nonlinear functional analysis of boundary value problems...

Guest Editors: Yong Hong Wu, Lishan Liu, Benchawan Wiwatanapataphee, and Shaoyong Lai

Nonlinear Functional Analysis of Boundary Value Problems: Novel Theory, Methods, and Applications

Abstract and Applied Analysis

Nonlinear Functional Analysis of BoundaryValue Problems: Novel Theory, Methods,and Applications

Abstract and Applied Analysis

Nonlinear Functional Analysis of BoundaryValue Problems: Novel Theory, Methods,and Applications

Guest Editors: Yong Hong Wu, Lishan Liu,Benchawan Wiwatanapataphee, and Shaoyong Lai

Copyright © 2013 Hindawi Publishing Corporation. All rights reserved.

This is a special issue published in “Abstract and Applied Analysis.” All articles are open access articles distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the originalwork is properly cited.

Editorial Board

Dirk Aeyels, BelgiumRavi P. Agarwal, Saudi ArabiaM. O. Ahmedou, GermanyNicholas D. Alikakos, GreeceDebora Amadori, ItalyPablo Amster, ArgentinaDouglas R. Anderson, USAJan Andres, Czech RepublicGiovanni Anello, ItalyStanislav Antontsev, PortugalMohamed Kamal Aouf, EgyptNarcisa C. Apreutesei, RomaniaNatig M. Atakishiyev, MexicoFerhan M. Atici, USAIvan G. Avramidi, USASoohyun Bae, KoreaZhanbing Bai, ChinaChuanzhi Bai, ChinaDumitru Baleanu, TurkeyJzef Bana, PolandGerassimos Barbatis, GreeceMartino Bardi, ItalyRoberto Barrio, SpainFeyzi Basar, TurkeyAbdelghani Bellouquid, MoroccoDaniele Bertaccini, ItalyMichiel Bertsch, ItalyLucio Boccardo, ItalyIgor Boglaev, New ZealandMartin J. Bohner, USAJulian F. Bonder, ArgentinaGeraldo Botelho, BrazilElena Braverman, CanadaRomeo Brunetti, ItalyJanusz Brzdek, PolandDetlev Buchholz, GermanySun-Sig Byun, KoreaFabio M. Camilli, ItalyAntonio Canada, SpainJinde Cao, ChinaAnna Capietto, ItalyKwang-chih Chang, ChinaJianqing Chen, ChinaWing-Sum Cheung, Hong KongMichel Chipot, Switzerland

Changbum Chun, KoreaSoon Y. Chung, KoreaJaeyoung Chung, KoreaSilvia Cingolani, ItalyJean M. Combes, FranceMonica Conti, ItalyDiego Cordoba, SpainJuan Carlos Cortes Lopez, SpainGraziano Crasta, ItalyGuillermo P. Curbera, SpainBernard Dacorogna, SwitzerlandVladimir Danilov, RussiaMohammad T. Darvishi, IranLuis F. Pinheiro de Castro, PortugalToka Diagana, USAJesus I. Dıaz, SpainJosef Diblık, Czech RepublicFasma Diele, ItalyTomas Dominguez, SpainAlexander I. Domoshnitsky, IsraelMarco Donatelli, ItalyBo-Qing Dong, ChinaOndrej Dosly, Czech RepublicWei-Shih Du, TaiwanLuiz Duarte, BrazilRoman Dwilewicz, USAPaul W. Eloe, USAAhmed El-Sayed, EgyptLuca Esposito, ItalyJose A. Ezquerro, SpainKhalil Ezzinbi, MoroccoDashan Fan, USAAngelo Favini, ItalyMarcia Federson, BrazilStathis Filippas, Equatorial GuineaAlberto Fiorenza, ItalyTore Flatten, NorwayIlaria Fragala, ItalyBruno Franchi, ItalyXianlong Fu, ChinaMassimo Furi, ItalyGiovanni P. Galdi, USAIsaac Garcia, SpainJesus Garcıa Falset, SpainJose A. Garcıa-Rodrıguez, Spain

Leszek Gasinski, PolandGyorgy Gat, HungaryVladimir Georgiev, ItalyLorenzo Giacomelli, ItalyJaume Gine, SpainValery Y. Glizer, IsraelLaurent Gosse, ItalyJean P. Gossez, BelgiumDimitris Goussis, GreeceJose L. Gracia, SpainMaurizio Grasselli, ItalyQian Guo, ChinaYuxia Guo, ChinaChaitan P. Gupta, USAUno Hamarik, EstoniaFerenc Hartung, HungaryBehnam Hashemi, IranNorimichi Hirano, JapanJiaxin Hu, ChinaChengming Huang, ChinaZhongyi Huang, ChinaGennaro Infante, ItalyIvan Ivanov, BulgariaHossein Jafari, IranJaan Janno, EstoniaAref Jeribi, TunisiaUn C. Ji, KoreaZhongxiao Jia, ChinaLucas Jodar, SpainJong Soo Jung, Republic of KoreaHenrik Kalisch, NorwayHamid Reza Karimi, NorwaySatyanad Kichenassamy, FranceTero Kilpelainen, FinlandSung Guen Kim, Republic of KoreaLjubisa Kocinac, SerbiaAndrei Korobeinikov, SpainPekka Koskela, FinlandVictor Kovtunenko, AustriaRen-Jieh Kuo, TaiwanPavel Kurasov, SwedenMiroslaw Lachowicz, PolandKunquan Lan, CanadaRuediger Landes, USAIrena Lasiecka, USA

Matti Lassas, FinlandChun-Kong Law, TaiwanMing-Yi Lee, TaiwanGongbao Li, ChinaPedro M. Lima, PortugalElena Litsyn, IsraelYansheng Liu, ChinaShengqiang Liu, ChinaCarlos Lizama, ChileMilton C. Lopes Filho, BrazilJulian Lopez-Gomez, SpainJinhu Lu, ChinaGrzegorz Lukaszewicz, PolandShiwang Ma, ChinaWanbiao Ma, ChinaEberhard Malkowsky, TurkeySalvatore A. Marano, ItalyCristina Marcelli, ItalyPaolo Marcellini, ItalyJesus Marın-Solano, SpainJose M. Martell, SpainMieczysław Mastyo, PolandMing Mei, CanadaTaras Mel’nyk, UkraineAnna Mercaldo, ItalyChangxing Miao, ChinaStanislaw Migorski, PolandMihai Mihailescu, RomaniaFeliz Minhos, PortugalDumitru Motreanu, FranceRoberta Musina, ItalyMaria Grazia Naso, ItalyGaston M. N’Guerekata, USASylvia Novo, SpainMicah Osilike, NigeriaMitsuharu Otani, JapanTurgut Ozis, TurkeyFilomena Pacella, ItalyNikolaos S. Papageorgiou, GreeceSehie Park, KoreaAlberto Parmeggiani, ItalyKailash C. Patidar, South AfricaKevin R. Payne, ItalyJosip E. Pecaric, CroatiaShuangjie Peng, ChinaSergei V. Pereverzyev, AustriaMaria Eugenia Perez, SpainDavid Perez-Garcia, SpainJosefina Perles, SpainAllan Peterson, USA

Andrew Pickering, SpainCristina Pignotti, ItalySomyot Plubtieng, ThailandMilan Pokorny, Czech RepublicSergio Polidoro, ItalyZiemowit Popowicz, PolandMaria M. Porzio, ItalyEnrico Priola, ItalyVladimir S. Rabinovich, MexicoIrena Rachunkova, Czech RepublicMaria Alessandra Ragusa, ItalySimeon Reich, IsraelWeiqing Ren, USAAbdelaziz Rhandi, ItalyHassan Riahi, MalaysiaJuan P. Rincon-Zapatero, SpainLuigi Rodino, ItalyYuriy Rogovchenko, NorwayJulio D. Rossi, ArgentinaWolfgang Ruess, GermanyBernhard Ruf, ItalyMarco Sabatini, ItalySatit Saejung, ThailandStefan G. Samko, PortugalMartin Schechter, USAJavier Segura, SpainSigmund Selberg, NorwayValery Serov, FinlandNaseer Shahzad, Saudi ArabiaAndrey Shishkov, UkraineStefan Siegmund, GermanyAbdel-Maksoud A. Soliman, EgyptPierpaolo Soravia, ItalyMarco Squassina, ItalySvatoslav Stanek, Czech RepublicStevo Stevic, SerbiaAntonio Suarez, SpainWenchang Sun, ChinaRobert Szalai, UKSanyi Tang, ChinaChun-Lei Tang, ChinaYoushan Tao, ChinaGabriella Tarantello, ItalyNasser-eddine Tatar, Saudi ArabiaSusanna Terracini, ItalyGerd Teschke, GermanyAlberto Tesei, ItalyBevan Thompson, AustraliaSergey Tikhonov, SpainClaudia Timofte, Romania

Thanh Dien Tran, AustraliaJuan J. Trujillo, SpainCiprian A. Tudor, FranceGabriel Turinici, FranceMilan Tvrdy, Czech RepublicMehmet Unal, TurkeyStephan A. van Gils, The NetherlandsCsaba Varga, RomaniaCarlos Vazquez, SpainGianmaria Verzini, ItalyJesus Vigo-Aguiar, SpainQing-Wen Wang, ChinaYushun Wang, ChinaShawn X. Wang, CanadaYouyu Wang, ChinaJing Ping Wang, UKPeixuan Weng, ChinaNoemi Wolanski, ArgentinaNgai-Ching Wong, TaiwanPatricia J. Y. Wong, SingaporeRoderick Wong, Hong KongZili Wu, ChinaYong Hong Wu, AustraliaShanhe Wu, ChinaTie-cheng Xia, ChinaXu Xian, ChinaYanni Xiao, ChinaFuding Xie, ChinaNaihua Xiu, ChinaDaoyi Xu, ChinaZhenya Yan, ChinaXiaodong Yan, USANorio Yoshida, JapanBeong In Yun, KoreaVjacheslav Yurko, RussiaAgacık Zafer, TurkeySergey V. Zelik, UKJianming Zhan, ChinaMeirong Zhang, ChinaWeinian Zhang, ChinaChengjian Zhang, ChinaZengqin Zhao, ChinaSining Zheng, ChinaTianshou Zhou, ChinaYong Zhou, ChinaQiji J. Zhu, USAChun-Gang Zhu, ChinaMalisa R. Zizovic, SerbiaWenming Zou, China

Contents

Nonlinear Functional Analysis of Boundary Value Problems: Novel Theory, Methods, and Applications,Yong Hong Wu, Lishan Liu, Benchawan Wiwatanapataphee, and Shaoyong LaiVolume 2013, Article ID 158358, 3 pages

Multiple Solutions for a Fractional Difference Boundary Value Problem via Variational Approach,Zuoshi Xie, Yuanfeng Jin, and Chengmin HouVolume 2012, Article ID 143914, 16 pages

On Spectral Homotopy Analysis Method for Solving Linear Volterra and Fredholm IntegrodifferentialEquations, Z. Pashazadeh Atabakan, A. Kılıcman, and A. Kazemi NasabVolume 2012, Article ID 960289, 16 pages

Multiple Solutions for a Class of Fractional Boundary Value Problems, Ge BinVolume 2012, Article ID 468980, 16 pages

On Impulsive Boundary Value Problems of Fractional Differential Equations with Irregular BoundaryConditions, Guotao Wang, Bashir Ahmad, and Lihong ZhangVolume 2012, Article ID 356132, 18 pages

Positive Solutions of Nonlinear Fractional Differential Equations with Integral Boundary ValueConditions, J. Caballero, I. Cabrera, and K. SadaranganiVolume 2012, Article ID 303545, 11 pages

Bifurcation Analysis for a Predator-Prey Model with Time Delay and Delay-Dependent Parameters,Changjin XuVolume 2012, Article ID 264870, 20 pages

A Galerkin Solution for Burgers’ Equation Using Cubic B-Spline Finite Elements, A. A. SolimanVolume 2012, Article ID 527467, 15 pages

Solutions of Sign-Changing Fractional Differential Equation with the Fractional Derivatives,Tunhua Wu, Xinguang Zhang, and Yinan LuVolume 2012, Article ID 797398, 16 pages

Existence of Positive Solution for Semipositone Fractional Differential Equations InvolvingRiemann-Stieltjes Integral Conditions, Wei Wang and Li HuangVolume 2012, Article ID 723507, 17 pages

A Generalization of Mahadevan’s Version of the Krein-Rutman Theorem and Applications to p-LaplacianBoundary Value Problems, Yujun Cui and Jingxian SunVolume 2012, Article ID 305279, 14 pages

On the Study of Local Solutions for a Generalized Camassa-Holm Equation, Meng WuVolume 2012, Article ID 164876, 17 pages

Existence of Solutions for Nonlinear Impulsive Fractional Differential Equations of Order α ∈ (2, 3]with Nonlocal Boundary Conditions, Lihong Zhang, Guotao Wang, and Guangxing SongVolume 2012, Article ID 717235, 26 pages

Existence of Solutions for Nonhomogeneous A-Harmonic Equations with Variable Growth,Yongqiang Fu and Lifeng GuoVolume 2012, Article ID 421571, 26 pages

Forward-Backward Splitting Methods for Accretive Operators in Banach Spaces, Genaro Lopez,Victoria Martın-Marquez, Fenghui Wang, and Hong-Kun XuVolume 2012, Article ID 109236, 25 pages

Blow-Up Analysis for a Quasilinear Degenerate Parabolic Equation with Strongly Nonlinear Source,Pan Zheng, Chunlai Mu, Dengming Liu, Xianzhong Yao, and Shouming ZhouVolume 2012, Article ID 109546, 19 pages

Positive Solutions for Sturm-Liouville Boundary Value Problems in a Banach Space, Hua Su, Lishan Liu,and Yonghong WuVolume 2012, Article ID 572172, 11 pages

Positive Solutions for Second-Order Singular Semipositone Differential Equations Involving StieltjesIntegral Conditions, Jiqiang Jiang, Lishan Liu, and Yonghong WuVolume 2012, Article ID 696283, 21 pages

The Asymptotic Solution of the Initial Boundary Value Problem to a Generalized Boussinesq Equation,Zheng Yin and Feng ZhangVolume 2012, Article ID 216320, 13 pages

Multiple Solutions for Degenerate Elliptic Systems Near Resonance at Higher Eigenvalues,Yu-Cheng An and Hong-Min SuoVolume 2012, Article ID 532430, 19 pages

The Solution of a Class of Singularly Perturbed Two-Point Boundary Value Problems by the IterativeReproducing Kernel Method, Zhiyuan Li, YuLan Wang, Fugui Tan, Xiaohui Wan, and Tingfang NieVolume 2012, Article ID 984057, 7 pages

Strong Global Attractors for 3D Wave Equations with Weakly Damping, Fengjuan MengVolume 2012, Article ID 469382, 12 pages

A Regularity Criterion for the Navier-Stokes Equations in the Multiplier Spaces, Xiang’ou ZhuVolume 2012, Article ID 682436, 7 pages

The Well-Posedness of Solutions for a Generalized Shallow Water Wave Equation,Shaoyong Lai and Aiyin WangVolume 2012, Article ID 872187, 15 pages

Positive Solutions of a Nonlinear Fourth-Order Dynamic Eigenvalue Problem on Time Scales,Hua Luo and Chenghua GaoVolume 2012, Article ID 798796, 17 pages

The Local Strong and Weak Solutions for a Generalized Pseudoparabolic Equation, Nan LiVolume 2012, Article ID 568404, 12 pages

Contents

The Cauchy Problem to a Shallow Water Wave Equation with a Weakly Dissipative Term,Ying Wang and YunXi GuoVolume 2012, Article ID 840919, 23 pages

Positive Solutions of Eigenvalue Problems for a Class of Fractional Differential Equations withDerivatives, Xinguang Zhang, Lishan Liu, Benchawan Wiwatanapataphee, and Yonghong WuVolume 2012, Article ID 512127, 16 pages

Positive Solutions of a Fractional Boundary Value Problem with Changing Sign Nonlinearity,Yongqing Wang, Lishan Liu, and Yonghong WuVolume 2012, Article ID 149849, 12 pages

The Local Strong and Weak Solutions for a Generalized Novikov Equation, Meng Wu and Yue ZhongVolume 2012, Article ID 158126, 14 pages

Regularity and Exponential Growth of Pullback Attractors for Semilinear Parabolic Equations Involvingthe Grushin Operator, Nguyen Dinh BinhVolume 2012, Article ID 272145, 20 pages

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2013, Article ID 158358, 3 pageshttp://dx.doi.org/10.1155/2013/158358

EditorialNonlinear Functional Analysis of Boundary Value Problems:Novel Theory, Methods, and Applications

Yong Hong Wu,1 Lishan Liu,2 Benchawan Wiwatanapataphee,3 and Shaoyong Lai4

1 Curtin University of Technology, Perth, WA 6845, Australia2 Qufu Normal University, Qufu, Shandong 273165, China3Department of Mathematics, Faculty of Science, Mahidol University, Bangkok 10400, Thailand4 Southwestern University of Finance and Economics, Chengdu, Sichuan 610074, China

Correspondence should be addressed to Benchawan Wiwatanapataphee; [email protected]

Received 24 February 2013; Accepted 24 February 2013

Copyright © 2013 Yong Hong Wu et al.This is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Nonlinear boundary value problems often arise from sci-entific research, modelling of nonlinear phenomena, andoptimal control of complex systems. In some cases, it is notpossible to solve the underlying nonlinear boundary valueproblems in a closed form, and, for these cases, nonlinearfunctional analysis can play its role as a tool for obtainingqualitative information about the solution. The primaryproblems concerned in functional analysis are as follows:does a solution of the boundary value problem exist? If so,is it unique and stable? If the answers to these questions areaffirmative, the problem is well posed and the analysis canproceed in a firm base to the next step—construct a methodfor finding approximate solutions and evaluate the quality ofthe approximation.

Over the last few decades, intensive research has beencarried out worldwide to develop functional analysis theoriesand methods to tackle complex boundary value problemsarising continually from scientific research and modelling ofreal world phenomena. The papers selected for this specialissue represent a typical set of contributions in this fieldof research. Of course, the selected topics and papers arenot an exhaustive representation of the recent developmentin the field. Nevertheless, they provide novel functionaltheories and methods as well as rich analysis for manycomplex boundary value problems with different applicationbackgrounds and we have the pleasure of sharing theseresearch results with the readers. Here, wewould like to thankthe authors and reviewers of the papers for their excellent

contributions. The efforts made by staff of the editorial officeof the publishing corporation are also greatly acknowledged.

This special issue contains twenty-nine papers, coveringfunctional analysis of various types of complex bound-ary value problems for different differential equations andboundary conditions. The particular focus is on fractionalorder differential equations and partial differential equations,and the boundary conditions include Riemann-Stieltjes inte-gral boundary conditions and nonlocal boundary conditions.

The first set of papers, including nine papers, focus onthe study of various complex fractional order boundary valueproblems.

(i) In the paper titled “Existence of positive solution forsemipositone fractional differential equations involvingRiemann-Stieltjes integral conditions,” based on thefixed-point theorem in a cone, the existence of atleast one positive solution is established for a classof semipositone fractional differential equations withRiemann-Stieltjes integral boundary condition.

(ii) In the paper titled “Multiple solutions for a class of frac-tional boundary value problems,” the authors establishconditions for the existence of multiple solutionsfor a class of fractional bocundary value problemsinvolving left and right Riemann-Liouville fractionalintegrals of order 0 ≤ 𝛽 < 1 by the variationalmethod.

2 Abstract and Applied Analysis

(iii) In the paper titled “Positive solutions of eigenvalueproblems for a class of fractional differential equationswith derivatives,” the existence of positive solutions forthe eigenvalue problem of a class of fractional differ-ential equations is established through establishing amaximal principles and constructing upper and lowersolutions.

(iv) In the paper titled “Positive solutions of nonlinearfractional differential equations with integral boundaryvalue conditions,” by using the fixed-point theorem inpartially ordered set, the authors investigate the exis-tence and uniqueness of the solutions of a boundaryvalue problem involvingCaputo fractional derivativesand integral boundary conditions.

(v) In the paper titled “On impulsive boundary value prob-lems of fractional differential equations with irregularboundary conditions,” the existence and uniquenessof solutions are established for a class of boundaryvalue problems with irregular boundary conditionsand impulsive jumps in the system state and its first-order derivative.

(vi) In the paper titled “Existence of solutions for nonlinearimpulsive fractional differential equations of order𝛼 ∈ (2, 3] with nonlocal boundary conditions,” theauthors established the existence and uniqueness ofthe solution to a nonlocal boundary value problem fornonlinear impulsive fractional differential equationsof order 𝛼 ∈ (2, 3].

(vii) In the paper titled “Positive solutions of a fractionalboundary value problem with changing sign nonlinear-ity,” by investigating the properties of the associatedGreen function and utilizing the Krasnoselskii fixed-point theorem in a cone, the authors establish theexistence of positive solutions of a boundary valueproblem for nonlinear fractional differential equa-tions with changing sign nonlinearities.

(viii) In the paper titled “Multiple solutions for a fractionaldifference boundary value problem via variationalapproach,” the existence of multiple solutions for afractional difference boundary value problem witha parameter is established within the variationalframework.

(ix) In the paper titled “Solutions of sign-changing frac-tional differential equation with the fractional deriva-tives,” the existence of nontrivial solutions is estab-lished for a singular fractional order boundaryvalue problem with a sign changing nonlinear term,through computing the topological degree of a com-pletely continuous field.

The second set of papers, including thirteen papers, focuson the study of various complex partial differential equationboundary value problems.

(i) In the paper titled “On the study of local solutionsfor a generalized Camassa-Holm equation,” the localwell-posedness of strong solutions for a nonlinear

dispersive model is established in the Sobolev space𝐻𝑠

(𝑅) with 𝑠 > 3/2 by using the pseudoparabolicregularization technique.

(ii) In the paper titled “Thewell-posedness of solutions for ageneralized shallow water wave equation,” the authorsinvestigate a generalized shallow water wave equationand discuss the local well-posedness of solutions inthe Sobolev space.

(iii) In the paper titled “Blow-up analysis for a quasilineardegenerate parabolic equation with strongly nonlinearsource,” the authors investigate the properties of thepositive solution of the Cauchy problem for a quasi-linear degenerate parabolic equation with a stronglynonlinear source and show the existence of a single-point blow-up for a large class of radial decreasingsolutions.

(iv) In the paper titled “The cauchy problem to a shallowwater wave equation with a weakly dissipative term,” ashallowwater wave equationwith a weakly dissipativeterm is investigated, and the sufficient conditions forthe existence of global strong solution are established.

(v) In the paper titled ”The local strong and weak solutionsfor a generalized pseudo-parabolic equation,” the well-posedness of local strong solutions for the Cauchyproblem of a nonlinear generalized pseudoparabolicequation is established in the Sobolev space.

(vi) In the paper titled “The local strong and weak solutionsfor a generalized Novikov equation,” the Kato theoremfor abstract differential equations is applied to estab-lish the local well-posedness of the strong solution fora nonlinear generalized Novikov equation.

(vii) In the paper titled “The asymptotic solution of theinitial boundary value problem to a generalized Boussi-nesq equation,” the 𝐿2 solution of an initial boundaryproblem for a generalized damped Boussinesq equa-tion is constructed, and the large time asymptoticsolution is also obtained.

(viii) In the paper titled “Strong global attractors for 3Dwaveequations with weakly damping,” the authors prove theexistence of a global attractor for a 3Dweakly dampedwave equation in a strong topological space.

(ix) In the paper titled “Regularity and exponential growthof pullback attractors for semilinear parabolic equa-tions involving the Grushin operator,” the authorsconsider a first order nonlinear initial boundary valueproblem for a semilinear degenerate parabolic equa-tion involving the Grushin operator in a boundeddomain and prove the regularity and exponentialgrowth of a pullback attractor in certain functionspace for the nonautonomous dynamical system asso-ciated to the problem.

(x) In the paper titled “A generalization of Mahadevan’sversion of the Krein-Rutman theorem and applicationsto 𝑝-Laplacian boundary value problems,” the authorsestablish a generalization of Mahadevan’s version ofthe Krein-Rutman theorem for a compact, positively

Abstract and Applied Analysis 3

1-homogeneous operator in a Banach space and thendemonstrate its application in establishing the exis-tence of positive solutions for 𝑝-Laplacian boundaryvalue problems under certain conditions.

(xi) In the paper titled “Existence of solutions for nonhomo-geneous A-harmonic equations with variable growth,”the authors establish a theorem for the existenceof weak solutions for nonhomogeneous A-harmonicequations in subspace and then give three examplesto demonstrate its application.

(xii) In the paper titled “Multiple solutions for degenerateelliptic systems near resonance at higher eigenvalues,”the authors study the degenerate semilinear ellipticsystem in an open bounded domain with smoothboundary, and some multiplicity results of solutionsare obtained for the system near resonance at certaineigenvalues by the classical saddle point theorem anda local saddle point theorem in critical point theory.

(xiii) In the paper titled “A regularity criterion for theNavier-Stokes equations in the multiplier spaces,” theauthors establish a regularity criterion in terms of thepressure gradient for weak solutions to the Navier-Stokes equations in a special class.

The third set of papers, including four papers, deal withseveral boundary value problems for highly nonlinear ordi-nary differential equations.

(i) In the paper titled “Positive solutions for second-ordersingular semipositone differential equations involvingStieltjes integral conditions,” the authors investigate theexistence of positive solutions for second-order singu-lar differential equations with a negatively perturbedterm, by means of the fixed-point theory in cones.

(ii) In the paper titled “Positive solutions for Sturm-Liou-ville boundary value problems in a Banach Space,” thesufficient conditions for the existence of single andmultiple positive solutions for a second-order Sturm-Liouville boundary value problem are established ina Banach space, by using the fixed-point theorem ofstrict set contraction operators in the frame of theODE technique.

(iii) In the paper titled “Positive solutions of a nonlinearfourth-order dynamic eigenvalue problem on timescales,” the authors study a nonlinear fourth-orderdynamic eigenvalue problem on time scales andobtain the existence and nonexistence of positivesolutions when 0 < 𝜆 ≤ 𝜆∗ and 𝜆 > 𝜆∗, respectively,for some 𝜆∗, by using the Schauder fixed-pointtheorem and the upper and lower solution method.

(iv) In the paper titled “Bifurcation analysis for a predator-prey model with time delay and delay-dependentparameters,” a class of stage-structured predator-preymodel with time delay and delay-dependent parame-ters is considered. By using the normal form theoryand center manifold theory, some explicit formulaefor determining the stability and the direction of the

Hopf bifurcation periodic solutions bifur-cating fromHopf bifurcations are obtained.

The fourth set of papers focus on finding the approxi-mate and numerical solutions of various complex nonlinearboundary value problems.

(i) In the paper titled “On spectral homotopy analy-sis method for solving linear Volterra and Fredholmintegrodifferential equations,” a spectral homotopyanalysis method (SHAM) is proposed to solve lin-ear Volterra integrodifferential equations, and someexamples are given to test the efficiency and theaccuracy of the proposed method.

(ii) In the paper titled “The solution of a class of singu-larly perturbed two-point boundary value problems bythe iterative reproducing kernel method,” the authorsestablish an iterative reproducing kernel method(IRKM) for solving singular perturbation problemswith boundary layers and give two numerical exam-ples to demonstrate the effectiveness of the method.

(iii) In the paper titled “A Galerkin solution for Burgers’equation using cubic B-spline finite elements,” a Gal-erkin method using cubic B-splines is set up to findthe numerical solutions of Burgers’ equation, and themethod is shown to be capable of solving Burgers’equation accurately for values of viscosity rangingfrom very small to very large.

(iv) In the paper titled “Forward-backward splitting meth-ods for accretive operators in Banach spaces,” theauthors introduce two iterative forward-backwardsplitting methods with relaxations to find zeros ofthe sum of two accretive operators in Banach spacesand prove the weak and strong convergence of thesemethods under mild conditions, and also discussapplications of these methods to variational inequal-ities, the split feasibility problem, and a constrainedconvex minimization problem.

Yong Hong WuLishan Liu

Benchawan WiwatanapatapheeShaoyong Lai

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2012, Article ID 143914, 16 pagesdoi:10.1155/2012/143914

Research ArticleMultiple Solutions for a Fractional DifferenceBoundary Value Problem via Variational Approach

Zuoshi Xie,1 Yuanfeng Jin,2 and Chengmin Hou2

1 School of Economics and International Trade, Zhejiang University of Finance and Economics,Hangzhou, Zhejiang 310018, China

2 Department of Mathematics, Yanbian University, Yanji 133002, China

Correspondence should be addressed to Chengmin Hou, [email protected]

Received 28 April 2012; Revised 5 November 2012; Accepted 8 November 2012

Academic Editor: Lishan Liu

Copyright q 2012 Zuoshi Xie et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

By establishing the corresponding variational framework and using the mountain pass theorem,linking theorem, and Clark theorem in critical point theory, we give the existence of multiplesolutions for a fractional difference boundary value problem with parameter. Under some suitableassumptions, we obtain some results which ensure the existence of a well precise interval ofparameter for which the problem admits multiple solutions. Some examples are presented toillustrate the main results.

1. Introduction

Variational methods for dealing with difference equations have appeared as early as 1985 in[1] in which the positive definiteness of quadratic forms (which are functionals) is related tothe existence of “nodes” of solutions (or positive solutions satisfying “conjugate” boundaryconditions) of linear self-adjoint second-order difference equations of the form

Δ(p(k − 1)Δy(k − 1)

)+ q(k)y(k) = 0, k = 1, 2, . . . , n, (1.1)

where p(k) is real and positive for k = 0, 1, . . . , n and q(k) is real for k = 1, 2, . . . , n. Laterthere are interests in solutions of nonlinear difference equations under various types ofboundary or subsidiary conditions, and more sophisticated methods such as the mountainpass theorems are needed to handle the existence problem (see, e.g., [2–11]).

Recently, fractional differential and difference “operators” are found themselves inconcrete applications, and hence attention has to be paid to associated fractional difference


and differential equations under various boundary or side conditions. For example, a recentpaper by Atici and Eloe [12] explores some of the theories of a discrete conjugate fractionalBVP. Similarly, in [13], a discrete right-focal fractional BVP is analyzed. Other recent advancesin the theory of the discrete fractional calculus may be found in [14, 15]. In particular, aninteresting recent paper by Atici and Sengul [16] addressed the use of fractional differenceequations in tumor growth modeling. Thus, it seems that there exists some promise in usingfractional difference equations as mathematical models for describing physical problems inmore accurate manners.

In order to handle the existence problem for fractional BVPs, various methods (amongwhich are some standard fixed-point theorems) can be used. In this paper, however, we showthat variational methods can also be applied. A good reason for picking such an approach isthat, in Atici and Sengul [16], some basic fractional calculuses are developed and a simplevariational problem is demonstrated, and hence advantage can be taken in obvious manners.We remark, however, that fractional difference operators can be approached in differentmanners and one by means of operator convolution rings can be found in the book by Cheng[17, Chapter 3] published in 2003.

More specifically, in this paper, we are interested in the existence of multiple solutionsfor the following 2ν-order fractional difference boundary value problem

TΔνt−1

(tΔν

ν−1x(t))= λf(t + ν − 1, x(t + ν − 1)), t ∈ [0, T]

N0, (1.2)

x(ν − 2) =[tΔν

ν−1x(t)]t=T = 0, (1.3)

where ν ∈ (0, 1), tΔνν−1 and TΔν

t are, respectively, left fractional difference and the rightfractional difference operators (which will be explained in more detail later), t ∈ [0, T]

N0=

{0, 1, 2, . . . , T}, f(t + ν − 1, ·) : [ν − 1, T + ν − 1]Nν−1

× R → R is continuous, and λ is a positiveparameter.

By establishing the corresponding variational framework and using critical pointtheory, we will establish various existence results (which naturally depend on f , ν, and λ).

For convenience, throughout this paper, we arrange∑m

i=j x(i) = 0, for m < j.

2. Preliminaries

We first collect some basic lemmas for manipulating discrete fractional operators. These andother related results can be found in [14, 16].

First, for any integer β, we let Nβ = {β, β+1, β+2, . . .}. We define t(ν) := Γ(t+1)/Γ(t+1−ν),for any t and ν for which the right-hand side is defined. We also appeal to the convention that,if t + 1 − ν is a pole of the Gamma function and t + 1 is not a pole, then t(ν) = 0.

Definition 2.1. The νth fractional sum of f for ν > 0 is defined by

Δ−νa f(t) =

1Γ(ν)

t−ν∑

s=a(t − s − 1)(ν−1)f(s), (2.1)

for t ∈ Na−ν. We also define the νth fractional difference for ν > 0 by Δνf(t) := ΔNΔν−Nf(t),where t ∈ Na+N−ν and N ∈ N is chosen so that 0 ≤N − 1 < ν ≤N.


Definition 2.2. Let f be any real-valued function and ν ∈ (0, 1). The left discrete fractionaldifference and the right discrete fractional difference operators are, respectively, defined as

tΔνaf(t) = ΔtΔ

−(1−ν)a f(t) =

1Γ(1 − ν)Δ

t+ν−1∑

s=a(t − s − 1)(−ν)f(s), t ≡ a − ν + 1 (mod1),

bΔνt f(t) = −ΔbΔ

−(1−ν)t f(t) =

1Γ(1 − ν) (−Δ)

b∑

s=t+1−ν(s − t − 1)(−ν)f(s), t ≡ b + ν − 1 (mod1).

(2.2)

Definition 2.3. For I ∈ C1(E,R), we say I satisfies the Palais-Smale condition (henceforthdenoted by (PS) condition) if any sequence {xn} ⊂ E for which I(xn) is bounded and I ′(xn) →0 as n → +∞ possesses a convergent subsequence.

Lemma 2.4 (see [18]). A real symmetric matrixA is positive definite if there exists a real nonsingularmatrixM such that A =M†M, whereM† is the transpose.

Lemma 2.5 (see [9]: linking theorem). Let E be a real Banach space, and I ∈ C1(E,R) satisfies(PS) condition and is bounded from below. Suppose I has a local linking at the origin θ, namely, thereis a decomposition E = Y ⊕ W and a positive number ρ such that k = dim Y < ∞, I(y) < I(θ) fory ∈ Y with 0 < ‖y‖ ≤ ρ; I(y) ≥ I(θ) for y ∈ W with ‖y‖ ≤ ρ. Then I has at least three criticalpoints.

Lemma 2.6 (see [6]). Let E be a real reflexive Banach space, and let the functional I : E → R beweakly lower (upper) semicontinuous and coercive, that is, lim||x||→∞I(x) = ∞ (resp., anticoercive,i.e., lim||x||→∞I(x) = −∞). Then there exists x0 ∈ E such that I(x0) = infEI(x) (resp., I(x0) =sup

EI(x)). Moreover, if I ∈ C1(E,R), then x0 is a critical point of functional I.Recall that, in the finite dimensional setting, it is well known that a coercive functional satisfies

the (PS) condition.Let Br denote the open ball in a real Banach space of radius r about 0, and let ∂Br denote its

boundary. Now some critical point theorems needed later can be stated.

Lemma 2.7 (mountain pass theorem [8]). Let E be a real Banach space and I ∈ C1(E,R), satisfy-ing (PS) condition. Suppose I(θ) = 0 and

(I1) there are constants ρ, α > 0 such that I|∂Bρ ≥ α,(I2) there is e ∈ E \ Bρ such that I(e) ≤ 0.

Then I possesses a critical value c ≥ α. Moreover c can be characterized as

c = infg∈Γ

maxu∈g([0,1])

I(u), (2.3)

where

Γ ={g ∈ C([0, 1],E) | g(0) = θ, g(1) = e}. (2.4)


Lemma 2.8 (see [7]). Let E be a reflexive Banach space and I ∈ C1(E,R) with I(θ) = 0. Supposethat I is an even functional satisfying (PS) condition and the following conditions:

(I3) there are constants ρ, α > 0 and a closed linear subspaceX1 ofE such that codimX1 = l <∞and I|∂Bρ∩X1 ≥ α,

(I4) there is a finite dimensional subspace X2 of E with dim X2 = m, m > l, such that I(x) →−∞ as ||x|| → ∞, x ∈ X2. Then I possesses at leastm−l distinct pairs of nontrivial criticalpoints.

Lemma 2.9 (the Clark theorem [8]). Let E be a real Banach space, I ∈ C1(E,R) with I even,bounded from below, and satisfying (PS) condition. Suppose I(θ) = 0, there is a set K ⊂ E such thatK is homeomorphic to S

j−1 (the j − 1 dimensional unit sphere) by an odd map and supKI < 0. Then I

possesses at least j distinct pairs of critical points.

3. Main Results

Firstly, we establish variational framework. Let

Ω ={x = (x(ν − 1), x(ν), . . . , x(ν + T − 1))† | x(ν + i − 1) ∈ R, i = 0, 1, . . . , T

}(3.1)

be the T + 1-dimensional Hilbert space with the usual inner product and the usual norm

〈x, z〉 =T+ν−1∑

t=ν−1

x(t)z(t), ‖x‖ =

(T+ν−1∑

t=ν−1

|x(t)|2)1/2

, x, z ∈ Ω. (3.2)

For r > 1, we recall the r-norm on Ω: ‖x‖r = (∑T+ν−1

t=ν−1 |x(t)|r)1/r . We also recall the standardfact that there exist positive constants cr and cr , such that

cr ||x|| ≤ ‖x‖r ≤ cr‖x‖, x ∈ Ω. (3.3)

Define a functional on Ω by

I(x) =12

T∑

t=−1

(tΔν

ν−1x(t))2 − λ

T∑

t=−1

F(t + ν − 1, x(t + ν − 1)) (3.4)

for x = (x(ν − 1), x(ν), . . . , x(ν + T − 1))† ∈ Ω, where

F(t + ν − 1, x(t + ν − 1)) =∫x(t+ν−1)

0f(t + ν − 1, s)ds,

x(ν − 2) = 0,[tΔν

ν−1x(t)]t=T =

−νΓ(1 − ν)

T+ν∑

s=ν−1

(T − s − 1)(−ν−1)x(s) = 0.

(3.5)


Obviously, I(θ) = 0. Let

E ={χ = (x(ν − 2), x(ν − 1), . . . , x(ν + T))† ∈ R

T+3 | x(ν − 2) = 0,[tΔν

ν−1x(t)]t=T = 0

}.

(3.6)

Then by (1.3) it is easy to see that E is isomorphic to Ω. In the following, when we say x ∈Ω, we always imply that x can be extended to χ ∈ E if it is necessary. Now we claim thatif x = (x(ν − 1), x(ν), . . . , x(ν + T − 1))† ∈ Ω is a critical point of I, then χ = (x(ν − 2),x(ν − 1), . . . , x(ν + T))† ∈ E is precisely a solution of BVP (1.2) and (1.3). Indeed, since Ican be viewed as a continuously differentiable functional defined on the finite dimensionalHilbert space Ω, the Frechet derivative I ′(x) is zero if and only if ∂I(x)/∂x(i) = 0 for alli = ν − 1,ν, . . . , ν + T − 1.

By computation,

∂I(x)∂x(ν − 1)

=T∑

t=−1

(tΔν

ν−1x(t))∂tΔν

ν−1x(t)∂x(ν − 1)

− λf(ν − 1, x(ν − 1))

=−ν

Γ(1 − ν)T∑

t=−1

∂∑t+ν

s=ν−1 (t − s − 1)(−ν−1)x(s)∂x(ν − 1)

(tΔν

ν−1x(t)) − λf(ν − 1, x(ν − 1))

=ν2

Γ2(1 − ν)T∑

t=−1

(t − ν)(−ν−1)t+ν∑

s=ν−1

(t − s − 1)(−ν−1)x(s) − λf(ν − 1, x(ν − 1))

=ν2

Γ2(1 − ν)T∑

s=−1

(s − ν)(−ν−1)s+ν∑

u=ν−1

(s − u − 1)(−ν−1)x(u) − λf(ν − 1, x(ν − 1))

=ν2

Γ2(1 − ν)

[T∑

s=t−ν(s − t − 1)(−ν−1)

s+ν∑

u=ν−1

(s − u − 1)(−ν−1)x(u)

]

t=ν−1

− λf(ν − 1, x(ν − 1))

=−ν

Γ(1 − ν)

[T∑

s=t−ν(s − t − 1)(−ν−1)(

sΔνν−1

)x(s)

]

t=ν−1

− λf(ν − 1, x(ν − 1))

=−1

Γ(1 − ν)

[T∑

s=t−ν

((s − t − 1)(−ν) − (s − t)(−ν)

)(sΔν

ν−1

)x(s)

]

t=ν−1

− λf(ν − 1, x(ν − 1))

=1

Γ(1 − ν)

[

(−Δ)T∑

s=t−ν(s − t)(−ν)(sΔν

ν−1

)x(s)

]

t=ν−1

− λf(ν − 1, x(ν − 1))

=[TΔν

t−1

(tΔν

ν−1

)x(t)]t=ν−1 − λf(ν − 1, x(ν − 1)),


∂I(x)∂x(ν)

=T∑

t=−1

(tΔν

ν−1x(t))∂tΔν

ν−1x(t)∂x(ν)

− λf(ν, x(ν))

=−ν

Γ(1 − ν)T∑

t=−1

∂∑t+ν

s=ν−1 (t − s − 1)(−ν−1)x(s)∂x(ν)

(tΔν

ν−1x(t)) − λf(ν, x(ν))

=ν2

Γ2(1 − ν)T∑

t=0(t − ν − 1)(−ν−1)

t+ν∑

s=ν−1

(t − s − 1)(−ν−1)x(s) − λf(ν, x(ν))

=ν2

Γ2(1 − ν)T∑

s=0(s − ν − 1)(−ν−1)

s+ν∑

u=ν−1

(s − u − 1)(−ν−1)x(u) − λf(ν, x(ν))

=ν2

Γ2(1 − ν)

[T∑

s=t−ν(s − t − 1)(−ν−1)

s+ν∑

u=ν−1

(s − u − 1)(−ν−1)x(u)

]

t=ν

− λf(ν, x(ν))

=−ν

Γ(1 − ν)

[T∑

s=t−ν(s − t − 1)(−ν−1)(

sΔνν−1

)x(s)

]

t=ν

− λf(ν, x(ν))

=−1

Γ(1 − ν)

[T∑

s=t−ν

((s − t − 1)(−ν) − (s − t)(−ν)

)(sΔν

ν−1

)x(s)

]

t=ν

− λf(ν, x(ν))

=1

Γ(1 − ν)

[

(−Δ)T∑


ν−1

)x(s)

]

t=ν

− λf(ν, x(ν))

=[TΔν

t−1

(tΔν

ν−1

)x(t)]t=ν − λf(ν, x(ν)),

...

∂I(x)∂x(ν + T − 1)

=T∑

t=−1

(tΔν

ν−1x(t)) ∂tΔν

ν−1x(t)∂x(ν + T − 1)

− λf(ν + T − 1, x(ν + T − 1))

=−ν

Γ(1 − ν)T∑

t=−1

∂∑t+ν

s=ν−1 (t − s − 1)(−ν−1)x(s)∂x(ν + T − 1)

(tΔν

ν−1x(t))

− λf(ν + T − 1, x(ν + T − 1))

=ν2

Γ2(1 − ν)T∑

t=T−1

(t − ν − T)(−ν−1)t+ν∑

s=ν−1

(t − s − 1)(−ν−1)x(s)

− λf(ν + T − 1, x(ν + T − 1))

=ν2

Γ2(1 − ν)T∑

s=T−1

(s − ν − T)(−ν−1)s+ν∑

u=ν−1

(s − u − 1)(−ν−1)x(u)

− λf(ν + T − 1, x(ν + T − 1))


=ν2

Γ2(1 − ν)

[T∑

s=t−ν(s − t − 1)(−ν−1)

s+ν∑

u=ν−1

(s − u − 1)(−ν−1)x(u)

]

t=ν+T−1

− λf(ν + T − 1, x(ν + T − 1))

=−ν

Γ(1 − ν)

[T∑

s=t−ν(s − t − 1)(−ν−1)(

sΔνν−1

)x(s)

]

t=ν+T−1

− λf(ν + T − 1, x(ν + T − 1))

=−1

Γ(1 − ν)

[T∑

s=t−ν

((s − t − 1)(−ν) − (s − t)(−ν)

)(sΔν

ν−1

)x(s)

]

t=ν+T−1

− λf(ν + T − 1, x(ν + T − 1))

=1

Γ(1 − ν)

[

(−Δ)T∑


ν−1

)x(s)

]

t=ν+T−1

− λf(ν + T − 1, x(ν + T − 1))

=[TΔν

t−1

(tΔν

ν−1

)x(t)]t=ν+T−1 − λf(ν + T − 1, x(ν + T − 1)).

(3.7)

So to obtain the existence of solutions for problem (1.2) and (1.3), we just need to study theexistence of critical points, that is, x ∈ Ω such that I ′(x) = 0, of the functional I on Ω.

Next, observe by Definition 2.2 that, for t ∈ [−1, T]N−1

,

tΔνν−1x(t) = Δ

1Γ(1 − ν)

t−(1−ν)∑

t=ν−1

(t − s − 1)(−ν)x(s). (3.8)

We let

z(t + ν − 1) =1

Γ(1 − ν)t−(1−ν)∑

s=ν−1

(t − s − 1)(−ν)x(s), (3.9)

then

z(ν − 2) = 0,

z(ν − 1) =1

Γ(1 − ν)0−(1−ν)∑

s=ν−1

(−s − 1)(−ν)x(s) = x(ν − 1),

z(ν) =1

Γ(1 − ν)1−(1−ν)∑

s=ν−1

(1 − s − 1)(−ν)x(s) = (1 − ν)x(ν − 1) + x(ν),


z(ν + 1) =1

Γ(1 − ν)2−(1−ν)∑

s=ν−1

(2 − s − 1)(−ν)x(s)

=(2 − ν)(1 − ν)

2!x(ν − 1) + (1 − ν)x(ν) + x(ν + 1),

...

z(ν + T − 1) =1

Γ(1 − ν)T−(1−ν)∑

s=ν−1

(T − s − 1)(−ν)x(s)

=(T − ν)(T − 1 − ν) · · · (1 − ν)

(T)!x(ν − 1) +

(T − ν − 1)(T − 2 − ν) · · · (1 − ν)(T − 1)!

x(ν)

+ · · · + (1 − ν)x(ν + T − 2) + x(ν + T − 1),

(3.10)

that is, z = Bx, where z = (z(ν−1), z(ν), . . . , z(ν+T−1))†, x = (x(ν−1), x(ν), . . . , x(ν+T−1))†:

B =

⎛

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 0 0 · · · 01 − ν 1 0 · · · 0

(2 − ν)(1 − ν)2!

1 − ν 1 · · · 0...

......

...(T − ν)(T − 1 − ν) · · · (1 − ν)

T !(T − 1 − ν)(T − 2 − ν) · · · (1 − ν)

(T − 1)!· · · · · · 1

⎞

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

(T+1)×(T+1)

.

(3.11)

By Lemma 2.4, (B−1)†B−1 is a positive definite matrix. Let λmin and λmax denote,respectively, the minimum and the maximum eigenvalues of (B−1)†B−1.

Since x = B−1z, we may easily see that

λmin‖z‖2 ≤ ‖x‖2 =⟨z†(B−1)†, B−1z

⟩≤ λmax‖z‖2. (3.12)

Then ||x|| → ∞ if and only if ||z|| → ∞. Next, let

A =

⎛

⎜⎜⎜⎜⎜⎜⎜⎜⎝

2 −1 0 · · · 0 0−1 2 −1 · · · 0 00 −1 2 · · · 0 0...

......

......

0 0 0 · · · 2 −10 0 0 · · · −1 1

⎞

⎟⎟⎟⎟⎟⎟⎟⎟⎠

(T+1)×(T+1)

. (3.13)


By direct verifications, we may find that A is a positive definite matrix. Let η1, η2, . . . ,ηT+1 be the orthonormal eigenvectors corresponding to the eigenvalues λ1, λ2, . . . , λT+1 of A,where 0 < λ1 < λ2 < · · · < λT+1.

For convenience, we list the following assumptions.

(C1) There exists μ ∈ (0, 2) such that lim sup|x|→∞(F(t+ν−1, x)/|x|μ) < a for t ∈ [0, T]N0

,where a is a constant.

(C2) There is a constant μ > 2 such that lim inf|x|→∞(F(t+ν−1, x)/|x|μ) > 0 for t ∈ [0, T]N0

.

(C3) There exists a constant d > 0 such that lim sup|x|→ 0(F(t + ν − 1, x)/|x|2) < d fort ∈ [0, T]

N0.

(C4) F(t + ν − 1, x) satisfies limx→ 0(F(t + ν − 1, x)/|x|2) = q1 > 0 for t ∈ [0, T]N0

, where q1

is a constant.

(C5) f(t + ν − 1, x) is odd with respect to x, that is, f(t + ν − 1,−x) = −f(t + ν − 1, x), fort ∈ [0, T]

N0, and x ∈ R.

(C6) There is a positive constant p2 such that lim infx→ 0(F(t + ν − 1, x)/|x|2) > p2 fort ∈ [0, T]

N0.

Theorem 3.1. If (C1) holds, then for all λ > 0, BVP (1.2), (1.3) has at least one solution.

Proof. By (C1), we obtain

F(t + ν − 1, x) ≤ a|x|μ + b, t ∈ [0, T]N0, |x| ≥ ς, (3.14)

where ς is some sufficiently large numbers and b > 0. Thus, by the continuity of F(t+ν − 1, x)− a|x|μ on [0, T]

N0× [−ς, ς], there exists a′ > 0 such that

F(t + ν − 1, x) ≤ a|x|μ + a′, (t, s) ∈ [0, T]N0

× R. (3.15)

Combining with (3.3)–(3.15), we have

I(x) =12

T∑

t=−1

(tΔν

ν−1x(t))2 − λ

T∑

t=−1

F(t + ν − 1, x(t + ν − 1))

=12

T∑

t=−1

(Δz(t + ν − 1))2 − λT∑

t=−1

F(t + ν − 1, x(t + ν − 1))

=12

T−1∑

t=−1

(Δz(t + ν − 1))2 − λT∑

t=0

F(t + ν − 1, x(t + ν − 1))


≥ 12λ1‖z‖2 − λ

T∑

t=0

F(t + ν − 1, x(t + ν − 1))

≥ 12λ1‖z‖2 − λ|a|

T∑

t=0|x(t + ν − 1)|μ − a′λ(T + 1)

≥ 12λ1‖z‖2 − λ|a|(cμ

)μ‖x‖μ − a′λ(T + 1)

≥ 12λ1‖z‖2 − λ|a|(cμ

)μλ(1/2)μmax ‖z‖μ − a′λ(T + 1).

(3.16)

So, in view of our assumption μ ∈ (0, 2), we see that, for λ > 0, I(x) → ∞ as ‖x‖ → ∞,that is, I(x) is a coercive map. In view of Lemma 2.6, we know that there exists at least onex ∈ Ω such that I ′(x) = 0; hence BVP (1.2), (1.3) has at least one solution. The proof iscompleted.

Remark 3.2. If μ = 2 and a > 0, from the proof of Theorem 3.1, we can get that, for λ ∈(0, λ1/2|a|λmax), our functional I is also coercive.

Theorem 3.3. If (C2) holds, then for all λ > 0, BVP (1.2), (1.3) has at least one solution.

Proof. Similar to the proof Theorem 3.1, we have

I(x) ≤ 12λT+1‖z‖2 − λ

T∑

t=0

F(t + ν − 1, x(t + ν − 1)). (3.17)

By (C2), there exists ς > 0 and ς′ > 0 such that F(t + ν − 1, x) ≥ ς|x|μ for |x| > ς′ witht ∈ [0, T]

N0, so

F(t + ν − 1, x) ≥ ς|x|μ − c, (t, x) ∈ [0, T]N0

× R, (3.18)

where c > 0. Since F(t+ ν − 1, x)− ς|x|μ is continuous on [0, T]N0

× [−ς′, ς′], through (3.17), weobtain

I(x) ≤ 12λT+1‖z‖2 − λς(cμ

)μλμ/2min‖z‖μ + cλ(T + 1). (3.19)

Thus I(x) → −∞ as ||x|| → ∞ for μ > 2. That is, I(x) is an anticoercive. In view of Lemma 2.6,we know that there exists at least one x ∈ Ω such that I ′(x) = 0; hence BVP (1.2), (1.3) has atleast one solution. The proof is completed.

Theorem 3.4. Assume (C2) and (C3) hold. Then, for λ ∈ (0, λ1/2dλmax), the BVP (1.2), (1.3)possesses at least two nontrivial solutions.

Proof. First, we know from Theorem 3.3 that I(x) → −∞ as ||x|| → ∞. Clearly, Ω is a realreflexive finite dimensional Banach space and I ∈ C1(Ω,R), so functional I is weakly upper


semicontinuous. By Lemma 2.6, there exists x0 ∈ Ω such that I(x0) = supΩI and I ′(x0) = 0.Set c0 = supΩI. Let {xn} ⊂ Ω, such that there exists M > 0 and |I(xn)| ≤ M for n ∈ N. By (C2)and (3.19), we may see that

−M ≤ I(xn) ≤ 12λT+1‖zn‖2 − λμ/2

minλς(cμ)μ‖zn‖μ + cλ(T + 1), (3.20)

that is,

−M − cλ(T + 1) ≤ 12λT+1‖zn‖2 − λμ/2

minλς(cμ)μ‖zn‖μ. (3.21)

In view of μ > 2, we see that {zn} ⊂ Ω is bounded, and hence {xn} ⊂ Ω is bounded.Since Ω is finite dimensional, there is a subsequence of {xn}, which is convergent in Ω.Therefore, the (PS) condition is verified.

By (C3), there exists δ > 0, F(t + ν − 1, x) ≤ dx2 for |x| ≤ δ, t ∈ [0, T]N0

. Thus, for x ∈ Ωwith ||x|| ≤ δ, we have

I(x) =12

T−1∑

t=−1

(Δz(t + ν − 1))2 − λT∑

t=0

F(t + ν − 1, x(t + ν − 1))

≥ 12λ1‖z‖2 − λ

T∑

t=0

F(t + ν − 1, x(t + ν − 1))

≥(

12λ1 − λmaxλd

)‖z‖2.

(3.22)

For λ ∈ (0, λ1/2dλmax), we choose ρ = δ and γ = ((1/2)λ1 − λλmax)ρ2. Then we haveI|∂Bρ ≥ γ > 0, so that the condition (I1) in Lemma 2.7 holds.

Since I(x) → −∞ as ||x|| → ∞, we can find e ∈ Ω with sufficiently large norm ||e||such that I(e) < 0. Hence (I2) in Lemma 2.7 is satisfied. Thus, functional I has one criticalvalue

c = infg∈Γ

maxu∈g([0,1])

I(u), (3.23)

where Γ = {g ∈ C([0, 1],Ω) | g(0) = θ, g(1) = e}. If c0 > c, the proof is completed. It suffices toconsider the case c0 = c. Then

c0 = c = infg∈Γ

maxu∈g([0,1])

I(u), (3.24)

that is, c0 = maxu∈g([0,1])I(u) for each g ∈ Γ.Similarly, we can also choose −e ∈ Ω such that I(−e) < 0. Applying Lemma 2.7 again,

we obtain another critical value of the functional I,

c = infg∈Γ

maxu∈g([0,1])

I(u), (3.25)


where Γ = {g ∈ C([0, 1],Ω) | g(0) = θ, g(1) = −e}. If c0 > c, then the proof is completed. Itsuffices to consider the case where c0 = c. Then c0 = maxu∈g([0,1])I(u) for each g ∈ Γ. By thedefinitions of Γ and Γ, we may choose g0 ∈ Γ and g0 ∈ Γ such that g0([0, 1]) ∩ g0([0, 1]) ={θ}. Therefore, we get the maximum of the functional I on g0([0, 1]) \ {θ} and g0([0, 1]) \{θ}, respectively, that is, we find two distinct nontrivial critical points of the functional I.Therefore, our BVP (1.2), (1.3) possesses at least two nontrivial solutions.

Theorem 3.5. Assume that (C1) and (C4) hold and that there exists N ∈ [1, T + 1]N1

such thatλN < λN+1. Then, for λ ∈ (λN/2q1λmin, λN+1/2q1λmax), BVP (1.2), (1.3) has at least three solutions.

Proof. By (C1) and Theorem 3.1, we obtain lim||x||→∞I(x) = ∞, thus functional I is boundedfrom below. Similar to the proof of (PS) condition in Theorem 3.4, we can verify thatfunctional I satisfies (PS) condition in our hypothesis. In order to apply linking theorem, weprove functional I is local linking at origin θ as follows. Clearly, Ω = span{η1, η2, . . . , ηT+1}.Let X = span{η1, η2, . . . , ηN},W = span{ηN+1, ηN+2, . . . , ηT+1}, then Ω = X ⊕ W.

By (C4), for ε ∈ (0, q1), there exists ρ > 0, such that

(q1 − ε

)x2 ≤ F(t + ν − 1, x) ≤ (q1 + ε

)x2, |x| ≤ ρ, t ∈ [0, T]

N0. (3.26)

So, for x ∈ X with 0 < ||x|| ≤ ρ, such that

T−1∑

t=−1

(Δz(t + ν − 1))2 = z′Az ≤ λN‖z‖2 ≤ λN‖x‖2

λmin,

T∑

t=0

F(t + ν − 1, x(t + ν − 1)) ≥ (q1 − ε) T∑

t=0(x(t + ν − 1))2 =

(q1 − ε

)‖x‖2.

(3.27)

Since z = Bx ∈ X, we have

I(x) =12

T−1∑

t=−1

(Δz(t + ν − 1))2 − λT∑

t=0

F(t + ν − 1, x(t + ν − 1))

≤(

λN2λmin

− λ(q1 − ε))‖x‖2.

(3.28)

Thus, for λ > λN/2(q1 − ε)λmin, we have I(x) < 0 for x ∈ X with 0 < ||x|| ≤ ρ.Similarly, for x ∈ W with 0 < ||x|| ≤ ρ,

I(x) =12

T−1∑

t=−1

(Δz(t + ν − 1))2 − λT∑

t=0

F(t + ν − 1, x(t + ν − 1)) ≥(λN+1

2λmax− λ(q1 + ε

))‖x‖2,

(3.29)

then for λ < λN+1/2(q1 + ε)λmax, we have I(x) > 0 for x ∈ W with 0 < ||x|| ≤ρ. So, by Lemma 2.5, for ε ∈ (0, q1), if λ ∈ (λN/2(q1 − ε)λmin, λN+1/2(q1 + ε)λmax),functional I possesses at least three critical points. By the arbitrariness of ε, we get for λ ∈(λN/2q1λmin, λN+1/2q1λmax), the problem (1.2), (1.3) possesses at least three solutions.


Theorem 3.6. Assume (C2), (C3), and (C5) hold. Then, for each N ∈ [0, T]N0, if λ ∈ (0, λN+1/

2dλmax), then BVP (1.2),(1.3) possesses at least T + 1 −N pairs of solutions.

Proof. By (C5), functional I is even, and based on the proof of Theorem 3.4, we know thatI satisfies (PS) condition. In order to obtain our result, we need to verify (I3) and (I4) ofLemma 2.8.

First, in view of (C3), there exists ρ > 0 such that

F(t + ν − 1, x) ≤ dx2 for |x| ≤ ρ, t ∈ [0, T]N0. (3.30)

For N ∈ [1, T + 1]N1

, if we choose X1 = span{ηN+1, ηN+2, . . . , ηT+1}, then codim X1 =N.So for x ∈ X1 with ||x|| ≤ ρ, since z = Bx, we have

I(x) =12

T−1∑

t=−1

(Δz(t + ν − 1))2 − λT∑

t=0

F(t + ν − 1, x(t + ν − 1)) ≥(λN+1

2λmax− λd

)‖x‖2. (3.31)

Thus, for λ ∈ (0, λN+1/2dλmax), I|X1⋂∂Bρ ≥ β > 0, where β = (λN+1/2λmax − λd)ρ2, (I3)

of Lemma 2.8 holds.Next if we choose X2 = span{η1, η2, . . . , ηT+1}, then for x ∈ X2, in view of (C2) and

Theorem 3.3, we get I(x) → −∞ as ||x|| → ∞. (I4) of Lemma 2.8 is satisfied.Therefore, for λ ∈ (0, λN+1/2dλmax), functional I possesses at least T + 1 −N pair of

critical points in Ω, and problem (1.2),(1.3) has at least T + 1 −N pairs of solutions.

Remark 3.7. In Theorem 3.6, if we choose N = 0, then for λ ∈ (0, λ1/2dλmax), the BVP (1.2),(1.3) possesses at least T + 1 pairs of solutions.

Obviously, compared with Theorem 3.4, the even condition (C5) ensures that theproblem (1.2), (1.3) possesses more solutions.

Theorem 3.8. Suppose (C1), (C5), and (C6) hold. Then for everyN ∈ [1, T + 1]N1, when λ ∈ (λN/

2p2λmin,∞), problem (1.2), (1.3) possesses at leastN pairs of nontrivial solutions.

Proof. I(x) is an even functional on Ω by (C5). From (C1), we obtain I(x) → ∞ as ||x|| → ∞,so it is clear that I is bounded from below on Ω and satisfies the (PS) condition. For N ∈[1, T +1]

N1, if we choose X1 = span{η1, . . . , ηN} and set K = X1∩∂Bρ, then K is homeomorphic

to SN−1 by an odd map. By (C6), there exists ρ1 > 0 such that F(t + ν − 1, x) ≥ p2x

2 for |x| ≤ ρ1,t ∈ [0, T]

N0. So for x ∈ K1 = X1 ∩ ∂Bρ1 ,

I(x) =12

T∑

t=−1

(Δz(t + ν − 1))2 − λT∑

t=−1

F(t + ν − 1, x(t + ν − 1)) ≤ 12λN‖z‖2 − λp2‖x‖2

=(

λN2λmin

− λp2

)ρ2

1.

(3.32)

For λ ∈ (λN/2p2λminx,+∞), we have supK1I(x) < 0. Therefore, by Lemma 2.9, func-

tional I has at least N pairs of nontrivial solutions.


Remark 3.9. From Theorem 3.5, it is easy to see that, when f is odd about the second variable,we can obtain more solutions of the problem (1.2), (1.3), and the number of solutions dependson where λ lies.

4. Applications

In the final section, we apply the results developed in Section 3 to some examples.

Example 4.1. Consider the following problem

TΔνt−1

(tΔν

ν−1x(t))= λ4(t + ν − 1)x3(t + ν − 1)(sinx(t + ν − 1) + 2)

+ (t + ν − 1)x4(t + ν − 1) cos(x(t + ν − 1)), t ∈ [0, T]N0,

x(ν − 2) =[tΔν

ν−1x(t)]t=T = 0,

(4.1)

where f(t + ν − 1, x) = 4(t + ν − 1)x3(sinx + 2) + (t + ν − 1)x4 cosx. Choose μ = 4 and d = 1 in(C2) and (C3). Since

lim inf|x|→∞

F(t + ν − 1, x)

|x|4= lim inf

|x|→∞(t + ν − 1)(sinx + 2) = t + ν − 1 > 0, t ∈ [0, T]

N0,

lim sup|x|→ 0

F(t + ν − 1, x)x2

= lim sup|x|→∞

(t + ν − 1)x2(sinx + 2) = 0, t ∈ [0, T]N0,

(4.2)

we see that (C2) and (C3) hold. Thus, by Theorem 3.4, when λ ∈ (0, λ1/2λmax), problem (4.1)has at least two nontrivial solutions.

Example 4.2. Consider the problem

TΔνt−1

(tΔν

ν−1x(t))= λ4 sin(x(t + ν − 1)) cos(x(t + ν − 1))

− e−(t+ν−1)x(t+ν−1)

×(

2x(t + ν − 1) − (t + ν − 1)x2(t + ν − 1))

t ∈ [0, T]N0,

x(ν − 2) =[tΔν

ν−1x(t)]t=T = 0.

(4.3)

Suppose there exists N0 ∈ [0, T]N0

such that λmaxλN0 < λminλN0+1. If we choose μ = q1 = a = 1in (C1) and (C4), then for t ∈ [0, T]

N0, we have

lim sup|x|→∞

F(t + ν − 1, x)|x|μ = lim sup

|x|→∞

2sin2x − x2e−(t+ν−1)x

|x|μ < 1 = a,

limx→ 0

F(t + ν − 1, x)x2

= 1 = q1 > 0,

(4.4)


and hence (C1) and (C4) are satisfied. So, in view of Theorem 3.5, for λ ∈ (λN0/2λmin, λN0+1/2λmax), problem (4.3) has at least three solutions.

Example 4.3. Consider the problem

TΔνt−1

(tΔν

ν−1x(t))= λ4(t + ν − 1)x3(t + ν − 1)(cosx(t + ν − 1) + 2)

− (t + ν − 1)x4(t + ν − 1) sin(x(t + ν − 1)), t ∈ [0, T]N0,

x(ν − 2) =[tΔν

ν−1x(t)]t=T = 0.

(4.5)

Condition (C5) is satisfied. If we choose μ = 4 and d = 1 in (C2) and (C3), then by somesimple calculation, we may show that the hypotheses (C2) and (C3) are fulfilled. Therefore,by Theorem 3.6, for any N ∈ [0, T]

N0and λ ∈ (0, λN+1/2λmax), problem (4.5) has at least

T + 1 −N pairs of solutions.

Example 4.4. Finally, consider the problem

TΔνt−1

(tΔν

ν−1x(t))= λ

1t + ν − 1

sin((t + ν − 1)x(t + ν − 1))

+ x(t + ν − 1) cos((t + ν − 1)x(t + ν − 1)), t ∈ [0, T]N0,

x(ν − 2) =[tΔν

ν−1x(t)]t=T = 0,

(4.6)

where f(t + ν − 1, x) = (1/(t + ν − 1)) sin((t + ν − 1)x) + x cos((t + ν − 1)x). Let a = μ = 1 andp2 = 1/2. Then it is easy to verify that (C1), (C5), and (C6) hold. Thus, by Theorem 3.8, foreach N ∈ [0, T]

N0and λ ∈ (λN/λmin,+∞), problem (4.6) has at least N pairs of solutions.

Acknowledgment

This Project was supported by the National Natural Science Foundation of China (11161049).

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Research ArticleOn Spectral Homotopy Analysis Method forSolving Linear Volterra and FredholmIntegrodifferential Equations

Z. Pashazadeh Atabakan, A. Kılıcman, and A. Kazemi Nasab

Department of Mathematics, Universiti Putra Malaysia (UPM), Selangor, 43400 Serdang, Malaysia

Correspondence should be addressed to A. Kılıcman, [email protected]

Received 12 June 2012; Revised 19 September 2012; Accepted 4 October 2012


Copyright q 2012 Z. Pashazadeh Atabakan et al. This is an open access article distributed underthe Creative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

A modification of homotopy analysis method (HAM) known as spectral homotopy analysismethod (SHAM) is proposed to solve linear Volterra integrodifferential equations. Some examplesare given in order to test the efficiency and the accuracy of the proposed method. The SHAMresults show that the proposed approach is quite reasonable when compared to SHAM results andexact solutions.

1. Introduction

Functional equations such as partial differential equations, integral and integrodifferentialequations and others are widely employed to model complex real-life problems. Manyphysical events in different fields of sciences and engineering have been formulated usingintegrodifferential equations [1, 2]. Integrodifferential equations are usually difficult to solveanalytically so there is a need to obtain an efficient approximate solution [3, 4]. Homotopyanalysis method (HAM) was first proposed by Liao in [5]. it is based on the homotopyconcept in topology for solving nonlinear differential equations. Unlike the traditionalperturbation methods like Lyapunov’s artificial small parameter method [6], Adomiandecomposition method [7–10] and the δ-expansion method [11], which are the special casesof HAM, this approach does not need a small perturbation parameter. In the HAM, originalnonlinear problem is converted to infinite number of linear problems without using theperturbation techniques [12]. Homotopy analysis method is more powerful than traditionalperturbation methods since it is applicable for solving problems with strong nonlinearity[13–15], even if they do not have any small or large parameters.


We can adjust the convergence region and the rate of approximation series to giveus freedom to use different base function to approximate a nonlinear problem. He in [16]proposed the so called Homotopy perturbation method (HPM). Later it was pointed byHayat et al. [17] that HPM is only a special case of the HAM (h = −1). Many nonlinearproblems were solved by Homotopy perturbation method [18, 19]. The HPM does notprovide us with convergent series solution for strongly nonlinear problems as was indicatedby Abbasbandy in [20] using a simple problem in which the physical parameters werelarge. In 2010, Motsa et al. [21] employed Chebyshev polynomials to solve higher-orderdeformation equation and called his approach spectral homotopy analysis method (SHAM).The spectral homotopy analysis method has been used only for solving partial and ordinarydifferential equations [22, 23]. Motsa et al. [21–23] found that the Spectral homotopy analysismethod is more efficient than the homotopy analysis method as it does not depend on the ruleof solution expression and the rule of ergodicity. This method is more flexible than homotopyanalysis method since it allows for a wider range of linear and nonlinear operators and oneis not restricted to use the method of higher order differential mapping for solving boundaryvalue problems in bounded domains, unlike the homotopy analysis method. The range ofadmissible h values is much wider in spectral homotopy analysis method than in homotopyanalysis method. The main restriction of HAM in solving integral equations is to choose thebest initial guess as the series solution be convergent. In SHAM the initial approximation istaken to be the solution of the nonhomogeneous linear part of the given equation.

In this paper, we propose spectral homotopy analysis method (SHAM) to solve linearVolterra and Fredholm type of integrodifferential equations. Volterra integrodifferentialequation is given by

a1(x)u′′(x) + a2(x)u′(x) + a3(x)u(x) = g(x) +∫x

−1k(x, t)u(t)dt,

u(−1) = 0, u(1) = 0.

(1.1)

The paper is organized in the following way. Section 2 included a brief introductionin homotopy analysis method. Spectral homotopy analysis method for solving integralequations is presented in Section 3. We then propose the way to calculate Sm−1 which isneeded to obtain Ym in Section 4. Numerical examples are presented in Section 5. In Section 6,concluding remarks are given.

2. Homotopy Analysis Solution

In this section, we give a brief introduction to HAM. We consider the following differentialequation in a general form:

N[u(τ)] = 0, (2.1)

where N is nonlinear operator, τ denotes independent variables and u(τ) is an unknownfunction, respectively. For simplicity we disregard all initial and all boundary conditions


which can be dealt in similar way. The so-called zero-order deformation equation wasconstructed by Liao as follows:

(1 − p)L[φ(τ ; p

) − u0(τ)]= phH(τ)

(N[φ(τ ; p)])

, (2.2)

where p ∈ [0, 1] is the embedding parameter and, h is a nonzero convergence-parameter,H(τ) is an auxiliary function and u0(τ) is called an initial guess of u(τ) and φ(τ ; p) is anunknown function. In addition, L is an auxiliary linear operator and N is nonlinear operatoras follows:

L(φ(x; p))

= a1(x)∂2φ(x; p)

∂x2, (2.3)

with the property L(c1t + c2) = 0 where c1 and c2 are constants and

N[φ(x; p)]

= a1(x)∂2φ(x; p)

∂x2+ a2(x)

∂φ(x; p)

∂x+ a3(x)φ

(x; p) − g(x) −

∫x

−1k(x, t)φ(t)dt

(2.4)

is a nonlinear operator. Obviously, when p = 0 and p = 1, it holds φ(τ ; 0) = u0(τ) and φ(τ ; 1) =u(τ). In this way, as p increase from 0 to 1, φ(τ ; p) alter from initial guess u0(τ) to the solutionu(τ) and φ(τ ; p) is expanded in Taylor series with respect to p:

φ(τ ; p)= u0(τ) +

+∞∑

m=1

um(τ)pm, (2.5)

where

um(τ) = Dm

[φ(τ ; p)], (2.6)

Dmφ =1m!

∂m(φ)

∂pm

∣∣∣∣∣p=0

. (2.7)

The series (2.5) converges at p = 1 if the auxiliary linear operator, the initial guess, theconvergence parameter, and the auxiliary function are properly selected:

φ(τ) = u0(τ) ++∞∑

m=1

um(τ). (2.8)

The admissible and valid values of the convergence-parameter h are found from thehorizontal portion of the h-curves. Liao proved that u(τ) is one of the solutions of originalnonlinear equation. As H(τ) = 1 so (2.2) becomes

(1 − p)L[φ(τ ; p

) − u0(τ)]= ph

(N[φ(τ ; p)])

. (2.9)


Define the vector um = {u0(τ), u1(τ), . . . , um(τ)}. Operating on both sides with Dm, wehave the so called mth-order deformation equation

L[um(τ) − χmum−1(τ)

]= hH(τ)Rm(um−1(τ)), (2.10)

where

Rm(um−1) =1

(m − 1)!∂m−1N

[φ(τ ; p)]

∂pm−1

∣∣∣∣∣p=0

,

χm =

{0, m ≤ 11, otherwise.

(2.11)

um(τ) for m ≥ 0 is governed by the linear equation (2.10) and can be solved by symboliccomputation software such as Maple, Matlab, and so on.

3. Spectral Homotopy Analysis Solution

Consider the second-order Volterra integrodifferential equation

a1(x)u′′(x) + a2(x)u′(x) + a3(x)u(x) = g(x) +∫x

−1k(x, t)u(t)dt, u(−1) = 0, u(1) = 0.

(3.1)

To obtain initial approximation we solve the following two-point boundary valueproblem:

a1(x)u′′0(x) + a2(x)u′0(x) + a3(x)u0(x) = g(x), (3.2)

subject to boundary conditions

u0(−1) = 0, u0(1) = 0. (3.3)

We use the Chebyshev pseudospectral method to solve (3.2). At first we approximate u0(τ)by a truncated series of Chebyshev polynomial of the form

u0(τ) ≈ uN0(τj)=

N∑

k=0

ukTk(τj), j = 0, . . . ,N, (3.4)

where Tk is the kth Chebyshev polynomials, uk, are coefficients and τ0, τ1, . . . , τN are Gauss-Lobatto points which are the extrema of the Nth-order Chebyshev polynomial, defined by

τj = cos(πj

N

). (3.5)


Derivatives of the functions y0(τ) at the collocation points are represented as

dsu0(τk)dτs

=N∑

j=0

Dskju0(τj), k = 0, . . . ,N, (3.6)

where s is the order of differentiation and D is the Chebyshev spectral differentiation matrix.By following [24] we express the entries of the differentiation matrix D as

Dkj =(−1

2

)ckcj

(−1)k+j

sin(π(j + k

)/2N

)sin(π(j − k)/2N

) , j /= k,

Dkj =(−1

2

)xk

sin2(πk/N), k /= 0,N, k = j,

D00 = −DNN =2N2 + 1

6.

(3.7)

Substituting (3.4)–(3.6) into (3.2) results in

AU0 = G (3.8)

subject to the boundary conditions

u0(τ0) = u0(τN) = 0, (3.9)

where

A = a1D2 + a2D + a3,

U0 = [u0(τ0), u0(τ1), . . . , u0(τN)]T ,

G =[g(τ0), g(τ1), . . . , g(τN)

]T,

ar = diag(ar(τ0), ar(τ1), . . . ar(τN)), r = 1, 2, 3.

(3.10)

To satisfy the boundary conditions we delete the first and the last rows and columns ofA and the first and the last rows of Y0 and G. The values of u0(τi), i = 0, . . . ,N are determinedfrom the equation

U0 = A−1G, (3.11)

which is the initial approximation for the SHAM solution of the governing equation (3.1).The zeroth-order deformation equation is given by

(1 − p)L[φ(τ ; p

)u0(τ)

]= ph

(N[φ(τ ; p)] − g(τ)), (3.12)


where

L[φ(τ ; p)]

= a1∂2φ

∂τ2+ a2

∂φ

∂τ+ a3φ, (3.13)

h is the nonzero convergence controlling auxiliary parameter, and N is a nonlinear operatorgiven by

N[φ(τ ; p)]

= a1∂2φ

∂τ2+ a2

∂φ

∂τ+ a3φ −

∫ τ

−1k(τ, t)φ(t)dt. (3.14)

Differentiating (3.2) m times with respect to the embedding parameter p and thensetting p = 0 and finally dividing them by m!, we have the so-called mth-order deformationequation

L[um(τ) − χmum−1(τ)

]= hRm (3.15)

subject to boundary conditions

um(−1) = um(1) = 0, (3.16)

where

Rm(τ) = a1u′′m−1 + a2u

′m−1 + a3um−1 − g(τ)

(1 − χm

) −∫ τ

−1k(τ, t)um−1(t)dt. (3.17)

Applying the Chebyshev pseudospectral transformation on (3.15)–(3.17) results in

AUm =(χm + h

)AUm−1 − h

[Sm−1 +

(1 − χm

)G], (3.18)

subject to the boundary conditions

um(τ0) = um(τN) = 0, (3.19)

where A and G are as defined in (3.10) and

Um = [um(τ0), um(τ1), . . . , um(τN)]T , Sm−1 =∫ τ

−1k(τ, t)Um−1dt. (3.20)

To implement the boundary conditions we delete the first and the last rows of Sm−1 and thefirst and the last rows and columns of A.

Finally, this recursive formula can be written as following for m ≥ 1:

Um =(χm + h

)A−1AUm−1 − hA−1[Sm−1 +

(1 − χm

)G], (3.21)


so, starting from the initial approximation. We can obtain higher-order approximation Um form ≥ 1, recursively.

4. Calculation of Sm−1

Consider the well-known Chebyshev polynomials of first kind of degree n, defined by [25]

Tn(x) = cos(n cos−1(x)

), n ≥ 0. (4.1)

Also they are derived from the following recursive formula:

T0(x) = 1, T1(x) = x, Tn+1(x) = 2xTn(x) − Tn−1(x), n = 1, 2, 3, . . . (4.2)

In this section we approximate the integrand in (3.15) by letting

k(τm, t, ym−1(t)

)=

m−1∑

j=0

Tj(t)wmj, t = τ0, τ1, . . . , τm−1, m = 1, 2, . . . ,N, (4.3)

where Tj(t), j = 0, 1, . . . , m − 1 are the basis functions.Clearly, wmj , 0 ≤ j ≤ m − 1 can be obtained by solving the following matrix equation:

⎛

⎜⎜⎜⎝

T0(τ0) T1(τ0) . . . Tm−1(τ0)T0(τ1) T1(τ1) . . . Tm−1(τ1)

......

...T0(τm−1) T1(τm−1) . . . Tm−1(τm−1)

⎞

⎟⎟⎟⎠

⎛

⎜⎜⎜⎝

wm,0

wm,1...

wm,m−1

⎞

⎟⎟⎟⎠

=

⎛

⎜⎜⎜⎝

k(tm, τ0, y0

)

k(tm, τ1, y1

)

...k(tm, τm−1, ym−1

)

⎞

⎟⎟⎟⎠, (4.4)

where τj = tj , j = 0, 1, . . . , m − 1 are Chebyshev collocation points defined in (2.6). We canobtainwmj by solving the above system so the integral in (3.15) can be rewritten as following:

∫ τ

−1k(τ, t)ym−1(t)dt =

m−1∑

j=0

am,jwm,j , (4.5)

where

am,j =∫ τ

−1Tj(t)dt = PTj(τ), (4.6)


Table 1: Numerical result of Example 5.1 against the HAM and the SHAM solutions whit h = −0.85.

xSHAM HAM Numerical

2nd order 4th order 5th order 2nd order 5th order 11th order−0.8 −0.360376 −0.360001 −0.36000000 −0.370525 −0.360086 −0.36000000 −0.36000000−0.6 −0.640699 −0.640003 −0.64000000 −0.658320 −0.640123 −0.64000000 −0.64000000−0.4 −0.840976 −0.840005 −0.84000000 −0.860438 −0.840105 −0.84000000 −0.84000000−0.2 −0.961198 −0.960005 −0.96000000 −0.979684 −0.960078 −0.96000000 −0.960000000.0 −1.001354 −1.000005 −1.00000000 −1.018938 −1.000066 −1.00000000 −1.000000000.2 −0.961421 −0.960005 −0.96000000 −0.978828 −0.960067 −0.96000000 −0.960000000.4 −0.841370 −0.840005 −0.84000000 −0.857858 −0.840073 −0.84000000 −0.840000000.6 −0.641159 −0.640004 −0.64000000 −0.654041 −0.64048 −0.64000000 −0.640000000.8 −0.360727 −0.360002 −0.36000000 −0.367054 −0.360011 −0.36000000 −0.36000000

where P as in [26], that is the (N + 1) × (N + 1) operational matrix for integration as follows:

P =

⎛

⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

1 1 0 0 . . . 0 0

−14

0 −14

0 . . . 0 0

−13

−12

016. . . 0 0

18

0 −14

0 . . . 0 0...

......

.... . .

......

(−1)N

(N − 1)2 − 10 0 0 . . . 0

12N

(−1)N+1

N2 − 10 0 0 . . . − 1

2N − 20

⎞

⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

. (4.7)

5. Numerical Examples

In this section we apply the technique described in Section 3 to some illustrative examples ofsecond-order Volterra integrodifferential equations.

Example 5.1. Consider the second-order Volterra integrodifferential equation

y′′(x) − xy′(x) + xy(x) =x4

6+ x3 − 2x2 − x

3+

52+∫x

−1(x − 2t)y(t)dt (5.1)

subject to the boundary conditions y(−1) = y(1) = 0 with the exact solution y(x) = x2 − 1.

We employ HAM and SHAM to solve this example. In Table 1, there is a comparisonof the numerical result against the HAM and SHAM approximation solutions at differentorders. It is worth noting that the SHAM results become very highly accurate only with a fewiterations and fourth-order results become very close to the exact solution. Comparison of thenumerical solution with the 4th-order SHAM solution for h = −0.85 is made in Figure 1. It


−1 −0.5 0 0.5 1

−0.2

−0.4

−0.6

−0.8

−1

Exact solutiony4

x

Figure 1: Comparison of the numerical solution with the 4th-order SHAM solution for h = −0.85.

4

2

0

−2

−4

−1 −0.5 0 0.5 1

u′′(1)u′′′(1)

h

Figure 2: 4th-order SHAM h-curve.

can be seen from the h-curves (Figures 2 and 3) that if −1 ≤ h ≤ −0.2 and −1 ≤ h ≤ −0.4 thenthe SHAM an HAM approximate solutions are in well agreement with the exact solution. InTable 2, we present that the SHAM results for another value of h in h (h = −0.9), are alsovery highly accurate, and in sixth order we obtain the exact numerical solutions. In HAM wechoose y0(x) = (x2 − 1)/2 as initial guess.


4

2

0

−2

−4

u′′(1)u′′′(1)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

h

Figure 3: 11th-order HAM h-curve.

Table 2: Numerical result of Example 5.1 against the SHAM solutions whit h = −0.9.

x 5th order 6th order 7th order Numerical−0.8 −0.35999595 −0.36000000 −0.36000000 −0.36000000−0.6 −0.63999432 −0.64000000 −0.64000000 −0.64000000−0.4 −0.83999202 −0.84000000 −0.84000000 −0.84000000−0.2 −0.95999009 −0.96000000 −0.96000000 −0.960000000.0 −0.99998868 −1.00000000 −1.00000000 −1.000000000.2 −0.95998890 −0.96000000 −0.96000000 −0.960000000.4 −0.83998833 −0.84000000 −0.84000000 −0.840000000.6 −0.63999070 −0.64000000 −0.64000000 −0.640000000.8 −0.35999373 −0.36000000 −0.36000000 −0.36000000

Example 5.2. Consider the second-order Volterra integrodifferential equation

y′′(x) + y(x) = −4x6 + 6x4 + x3 + 5x − 2 +∫x

−124t2y(t)dt (5.2)

subject to the boundary conditions y(−1) = y(1) = 0 with the exact solution y(x) = x3 − x.

Similar to the previous example, we employ HAM and SHAM to solve this example.In Table 3, there is a comparison of the numerical results against the HAM and SHAMapproximation solutions (5.2) at different orders. It is worth noting that the rate ofconvergence in SHAM results is higher than the HAM results. Comparison of the numericalsolution with the 16th-order SHAM solution for h = −0.4 is made in Figure 4. It can beseen from the h-curves (Figures 5 and 6) that if −0.4 ≤ h ≤ 0 and −0.6 ≤ h ≤ −0.3 then


Table 3: Numerical result of Example 5.2 against the HAM and the SHAM solutions whit h = −0.4.

xSHAM Numerical HAM

8th order 14th order 16th order 8th order 23th order−0.8 0.28818730 0.28800272 0.28800000 0.28800000 0.28786022 0.28800097−0.6 0.38426358 0.38400174 0.38400000 0.38400000 0.38548670 0.3840003−0.4 0.33618044 0.33599640 0.33600000 0.33600000 0.34009630 0.33600088−0.2 0.19199056 0.19198946 0.19200000 0.19200000 0.19905046 0.19200140.0 −0.00024441 −0.00000165 0.00000000 0.00000000 −0.00041662 0.000001660.2 −0.19249771 −0.19202092 −0.19200000 −0.19200000 −0.18206491 −0.192098840.4 −0.33673440 −0.33602266 −0.33600000 −0.33600000 −0.32825115 −0.336001190.6 −0.36072793 −0.38401942 −0.38400000 −0.38400000 −0.38049994 −0.384001530.8 −0.28863110 −0.28801009 −0.28800000 −0.28800000 −0.28821582 −0.28800159

the approximate solution is in well agreement with the exact solution. In HAM, we choosey0(x) = −x2 + 1 as initial guess.

Example 5.3. Consider the first-order Fredholm integrodifferential equation [27]

y′(x) − y(x) = sin(4πx) − cos(2πx) − 2π sin(2πx) −∫1

−1sin(4πx + 2πt)y(t)dt (5.3)

subject to the boundary conditions y(−1) = y(1) = 1.

In this example, by a change of variables, the Fredholm-integro differential equationwith nonhomogeneous conditions are transformed to the equation with homogeneousconditions and then apply the spectral homotopy analysis method for this problem.According to governing equation and the boundary condition it is reasonable to set thetransformation

y(x) = cos(2πx) + Y (x) (5.4)

to make the governing boundary conditions homogeneous. Substituting (5.4) in (5.3) andboundary conditions gives

Y ′(x) − Y (x) =∫1

−1sin(4πx + 2πt)Y (t)dt (5.5)

subject to the boundary condition

Y (−1) = Y (1) = 0. (5.6)

Comparison between absolute errors in solutions by SHAM and Legendre collocationmatrix method (LCMM) is tabulated in Table 4. It is worth noting that the SHAM resultsbecome very highly accurate only with four iterations.


0.3

0.2

0.1

−0.1

−0.2

−0.3

−1 −0.5 0 0.5 1

Exact solutiony16

x

Figure 4: Comparison of the numerical solution with the 16th-order SHAM solution for h = −0.4.

8

6

4

2

0

−2

−4

−6

−8−0.6 −0.4 −0.2 0 0.2 0.4 0.6

h

u′′(1)u′′′(1)

Figure 5: 16th-order SHAM h-curve3.

Example 5.4. Consider the following second-order Fredholm integrodifferential equation [28]:

y′′(x) + xy′(x) − xy(x) = ex − 2 sin(x) +∫1

−1sin(x)e−ty(t)dt, (5.7)

where y(−1) = e−1, y(1) = e.

In this example, by a change of variables, the Fredholm-integro differential equationwith nonhomogeneous conditions are transformed to the equation with homogeneousconditions and then we apply the spectral homotopy analysis method for this problem.


−0.6 −0.4 −0.2 0 0.2 0.4 0.6

h

10

5

0

−5

−10

u′′(1)u′′′(1)

Figure 6: 23th-order HAM h-curve.

Table 4: Absolute errors for Example 5.3.

x SHAM LCMM−1.0 3.30 × 10−19 2.49 × 100

−0.8 1.73 × 10−9 3.56 × 10−1

−0.6 6.69 × 10−9 2.66 × 10−1

−0.4 2.91 × 10−10 5.85 × 10−2

−0.2 6.93 × 10−9 1.74 × 10−2

0.0 9.16 × 10−9 0.000.2 6.93 × 10−9 1.97 × 10−2

0.4 2.91 × 10−10 7.50 × 10−2

0.6 6.69 × 10−9 2.59 × 10−2

0.8 1.73 × 10−9 4.16 × 10−1

1.0 6.38 × 10−17 2.58 × 100

According to governing equation and the boundary condition it is reasonable to set thetransformation

y(x) = ex + Y (x) (5.8)

to make the governing boundary conditions homogeneous. Substituting (5.8) in (5.7) andboundary conditions gives

Y ′′(x) + xY ′(x) − xY (x) =∫1

−1sin(x)e−tY (t)dt (5.9)


Table 5: Absolute errors for Example 5.4.

xSHAM HAM

4th order 20th order0.0 0 00.2 0 2.2760 × 10−14

0.4 0 3.5971 × 10−13

0.6 1 × 10−19 1.8209 × 10−12

0.8 1 × 10−19 5.7547 × 10−12

1.0 0 1.4550 × 10−11

1.2 1.2241 × 10−15 2.9134 × 10−11

1.4 1.1239 × 10−13 5.3974 × 10−11

subject to boundary condition

Y (−1) = Y (1) = 0. (5.10)

Comparison between absolute errors in solutions by SHAM and HAM is tabulated inTable 5. It is worth noting that the SHAM results become very highly accurate only with fouriterations.

6. Conclusion

In this paper for the first time in the literature we presented the application of spectralhomotopy analysis method (SHAM) in solving linear integro differential equations. Acomparison was made between exact analytic solutions and numerical results from thespectral homotopy analysis method and homotopy analysis method (HAM) solutions. Thenumerical results indicate that in SHAM the rate of convergence is faster than HAM. Forexample, in Example 5.1 we found that the fourth-order SHAM approximation sufficientlygives a match with the numerical results up to twelve decimal places.In contrast, HAMsolutions have a good agreement with the numerical results in 11th-order. In this paperwe choose the admissible value of h from h-curve for y′′(1) and y′′′(1). As it is shown inFigures 2 and 3 the range of admissible h values is much wider in SHAM than in HAM. Inusing HAM there is restriction on choosing an initial approximation that the higher-orderdeformation can be integrated whereas in SHAM the initial approximation is the solutionof the nonhomogeneous linear part of (3.1). The spectral homotopy analysis method is moreefficient as it does not depend on the rule of coefficient ergodicity unlike the HAM. With somechanges we can apply this method on Fredholm integrodifferential equations. In Example 5.3[27], we compare the SHAM solution with Legendre collocation matrix method (LCCM)to solve linear Fredholm integrodifferential equation. As can be seen in Table 4 SHAM ismore accurate and efficient than LCCM. In this work we obtained the numerical results up totwelve decimal places having well agreement with exact solution but reported them only upto six decimal places.

In this paper, we described the spectral homotopy analysis method to solve linearVolterra and Fredholm integrodifferential equations; however, it remains to be generalizedand verified for more complicated integral equations that we consider it as future works.


Acknowledgment

The authors gratefully acknowledge that this research was partially supported by theUniversiti Putra Malaysia under the ERGS Grant Scheme having project number 5527068.

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Research ArticleMultiple Solutions for a Class of FractionalBoundary Value Problems

Ge Bin

Department of Mathematics, Harbin Engineering University, Harbin 150001, China

Correspondence should be addressed to Ge Bin, [email protected]

Received 6 March 2012; Accepted 19 September 2012

Academic Editor: Yong H. Wu

Copyright q 2012 Ge Bin. This is an open access article distributed under the Creative CommonsAttribution License, which permits unrestricted use, distribution, and reproduction in anymedium, provided the original work is properly cited.

We study the multiplicity of solutions for the following fractional boundary value problem:(d/dt)((1/2) 0D

−βt (u′(t)) + (1/2) 0D

−βT (u′(t))) + λ∇F(t, u(t)) = 0, a.e. t ∈ [0, T], u(0) = u(T) = 0,

where 0D−βt and 0D

−βT are the left and right Riemann-Liouville fractional integrals of order

0 ≤ β < 1, respectively, λ > 0 is a real number, F : [0, T] × RN → R is a given function, and

∇F(t, x) is the gradient of F at x. The approach used in this paper is the variational method. Moreprecisely, the Weierstrass theorem and mountain pass theorem are used to prove the existence ofat least two nontrivial solutions.

1. Introduction

In this paper, we consider the fractional boundary value problem of the following form:

d

dt

(12 0D

−βt

(u′(t)

)+

12 0D

−βT

(u′(t)

))+ λ∇F(t, u(t)) = 0, a.a. t ∈ [0, T],

u(0) = u(T) = 0,

(P)

where 0D−βt and 0D

−βT are the left and right Riemann-Liouville fractional integrals of order

0 ≤ β < 1, respectively, λ > 0 is a real number, F : [0, T] × RN → R is a given function, and

∇F(t, x) is the gradient of F at x.


In particular, if λ = 1, the problem (P1) reduces to the standard second-order boundaryvalue problem of the following form:

d

dt

(12 0D

−βt

(u′(t)

)+

12 0D

−βT

(u′(t)

))+∇F(t, u(t)) = 0, a.a. t ∈ [0, T],

u(0) = u(T) = 0.(P1)

Fractional calculus and fractional differential equations can find many applicationsin various fields of physical science such as viscoelasticity, diffusion, control, relaxationprocesses, and modeling phenomena in engineering, see [1–12]. Recently, many results wereobtained dealing with the existence and multiplicity of solutions of nonlinear fractionaldifferential equations by use of techniques of nonlinear analysis, such as fixed-point theory(including Leray-Schauder nonlinear alternative) (see [13, 14]), topological degree theory(including coincidence degree theory) (see [15, 16]), and comparison method (includingupper and lower solutions methods and monotone iterative method) (see [17, 18]). However,it seems that the popular methods mentioned above are not appropriate for discussing (P)and (P1), as the equivalent integral equation is not easy to be obtained.

In the past, there were investigations of the eigenvalue problems for fractionaldifferential equations. For more detailed information on this topic, we refer to Zhang et al.[19–21], Wang et al. [22, 23], and Jiang et al. [24].

Recently, there are many papers dealing with the existence of solutions for problem(P1). In [25], Jiao and Zhou obtained the existence of solutions for (P1) by mountainpass theorem under the Ambrosetti-Rabinowitz condition. Chen and Tang [26] studied theexistence and multiplicity of solutions for the system (P1) when the nonlinearity F(t, ·) issuperquadratic, asymptotically quadratic, and subquadratic, respectively.

But so far, few papers discuss the two solutions of the system (P) via critical pointtheory. The aim of the present paper is to study the existence of at least two solutions for thesystem (P) as the parameter λ > λ0 for some constant λ0.

The paper is organized as follows. We first introduce some basic preliminary resultsand a well-known lemma in Section 2, including the fractional derivative space Eα, whereα ∈ (1/2, 1]. In Section 3, we give the main result and its proof. In Section 4, we give thesummary of this paper.

2. Preliminary

In this section, we recall some related preliminaries and display the variational setting whichhas been established for our problem.

Definition 2.1 (see [8]). Let f(t) be a function defined on [a, b] and τ > 0. The left and rightRiemann-Liouville fractional integrals of order τ for function f(t) denoted by aD

−τt f(t) and

tD−τb f(t), respectively, are defined by

aD−τt f(t) =

1Γ(τ)

∫ t

a

(t − s)τ−1f(s)ds, t ∈ [a, b],

tD−τb f(t) =

1Γ(τ)

∫b

t

(t − s)τ−1f(s)ds, t ∈ [a, b],

(2.1)

provided the right-hand sides are pointwise defined on [a, b], where Γ is the gamma function.


Definition 2.2 (see [8]). Let f(t) be a function defined on [a, b]. The left and right Riemann-Liouville fractional derivatives of order τ for function f(t) denoted by aD

τbf(t) and tD

τbf(t),

respectively, are defined by

aDτt f(t) =

dn

dtnaD

τ−nt f(t) =

1Γ(n − τ)

dn

dtn

(∫ t

a

(t − s)n−τ−1f(s)ds

)

,

tDτbf(t) = (−1)n

dn

dtntD

τ−nb f(t) =

1Γ(n − τ)

dn

dtn

(∫b

t

(t − s)n−τ−1f(s)ds

)

,

(2.2)

where t ∈ [a, b], n − 1 ≤ τ < n, and n ∈ N.

The left and right Caputo fractional derivatives are defined via the above Riemann-Liouville fractional derivatives. In particular, they are defined for the function belong-ing to the space of absolutely continuous functions, which we denote by AC([a, b],RN).ACk([a, b],RN) (k = 1, 2, . . .) is the space of functions f such that f ∈ Ck([a, b],RN). Inparticular, AC([a, b],RN) = AC1([a, b],RN).

Definition 2.3 (see [8]). Let τ ≥ 0 and n ∈ N. If τ ∈ [n − 1, n) and f(t) ∈ ACn([a, b],RN),then the left and right Caputo fractional derivatives of order τ for function f(t) denotedby c

aDτt f(t) and c

tDτbf(t), respectively, which exist a.e. on [a, b]. c

aDτt f(t) and c

tDτbf(t) are

represented by

caD

τt f(t) = aD

τ−nt f (n)(t) =

1Γ(n − τ)

(∫ t

a

(t − s)n−τ−1f (n)(s)ds

)

,

ctD

τbf(t) = (−1) nt D

τ−nb f (n)(t) =

1Γ(n − τ)

(∫b

t

(t − s)n−τ−1f (n)(s)ds

)

,

(2.3)

respectively, where t ∈ [a, b].

Definition 2.4 (see [25]). Define 0 < α ≤ 1 and 1 < p < ∞. The fractional derivative space Eα,p0is defined by the closure of C∞

0 ([0, T],RN) with respect to the norm

‖u‖α,p =(∫T

0|u(t)|pdt +

∫T

0

∣∣ c0D

αt u(t)

∣∣pdt

)1/p

, ∀u ∈ Eα,p0 , (2.4)

where C∞0 ([0, T],RN) denotes the set of all functions u ∈ C∞([0, T],RN) with u(0) =

u(T) = 0. It is obvious that the fractional derivative space Eα,p

0 is the space of functionsu ∈ Lp([0, T],RN) having an α-order Caputo fractional derivative c

0Dαt u ∈ Lp([0, T],RN)

and u(0) = u(T) = 0.

Proposition 2.5 (see [25]). Let 0 < α ≤ 1 and 1 < p < ∞. The fractional derivative space Eα,p0 is areflexive and separable space.


Proposition 2.6 (see [25]). Let 0 < α ≤ 1 and 1 < p <∞. For all u ∈ Eα,p0 , one has

‖u‖Lp ≤Tα

Γ(α + 1)

∥∥ c

0Dαt u∥∥Lp. (2.5)

Moreover, if α > 1/p and 1/p + 1/q = 1, then

‖u‖∞ ≤ T (α−1)/p

Γ(α)((α − 1)q + 1

)1/q

∥∥ c

0Dαt u∥∥Lp. (2.6)

According to (2.6), one can consider Eα,p0 with respect to the norm

‖u‖α,p =∥∥ c

0Dαt u∥∥Lp

=

(∫T

0

∣∣ c

0Dαt u∣∣pdt

)1/p

. (2.7)

Proposition 2.7 (see [25]). Define 0 < α ≤ 1 and 1 < p < ∞. Assume that α > 1/p, and thesequence uk converges weakly to u ∈ E

α,p

0 , that is, uk ⇀ u. Then uk → u in C([0, T],RN), that is,‖uk − u‖∞ → 0, as k → ∞.

Making use of Definition 2.3, for any u ∈ AC([0, T],RN), problem (P) is equivalent tothe following problem:

d

dt

(12 0D

α−1t

( c0D

αt u(t)

) − 12 tD

α−1T

( ctD

αTu(t)

))+ λ∇F(t, u(t)) = 0, a.e. t ∈ [0, T],

u(0) = u(T) = 0,

(P2)

where α = 1 − β/2 ∈ (1/2, 1].In the following, we will treat problem (P2) in the Hilbert space Eα = Eα,20 with the

corresponding norm ‖u‖α = ‖u‖α,2. It follows from [25, Theorem 4.1] that the functional ϕgiven by

ϕ(u) =∫T

0

[−1

2( c

0Dαt u(t),

ctD

αTu(t)

)]dt − λ

∫T

0F(t, u(t))dt (2.8)

is continuously differentiable on Eα. Moreover, for u, v ∈ Eα, we have

⟨ϕ′(u), v

⟩= −

∫T

0

12[( c

0Dαt u(t),

ctD

αTv(t)

)+( ctD

αTu(t),

c0D

αt v(t)

)]dt

− λ∫T

0(∇F(t, u(t)), v(t)]dt.

(2.9)

Definition 2.8 (see [25]). A function u ∈ AC([0, T],RN) is called a solution of (P2) if(i) Dα(u(t)) is derivative for a.e. t ∈ [0, T],(ii) u satisfies (P2), where Dα(u(t)) := (1/2) 0D

α−1t (c0D

αt u(t)) − (1/2) tD

α−1T (ctD

αTu(t)).


Proposition 2.9 (see [25]). If 1/2 < α ≤ 1, then for any u ∈ Eα, one has

|cos(πα)|‖u‖2α ≤ −

∫T

0

( c0D

αt u(t),

ctD

αTu(t)

)dt ≤ 1

|cos(πα)| ‖u‖2α. (2.10)

Proposition 2.10 (see [25]). Let 1/2 < α ≤ 1 be satisfied. If u ∈ Eα, then the functional J : Eα → R

denoted by

J(u) = −12

∫T

0

( c0D

αt u(t),

ctD

αTu(t)

)dt (2.11)

is convex and continuous on Eα.

In order to prove the existence of two solutions for problem (P), firstly, we recall somewell-known results. Their proofs can be found in many books. Please refer to the referencesand its references therein.

Lemma 2.11 (see [27]). If X is a Banach space, ϕ ∈ C1(X,R), e ∈ X, and r > 0, such that ‖e‖ > rand

b := inf‖u‖=r

ϕ(u) > ϕ(0) ≥ ϕ(e), (2.12)

and if ϕ satisfies the PS condition, with

c := infγ∈Γ

maxt∈[0,1]

ϕ(γ(t)), Γ :=

{γ ∈ C([0, 1], X) : γ(0) = 0, γ(1) = e

}, (2.13)

then c is a critical value of ϕ.

3. The Main Result and Proof of the Theorem

In this part, we will prove that for (P) there also exist two solutions for the general case.Our hypotheses on nonsmooth potential F(t, x) are as follows.H(F)1: F : [0, T] ×R

N → R is a function such that F(t, 0) = 0 a.e. on [0, T] and satisfiesthe following facts:

(i) for all x ∈ RN , t �→ F(t, x) is measurable,

(ii) for a.a. t ∈ [0, T], x �→ F(t, x) is continuously differentiable,

(iii) there exist c ∈ C([0, T],R) and 0 < α0 < 2, such that

|∇F(t, x)|, |F(t, x)| ≤ c(t)(1 + |x|α0), for a.a. t ∈ [0, T], all x ∈ R

N, (3.1)

(iv) there exist γ > 2 and μ ∈ L∞([0, T]), such that

lim sup|x|→ 0

(∇F(t, x), x)|x|γ < μ(t), uniformly for a.a. t ∈ [0, T], (3.2)


(v) there exist ξ0 ∈ RN , t0 ∈ (0, T), and r0 > 0, such that

F(t, ξ0) > δ0 > 0, a.e. t ∈ Br0(t0), (3.3)

where Br0(t0) := {t ∈ [0, T] : |t − t0| ≤ r0} ⊂ [0, T].

Remark 3.1. It is easy to verify that F : [0, T]×RN → R satisfies the whole assumption in [25,

Theorem 4.1]. So, u ∈ Eα is a solution of the corresponding Euler equation ϕ′(u) = 0, then u isa solution of problem (P2) which, of course, corresponds to the solution of problem (P). Forgreat details, please see [25, Theorem 4.2].

Theorem 3.2. Suppose that H(F)1 holds. Then there exists λ0 > 0 such that for each λ > λ0, the pro-blem (P2) has at least two nontrivial solutions, which correspond to the two solutions of problem (P).

Proof. The proof is divided into four steps as follows.Step 1. We will show that ϕ is coercive in this step.

Firstly, by H(F)1(iii), (2.6), and (2.10), we have

ϕ(u) =∫T

0

[−1

2( c

0Dαt u(t),

ctD

αTu(t)

)]dt − λ

∫T

0F(t, u(t))dt

≥ |cos(πα)|2

‖u‖2α − c0λT − c0λ

∫T

0|u(t)|α0dt

≥ |cos(πα)|2

‖u‖2α − c0λT − c0λT

(T (α−1)/2

√2α − 1Γ(α)

)α0

‖u‖α0α

−→ ∞, as ‖u‖α −→ ∞,

(3.4)

where c0 = maxt∈[0,T]|c(t)|.Step 2. We will show that the ϕ is weakly lower semicontinuous.

Let un ⇀ u weakly in Eα, and by Proposition 2.7, we obtain the following results:

Eα ↪→ C([0, T],RN

),

un(t) −→ u(t) for a.e. t ∈ [0, T],

F(t, un(t)) −→ F(t, u(t)) for a.e. t ∈ [0, T].

(3.5)

By Fatou’s lemma,

lim supn→∞

∫T

0F(t, un(t))dt ≤

∫T

0F(t, u(t))dt. (3.6)

On the other hand, by Proposition 2.10, we have limn→∞J(un) = J(u), that is,

limn→∞

∫T

0

[−1

2( c

0Dαt un(t),

ctD

αTun(t)

)]dt =

∫T

0

[−1

2( c

0Dαt u(t),

ctD

αTu(t)

)]dt. (3.7)


Thus,

lim infn→∞

ϕ(un) = lim infn→∞

∫T

0

[−1

2( c

0Dαt un(t),

ctD

αTun(t)

)]dt − lim sup

n→∞λ

∫T

0F(t, un(t))dt

≥∫T

0

[−1

2( c

0Dαt u(t),

ctD

αTu(t)

)]dt − λ

∫T

0F(t, u(t))dt = ϕ(u).

(3.8)

Hence, by the Weierstrass theorem, we deduce that there exists a global minimizeru0 ∈ Eα such that

ϕ(u0) = minu∈Eα

ϕ(u). (3.9)

Step 3. We will show that there exists λ0 > 0 such that for each λ > λ0, ϕ(u0) < 0.By the condition by H(F)1 (v), there exists ξ0 ∈ R

N such that F(t, ξ0) > δ0 > 0, a.e.t ∈ Br0(t0). It is clear that

0 < M1 := max|x|≤|ξ0|

{c0 + c0|x|α0

}< +∞. (3.10)

Now we denote

κ0 =M1

δ0 +M1,

λ0 = maxκ∈[κ1,κ2]

|ξ0|2T3−2α

2Γ2(1 − α)r30(1 − κ)2(1 − α)2(3 − 2α)(δ0κ −M1 + κM1)|cos(πα)|

,

(3.11)

where κ0 < κ1 < κ2 < 1, and δ0 is given in the condition H(F)1 (v). A simple calculation showsthat the function κ �→ δ0κ−M1 +κM1 is positive whenever κ > κ0 and δ0κ0 −M1 +κ0M1 = 0.Thus, λ0 is well defined and λ0 > 0.

We will show that for each λ > λ0, the problem (P2) has two nontrivial solutions. Inorder to do this, for t ∈ [t1, t2], let us define

ηκ(t) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

0 if t ∈ [0, T]/Br0(t0),ξ0 if t ∈ Bκr0(t0),

ξ0

r0(1 − κ) (r0 − |t − t0|) if t ∈ Br0(t0)/Bκr0(t0).(3.12)

Then |η′κ(t)| ≤ |ξ0|/r0(1 − κ) and

∣∣ c0D

αt ηκ(t)

∣∣ ≤ 1Γ(1 − α)

∫ t

0(t − s)−α∣∣η′κ(s)

∣∣ds ≤ 1Γ(1 − α)

|ξ0|r0(1 − κ)

t1−α

1 − α. (3.13)

Hence,

∥∥ηι∥∥2α =∫T

0

∣∣ c0D

αt ηκ(t)

∣∣2dt ≤ |ξ0|2T3−2α

Γ2(1 − α)r20(1 − κ)2(1 − α)2(3 − 2α)

. (3.14)


By H(F)1 (iii) and (v), we have

∫T

0F(t, ηκ(t)

)dt =

∫

Bκr0 (t0)F(t, ηκ(t)

)dt +

∫

Br0 (t0)\Bκr0 (t0)F(t, ηκ(t)

)dt

≥ 2κr0δ0 −M1(2r0 − 2κr0)

= 2r0(δ0κ −M1 + κM1).

(3.15)

For κ ∈ [κ1, κ2], by Proposition 2.9, we have

ϕ(ηκ)=∫T

0

[−1

2( c

0Dαt ηκ(t),

ctD

αTηκ(t)

)]dt − λ

∫T

0F(t, ηκ(t)

)dt

≤ 1|cos(πα)|

∥∥ηκ∥∥2α − 2λr0(δ0κ −M1 + κM1)

≤ 1|cos(πα)|

|ξ0|2T3−2α

Γ2(1 − α)r20(1 − κ)2(1 − α)2(3 − 2α)

− 2λr0(δ0κ −M1 + κM1),

(3.16)

so that ϕ(ηt) < 0 whenever λ > λ0.Step 4. We will check the PS condition in the following.

Suppose that {un}n≥1 ⊆ Eα such that ϕ(un) → c and ‖ϕ′(un)‖α → 0.Since ϕ is coercive and {un}n≥1 is bounded in Eα and passed to a subsequence, which

still denote {un}n≥1, we may assume that there exists u ∈ Eα, such that un ⇀ u weakly in Eα;thus, we have

⟨ϕ′(un) − ϕ′(u), un − u

⟩=⟨ϕ′(un), un − u

⟩ − ⟨ϕ′(u), un − u⟩

≤ ∥∥ϕ′(un)∥∥‖un‖α −

⟨ϕ′(u), un − u

⟩ −→ 0,(3.17)

as n → ∞. Moreover, according to Proposition 2.7, we have ‖un − u‖∞ → 0, as n → ∞.Observing that

⟨ϕ′(un) − ϕ′(u), un − u

⟩

= −∫T

0

( c0D

αt (un(t) − u(t)), ctDα

T (un(t) − u(t)))dt

−∫T

0(∇F(t, un(t)) − F(t, u(t)), un(t) − u(t))dt

≥ |cos(πα)|‖un − u‖2α −∫T

0|∇F(t, un(t)) − F(t, u(t))|dt‖un − u‖∞,

(3.18)

combining this with (3.17), it is easy to verify that ‖un − u‖α → 0, as n → ∞, and hence thatun → u in Eα. Thus, ϕ satisfies the PS condition.


Step 5. We will show that there exists another nontrivial weak solution of problem (P2).From the mean value theorem and H(F)1 (v), we have

F(t, x) − F(t, 0) =∫1

0(∇F(t, θx), x)dθ

=∫1

0

(∇F(t, θx), θx)θ

dθ

≤∫1

0

μ(t)θ

|θx|γdθ

= μ(t)|x|γ∫1

0θγ−1dθ

=μ(t)|x|γ

γ,

(3.19)

for all |x| < β, a.e. t ∈ [0, T].It follows from the conditions H(F)1 (iii) and all |x| ≥ β and a.e. t ∈ [0, T] that

|F(t, x)| ≤ c0|t| + c0|x|α0

≤ c0

∣∣∣∣x

β

∣∣∣∣ + c0|x|α0

≤(c0

βα0+ c0

)|x|α0−γ |x|γ

≤(c0

βγ+

c0

βγ−α0

)|x|γ ,

(3.20)

and this together with (3.19) yields that for all x ∈ RN and a.e. t ∈ [0, T],

|F(t, x)| ≤(μ(t)γ

+c0

βγ+

c0

βγ−α0

)|x|γ(x) ≤ c1|x|γ , (3.21)

for some positive constant c1.For all λ > λ0, ‖u‖α < 1, and |u|∞ < 1, we have

ϕ(u) =∫T

0

[−1

2( c

0Dαt u(t),

ctD

αTu(t)

)]dt − λ

∫T

0F(t, u(t))dt

≥ |cos(πα)|‖u‖2α − λc1

∫T

0|u(t)|γdt


[T (α−1)/2

Γα(α)√

2α − 1

]γ| ‖u‖γα.

(3.22)


So, for ρ > 0 small enough, there exists a ν > 0 such that

ϕ(u) > ν, for ‖u‖α = ρ , (3.23)

and ‖u0‖α > ρ. So by the mountain pass theorem (cf. Lemma 2.11), we can get u1 ∈ Eα whichsatisfies

ϕ(u1) = c > 0, ϕ′(u1) = 0. (3.24)

Therefore, u1 is another nontrivial critical point of ϕ.

Remark 3.3. We can find a potential function satisfying the hypothesis of our Theorem 3.2. Forgreat details, please see Section 4(B) in Summary.

So far, the results involved potential functions exhibiting sublinear. The next theoremconcerns problems where the potential function is superlinear.

Our hypotheses on nonsmooth potential F(t, x) are as follows.H(F)2: F : [0, T] ×R

N → R is a function such that F(t, 0) = 0 a.e. on [0, T] and satisfiesthe following facts:

(i) for all x ∈ RN , t �→ F(t, x) is measurable,

(ii) for a.a. t ∈ [0, T], x �→ F(t, x) is continuously differentiable,

(iii) there exist c ∈ C([0, T],R) and α0 > 2, such that

|∇F(t, x)|, |F(t, x)| ≤ c(t)(1 + (t)|x|α0), for a.a. t ∈ [0, T], allx ∈ R

N, (3.25)

(iv) there exist γ > 2 and μ ∈ L∞([0, T]), such that

lim sup|x|→ 0

(∇F(t, x), x)|x|γ < μ(t), uniformly for a.a. t ∈ [0, T], (3.26)

(v) there exist ξ0 ∈ RN , t0 ∈ (0, T), and r0 > 0, such that F(t, ξ0) > δ0 > 0, a.e. t ∈

Br0(t0),where Br0(t0) := {t ∈ [0, T] : |t − t0| ≤ r0} ⊂ [0, T],

(vi) for a.a. t ∈ [0, T] and all x ∈ RN , we have

F(t, x) ≤ ν(t)with ν ∈ Lβ([0, T],R), 1 ≤ β < 2. (3.27)

Theorem 3.4. Suppose that H(F)2 holds. Then there exists a λ0 > 0 such that for each λ > λ0, theproblem (P2) has at least two nontrivial solutions, which correspond to the two solutions of problem(P).

Proof. The steps are similar to those of Theorem 3.2. In fact, we only need to modify Step 1and Step 4 as follows: 1′ shows that ϕ is coercive under the condition H(F)2 (vi); 4′ shows that


there exists a second nontrivial solution under the conditions H(F)2 (iii) and H(F)2 (iv). Thenfrom Steps 1′, 2, 3, and 4′ above, problem (P) has at least two nontrivial solutions.Step 1′ . By H(F)2 (vi), for all u ∈ Eα, ‖u‖α > 1, we have

ϕ(u) =∫T

0

[−1

2( c

0Dαt u(t),

ctD

αTu(t)

)]dt − λ

∫T

0F(t, u(t))dt


∫T

0ν(t)dt −→ ∞, as ‖u‖α −→ ∞,

(3.28)

where c0 = maxt∈[0,T]|c(t)|.Step 4′. Because of hypothesis H(F)2 (iii), we have

F(t, x) ≤ c0 + c0|x|α0

≤ c0

∣∣∣∣x

β

∣∣∣∣

α0

+ c0|x|α0

= c0

∣∣∣∣1β

∣∣∣∣

α0

|x|α0 + c0|x|α0

= c2|x|α0 ,

(3.29)

for a.e. t ∈ [0, T] and all |x| ≥ β with c2 > 0.Combining (3.19) and (3.29), it follows that

|F(t, x)| ≤ μ(t)γ

|x|γ + c2|x|α0 , (3.30)

for a.e. t ∈ [0, T] and all x ∈ RN .

Thus, for all λ > λ0, ‖u‖α < 1 and |u|∞ < 1, we have

ϕ(u) =∫T

0

[−1

2( c

0Dαt u(t),

ctD

αTu(t)

)]dt − λ

∫T

0F(t, u(t))dt

≥ |cos(πα)|‖u‖2α − λ

∫T

0

μ(t)γ

|u(t)|γdt − λc2

∫T

0|u(t)|α0dt


[T (α−1)/2

Γ(α)√

2α − 1

]γ ∣∣∣∣∣‖u‖γα − λc4

[T (α−1)/2

Γ(α)√

2α − 1

]α0∣∣∣∣∣‖u‖α0

α ,

(3.31)

where c3 and c4 are positive constants.So, for ρ > 0 small enough, there exists a ν > 0 such that

ϕ(u) > ν, for ‖u‖α = ρ , (3.32)

and ‖u0‖α > ρ.Arguing as in proof of Step 4 of Theorem 3.2, we conclude that ϕ satisfies the PS

condition. So by the mountain pass theorem (cf. Lemma 2.11), we can get that u1 ∈ Eα satisfies

ϕ(u1) = c > 0, ϕ′(u1) = 0. (3.33)


Therefore, u1 is another nontrivial critical point of ϕ.

Remark 3.5. We will give some examples, which satisfy the hypothesis of our Theorem 3.4.For great details, please see Section 4(C) in Summary.

4. Summary

(A) If β = 0, then α = 1 − β/2 = 1. Therefore, by Theorems 3.2 and 3.4, we actually obtain theexistence of two weak solutions of the following eigenvalue problem:

u′′(t) + λ∇F(t, u(t)) = 0, a.a. t ∈ [0, T],

u(0) = u(T) = 0,(P3)

where λ > 0 is a real number, F : [0, T] × RN → R is a given function, and ∇F(t, x) is the

gradient of F at x. Although many excellent results have been worked out on the existenceof solutions for second-order system (P3) (e.g., [28, 29]), it seems that no similar results wereobtained in the literature for fractional system (P).

(B) We give an example in the following to illustrate our viewpoint in Remark 3.1. Weconsider

F(t, x) =

⎧⎪⎨

⎪⎩

a(t) sin(π

2|x|γ), 0 ≤ |x| < 1,

a(t), |x| ≥ 1,(4.1)

where γ > 2, a ∈ C([0, T],R), a(t) > 0 for all t ∈ [0, T].Obviously, hypotheses H(F)1 (i), (ii), and (v) are satisfied. Moreover,

|F(t, x)| ≤

⎧⎪⎪⎨

⎪⎪⎩

a(t)π

2|x|γ ≤ a(t)π

2|x|γ−α0 |x|α0 ≤ a(t)π

2|x|α0 ≤ a(t)πγ

2|x|α0 , 0 ≤ |x| < 1,

a(t) ≤ a(t)πγ2, |x| ≥ 1,

|∇F(t, x)| =

⎧⎪⎨

⎪⎩

∣∣∣a(t)πγ

2cos(π

2|x|γ)|x|γ−2x

∣∣∣ ≤ a(t)πγ2

|x|γ−1 ≤ a(t)πγ2, 0 ≤ |x| < 1,

0, |x| ≥ 1.

(4.2)


Therefore,

|∇F(t, x)|, |F(t, x)| ≤ πγa(t)2

+πγa(t)

2|x|α0 . (4.3)

So, condition H(F)1(iii) holds.On the other hand,

lim sup|x|→ 0

〈∇F(t, x), x〉|x|γ = lim sup

|x|→ 0

γ(π/2)a(t) cos((π/2)|x|γ)|x|γ

|x|γ =π

2γa(t), (4.4)

uniformly for a.a. t ∈ [0, T], so condition H(F)1 (iv) holds.(C) We can find the following potential functions satisfying the conditions stated in

Theorem 3.4:

F1(t, x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

−a(t)γ

|x|γ , |x| < 1,

a(t)π

cos(π

2|x|2)− a(t)

γ, |x| ≥ 1,

F2(t, x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

a(t) sin(π

2|x|γ), |x| < 1,

a(t)[

1|x| +

12|x|2 − 1

2

], |x| ≥ 1,

F3(t, x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

a(t)γ

|x|γ , |x| < 1,

−2a(t)√|x|

+ a(t)(

2 +1γ

), |x| ≥ 1,

(4.5)

where a ∈ C([0, T],R), a(t) > 0 for all t ∈ [0, T].It is clear that Fi(x, 0) = 0 (i = 1, 2, 3) for a.e. t ∈ [0, T], and hypotheses H(F)2 (i) and

H(F)2 (ii) are satisfied. A direct verification shows that conditions H(F)2 (v) and H(F)2 (vi)are satisfied. Note that

|∇F1(t, x)| =

⎧⎪⎪⎨

⎪⎪⎩

∣∣∣−a(t)|x|γ−2x∣∣∣ ≤ a(t)|x|γ−1 ≤ 2a(t), 0 ≤ |x| < 1,

∣∣∣−a(t) sin(π

2|x|2)x∣∣∣ ≤ a(t)|x| ≤ 2a(t)|x|γ , |x| ≥ 1,


|∇F2(t, x)| =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

∣∣∣γa(t) cos

(π2|x|γ)π

2|x|γ−2x

∣∣∣ ≤ γa(t)π

2|x|γ−1 ≤ a(t)πγ

2, 0 ≤ |x| < 1,

∣∣∣∣∣a(t)

(

x − x

|x|3)∣∣∣∣∣≤ a(t)

(

|x| + 1

|x|2)

≤ a(t) + a(t)|x|γ , |x| ≥ 1,

|∇F3(t, x)| =

⎧⎪⎨

⎪⎩

∣∣∣a(t)|x|γ−2x

∣∣∣ ≤∣∣∣a(t)|x|2(5/2)

∣∣∣ ≤ a(t), 0 ≤ |x| < 1,

∣∣∣a(t)|x|−5/2x

∣∣∣ ≤ a(t)|x| ≤ a(t)|x|γ , |x| ≥ 1,

|F1(t, x)| ≤ a(t)|x|γ + a(t)(1 + 1) = 2a(t) + 2a(t)|x|γ ,

|F2(t, x)| ≤ a(t)∣∣∣sin(π

2|x|γ)∣∣∣ + a(t)

[1 +

12|x|γ − 1

2

]≤ 3a(t) + a(t)|x|γ

2,

|F3(t, x)| ≤ a(t)1γ|x|γ + 2a(t)

√|x|

+ a(t)(

2 +1γ

)≤ 5a(t) + a(t)|x|γ .

(4.6)

So,

|∇F1(t, x)|, |F1(t, x)| ≤ 2a(t)(1 + |x|γ),

|∇F2(t, x)|, |F2(t, x)| ≤ a(t)πγ

2(1 + |x|γ),

|∇F3(t, x)|, |F3(t, x)| ≤ 5a(t)(1 + |x|γ),

lim sup|x|→ 0

(∇F1(t, x), x)|x|γ = lim

|x|→ 0

−a(t)|x|γ|x|γ = −a(t),

lim sup|x|→ 0

(∇F2(t, x), x)|x|γ = lim

|x|→ 0

a(t) cos((π/2)|x|γ)(π/2)γ |x|γ

|x|γ

= lim|x|→ 0

a(t) cos(π

2|x|γ)π

2γ =

γπa(t)2

,

lim sup|x|→ 0

(∇F3(t, x), x)|x|γ = lim

|x|→ 0

a(t)|x|γ|x|γ = a(t),

(4.7)

which shows that assumptions H(F)2 (iii) and H(F)2 (iv) are fulfilled.

Acknowlegments

This work was supported by the National Natural Science Foundation of China (nos.11126286, 11001063, and 11201095), the Fundamental Research Funds for the Central Univer-sities (no. 2012), China Postdoctoral Science Foundation funded project (no. 20110491032),and China Postdoctoral Science (Special) Foundation (no. 2012T50325).


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Research ArticleOn Impulsive Boundary Value Problems ofFractional Differential Equations with IrregularBoundary Conditions

Guotao Wang,1 Bashir Ahmad,2 and Lihong Zhang1

1 School of Mathematics and Computer Science, Shanxi Normal University, Linfen, Shanxi 041004, China2 Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203,Jeddah 21589, Saudi Arabia

Correspondence should be addressed to Lihong Zhang, [email protected]

Received 25 February 2012; Accepted 4 September 2012


Copyright q 2012 Guotao Wang et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We study nonlinear impulsive differential equations of fractional order with irregular boundaryconditions. Some existence and uniqueness results are obtained by applying standard fixed-pointtheorems. For illustration of the results, some examples are discussed.

1. Introduction

Boundary value problems of nonlinear fractional differential equations have recently beenstudied by several researchers. Fractional differential equations appear naturally in variousfields of science and engineering and constitute an important field of research. As a matterof fact, fractional derivatives provide an excellent tool for the description of memoryand hereditary properties of various materials and processes [1–4]. Some recent work onboundary value problems of fractional order can be found in [5–23] and the referencestherein. In [24], some existence and uniqueness results were obtained for an irregularboundary value problem of fractional differential equations.

Dynamical systems with impulse effect are regarded as a class of general hybridsystems. Impulsive hybrid systems are composed of some continuous variable dynamicsystems along with certain reset maps that define impulsive switching among them. It isthe switching that resets the modes and changes the continuous state of the system. There arethree classes of impulsive hybrid systems, namely, impulsive differential systems [25, 26],sampled data or digital control system [27, 28], and impulsive switched system [29, 30].


Applications of such systems include air traffic management [31], automotive control [32, 33],real-time software verification [34], transportation systems [35, 36], manufacturing [37],mobile robotics [38], and process industry [39]. In fact, hybrid systems have a central rolein embedded control systems that interact with the physical world. Using hybrid models,one may represent time and event-based behaviors more accurately so as to meet challengingdesign requirements in the design of control systems for problems such as cut-off control andidle speed control of the engine. For more details, see [40] and the references therein.

The theory of impulsive differential equations of integer order has found its extensiveapplications in realistic mathematical modelling of a wide variety of practical situations andhas emerged as an important area of investigation. The impulsive differential equations offractional order have also attracted a considerable attention and a variety of results can befound in the papers [41–50].

In this paper, motivated by [24], we study a nonlinear impulsive hybrid system offractional differential equations with irregular boundary conditions given by

CDαu(t) = f(t, u(t)), 1 < α ≤ 2, t ∈ J ′,Δu(tk) = Ik(u(tk)), Δu′(tk) = I∗k(u(tk)), k = 1, 2, . . . , p,

u′(0) + (−1)θu′(T) + bu(T) = 0, u(0) + (−1)θ+1u(T) = 0, θ = 1, 2,

(1.1)

where CDα is the Caputo fractional derivative, f ∈ C(J × R,R), Ik, I∗k ∈ C(R,R), b ∈R, b /= 0, J = [0, T](T > 0), 0 = t0 < t1 < · · · < tk < · · · < tp < tp+1 = T, J ′ =J \ {t1, t2, . . . , tp},Δu(tk) = u(t+

k) − u(t−

k), where u(t+

k) and u(t−

k) denote the right and the left

limits of u(t) at t = tk (k = 1, 2, . . . , p), respectively. Δu′(tk) have a similar meaning for u′(t).Here, we remark that irregular boundary value problems for ordinary and partial

differential equations occur in scientific and engineering disciplines and have been addressedby many authors, for instance, see [24] and the references.

The paper is organized as follows. Section 2 deals with some definitions andpreliminary results, while the main results are presented in Section 3.

2. Preliminaries

Let us fix J0 = [0, t1], Jk−1 = (tk−1, tk], k = 2, . . . , p + 1 with tp+1 = T and introduce the spaces:

PC(J,R) ={u : J −→ R | u ∈ C(Jk), k = 0, 1, . . . , p, andu

(t+k)

exist, k = 1, 2, . . . , p}, (2.1)

with the norm ‖u‖ = supt∈J |u(t)|, and

PC1(J,R) ={u : J → R | u ∈ C1(Jk), k = 0, 1, . . . , p,and u

(t+k

), u′(t+k

)exist, k = 1, 2, . . . , p

}, (2.2)

with the norm ‖u‖PC1 = max{‖u‖, ‖u′‖}. Obviously, PC(J,R) and PC1(J,R) are Banach spaces.

Definition 2.1. A function u ∈ PC1(J,R) with its Caputo derivative of order α existing on J isa solution of (1.1) if it satisfies (1.1).


To prove the existence of solutions of problem (1.1), we need the following fixed-pointtheorems.

Theorem 2.2 (see [51]). Let E be a Banach space. Assume that Ω is an open bounded subset of Ewith θ ∈ Ω and let T : Ω → E be a completely continuous operator such that

‖Tu‖ ≤ ‖u‖, ∀u ∈ ∂Ω. (2.3)

Then T has a fixed point in Ω.

Lemma 2.3 (see [1]). For α > 0, the general solution of fractional differential equation CDαu(t) = 0is

u(t) = C0 + C1t + C2t2 + · · · + Cn−1t

n−1, (2.4)

where Ci ∈ R, i = 0, 1, 2, . . . , n − 1, n = [α] + 1 ([α] denotes integer part of α).

Lemma 2.4 (see [1]). Let α > 0. Then

Iα CDαu(t) = u(t) + C0 + C1t + C2t2 + · · · + Cn−1t

n−1 (2.5)

for some Ci ∈ R, i = 1, 2, . . . , n − 1, n = [α] + 1.

Lemma 2.5. For a given y ∈ C[0, T], a function u is a solution of the following impulsive irregularboundary value problem

CDαu(t) = y(t), 1 < α ≤ 2, t ∈ J ′,Δu(tk) = Ik(u(tk)), Δu′(tk) = I∗k(u(tk)), k = 1, 2, . . . , p,

u′(0) + (−1)θu′(T) + bu(T) = 0, u(0) + (−1)θ+1u(T) = 0, θ = 1, 2, b /= 0,

(2.6)


if and only if u is a solution of the impulsive fractional integral equation

u(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

∫ t

0

(t − s)α−1

Γ(α)y(s)ds +

1 − (−1)θ+1

bT

∫T

tp

(T − s)α−1

Γ(α)y(s)ds

+

[1 + (−1)θ+1

]t

bT

∫T

tp

(T − s)α−2

Γ(α − 1)y(s)ds

−1b

∫T

tp

(T − s)α−2

Γ(α − 1)y(s)ds − t

T

∫T

tp

(T − s)α−1

Γ(α)y(s)ds +A, t ∈ J0;

∫ t

tk

(t − s)α−1

Γ(α)y(s)ds +

1 − (−1)θ+1

bT

∫T

tp

(T − s)α−1

Γ(α)y(s)ds

+

[1 + (−1)θ+1

]t

bT

∫T

tp

(T − s)α−2

Γ(α − 1)y(s)ds

−1b

∫T

tp

(T − s)α−2

Γ(α − 1)y(s)ds − t

T

∫T

tp

(T − s)α−1

Γ(α)y(s)ds

+k∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)y(s)ds + Ii(u(ti))

]

+k−1∑

i=1

(tk − ti)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+k∑

i=1

(t − tk)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+A, t ∈ Jk, k = 1, 2, . . . , p,

(2.7)

where

A =

[1 + (−1)θ+1

]t − T

bT

p∑

i=1

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+1 − (−1)θ+1 − bt

bT

{p∑

i=1

[∫ ti

ti−1

(ti − s)α−1


]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]}

.

(2.8)

Proof. Let u be a solution of (2.6). Then, by Lemma 2.4, we have

u(t) = Iαy(t) − c1 − c2t =1

Γ(α)

∫ t

0(t − s)α−1y(s)ds − c1 − c2t, t ∈ J0, (2.9)


for some c1, c2 ∈ R. Differentiating (2.9), we get

u′(t) =1

Γ(α − 1)

∫ t

0(t − s)α−2y(s)ds − c2, t ∈ J0. (2.10)

If t ∈ J1, then

u(t) =1

Γ(α)

∫ t

t1

(t − s)α−1y(s)ds − d1 − d2(t − t1),

u′(t) =1

Γ(α − 1)

∫ t

t1

(t − s)α−2y(s)ds − d2,

(2.11)

for some d1, d2 ∈ R. Thus,

u(t−1)=

1Γ(α)

∫ t1

0(t1 − s)α−1y(s)ds − c1 − c2t1, u

(t+1)= −d1,

u′(t−1)=

1Γ(α − 1)

∫ t1

0(t1 − s)α−2y(s)ds − c2, u′

(t+1)= −d2.

(2.12)

Using the impulse conditions

Δu(t1) = u(t+1) − u(t−1

)= I1(u(t1)), Δu′(t1) = u′

(t+1) − u′(t−1

)= I∗1(u(t1)), (2.13)

we find that

−d1 =1

Γ(α)

∫ t1

0(t1 − s)α−1y(s)ds − c1 − c2t1 + I1(u(t1)),

−d2 =1

Γ(α − 1)

∫ t1

0(t1 − s)α−2y(s)ds − c2 + I∗1(u(t1)).

(2.14)

Consequently, we obtain

u(t) =1

Γ(α)

∫ t

t1

(t − s)α−1y(s)ds +1

Γ(α)

∫ t1

0(t1 − s)α−1y(s)ds

+t − t1

Γ(α − 1)

∫ t1

0(t1 − s)α−2y(s)ds + I1(u(t1)) + (t − t1)I∗1(u(t1)) − c1 − c2t, t ∈ J1.

(2.15)


By a similar process, we get

u(t) =∫ t

tk

(t − s)α−1

Γ(α)y(s)ds +

k∑

i=1

[∫ ti

ti−1

(ti − s)α−1


]

+k−1∑

i=1

(tk − ti)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+k∑

i=1

(t − tk)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

− c1 − c2t, t ∈ Jk, k = 1, 2, . . . , p.

(2.16)

Applying the boundary conditions u′(0) + (−1)θu′(T) + bu(T) = 0 and u(0) +(−1)θ+1u(T) = 0, we find that

c1 = − 1 − (−1)θ+1

bT

∫T

tp

(T − s)α−1

Γ(α)y(s)ds +

1b

∫T

tp

(T − s)α−2

Γ(α − 1)y(s)ds

+p∑

i=1

1b

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

− 1 − (−1)θ+1

bT

{p∑

i=1

[∫ ti

ti−1

(ti − s)α−1


]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]}

,

c2 = − 1 + (−1)θ+1

bT

∫T

tp

(T − s)α−2

Γ(α − 1)y(s)ds +

1T

∫T

tp

(T − s)α−1

Γ(α)y(s)ds

− 1 + (−1)θ+1

bT

p∑

i=1

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+1T

{p∑

i=1

[∫ ti

ti−1

(ti − s)α−1


]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)y(s)ds + I∗i (u(ti))

]}

.

(2.17)


Substituting the value of ci (i = 1, 2) in (2.9) and (2.16), we obtain (2.7). Conversely, assumethat u is a solution of the impulsive fractional integral equation (2.7), then by a directcomputation, it follows that the solution given by (2.7) satisfies (2.6). This completes theproof.

Remark 2.6. With T = π , the first five terms of the solution (2.7) correspond to the solution forthe problem without impulses [24].

3. Main Results

Define an operator G : PC(J,R) → PC(J,R) by

Gu(t) =∫ t

tk

(t − s)α−1

Γ(α)f(s, u(s))ds +

k∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)f(s, u(s))ds + Ii(u(ti))

]

+k−1∑

i=1

(tk − ti)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)f(s, u(s))ds + I∗i (u(ti))

]

+k∑

i=1

(t − tk)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)f(s, u(s))ds + I∗i (u(ti))

]

+1 − (−1)θ+1

bT

∫T

tp

(T − s)α−1

Γ(α)f(s, u(s))ds +

[1 + (−1)θ+1

]t

bT

∫T

tp

(T − s)α−2

Γ(α − 1)f(s, u(s))ds

− 1b

∫T

tp

(T − s)α−2

Γ(α − 1)f(s, u(s))ds − t

T

∫T

tp

(T − s)α−1

Γ(α)f(s, u(s))ds

+

[1 + (−1)θ+1

]t − T

bT

p∑

i=1

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)f(s, u(s))ds + I∗i (u(ti))

]

+1 − (−1)θ+1 − bt

bT

{p∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)f(s, u(s))ds + Ii(u(ti))

]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)f(s, u(s))ds + I∗i (u(ti))

]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)f(s, u(s))ds + I∗i (u(ti))

]}

.

(3.1)

Notice that problem (1.1) has a solution if and only if the operator G has a fixed point.


For the sake of convenience, we set the following notations:

μ = 2(1 + p

)(

1 +1

|b|T)Iαa(T) +

[2(2p − 1

)T +

5p + 1|b|

]Iα−1

+ 2(

1 +1

|b|T)pL2 +

[2(2p − 1

)T +

5p − 2|b|

]L3,

ν = 2(1 + p

)(

1 +1

|b|T)

Tα

Γ(α + 1)+[

2(2p − 1

)T +

5p + 1|b|

]Tα−1

Γ(α).

(3.2)

Theorem 3.1. Assume that

(H1) there exists a nonnegative function a(t) ∈ L(0, T) such that

∣∣f(t, u)∣∣ ≤ a(t) + ξ|u|ρ, 0 < ρ < 1, (3.3)

where ξ is a nonnegative constant;

(H2) there exist positive constants L2 and L3 such that

|Ik(u)| ≤ L2,∣∣I∗k(u)

∣∣ ≤ L3, for t ∈ J, u ∈ R , k = 1, 2, . . . , p. (3.4)

Then problem (1.1) has at least one solution.

Proof. As a first step, we show that the operator G : PC(J,R) → PC(J,R) is completelycontinuous. Observe that continuity of G follows from the continuity of f, Ik and I∗

k.

Let Ω ⊂ PC(J,R) be bounded. Then, there exist positive constants Li > 0 (i = 1, 2, 3)such that |f(t, u)| ≤ L1, |Ik(u)| ≤ L2, and |I∗

k(u)| ≤ L3, for all u ∈ Ω. Thus, for all u ∈ Ω, we

have

|Gu(t)| ≤∫ t

tk

(t − s)α−1

Γ(α)

∣∣f(s, u(s))∣∣ds +

k∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)

∣∣f(s, u(s))∣∣ds + |Ii(u(ti))|

]

+k−1∑

i=1

(tk − ti)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

∣∣I∗i (u(ti))∣∣]

+k∑

i=1

(t − tk)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

∣∣I∗i (u(ti))∣∣]


+1 − (−1)θ+1

|b|T∫T

tp

(T − s)α−1

Γ(α)

∣∣f(s, u(s))

∣∣ds

+

[1 + (−1)θ+1

]t

|b|T∫T

tp

(T − s)α−2

Γ(α − 1)

∣∣f(s, u(s))

∣∣ds

+1|b|∫T

tp

(T − s)α−2

Γ(α − 1)

∣∣f(s, u(s))

∣∣ds +

t

T

∫T

tp

(T − s)α−1

Γ(α)

∣∣f(s, u(s))

∣∣ds

+

∣∣∣∣∣∣∣

[1 + (−1)θ+1

]t − T

bT

∣∣∣∣∣∣∣

p∑

i=1

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))

∣∣ds +

∣∣I∗i (u(ti))

∣∣]

+

∣∣∣∣∣

1 − (−1)θ+1 − btbT

∣∣∣∣∣

{p∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)

∣∣f(s, u(s))

∣∣ds + |Ii(u(ti))|

]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

∣∣I∗i (u(ti))∣∣]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

∣∣I∗i (u(ti))∣∣]}

≤ L1

∫ t

tk

(t − s)α−1

Γ(α)ds +

p∑

i=1

[

L1

∫ ti

ti−1

(ti − s)α−1

Γ(α)ds + L2

]

+p−1∑

i=1

T

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]

+p∑

i=1

T

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]

+2L1

|b|T∫T

tp

(T − s)α−1

Γ(α)ds +

2L1

|b|∫T

tp

(T − s)α−2

Γ(α − 1)ds

+L1

|b|∫T

tp

(T − s)α−2

Γ(α − 1)ds + L1

∫T

tp

(T − s)α−1

Γ(α)ds +

1|b|

p∑

i=1

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]

+2 + |b|T|b|T

{p∑

i=1

[

L1

∫ ti

ti−1

(ti − s)α−1

Γ(α)ds + L2

]

+p−1∑

i=1

T

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]

+p∑

i=1

T

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]}

≤ 1|b|

{2(1 + p

)(|b|T + 1)Tα−1L1

Γ(α + 1)+

[5p + 1 + 2|b|T(2p − 1

)]Tα−1L1

Γ(α)

+2(1 + |b|T)pL2

T+[5p − 2 + 2|b|T(2p − 1

)]L3

},

(3.5)


which implies

‖Gu‖ ≤ 1|b|

{2(1 + p

)(|b|T + 1)Tα−1L1

Γ(α + 1)+

[5p + 1 + 2|b|T(2p − 1

)]Tα−1L1

Γ(α)+

2(1 + |b|T)pL2

T

+[5p − 2 + 2|b|T(2p − 1

)]L3

}

:= L.

(3.6)

On the other hand, for any t ∈ Jk, 0 ≤ k ≤ p, we get

∣∣(Gu)′(t)∣∣ ≤

∫ t

tk

(t − s)α−2

Γ(α − 1)

∣∣f(s, u(s))

∣∣ds +

k∑

i=1

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))

∣∣ds +

∣∣I∗i (u(ti))

∣∣]

+

[1 + (−1)θ+1

]

|b|T∫T

tp

(T − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

1T

∫T

tp

(T − s)α−1

Γ(α)

∣∣f(s, u(s))∣∣ds

+

[1 + (−1)θ+1

]

|b|Tp∑

i=1

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

∣∣I∗i (u(ti))∣∣]

+1T

{p∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)

∣∣f(s, u(s))∣∣ds + |Ii(u(ti))|

]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

∣∣I∗i (u(ti))∣∣]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s))∣∣ds +

∣∣I∗i (u(ti))∣∣]}

≤ L1

∫ t

tk

(t − s)α−2

Γ(α − 1)ds +

p∑

i=1

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]

+2L1

|b|T∫T

tp

(T − s)α−2

Γ(α − 1)ds

+L1

T

∫T

tp

(T − s)α−1

Γ(α)ds +

2|b|T

p∑

i=1

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]

+1T

{p∑

i=1

[

L1

∫ ti

ti−1

(ti − s)α−1

Γ(α)ds + L2

]

+p−1∑

i=1

T

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]

+p∑

i=1

T

[

L1

∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)ds + L3

]}

≤(1 + p

)Tα−1L1

Γ(α + 1)+

3pTα−1L1

Γ(α)+

2(1 + 2p

)Tα−2L1

|b|Γ(α) +pL2

T+

2pL3

|b|T +(3p − 1

)L3 := L.

(3.7)


Hence, for t1, t2 ∈ Jk with t1 < t2, 0 ≤ k ≤ p, we have

|(Gu)(t2) − (Gu)(t1)| ≤∫ t2

t1

∣∣(Gu)′(s)∣∣ds ≤ L(t2 − t1). (3.8)

This implies that G is equicontinuous on all Jk, k = 0, 1, 2, . . . , p and hence, by the Arzela-Ascoli theorem, the operator G : PC(J,R) → PC(J,R) is completely continuous.

Next, we prove that G : B → B. For that, let us choose R ≥ max{2μ, (2νξ)1/(1−ρ)} anddefine a ball B = {u ∈ PC(J,R) : ‖u‖ ≤ R}. For any u ∈ B, by the assumptions (H1) and (H2),we have

|Gu(t)| ≤∫ t

tk

(t − s)α−1

Γ(α)[a(s) + ξ|u(s)|ρ]ds +

k∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)[a(s) + ξ|u(s)|ρ]ds + |Ii(u(ti))|

]

+k−1∑

i=1

(tk − ti)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)[a(s) + ξ|u(s)|ρ]ds + ∣∣I∗i (u(ti))

∣∣]

+k∑

i=1

(t − tk)[∫ ti

ti−1

(ti − s)α−2


∣∣]

+1 − (−1)θ+1

|b|T∫T

tp

(T − s)α−1

Γ(α)[a(s) + ξ|u(s)|ρ]ds

+

[1 + (−1)θ+1

]t

|b|T∫T

tp

(T − s)α−2

Γ(α − 1)[a(s) + ξ|u(s)|ρ]ds

+1|b|∫T

tp

(T − s)α−2

Γ(α − 1)[a(s) + ξ|u(s)|ρ]ds + t

T

∫T

tp

(T − s)α−1

Γ(α)[a(s) + ξ|u(s)|ρ]ds

+

∣∣∣∣∣∣∣

[1 + (−1)θ+1

]t − T

bT

∣∣∣∣∣∣∣

p∑

i=1

[∫ ti

ti−1

(ti − s)α−2


∣∣]

+

∣∣∣∣∣1 − (−1)θ+1 − bt

bT

∣∣∣∣∣

{p∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)[a(s) + ξ|u(s)|ρ]ds + |Ii(u(ti))|

]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2


∣∣]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2


∣∣]}


≤ 2(1 + p

)(

1 +1

|b|T)Iαa(T) +

[2(2p − 1

)T +

5p + 1|b|

]Iα−1a(T) + 2

(1 +

1|b|T)pL2

+[

2(2p − 1

)T +

5p − 2|b|

]L3 + 2

(1 + p

)(

1 +1

|b|T)

TαξRρ

Γ(α + 1)

+[

2(2p − 1

)T +

5p + 1|b|

]Tα−1ξRρ

Γ(α).

(3.9)

Thus,

‖Gu‖ ≤ μ + νξRρ ≤ R

2+R

2= R, (3.10)

where μ and ν are given by (3.2). This implies G : B → B. Hence, G : B → B is completelycontinuous. Therefore, by the Schauder fixed-point theorem, the operator G has at least onefixed point. Consequently, problem (1.1) has at least one solution in B.

Remark 3.2. For ρ = 1 in (H1), if νξ < 1, we can take R ≥ μ/(1 − νξ), then the conclusion ofTheorem 3.1 holds.

Theorem 3.3. Suppose that there exist a nonnegative functions a1 ∈ L(0, 1) and a nonnegativenumber ξ1 such that |f(t, u)| ≤ a1(t) + ξ1|u|ρ for ρ > 1. Furthermore, the assumption (H2) holds.Then problem (1.1) has at least one solution.

Proof. The proof is similar to that of Theorem 3.1, so we omit it.

Theorem 3.4. Suppose that

limu→ 0

f(t, u)u

= 0, limu→ 0

Ik(u)u

= 0, limu→ 0

I∗k(u)u

= 0. (3.11)


Proof. By Theorem 3.1, we know that the operator G : PC(J,R) → PC(J,R) is completelycontinuous. In view of (3.11), we can find a constant r > 0 such that |f(t, u)| ≤ δ1|u|, |Ik(u)| ≤δ2|u| and |I∗k(u)| ≤ δ3|u| for 0 < |u| < r, where δi > 0 (i = 1, 2, 3) satisfy

2(1 + p

)(|b|T + 1)Tα−1δ1

Γ(α + 1)+

[5p + 1 + 2|b|T(2p − 1

)]Tα−1δ1

Γ(α)+

2(1 + |b|T)pδ2

T

+[5p − 2 + 2|b|T(2p − 1

)]δ3 ≤ |b|.

(3.12)


Let Ω = {u ∈ PC(J,R) | ‖u‖ < r}. Take u ∈ PC(J,R) such that ‖u‖ = r, which meansu ∈ ∂Ω. Then, as in the proof of Theorem 3.1, we have

|Gu(t)| ≤ 1|b|

{2(1 + p

)(|b|T + 1)Tα−1δ1

Γ(α + 1)+

[5p + 1 + 2|b|T(2p − 1

)]Tα−1δ1

Γ(α)

+2(1 + |b|T)pδ2

T+[5p − 2 + 2|b|T(2p − 1

)]δ3

}

‖u‖,(3.13)

which, in view of (3.12), implies that ‖Gu‖ ≤ ‖u‖, u ∈ ∂Ω. Therefore, by Theorem 2.2, theoperator G has at least one fixed point. Thus we conclude that problem (1.1) has at least onesolution u ∈ Ω.


(H3) there exist positive constants Ki (i = 1, 2, 3) such that

∣∣f(t, u) − f(t, v)∣∣ ≤ K1|u − v|, |Ik(u) − Ik(v)| ≤ K2|u − v|, ∣∣I∗k(u) − I∗k(v)∣∣ ≤ K3|u − v|,

(3.14)

for t ∈ J, u, v ∈ R and k = 1, 2, . . . , p.

Then problem (1.1) has a unique solution if

Λ =2(1 + p

)(|b|T + 1)Tα−1K1

Γ(α + 1)+

[5p + 1 + 2|b|T(2p − 1

)]Tα−1K1

Γ(α)+

2(1 + |b|T)pK2

T

+[5p − 2 + 2|b|T(2p − 1

)]K3 < |b|.

(3.15)

Proof. For u, v ∈ PC(J,R), we have

|(Gu)(t) − (Gv)(t)| ≤∫ t

tk

(t − s)α−1

Γ(α)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+k∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds + |Ii(u(ti)) − Ii(v(ti))|]

+k−1∑

i=1

(tk − ti)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣]


+k∑

i=1

(t − tk)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣]

+1 − (−1)θ+1

|b|T∫T

tp

(T − s)α−1

Γ(α)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+

[1 + (−1)θ+1

]t

|b|T∫T

tp

(T − s)α−2

Γ(α − 1)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+1|b|∫T

tp

(T − s)α−2

Γ(α − 1)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+t

T

∫T

tp

(T − s)α−1

Γ(α)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+

∣∣∣∣∣∣∣

[1 + (−1)θ+1

]t − T

bT

∣∣∣∣∣∣∣

p∑

i=1

[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣]

+

∣∣∣∣∣1 − (−1)θ+1 − bt

bT

∣∣∣∣∣

×{

p∑

i=1

[∫ ti

ti−1

(ti − s)α−1

Γ(α)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds + |Ii(u(ti)) − Ii(v(ti))|]

+p−1∑

i=1

(tp − ti

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣]

+p∑

i=1

(T − tp

)[∫ ti

ti−1

(ti − s)α−2

Γ(α − 1)

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣]}


≤ 1|b|

{2(1 + p

)(|b|T + 1)Tα−1K1

Γ(α + 1)+

[5p + 1 + 2|b|T(2p − 1

)]Tα−1K1

Γ(α)

+2(1 + |b|T)pK2

T+[5p − 2 + 2|b|T(2p − 1

)]K3

}

‖u − v‖

=Λ|b| ‖u − v‖,

(3.16)

which, by (3.15), yields ‖Tu − Tv‖ < ‖u − v‖. So, G is a contraction. Therefore, by the Banachcontraction mapping principle, problem (1.1) has a unique solution.

Example 3.6. Consider the following fractional impulsive irregular boundary value problem

CDαu(t) =e3tcos5[u(t) + eu(t)

]

1 + u4(t)+

sin(t + 1)√

5 + u2(t)|u|ρ, 0 < t < 1, t /=

14,

Δu(

14

)= 2 + 3sin2

[ln(

1 + 2u2(

14

))], Δu′

(14

)=

7 + 2u2(1/4)2 + u2(1/4)

,

u′(0) + (−1)θu′(1) + bu(1) = 0, u(0) + (−1)θ+1u(1) = 0, θ = 1, 2, b /= 0,

(3.17)

where 1 < α ≤ 2 and p = 1.

Observe that

∣∣f(t, u)∣∣ =

∣∣∣∣∣e3tcos5[u(t) + eu(t)

]

1 + u4(t)+

sin(t + 1)√

5 + u2(t)|u|ρ∣∣∣∣∣≤ e3t + |u|ρ. (3.18)

Clearly, a(t) = e3t, ξ = 1, L2 = 5, L3 = 7/2, and the conditions of Theorem 3.1 hold for 0 < ρ < 1.Thus, by Theorem 3.1, problem (3.17) has at least one solution. In a similar way, for ρ > 1,the impulsive irregular fractional boundary value problem (3.17) has at least one solution bymeans of Theorem 3.3.

Example 3.7. Consider the impulsive fractional irregular boundary value problem given by

CDαu(t) = t2(1 − cosu(t)) + e(3+t)u4(t), 0 < t < 1, t /=15,

Δu(

15

)=

arctan2u(1/5)5

, Δu′(

15

)= eu

3(1/5) − 1,

u′(0) + (−1)θu′(1) + bu(1) = 0, u(0) + (−1)θ+1u(1) = 0, θ = 1, 2, b /= 0,

(3.19)

where 1 < α ≤ 2 and p = 1.


It can easily be verified that all the assumptions of Theorem 3.4 are satisfied. Thus, bythe conclusion of Theorem 3.4, we deduce that the problem (3.19) has at least one solution.

Example 3.8. Consider

CD7/4u(t) = 100t5 +t2

200e−cos2u(t), 0 < t < 1, t /=

34,

Δu(

34

)=

19

cosu(

34

), Δu′

(34

)=

|u(3/4)|5(1 + |u(3/4)|) ,

u′(0) + (−1)θu′(1) + 8u(1) = 0, u(0) + (−1)θ+1u(1) = 0, θ = 1, 2.

(3.20)

Here q = 7/4, b = 8, T = 1, and p = 1. With

K1 =1

200, K2 =

19, K3 =

15, (3.21)

we find that

Λ =2(1 + p

)(|b|T + 1)Tα−1K1

Γ(α + 1)+

[5p + 1 + 2|b|T(2p − 1

)]Tα−1K1

Γ(α)+

2(1 + |b|T)pK2

T

+[5p − 2 + 2|b|T(2p − 1

)]K3 < 6.031587 < |b| = 8.

(3.22)

Thus, all the conditions of Theorem 3.5 are satisfied. Consequently, the conclusion ofTheorem 3.5 applies and the fractional order impulsive irregular boundary value problem(3.20) has a unique solution on [0, 1].

Acknowledgment

The research of G. Wang and L. Zhang was supported by the Natural Science Foundation forYoung Scientists of Shanxi Province (2012021002-3), China.

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Research ArticlePositive Solutions of Nonlinear FractionalDifferential Equations with Integral BoundaryValue Conditions

J. Caballero, I. Cabrera, and K. Sadarangani

Departamento de Matematicas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja,35017 Las Palmas de Gran Canaria, Spain

Correspondence should be addressed to K. Sadarangani, [email protected]

Received 28 February 2012; Accepted 7 September 2012


Copyright q 2012 J. Caballero et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We investigate the existence and uniqueness of positive solutions of the following nonlinearfractional differential equation with integral boundary value conditions CDαu(t) + f(t, u(t)) = 0,0 < t < 1, u(0) = u′′(0) = 0, u(1) = λ

∫10 u(s)ds, where 2 < α < 3, 0 < λ < 2 and CDα is the Caputo

fractional derivative and f : [0, 1] × [0,∞) → [0,∞) is a continuous function. Our analysis relieson a fixed point theorem in partially ordered sets. Moreover, we compare our results with othersthat appear in the literature.

1. Introduction

Many papers and books on fractional differential equations have appeared recently (see, forexample, [1–22]). The interest of the study of fractional-order differential equations lies in thefact that fractional-order models are more accurate than integer-order models, that is, thereare more degrees of freedom in the fractional-order models.

Integral boundary conditions have various applications in chemical engineering,thermo-elasticity, population dynamics, and so forth. For a detailed description of the integralboundary conditions, we refer the reader to some recent papers (see, [23–30]) and thereferences therein. Recently, Cabada and Wang in [31] investigated the existence of positivesolutions for the fractional boundary value problem

CDαu(t) + f(t, u(t)) = 0, 0 < t < 1,

u(0) = u′′(0) = 0, u(1) = λ∫1

0u(s)ds,

(1.1)


where 2 < α < 3, 0 < λ < 2, CDα is the Caputo fractional derivative and f : [0, 1] × [0,∞) →[0,∞) is a continuous function.

The main tool used in [31] is the well-known Guo-Krasnoselskii fixed point theoremand the question of uniqueness of solutions is not treated. We consider our paper as analternative answer to the results of [31]. The fixed point theorem in partially ordered setsis the main tool used in our results. The existence of fixed points in partially ordered sets hasbeen considered recently (see, e.g. [32–34]).

2. Preliminaries and Basic Facts

For the convenience of the reader, we present in this section some notations and lemmaswhich will be used in the proofs of our results. For details, see [35, 36].

Definition 2.1. The Caputo derivative of fractional order α > 0 of a function f : [0,∞) → R isdefined by

CDαf(t) =1

Γ(n − α)∫ t

0(t − s)n−α−1f (n)(s)ds, (2.1)

where n = [α] + 1 and [α] denotes the integer part of α.

Definition 2.2. The Riemman-Liouville fractional integral of order α > 0 of a function f :(0,∞) → R is defined by

Iαf(t) =1

Γ(α)

∫ t

0(t − s)α−1f(s)ds, (2.2)

provided that such integral exists.

Definition 2.3. The Riemman-Liouville fractional derivative of order α > 0 of a function f :(0,∞) → R is given by

Dαf(t) =1

Γ(n − α)(d

dt

)n ∫ t

0(t − s)n−α−1f(s)ds, (2.3)

where n = [α] + 1, provided that the right hand side is pointwise defined on (0,∞).

Lemma 2.4. Let α > 0 then the fractional differential equation

CDαu(t) = 0, (2.4)

has

u(t) =[α]∑

j=0

u(j)(0)j!

tj , (2.5)

as unique solution.


Lemma 2.5. Let α > 0 then

Iα CDαu(t) = u(t) −[α]∑

j=0

u(j)(0)j!

tj . (2.6)

In [31], the authors obtain the Green’s function associated with Problem (1.1). Moreprecisely, they proved the following result.

Theorem 2.6 (see [31]). Let 2 < α < 3 and λ/= 2. Suppose that f ∈ C[0, 1] then the unique solutionof

CDαu(t) + f(t) = 0, 0 < t < 1,

u(0) = u′′(0) = 0, u(1) = λ∫1

0u(s)ds

(2.7)

is u(t) =∫1

0 G(t, s)f(s)ds, where

G(t, s) =1

(2 − λ)Γ(α + 1)

{2t(1 − s)α−1(α − λ + λs) − (2 − λ)α(t − s)α−1, 0 ≤ s ≤ t ≤ 1,2t(1 − s)α−1(α − λ + λs), 0 ≤ t ≤ s ≤ 1.

(2.8)

In [31], the following lemma is proved.

Lemma 2.7. Let G(t, s) be the Green’s function associated to Problem (2.7), which has the expression(2.8). Then:

(i) G(t, s) > 0 for all t, s ∈ (0, 1) if and only if λ ∈ [0, 2).

(ii) G(t, s) ≤ 2/(2 − λ)Γ(α) for all t, s ∈ [0, 1] and λ ∈ [0, 2).

(iii) For 2 < α < 3 and λ/= 2G(t, s) is a continuous function on [0, 1] × [0, 1].

In the sequel, we present the fixed point theorem which we will be use later. This resultappears in [32].

Theorem 2.8 (see [32]). Let (X,≤) be a partially ordered set and suppose that there exists a metricd in X such that (X, d) is a complete metric space. Let T : X → X be a nondecreasing mapping suchthat there exists an element x0 ∈ X with x0 ≤ Tx0.

Suppose that

d(Tx, Ty

) ≤ d(x, y) − ψ(d(x, y)) for x, y ∈ X with x ≥ y, (2.9)

where ψ : [0,∞) → [0,∞) is a continuous and nondecreasing function, ψ is positive on (0,∞) andψ(0) = 0.

Assume that either T is continuous or X is such that

if {xn} is a nondecreasing sequencein X such that xn → x then xn ≤ x, ∀n ∈ N. (2.10)


Besides, if

for each x, y ∈ X there exists z ∈ X which is comparable to x and y, (2.11)

then T has a unique fixed point.

Remark 2.9. Notice that the condition limt→∞ψ(t) = ∞ is superfluous in Theorem 2 of [32].

Remark 2.10. If we look at the proof of Theorem 2.2 in [32] we notice that the condition aboutthe continuity of ψ is redundant.

In fact, from x0 ≤ Tx0 the authors generate the sequence {Tnx0} and if we put xn+1 =Tnx0 it is proved that

d(xn+1, xn) = d(Txn, Txn−1) ≤ d(xn, xn−1) − ψ(d(xn, xn−1)) ≤ d(xn, xn−1). (2.12)

Consequently, {d(xn+1, xn)} is a nonnegative decreasing sequence of real numbers andhence {d(xn+1, xn)} posseses a limit ρ∗.

Taking limit when n → ∞ in the last inequality, we obtain

ρ∗ ≤ ρ∗ − limn→∞

ψ(d(xn, xn−1)) ≤ ρ∗, (2.13)

and, therefore,

limn→∞

ψ(d(xn, xn−1)) = 0. (2.14)

Suppose that ρ∗ > 0, since {d(xn+1, xn)} is a decreasing sequence ρ∗ ≤ d(xn+1, xn) for alln ∈ N, and, since ψ is a nondecreasing function, we have ψ(ρ∗) ≤ ψ(d(xn+1, xn)) for all n ∈ N.

As ψ is positive on (0,∞), 0 < ψ(ρ∗) ≤ ψ(d(xn+1, xn)) for all n ∈ N and, therefore,

0 < ψ(ρ∗) ≤ limψ(d(xn+1, xn)). (2.15)

This contradicts to (2.14). Consequently, ρ∗ = 0.The rest of the proof works well and the condition about the continuity of ψ is not

used.

Theorem 2.11. Theorem 2.8 is valid without the assumption ψ : [0,∞) → [0,∞) is continuous.

In our considerations, we will work in the Banach space C[0, 1] = {x : [0, 1] →R, continuous} with the classical metric given by d(x, y) = sup0≤t≤1{|x(t) − y(t)|}.

Notice that this space can be equipped with a partial order given by

x, y ∈ C[0, 1], x ≤ y ⇐⇒ x(t) ≤ y(t) for t ∈ [0, 1]. (2.16)

In [33] it is proved that (C[0, 1],≤) satisfies condition (2.10) of Theorem 2.8. Moreover,for x, y ∈ C[0, 1], as the function max{x, y} ∈ C[0, 1], (C[0, 1],≤) satisfies condition (2.11) ofTheorem 2.8.


3. Main Result

Our starting point in this section is to present the class of function A which we will use later.By A we will denote the class of those functions φ : [0,∞) → [0,∞) which are nondecreasingand such that if ϕ(x) = x − φ(x) then the following conditions are satisfied:

(a) ϕ : [0,∞) → [0,∞) and ϕ is nondecreasing.

(b) ϕ(0) = 0.

(c) ϕ is positive on (0,∞).

Examples of such functions are φ(x) = arctanx and φ(x) = x/(1 + x).In what follows, we formulate our main result.

Theorem 3.1. Suppose that 2 < α < 3, 0 < λ < 2 and f : [0, 1] × [0,∞) → [0,∞) satisfies thefollowing assumptions:

(i) f is continuous.

(ii) f(t, x) is nondecreasing respect to the second argument for each t ∈ [0, 1].

(iii) There exist 0 < ρ ≤ (2 − λ)Γ(α)/2 and φ ∈ A such that

f(t, y

) − f(t, x) ≤ ρφ(y − x), (3.1)

for x, y ∈ [0,∞) with y ≥ x and t ∈ [0, 1].

Then Problem (1.1) has a unique nonnegative solution.

Proof. Consider the cone

P = {u ∈ C[0, 1] : u ≥ 0}. (3.2)

Notice that, as P is a closed set ofC[0, 1], P is a complete metric with the distance givenby d(x, y) = sup0≤t≤1{|x(t) − y(t)|} satisfying conditions (2.10) and (2.11) of Theorem 2.8.

Now, for u ∈ P we define the operator T by

(Tu)(t) =∫1

0G(t, s)f(s, u(s))ds, (3.3)

where G(t, s) is the Green’s function defined by (2.8).By Lemma 2.7 and assumption (i), it is clear that T applies P into itself.In the sequel, we will check that the assumptions of Theorem 2.11 are satisfied.Firstly, the operator T is nondecreasing. In fact, by (ii), for u ≥ v we have

(Tu)(t) =∫1

0G(t, s)f(s, u(s))ds ≥

∫1

0G(t, s)f(s, v(s))ds = (Tv)(t). (3.4)


Besides, for u ≥ v and taking into account our assumptions, we can obtain

d(Tu, Tv) = max0≤t≤1

{|(Tu)(t) − (Tv)(t)|}

= max0≤t≤1

{Tu(t) − Tv(t)}

= max0≤t≤1

[∫1

0G(t, s)

(f(s, u(s)) − f(s, v(s)))ds

]

≤ max0≤t≤1

[∫1

0G(t, s)ρφ(u(s) − v(s))ds

]

.

(3.5)

Since φ is nondecreasing and u ≥ v, we have

φ(u(s) − v(s)) ≤ φ(d(u, v)), (3.6)

and, from the last inequality it follows

d(Tu, Tv) ≤ ρφ(d(u, v))max0≤t≤1

∫1

0G(t, s)ds. (3.7)

By Lemma 2.7(ii) and since ρ ≤ (2 − λ)Γ(α)/2, we can obtain

d(Tu, Tv) ≤ φ(d(u, v)) = d(u, v) − [d(u, v) − φ(d(u, v))]. (3.8)

Since φ ∈ A, ψ(x) = x − φ(x) satifies the properties (a), (b), and (c) mentioned at thebeginning of this section, for u ≥ v we have

d(Tu, Tv) ≤ d(u, v) − ψ(d(u, v)). (3.9)

Finally, taking into account that the zero function, 0 ≤ T0 (where T0 = Tu with u(t) = 0for all t ∈ [0, 1]), by Theorem 2.11, Problem (1.1) has a unique nonnegative solution.

Now, we present a sufficient condition for the existence and uniqueness of a positivesolution for Problem (1.1) (positive solution means a solution satisfying x(t) > 0 for t ∈ (0, 1)).

Theorem 3.2. Under assumptions of Theorem 3.1 and adding the following condition

f(t0, 0)/= 0 for certain t0 ∈ [0, 1], (3.10)

we obtain existence and uniqueness of a positive solution for Problem (1.1).

Proof. Consider the nonnegative solution x(t) for Problem (1.1) whose existence is guaran-teed by Theorem 3.1.


Notice that x(t) satisfies

x(t) =∫1

0G(t, s)f(s, x(s))ds. (3.11)

In what follows, we will prove that x(t) > 0 for t ∈ (0, 1).In fact, in contrary case we can find 0 < t∗ < 1 such that x(t∗) = 0, and therefore,

x(t∗) =∫1

0G(t∗, s)f(s, x(s))ds = 0. (3.12)

Since x ≥ 0 and G(t, s) ≥ 0 (see, Lemma 2.7) and, taking into account the nondecreas-ing character with respect to the second argument of the function f , from the last inequalityit follows

0 = x(t∗) =∫1

0G(t∗, s)f(s, x(s))ds ≥

∫1

0G(t∗, s)f(s, 0)ds ≥ 0. (3.13)

Thus,∫1

0 G(t∗, s)f(s, 0)ds = 0. This fact and the nonnegative character of the functions

G(t∗, s) and f(s, 0) give us

G(t∗, s)f(s, 0) = 0 a.e. (s). (3.14)

Since G(t∗, s) > 0 for s ∈ (0, 1), we get

f(s, 0) = 0 a.e. (s). (3.15)

By (3.10), since f(t0, 0)/= 0 for certain t0 ∈ [0, 1], this means that f(t0, 0) > 0, and takinginto account the continuity of f , we can find a set Ω ⊂ [0, 1] with t0 ∈ Ω and μ(Ω) > 0 (whereμ is the Lebesgue measure) such that f(t, 0) > 0 for any t ∈ Ω. This contradicts to (3.15).

Therefore, x(t) > 0.

Remark 3.3. In Theorem 3.2, the condition f(t0, 0)/= 0 for certain t0 ∈ [0, 1] seems to be a strongcondition in order to obtain a positive solution for Problem (1.1), but when the solution isunique we will see that the condition is very adjusted one. In fact, under the assumption thatProblem (1.1) has a unique nonnegative solution x(t) we have that

f(t, 0) = 0 for any t ∈ [0, 1] iff x(t) ≡ 0. (3.16)

Indeed, if f(t, 0) = 0 for any t ∈ [0, 1] then it is easily seen that the zero function is asolution for Problem (1.1) and the uniqueness of solution gives us x(t) ≡ 0.

The reverse implication is obvious.

Remark 3.4. Notice that assumptions in Theorem 3.1 are invariant by nonnegative andcontinuous perturbations. More precisely, if f(t, 0) = 0 for any t ∈ [0, 1] and f satisfies


conditions (i), (ii), and (iii) of Theorem 3.1 then g(t, x) = f(t, x) + a(t), where a : [0, 1] →[0,∞) continuous and a/= 0 satisfies assumptions of Theorem 3.2 and, consequently, theboundary value problem

CDαu(t) + g(t, u(t)) = 0, 0 < t < 1, 2 < α < 3

u(0) = u′′(0) = 0, u(1) = λ∫1

0u(s)ds,

(3.17)

with 0 < λ < 2, has a unique positive solution.

Example 3.5. Now, consider the following boundary value problem

CDαu(t) + t +

γu(t)1 + u(t)

= 0, 0 < t < 1, 2 < α < 3, γ > 0

u(0) = u′′(0) = 0, u(1) =sin 1

1 − cos 1

∫1

0u(s)ds.

(3.18)

In this case, f(t, u) = t + (γu/(1 + u)). Obviously, f : [0, 1] × [0,∞) → [0,∞) and f iscontinuous. Since ∂f/∂u = γ/(1 + u)2 > 0, f satisfies condition (ii) of Theorem 3.1.

Moreover, for u ≥ v and t ∈ [0, 1] we have

f(t, u) − f(t, v) = t + γu

1 + u− t − γv

1 + v

= γ( u

1 + u− v

1 + v

)= γ

(u − v

(1 + u)(1 + v)

)

≤ γ u − v1 + u − v = γφ(u − v),

(3.19)

where φ(x) = x/(1+x). It is easily seen that φ belongs to the class A. In this case, λ = sin 1/(1−cos 1) ≈ 1, 83048, consequently 0 < λ < 2, and for 0 < γ ≤ (2 − λ)/2 · Γ(α) ≈ 0, 08476Γ(α),Problem (3.18) satisfies (iii) of Theorem 3.1. Since f(t, 0) = t /= 0 for t /= 0, Theorem 3.2 saysthat Problem (3.18) has a unique positive solution for 0 < γ ≤ 0, 08476Γ(α), where 2 < α < 3.

4. Some Remarks and Examples

In [31] the authors consider Problem (1.1).In order to present the main result of [31] we need the following notation. Denote by

f0 and f∞ the following limits:

f0 = limu→ 0+

{mint∈[0,1]

f(t, u)u

}, f∞ = lim

u→∞

{maxt∈[0,1]

f(t, u)u

}. (4.1)

The main result of [31] is the following theorem.


Theorem 4.1. Assume that one of the two following conditions is fulfilled:

(i) (sublinear case) f0 = ∞ and f∞ = 0,

(ii) (superlinear case) f0 = 0, f∞ = ∞ and there exist μ > 0 and θ > 0 for which f(t, ρx) ≥μρθf(t, x) for all ρ ∈ (0, 1].

Then, Problem (1.1) has at least one positive solution that belongs to

P ={u ∈ C[0, 1] :

u

id[0,1]∈ C[0, 1], u(t) ≥ tλ(α − 2)

2α‖u‖ ∀t ∈ [0, 1]

}, (4.2)

where id[0,1] is the identity mapping on [0, 1].

Notice that in Example 3.5, f(t, u) = t + (γu/(1 + u)) and in this case we have

mint∈[0,1]

f(t, u)u

=γ

1 + u. (4.3)

Consequently,

f0 = limu→ 0+

{mint∈[0,1]

f(t, u)u

}= lim

u→ 0+

γ

1 + u= γ. (4.4)

Therefore, for 0 < γ ≤ 0, 08476Γ(α) Example 3.5 cannot be treated by Theorem 4.1.

Example 4.2. Consider the following boundary value problem

CDαu(t) + c + γ arctan u(t) = 0, 0 < t < 1, c > 0, γ > 0, 2 < α < 3,

u(0) = u′′(0) = 0, u(1) =12

∫1

0u(s)ds,

(4.5)

In this case, 0 < λ = (1/2) < 2, f(t, u) = c + γ arctanu.It is easily proved that f satisfies condition (i) and (ii) of Theorem 3.1. In [37], it is

proved that if u ≥ v ≥ 0

arctan u − arctan v ≤ arctan(u − v). (4.6)

Using this fact, for u ≥ v ≥ 0 and t ∈ [0, 1], we have

f(t, u) − f(t, v) = γ(arctan u − arctan v) ≤ γ arctan(u − v) = γφ(u − v), (4.7)

where φ(x) = arctanx. It is easily proved that φ ∈ A.Then, for 0 < γ ≤ (3/4)Γ(α), the function f satisfies condition (iii) of Theorem 3.1.

Moreover, since f(t, 0) = c > 0, Theorem 3.2 gives us the existence and uniqueness of apositive solution for Problem (4.5) when 0 < γ ≤ (3/4)Γ(α).


On the other hand, we have

maxt∈[0,1]

f(t, u)u

= mint∈[0,1]

f(t, u)u

=c + γ arctan u

u,

f0 = limu→ 0+

c + γ arctan u

u= ∞, f∞ = lim

u→∞c + γ arctan u

u= 0.

(4.8)

Consequently, this example corresponds to the sublinear case of Theorem 4.1.Therefore, Theorem 4.1 gives us the existence of at least one positive solution for 0 < γ ≤(3/4)Γ(α). The question of uniqueness of solutions is not treated in [31].

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Research ArticleBifurcation Analysis for a Predator-Prey Modelwith Time Delay and Delay-Dependent Parameters

Changjin Xu

Guizhou Key Laboratory of Economics System Simulation, School of Mathematics and Statistics,Guizhou University of Finance and Economics, Guiyang 550004, China

Correspondence should be addressed to Changjin Xu, [email protected]

Received 13 January 2012; Accepted 20 August 2012

Academic Editor: Benchawan Wiwatanapataphee

Copyright q 2012 Changjin Xu. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

A class of stage-structured predator-prey model with time delay and delay-dependent parametersis considered. Its linear stability is investigated and Hopf bifurcation is demonstrated. Usingnormal form theory and center manifold theory, some explicit formulae for determining thestability and the direction of the Hopf bifurcation periodic solutions bifurcating from Hopfbifurcations are obtained. Finally, numerical simulations are performed to verify the analyticalresults.

1. Introduction

Over the past decade, a great many predator-prey models have been developed todescribe the interaction between predator and prey. Their dynamical phenomena havebeen extensively studied because of the wide application in the field of biomathematics. Inparticular, the appearance of a cycle bifurcating from the equilibrium of an ordinary or adelayed predator-prey model with a single parameter, which is known as a Hopf bifurcation,has attracted much attention due to its theoretical and practical significance [1–5]. But mostof the research literature on these models are only connected with parameters which areindependent of time delay; thus, the corresponding characteristic equations are easy to dealwith. While in most applications of delay differential equations in population dynamics, theneed of incorporation of a time delay is often the result of existence of some stage structure [6–8]. Indeed, every population goes through some distinct life stages [9, 10]. Since the through-stage survival rate is often a function of time delay, it is easy to conceive that thesemodels willinevitably involve some delay-dependent parameters. Thus, the corresponding characteristicequations dependent on the delay τ become more complicated. In view of the fact that it is


often difficult to analytically study models with delay-dependent parameters even if only asingle discrete delay is present, we resort to the help of computer programs.

In 2008, Wang et al. [11] introduced and investigated the following predator-preyinteraction model with time delay and delay-dependent parameters:

x(t) = x(t)[a − bx(t)] − cx2(t)y(t)1 +�x2(t)

,

y(t) =cx2(t − τ)y(t − τ)1 +�x2(t − τ) e−dτ − dy(t),

(1.1)

where x(t) and y(t) stand for prey and predator density at time t, respectively. a, b, c,� arereal positive parameters and the time delay τ is a positive constant. Wang et al. [11] obtainedthe conditions that guarantee the system asymptotically stable and permanent. For moreknowledge about the model, one can see [11].

It is well known that time delays which occur in the interaction between predator-preywill affect the stability of a model by creating instability, oscillation, and chaos phenomena.Based on the discussion above, the main purpose of this paper is to investigate the stabilityand the properties of Hopf bifurcation of the model (1.1) which involves some delay-dependent parameters. Recently, there are few papers on the topic that involves some delay-dependent parameters, for example, Liu and Zhang [12] investigated the stability and Hopfbifurcation of the following SIS model with nonlinear birth rate:

I(t) = β(N(t) − I(t)) I(t)N(t)

− (d + ε + γ)I(t),

N(t) =PN(t − τ)

1 + qN3(t − τ)e−d1τ − dN(t) − εI(t).

(1.2)

Jiang and Wei [13] studied the stability and Hopf bifurcation of the following SIR model:

S(t) = μ − μS(t) − φI(t)S(t)1 + I(t)

+ γI(t − τ)e−μτ ,

I(t) =φI(t)S(t)1 + I(t)

− (μ + γ)I(t).

(1.3)

It worth pointing out that Liu and Zhang [12] investigated the Hopf bifurcation of system(1.2) by choosing p (not delay τ) as the bifurcation parameters and Jiang and Wei [13]studied the Hopf bifurcation of system (1.3) by choosing φ (not delay τ) as the bifurcationparameters. In this paper, we will investigate the Hopf bifurcation by regarding the delay τas the bifurcation parameter which is different from the papers [12, 13]. To the best of ourknowledge, it is the first time to deal with the stability and Hopf bifurcation of system (1.1).

This paper is organized as follows. In Section 2, the stability of the equilibrium andthe existence of Hopf bifurcation at the equilibrium are studied. In Section 3, the direction ofHopf bifurcation and the stability and periodic of bifurcating periodic solutions on the centermanifold are determined. In Section 4, numerical simulations are carried out to illustrate thevalidity of the main results. Some main conclusions are drawn in Section 5.


2. Stability of the Equilibrium and Local Hopf Bifurcations

Throughout this paper, we assume that the following condition

(H1) ce−dτ > d�, a2(ce−dτ − d�) > b2d holds.

The hypothesis (H1) implies that system (1.1) has a unique positive equilibriumE∗(x∗, y∗), where

x∗ =

√d

ce−dτ − d� , y∗ =(a − bx∗)

(1 +�x∗2)

cx∗ . (2.1)

The linearized system of (1.1) around E∗(x∗, y∗) takes the form

dx(t)dt

= a1x(t) − b1y(t),

dy(t)dt

= −dy(t) + c1x(t − τ) + d1y(t − τ),(2.2)

where

a1 = a − 2bx∗ − 2cx∗y∗

1 +�x∗2 +2x∗3y∗�

(1 +�x∗)2, b1 = − cx∗2

1 +�x∗2 ,

c1 = e−dτ[

2cx∗y∗

1 +�x∗2 − 2x∗3y∗�

(1 +�x∗)2

]

, d1 = e−dτcx∗2

1 +�x∗2 .

(2.3)

The associated characteristic equation of (2.2) is

P(λ, τ) +Q(λ, τ)e−λτ = 0, (2.4)

where

P(λ, τ) = λ2 + (d − a1)λ − a1d,Q(λ, τ) = − (d1λ − a1d1 − b1c1).

(2.5)

When τ = 0, then (2.4) becomes

λ2 +(d − a1 − d0

1

)λ + b1c01 − a1d − a1d0

1 = 0, (2.6)

where

c01 =

[2cx∗y∗

1 +�x∗2 − 2x∗3y∗�

(1 +�x∗)2

]

, d01 =

cx∗

1 +�x∗2 . (2.7)

It is easy to obtain the following result.


Lemma 2.1. If the condition(H2) d − a1 − d0

1 > 0, b1c01 − a1d − a1d01 > 0,

holds, then the positive equilibrium E∗(x∗, y∗) of system (1.1) is asymptotically stable.

In the following, one investigates the existence of purely imaginary roots λ = iω (ω >0) of (2.4). Equation (2.4) takes the form of a second-degree exponential polynomial in λ,which some of the coefficients of P and Q depend on τ . Beretta and Kuang [14] establisheda geometrical criterion which gives the existence of purely imaginary roots of a characteristicequation with delay-dependent coefficients. In order to apply the criterion due to Beretta andKuang [14], one needs to verify the following properties for all τ ∈ [0, τmax), where τmax is themaximum value which E∗(x∗, y∗) exists.

(a) P(0, τ) +Q(0, τ)/= 0;

(b) P(iω, τ) +Q(iω, τ)/= 0;

(c) lim sup{|Q(λ, τ)/P(λ, τ)| : |λ| → ∞,Reλ ≥ 0} < 1;

(d) F(ω, τ) = |P(iω, τ)|2 − |Q(iω, τ)|2 has a finite number of zeros;

(e) Each positive root ω(τ) of F(ω, τ) = 0 is continuous and differentiable in τwhenever it exists.

Here, P(λ, τ) and Q(λ, τ) are defined as in (2.5), respectively.Let τ ∈ [0, τmax). It is easy to see that

P(0, τ) +Q(0, τ) = −a1d + a1d1 + b1c1 /= 0, (2.8)

which implies that (a) is satisfied, and (b)

P(iω, τ) +Q(iω, τ) = −ω2 + iω(d − a1) − a1d − iωd1 + a1d1 + b1c1= −ω2 − a1d + a1d1 + b1c1 + iω(d − a1 − d1)/= 0.

(2.9)

From (2.4), one has

lim|λ|→+∞

∣∣∣∣Q(λ, τ)P(λ, τ)

∣∣∣∣ = lim|λ|→+∞

∣∣∣∣−(d1λ + a1d1 − b1c1)λ2 + (d − a1)λ − a1d

∣∣∣∣ = 0. (2.10)

Therefore, (c) follows. Let F be defined as in (d). From

|P(iω, τ)|2 =(ω2 + a1d

)2+ (d − a1)2ω2

= ω4 +(d2 + a21

)ω2 + a21d

2,

|Q(iω, τ)|2 = d21ω

2 + (a1d1 + b1c1)2,

(2.11)

one obtain

F(ω, τ) = ω4 +(d2 + a21 − d2

1

)ω2 + a1d2 − (a1d1 + b1c1)

2. (2.12)

Obviously, property (d) is satisfied, and by implicit function theorem, (e) is also satisfied.


Now let λ = iω (ω > 0) be a root of (2.4). Substituting it into (2.4) and separating thereal and imaginary parts yields

(a1d1 + b1c1) cosωτ − d1ω sinωτ = ω2 + a1d,

d1ω cosωτ + (a1d1 + b1c1) sinωτ = (d − a1)ω.(2.13)

From (2.13), it follows that

sinωτ = −(ω2 + a1d

)d1ω − (d − a1)ω(a1d1 + b1c1)

d21ω

2 + (a1d1 + b1c1)2

,

cosωτ =

(ω2 + a1d

)(a1d1 + b1c1) + (d − a1)ωd1ω

d21ω

2 + (a1d1 + b1c1)2

.

(2.14)

By the definitions of P and Q as in (2.5), respectively, and applying the property (a), then(2.14) can be written as

sinωτ = Im[P(iω, τ)Q(iω, τ)

],

cosωτ = −Re[P(iω, τ)Q(iω, τ)

],

(2.15)

which yields |P(iω, τ)|2 = |Q(iω, τ)|2. Assume that I ∈ R+0 is the set where ω(τ) is a positiveroot of

F(ω, τ) = |P(iω, τ)|2 − |Q(iω, τ)|2, (2.16)

and for τ ∈ I, ω(τ) is not definite. Then for all τ in I, ω(τ) satisfied F(ω, τ) = 0. The polynomialfunction F can be written as

F(ω, τ) = h(ω2, τ

), (2.17)

where h is a second degree polynomial, defined by

h(z, τ) = z2 +(d2 + a21 − d2

1

)z + a1d2 − (a1d1 + b1c1)

2. (2.18)

It is easy to see that

h(z, τ) = z2 +(d2 + a21 − d2

1

)z + a1d2 − (a1d1 − b1c1)2 = 0 (2.19)

has only one positive real root if the following condition (H3) holds:

(H3) a1d2 < (a1d1 + b1c1)2.


One denotes this positive real root by z+. Hence, (2.17) has only one positive real rootω =

√z+. Since the critical value of τ and ω(τ) are impossible to solve explicitly, so one

will use the procedure described in Beretta and Kuang [14]. According to this procedure,one defines θ(τ) ∈ [0, 2π) such that sin θ(τ) and cos θ(τ) are given by the righthand sides of(2.14), respectively, with θ(τ) given by (2.19). This define θ(τ) in a form suitable for numericalevaluation using standard software. And the relation between the argument θ and ωτ in(2.18) for τ > 0 must be ωτ = θ + 2nπ , n = 1, 2, . . ..

Hence, one can define the maps: τn : I → R+0 given by

τn(τ) :=θ(τ) + 2nπ

ω(τ), τn > 0, n = 0, 1, 2, . . . , (2.20)

where a positive root ω(τ) of F(ω, τ) = 0 exists in I. Let us introduce the functions Sn(τ) :I → R,

Sn(τ) = τ − θ(τ) + 2nπω(τ)

, n = 0, 1, 2, . . . , (2.21)

which are continuous and differentiable in τ . Thus, one gives the following theorem which isdue to Beretta and Kuang [14].

Theorem 2.2. Assume thatω(τ) is a positive root of (2.4) defined for τ ∈ I, I ⊆ R+0, and at some τ0 ∈I, Sn(τ0) = 0 for some n ∈ N0. Then, a pair of simple conjugate pure imaginary roots λ = ±iω existsat τ = τ0 which crosses the imaginary axis from left to right if δ(τ0) > 0 and crosses the imaginaryaxis from right to left if δ(τ0) < 0, where δ(τ0) = sign[F ′

ω(ωτ0, τ0)] sign[(dSn(τ))/dτ |τ=τ0].

Applying Lemma 2.1 and the Hopf bifurcation theorem for functional differentialequation [5], we can conclude the existence of a Hopf bifurcation as stated in the followingtheorem.

Theorem 2.3. For system (1.1), if (H1)–(H3) hold, then there exists s τ0 ∈ I such that the positiveequilibrium E∗(x∗, y∗) is asymptotically stable for 0 ≤ τ < τ0 and becomes unstable for τ staying insome right neighborhood of τ0, with a Hopf bifurcation occurring when τ = τ0.

3. Direction and Stability of the Hopf Bifurcation

In the previous section, we obtained some conditions which guarantee that the stage-structured predator-prey model with time delay undergoes the Hopf bifurcation at somevalues of τ = τ0. In this section, we will derive the explicit formulae determining thedirection, stability, and period of these periodic solutions bifurcating from the positiveequilibrium E∗(x∗, y∗) at these critical value of τ , by using techniques from normal formand center manifold theory [15]. Throughout this section, we always assume that system(1.1) undergoes Hopf bifurcation at the positive equilibrium E∗(x∗, y∗) for τ = τ0, and then±iω0 is corresponding purely imaginary roots of the characteristic equation at the equilibriumE∗(x∗, y∗).

For convenience, let τ = τ0 +μ, μ ∈ R. Then μ = 0 is the Hopf bifurcation value of (1.1).Thus, one will study Hopf bifurcation of small amplitude periodic solutions of (1.1) from thepositive equilibrium point E∗(x∗, y∗) for μ close to 0.


Let u1(t) = x(t) − x∗, u2(t) = y(t) − y∗, xi(t) = ui(τt), (i = 1, 2), τ = τ0 + μ, then system(1.1) can be transformed into an functional differential equation (FDE) in (C = C(−1, 0], R2)as

du

dt= Lμ(ut) + f

(μ, ut

), (3.1)

where u(t) = (x1(t), x2(t))T ∈ R2 and Lμ : C → R, f : R × C → R are given, respectively, by

Lμφ =(τ0 + μ

)Bφ(0) +

(τ0 + μ

)Gφ(−1), (3.2)

where

B =(a1 −b10 −d

), C =

(0 0c1 d1

),

a1 = a − 2bx∗ − 2cx∗y∗

1 +�x∗2 +2x∗3y∗�

(1 +�x∗)2, b1 = − cx∗

1 +�x∗2 ,

c1 = e−dτ[

2cx∗y∗

1 +�x∗2 − 2x∗3y∗�

(1 +�x∗)2

]

, d1 = e−dτcx∗

1 +�x∗2

f(μ, φ)=

(f1(μ, φ)

f2(μ, φ)

)

,

(3.3)

where

f1(μ, φ)=(τ0 + μ

)[m1φ

21(0) +m2φ1(0)φ2(0) +m3φ

31(0) +m4φ

21(0)φ2(0) + h.o.t.

],

f2(μ, φ)=(τ0 + μ

)[n1φ

21(−1) + n2φ1(−1)φ2(−1) + n3φ3

1(−1) +m4φ21(−1)φ2(−1) + h.o.t.

],

(3.4)

where

m1 =4x∗2y∗�

(1 +�x∗)2− cy∗

1 +�x∗2 − cx∗2y∗(4�2x∗ −�x∗2 −�)

(1 +�x∗)4,

m2 =2c�x∗3

(1 +�x∗2)2− 2cx∗

1 +�x∗2 ,


m3 =2c�x∗y∗

(1 +�x∗2)2− cx∗2y∗g

6− 2cx∗y∗(4�2x∗ −�2x∗2 −�)

(1 +�x∗2)4,

m4 =4c�x∗2

(1 +�x∗2)2− c

1 +�x∗ − cx∗2(4�2x∗ −�2x∗2 −�)

(1 +�x∗2)4,

n1 = e−dτ0[

cy∗

1 +�x∗2 +cx∗2y∗(4�2x∗ −�x∗2 −�)

(1 +�x∗)4− 4x∗2y∗�

(1 +�x∗)2

]

,

n2 = e−dτ0[

2cx∗

1 +�x∗2 − 2c�x∗3

(1 +�x∗2)2

]

,

n3 = e−dτ0[cx∗2y∗g1

6+2cx∗y∗(4�2x∗ −�2x∗2 −�)

(1 +�x∗2)4− 2c�x∗y∗

(1 +�x∗2)2

]

,

n4 = e−dτ0[

c

1 +�x∗ +cx∗2(4�2x∗ −�2x∗2 −�)

(1 +�x∗2)4− 4c�x∗2

(1 +�x∗2)2

]

,

(3.5)

where

g =18�2x∗(1 +�2x∗2)3(�2x∗4 − 4�2x∗3 + 2�x∗2 − 4�x∗ + 1

)

(1 +�x∗2)8

− 8�2(x∗ + 4x∗3 − 3�x∗2 − 4�)

(1 +�x∗2)4.

(3.6)

Clearly, Lμ is a linear continuous operator from C to R2. By the Riesz representationtheorem, there exists a matrix function with bounded variation components η(θ, μ), θ ∈[−1, 0] such that

Lμφ =∫0

−1dη(θ, μ)φ(θ), for φ ∈ C. (3.7)

In fact, we can choose

η(θ, μ)=(τ0 + μ

)(a1 −b10 −d

)δ(θ) − (τ0 + μ

)(0 0c1 d1

)δ(θ + 1), (3.8)

where δ is the Dirac delta function.


For φ ∈ C([−1, 0], R2), define

A(μ)φ =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

dφ(θ)dθ

, −1 ≤ θ < 0,

∫0

−1dη(s, μ)φ(s), θ = 0,

R(μ)φ =

{0, −1 ≤ θ < 0,f(μ, φ), θ = 0.

(3.9)

Then (1.1) is equivalent to the abstract differential equation

ut = A(μ)ut + R

(μ)ut, (3.10)

where u = (x1, x2)T , ut(θ) = u(t + θ), θ ∈ [−1, 0].

For ψ ∈ C([0, 1], (R2)∗), define

A∗ψ(s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

−dψ(s)ds

, s ∈ (0, 1],

∫0

−1dηT (t, 0)ψ(−t), s = 0.

(3.11)

For φ ∈ C([−1, 0], R2) and ψ ∈ C([0, 1], (R2)∗), define the bilinear form

⟨ψ, φ⟩= ψ(0)φ(0) −

∫0

−1

∫θ

ξ=0ψ(ξ − θ)dη(θ)φ(ξ)dξ, (3.12)

where η(θ) = η(θ, 0). We have the following result on the relation between the operatorsA = A(0) and A∗.

Lemma 3.1. A = A(0) and A∗ are adjoint operators.

The proof is easy from (3.12), so we omit it.By the discussions in Section 2, we know that ±iω0τ0 are eigenvalues of A(0), and

they are also eigenvalues of A∗ corresponding to iω0τ0 and −iω0τ0, respectively. We have thefollowing result.

Lemma 3.2. The vector

q(θ) =(1, γ)Teiω0τ0θ, θ ∈ [−1, 0], (3.13)

where

γ =iω0 − a1

b1(3.14)


is the eigenvector of A(0) corresponding to the eigenvalue iω0τ0, and

q∗(s) = D(1, γ∗

)eiω0τ0s, s ∈ [0, 1], (3.15)

where

γ∗ = − iω0a1c1e−iωτ0

(3.16)

is the eigenvector of A∗ corresponding to the eigenvalue −iω0τ0; moreover, 〈q∗(s), q(θ)〉 = 1, where

D = 1 + γγ∗ + c1γ∗eiω0τ0 + d1γγ∗eiω0τ0 . (3.17)

Proof. Let q(θ) be the eigenvector of A(0) corresponding to the eigenvalue iω0τ0 and q∗(s) bethe eigenvector of A∗ corresponding to the eigenvalue −iω0τ0, namely, A(0)q(θ) = iω0τ0q(θ)and A∗q∗T (s) = −iω0τ0q

∗T (s). From the definitions of A(0) and A∗, we have A(0)q(θ) =dq(θ)/dθ and A∗q∗T (s) = −dq∗T (s)/ds. Thus, q(θ) = q(0)eiω0τ0θ and q∗(s) = q∗(0)eiω0τ0s. Inaddition,

∫0

−1dη(θ)q(θ) = τ0Bq(0) + τ0Gq(−1) = τ0A(0)q(0) = iω0τ0q(0). (3.18)

That is,

(iω0 − a1 −b1−c1e−iω0τ0 iω0 + d − d1e−iω0τ0

)q(0) =

(00

). (3.19)

Therefore, we can easily obtain

γ =iω0 − a1

b1. (3.20)

And so

q(0) =(1,iω0 − a1

b1

)T, (3.21)

and hence

q(θ) =(1,iω0 − a1

b1

)Teiω0τ0θ. (3.22)

On the other hand,

∫0

−1q∗(−t)dη(t) = τ0BTq∗T (0) + τ0GTq∗T (−1) = τ0A∗q∗T (0) = −iω0τ0q

∗T (0). (3.23)


Namely,

(−iω0 − a1 −c1e−iω0τ0

b1 −iω0 + d − d1e−iω0τ0

)q∗T (0) =

(00

). (3.24)

Therefore, we can easily obtain

γ∗ = − iω0a1c1e−iω0τ0

, (3.25)

and so

q∗(0) =(1,− iω0a1

c1e−iωτ0

), (3.26)

and hence

q∗(s) =(1,− iω0a1

c1e−iωτ0

)eiω0τ0s. (3.27)

In the sequel, one will verify that 〈q∗(s), q(θ)〉 = 1. In fact, from (3.12), we have

⟨q∗(s), q(θ)

⟩= D(1, γ∗

)(1, γ)T

−∫0

−1

∫0

ξ=0D(1, γ∗

)e−iω0(ξ−θ) dη(θ)

(1, γ)Teiω0ξ dξ

= D

[

1 + γγ∗ −∫0

−1

(1, γ∗

)θeiω0θ dη(θ)

(1, γ)T]

= D{1 + γγ∗ −

(1, γ∗

)[−τ0Ce−iω0τ0

](1, γ)T}

= D[1 + γγ∗ + c1γ

∗e−iω0τ0 + d1γγ∗e−iω0τ0]= 1.

(3.28)

Next, we use the same notations as those in Hassard et al. [15], and we first computethe coordinates to describe the center manifold C0 at μ = 0. Let ut be the solution of (1.1)when μ = 0.

Define

z(t) =⟨q∗, ut

⟩, W(t, θ) = ut(θ) − 2Re

{z(t)q(θ)

}, (3.29)

on the center manifold C0, and we have

W(t, θ) =W(z(t), z(t), θ), (3.30)


where

W(z(t), z(t), θ) =W(z, z) =W20z2

2+W11zz +W02

z2

2+ · · · , (3.31)

and z and z are local coordinates for center manifold C0 in the direction of q∗ and q∗. NotingthatW is also real if ut is real, we consider only real solutions. For solutions ut ∈ C0 of (1.1) ,

z(t) =⟨q∗(s), ut

⟩=⟨q∗(s), A(0)ut + R(0)ut

⟩

=⟨q∗(s), A(0)ut

⟩+⟨q∗(s), R(0)ut

⟩

=⟨A∗q∗(s), ut

⟩+ q∗(0)R(0)ut −

∫0

−1

∫θ

ξ=0q∗(ξ − θ)dη(θ)A(0)R(0)ut(ξ)dξ

=⟨iω0q

∗(s), ut⟩+ q∗(0)f(0, ut(θ))

def= iω0z(t) + q∗(0)f0(z(t), z(t)).

(3.32)

That is,

z(t) = iω0z + g(z, z), (3.33)

where

g(z, z) = g20z2

2+ g11zz + g02

z2

2+ g21

z2z

2+ · · · . (3.34)

Hence, we have

g(z, z) = q∗(0)f0(z, z) = f(0, ut) = D(1, γ∗

)(f1(0, ut), f2(0, ut)

)T, (3.35)

where

f1(0, ut) = τ0[m1x

2t (0) +m2xt(0)yt(0) +m3x

3t (0) +m4x

2t (0)yt(0) + h.o.t.

],

f2(0, ut) = τ0[n1x

2t (−1) + n2xt(−1)yt(−1) + n3x3

t (−1) +m4x2t (−1)yt(−1) + h.o.t.

].

(3.36)


Noticing ut(θ) = (xt(θ), yt(θ))T =W(t, θ) + zq(θ) + zq(θ) and q(θ) = (1, γ)Teiω0τ0θ, we have

xt(0) = z + z +W(1)20 (0)

z2

2+W (1)

11 (0)zz +W(1)02 (0)

z2

2+ · · · ,

yt(0) = γz + γ z +W(2)20 (0)

z2

2+W (2)

11 (0)zz +W(2)02 (0)

z2

2+ · · · ,

xt(−1) = e−iω0τ0z + eiω0τ0z +W (1)20 (−1)

z2

2+W (1)

11 (−1)zz +W(1)02 (−1)

z2

2+ · · · ,

yt(−1) = γe−iω0τ0βz + γeiω0τ0z +W (2)20 (−1)

z2

2+W (2)

11 (−1)zz +W(2)02 (−1)

z2

2+ · · · .

(3.37)

From (3.34) and (3.35), we have

g(z, z) = q∗(0)f0(z, z) = D[f1(0, xt) + γ∗ f2(0, xt)

]

= Dτ0[(m1 +m2γ

)+ γ∗(n1 + n2γ

)]z2

+Dτ0[2m1 +m2

(γ + γ∗

)(2n1 + n2

(γ + γ∗

))]zz

+Dτ0[m1 +m2γ + γ∗

(n1 + n2γ

)]z2

+Dτ0

[

m1

(W

(1)20 (0) + 2W (1)

11 (0))

+m2

(γW

(1)20 (0)2

+W (2)20 (0) + γW

(1)11 (0) +W

(2)11 (0) + 3m3 +m4

(γ + 2γ

))

+ n1(W

(1)20 (−1) + 2W (1)

11 (−1))

+ n2

(γW

(1)20 (−1)2

+W (2)20 (−1) + γW

(1)11 (−1)

+W (2)11 (−1) + 3n3 + n4

(γ + 2γ

))]

z2z + h.o.t,

(3.38)

and we obtain

g20 = 2Dτ0[(m1 +m2γ

)+ γ∗

(n1 + n2γ

)],

g11 = Dτ0[2m1 +m2

(γ + γ∗

)+ γ∗

(2n1 + n2

(γ + γ∗

))],

g02 = 2Dτ0[m1 +m2γ + γ∗

(n1 + n2γ

)],


g21 = 2Dτ0

[

m1

(W

(1)20 (0) + 2W (1)

11 (0))

+m2

(γW

(1)20 (0)2

+W (2)20 (0) + γW

(1)11 (0) +W

(2)11 (0) + 3m3 +m4

(γ + 2γ

))

+ n1(W

(1)20 (−1) + 2W (1)

11 (−1))

+ n2

(γW

(1)20 (−1)2

+W (2)20 (−1) + γW

(1)11 (−1) +W

(2)11 (−1) + 3n3 + n4

(γ + 2γ

))]

.

(3.39)

For unknown

W(1)20 (0),W

(1)20 (−1),W

(1)11 (0),W

(2)11 (−1),W

(2)11 (0),W

(2)11 (−1), (3.40)

in g21, we still need to compute them.From (3.10) and (3.29), we have

W ′ =

⎧⎨

⎩

W − 2Re{q∗(0)fq(θ)

}, −1 ≤ θ < 0,

W − 2Re{q∗(0)fq(θ)

}+ f, θ = 0 .

def= AW +H(z, z, θ),

(3.41)

where

H(z, z, θ) = H20(θ)z2

2+H11(θ)zz +H02(θ)

z2

2+ · · · . (3.42)

Comparing the coefficients, we obtain

(A − 2iω0τ0)W20 = −H20(θ), (3.43)

AW11(θ) = −H11(θ), (3.44)

and we know that for θ ∈ [−1, 0),

H(z, z, θ) = −q∗(0)f0q(θ) − q∗(0)f0q(θ) = −g(z, z)q(θ) − g(z, z)q(θ). (3.45)

Comparing the coefficients of (3.42) with (3.45) gives that

H20(θ) = −g20q(θ) − g02q(θ), (3.46)

H11(θ) = −g11q(θ) − g11 q(θ). (3.47)


From (3.43) and (3.46) and the definition of A, we get

W20(θ) = 2iω0τ0W20(θ) + g20q(θ) + g02 q(θ). (3.48)

Noting that q(θ) = q(0)eiω0τ0θ, we have

W20(θ) =ig20ω0τ0

q(0)eiω0τ0θ +ig02

3ω0τ0q(0)e−iω0τ0θ + E1e

2iω0τ0θ, (3.49)

where E1 = (E(1)1 , E

(2)1 )T is a constant vector.

Similarly, from (3.44) and (3.47) and the definition of A, we have

W11(θ) = g11q(θ) + g11 q(θ), (3.50)

W11(θ) = − ig11ω0τ0

q(0)eiω0τ0θ +ig11

ω0τ0q(0)e−iω0θ + E2, (3.51)

where E2 = (E(1)2 , E

(2)2 )T is a constant vector.

In what follows, one will seek appropriate E1,E2 in (3.49) and (3.51), respectively. Itfollows from the definition of A and (3.46) and (3.47) that

∫0

−1dη(θ)W20(θ) = 2iω0τ0W20(0) −H20(0), (3.52)

∫0

−1dη(θ)W11(θ) = −H11(0), (3.53)

where η(θ) = η(0, θ).From (3.43), we have

H20(0) = −g20q(0) − g02 q(0) + τ0(M1,M2)T , (3.54)

where

M1 = m1 +m2γ,

M2 = n1 + n2γ.(3.55)

From (3.44), we have

H11(0) = −g11q(0) − g11 (0)q(0) + τ0(N1,N2)T , (3.56)

where

N1 = 2m1 +m2(γ + γ

),

N2 = 2n1 + n2(γ + γ

).

(3.57)


Noting that

(

iω0τ0I −∫0

−1eiω0τ0θdη(θ)

)

q(0) = 0,

(

−iω0τ0I −∫0

−1e−iω0τ0θdη(θ)

)

q(0) = 0,

(3.58)

and substituting (3.49) and (3.54) into (3.52), we have

(

2iω0τ0I −∫0

−1e2iω0τ0θdη(θ)

)

E1 = τ0(M1,M2)T . (3.59)

That is,

(2iω0τ0I − τ0B − τ0Ge−2iω0τ0

)E1 = τ0(M1,M2)T , (3.60)

then

(2iω0 − a1 b1c1e

−2iω0τ0 2iω0 + d − d1e−2iω0τ0

)(E(1)1

E(2)1

)

=(m1 +m2γn1 + n2γ

). (3.61)

Hence,

E(1)1 =

Δ11

Δ1, E

(2)1 =

Δ12

Δ1, (3.62)

where

Δ1 = det(2iω0 − a1 b1c1e

−2iω0τ0 2iω0 + d − d1e−2iω0τ0

),

Δ11 = det(m1 +m2γ b1n1 + n2γ 2iω0 + d − d1e−2iω0τ0

),

Δ12 = det(2iω0 − a1 m1 +m2γc1e

−2iω0τ0 n1 + n2γ

).

(3.63)

Similarly, substituting (3.51) and (3.56) into (3.53), we have

(∫0

−1dη(θ)

)

E2 = τ0(N1,N2)T . (3.64)


Then,

(B +G)E2 = (−N1,−N2)T . (3.65)

That is,

(a1 −b1c1 D1 − d

)(E(1)2

E(2)2

)

=(−N1

−N2

). (3.66)

Hence,

E(1)2 =

Δ21

Δ2, E

(2)2 =

Δ22

Δ2, (3.67)

where

Δ2 = det(a1 −b1c1 d1 − d

),

Δ21 = det(−2m1 −m2

(γ + γ

) −b1−2n1 − n2

(γ + γ

)d1 − d

),

Δ22 = det(a1 −2m1 −m2

(γ + γ

)

c1 −2n1 − n2(γ + γ

)).

(3.68)

From (3.49) and (3.51), we can calculate g21 and derive the following values:

c1(0) =i

2ω0τ0

(

g20g11 − 2∣∣g11∣∣2 −

∣∣g02∣∣2

3

)

+g212,

μ2 = − Re{c1(0)}Re{λ′(τ0)} ,

β2 = 2Re (c1(0)),

T2 = − Im {c1(0)} + μ2 Im {λ′(τ0)}ω0τ0

.

(3.69)

These formulae give a description of the Hopf bifurcation periodic solutions of (1.1) at τ = τ0on the center manifold. From the discussion above, we have the following result.

Theorem 3.3. The periodic solution is supercritical (subcritical) if μ2 > 0 (μ2 < 0). The bifurcatingperiodic solutions are orbitally asymptotically stable with asymptotical phase (unstable) if β2 < 0(β2 > 0). The periods of the bifurcating periodic solutions increase (decrease) if T2 > 0 (T2 < 0).


0 100 200 300 400 5000.20.30.40.50.60.70.80.9

1

x(t)

(t)

(a)

0 100 200 300 400 5000.8

1

1.2

1.4

1.6

1.8

2

2.2

y(t)

(t)

(b)

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1x (t)

0.8

1

1.2

1.4

1.6

1.8

2

2.2

y(t)

(c)

0 200 400 60080010001200

0.20.4

0.60.8

10.5

1

1.5

2

2.5

(t)x (t)

z(t)

(d)

Figure 1: The time histories and phase portrait of system (4.1) with the following parameters: a = 0.5, b =0.5, c = 2, � = 2, d = 0.2, and τ = 1.7 < τ0 ≈ 1.7355. The positive equilibrium E∗(x∗, y∗) = (0.4928, 1.3217) isasymptotically stable. The initial value is (0.2, 2).


In this section, we present some numerical results to verify the analytical predictions obtainedin the previous section. As an example, we consider the following special case of system (1.1)with the parameters a = 0.5, b = 0.5, c = 2, � = 2, and d = 0.2. Then system (1.1) becomes

x(t) = x(t)[1 − 0.5x(t)] − 2x2(t)y(t)1 + 3x2(t)

,

y(t) =2x2(t − τ)y(t − τ)1 + 3x2(t − τ) e−0.2τ − 0.2y(t),

(4.1)

which has a positive equilibrium E∗(x∗, y∗) = (1.4928, 1.3217). By some complicatedcomputation by means of Matlab 7.0, we get only one critical values of the delay τ0 ≈1.7355, λ′(τ0) ≈ 0.2035 − 0.5423i. Thus, we derive c1(0) ≈ −1.3122 − 5.0131i, μ2 ≈ 0.6177, β2 ≈−3.3326, T2 ≈ 9.3042. We obtain that the conditions indicated in Theorem 2.3 are satisfied.Furthermore, it follows that μ2 > 0 and β2 < 0. Thus, the positive equilibrium E∗(x∗, y∗)is stable when τ < τ0 which is illustrated by the computer simulations (see Figures 1(a)–1(d)). When τ passes through the critical value τ0, the positive equilibrium E∗(x∗, y∗) losesits stability and a Hopf bifurcation occurs, that is, a family of periodic solutions bifurcations


0 100 200 300 400 500 600 700 800 9000.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

x(t)

(t)

(a)

0 100 200 300 400 500 600 700 800 9000.8

1

1.2

1.4

1.6

1.8

2

2.2

y(t)

(t)

(b)

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1x (t)

0.8

1

1.2

1.4

1.6

1.8

2

2.2

y(t)

(c)

0 200 400 600 800 1000

0.20.4

0.60.8

10.5

1

1.5

2

2.5

(t)x (t)

y(t)

(d)

Figure 2: The time histories and phase portrait of system (4.1) with the following parameters: a = 0.5, b =0.5, c = 2, � = 2, d = 0.2, and τ = 1.8 > τ0 ≈ 1.7355. Hopf bifurcation occurs from the positive equilibriumE∗(x∗, y∗) = (0.5014, 1.3108). The initial value is (0.2, 2).

from the positive equilibrium E∗(x∗, y∗). Since μ2 > 0 and β2 < 0, the direction of the Hopfbifurcation is τ > τ0, and these bifurcating periodic solutions from E∗(x∗, y∗) at τ0 are stable,which are depicted in Figures 2(a)–2(d).

5. Conclusions

In this paper, the main object is to investigate the local stability and Hopf bifurcation and alsoto study the stability of bifurcating periodic solutions and some formulae for determiningthe direction of Hopf bifurcation for a stage-structured predator-prey model with time delayand delay. By choosing the delay as a bifurcation parameter, It is shown that under certaincondition, the positive equilibrium E∗(x∗, y∗) of system (1.1) is asymptotically stable forall τ ∈ [0, τ0) and unstable for τ > τ0 and under another condition; when the delay τincreases, the equilibrium loses its stability and a sequence of Hopf bifurcations occur atthe positive equilibrium E∗(x∗, y∗), that is, a family of periodic orbits bifurcate from thepositive equilibrium E∗(x∗, y∗). At the same time, using the normal form theory and thecenter manifold theorem, the direction of Hopf bifurcation and the stability of the bifurcatingperiodic orbits are discussed. Finally, numerical simulations are carried out to validate thetheorems obtained.


Acknowledgments

This work is supported by The National Natural Science Foundation of China (no.11261010), Soft Science and Technology Program of Guizhou Province (no. 2011LKC2030),Natural Science and Technology Foundation of Guizhou Province (J (2012) 2100), GovernorFoundation of Guizhou Province ((2012)53), and Doctoral Foundation of Guizhou Universityof Finance and Economics (2010).

References

[1] Z.-P. Ma, H.-F. Huo, and C.-Y. Liu, “Stability and Hopf bifurcation analysis on a predator-prey modelwith discrete and distributed delays,” Nonlinear Analysis. Real World Applications, vol. 10, no. 2, pp.1160–1172, 2009.

[2] Y. Song, M. Han, and Y. Peng, “Stability and Hopf bifurcations in a competitive Lotka-Volterra systemwith two delays,” Chaos, Solitons & Fractals, vol. 22, no. 5, pp. 1139–1148, 2004.

[3] X.-P. Yan and W.-T. Li, “Bifurcation and global periodic solutions in a delayed facultative mutualismsystem,” Physica D, vol. 227, no. 1, pp. 51–69, 2007.

[4] X.-P. Yan and C.-H. Zhang, “Hopf bifurcation in a delayed Lokta-Volterra predator-prey system,”Nonlinear Analysis. Real World Applications, vol. 9, no. 1, pp. 114–127, 2008.

[5] C. Yu, J. Wei, and X. Zou, “Bifurcation analysis in an age-structured model of a single species livingin two identical patches,” Applied Mathematical Modelling, vol. 34, no. 4, pp. 1068–1077, 2010.

[6] W. G. Aiello and H. I. Freedman, “A time-delay model of single-species growth with stage structure,”Mathematical Biosciences, vol. 101, no. 2, pp. 139–153, 1990.

[7] J. R. Bence and R. M. Nisbet, “Space-limited recruitment in open systems: the importance of timedelays,” Ecology, vol. 70, no. 5, pp. 1434–1441, 1989.

[8] K. L. Cooke and P. van den Driessche, “On zeroes of some transcendental equations,” FunkcialajEkvacioj, vol. 29, no. 1, pp. 77–90, 1986.

[9] V. A. A. Jansen, R. M. Nisbet, Gurney, and W. S. C, “Generation cycles in stage structuredpopulations,” Bulletin of Mathematical Biology, vol. 52, no. 3, pp. 375–396, 1990.

[10] R. M. Nisbet, W. S. C. Gurney, and J. A. J. Metz, “Stage structure models applied in evolutionaryecology,” in Applied Mathematical Ecology, Biomathematics, pp. 428–449, Springer, Berlin, Germany,1989.

[11] X. Wang, Z. L. Xiong, and S. Y. Li, “Dynamics of predator-prey model with impulses and time delay,”Mathematics in Practice and Theory, vol. 38, no. 12, pp. 80–87, 2008.

[12] J. Liu and T. Zhang, “Bifurcation analysis of an SIS epidemic model with nonlinear birth rate,” Chaos,Solitons and Fractals, vol. 40, no. 3, pp. 1091–1099, 2009.

[13] Z. Jiang and J. Wei, “Stability and bifurcation analysis in a delayed SIR model,” Chaos, Solitons andFractals, vol. 35, no. 3, pp. 609–619, 2008.

[14] E. Beretta and Y. Kuang, “Geometric stability switch criteria in delay differential systems with delaydependent parameters,” SIAM Journal on Mathematical Analysis, vol. 33, no. 5, pp. 1144–1165, 2002.

[15] B. D. Hassard, N. D. Kazarinoff, and Y. H. Wan, Theory and Applications of Hopf Bifurcation, CambridgeUniversity Press, Cambridge, UK, 1981.


Research ArticleA Galerkin Solution for Burgers’ Equation UsingCubic B-Spline Finite Elements

A. A. Soliman

Department of Mathematics, Faculty of Education, Suez Canal University, Al-Arish 45111, Egypt

Correspondence should be addressed to A. A. Soliman, asoliman [email protected]

Received 13 May 2012; Accepted 24 July 2012


Copyright q 2012 A. A. Soliman. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

Numerical solutions for Burgers’ equation based on the Galerkins’ method using cubic B-splinesas both weight and interpolation functions are set up. It is shown that this method is capable ofsolving Burgers’ equation accurately for values of viscosity ranging from very small to large. Threestandard problems are used to validate the proposed algorithm. A linear stability analysis showsthat a numerical scheme based on a Cranck-Nicolson approximation in time is unconditionallystable.

1. Introduction

A study of Burgers’ equation is important since it arises in the approximate theory offlow through a shock wave propagating in a viscous fluid [1] and in the modeling ofturbulence [2]. Burgers’ equation and the Navier-Stokes equation are similar in the formof their nonlinear terms and in the occurrence of higher order derivatives with smallcoefficients in both [3]. The applications have been found in field as diverse as numbertheory, gas dynamics, heat conduction, elasticity, and so forth [4]. The exact solutions ofthe one-dimensional Burgers’ equation have been surveyed by Benton and Platzman [5].However, difficulties arise in the numerical solution of Burgers’ equation for small valuesof the viscosity coefficient, that is large Reynolds numbers, which correspond to steepwave fronts [6]. In these cases, numerical methods are likely to produce results whichinclude large nonphysical oscillations unless the size of the elements in unrealistically small.Many studies have been done on the numerical solutions of Burgers’ equation to deal withsolutions for the large Reynolds number values. Some of the earlier numerical studies aredocumented as follows: a finite element method has been given by Caldwell et al. [7] to solveBurgers’ equation by altering the size of the element at each stage using information from


three previous steps. Galerkin and Petrov-Galerkin finite element methods involving a timedependent grid [8, 9], and product approximation methods [10], have been used successfullyto obtain accurate numerical solutions even for small viscosity coefficients.

Rubin and Graves Jr. [11] have used the spline function technique and quasilin-earization for the numerical solution of Burgers’ equation in one space variable. A cubicspline collocation procedures were developed for Burgers’ equation in the papers [12–15].Soliman [16, 17] has solved Burgers’ equation numerically using a new similarity technique,and analytically using the modified extended tanh-function method. Raslan [18] has solvedBurgers’ equation using collocation method based on quadratic B-spline finite elementmethod. In this paper we set up the finite element method using Galerkins’ method withcubic splines as shape functions to obtain a numerical solution for Burgers’ equation. Theresulting system will be a system of ordinary differential equations which can be solved usinga Crank-Nicolson approximation in time.

2. The Governing Equation

Consider Burgers’ equation in the form

ut + εuux − νuxx=0, a ≤ x ≤ b, (2.1)

where ε and ν are positive parameters and the subscripts t and x denote temporal and spatialdifferentiation, respectively. Appropriate boundary conditions will be chosen from

u(a, t) = α,

u(b, t) = β,

ux(a, t) = ux(b, t) = 0,

uxx(a, t) = uxx(b, t) = 0.

(2.2)

Applying the Galerkin approach [19–21] to (2.1) with weight function v, integratingby parts, and choosing the boundary conditions from (2.2) lead to the weak form

∫b

a

v(ut + εuux)dx + ν∫b

a

vxuxdx = ν[vux]ba, (2.3)

where the right-hand side of (2.3) is evaluated only at the boundaries. The conditions on theinterpolation functions are now simply that only the functions and their first derivative needto be continuous throughout the region. However, we have chosen to use, as trial functions,the very adaptable cubic splines with their well-known advantages.

3. The B-Spline Finite Element Solution

The region [a, b] is partitioned into N finite elements of equal length h by knots xi such thata = x0 < x1 · · · < xN = b, and let φi(x) be the cubic B-splines with knots at the points xi. The


Table 1

x x�−2 x�−1 x� x�+1 x�+2

φ� 0 1 4 1 0φ

′�

0 3/h 0 −3/h 0φ

′′�

0 6/h2 −12/h2 6/h2 0

set of splines {φ−1, φ0, . . . , φN, φN+1} form a basis for functions defined over [a, b]. We seekthe approximation uN(x, t) to the solution u(x, t), which uses these splines as trial functions

uN(x, t) =N+1∑

�=−1

δ�(t)φ�(x), (3.1)

where the δ� are time-dependent quantities to be determined from (2.3).Defining a local coordinate system for the finite element [x�, x�+1] with nodes at x�

and x�+1, each cubic B-spline covers four elements [22]; consequently each element [x�, x�+1]is covered by four splines (φ�−1, φ�, φ�+1, φ�+2) which are given in terms of a local coordinatesystem η, where η = x − x� and 0 ≤ η ≤ h, by

⎧⎪⎪⎨

⎪⎪⎩

φ�−1

φ�φ�+1

φ�+2

=1h3

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(h − η)3

,

h3 + 3h2(h − η) + 3h(h − η)2 − 3

(h − η)3

,

h3 + 3h2η + 3hη2 − 3η3,

η3, 0 ≤ η ≤ h.

(3.2)

It is the representation of cubic B-spline over a single element which is most appropriate forthe finite element approach where all other splines are zero over this element.

The spline φ�(x) and its two principle derivatives vanish outside the interval[x�−2, x�+2]. In Table 1 we list for convenience the values of φ�(x) and its derivatives φ′

�(x),

φ′′�(x) at the relevant knots.

We now identify the finite elements for the problem with the intervals [x�, x�+1], andthe element nodes with the knots x�, x�+1.

Using (3.1) and Table 1, we see that the nodal parameters u� are given in terms of theparameters δ� by

u� = u(x�) = δ�−1 + 4δ� + δ�+1,

u�+1 = u(x�+1) = δ� + 4δ�+1 + δ�+2,(3.3)

and the variation of a function u over the element [x�, x�+1] is given by

u =�+2∑

j=�−1

δjφj . (3.4)


In addition, we have the valuable property that δ�−1 − δ�+2 determine also the first andsecond derivatives at the nodes (element boundaries), and these are also continuous

u′ =3h(−δ�−1 + δ�+1), (3.5)

u′′ =6h2 (δ�−1 − 2δ� + δ�+1). (3.6)

The finite element equations we will set up will not be expressed in terms of the nodalparameters u�, u′� but in terms of element parameters δ� , so we shall not directly determinethe nodal values as is the case with the usual finite element formulations, but these can alwaysbe recovered using (3.3) and (3.5).

We will now set up the element matrices relevant to (2.3). For a typical element[x�, x�+1], we have the contribution

∫x�+1

x�

v(ut + εuux)dx + ν∫x�+1

x�

vxuxdx = 0. (3.7)

Using (3.4) in (3.7) and identifying the weight function v(η) with the cubic splines weobtain

�+2∑

j=�−1

[∫h

0φiφjdη

]dδej

dt+ ε

�+2∑

j=�−1

�+2∑

k=�−1

[∫h

0φiφ

′jφkdη

]

δekδej + ν

�+2∑

j=�−1

[∫h

0φ′iφ

′jdη

]

δej = 0, (3.8)

which can be written in matrix form as

Ae dδe

dt+ εδe

T

Leδe + νCeδe = 0, (3.9)

where δe = (δe�−1, δe�, δ

e�+1, δ

e�+2)

T .The element matrices are given by the integrals

Aeij =

∫h

0φiφjdη,

Ceij =

∫h

0φ′iφ

′jdη,

Leijk =∫h

0φiφ

′jφkdη,

(3.10)


where i, j, k take only the values � − 1, �, � + 1, � + 2, for this typical element [x�, x�+1]. Thematrices Ae,Ce are therefore 4 × 4 and the matrix Le is 4 × 4 × 4. An associated 4 × 4 matrixBe is given by

Beij =�+2∑

k=�−1

Leijkδek, (3.11)

which also depends on the parameters δek and will be used in the following theoreticaldiscussions.

The element matrices Ae,Ce, Le, and Be can be determined algebraically from (3.2).Then, we obtain

Ae =h

140

⎛

⎜⎜⎝

20 129 60 1129 1188 933 6060 933 1188 1291 60 129 20

⎞

⎟⎟⎠,

Ce =1

10h

⎛

⎜⎜⎝

18 21 −36 −321 102 −87 −36−36 −87 102 21−3 −36 21 18

⎞

⎟⎟⎠,

(3.12)

Be11 = − 1840

(280, 1605, 630, 5)δe,

Be12 = − 1840

(150, 1305, 792, 21)δe,

Be13 =1

840(420, 2781, 1314, 21)δe,

Be14 =1

840(10, 129, 108, 5)δe,

Be21 = − 1840

(1605, 10830, 5349, 108)δe,

Be22 = − 1840

(1305, 17640, 17541, 1314)δe,

Be23 =1

840(2781, 25002, 17541, 792)δe,

Be24 =1

840(129, 3468, 5349, 630)δe,

Be31 = − 1840

(630, 5349, 3468, 129)δe,

Be32 = − 1840

(792, 17541, 25002, 2781)δe,


Be33 =1

840(1314, 17541, 17640, 1305)δe,

Be34 =1

840(108, 5349, 10830, 1605)δe,

Be41 = − 1840

(5, 108, 129, 10)δe,

Be42 = − 1840

(21, 1314, 2781, 420)δe,

Be43 =1

840(21, 792, 1305, 150)δe,

Be44 =1

840(5, 630, 1605, 280)δe.

(3.13)

Combining contributions from all elements leads to the matrix equation

Adδ

dt+ Bδ + νCδ = 0, (3.14)

where

δ = (δ−1, δ0, δ1, . . . , δN+1)T , (3.15)

and A,B,C assembled from the element matrices Ae, Be, Ce in the usual way, areseptadiagonal in form. We have assumed homogeneous boundary conditions on u or u′ andso the term on the right-hand side of (2.3) is zero. The general row for each matrix has thefollowing form:

A:h

140(1, 120, 1191, 2416, 1191, 120, 1),

C:−110h

(3, 72, 45,−240, 45, 72, 3),

B:1

840(−(5, 108, 129, 10, 0, 0, 0)δ,−(21, 1944, 8130, 3888, 129, 0, 0)δ,

− (−21, 0, 17841, 35682, 8130, 108, 0)δ, (5, 1944, 17841, 0,−17841,−1944,−5)δ,

(0, 108, 8130, 35682, 17841, 0,−21)δ, (0, 0, 129, 3888, 8130, 1944, 21)δ,

(0, 0, 0, 10, 129, 108, 5)δ),

(3.16)

where δ = (δ�−3, δ�−2, δ�−1, δ�, δ�+1, δ�+2, δ�+3)T , for row �. The matricesA andC are symmetric

while the form of the matrix B has a somewhat more complex structure.


We will time centre on t = (n + 1/2)Δt and write

δ =12

(δn + δn+1

),

dδ

dt=

1Δt

(δn+1 − δn

), (3.17)

where the δn are the parameters at time nΔt, and Δt is the timestep. Then (3.14), can bewritten as

[A +

εΔt2B +

νΔt2C

]δn+1 =

[A − εΔt

2B − νΔt

2C

]δn. (3.18)

The matrices A, C are independent of the time; so, they will remain constantthroughout the calculations. While the matrix B is dependent on the time, it must thereforebe recalculated at each step. Apply the boundary conditions and eliminate δ−1, δN+1 from(3.18) which then becomes (N + 1) × (N + 1) septadiagonal system and can be solved usingseptadiagonal algorithm.

The time evolution of the approximate solution uN(x, t) is determined by the timeevolution of the vector δn. This is found by repeatedly solving the system (3.18), oncethe initial vector δ0 has been computed from the initial conditions. The system (3.18) isseptadiagonal and so can be solved using the septadiagonal algorithm, but an inner iterationis needed at each timestep to scope with the nonlinear terms in which the matrix B dependson δ = (1/2)(δn + δn+1). The following solution procedure is followed.

(i) At time t = 0, for the initial step of the inner iteration, we approximate B by B∗

calculated from δ0 only and obtain a first approximation to δ1 from (3.18). We theniterate, using (3.18) with matrix B calculated from δ = (1/2)(δ0 + δ1), for up to10-times to refine the approximation to δ1.

(ii) At the other timesteps we use for the matrix B, at the first step of inner iteration,B∗ derived from δ∗ = δn + (1/2)(δn − δn−1), to obtain a first approximation to δn+1,by solving (3.18). We then iterate, using (3.18) with matrix B calculated from δ =(1/2)(δn + δn+1), two- or three-times to refine the approximation to δn+1.

4. The Stability Analysis

An investigation into the stability of the numerical scheme (3.18) is based on the VonNeumann theory in which the growth factor of a typical Fourier mode defined as

δnj = δneijkh, (4.1)

where k is the mode number, and h is the element size, is determined for a linearization ofthe numerical scheme (3.18).

In this linearization, we assume that the quantity u in the nonlinear term uux is locallyconstant. This is equivalent to assuming that the corresponding values of δj are also constantand equal to d.


A linearized recurrence relationship corresponding to (3.18) is then given by

α1δn+1�−3 + α2δ

n+1�−2 + α3δ

n+1�−1 + α4δ

n+1� + α5δ

n+1�+1 + α6δ

n+1�+2 + α7δ

n+1�+3

= α7δn�−3 + α6δ

n�−2 + α5δ

n�−1 + α8δ

n� + α3δ

n�+1 + α2δ

n�+2 + α1δ

n�+3,

(4.2)

where

α1 = r1 − 3r2 − 3r3, α2 = 120r1 − 168r2 − 72r3,

α3 = 1191r1 − 735r2 − 45r3, α4 = 2416r1 + 240r3,

α5 = 1191r1 + 735r2 + 45r3, α6 = 120r1 + 168r2 + 72r3,

α7 = r1 + 3r2 + 3r3, α8 = 2416r1 − 240r3,

r1 =h

140, r2 =

εdΔt20

, r3 =νΔt20

.

(4.3)

Substituting (4.1) into (4.2) we obtain

(a + 120 + ib)δn+1 = (a − 120 − ib)δn, (4.4)

where

i =√−1,

a = r1 cos 3kh + 120r1 cos 2kh + 1191r1 cos kh + 1208r1,

b = (3r2 + 3r3) sin 3kh + (168r2 + 72r3) sin 2kh + (1470r2 + 45r3) sin kh.

(4.5)

Let δn+1 = gδn and substitute in (4.4) to give

g =a − 120 − iba + 120 + ib

, (4.6)

where g is the growth factor for the mode. Then, the modulus of the growth factor is

∣∣g∣∣ =

√gg =

√√√√ (a − 120)2 + b2

(a + 120)2 + b2< 1. (4.7)

Therefore the linearized scheme is unconditionally stable.

5. The Initial State

From the initial condition u(x, 0) on the function u(x, t), we must determine the initial vectorδ0 so that the time evolution of δ, using (3.18), can be started.


First rewrite (3.1) for the initial condition as

uN(x, 0) =N+1∑

j=−1

δ0j φj(x), (5.1)

where δ0j are unknown parameters to be determined. To do this we require uN(x, 0) to satisfy

the following constraints:

(a) it must agree with the initial condition u(x, 0) at the knots; (3.3) leads to N + 1conditions, that is, uN(x�, 0) = u(x�, 0), � = 0, . . . ,N,

(b) the first or second derivatives of the approximate initial condition shall agree withthose of the exact initial condition at both ends of the range; (3.5) or (3.6) producestwo further equations, that is, u′(x0, 0) = u′(xN, 0) = 0.

The initial vector δ0 is then determined as the solution of a matrix equation derivedfrom (3.3)–(3.6)

Mδ0 = b. (5.2)

This system can be solved using the Thomas algorithm.

6. The Test Problems

Numerical solutions of Burgers’ equation for three test problems will now be considered. Tomeasure the accuracy of the numerical algorithm, we compute the difference between theanalytic and numerical solutions at each mesh point after specified timesteps and use theseto calculate the discrete L2- and L∞- error norms.

(a) Burgers’ equation has the analytic solution [23]

u(x, t) =x/t

1 +√(t/t0) exp(x2/4νt)

, 0 ≤ x ≤ 1, t ≥ 1, (6.1)

where t0 = exp(1/8ν). With (6.1), at time t = 1 as initial condition, numerical solutions aredetermined up to a time of t = 4, and use as boundary conditions u(0, t) = u(1, t) = 0. Wediscuss the three cases.

(i) Figures 1 and 2 shows us the behavior of the numerical solutions for viscositycoefficients ν = 0.005 and 0.0005 at times from t = 1 to 3.1 and from t = 1 to 3.25,respectively. It is seen that for the smaller viscosity value the propagation frontis steeper. These graphs agree with those reported by Nguyen and Reynen [23]to within plotting accuracy. In both cases, if the exact solutions are plotted on thesame diagram, the curves are indistinguishable. The L2- and L∞- error norms arecompared with results presented in the papers [9, 14], which are given in Tables 1, 2,and 3. From these tables, we find that the errors obtained by the present algorithmare smaller than the errors calculated by [9, 14].


0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.5

t = 1

t = 1.7t = 2.4

t = 3.1

uax

is

X axis

Figure 1: Example (a): v = 0.005, h = 0.02,Δt = 0.1.

0 0.2 0.4 0.6 0.8 10

0.1

0.2

0.3

0.4

0.6

0.5 t = 1

t = 1.75t = 2.5

t = 3.25

uax

is

X axis

Figure 2: Example (a): v = 0.0005, h = 0.005,Δt = 0.01.

(ii) The case with ν = 0.0005, h = 0.005, Δt = 0.01, at different times t = 1.7, 2.5,and 3.25, respectively. The numerical and the analytical solutions computed by thepresent approach are given in Table 4, with corresponding previous results [15].From Table 4, we deduce that the numerical solutions computed by the presentalgorithm are more accurate than those evaluated by [15], at different times.

(iii) In this case, we take h = 0.01, Δt = 0.01, and viscosity coefficients ν = 0.005, ν = 0.01.The results given in Tables 5 and 6 show us that the computed errors using thepresent algorithm are smaller than the corresponding errors obtained by Raslan[18], even, when we chose large timestep.

(b) A second analytic solution is [11]

u(x, t) =12

[1 − tanh

{1

4ν

(x − 1

2t − 15

)}]. (6.2)

We take as initial condition (6.2), at t = 0, over the range 0 ≤ x ≤ 30, with boundaryconditions

u(0, t) = 1, u(30, t) = 0, t > 0. (6.3)


0 5 10 15 20 25 300

0.4

0.8

uax

is t = 0 t = 10ν = 1/2

X axis

(a)

0 5 10 15 20 25 300

0.4

0.8u

axis t = 0 t = 10ν = 1/8

X axis

(b)

0 5 10 15 20 25 300

0.4

0.8

uax

is

X axis

t = 0 t = 10ν = 1/24

(c)

Figure 3: The behavior of the numerical solution with different viscosity at different times.

0.2

0.4

0.6

0.8

1

0 0.2 0.4 0.6 0.8 1

uax

is

X axis

t = 0.9 t = 0.5 t = 1.3

Figure 4: Example (c): the numerical solution with v = 0.01 at different times.

Table 2: Error norm L2 × 103 for problem (a).

ν = 0.005 ν = 0.0005t = 1.7 t = 2.4 t = 3.1 t = 1.75 t = 2.5 t = 3.25

Present 1.206 0.653 0.434 0.581 0.252 0.179[14] 1.003 0.598 0.522 0.567 0.308 0.291[9] 2.281 1.175 0.808 2.023 2.094 0.933

Table 3: Error norm L∞ × 103 for problem (a).

ν = 0.005 ν = 0.0005t = 1.7 t = 2.4 t = 3.1 t = 1.75 t = 2.5 t = 3.25

Present 3.868 1.679 1.295 5.371 1.619 0.821[14] 3.730 2.029 1.537 5.880 2.705 2.291[9] 7.150 2.921 2.297 14.033 12.615 8.394


Table 4: Comparison of results at different times for ν = 0.0005 and 0 ≤ x ≤ 1, with h = 0.005 and Δt = 0.01,problem (a).

x

t = 1.7 t = 1.7 t = 1.7 t = 2.5 t = 2.5 t = 2.5 t = 3.25 t = 3.25 t = 3.25Exact Numer Numer Exact Numer Numer Exact Numer Numer

Present [15] Present [15] Present [15]0.1 0.05882 0.05882 0.05883 0.04000 0.04003 0.04000 0.03077 0.03082 0.030770.2 0.11765 0.11765 0.11765 0.08000 0.08000 0.08000 0.06154 0.06154 0.061540.3 0.17647 0.17647 0.17648 0.12000 0.12000 0.12001 0.09231 0.09231 0.092310.4 0.23529 0.23529 0.23531 0.16000 0.16000 0.16001 0.12308 0.12308 0.123080.5 0.29412 0.29412 0.29414 0.20000 0.20000 0.20001 0.15385 0.15385 0.153850.6 0.35294 0.35287 0.35296 0.24000 0.24000 0.24001 0.18462 0.18462 0.184620.7 0.00000 0.00000 0.00000 0.28000 0.28000 0.28001 0.21538 0.21538 0.215390.8 0.00000 0.00000 0.00000 0.00977 0.00994 0.00811 0.24615 0.24615 0.246160.9 0.00000 0.00000 0.00000 0.00000 0.00000 0.00000 0.12433 0.12401 0.12358

Table 5: Comparison of errors at different times for ν = 0.005 and 0 ≤ x ≤ 1, with h = 0.01, problem (a).

TimeL2-error L2-error L∞-error L∞-errorPresent [18] Present [18]Δt = 0.01 Δt = 0.001 Δt = 0.01 Δt = 0.001

1.5 0.00019 0.00770 0.00132 0.042102.0 0.00018 0.01154 0.00102 0.051662.5 0.00017 0.01431 0.00082 0.056233.0 0.00016 0.01654 0.00069 0.058823.5 0.00019 0.01872 0.00109 0.060464.0 0.00035 0.02255 0.00268 0.07691

Table 6: Comparison of errors at different times for ν = 0.01 and 0 ≤ x ≤ 1, with h = 0.01, problem (a).

TimeL2-error L2-error L∞-error L∞-errorPresent [18] Present [18]Δt = 0.01 Δt = 0.001 Δt = 0.01 Δt = 0.001

1.07 0.00019 0.00207 0.00168 0.016601.14 0.00021 0.00333 0.00166 0.024131.35 0.00023 0.00613 0.00147 0.036671.70 0.00022 0.00938 0.00120 0.046772.40 0.00020 0.01376 0.00086 0.055373.10 0.00021 0.01689 0.00074 0.059063.24 0.00022 0.01746 0.00089 0.059553.31 0.00023 0.01775 0.00096 0.05977

Table 7: Comparison of L2 × 103 at different times for 0 ≤ x ≤ 30, with h = 0.1, and Δt = 0.025, problem (b).

t = 2 t = 4 t = 6 t = 8 t = 10 ν

Present 0.002 0.003 0.003 0.004 0.004 1/2[9] 0.006 0.010 0.014 0.022 0.043 1/2Present 0.032 0.036 0.038 0.038 0.038 1/8[9] 0.228 0.227 0.227 0.227 0.226 1/8Present 0.431 0.433 0.433 0.433 0.433 1/24[9] 3.431 3.426 3.426 3.426 3.429 1/24


Table 8: Comparison of L∞ × 103 at different times for 0 ≤ x ≤ 30, with h = 0.1, and Δt = 0.025, problem(b).

t = 2 t = 4 t = 6 t = 8 t = 10 ν

Present 0.001 0.002 0.002 0.003 0.003 1/2[9] 0.004 0.006 0.007 0.017 0.045 1/2Present 0.044 0.049 0.050 0.051 0.051 1/8[9] 0.346 0.339 0.340 0.347 0.356 1/8Present 0.937 0.939 0.939 0.939 0.939 1/24[9] 4.340 4.336 4.363 4.402 4.432 1/24

Table 9: Comparison of results for various numerical schemes with exact at time t = 0.5, for ν = 0.01.

x

SGA [10] PAG [10] ML [9] CSCM [14] CSCM [15] PresentExacth = 1/18 h = 1/18 h = 1/36 h = 1/36 h = 1/36 h = 1/36

Δt = .001 Δt = .001 Δt = .025 Δt = .025 Δt = .025 Δt = .025.000 1.000 1.000 1.000 1.000 1.000 1.000 1.000.056 1.000 1.000 1.000 1.000 1.000 1.000 1.000.111 1.000 1.000 1.000 1.000 1.000 1.000 1.000.167 1.000 1.000 1.000 1.000 1.000 1.000 1.000.222 0.998 0.999 1.000 1.000 1.000 1.000 1.000.278 0.991 0.997 0.998 0.999 0.999 0.999 0.998.333 0.970 0.982 0.985 0.985 0.986 0.986 0.980.389 0.862 0.850 0.851 0.847 0.850 0.850 0.847.444 0.461 0.444 0.447 0.452 0.448 0.448 0.452.500 0.159 0.171 0.238 0.238 0.236 0.238 0.238.556 0.300 0.286 0.205 0.204 0.204 0.204 0.204.611 0.194 0.197 0.201 0.200 0.200 0.200 0.200.667 0.213 0.211 0.200 0.200 0.200 0.200 0.200.722 0.211 0.210 0.200 0.200 0.200 0.200 0.200.778 0.188 0.190 0.200 0.200 0.200 0.200 0.200.833 0.201 0.207 0.200 0.200 0.200 0.200 0.200.889 0.191 0.193 0.200 0.200 0.200 0.200 0.200.944 0.203 0.202 0.200 0.200 0.200 0.200 0.2001.00 0.200 0.200 0.200 0.200 0.200 0.200 0.200

For ν = 1/2, a very weak shock wave develops, when ν = 1/8, we obtain a moderateshock wave and when ν = 1/24, a strong shock wave is produced. As the value of ν isdecreased the propagation front becomes steeper. The L2- and L∞- error norms for thesesimulations with the corresponding errors obtained by [9], are given in Tables 7 and 8. Forlarge values of ν the errors are small and as the value of ν is decreased the errors tend toincrease, but for the values of ν used here, the errors are still acceptable. From Tables 7 and 8,we deduce that the errors computed by the present algorithm are more accurate than thoseevaluated by [9] at different times. Figure 3 shows us the behavior of the numerical solutionsfor viscosity coefficients ν = 1/24, at different times from t = 0 to 10.


(c) A third test example, consider the particular analytic solution of Burgers’ equation[8, 10]

u(x, t) =

[α + μ +

(α − μ) exp

(η)]

1 + exp(η) , 0 ≤ x ≤ 1, t ≥ 0, (6.4)

where η = α(x − μt − β)/ν, α, μ, and β are constants.The comparison can be made with [8–10, 14, 15], these constants are chosen to have

the values α = 0.4, β = 0.125, μ = 0.6, and ν = 0.01. The initial condition is obtained from (6.4),when t = 0. The boundary conditions are taken as

u(0, t) = 1, (1, t) = 0.2, t ≥ 0. (6.5)

From Figure 4, it can be seen that the solution represents a travelling wave, initiallysituated at x = β, moving to the right with speed μ. The present method behaves verywell until the wave encounters the right-hand boundary. The numerical results at t = 0.5are compared with the analytic solution, and the results presented in papers [8–10, 14, 15],known as methods of method of lines (ML), standard Galerkin approach (SGA), productapproximation Galerkin method (PAG), cubic B-spline collocation method (CSCM), and anumerical solution of Burgers’ equation using cubic B-spline (CSCM), are given in Table 9. Itis seen from Table 9 that the agreement between the numerical and the exact solution appearsvery satisfactorily.

7. Conclusions

We have seen that the algorithm proposed here using Galerkins’ method with cubic splineshape functions compares well in the accuracy with the other published methods, such asmethods of lines, standard Galerkin approach, product approximation Galerkin method,and cubic B-spline collocation method. The L2- and L∞- error norms are satisfactorily smallduring the simulations. A linear stability analysis based on the Von Neumann theory showsthat the numerical scheme is unconditionally stable. We conclude that a finite elementapproach based on Galerkins’ method with cubic spline shape functions is eminently suitablefor the computation of solutions to Burgers’ equation. We believe that the approach shouldbe suitable for other applications where the continuity of derivatives is essential.

References

[1] J. D. Cole, “On a quasi-linear parabolic equation occurring in aerodynamics,” Quarterly of AppliedMathematics, vol. 9, pp. 225–236, 1951.

[2] J. M. Burgers, “A mathematical model illustrating the theory of turbulence,” in Advances in AppliedMechanics, pp. 171–199, Academic Press, New York, NY, USA, 1948.

[3] E. Varoglu and W. D. L. Finn, “Space-time finite elements incorporating characteristics for the Burgersequation,” International Journal for Numerical Methods in Engineering, vol. 16, pp. 171–184, 1980.

[4] E. Hopf, “The partial differential equation ut + uuxx = vuxx,” Communications on Pure and AppliedMathematics, vol. 3, pp. 201–230, 1950.

[5] E. R. Benton and G. W. Platzman, “A table of solutions of the one-dimensional Burgers equation,”Quarterly of Applied Mathematics, vol. 30, pp. 195–212, 1972.


[6] D. U. Rosenberg, Methods for Solution of Partial Differntial Equations, vol. 113, American Elsevier, NewYork, NY, USA, 1969.

[7] J. Caldwell, P. Wanless, and A. E. Cook, “A finite element approach to Burgers’ equation,” AppliedMathematical Modelling, vol. 5, no. 3, pp. 189–193, 1981.

[8] B. M. Herbst, S. W. Schoombie, and A. R. Mitchell, “A moving Petrov-Galerkin method for transportequations,” International Journal for Numerical Methods in Engineering, vol. 18, no. 9, pp. 1321–1336,1982.

[9] L. R. T. Gardner, G. A. Gardner, and A. H. A. Ali, “A method of lines solutions for Burgers’ equation,”in Computer Mech, Y. K. Cheung et al., Ed., pp. 1555–1561, Balkema, Rotterdam, 1991.

[10] I. Christie, D. F. Griffiths, A. R. Mitchell, and J. M. Sanz-Serna, “Product approximation for nonlinearproblems in the finite element method,” Journal of Numerical Analysis, vol. 1, no. 3, pp. 253–266, 1981.

[11] S. G. Rubin and R. A. Graves, Jr., “Viscous flow solutions with a cubic spline approximation,”Computers & Fluids, vol. 3, pp. 1–36, 1975.

[12] S. G. Rubin and P. K. Khosla, “Higher-order numerical solutions using cubic splines,” AmericanInstitute of Aeronautics and Astronautics, vol. 14, no. 7, pp. 851–858, 1976.

[13] J. Caldwell, “Application of cubic splines to the nonlinear Burgers’ equation,” in Numerical Methodsfor Nonlinear Problems, vol. 3, pp. 253–261, Pineridge Press, 1986.

[14] A. H. A. Ali, G. A. Gardner, and L. R. T. Gardner, “A collocation solution for Burgers’ equation usingcubic Cubic-spline finite elements,” Computer Methods in Applied Mechanics and Engineering, vol. 100,no. 3, pp. 325–337, 1992.

[15] I. Dag, D. Irk, and B. Saka, “A numerical solution of the Burgers’ equation using cubic B-splines,”Applied Mathematics and Computation, vol. 163, no. 1, pp. 199–211, 2005.

[16] A. A. Soliman, “Numerical technique for Burgers’ equation based on similarity reductions,” inInternational Conference on Applied Computational Fluid Dynamics, pp. 559–566., Bejing, China, October2000.

[17] A. A. Soliman, “The modified extended tanh-function method for solving Burgers-type equations,”Physica A, vol. 361, no. 2, pp. 394–404, 2006.

[18] K. R. Raslan, “A collocation solution for Burgers equation using quadratic B-spline finite elements,”International Journal of Computer Mathematics, vol. 80, no. 7, pp. 931–938, 2003.

[19] G. A. Gardner, A. H. A. Ali, and L. R. T. Gardner, “A finite element solution for the korteweg-de vriesequation using cubic B-spline shape function,” in Proceedings of the Information Security Policy MadeEasy (ISNME ’89), R. Gruber, J. Periaux, and R. P. Shaw, Eds., vol. 2, pp. 565–570, Springer, 1989.

[20] L. R. T. Gardner and G. A. Gardner, “Solitary waves of the regularised long-wave equation,” Journalof Computational Physics, vol. 91, no. 2, pp. 441–459, 1990.

[21] O. C. Zienkiewicz, The Finite Elment Method, McGraw Hill, London, UK, 3th edition, 1977.[22] P. M. Prenter, Splines and Variational Methods, John Wiley & Sons, New York, NY, USA, 1989.[23] H. Nguyen and J. Reynen, “A space-time finite element approach to Burgers’ equation,” in Numerical

Methods for Nonlinear Problems, E. Hinton, Ed., vol. 2, pp. 718–728, Pineridge Press, 1984.


Research ArticleSolutions of Sign-Changing Fractional DifferentialEquation with the Fractional Derivatives

Tunhua Wu,1 Xinguang Zhang,2 and Yinan Lu3

1 School of Information and Engineering, Wenzhou Medical College, Zhejiang,Wenzhou 325035, China

2 School of Mathematical and Informational Sciences, Yantai University, Shandong,Yantai 264005, China

3 Information Engineering Department, Anhui Xinhua University, Anhui, Hefei 230031, China

Correspondence should be addressed to Xinguang Zhang, [email protected]

Received 24 April 2012; Accepted 29 May 2012

Academic Editor: Yonghong Wu

Copyright q 2012 Tunhua Wu et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We study the singular fractional-order boundary-value problem with a sign-changing nonlinearterm −Dαx(t) = p(t)f(t, x(t),Dμ1x(t),Dμ2x(t), . . . ,Dμn−1x(t)), 0 < t < 1, Dμix(0) = 0, 1 ≤ i ≤ n −1,Dμn−1+1x(0) = 0,Dμn−1x(1) =

∑p−2j=1 ajDμn−1x(ξj), where n − 1 < α ≤ n, n ∈ N and n ≥ 3 with

0 < μ1 < μ2 < · · · < μn−2 < μn−1 and n − 3 < μn−1 < α − 2, aj ∈ R, 0 < ξ1 < ξ2 < · · · < ξp−2 < 1 satisfying

0 <∑p−2

j=1 ajξα−μn−1−1j < 1, Dα is the standard Riemann-Liouville derivative, f : [0, 1] × R

n → R isa sign-changing continuous function and may be unbounded from below with respect to xi, andp : (0, 1) → [0,∞) is continuous. Some new results on the existence of nontrivial solutions forthe above problem are obtained by computing the topological degree of a completely continuousfield.

1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines, andparticularly in the mathematical modeling of systems and processes in physics, chemistry,aerodynamics, electrodynamics of complex medium, and polymer rheology [1–6]. Fractional-order models have proved to be more accurate than integer-order models, that is, thereare more degrees of freedom in the fractional-order models. Hence fractional differentialequations have attracted great research interest in recent years, and for more details we referthe reader to [7–16] and the references cited therein.


In this paper, we consider the existence of nontrivial solutions for the followingsingular fractional-order boundary-value problem with a sign-changing nonlinear term andfractional derivatives:

−Dαx(t) = p(t)f(t, x(t),Dμ1x(t),Dμ2x(t), . . . ,Dμn−1x(t)), 0 < t < 1,

Dμix(0) = 0, 1 ≤ i ≤ n − 1, Dμn−1+1x(0) = 0, Dμn−1x(1) =p−2∑

j=1

ajDμn−1x(ξj),

(1.1)

where n − 1 < α ≤ n, n ∈ N and n ≥ 3 with 0 < μ1 < μ2 < · · · < μn−2 < μn−1 and n − 3 < μn−1 <

α − 2, aj ∈ R, 0 < ξ1 < ξ2 < · · · < ξp−2 < 1 satisfying 0 <∑p−2

j=1 ajξα−μn−1−1j < 1, Dα is the standard

Riemann-Liouville derivative, f : [0, 1]×Rn → R is a sign-changing continuous function and

may be unbounded from below with respect to xi, and p : (0, 1) → [0,∞) is continuous.In this paper, we assume that f : [0, 1] × R

n → R, which implies that the problem(1.1) is changing sign (or semipositone particularly). Differential equations with changing-sign arguments are found to be important mathematical tools for the better understanding ofseveral real-world problems in physics, chemistry, mechanics, engineering, and economics[17–19]. In general, the cone theory is difficult to handle this type of problems since theoperator generated by f is not a cone mapping. So to find a new method to solve changing-sign problems is an interesting, important, and difficult work. An effective approach to thisproblem was recently suggested by Sun [20] based on the topological degree of a completelycontinuous field. Then, Han and Wu [21, 22] obtained a new Leray-Schauder degree theoremby improving the results of Sun [20]. In [22], Han et al. also investigated a kind of singulartwo-point boundary-value problems with sign-changing nonlinear terms by applying thenew Leray-Schauder degree theorem obtained in [22].

To our knowledge, very few results have been established when f is changing sign[20–24]. In [20, 21, 23], f permits sign changing but required to be bounded from below. In[22, Theorem 1.1], f may be a sign-changing and unbounded function, but the Green functionmust be symmetric and f is controlled by a special function h(u) = −b−c|u|μ, where b > 0, c >0 and μ ∈ (0, 1). Recently, by improving and generalizing the main results of Sun [20] andHan et al. [21, 22], Liu et al. [24] established a generalized Leray-Schauder degree theoremof a completely continuous field for solving m-point boundary-value problems for singularsecond-order differential equations.

Motivated by [20–24], we established some new results on the existence of nontrivialsolutions for the problem (1.1) by computing the topological degree of a completelycontinuous field. The conditions used in the present paper are weaker than the conditionsgiven in previous works [20–24], and particularly we drop the assumption of even functionin [24]. The new features of this paper mainly include the following aspects. Firstly, thenonlinear term f(t, x1, x2, . . . , xn) in the BVP (1.1) is allowed to be sign changing andunbounded from below with respect to xi. Secondly, the nonlinear term f involves fractionalderivatives of unknown functions. Thirdly, the boundary conditions involve fractionalderivatives of unknown functions which is a more general case, and include the two-point,three-point, multipoint, and some nonlocal problems as special cases of (1.1).

2. Preliminaries and Lemmas

In this section, we give some preliminaries and lemmas.


Definition 2.1. Let E be a real Banach space. A nonempty closed convex set P ⊂ E is called acone of E if it satisfies the following two conditions:

(1) x ∈ P, σ > 0 implies σx ∈ P ;(2) x ∈ P,−x ∈ P implies x = θ.

Definition 2.2. An operator is called completely continuous if it is continuous and mapsbounded sets into precompact sets.

Let E be a real Banach space, E∗ the dual space of E, P a total cone in E, that is, E =P − P, and P ∗ the dual cone of P .

Lemma 2.3 (Deimling [25]). Let L : E → E be a continuous linear operator, P a total cone, andL(P) ⊂ P . If there exist ψ ∈ E \ (−P) and a positive constant c such that cL(ψ) ≥ ψ, then the spectralradius r(L)/= 0 and has a positive eigenfunction corresponding to its first eigenvalue λ = r(L)−1.

Lemma 2.4 (see [25]). Let P be a cone of the real Banach space E, and Ω a bounded open subsets ofE. Suppose that T : Ω ∩ P → P is a completely continuous operator. If there exists x0 ∈ P \ {θ} suchthat x − Tx /=μx0, ∀x ∈ ∂Ω ∩ P, μ ≥ 0, then the fixed-point index i(T,Ω ∩ P, P) = 0.

Let L : E → E be a completely continuous linear positive operator with the spectral radiusr1 /= 0. On account of Lemma 2.3, there exist ϕ1 ∈ P \ {θ} and g1 ∈ P ∗ \ {θ} such that

Lϕ1 = r1ϕ1,

r1L∗g1 = g1,

(2.1)

where L∗ is the dual operator of L. Choose a number δ > 0 and let

P(g1, δ

)={u ∈ P : g1(u) ≥ δ||u||

}, (2.2)

then P(g1, δ) is a cone in E.

Lemma 2.5 (see [24]). Suppose that the following conditions are satisfied.

(A1) T : E → P is a continuous operator satisfying

lim‖u‖→+∞

‖Tu‖‖u‖ = 0; (2.3)

(A2) F : E → E is a bounded continuous operator and there exists u0 ∈ E such that Fu + u0 +Tu ∈ P , for all u ∈ E;

(A3) r1 > 0 and there exist v0 ∈ E and η > 0 such that

LFu ≥ r1(1 + η

)Lu − LTu − v0, ∀u ∈ E. (2.4)

Let A = LF, then there exists R > 0 such that

deg(I −A,BR, θ) = 0, (2.5)

where BR = {u ∈ E : ‖u‖ < R} is the open ball of radius R in E.


Remark 2.6. If the operator T which satisfies the conditions of Lemma 2.5 is a null operator,then Lemma 2.5 turns into Theorem 1 in [20]. On the other hand, if the operator T inLemma 2.5 is such that there exist constants α ∈ (0, 1) and N > 0 satisfying ||Tu|| ≤N||u||α forall u ∈ E, then Lemma 2.5 turns into Theorem 2.1 in [22] or Theorem 1 in [21]. So Lemma 2.5is an improvement of the results of paper [20–22].

Now we present the necessary definitions from fractional calculus theory. Thesedefinitions can be found in some recent literatures, for example, [26, 27].

Definition 2.7 (see [26, 27]). The Riemann-Liouville fractional integral of order α > 0 of afunction x : (0,+∞) → R is given by

Iαx(t) =1

Γ(α)

∫ t

0(t − s)α−1x(s)ds (2.6)

provided that the right-hand side is pointwisely defined on (0,+∞).

Definition 2.8 (see [26, 27]). The Riemann-Liouville fractional derivative of order α > 0 of afunction x : (0,+∞) → R is given by

Dαx(t) =1

Γ(n − α)(d

dt

)n ∫ t

0(t − s)n−α−1x(s)ds, (2.7)

where n = [α] + 1, and [α] denotes the integer part of the number α, provided that the right-hand side is pointwisely defined on (0,+∞).

Remark 2.9. If x, y : (0,+∞) → R with order α > 0, then

Dα(x(t) + y(t))= Dαx(t) +Dαy(t). (2.8)

Lemma 2.10 (see [27]). (1) If x ∈ L1(0, 1), ρ > σ > 0, then

IρIσx(t) = Iρ+σx(t), DσIρx(t) = Iρ−σx(t), DσIσx(t) = x(t). (2.9)

(2) If ρ > 0, σ > 0, then

Dρtσ−1 =Γ(σ)

Γ(σ − ρ) t

σ−ρ−1. (2.10)

Lemma 2.11 (see [27]). Assume that x ∈ C(0, 1) ∩ L1(0, 1) with a fractional derivative of orderα > 0. Then

IαDαx(t) = x(t) + c1tα−1 + c2t

α−2 + · · · + cntα−n, (2.11)

where ci ∈ R (i = 1, 2, . . . , n), n is the smallest integer greater than or equal to α.


Noticing that 2 < α − μn−1 ≤ n − μn−1 < 3, let

k(t, s) =

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(t(1 − s))α−μn−1−1 − (t − s)α−μn−1−1

Γ(α − μn−1

) , 0 ≤ s ≤ t ≤ 1,

(t(1 − s))α−μn−1−1

Γ(α − μn−1

) , 0 ≤ t ≤ s ≤ 1,

(2.12)

by [28], for t, s ∈ [0, 1], one has

tα−μn−1−1(1 − t)s(1 − s)α−μn−1−1

Γ(α − μn−1

) ≤ k(t, s) ≤ s(1 − s)α−μn−1−1

Γ(α − μn−1

) . (2.13)

Lemma 2.12. If 2 < α − μn−1 < 3 and ρ ∈ L1[0, 1], then the boundary-value problem

Dα−μn−1w(t) + ρ(t) = 0,

w(0) = w′(0) = 0, w(1) =p−2∑

j=1

ajw(ξj),

(2.14)

has the unique solution

w(t) =∫1

0K(t, s)ρ(s)ds, (2.15)

where

K(t, s) = k(t, s) +tα−μn−1−1

1 −∑p−2j=1 ajξ

α−μn−1−1j

p−2∑

j=1

ajk(ξj , s), (2.16)

is the Green function of the boundary-value problem (2.14).

Proof. By applying Lemma 2.11, we may reduce (2.14) to an equivalent integral equation:

w(t) = −Iα−μn−1ρ(t) + c1tα−μn−1−1 + c2t

α−μn−1−2 + c3tα−μn−1−3, c1, c2, c3 ∈ R. (2.17)

Note thatw(0) = w′(0) = 0 and (2.17), we have c2 = c3 = 0. Consequently the general solutionof (2.14) is

w(t) = −Iα−μn−1ρ(t) + c1tα−μn−1−1. (2.18)

By (2.18) and Lemma 2.10, we have

w(t) = − Iα−μn−1ρ(t) + c1tα−μn−1−1

= −∫ t

0

(t − s)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds + c1tα−μn−1−1.

(2.19)


So,

w(1) = −∫1

0

(1 − s)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds + c1, (2.20)

and for j = 1, 2, . . . , p − 2,

w(ξj)= −∫ ξj

0

(ξj − s

)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds + c1ξα−μn−1−1j . (2.21)

By w(1) =∑p−2

j=1 ajw(ξj), combining with (2.20) and (2.21), we obtain

c1 =

∫10 (1 − s)α−μn−1−1ρ(s)ds −∑p−2

j=1 aj∫ ξj

0

(ξj − s

)α−μn−1−1ρ(s)ds

Γ(α − μn−1

)(1 −∑p−2

j=1 ajξα−μn−1−1j

) . (2.22)

So, the unique solution of problem (2.14) is

w(t) = −∫ t

0

(t − s)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds

+tα−μn−1−1

1 −∑p−2j=1 ajξ

α−μn−1−1j

⎧⎨

⎩

∫1

0

(1 − s)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds −p−2∑

j=1

aj

∫ ξj

0

(ξj − s

)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds

⎫⎬

⎭

= −∫ t

0

(t − s)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds +∫1

0

(1 − s)α−μn−1−1tα−μn−1−1

Γ(α − μn−1

) ρ(s)ds

+tα−μn−1−1

1 −∑p−2j=1 ajξ

α−μn−1−1j

p−2∑

j=1

aj

∫1

0

(1 − s)α−μn−1−1ξα−μn−1−1j

Γ(α − μn−1

) ρ(s)ds

− tα−μn−1−1

1 −∑p−2j=1 ajξ

α−μn−1−1j

p−2∑

j=1

aj

∫ ξj

0

(ξj − s

)α−μn−1−1

Γ(α − μn−1

) ρ(s)ds

=∫1

0

⎛

⎝k(t, s) +tα−μn−1−1

1 −∑p−2j=1 ajξ

α−μn−1−1j

p−2∑

j=1

ajk(ξj , s)⎞

⎠ρ(s)ds

=∫1

0K(t, s)ρ(s)ds.

(2.23)

The proof is completed.


Lemma 2.13. The function K(t, s) has the following properties:(1) K(t, s) > 0, for t, s ∈ (0, 1);(2) K(t, s) ≤Ms(1 − s)α−μn−1−1, for t, s ∈ [0, 1], where

M =1

Γ(α − μn−1

)

⎛

⎝1 +

∑p−2j=1 aj

1 −∑p−2j=1 ajξ

α−μn−1−1j

⎞

⎠. (2.24)

Proof. It is obvious that (1) holds. In the following, we will prove (2). In fact, by (2.13), wehave

K(t, s) = k(t, s) +tα−μn−1−1

1 −∑p−2j=1 ajξ

α−μn−1−1j

p−2∑

j=1

ajk(ξj , s)

≤ s(1 − s)α−μn−1−1

Γ(α − μn−1

) +

∑p−2j=1 aj

1 −∑p−2j=1 ajξ

α−μn−1−1j

s(1 − s)α−μn−1−1

Γ(α − μn−1

)

≤⎛

⎝1 +

∑p−2j=1 aj

1 −∑p−2j=1 ajξ

α−μn−1−1j

⎞

⎠s(1 − s)α−μn−1−1

Γ(α − μn−1

) .

(2.25)

This completes the proof.Now let us consider the following modified problems of the BVP (1.1):

−Dα−μn−1u(t) = p(t)f(t, Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

),

u(0) = 0, u′(0) = 0, Dμ−μn−1u(1) =p−2∑

j=1

ajDμ−μn−1u(ξj).

(2.26)

Lemma 2.14. Let x(t) = Iμn−1u(t), u(t) ∈ C[0, 1], Then (1.1) can be transformed into (2.26).Moreover, if u ∈ C([0, 1], [0,+∞) is a solution of problem (2.26), then, the function x(t) = Iμn−1u(t)is a positive solution of problem (1.1).

Proof. Substituting x(t) = Iμn−1u(t) into (1.1), by the definition of the Riemann-Liouvillefractional derivative and Lemmas 2.10 and 2.11, we obtain that

Dαx(t) =dn

dtnIn−αx(t) =

dn

dtnIn−αIμn−1u(t)

=dn

dtnIn−α+μn−1u(t) = Dα−μn−1u(t),

Dμ1x(t) = Dμ1Iμn−1u(t) = Iμn−1−μ1u(t),

Dμ2x(t) = Dμ2Iμn−1u(t) = Iμn−1−μ2u(t),

......

Dμn−2x(t) = Dμn−2Iμn−1u(t) = Iμn−1−μn−2u(t),

Dμn−1x(t) = Dμn−1Iμn−1u(t) = u(t),

Dμn−1+1x(t) = [Dμn−1Iμn−1u(t)]′ = u′(t).

(2.27)


Also, we have Dμn−1x(0) = u(0) = 0,Dμn−1+1x(0) = u′(0) = 0, and it follows from Dμn−1x(t) =u(t) that u(1) =

∑p−2j=1 aju(ξj). Hence, by x(t) = Iμn−1u(t), u ∈ C[0, 1], (1.1) is transformed into

(2.14).Now, let u ∈ C([0, 1, 0,+∞)) be a solution of problem (2.26). Then, by Lemma 2.10,

(2.26), and (2.27), one has

−Dαx(t) = − dn

dtnIn−αx(t) = − d

n

dtnIn−αIμn−1u(t)

= − dn

dtnIn−α+μn−1u(t) = −Dα−μn−1u(t)

= p(t)f(t, Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

)

= p(t)f(t, x(t),Dμ1x(t),Dμ2x(t), . . . ,Dμn−1x(t)), 0 < t < 1.

(2.28)

Noticing that

Iαu(t) =1

Γ(α)

∫ t

0(t − s)α−1u(s)ds, (2.29)

which implies that Iαu(0) = 0, from (2.27), for i = 1, 2, . . . , n − 1, we have

Dμix(0) = 0, Dμn−1+1x(0) = 0, Dμn−1x(1) =p−2∑

j=1

ajDμn−1x(ξj). (2.30)

Moreover, it follows from the monotonicity and property of Iμn−1 that

Iμn−1u ∈ C([0, 1], [0,+∞)). (2.31)

Consequently, x(t) = Iμn−1u(t) is a positive solution of problem (1.1).

In the following let us list some assumptions to be used in the rest of this paper.(H1) p : (0, 1) → [0,+∞) is continuous, p(t)/≡ 0 on any subinterval of (0,1), and

∫1

0p(s)ds < +∞. (2.32)

(H2) f : [0, 1] × Rn → (−∞,+∞) is continuous.

In order to use Lemma 2.5, let E = C[0, 1] be our real Banach space with the norm‖u‖ = maxt∈[0,1]|u(t)| and P = {u ∈ C[0, 1] : u(t) ≥ 0, for all t ∈ [0, 1]}, then P is a total conein E.

Define two linear operators L, J : C[0, 1] → C[0, 1] by

(Lu)(t) =∫1

0K(t, s)p(s)u(s)ds, t ∈ [0, 1], u ∈ C[0, 1], (2.33)

(Ju)(t) =∫1

0K(s, t)p(s)u(s)ds, t ∈ [0, 1], u ∈ C[0, 1], (2.34)


and define a nonlinear operator A : C[0, 1] → C[0, 1] by

(Au)(t) =∫1

0K(t, s)p(s)f

(t, Iμn−1u(s), Iμn−1−μ1u(s), . . . , Iμn−1−μn−2u(s), u(s)

)ds, t ∈ [0, 1].

(2.35)

Lemma 2.15. Assume (H1) holds. Then

(i) L, J : C[0, 1] → C[0, 1] are completely continuous positive linear operators with the firsteigenvalue r > 0 and λ > 0, respectively.

(ii) L satisfies L(P) ⊂ P(g1, δ).

Proof. (i) By using the similar method of paper [22], it is easy to know that L, J : C[0, 1] →C[0, 1] are completely continuous positive linear operators. In the following, by using theKrein-Rutmann’s theorem, we prove that L, J have the first eigenvalue r > 0 and λ > 0,respectively.

In fact, it is obvious that there is t1 ∈ (0, 1) such that K(t1, t1)p(t1) > 0. Thus there exists[a, b] ⊂ (0, 1) such that t1 ∈ (a, b) and K(t, s)p(s) > 0 for all t, s ∈ [a, b]. Choose ψ ∈ P suchthat ψ(t1) > 0 and ψ(t) = 0 for all t /∈ [a, b]. Then for t ∈ [a, b],

(Lψ)(t) =

∫1

0K(t, s)p(s)ψ(s)ds ≥

∫b

a

K(t, s)p(s)ψ(s)ds > 0. (2.36)

So there exists ν > 0 such that ν(Lψ)(t) ≥ ψ(t) for t ∈ [0, 1]. It follows from Lemma 2.3that the spectral radius r1 /= 0. Thus corresponding to r = r−1

1 , the first eigenvalue of L, and Lhas a positive eigenvector ϕ1 such that

rLϕ1 = ϕ1. (2.37)

In the same way, J has a positive first eigenvalue λ and a positive eigenvector ϕ2

corresponding to the first eigenvalue λ, which satisfy

λJϕ2 = ϕ2. (2.38)

(ii) Notice that K(t, 0) = K(t, 1) ≡ 0 for t ∈ [0, 1], by λJϕ2 = ϕ2 and (2.12)–(2.16), wehave ϕ2(0) = ϕ2(1) = 0. This implies that ϕ′

2(0) > 0 and ϕ′2(1) < 0 (see [29]). Define a function

χ on [0, 1] by

χ(s) =

⎧⎪⎪⎨

⎪⎪⎩

ϕ′2(0), s = 0,ϕ2(s)

(1 − s)s , 0 < s < 1,

−ϕ′2(1), s = 1.

(2.39)

Then χ is continuous on [0, 1] and χ(s) > 0 for all s ∈ [0, 1]. So, there exist δ1, δ2 > 0such that δ1 ≤ χ(s) ≤ δ2 for all s ∈ [0, 1]. Thus

δ1(1 − s)s ≤ ϕ2(s) ≤ δ2(1 − s)s, (2.40)

for all s ∈ [0, 1].


Let L∗ be the dual operator of L, we will show that there exists g1 ∈ P ∗ \ {θ} such that

λL∗g1 = g1. (2.41)

In fact, let

g1(u) =∫1

0p(t)ϕ2(t)u(t)dt ∀u ∈ E. (2.42)

Then by (H1) and (2.40), we have

∫1

0p(t)ϕ2(t)u(t)dt ≤ δ2‖u‖

∫1

0t(1 − t)p(t)dt ≤ δ2‖u‖

∫1

0p(t)dt < +∞, (2.43)

which implies that g1 is well defined. We state that g1 of (2.42) satisfies (2.41). In fact, by(2.40), (2.41), and interchanging the order of integration, for any s, t ∈ [0, 1], we have

λ−1g1(u) =∫1

0p(t)(λ−1ϕ2(t)

)u(t)dt =

∫1

0p(t)(Jϕ2)(t)u(t)dt

=∫1

0p(t)u(t)

∫1

0K(s, t)p(s)ϕ2(s)dsdt

=∫1

0p(s)ϕ2(s)

∫1

0K(s, t)p(t)u(t)dtds

=∫1

0p(s)ϕ2(s)(Lu)(s)ds

= g1(Lu) =(L∗g1

)(u) ∀u ∈ E.

(2.44)

So (2.41) holds.In the following we prove that L(P) ⊂ P(g1, δ). In fact, by (2.41) and α − μn−1 − 1 > 1,

we have

ϕ2(s) ≥ δ1(1 − s)s ≥ δ1(1 − s)α−μn−1−1s ≥ δ1M−1K(t, s), t, s ∈ [0, 1]. (2.45)

Take δ = δ1M−1λ−1 > 0 in (2.2). For any u ∈ P , by (2.44), (2.45), we have

g1(Lu) = λ−1g1(u) = λ−1∫1

0p(s)ϕ2(s)u(s)ds

≥ δ1M−1λ−1

∫1

0K(t, s)p(s)u(s)ds = δ(Lu)(t) ∀t ∈ [0, 1].

(2.46)

Hence, g1(Lu) ≥ δ||Lu||, that is, L(P) ⊂ P(g1, δ). The proof is completed.


3. Main Result

Theorem 3.1. Assume that (H1)(H2) hold, and the following conditions are satisfied.(H3) There exist nonnegative continuous functions b, c : [0, 1] → (0,+∞) and a nondecreas-

ing continuous function h : Rn → [0,+∞) satisfying

lim∑ni=1|xi|→+∞

h(x1, x2, . . . , xn)∑ni=1|xi|

= 0,

f(t, x1, x2, . . . , xn) ≥ −b(t) − c(t)h(x1, x2, . . . , xn), ∀xi ∈ R.

(3.1)

(H4) f also satisfies

lim inf∑ni=1 xi→+∞∑n−1

i=1 xi≥0

f(t, x1, x2, . . . , xn)∑n

i=1 xi> λ lim sup

∑ni=1 |xi|→ 0

∣∣f(t, x1, x2, . . . , xn)∣∣

∑ni=1|xi|

< λ, (3.2)

uniformly on t ∈ [0, 1], where λ is the first eigenvalue of the operator J defined by (2.34).Then the singular fractional-order boundary-value problem (1.1) has at least one nontrivial

solution.

Proof. According to Lemma 2.15, L satisfies L(P) ⊂ P(g1, δ). Let

(Tu)(t) = ch(Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

)for u ∈ E, (3.3)

where c = maxt∈[0,1]c(t). It follows from (H3) that T : E → P is a continuous operator. Notethat

Iμn−1u(t) =∫ t

0

(t − s)μn−1−1u(s)Γ(μn−1

) ds ≤ ‖u‖Γ(μn−1

) ,

Iμn−1−μiu(t) =∫ t

0

(t − s)μn−1−μi−1u(s)Γ(μn−1 − μi

) ds ≤ ‖u‖Γ(μn−1 − μi

) , i = 1, 2, . . . , n − 2.

(3.4)

Thus from the monotone assumption of h on xi, we have

h(Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

)

≤ h(

‖u‖Γ(μn−1

) ,‖u‖

Γ(μn−1 − μ1

) , . . . ,‖u‖

Γ(μn−1 − μn−2

) , ‖u‖)

, for any u ∈ E, t ∈ [0, 1],(3.5)


which implies that

‖Tu‖ = maxt∈[0,1]

{ch(Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

)}

≤ ch(

‖u‖Γ(μn−1

) ,‖u‖

Γ(μn−1 − μ1

) , . . . ,‖u‖

Γ(μn−1 − μn−2

) , ‖u‖)

for any u ∈ E.(3.6)

Let

τ =1

Γ(μn−1

) +1

Γ(μn−1 − μ1

) + · · · + 1Γ(μn−1 − μn−2

) + 1, (3.7)

then

lim‖u‖→+∞

‖Tu‖‖u‖ ≤ lim

‖u‖→+∞ch(‖u‖/Γ(μn−1

), ‖u‖/Γ(μn−1 − μ1

), . . . , ‖u‖/Γ(μn−1 − μn−2

), ‖u‖)

‖u‖

= lim‖u‖→+∞

cτh(‖u‖/Γ(μn−1

), ‖u‖/Γ(μn−1 − μ1

), . . . , ‖u‖/Γ(μn−1 − μn−2

), ‖u‖)

τ‖u‖

≤ lim‖u‖→+∞

cτh(‖u‖/Γ(μn−1

), ‖u‖/Γ(μn−1 − μ1

), . . . , ‖u‖/Γ(μn−1 − μn−2

), ‖u‖)

‖u‖/Γ(μn−1)

+ ‖u‖/Γ(μn−1 − μ1)+ · · · + ‖u‖/Γ(μn−1 − μn−2

)+ ‖u‖

= cτ lim∑ni=1|xi|→+∞

h(x1, x2, . . . , xn)∑ni=1|xi|

= 0,

(3.8)

that is,

lim‖u‖→+∞

‖Tu‖‖u‖ = 0. (3.9)

Hence T satisfies condition (A1) in Lemma 2.5.Next take u0(t) ≡ b(t), and (Fu)(t) = f(t, Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t))

for u ∈ E. Then it follows from (H3) that

Fu + u0 + Tu = f(t, Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

)

+ b(t) + ch(Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

) ≥ 0,(3.10)

that yields

Fu + u0 + Tu ∈ P, ∀u ∈ E, (3.11)

namely, condition (A2) in Lemma 2.5 holds.


From (H4), there exists ε > 0 and a sufficiently large l1 > 0 such that, for any t ∈ [0, 1],

f(t, x1, x2, . . . , xn) ≥ λ(1 + ε)(x1 + x2 + · · · + xn)≥ λ(1 + ε)xn, x1 + x2 + · · · + xn > l1.

(3.12)

Combining (H3) with (3.12), there exists b1 ≥ 0 such that

Fu = f(t, Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

)

≥ λ(1 + ε)u(t) − b1 − ch(Iμn−1u(t), . . . , Iμn−1−μn−2u(t), u(t)

), ∀u ∈ E,

(3.13)

that is,

Fu ≥ λ(1 + ε)u − b1 − Tu ∀u ∈ E. (3.14)

As L is a positive linear operator, it follows from (3.14) that

(LFu)(t) ≥ λ(1 + ε)(Lu)(t) − Lb1 − (LTu)(t), ∀t ∈ [0, 1]. (3.15)

So condition (A3) in Lemma 2.5 holds. According to Lemma 2.5, there exists a sufficientlylarge number R > 0 such that

deg(I −A,BR, θ) = 0. (3.16)

On the other hand, it follows from (H4) that there exist 0 < ε < 1 and 0 < r < R, forany t ∈ [0, 1], such that

∣∣f(t, x1, x2, . . . , xn)∣∣ ≤ 1 − ε

Mτ∫1

0 p(s)ds(|x1| + |x2| + · · · + |xn|),

|x1| + |x2| + · · · + |xn| < r.(3.17)

Thus for any u ∈ E with ||u|| ≤ r/τ ≤ r ≤ R, we have

|Iμn−1u(t)| + ∣∣Iμn−1−μ1u(t)∣∣ + · · · + ∣∣Iμn−1−μn−2u(t)

∣∣ + |u(t)|

≤ ‖u‖Γ(μn−1

) +‖u‖

Γ(μn−1 − μ1

) + · · · + ‖u‖Γ(μn−1 − μn−2

) + ‖u‖

≤ τ‖u‖ ≤ r.

(3.18)


By (3.17), for any t ∈ [0, 1], we have

∣∣f(t, Iμn−1u(t), Iμn−1−μ1u(t), . . . , Iμn−1−μn−2u(t), u(t)

)∣∣

≤ 1 − εMτ

∫10 p(s)ds

(|Iμn−1u(t)| + ∣∣Iμn−1−μ1u(t)∣∣ + · · · + ∣∣Iμn−1−μn−2u(t)

∣∣ + |u(t)|).

(3.19)

Thus if there exist u1 ∈ ∂Br/τ and μ1 ∈ [0, 1] such that u1 = μ1Au1, then by (3.19), we have

g1(‖u1‖) = g1(μ1‖Au1‖

)= μ1g1(‖Au1‖)

= μ1g1

(

max0≤t≤1

∣∣∣∣∣

∫1

0K(t, s)p(s)f

(s, Iμn−1u1(s), . . . , Iμn−1−μn−2u1(s), u1(s)

)ds

∣∣∣∣∣

)

≤ 1 − εMτ

∫10 p(s)ds

g1

(∫1

0Mp(s)

(|Iμn−1u1(s)| + · · · + ∣∣Iμn−1−μn−2u1(s)∣∣ + |u1(s)|

)ds

)

≤ 1 − εMτ

∫10 p(s)ds

Mτ

∫1

0p(s)dsg1(‖u1‖) = (1 − ε)g1(‖u1‖).

(3.20)

Therefore, g1(‖u1‖) ≤ 0.But ϕ2(t) > 0 for all t ∈ (0, 1) by the maximum principle, and u1(t) attains zero on

isolated points by the Sturm theorem. Hence, from (2.42),

g1(‖u1‖) =∫1

0p(t)ϕ2(t)‖u1‖dt > 0. (3.21)

This is a contradiction. Thus

u/=μAu, ∀u ∈ ∂Br, μ ∈ [0, 1]. (3.22)

It follows from the homotopy invariance of the Leray-Shauder degree that

deg(I −A,Br/τ , θ) = 1. (3.23)

By (3.16), (3.23), and the additivity of Leray-Shauder degree, we obtain

deg(I −A,BR \ Br/τ , θ) = deg(I −A,BR, θ) − deg(I −A,Br/τ , θ) = −1. (3.24)

As a result, A has at least one fixed point u on BR \ Br/τ , namely, the BVP (1.1) has at leastone nontrivial solution x(t) = Iμn−1u(t).

Corollary 3.2. Assume that (H1), (H2), and (H4) hold. If the following condition is satisfied.


(H∗3) There exists a nonnegative continuous functions b : [0, 1] → (0,+∞) such that

f(t, x1, x2, . . . , xn) ≥ −b(t), for any t ∈ [0, 1], ∀xi ∈ R, (3.25)

then the singular higher multipoint boundary-value problems (1.1) have at least one nontrivial solu-tion.

Corollary 3.3. Assume that (H1), (H2), and (H4) hold. If the following condition is satisfied.(H3∗∗) There exist three constants b > 0, c > 0, and αi ∈ (0, 1) such that

f(t, x1, x2, . . . , xn) ≥ −b − cn∑

i=1

|xi|αi , for any t ∈ [0, 1], ∀xi ∈ R, (3.26)

then the singular higher multipoint boundary-value problems (1.1) have at least one nontrivial solu-tion.

Remark 3.4. Noticing that the Green function of the BVP (1.1) is not symmetrical, whichimplies that the existence of nontrivial solutions of the BVP (1.1) cannot be obtained byTheorem 2.1 in [22] and Theorem 1 in [20]. It is interesting that we construct a new linearoperator J instead of K in paper [22] and use its first eigenvalue and its correspondingeigenfunction to seek a linear continuous functional g of P . As a result, we overcome thedifficulty caused by the nonsymmetry of the Green function. In [24], the nonlinearity does notcontain derivatives and a stronger condition is required, that is, h must be an even function;here we omit this stronger assumption.

Remark 3.5. The results of [20–22] is a special case of the Corollary 3.2 and Corollary 3.3 whenα1 (i = 1, 2, . . . , n) are integer and the nonlinear term does not involve derivatives of unknownfunctions.

Acknowledgments

This project is supported financially by Scientific Research Project of Zhejiang EducationDepartment (no. Y201016244), also by Scientific Research Project of Wenzhou (no.G20110004), Natural Science Foundation of Zhejiang Province (No. 2012C31025) and theNational Natural Science Foundation of China (11071141, 11126231, 21207103), and theNatural Science Foundation of Shandong Province of China (ZR2010AM017).

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Research ArticleExistence of Positive Solution forSemipositone Fractional Differential EquationsInvolving Riemann-Stieltjes Integral Conditions

Wei Wang1 and Li Huang2

1 Water Transportation Planning & Logistics Engineering Institute, College of Harbor,Coastal and Offshore Engineering, Hohai University, Jiangsu, Nanjing 210098, China

2 National Research Center for Resettlement, Hohai University, Jiangsu, Nanjing 210098, China

Correspondence should be addressed to Wei Wang, [email protected]

Received 13 May 2012; Accepted 11 July 2012

Academic Editor: Yong Hong Wu

Copyright q 2012 W. Wang and L. Huang. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

The existence of at least one positive solution is established for a class of semipositone fractionaldifferential equations with Riemann-Stieltjes integral boundary condition. The technical approachis mainly based on the fixed-point theory in a cone.

1. Introduction

In this paper, we discuss the existence of positive solutions for the following singular semi-positone fractional differential equation with nonlocal condition:

−Dtαx(t) = f

(t, x(t),−Dt

βx(t))+ q(t), t ∈ (0, 1),

Dtβx(0) = Dt

β+1x(0) = 0, Dtβx(1) =

∫1

0Dt

βx(s)dA(s),(1.1)

where 2 < α ≤ 3, 0 < β < 1, and α − β > 2, Dt is the standard Riemann-Liouville derivative.∫1

0 Dtβx(s)dA(s) denotes the Riemann-Stieltjes integral, and A is a function of bounded

variation. f : (0, 1) × [0,+∞) × (−∞, 0] → [0,+∞) is continuous, q : (0, 1) → (−∞,+∞)is Lebesgue integrable.

Differential equations of fractional order have been recently proved to be valuabletools in the modeling of many phenomena in various fields of science and engineering.


Indeed, we can find numerous applications in physics, engineering (like traffic, trans-portation, logistics, etc.), mechanics, chemistry, and so forth, (see [1–5]). There has been asignificant development in the study of fractional differential equations in recent years, seethe monographs of Kilbas et al. [6], Lakshmikantham et al. [7], Podlubny [4], Samko et al.[8], the survey by Agarwal et al. [9] and some recent results [10–14].

On the other hand, the nonlocal condition given by a Riemann-Stieltjes integral isdue to Webb and Infante in [15–17] and gives a unified approach to many BVPs in [15, 16].Motivated by [15–17], Hao et al. [18] studied the existence of positive solutions for nth-ordersingular nonlocal boundary value problem:

x(n)(t) + a(t)f(t, x(t)) = 0, 0 < t < 1,

x(k)(0) = 0, 0 ≤ k ≤ n − 2, x(1) =∫1

0x(s)dA(s),

(1.2)

where a can be singular at t = 0, 1, f also can be singular at x = 0, but there is no singularity att = 0, 1. The existence of positive solutions of the BVP (1.2) is obtained by means of the fixedpoint index theory in cones.

More recently, Zhang [19] considered the following problem whose nonlinear termand boundary condition contain integer order derivatives of unknown functions

Dαx(t) + q(t)f(x, x′, . . . , x(n−2)

)= 0, 0 < t < 1, n − 1 < α ≤ n,

x(0) = x′(0) = · · · = x(n−2)(0) = x(n−2)(1) = 0,(1.3)

where Dα is the standard Riemann-Liouville fractional derivative of order α, q may besingular at t = 0 and f may be singular at x = 0, x′ = 0, . . . , x(n−2) = 0. By using fixed pointtheorem of the mixed monotone operator, the unique existence result of positive solutionto problem (1.3) was established. And then, Goodrich [20] was concerned with a partialextension of the problem (1.3) by extending boundary conditions

−Dαx(t) = f(t, x(t)), 0 < t < 1, n − 1 < α ≤ n,

x(i)(0) = 0, 0 ≤ i ≤ n − 2, Dαx(1) = 0, 1 ≤ α ≤ n − 2,(1.4)

and the author derived the Green’s function for the problem (1.4) and showed that it satisfiescertain properties, then by using cone theoretic techniques, a general existence theorem for(1.4) was obtained when f(t, x) satisfies some growth conditions.

Recently, Rehman and Khan [21] investigated the multipoint boundary value prob-lems for fractional differential equations of the form:

Dtαy(t) = f

(t, y(t),Dt

βy(t)), t ∈ (0, 1),

y(0) = 0, Dtβy(1) −

m−2∑

i=1

ζiDtβy(ξi) = y0,

(1.5)


where 1 < α ≤ 2, 0 < β < 1, 0 < ξi < 1, ζi ∈ [0,+∞) with∑m−2

i=1 ζiξα−β−1i < 1. By using the

Schauder fixed point theorem and the contraction mapping principle, the authors establishedthe existence and uniqueness of nontrivial solutions for the BVP (1.5) provided that thenonlinear function f : [0, 1]×R×R is continuous and satisfies certain growth conditions. Since∫1

0 Dtβx(s)dA(s) covers the multipoint BVP and integral BVP as special case, the fractional

differential equations with the Riemann-Stieltjes integral condition also were extensivelystudied by many authors, see [22, 23]. In [23], Zhang and Han considered the existence ofpositive solution of the following singular fractional differential equation:

Dtαx(t) + f(t, x(t)) = 0, 0 < t < 1, n − 1 < α ≤ n,

x(k)(0) = 0, 0 ≤ k ≤ n − 2, x(1) =∫1

0x(s)dA(s),

(1.6)

where α ≥ 2 and dA(s) can be a signed measure. Some growth conditions were adoptedto guarantee that (1.6) has an unique positive solution, moreover, the authors also gave theiterative sequence of the solution, an error estimation, and the convergence rate of the positivesolution.

Fractional differential equations like (1.1), with nonlinearities which are allowed tochange sign and boundary conditions which contain nonlocal condition given by a Riemann-Stieltjes integral with a signed measure, are rarely studied. This type problems are referred toas semipositone problems in the literature, which arise naturally in chemical reactor theory[24]. In the recent work [25], by constructing a modified function, Zhang and Liu studiedthe existence of positive solution of a class of semipositone singular second-order Dirichletboundary value problem, and when f is superlinear, a sufficient condition for the existenceof positive solution is obtained under the simple assumptions.

Motivated by the above work, in this paper, we establish the existence of positivesolutions for the semipositone fractional differential equations (1.1) when f is superlinearand involves fractional derivatives of unknown functions.


Definition 2.1 (see [4, 5]). The Riemann-Liouville fractional integral of order α > 0 of afunction x : (0,+∞) → R is given by

Iαx(t) =1

Γ(α)

∫ t

0(t − s)α−1x(s)ds, (2.1)

provided that the right-hand side is pointwise defined on (0,+∞).

Definition 2.2 (see [4, 5]). The Riemann-Liouville fractional derivative of order α > 0 of afunction x : (0,+∞) → R is given by

Dtαx(t) =

1Γ(n − α)

(d

dt

)n ∫ t

0(t − s)n−α−1x(s)ds, (2.2)


where n = [α]+1, [α] denotes the integer part of number α, provided that the right-hand sideis pointwise defined on (0,+∞).


Dtα(x(t) + y(t)

)= Dt

αx(t) +Dtαy(t). (2.3)

Proposition 2.4 (see [4, 5]). (1) If x ∈ L1(0, 1), ν > σ > 0, then

IνIσx(t) = Iν+σx(t), DtσIνx(t) = Iν−σx(t), Dt

σIσx(t) = x(t). (2.4)

(2) If ν > 0, σ > 0, then

Dtνtσ−1 =

Γ(σ)Γ(σ − ν) t

σ−ν−1. (2.5)

Proposition 2.5 (see [4, 5]). Let α > 0, and f(x) is integrable, then

IαDtαx(t) = f(x) + c1x

α−1 + c2xα−2 + · · · + cnxα−n, (2.6)

where ci ∈ R (i = 1, 2, . . . , n), n is the smallest integer greater than or equal to α.Let x(t) = Iβy(t), y(t) ∈ C[0, 1], by standard discuss, one easily reduces the BVP (1.1) to the

following modified problems,

− Dtα−βy(t) = f

(t, Iβy(t),−y(t)

)+ q(t),

y(0) = y′(0) = 0, y(1) =∫1

0y(s)dA(s),

(2.7)

and the BVP (2.7) is equivalent to the BVP (1.1).

Lemma 2.6 (see [26]). Given y ∈ L1(0, 1), then the problem,

Dtα−βx(t) + y(t) = 0, 0 < t < 1,

x(0) = x′(0) = 0, x(1) = 0,(2.8)

has the unique solution

x(t) =∫1

0G(t, s)y(s)ds, (2.9)

where G(t, s) is the Green function of the BVP (2.8) and is given by

G(t, s) =1

Γ(α − β)

{[t(1 − s)]α−β−1, 0 ≤ t ≤ s ≤ 1,[t(1 − s)]α−β−1 − (t − s)α−β−1, 0 ≤ s ≤ t ≤ 1.

(2.10)


Lemma 2.7 (see [26]). For any t, s ∈ [0, 1], G(t, s) satisfies:

tα−β−1(1 − t)s(1 − s)α−β−1

Γ(α − β) ≤ G(t, s)

≤ s(1 − s)α−β−1

Γ(α − β − 1

) , or

(tα−β−1(1 − t)Γ(α − β − 1

)

)

.

(2.11)

By Lemma 2.6, the unique solution of the problem,

Dtαx(t) = 0, 0 < t < 1,

x(0) = x′(0) = 0, x(1) = 1,(2.12)

is tα−β−1. Let

C =∫1

0tα−β−1dA(t), B =

∫1

0tα−β−1(1 − t)dA(t) (2.13)

and define

GA(s) =∫1

0G(t, s)dA(t), (2.14)

as in [25], one can get that the Green function for the nonlocal BVP (2.7) is given by

H(t, s) =tα−β−1

1 − C GA(s) +G(t, s). (2.15)

Throughout paper one always assumes the following holds.(H0)A is a increasing function of bounded variation such that GA(s) ≥ 0 for s ∈ [0, 1]

and 0 ≤ C < 1, where C is defined by (2.13).Define

q+(t) = max{q(t), 0

}, q−(t) = max

{−q(t), 0}. (2.16)

One has the following Lemma.

Lemma 2.8. Let 1 < α − β ≤ 2 and (H0) hold, then the unique solution w(t) of the linear problem,

−Dtα−βw(t) = q−(t), t ∈ (0, 1),

w(0) = w′(0) = 0, w(1) =∫1

0w(s)dA(s),

(2.17)


satisfies

w(t) ≤ ηγ(t), (2.18)

where

η =1

Γ(α − β − 1

)(

1 +CB)∫1

0q−(s)ds, γ(t) = tα−β−1(1 − t) + B

1 − C tα−β−1. (2.19)

Proof. By (2.11) and that A(t) is a increasing function of bounded variation, we have

GA(s) =∫1

0G(t, s)dA(t) ≤

∫1

0

tα−β−1

Γ(α − β − 1

)dA(t) =C

Γ(α − β − 1

) . (2.20)

Consequently,

w(t) =∫1

0H(t, s)q−(t)ds =

∫1

0

(tα−β−1

1 − C GA(s) +G(t, s)

)

q−(s)ds

≤ tα−β−1(1 − t)Γ(α − β − 1

)∫1

0q−(s)ds +

Ctα−β−1

Γ(α − β − 1

)(1 − C)

∫1

0q−(s)ds

≤ 1Γ(α − β − 1

)(

1 +CB)(

tα−β−1(1 − t) + B1 − C t

α−β−1)∫1

0q−(s)ds

= ηγ(t).

(2.21)

Remark 2.9. (1) γ(t) satisfies

(1 − C)γ(t)2

≤ (1 − C)γ(t)α − β − 1

≤ 1, (2.22)

(2)

H(t, s) ≤ GA(s)1 − C +

s(1 − s)α−β−1

Γ(α − β − 1

) = Φ(s). (2.23)

In fact, since 1 < α − β − 1 < 2, the left side of (1) clearly holds. For right side of (1),from B ≤ C, one gets

γ(t) ≤ 1 +B

1 − C ≤ 1 +C

1 − C =1

1 − C , (2.24)


thus we have

(1 − C)γ(t)α − β − 1

≤ 1. (2.25)

(2) is obvious from (2.11).Now define a function [·]∗ for any z ∈ C[0, 1] by

[z(t)]∗ =

{z(t), z(t) ≥ 0,0, z(t) < 0

(2.26)

and consider the following approximate problem of the BVP (2.7):

−Dtα−βv(t) = f

(t, Iβ[v(t) −w(t)]∗,−[v(t) −w(t)]∗

)+ q+(t),

v(0) = v′(0) = 0, v(1) =∫1

0v(s)dA(s).

(2.27)

Lemma 2.10. Suppose v is a positive solution of the problem (2.27) and satisfies v(t) ≥ w(t), t ∈[0, 1], then v − w is a positive solution of the problem (2.7), consequently, Iβ[v(t) − w(t)] also is apositive solution of the BVP (1.1).

Proof. In fact, if v is a positive solution of the BVP (2.27) such that v(t) ≥ w(t) for any t ∈ [0, 1],then, from (2.27) and the definition of [z(t)]∗, we have

−Dtα−βv(t) = f

(t, Iβ(v(t) −w(t)),−(v(t) −w(t))

)+ q+(t),

v(0) = v′(0) = 0, v(1) =∫1

0v(s)dA(s).

(2.28)

Let y = v −w, then we have

w(0) = w′(0) = 0, w(1) =∫1

0w(s)dA(s), (2.29)

and Dtα−βy(t) = Dt

α−βv(t) − Dtα−βw(t), which implies that

−Dtα−βv(t) = −Dt

α−βy(t) + q−(t). (2.30)

Since q(t) = q+(t)−q−(t), then (2.27) is transformed to (2.7), that is, v−w is a positive solutionof the BVP (2.27). By (2.7), Iβ[v(t) −w(t)] is a positive solution of the BVP (1.1).


It is well known that the BVP (2.27) is equivalent to the fixed points of the mapping Tgiven by

(Tv)(t) =∫1

0H(t, s)

(f(s, Iβ[v(s) −w(s)]∗,−[v(s) −w(s)]∗

)+ q+(s)

)ds. (2.31)

The basic space used in this paper is E = C([0, 1]; R), where R is a real number set.Obviously, the space E is a Banach space if it is endowed with the norm as follows:

‖v‖ = maxt∈[0,1]

|v(t)|, (2.32)

for any u ∈ E. Let

P ={v ∈ E : v(t) ≥ (1 − C)γ(t)

4‖v‖}, (2.33)

then P is a cone of E.For the convenience in presentation, we now present some assumptions to be used in

the rest of the paper.(H1) For any fixed t ∈ (0, 1) and for any c ∈ (0, 1], there exist constants μi > 1, i =

1, 2, 3, 4, such that, for all (t, u, v) ∈ (0, 1) × [0,+∞) × (−∞,+0],

cμ1f(t, u, v) ≤ f(t, cu, v) ≤ cμ3f(t, u, v),

cμ2f(t, u, v) ≤ f(t, u, cv) ≤ cμ4f(t, u, v).(2.34)

(H2)∫1

0 q−(s)ds > 0, for any s ∈ (0, 1), f(s, 1,−1) > 0, and

∫1

0Φ(s)

[f(s, 1,−1) + q+(s)

]ds ≤ r

2(1 + r)μ1+μ2, (2.35)

where

r =8η

1 − C , (2.36)

and C, η,Φ are defined by (2.13), (2.19), and (2.23), respectively.

Remark 2.11. If c > 1, the reversed inequality of (2.34) holds, that is, for all (t, u, v) ∈ (0, 1) ×[0,+∞) × (−∞,+0], one has

cμ3f(t, u, v) ≤ f(t, cu, v) ≤ cμ1f(t, u, v), cμ4f(t, u, v) ≤ f(t, u, cv) ≤ cμ2f(t, u, v). (2.37)


Lemma 2.12. Suppose (H1)–(H2) holds, for any fixed t ∈ (0, 1), f(t, u, v) is nondecreasing in u on[0,+∞) and nonincreasing in v on (−∞, 0]; and for any [a, b] ⊂ (0, 1),

lim(u,v)→ (+∞,−∞)

mint∈[a,b]

f(t, u, v)|u||v| = +∞. (2.38)

Proof. Let 0 ≤ u1 ≤ u2. If u2 = 0, obviously f(t, u1, v) ≤ f(t, u2, v) holds. If u2 /= 0, let c0 = u1/u2,then 0 < c0 ≤ 1. It follows from (H1) that

f(t, u1, v) = f(t, c0u2, v) ≤ cμ3

0 f(t, u2, v) ≤ f(t, u2, v). (2.39)

Thus f(t, u, v) is nondecreasing in u on [0,∞).On the other hand, for any v1 ≤ v2 ≤ 0, if v1 = 0, obviously f(t, u, v1) ≥ f(t, u, v2)

holds. If v1 /= 0, let c1 = v2/v1, then 0 < c1 ≤ 1. By (H1), we have

f(t, u, v2) = f(t, u, c1v1) ≤ cμ4

1 f(t, u, v1) ≤ f(t, u, v1). (2.40)

Thus f(t, u, v) is nonincreasing in v on (−∞, 0].Now choose u > 1 and v < −1, then by Remark 2.11, we have

f(t, u, v) = f(t, |u|,−|v|) ≥ |u|μ3f(t, 1,−|v|) ≥ |u|μ3 |v|μ4f(t, 1,−1). (2.41)

Thus for any [a, b] ⊂ (0, 1), and any t ∈ [a, b], we have

mint∈[a,b]

f(t, u, v)|u||v| ≥ |u|μ3−1|v|μ4−1 min

t∈[a,b]f(t, 1,−1) > 0. (2.42)

Therefore

lim(u,v)→ (+∞,−∞)

mint∈[a,b]

f(t, u, v)|u||v| = +∞. (2.43)

Lemma 2.13. Assume that (H0)-(H2) holds. Then T : P → P is well defined. Furthermore, T :P → P is a completely continuous operator.

Proof. For any fixed v ∈ P , there exists a constant L > 0 such that ‖v‖ ≤ L. And then,

0 ≤ [v(s) −w(s)]∗ ≤ v(s) ≤ ‖v‖ ≤ L,

0 ≤ Iβ[v(s) −w(s)]∗ =∫ t

0

(t − s)β−1[v(s) −w(s)]∗

Γ(β) ds ≤ L

Γ(β) .

(2.44)


By (2.44) and (H1)-(H2), we have

(Tv)(t) =∫1

0H(t, s)

(f(s, Iβ[v(s) −w(s)]∗,−[v(s) −w(s)]∗

)+ q+(s)

)ds

≤∫1

0H(t, s)

(

f

(

s,L

Γ(β) + 1,−(L + 1)

)

+ q+(s)

)

ds

≤(

L

Γ(β) + 1

)μ1

(L + 1)μ2

∫1

0Φ(s)

[f(s, 1,−1) + q+(s)

]ds < +∞,

(2.45)

which implies that the operator T : P → E is well defined.Next let f∗(s) = f(s, Iβ[v(s)−w(s)]∗,−[v(s)−w(s)]∗) + q+(s), for any v ∈ P , by (2.11),

we have

‖Tv‖ ≤∫1

0

s(1 − s)α−β−1

Γ(α − β − 1

) f∗(s)ds +tα−β−1

1 − C∫1

0GA(s)f∗(s)ds

≤ 11 − C

∫1

0

s(1 − s)α−β−1

Γ(α − β − 1


1 − C∫1

0GA(s)f∗(s)ds.

(2.46)

On the other hand, by (2.11), (2.22), and (2.46), we also have

(Tu)(t) ≥ tα−β−1(1 − t)∫1

0

s(1 − s)α−β−1

Γ(α − β) f∗(s)ds +

tα−β−1

1 − C∫1

0GA(s)f∗(s)ds

≥ 12tα−β−1(1 − t)

∫1

0

s(1 − s)α−β−1

Γ(α − β) f∗(s)ds +

tα−β−1

1 − C∫1

0GA(s)f∗(s)ds

=12

{

tα−β−1(1 − t)∫1

0

s(1 − s)α−β−1

Γ(α − β) f∗(s)ds +

tα−β−1

1 − C∫1

0GA(s)f∗(s)ds

}

+tα−β−1

2(1 − C)∫1

0GA(s)f∗(s)ds

≥ 12

{

tα−β−1(1 − t)∫1

0

s(1 − s)α−β−1

Γ(α − β) f∗(s)ds +

Btα−β−1

1 − C∫1

0

s(1 − s)α−β−1

Γ(α − β) f∗(s)ds

}

+tα−β−1

2(1 − C)∫1

0GA(s)f∗(s)ds


=(1 − C)γ(t)

2(α − β − 1

) × 11 − C

∫1

0

s(1 − s)α−β−1

Γ(α − β − 1


2(1 − C)∫1

0GA(s)f∗(s)ds

≥ (1 − C)γ(t)2(α − β − 1

)

{1

1 − C∫1

0

s(1 − s)α−β−1

Γ(α − β − 1


1 − C∫1

0GA(s)f∗(s)ds

}

≥ (1 − C)γ(t)4

{1

1 − C∫1

0

s(1 − s)α−β−1

Γ(α − β − 1


1 − C∫1

0GA(s)f∗(s)ds

}

.

(2.47)

So we have

(Tv)(t) ≥ (1 − C)γ(t)4

‖Tv‖, t ∈ [0, 1]. (2.48)

which yields that T(P) ⊂ P .At the end, using standard arguments, according to the Ascoli-Arzela Theorem, one

can show that T : P → P is continuous. Thus T : P → P is a completely continuousoperator.

Lemma 2.14 (see [27]). Let E be a real Banach space, P ⊂ E be a cone. Assume Ω1,Ω2 are twobounded open subsets of E with θ ∈ Ω1, Ω1 ⊂ Ω2, and let T : P ∩ (Ω2 \ Ω1) → P be a completelycontinuous operator such that either

(1) ‖Tx‖ ≤ ‖x‖, x ∈ P ∩ ∂Ω1, and ‖Tx‖ ≥ ‖x‖, x ∈ P ∩ ∂Ω2 or

(2) ‖Tx‖ ≥ ‖x‖, x ∈ P ∩ ∂Ω1, and ‖Tx‖ ≤ ‖x‖, x ∈ P ∩ ∂Ω2.

Then T has a fixed point in P⋂(Ω2 \Ω1).

3. Main Results

Theorem 3.1. Suppose (H0)–(H2) hold. Then the BVP (1.1) has at least one positive solutions x(t),and x(t) satisfies

x(t) ≥ ηΓ(β)Γ(α − β)

Γ(α)

[(1 +

B1 − C

)tα−1 − α − β

αtα], t ∈ [0, 1]. (3.1)

Proof. Let

r =8η

1 − C , (3.2)

and Ω1 = {v ∈ P : ‖v‖ < r}. Then, for any v ∈ ∂Ω1, s ∈ [0, 1], notice that 0 < β < 1, we have

[v(s) −w(s)]∗ ≤ v(s) ≤ ‖v‖ ≤ r,

Iβ[v(s) −w(s)]∗ =1

Γ(β)∫ s

0(s − t)β−1[v(t) −w(t)]∗dt ≤ r

Γ(β) ≤ r.

(3.3)


By (H1) and (H2), one has

‖Tv‖ = max0≤t≤1

∫1

0H(t, s)

[f(s, Iβ[v(s) −w(s)]∗,−[v(s) −w(s)]∗

)+ q+(s)

]ds

≤∫1

0Φ(s)

[

f

(

s,r

Γ(β) ,−r

)

+ q+(s)

]

ds

≤∫1

0Φ(s)

[f(s, r + 1,−(r + 1)) + q+(s)

]ds

≤ 2(r + 1)μ1+μ2

∫1

0Φ(s)

[f(s, 1,−1) + q+(s)

]ds

≤ r = ‖v‖.

(3.4)

Therefore,

‖Tv‖ ≤ ‖v‖, v ∈ P ∩ ∂Ω1. (3.5)

On the other hand, choose a real number M > 0 such that

aα−β(1 − b)α−βΓ(α − β) × M(1 − C)2aβ

64Γ(β + 1

)(aα−β−1(1 − b) + B

1 − Caα−β−1

)2

≥ 1. (3.6)

From (2.38), there exists N > r such that, for any t ∈ [a, b],

f(t, u, v) ≥M|u||v|, |u|, |v| ≥N. (3.7)

Take

R

> max

⎧⎪⎪⎨

⎪⎪⎩1,

8N

(1−C)(aα−β−1(1−b)+ B

1 − Caα−β−1

) ,8NΓ

(β + 1

)

(1−C)aβ(aα−β−1(1−b) + B

1 − Caα−β−1

) , r

⎫⎪⎪⎬

⎪⎪⎭,

(3.8)


then R > r. Let Ω2 = {v ∈ P : ‖v‖ < R}, for any v ∈ P ∩ ∂Ω2 and for any t ∈ [a, b], we have

v(t) −w(t) ≥ v(t) − ηγ(t) ≥ v(t) − 4(1 − C)Rv(t) ≥

12v(t)

≥ (1 − C)γ(t)8

R ≥ (1 − C)8

(aα−β−1(1 − b) + B

1 − Caα−β−1

)R ≥N > 0,

Iβ[v(t) −w(t)] ≥ (1 − C)8

(aα−β−1(1 − b) + B

1 − Caα−β−1

)R

∫ t

0

(t − s)β−1

Γ(β) ds

≥ (1 − C)aβ8Γ(β + 1

)(aα−β−1(1 − b) + B

1 − Caα−β−1

)R ≥N > 0.

(3.9)

So for any v ∈ P ∩ ∂Ω2, t ∈ [a, b], by (3.7)-(3.9), we have

‖Tv‖ ≥∫1

0H(t, s)

(f(s, Iβ[v(s) −w(s)]∗,−[v(s) −w(s)]∗

)+ q+(s)

)ds

≥∫1

0G(t, s)f

(s, Iβ[v(s) −w(s)]∗,−[v(s) −w(s)]∗

)dss

≥ tα−β−1(1 − t)Γ(α − β)

∫b

a

s(1 − s)α−β−1f(s, Iβ[v(s) −w(s)]∗,−[v(s) −w(s)]∗

)ds

≥ aα−β−1(1 − b)Γ(α − β)

∫b

a

s(1 − s)α−β−1M∣∣∣Iβ[v(s) −w(s)]

∣∣∣|v(s) −w(s)|ds

≥ aα−β(1 − b)α−βΓ(α − β) × M(1 − C)2aβ

64Γ(β + 1

)(aα−β−1(1 − b) + B

1 − Caα−β−1

)2

R2

≥ R2 ≥ R = ‖x‖.

(3.10)

Thus

‖Tv‖ ≥ ‖v‖, v ∈ P ∩ ∂Ω2. (3.11)

By Lemma 2.14, T has at least one fixed points v such that r ≤ ||v|| ≤ R.In the end,

v(t) −w(t) ≥ (1 − C)γ(t)4

‖v‖ − ηγ(t) ≥ (1 − C)γ(t)4

r − ηγ(t)

≥ ηγ(t) > 0, t ∈ (0, 1).

(3.12)


Equation (3.12) implies that v(t) > w(t), t ∈ (0, 1), and

x(t) = Iβ(v(t) −w(t)) ≥ ηIβγ(t) = η∫ t

0(t − s)β−1γ(s)ds

=ηΓ(β)Γ(α − β)

Γ(α)

[(1 +

B1 − C

)tα−1 − α − β

αtα]> 0, t ∈ (0, 1).

(3.13)

So by Lemma 2.10, the BVP (1.1) has at least one positive solution x, and x satisfies (3.13).

Example 3.2. Consider the following semipositone boundary value problem with fractionalorder α = 21/8:

−D21/8x(t) =x2(t) +

(D1/8x(t))3

1012 × 4√(1 − t)

− 1√t, 0 < t < 1,

D1/8x(0) = D9/8x(0) = 0, D1/8x(1) =∫1

0D1/8x(s)dA(s).

(3.14)

Let

A(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

0, t ∈[

0,14

),

1, t ∈[

14,

34

),

2, t ∈[

34, 1],

(3.15)

then the BVP (3.14) becomes the 4-Point BVP with coefficients

−D21/8x(t) =x2(t) +

(D1/8x(t))3

1012 × 4√(1 − t)

− 1√t, 0 < t < 1,

D1/8x(0) = D9/8x(0) = 0, D1/8x(1) = D1/8x

(14

)+D1/8x

(34

).

(3.16)

Then the BVP (3.14) has at least one positive solution x(t), and x(t) satisfies

x(t) ≥ 62.4289[

1.1357t5/2 − 1021t21/8], t ∈ [0, 1]. (3.17)


Proof. Obviously, α = 21/8, β = 1/8, and

0 ≤ C =∫1

0t3/2dA(t) =

(14

)3/2

+(

34

)3/2

≈ 0.7745 < 1,

B =∫1

0t3/2(1 − t)dA(t) =

34

(14

)3/2

+14

(34

)3/2

≈ 0.2561.

(3.18)

On the other hand, we have

G(t, s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

G1(t, s) =[t(1 − s)]3/2

Γ(5/2), 0 ≤ t ≤ s ≤ 1,

G2(t, s) =[t(1 − s)]3/2 − (t − s)3/2

Γ(5/2), 0 ≤ s ≤ t ≤ 1,

GA(s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

G2

(14, s

)+ 2G2

(34, s

), 0 ≤ s < 1

4,

G1

(14, s

)+ 2G2

(34, s

),

14≤ s < 3

4,

G1

(14, s

)+ 2G1

(34, s

),

34≤ s ≤ 1.

(3.19)

Thus GA(s) ≥ 0, 0 ≤ C < 1 and A(s) is increasing. So (H0) holds.Take

f(t, u, v) =u2 − v3

1012 × 4√(1 − t)

, (t, u, v) ∈ (0, 1) × [0,+∞) × (−∞, 0], q(t) = − 1√t, (3.20)

then

f(t, x(t),−Dt

1/8x(t))=x2(t) +

(D1/8x(t))3

1012 × 4√(1 − t)

, q+(t) = 0, q−(t) =1√t. (3.21)

For any 0 < c ≤ 1, we also have

c3f(t, u, v) ≤ f(t, cu, v) ≤ c3/2f(t, u, v),

c2f(t, u, v) ≤ f(t, cu, v) ≤ c4f(t, u, v).(3.22)

Then (H1) holds.


In the end, we notice

Φ(s) =GA(s)1 − C +

s(1 − s)α−β−1

Γ(α − β − 1

) ≤ CΓ(α − β − 1

)(1 − C) +

1Γ(α − β − 1

)

=1

Γ(α − β − 1

)(1 − C) ≈ 5.0041,

(3.23)

then

∫1

0Φ(s)

[f(s, 1,−1) + q+(s)

]ds ≤ 5.0041

∫1

0

2

1012 × 4√(1 − t)

ds ≈ 1.33443 × 1011. (3.24)

Moreover,

η =1

Γ(α − β − 1

)(

1 +CB)∫1

0q−(s)ds ≈ 9.0817, r =

8η1 − C = 322.1889,

r

2(1 + r)μ1+μ2=

322.1889

2(1 + 322.1889)5≈ 4.5688 × 10−11,

(3.25)

which implies that (H2) holds.According to Theorem 3.1, the BVP (3.14) has at least one positive solution x(t), and

x(t) satisfies

x(t) ≥ 62.4289[

1.1357t5/2 − 1021t21/8], t ∈ [0, 1]. (3.26)

Acknowledgments

The paper is Funded by the Humanities and Social Sciences Foundation of Ministryof Education (no. 09YJC630056), the National Natural Science Foundation of China (no.51009060, no. 50909042), the Fundamental Research Funds for the Central Universities (no.2009B13414), and the Priority Academic Program Development of Jiangsu Higher EducationInstitutions(Coastal Development Conservancy).

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Research ArticleA Generalization of Mahadevan’s Version ofthe Krein-Rutman Theorem and Applications top-Laplacian Boundary Value Problems

Yujun Cui1 and Jingxian Sun2

1 Department of Applied Mathematics, Shandong University of Science and Technology,Qingdao 266590, China

2 Department of Mathematics, Xuzhou Normal University, Xuzhou 221116, China

Correspondence should be addressed to Yujun Cui, [email protected]

Received 12 January 2012; Revised 24 March 2012; Accepted 13 July 2012


Copyright q 2012 Y. Cui and J. Sun. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We will present a generalization of Mahadevan’s version of the Krein-Rutman theorem for acompact, positively 1-homogeneous operator on a Banach space having the properties of beingincreasing with respect to a cone P and such that there is a nonzero u ∈ P\{θ}−P for whichMTpu ≥u for some positive constant M and some positive integer p. Moreover, we give some new resultson the uniqueness of positive eigenvalue with positive eigenfunction and computation of the fixedpoint index. As applications, the existence of positive solutions for p-Laplacian boundary-valueproblems is considered under some conditions concerning the positive eigenvalues correspondingto the relevant positively 1-homogeneous operators.

1. Introduction

The Krein-Rutman theorem [1, 2] plays a very important role in nonlinear differentialequations, as it provides the abstract basis for the proof of the existence of various principaleigenvalues, which in turn are crucial in bifurcation theory, in topological degree calculation,and in the stability analysis of solutions to elliptic equations. Owing to its importance, muchattention has been given to the most general versions of the linear Krein-Rutman theoremby a number of authors, see [3–7]. For example, Krasnosel’skiı [3] introduced the concept ofthe e-positive linear operator and then used it to prove the following results concerning theeigenvalues of positive linear compact operator.

Theorem 1.1. Let X, a Banach space, P ⊂ X a cone in X. Let T : X → X be a linear, positive, andcompact operator. Suppose that for some non-zero element u = v −w, where v,w ∈ P and −u /∈ P ,


the following relation is satisfied:

MTpu ≥ u, for some M > 0, (1.1)

where p is some positive integer. Then T has a non-zero eigenvector x0 in P :

Tx0 = λ0x0, (1.2)

where the positive eigenvalue λ0 satisfies the inequality λ0p√M ≥ 1.

Furthermore, if P is a reproducing cone and T is e-positive for some e ∈ P \ {θ}, then

(1) the positive eigenvalue λ0 of T is simple;

(2) the operator T has a unique positive eigenvector upto a multiplicative constant.

Recently, the nonlinear version of the Krein-Rutman theorem has been extended topositive eigenvalue problem for increasing, positively 1-homogeneous, compact, continuousoperators by Mallet-Paret and Nussbaum [8, 9], Mahadevan [10], and Chang [11].

The following nonlinear Krein-Rutman theorem has been established in [10].

Theorem 1.2. Let X be a Banach space, P ⊂ X be a cone in X. Let T : X → X be an increasing,positively 1-homogeneous, compact, continuous operator for which there exists a non-zero u ∈ P andM > 0 such that

MTu ≥ u, (1.3)

Then T has a non-zero eigenvector x0 in P .

Compared with Theorem 1.1, we note that the element u, appeared in Theorem 1.2,belongs to P . Consequently we put forward a problem: are the results in Theorem 1.2 validif the condition u ∈ P is replaced with that in Theorem 1.1. The purpose of this studyis to solve the above problem. By means of global structure of the positive solution set,we present a generalization of Mahadevan’s version of the Krein-Rutman Theorem for acompact, positively 1-homogeneous operator on a Banach space having the properties ofbeing increasing with respect to a convex cone P and such that there is a non-zero u ∈P \ {θ} − P for which MTpu ≥ u for some positive constant M and some positive integerp. The method in this paper is somewhat different from that in [10].

The paper is organized as follow. In Section 2, we give some basic definitions andstate three lemmas which are needed later. In Section 3, we establish some results for theexistence of the eigenvalues of positively compact, 1-homogeneous operator and deducesome results on the uniqueness of positive eigenvalue with positive eigenfunction. InSection 4, we present some new methods of computation of the fixed point index forcone mapping. The final section is concerned with applications to the existence of positivesolutions for p-Laplacian boundary-value problems under some conditions concerning thepositive eigenvalues corresponding to the relevant positively 1-homogeneous operators.


2. Preliminaries

Let X a Banach space, P ⊂ X be a cone in X. A cone P is called solid if it contains interior

points, that is,o

P /= ∅. A cone is said to be reproducing if X = P − P . Every cone P in E definesa partial ordering in E given by x ≤ y if and only if y − x ∈ P . If x ≤ y and x /=y, we write

x < y; if cone P is solid and y − x ∈ o

P , we write x y. For the concepts and the propertiesabout the cone we refer to [12, 13].

A mapping T : X → X is said to be increasing if x ≤ y implies Tx ≤ Ty and it is saidto be strictly increasing if x < y implies Tx < Ty. The mapping is said to be compact if ittakes bounded subsets of X into relatively compact subsets of X. We say that the mapping ispositively 1-homogeneous if it satisfies the relation

T(tx) = tT(x) ∀x ∈ X, t ∈ R+0 = [0,+∞). (2.1)

We say that a real number λ is an eigenvalue of the operator if there exists a non-zero x ∈ Xsuch that Tx = λx.

Definition 2.1. Let e ∈ P \ {θ}, a mapping T : P → P is called e−positive if for every non-zerox ∈ P a natural number n = n(x) and two positive number c(x), d(x) can be found such that

c(x)e ≤ Tnx ≤ d(x)e. (2.2)

This is stronger than requiring that T is positive, that is, T(P) ⊂ P . It is always satisfied

if P is a solid cone and T is strongly positive, that is, T(P) ⊂ o

P , with any e ∈ P \ {θ}, but it canbe satisfied more generally.

For the application in the sequel, we state the following three lemmas which can befound in [14, Theorem 17.1] [3, Lemma 1.2] [15, Theorem 1.1]. The first one involves theglobal structure of the positive solution set for completely continuous map, the second oneinvolve cones, and the last one involves the computation of fixed-point index.

Lemma 2.2. Let F : R+0 × P → P be a compact, continuous map and such that F(0, x) = θ for all

x ∈ P . Then, F(λ, x) = x has a nontrivial connected unbounded component of solutions C+ ⊂ R+0 × P

containing the point (0, θ).

Lemma 2.3. Let x ∈ P . For an element y ∈ X, suppose a δ1 can be found such that y ≤ δ1x. Then asmall δx(y) exists for which y ≤ δx(y)x.

Lemma 2.4. Let Ω be a bounded open set in X, let P be a cone in X, and let A : P → P be acompletely continuous map. Suppose that there is an increasing, positively 1-homogeneous mappingT and u∗ ∈ P \ {θ} such that Tu∗ ≥ u∗, and that

Au ≥ Tu, Au/=u, ∀u ∈ ∂Ω ∩ P. (2.3)

Then the fixed-point index i(A,Ω ∩ P, P) = 0.


3. Main Results

Theorem 3.1. Let T : X → X be an increasing, positively 1-homogeneous, compact, continuousmapping. Suppose that for some non-zero element u = v − w, where v,w ∈ P and −u /∈ P , thefollowing relation is satisfied:

MTpu ≥ u (for some M > 0), (3.1)

where p is some positive integer. Then T has a non-zero eigenvector x0 in P :

Tx0 = λ0x0, (3.2)

where the positive eigenvalue λ0 satisfies the inequality

λ0p√M ≥ 1. (3.3)

Proof. Let v ∈ P (v /= θ) be as in the hypothesis of the theorem. For every positive integern > 0, define Fn : R

+0 × P → P by

Fn(λ, x) := λTx +1nλv. (3.4)

Since T is compact and continuous, each of these operators Fn is clearly compact andcontinuous on R

+0 × P . Also they map R

+0 × P into P since T maps P into itself, which follows

from the fact that T is increasing and Tθ = θ. Let, by Lemma 2.2, C+n ⊂ R

+0 × P → P be a

connected unbounded branch of solutions to the equation

Fn(λ, x) = x. (3.5)

First we show that C+n ⊂ [0, p

√M] × P for all n > 0. Indeed, suppose that x is a fixed

point of Fn(λ, ·) for some λ ≥ 1. Then x = Fn(λ, x) = λTx + (1/n)λv and we obtain, from theproperties of T and the inequalities v ≥ θ, v ≥ u, respectively, that

x ≥ λTx, x ≥ 1nλu. (3.6)

Let τn = sup{τ | x ≥ τu}. Obviously, τn ≥ (1/n)λ > 0. Since T is increasing and 1-homogeneous, by (3.4), we have

x = λTx +1nλv ≥ λ2T2x +

1nλv

≥ · · · ≥ λpTpx +1nλv ≥ λpTpx

≥ λpTp(τnu) ≥ τn λp

Mu.

(3.7)


Consequently, by the definition of τn,

λp

M≤ 1. (3.8)

In other words, if λ > p√M, then Fn(λ, ·) has no fixed point. This implied thatC+

n ⊂ [0, p√M]×P

for every n > 0.Notice that the branch C+

n is connected and unbounded starting from (0, θ), there mustnecessarily exist xn with ‖xn‖ = 1 and λn ∈ [0, p

√M] such that (λn, xn) ∈ C+

n . That is,

xn = λnTxn +1nλnv, λn ∈

[0, p√M

], ‖xn‖ = 1. (3.9)

Since the operator T is compact, a subsequence of indices ni (i = 1, 2, . . .) can be chosen suchthat the sequence Txn strongly converges to some element y∗ ∈ P . By virtue of (3.9), withthis choice of the sequence ni, the convergence of the number λni to some λ∗ which satisfiesthe inequality (3.3) can be guaranteed simultaneously. Then xni will converge in norm to theelement x0 = λ∗y∗ with ‖x0‖ = 1. Further, it follows from the fact ‖x0‖ = 1 that λ∗ /= 0. Letλ0 = λ−1

∗ . To obtain the equality (3.2), it suffices to pass to the limits in the equality:

xni = λniTxni +1niλniv (i = 1, 2, . . .). (3.10)

This completes the proof of the theorem.

Example 3.2. Consider the positive 1-homogeneous map T :

(Tx)(t) =(∫

G

K(t, s)xp(s)ds)1/p

, (3.11)

where G is a bounded closed set in a finite-dimensional space, the kernel K(t, s) isnonnegative, and p = 2n + 1 for some n ∈ N.

If there exists a system of points s1, s2, . . . , sp such that

K(s1, s2)K(s2, s3) · · ·K(sp, s1

)> 0. (3.12)

Then the map T defined by (3.11) has a nonnegative eigenfunction. In fact, it is easy to seethat

(Tpx)(t) =(∫

G

K(p)(t, s)xp(s)ds)1/p

, (3.13)

where

K(p)(t, s) =∫

G

· · ·∫

G

K(t, t1) · · ·K(tp−1, s

)dt1 · · ·dtp−1 (3.14)


is positive at the point (s1, s1) of the topological product G × G. We denoted by G1 ⊂ G aclosed neighborhood of the point s1 ∈ G1 such that K(p)(t, s) > 0 when t, s ∈ G1. We denoteby y(t) a continuous nonnegative function such that y(s1) > 0, y(t) = 0 when t /∈ G1. Then

(Tpy

)(t) =

(∫

G

K(p)(t, s)yp(s)ds)1/p

> 0 (3.15)

when t ∈ G1. From (3.15) it follows that there exists a number M > 0 such that

MTpy ≥ y. (3.16)

This inequality is just the condition of Theorem 3.1.

Remark 3.3. Positive 1-homogeneous maps are usually only defined on a cone. In this case,Theorem 3.1 remains valid provided u ∈ P \ {θ}. Moreover, Theorem 2.1 and Corollary 2.1 of[5] already give a general result for a k-set contraction, positive 1-homogeneous maps.

Theorem 3.4. Suppose that T is an increasing, positively 1-homogeneous, e-positive mapping. If thereexist

u1 ∈ P \ {θ}, λ1 > 0 such that λ1u1 ≤ Tu1,

u2 ∈ P \ {θ}, λ2 > 0 such that λ2u2 ≥ Tu2,(3.17)

then λ2 ≥ λ1. Furthermore, if for x > y > θ, a positive number c(x, y) can be found such that

Tx − Ty ≥ c(x, y)e, (3.18)

then λ1 = λ2 implies that u1 is a scalar multiple of u2.

Proof. It follows from the e-positiveness of T that there exist m,n such that

Tnu1 ≤ d(u1)e, c(u2)e ≤ Tmu2, (c(u1), d(u2) > 0). (3.19)

Then for t > 0, we have

(tc(u2) − d(u1))e ≤ tTmu2 − Tnu1 ≤ tλm2 u2 − λn1u1. (3.20)

From this and the fact that u1 ∈ P \ {θ} we deduce that δ � δλm2 u2(λn1u1) > 0, that is, δλm2 u2 −

λn1u1 ∈ P . Since T is increasing, T(δλm2 u2) ≥ T(λn1u1), from which, by virtue of (3.17), it followsthat

λ2δλm2 u2 ≥ T(δλm2 u2

) ≥ T(λn1u1) ≥ λ1λ

n1u1. (3.21)

By Lemma 2.3, we obtain that λ2 ≥ λ1.


Now suppose that λ1 = λ2. According to the above proof, the number δ > 0. Sinceδλm2 u2 − λn1u1 ∈ P and (3.18), if δλm2 u2 /=λn1u1, then there exists c > 0 such that

T(δλm2 u2

) − T(λn1u1) ≥ ce. (3.22)

Therefore, by (3.19)

λ2δλm2 u2 ≥ T(δλm2 u2

) ≥ T(λn1u1)+ ce ≥ λ1λ

n1u1 +

c

d(u1)Tn(u1)

≥ λn+11 u1 +

c

d(u1)λn1u1 =

(1 +

c

λ1d(u1)

)λ2λ

n1u1.

(3.23)

This contradicts with the definition of δ. This shows that we must have δλm2 u2 = λn1u1. Thiscompletes the proof of the theorem.

4. Computation for the Fixed-Point Index

We illustrate how e-positivity can be used to prove some fixed-point index results which canthen be used to prove existence results for nonlinear equations. When Ω is a bounded openset in a Banach space X, we write ΩP := Ω ∩ P and ∂ΩP for its boundary relative to P .

Theorem 4.1. LetΩ be a bounded open set inX containing θ, let P be a cone inX, and letA : P → Pbe a completely continuous map. Suppose that there is an increasing, positively 1-homogeneous, e-positive mapping T such that Te ≤ e, and that

Au ≤ Tu, Au/=u ∀u ∈ ∂ΩP . (4.1)

Then the fixed-point index i(A,ΩP , P) = 1.

Proof. We show that Au/=μu for all u ∈ ∂ΩP and all μ ≥ 1, from which the result follows bystandard properties of fixed-point index (see, e.g., [12–14]). Suppose that there exist u0 ∈ ∂ΩP

and μ0 ≥ 1 such that Au0 = μ0u0, then μ0 > 1. It follows from the e-positiveness of T that thereexists a natural number n such that

Tnu0 ≤ d(u0)e, d(u0) > 0. (4.2)

So, by induction, for all m ∈N, we have

μmn0 u0 ≤ Tmnu0 ≤ d(u0)e (4.3)

which implies u0 = θ. This contradicts u0 ∈ ∂ΩP .


Theorem 4.2. Let P be a normal cone in a real Banach space X and let A : P → P be a completelycontinuous map. Suppose that there is an increasing, positively 1-homogeneous, e-positive mapping T(with n = 1 in Definition 2.1) which satisfies the following conditions:

(1) there exists k ∈ [0, 1) such that

Te ≤ ke, (4.4)

(2) there existsM > 0 such that

Au ≤ Tu +Me, u ∈ P. (4.5)

Then there exists R1 > 0 such that for any R > R1, the fixed-point index i(A,BR ∩ P, P) = 1, whereBR = {x ∈ X | ‖x‖ ≤ R}.

Proof. Let

W ={u ∈ P | u = μAu, 0 ≤ μ ≤ 1

}. (4.6)

In the following, we prove that W is bounded.For any u ∈W \ {θ}, using the e-positiveness of T , we have

u = μAu ≤ Tu +Me ≤ (d(u) +M)e. (4.7)

Let

μ0 = inf{μ | u ≤ μe}. (4.8)

It is easy to see that 0 < μ0 < +∞ and u ≤ μ0e. We now have

u ≤ Tu +Me ≤ (kμ0 +M

)e, (4.9)

which, by the definition of μ0, implies that μ0 ≤M/(1−k). So we know that u ≤ (M/(1−k))eand W is bounded by the normality of the cone P .

Select R1 > sup{‖x‖ | x ∈ W}. Then from the homotopy invariance property of fixed-point index we have

i(A,BR ∩ P, P) = i(θ, BR ∩ P, P) = 1, ∀R > R1. (4.10)

This completes the proof of the theorem.


5. Applications

In the following, we will apply the results in this paper to the existence of positive solutionfor two-point boundary-value problems for one-dimensional p-Laplacian:

(φp

(v′))′ + a(t)f(v(t)) = 0, t ∈ (0, 1),

v′(0) = 0, v(1) = 0,(5.1)

where φp(s) = |s|p−2s, p ≥ 2, and (φp)−1 = φq = |s|q−2s, (1/p) + (1/q) = 1.

We make the following assumptions:

(H1) f : [0,+∞) → [0,+∞) is continuous;

(H2) a : [0, 1] → (0,+∞) is continuous.

For each v ∈ E := C[0, 1], we write ‖v‖ = max{|v(t)| : t ∈ [0, 1]}. Define

K = {v ∈ E : v(t) ≥ (1 − t)‖v‖}. (5.2)

Clearly, (E, ‖ · ‖) is a Banach space and K is a cone of E. For any real constant r > 0, defineBr = {v ∈ E : ‖x‖ < r}.

Let the operators T and A be defined by:

(Tv)(t) =∫1

t

φq

(∫s

0a(τ)vp−1(τ)dτ

)ds,

(Av)(t) =∫1

t

φq

(∫s

0a(τ)f(v(τ))dτ

)ds,

(5.3)

respectively.Under (H1) and (H2), it is not difficult to verify that the non-zero fixed points of the

operator A are the positive solutions of boundary-value problem (5.1). In addition, we havefrom (H2) that T : K → E is a completely continuous, positively 1-homogeneous operatorand T(K) ⊂ K.

Lemma 5.1. Suppose that (H2) holds. Then for the operator T defined by (5.3), there is a uniquepositive eigenvalue λ1 of T with its eigenfunction in K.

Proof. First, we show that T is e-positive with e = 1 − t, that is, for any v > θ from K, thereexist α, β > 0 such that

αe ≤ Tv ≤ βe. (5.4)

Let M1 = maxt∈[0,1]a(t). Then

(Tv)(t) ≤∫1

t

φq(M1‖v‖p−1

)ds = φq

(M1‖v‖p−1

)(1 − t). (5.5)

So, we may take β = φq(M1‖v‖p−1).


Clearly, we may take α = ‖Tv‖ = (Tv)(0) since T(K) ⊂ K. So (5.4) is proved.Now we need to show that for any u > v > θ, there always exists some c > 0 such that

Tu − Tv ≥ ce. (5.6)

In fact, we note that φq is increasing, there exists an η ∈ (0, 1) such that

m = mins∈[η,1]

φq

(∫s

0a(τ)up−1(τ)dτ

)− φq

(∫s

0a(τ)vp−1(τ)dτ

)> 0. (5.7)

Then for all t ∈ [η, 1], we have

(Tu)(t) − (Tv)(t) ≥∫1

η

φq

(∫s

0a(τ)up−1(τ)dτ

)ds

−∫1

η

φq

(∫ s

0a(τ)vp−1(τ)dτ

)ds ≥ m(1 − t).

(5.8)

Since (Tu)(t) − (Tv)(t) ≥ (Tu)(η) − (Tv)(η) ≥ m(1 − η) for all t ∈ [0, η], we have

(Tu)(t) − (Tv)(t) ≥ m(1 − η)(1 − t), (5.9)

for all t ∈ [0, 1]. Therefore, the proof is complete and follows from Theorems 3.1 and 3.4.

Remark 5.2. Let ϕ∗ be the positive eigenfunction of T corresponding to λ1, thus λ1Tϕ∗ = ϕ∗.

Then by Lemma 5.1, there exist α, β > 0 such that

αe ≤ Tϕ∗ = λ1ϕ∗ ≤ βe. (5.10)

Hence we obtained that T is ϕ∗-positive operator.

Theorem 5.3. Suppose that the conditions (H1) and (H2) are satisfied, and

lim infu→ 0+

f(u)up−1

> λ1−p1 , (5.11)

lim supu→+∞

f(u)up−1

< λ1−p1 , (5.12)

where λ1 is given in Lemma 5.1, then the boundary-value problem (5.1) has at least one positivesolution.


Proof. It follows from (5.11) that there exists r > 0 such that

f(u) ≥ λ1−p1 up−1, ∀0 ≤ u ≤ r. (5.13)

We may suppose that A has no fixed point on ∂Br ∩ K (otherwise, the proof is finished).Therefore by (5.13),

(Av)(t) ≥∫1

t

φq

(∫s

0a(τ)λ1−p

1 up−1(τ)dτ)ds =

1λ1

(Tv)(t), v ∈ ∂Br ∩K. (5.14)

Hence we have from Lemma 2.4 and Remark 5.2 that

i(A,Br ∩K,K) = 0. (5.15)

It follows from (5.12) that there exist 0 < σ < 1 and M1 > 0 such that

f(u) ≤ σp−1λ1−p1 up−1 +M2, ∀u ≥ 0. (5.16)

Thus, we have

(Av)(t) ≤∫1

t

φq

(∫s

0a(τ)

[σp−1λ

1−p1 up−1(τ) +M2

]dτ

)ds

≤ σ

λ1

∫1

t

φq

(∫s

0a(τ)up−1(τ)dτ

)ds +

∫1

t

φq(M1M2)ds

=σ

λ1(Tv)(t) + φq(M1M2)(1 − t), v ∈ K.

(5.17)

Here we have used the following inequality:

φq(a + b) ≤ φq(a) + φq(b), a, b > 0, 1 < q ≤ 2. (5.18)

Thus by Theorem 4.2 and Remark 5.2, there exists R1 > r such that

i(A,BR ∩K,K) = 0, R > R1, (5.19)


and hence we obtained

i(A,

(BR \ Br

)∩K,K

)= 1. (5.20)

Thus, A has a fixed point in (BR \ Br) ∩K. Consequently, (5.1) has a positive solution.

Theorem 5.4. Suppose that the conditions (H1) and (H2) are satisfied, and

lim supu→ 0+

f(u)up−1

< λ1−p1 , (5.21)

lim infu→+∞

f(u)up−1

> λ1−p1 , (5.22)

where λ1 is given in Lemma 5.1, then the boundary-value problem (5.1) has at least one positivesolution.

Proof. It follows from (5.21) that there exists r2 > 0 such that

f(u) ≤ λ1−p1 up−1, ∀0 ≤ u ≤ r2. (5.23)

We may suppose that A has no fixed point on ∂Br2 ∩ K (otherwise, the proof is finished).Therefore by (5.23),

(Av)(t) ≤∫1

t

φq

(∫ s

0a(τ)λ1−p


1λ1

(Tv)(t), v ∈ ∂Br2 ∩K. (5.24)

Hence we have from Theorem 4.1 and Remark 5.2 that

i(A,Br2 ∩K,K) = 1. (5.25)

It follows from (5.22) that there exists ε > 0 such that f(u) ≥ (λ1−p1 + ε)up−1 when u is

sufficiently large. We know from the continuity of f that there exists b ≥ 0 such that

f(u) ≥(λ

1−p1 + ε

)up−1 − b, ∀u ≥ 0. (5.26)

Take

R2 > max

{

r2,

(M1b2p

m0ε

)1/(p−1)}

, (5.27)

where m0 = mint∈[0,1]a(t), M1 = maxt∈[0,1]a(t).


For v ∈ ∂BR2 ∩K, we have

∫s

0a(τ)f(v(τ))dτ ≥

∫s

0a(τ)

[(λ

1−p1 + ε

)vp−1(τ) − b

]dτ

≥ λ1−p1

∫s

0a(τ)vp−1(τ)dτ +m0ε

∫s

0(1 − τ)p−1‖v‖p−1dτ −M1bs

= λ1−p1

∫s

0a(τ)vp−1(τ)dτ +m0εR

p−12

∫s

0(1 − τ)p−1dτ −M1bs

≥ λ1−p1

∫s


p−12

∫min{s,1/2}

0(1 − τ)p−1dτ −M1bs

≥ λ1−p1

∫s


p−12

∫min{s,1/2}

0

(1 − 1

2

)p−1

dτ −M1bs

≥ λ1−p1

∫s

0a(τ)vp−1(τ)dτ +

(m0εR

p−12

12p

−M1b

)s

≥ λ1−p1

∫s

0a(τ)vp−1(τ)dτ.

(5.28)

Thus, we have

(Av)(t) ≥∫1

t

φq

(∫s

0a(τ)λ1−p


1λ1

(Tv)(t), v ∈ ∂BR2 ∩K. (5.29)

It follows from Lemma 2.4 that

i(A,BR2 ∩K,K) = 0, (5.30)

and hence we obtained

i(A,

(BR2 \ Br2

)∩K,K

)= −1. (5.31)

Thus, A has a fixed point in (BR2 \ Br2) ∩K. Consequently, (5.1) has a positive solution.

Remark 5.5. p-Laplacian boundary-value problems have been studied by some authors([16, 17] and references therein). In preceding works mentioned, they study the existenceof positive solutions by the shooting method, fixed-point theorem, or the fixed-point indexunder some different conditions. It is known that, when p = 2, there are very good conditionsimposed on f that ensure the existence of positive solution for two-point boundary-valueproblems (5.1). In particular, some of those involving the first eigenvalues correspondingto the relevant linear operator are sharp conditions. So, Theorems 5.3 and 5.4 generalizea number of recent works about the existence of solutions for p-Laplacian boundary-valueproblems.


Acknowledgments

The authors sincerely thank the reviewers for their valuable suggestions and usefulcomments that have led to the present improved version of the original paper. The paperis supported by the National Science Foundation of China (10971179) and Research AwardFund for Outstanding Young Scientists of Shandong Province (BS2010SF023).

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Research ArticleOn the Study of Local Solutions fora Generalized Camassa-Holm Equation

Meng Wu

School of Mathematics, Southwestern University of Finance and Economics, Chengdu 610074, China

Correspondence should be addressed to Meng Wu, [email protected]

Received 23 May 2012; Revised 30 June 2012; Accepted 18 July 2012


Copyright q 2012 Meng Wu. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

The pseudoparabolic regularization technique is employed to study the local well-posedness ofstrong solutions for a nonlinear dispersive model, which includes the famous Camassa-Holmequation. The local well-posedness is established in the Sobolev space Hs(R) with s > 3/2 viaa limiting procedure.

1. Introduction

In recent years, extensive research has been carried out worldwide to study highly nonlinearequations including the Camassa-Holm (CH) equation and its various generalizations [1–6].It is shown in [7–9] that the inverse spectral or scattering approach is a powerful techniqueto handle the Camassa-Holm equation and analyze its dynamics. It is pointed out in [10–12]that the CH equation gives rise to geodesic flow of a certain invariant metric on the Bott-Virasoro group, and this geometric illustration leads to a proof that the Least Action Principleholds. Li and Olver [13] established the local well-posedness to the CH model in the Sobolevspace Hs(R) with s > 3/2 and gave conditions on the initial data that lead to finite timeblow-up of certain solutions. Constantin and Escher [14] proved that the blow-up occurs inthe form of breaking waves, namely, the solution remains bounded but its slope becomesunbounded in finite time. Hakkaev and Kirchev [15] investigated a generalized form of theCamassa-Holm equation with high order nonlinear terms and obtained the orbit stabilityof the traveling wave solutions under certain assumptions. Lai and Wu [16] discussed ageneralized Camassa-Holm model and acquired its local existence and uniqueness. Recently,Li et al. [17] investigated the generalized Camassa-Holm equation

ut − utxx + kumux + (m + 3)um+1ux = (m + 2)umuxuxx + um+1uxxx, (1.1)


where m ≥ 0 is a natural number and k ≥ 0. The authors in [17] assume that the initial valuesatisfies the sign condition and establish the global existence of solutions for (1.1).

In this paper, we will study the following generalization of (1.1):

ut − utxx + kumux + (m + 3)um+1ux = (m + 2)umuxuxx + um+1uxxx + λ(u − uxx), (1.2)

where m ≥ 0 is a natural number, k ≥ 0, and λ is a constant.The objective of this paper is to study the local well-posedness of (1.2). Its local well-

posedness of strong solutions in the Sobolev space Hs(R) with s > 3/2 is investigated byusing the pseudoparabolic regularization method. Comparing with the work by Li et al.[17], (1.2) considered in this paper possesses a conservation law different to that in [17](see Lemma 3.2 in Section 3). Also (1.2) contains a dissipative term λ(u − uxx), whichcauses difficulty to establish its local and global existence in the Sobolev space. It shouldbe mentioned that the existence and uniqueness of local strong solutions for the generalizednonlinear Camassa-Holm models like (1.2) have never been investigated in the literatures.

The organization of this work is as follows. The main result is given in Section 2.Section 3 establishes several lemmas, and the last section gives the proof of the main result.

2. Main Result

Firstly, we introduce several notations.Lp = Lp(R) (1 ≤ p < +∞) is the space of all measurable functions h such that

‖h‖pLp =∫R |h(t, x)|pdx < ∞. We define L∞ = L∞(R) with the standard norm ‖h‖L∞ =

infm(e)=0 supx∈R\e|h(t, x)|. For any real number s, Hs = Hs(R) denotes the Sobolev space withthe norm defined by

‖h‖Hs =(∫

R

(1 + |ξ|2

)s∣∣∣h(t, ξ)∣∣∣

2dξ

)1/2

<∞, (2.1)

where h(t, ξ) =∫R e

−ixξh(t, x)dx.For T > 0 and nonnegative number s, C([0, T);Hs(R)) denotes the Frechet space of all

continuous Hs-valued functions on [0, T). We set Λ = (1 − ∂2x)

1/2. For simplicity, throughoutthis paper, we let c denote any positive constant that is independent of parameter ε.

We consider the Cauchy problem of (1.2)

ut − utxx = − k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

− (m + 1)∂x(umu2

x

)+ umuxuxx + λ(u − uxx), k ≥ 0, m ≥ 0,

u(0, x) = u0(x).

(2.2)

Now, we give our main results for problem of (2.2).


Theorem 2.1. Suppose that the initial function u0(x) belongs to the Sobolev space Hs(R) with s >3/2 and λ is a constant. Then, there is a T > 0, which depends on ‖u0‖Hs , such that problem (2.2) hasa unique solution u(t, x) satisfying

u(t, x) ∈ C([0, T);Hs(R))⋂C1([0, T);Hs−1(R)

). (2.3)

3. Local Well-Posedness

In order to prove Theorem 2.1, we consider the associated regularized problem

ut − utxx + εutxxxx = − k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

− (m + 1)∂x(umu2

x

)+ umuxuxx + λ(u − uxx),

u(0, x) = u0(x),

(3.1)

where the parameter ε satisfies 0 < ε < 1/4.

Lemma 3.1. For s ≥ 1 and f(x) ∈ Hs(R) and letting k1 > 0 be an integer such that k1 ≤ s − 1,f, f ′, . . . , fk1 are uniformly continuous bounded functions that converge to 0 at x = ±∞.

The proof of Lemma 3.1 was stated on page 559 by Bona and Smith [18].

Lemma 3.2. If u(t, x) ∈ Hs(s > 7/2) is a solution to problem (3.1), it holds that

∫

R

(u2 + u2

x + εu2xx

)dx =

∫

R

(u2

0 + u20x + εu

20xx

)dx + 2λ

∫ t

0

∫

R

(u2 + u2

x

)dx. (3.2)

Proof. Using Lemma 3.1, we have u(t,±∞) = ux(t,±∞) = uxx(t,±∞) = uxxx(t,±∞) = 0. Theintegration by parts results in

∫

R

um+2uxxxdx =∫

R

um+2duxx = um+2uxx|+∞−∞ − (m + 2)∫

R

um+1uxuxxdx

= −(m + 2)∫

R

um+1uxuxxdx.

(3.3)


Direct calculation and integration by parts give rise to

12d

dt

∫

R

(u2 + u2

x + εu2xx

)dx

=∫

R

(uut + uxutx + εuxxutxx)dx

=∫

R

u(ut − utxx + εutxxxx)dx

=∫

R

u

[− k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

−(m + 1)∂x(umu2

x

)+ umuxuxx + λ(u − uxx)

]dx

=∫

R

u[(m + 2)umuxuxx + um+1uxxx

]dx + λ

∫

R

(u2 − uuxx

)dx

=∫

R

[(m + 2)um+1uxuxx + um+2uxxx

]dx + λ

∫

R

(u2 + u2

x

)dx

= λ∫

R

(u2 + u2

x

)dx,

(3.4)

in which we have used (3.3). From (3.4), we obtain the conservation law (3.2).

Lemma 3.3. Let s ≥ 7/2. The function u(t, x) is a solution of problem (3.1) and the initial valueu0(x) ∈ Hs. Then, the following inequality holds:

‖u‖2H1 ≤

∫

R

(u2

0 + u20x + εu

20xx

)dx, if λ ≤ 0,

‖u‖2H1 ≤ e2λt

∫

R

(u2

0 + u20x + εu

20xx

)dx, if λ > 0.

(3.5)

For q ∈ (0, s − 1], there is a constant c independent of ε such that

∫

R

(Λq+1u

)2dx ≤

∫

R

[(Λq+1u0

)2+ ε(Λqu0xx)

2]dx

+ c∫ t

0‖u‖2

Hq+1

(|λ| +

(‖u‖m−1

L∞ + ‖u‖mL∞

)‖ux‖L∞ + ‖u‖m−1

L∞ ‖ux‖2L∞

)dτ.

(3.6)

For q ∈ [0, s − 1], there is a constant c independent of ε such that

(1 − 2ε)‖ut‖Hq ≤ c‖u‖Hq+1

(|λ| +

(‖u‖m−1

L∞ + ‖u‖mL∞

)‖u‖H1 +‖u‖mL∞‖ux‖L∞ + ‖u‖m−1

L∞ ‖ux‖2L∞

).

(3.7)


The proof of this lemma is similar to that of Lemma 3.5 in [17]. Here we omit it.

Lemma 3.4. Let r and q be real numbers such that −r < q ≤ r. Then,

‖uv‖Hq ≤ c‖u‖Hr‖v‖Hq , if r >12,

‖uv‖Hr+q−1/2 ≤ c‖u‖Hr‖v‖Hq , if r <12.

(3.8)

This lemma can be found in [19] or [20].

Lemma 3.5. Let u0(x) ∈ Hs(R) with s > 3/2. Then, the Cauchy problem (3.1) has a uniquesolution u(t, x) ∈ C([0, T];Hs(R)), where T > 0 depends on ‖u0‖Hs(R). If s ≥ 7/2, the solutionu ∈ C([0,+∞);Hs) exists for all time.

Proof. Letting D = (1 − ∂2x + ε∂

4x)

−1, we know that D : Hs → Hs+4 is a bounded linearoperator. Applying the operator D on both sides of the first equation of system (3.1) andthen integrating the resultant equation with respect to t over the interval (0, t), we get

u(t, x) = u0(x) +∫ t

0D

[− k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

−(m + 1)∂x(umu2

x


]dt.

(3.9)

Suppose that both u and v are in the closed ball BM0(0) of radiusM0 about the zero function inC([0, T];Hs(R)) and A is the operator in the right-hand side of (3.9). For any fixed t ∈ [0, T],we obtain

∥∥∥∥∥

∫ t

0D

[− k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

−(m + 1)∂x(umu2

x


]dt

−∫ t

0D

[− k

m + 1

(vm+1

)

x− m + 3m + 2

(vm+2

)

x+

1m + 2

∂3x

(vm+2

)

−(m + 1)∂x(vmv2

x

)+ vmvxvxx + λ(v − vxx)

]dt

∥∥∥∥∥Hs

≤ TC1

(

sup0≤t≤T

‖u − v‖Hs + sup0≤t≤T

∥∥∥um+1 − vm+1∥∥∥Hs

+ sup0≤t≤T

∥∥∥um+2 − vm+2∥∥∥Hs

+ sup0≤t≤T

∥∥∥D∂x[umu2

x − vmv2x

]∥∥∥Hs

+ sup0≤t≤T

‖D[umuxuxx − vmvxvxx]‖Hs

)

,

(3.10)


where C1 may depend on ε. The algebraic property of Hs0(R) with s0 > 1/2 derives

∥∥∥um+2 − vm+2

∥∥∥Hs

=∥∥∥(u − v)

(um+1 + umv + · · · + uvm + vm+1

)∥∥∥Hs

≤ ‖(u − v)‖Hs

m+1∑

j=0‖u‖m+1−j

Hs ‖v‖jHs ,

≤Mm+10 ‖u − v‖Hs,

∥∥∥um+1 − vm+1

∥∥∥Hs

≤Mm0 ‖u − v‖Hs,

∥∥∥D∂x

(umu2

x − vmv2x

)∥∥∥Hs

≤∥∥∥D∂x

[um(u2x − v2

x

)]∥∥∥Hs

+∥∥∥D∂x

[v2x(u

m − vm)]∥∥∥Hs

≤ C(∥∥∥um

(u2x − v2

x

)∥∥∥Hs−1

+∥∥∥v2

x(um − vm)

∥∥∥Hs−1

)

≤ CMm+10 ‖u − v‖Hs.

(3.11)

Using the first inequality of Lemma 3.4 gives rise to

‖D[umuxuxx − vmvxvxx]‖Hs =∥∥∥∥

12D[um(u2x

)

x− vm

(v2x

)

x

]∥∥∥∥Hs

≤ 12

(∥∥∥D[um(u2x − v2

x

)

x

]∥∥∥Hs

+∥∥∥D[(v2x

)

x(um − vm)

] ∥∥∥Hs

)

≤ C(∥∥∥um

(u2x − v2

x

)

x

∥∥∥Hs−2

+∥∥∥(v2x

)

x(um − vm)

∥∥∥Hs−2

)

≤ C(‖um‖Hs

∥∥∥u2x − v2

x

∥∥∥Hs−1

+∥∥∥v2

x

∥∥∥Hs−1

‖um − vm‖Hs

)

≤ CMm+10 ‖u − v‖Hs,

(3.12)

where C may depend on ε. From (3.11)-(3.12), we obtain

‖Au −Av‖Hs ≤ θ‖u − v‖Hs, (3.13)

where θ = TC2(Mm0 + Mm+1

0 ) and C2 is independent of 0 < t < T . Choosing T sufficientlysmall such that θ < 1, we know that A is a contraction. Similarly, it follows from (3.10) that

‖Au‖Hs ≤ ‖u0‖Hs + θ‖u‖Hs. (3.14)

Choosing T sufficiently small such that θM0 + ‖u0‖Hs < M0, we deduce that A maps BM0(0)to itself. It follows from the contraction-mapping principle that the mapping A has a uniquefixed point u in BM0(0). It completes the proof.


From the above and Lemma 3.2, we have

∫

R

(u2 + u2

x + εu2xx

)dx ≤ e2|λ|t

∫

R

(u2

0 + u20x + εu

20xx

)dx. (3.15)

Therefore,

‖ux‖L∞ ≤ Cεe2|λ|t∫

R

(u2

0 + u20x + εu

20xx

)dx, (3.16)

which together with Lemma 3.3 completes the proof of the global existence.Setting φε(x) = ε−1/4φ(ε−1/4x) with 0 < ε < 1/4 and uε0 = φε u0, we know that

uε0 ∈ C∞ for any u0 ∈ Hs, s > 0. From Lemma 3.5, it derives that the Cauchy problem

ut − utxx + εutxxxx = − k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

− (m + 1)∂x(umu2

x

)+ umuxuxx + λ(u − uxx),

u(0, x) = uε0(x), x ∈ R,

(3.17)

has a unique solution uε(t, x) ∈ C∞([0,∞);H∞) .Furthermore, we have the following.

Lemma 3.6. For s > 0, u0 ∈ Hs, it holds that

‖uε0x‖L∞ ≤ c‖u0x‖L∞ , (3.18)

‖uε0‖Hq ≤ c, if q ≤ s, (3.19)

‖uε0‖Hq ≤ cε(s−q)/4, if q > s, (3.20)

‖uε0 − u0‖Hq ≤ cε(s−q)/4, if q ≤ s, (3.21)

‖uε0 − u0‖Hs = o(1), (3.22)

where c is a constant independent of ε.

The proof of Lemma 3.6 can be found in [16].


Remark 3.7. For s ≥ 1, using ‖uε‖L∞ ≤ c‖uε‖H1/2+ ≤ c‖uε‖H1 , ‖uε‖2H1 ≤ c

∫R(u

2ε + u

2εx)dx, (3.5),

(3.19), and (3.20), we know that,

‖uε‖2L∞ ≤ c‖uε‖2

H1 ≤ ce2|λ|t∫

R

(u2ε0 + u

2ε0x + εu

2ε0xx

)dx

≤ ce2|λ|t(‖uε0‖2

H1 + ε‖uε0‖2H2

)

≤ ce|2λ|t(c + cε × ε(s−2)/2

)

≤ c0e2|λ|t,

(3.23)

where c0 is independent of ε and t.

Lemma 3.8. Suppose u0(x) ∈ Hs(R) with s ≥ 1 such that ‖u0x‖L∞ < ∞. Let uε0 be defined as insystem (3.17). Then, there exist two positive constants T and c, which are independent of ε, such thatthe solution uε of problem (3.17) satisfies ‖uεx‖L∞ ≤ c for any t ∈ [0, T).

Here we omit the proof of Lemma 3.8 since it is similar to Lemma 3.9 presented in[17].

Lemma 3.9 (see Li and Olver [13]). If u and f are functions inHq+1 ∩ {‖ux‖L∞ <∞}, then

∣∣∣∣

∫

R

ΛquΛq(uf)xdx

∣∣∣∣ ≤

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

cq∥∥f∥∥Hq+1‖u‖2

Hq , q ∈(

12, 1],

cq(∥∥f∥∥Hq+1‖u‖Hq‖u‖L∞

+∥∥fx∥∥L∞

‖u‖2Hq +

∥∥f∥∥Hq‖u‖Hq‖ux‖L∞

),

q ∈ (0,∞).

(3.24)

Lemma 3.10 (see Lai and Wu [16]). For u, v ∈ Hs(R) with s > 3/2, w = u − v, q > 1/2, and anatural number n, it holds that

∣∣∣∣

∫

R

ΛswΛs(un+1 − vn+1

)

xdx

∣∣∣∣ ≤ c(‖w‖Hs‖w‖Hq‖v‖Hs+1 + ‖w‖2

Hs

). (3.25)

Lemma 3.11 (see Lai and Wu [16]). If 1/2 < q < min{1, s−1} and s > 3/2, then for any functionsw, f defined on R, it holds that

∣∣∣∣

∫

R

ΛqwΛq−2(wf)xdx

∣∣∣∣ ≤ c‖w‖2Hq

∥∥f∥∥Hq , (3.26)

∣∣∣∣

∫

R

ΛqwΛq−2(wxfx)xdx

∣∣∣∣ ≤ c‖w‖2Hq

∥∥f∥∥Hs. (3.27)


Lemma 3.12. For problem (3.17), s > 3/2, and u0 ∈ Hs(R), there exist two positive constants c andM, which are independent of ε, such that the following inequalities hold for any sufficiently small εand t ∈ [0, T):

‖uε‖Hs ≤ Mect,

‖uε‖Hs+k1 ≤ ε−k1/4Mect, k1 > 0,

‖uεt‖Hs+k1 ≤ ε−(k1+1)/4Mect, k1 > −1.

(3.28)

Slightly modifying the methods presented in [16] can complete the proof ofLemma 3.12.

Our next step is to demonstrate that uε is a Cauchy sequence. Let uε and uδ be solutionsof problem (3.17), corresponding to the parameters ε and δ, respectively, with 0 < ε < δ < 1/4,and let w = uε − uδ. Then, w satisfies the problem

(1 − ε)wt − εwxxt + (δ − ε)(uδt + uδxxt)

=(

1 − ∂2x

)−1[−εwt + (δ − ε)uδt − k

m + 1∂x(um+1ε − um+1

δ

)− ∂x

(um+2ε − um+2

δ

)

− ∂x[∂x(um+1ε

)∂xw + ∂x

(um+1ε − um+1

ε

)∂xuδ

]

+[umε uεxuεxx − umδ uδxuδxx

]]− 1m + 2

∂x(um+2ε − um+2

δ

)+ λw,

(3.29)

w(x, 0) = w0(x) = uε0(x) − uδ0(x). (3.30)

Lemma 3.13. For s > 3/2, u0 ∈ Hs(R), there exists T > 0 such that the solution uε of (3.17) is aCauchy sequence in C([0, T];Hs(R))

⋂C1([0, T];Hs−1(R)).

Proof. For q with 1/2 < q < min{1, s−1}, multiplying both sides of (3.29) by ΛqwΛq and thenintegrating with respect to x give rise to

12d

dt

∫

R

[(1 − ε)(Λqw)2 + ε(Λqwx)

2]dx

= (ε − δ)∫

R

(Λqw)[(Λquδt) + (Λquδxxt)]dx − ε∫

R

ΛqwΛq−2wtdx

+ (δ − ε)∫

R

ΛqwΛq−2uδtdx − 1m + 2

∫

R

(Λqw)Λq(um+2ε − um+2

δ

)

xdx


− k

m + 1

∫

R

ΛqwΛq−2(um+1ε − um+1

δ

)

xdx −

∫

R


δ

)

xdx

−∫

R

ΛqwΛq−2[∂x(um+1ε

)∂xw

]

xdx −

∫

R

ΛqwΛq−2[∂x(um+1ε − um+1

δ

)∂xuδ

]

xdx

+∫

R

ΛqwΛq−2[umε uεxuεxx − umδ uδxuδxx]dx + λ

∫

R

ΛqwΛqwdx.

(3.31)

It follows from the Schwarz inequality that

d

dt

∫[(1 − ε)(Λqw)2 + ε(Λqwx)

2]dx

≤ c{‖Λqw‖L2

[(δ − ε)(‖Λquδt‖L2 + ‖Λquδxxt‖L2) + ε

∥∥∥Λq−2wt

∥∥∥L2

+ (δ − ε)∥∥∥Λq−2uδt

∥∥∥L2

]

+ |λ|∫

R

(Λqw)2 dx +∣∣∣∣

∫

R

ΛqwΛq(um+2ε − um+2

δ

)

xdx

∣∣∣∣

∣∣∣∣

∫ΛqwΛq−2

(um+1ε − um+1

δ

)

xdx

∣∣∣∣ +∣∣∣∣

∫ΛqwΛq−2

(um+2ε − um+2

δ

)

xdx

∣∣∣∣

+∣∣∣∣

∫

R

ΛqwΛq−2[∂x(um+1ε

)∂xw

]

xdx

∣∣∣∣ +∣∣∣∣

∫

R

ΛqwΛq−2[∂x(um+1ε − um+1

δ

)∂xuδ

]

xdx

∣∣∣∣

+∣∣∣∣

∫

R

ΛqwΛq−2[umε uεxuεxx − umδ uδxuδxx]dx

∣∣∣∣

}.

(3.32)

Using the first inequality in Lemma 3.9, we have

∣∣∣∣

∫

R

ΛqwΛq(um+2ε − um+2

δ

)

xdx

∣∣∣∣ =∣∣∣∣

∫

R

ΛqwΛq(wgm+1)xdx

∣∣∣∣

≤ c‖w‖2Hq

∥∥gm+1∥∥Hq+1 ,

(3.33)

where gm+1 =∑m+1

j=0 um+1−jε u

j

δ. For the last three terms in (3.32), using Lemmas 3.4 and 3.12,1/2 < q < min{1, s − 1}, s > 3/2, the algebra property of Hs0 with s0 > 1/2, and (3.23),


we have

∣∣∣∣

∫

R

ΛqwΛq−2(∂x(um+1ε

)∂xw

)

xdx

∣∣∣∣ ≤ c‖w‖2

Hq‖uε‖m+1Hs , (3.34)

∣∣∣∣

∫

R

ΛqwΛq−2(∂x(um+1ε − um+1

δ

)∂xuδ

)

xdx

∣∣∣∣

≤ c‖w‖Hq‖uδ‖Hs

∥∥∥um+1

ε − um+1δ

∥∥∥Hq

≤ c‖w‖2Hq‖uδ‖Hs,

(3.35)

∣∣∣∣

∫

R

ΛqwΛq−2[umε uεxuεxx − umδ uδxuδxx]dx

∣∣∣∣

≤ c‖w‖Hq

∥∥∥(umε − umδ

)(u2εx

)

x+ umδ

[u2εx − u2

δx

]

x

∥∥∥Hq−2

≤ c‖w‖Hq

(∥∥∥(umε − umδ

)(u2εx

)

x

∥∥∥Hq−1

+∥∥∥umδ

[u2εx − u2

δx

]

x

∥∥∥Hq−2

)

≤ c‖w‖Hq

(∥∥umε − umδ∥∥Hq

∥∥∥(u2εx

)

x

∥∥∥Hq−1

+∥∥umδ

∥∥Hs

∥∥∥[u2εx − u2

δx

]

x

∥∥∥Hq−2

)

≤ c‖w‖Hq

(‖w‖Hq

∥∥gm−1∥∥Hq‖u‖2

Hs +∥∥umδ

∥∥Hs‖uεx + uδx‖Hq‖w‖Hq

)

≤ c‖w‖2Hq

(∥∥gm−1∥∥Hq‖u‖2

Hs +∥∥umδ

∥∥Hs‖uεx + uδx‖Hq

).

(3.36)

Using (3.26), we derive that the inequality

∣∣∣∣

∫

R


δ

)

xdx

∣∣∣∣ =∣∣∣∣

∫

R

ΛqwΛq−2(wgm+1)xdx

∣∣∣∣

≤ c∥∥gm+1∥∥Hq‖w‖2

Hq

(3.37)

holds for some constant c, where gm+1 =∑m+1

j=0 um+1−jε u

j

δ. Using the algebra property of Hq

with q > 1/2, q + 1 < s and Lemma 3.11, we have ‖gm‖Hq+1 ≤ c for t ∈ (0, T]. Then, it followsfrom (3.28) and (3.33)–(3.37) that there is a constant c depending on T such that the estimate

d

dt

∫

R

[(1 − ε)(Λqw)2 + ε(Λqwx)

2]dx ≤ c

(δγ‖w‖Hq + ‖w‖2

Hq

)(3.38)


holds for any t ∈ [0, T), where γ = 1 if s ≥ 3 + q and γ = (1 + s − q)/4 if s < 3 + q. Integrating(3.38) with respect to t, one obtains the estimate

12‖w‖2

Hq =12

∫

R

(Λqw)2dx

≤∫

R

[(1 − ε)(Λqw)2 + ε(Λqw)2

]dx

≤∫

R

[(Λqw0)

2 + ε(Λqw0x)2]dx + c

∫ t

0

(δγ‖w‖Hq + ‖w‖2

Hq

)dτ.

(3.39)

Applying the Gronwall inequality and using (3.20) and (3.22) yield

‖u‖Hq ≤ cδ(s−q)/4ect + δγ(ect − 1

)(3.40)

for any t ∈ [0, T).Multiplying both sides of (3.29) by ΛswΛs and integrating the resultant equation with

respect to x, one obtains

12d

dt

∫

R

[(1 − ε)(Λsw)2 + ε(Λswx)

2]dx

= (ε − δ)∫

R

(Λsw)[(Λsuδt) + (Λsuδxxt)]dx − ε∫

R

ΛswΛs−2wtdx

+ (δ − ε)∫

R

ΛswΛs−2uδtdx − k

m + 1

∫

R

(Λsw)Λs(um+1ε − um+1

δ

)

xdx

− 1m + 2

∫

R

(Λsw)Λs(um+2ε − um+2

δ

)

xdx −

∫

R

ΛswΛs−2(um+2ε − um+2

δ

)

xdx

−∫

R

ΛswΛs−2[∂x(um+1ε

)∂xw

]

xdx −

∫

R

ΛswΛs−2[∂x(um+1ε − um+1

δ

)∂xuδ

]

xdx

+∫

R

ΛswΛs−2[umε uεxuεxx − umδ uδxuδxx]dx + λ

∫

R

(Λsw)2dx.

(3.41)

From Lemma 3.12, we have

∣∣∣∣

∫

R

ΛswΛs−2(um+2ε − um+2

δ

)

xdx

∣∣∣∣ ≤ c3∥∥gm+1

∥∥Hs‖w‖2

Hs. (3.42)

From Lemma 3.10, it holds that

∣∣∣∣

∫

R

ΛswΛs(um+2ε − um+2

δ

)

xdx

∣∣∣∣ ≤ c(‖w‖Hs‖w‖Hq‖uδ‖Hs+1 + ‖w‖2

Hs

). (3.43)


Using the Cauchy-Schwartz inequality and the algebra property of Hs0 with s0 > 1/2, fors > 3/2, we have

∣∣∣∣

∫

R

ΛswΛs−2[∂x(um+1ε

)∂xw

]

xdx

∣∣∣∣

=∣∣∣∣

∫

R

ΛqwΛs−2[∂x(um+1ε

)∂xw

]

xdx

∣∣∣∣

≤ c‖Λsw‖L2

∥∥∥Λs−2

[∂x(um+1ε

)∂xw

]

x

∥∥∥L2

≤ c‖w‖Hq

∥∥∥∂x(um+1ε

)∂xw

∥∥∥Hs−1

≤ c∥∥∥uεm+1

Hs

∥∥∥‖w‖2

Hs,

(3.44)

∣∣∣∣

∫

R

ΛswΛs−2[∂x(um+1ε − um+1

δ

)∂xuδ

]

xdx

∣∣∣∣

≤ c‖w‖Hs

∥∥∥Λs−2[∂x(um+1ε − um+1

δ

)∂xuδ

]

x

∥∥∥L2

≤ c‖uδ‖Hs

∥∥gm∥∥Hs‖w‖2

Hs,

(3.45)

∣∣∣∣

∫

R

ΛswΛs−2[umε uεxuεxx − umδ uδxuδxx]dx

∣∣∣∣

≤ c‖w‖Hs

(∥∥∥(umε − umδ

)(u2εx

)

x

∥∥∥Hs−2

+∥∥∥umδ

[u2εx − u2

δx

]

x

∥∥∥Hs−2

)

≤ c‖w‖Hs

(∥∥umε − umδ∥∥Hs

∥∥∥(u2εx

)

x

∥∥∥Hs−2

+∥∥umδ

∥∥Hs

∥∥∥[u2εx − u2

δx

]

x

∥∥∥Hs−2

)

≤ c‖w‖2Hs,

(3.46)

in which we have used Lemma 3.4 and the bounded property of ‖uε‖Hs and ‖uδ‖Hs (seeRemark 3.7). It follows from (3.41)–(3.46) and the inequalities (3.28) and (3.40) that thereexists a constant c depending on m such that

d

dt

∫

R

[(1 − ε)(Λsw)2 + ε(Λswx)

2]dx

≤ 2δ(‖uδt‖Hs + ‖uδxxt‖Hs +

∥∥∥Λs−2wt

∥∥∥L2

+∥∥∥Λs−2uδt

∥∥∥)‖w‖Hs

+ c(‖w‖2

Hs + ‖w‖Hq‖w‖Hs‖uδ‖Hs+1

)

≤ c(δγ1‖w‖Hs + ‖w‖2

Hs

),

(3.47)


where γ1 = min(1/4, (s−q−1)/4) > 0. Integrating (3.47) with respect to t leads to the estimate

12‖w‖2

Hs ≤∫

R

[(1 − ε)(Λsw)2 + ε(Λswx)

2]dx

≤∫

R

[(Λsw0)

2 + ε(Λsw0x)2]dx + c

∫ t

0

(δγ1‖w‖Hs + ‖w‖2

Hs

)dτ.

(3.48)

It follows from the Gronwall inequality and (3.48) that

‖w‖Hs ≤(

2∫

R

[(Λsw0)

2 + ε(Λsw0x)2]dx

)1/2

ect + δγ1(ect − 1

)

≤ c1

(‖w0‖Hs + δ3/4

)ect + δγ1

(ect − 1

),

(3.49)

where c1 is independent of ε and δ.Then, (3.22) and the above inequality show that

‖w‖Hs −→ 0 as ε −→ 0, δ −→ 0. (3.50)

Next, we consider the convergence of the sequence {uεt}. Multiplying both sides of (3.29) byΛs−1wtΛs−1 and integrating the resultant equation with respect to x, we obtain

(1 − ε)‖wt‖2Hs−1 +

1m + 2

∫

R

(Λs−1wt

)Λs−1

(um+2ε − um+2

δ

)

xdx

+∫

R

[−ε(Λs−1wt

)(Λs−1wxxt

)+ (δ − ε)

(Λs−1wt

)Λs−1(uδt + uδxxt)

]dx

=∫

R

(Λs−1wt

)Λs−3

[−εwt + (δ − ε)uδt − k

m + 1∂x(um+1ε − um+1

δ

)− ∂x

(um+2ε − um+2

δ

)

− ∂x[∂x(um+1ε

)∂xw + ∂x

(um+1ε − um+1

ε

)∂xuδ

]

+[umε uεxuεxx − umδ uδxuδxx

]]dx + λ

∫

R

Λs−1wtΛs−1wdx.

(3.51)

It follows from inequalities (3.28) and the Schwartz inequality that there is a constantc depending on T and m such that

(1 − ε)‖wt‖2Hs−1 ≤ c

(δ1/2 + ‖w‖Hs + ‖w‖s−1

)‖wt‖Hs−1 + ε‖wt‖2

Hs−1 (3.52)


Hence

12‖wt‖2

Hs−1 ≤ (1 − 2ε)‖wt‖2Hs−1

≤ c(δ1/2 + ‖w‖Hs + ‖w‖Hs−1

)‖wt‖Hs−1 ,

(3.53)

which results in

12‖wt‖Hs−1 ≤ c

(δ1/2 + ‖w‖Hs + ‖w‖Hs−1

). (3.54)

It follows from (3.40) and (3.50) that wt → 0 as ε, δ → 0 in the Hs−1 norm. Thisimplies that uε is a Cauchy sequence in the spaces C([0, T);Hs(R)) and C([0, T);Hs−1(R)),respectively. The proof is completed.

4. Proof of the Main Result

We consider the problem

(1 − ε)ut − εutxx =(

1 − ∂2x

)−1[− k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

−(m + 1)∂x(umu2

x

)+ umuxuxx

]+ λu,

u(0, x) = uε0(x).

(4.1)

Letting u(t, x) be the limit of the sequence uε and taking the limit in problem (4.1) as ε → 0,from Lemma 3.13, we know that u is a solution of the problem

ut =(

1 − ∂2x

)−1[− k

m + 1

(um+1

)

x− m + 3m + 2

(um+2

)

x+

1m + 2

∂3x

(um+2

)

−(m + 1)∂x(umu2

x

)+ umuxuxx

]+ λu,

u(0, x) = u0(x),

(4.2)

and hence u is a solution of problem (4.2) in the sense of distribution. In particular, if s ≥ 4,u is also a classical solution. Let u and v be two solutions of (4.2) corresponding to the same


initial value u0 such that u, v ∈ C([0, T);Hs(R)). Then,w = u−v satisfies the Cauchy problem

wt =(

1 − ∂2x

)−1{∂x

[− k

m + 1wgm − m + 3

m + 2wgm+1 +

1m + 2

∂2x

(wgm+1

)

−∂x(um+1

)∂xw − ∂x

(um+1 − vm+1

)∂xv

]+ umuxuxx − vmvxvxx

}+ λw,

w(0, x) = 0.(4.3)

For any 1/2 < q < min{1, s − 1}, applying the operator ΛqwΛq to both sides of (4.3) andintegrating the resultant equation with respect to x, we obtain the equality

12d

dt‖w‖2

Hq =∫

R

(Λqw)Λq−2{∂x

[− k

m + 1wgm − m + 3

m + 2wgm+1 +

1m + 2

∂2x

(wgm+1

)

−∂x(um+1

)∂xw − ∂x

(um+1 − vm+1

)∂xv

]+ umuxuxx

−vmvxvxx}dx + |λ|‖w‖2

Hq .

(4.4)

By the similar estimates presented in Lemma 3.13, we have

d

dt‖w‖2

Hq ≤ c‖w‖2Hq . (4.5)

Using the Gronwall inequality leads to the conclusion that

‖w‖Hq ≤ 0 × ect = 0 (4.6)

for t ∈ [0, T). This completes the proof.

Acknowledgments

Thanks are given to the referees whose suggestions were very helpful in improving the paper.This work is supported by the Fundamental Research Funds for the Central Universities(JBK120504).

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Research ArticleExistence of Solutions for Nonlinear ImpulsiveFractional Differential Equations of Order α ∈ (2,3]with Nonlocal Boundary Conditions

Lihong Zhang,1 Guotao Wang,1 and Guangxing Song2

1 School of Mathematics and Computer Science, Shanxi Normal University, Shanxi, Linfen 041004, China2 Department of Mathematics, China University of Petroleum, Shandong, Qingdao 266555, China

Correspondence should be addressed to Guotao Wang, [email protected]

Received 2 February 2012; Revised 13 April 2012; Accepted 28 April 2012


Copyright q 2012 Lihong Zhang et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We investigate the existence and uniqueness of solutions to the nonlocal boundary value problemfor nonlinear impulsive fractional differential equations of order α ∈ (2, 3]. By using some well-known fixed point theorems, sufficient conditions for the existence of solutions are established.Some examples are presented to illustrate the main results.

1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines suchas the mathematical modeling of systems and processes in the fields of physics, chemistry,aerodynamics, control theory, signal and image processing, biophysics, electrodynamics ofcomplex medium, polymer rheology, and fitting of experimental data [1–6]. For example,one could mention the problem of anomalous diffusion [7–9], the nonlinear oscillation ofearthquake can be modeled with fractional derivative [10], and fluid-dynamic traffic modelwith fractional derivatives [11] can eliminate the deficiency arising from the assumption tocontinuum traffic flow and many other [12, 13] recent developments in the description ofanomalous transport by fractional dynamics. For some recent development on nonlinear frac-tional differential equations, see [14–29] and the references therein.

In this paper, we investigate a three-point boundary value problems for nonlinearimpulsive fractional differential equations of order α ∈ (2, 3]:

CDαu(t) = f(t, u(t)), t ∈ J ′,

Δu′(tk) = Ik(u(tk)), k = 1, 2, . . . , p,


Δu′′(tk) = I∗k(u(tk)), k = 1, 2, . . . , p,

u(0) = u′(0) = 0, βu(η)= u(1),

(1.1)

where CDα is the Caputo fractional derivative, 2 < α ≤ 3, f ∈ C(J × R,R), Ik, I∗k ∈ C(R,R),J = [0, 1], 0 = t0 < t1 < · · · < tk < · · · < tp < tp+1 = 1, J ′ = J \ {t1, t2, . . . , tp}, 0 < η < 1, η /= tk(k =1, 2, . . . , p), 0 < β < 1/η2, Δu′(tk) = u′(t+k)−u′(t−k), where u′(t+

k) and u′(t−

k) denote the right and

the left limits of u′(t) at t = tk (k = 1, 2, . . . , p), respectively. Δu′′(tk) has a similar meaning foru′′(t).

Impulsive differential equations arise in many engineering and scientific disciplinesas the important mathematical modeling of systems and processes in the fields of biology,physics, engineering, and so forth. Due to their significance, it is important to study thesolvability of impulsive differential equations. The theory of impulsive differential equationsof integer order has emerged as an important area of investigation. Recently, the impulsivedifferential equations of fractional order have also attracted a considerable attention, and avariety of results can be found in the papers [30–42].

The study of multipoint boundary-value problems was initiated by Bicadze andSamarskiı in [43]. Many authors since then considered nonlinear multipoint boundary-value problems, see [44–51] and the references therein. The multipoint boundary conditionsare important in various physical problems of applied science when the controllers at theend points of the interval (under consideration) dissipate or add energy according to thesensors located at intermediate points. For example, the vibrations of a guy wire of uniformcross-section and composed of N parts of different densities can be set up as a multipointboundary-value problem.

To our knowledge, no paper has considered nonlinear impulsive fractional differentialequations of order α ∈ (2, 3] with nonlocal boundary conditions, that is, problem (1.1). Thispaper fills this gap in the literature. Our purpose here is to give the existence and uniquenessof solutions for nonlinear impulsive fractional differential equations (1.1). Our results arebased on some well-known fixed point theorems.

2. Preliminaries

Let J0 = [0, t1], J1 = (t1, t2], . . . , Jp−1 = (tp−1, tp], Jp = (tp, 1].We introduce the space:

PC2(J, R) ={u : J −→ R | u ∈ C2(Jk), k = 0, 1, . . . , p, u′

(t+k), u′′(t+k)

exist, k = 1, 2, . . . , p,},

(2.1)

with the norm:

‖u‖ = supt∈J

|u(t)|, ‖u‖PC2 = max{‖u‖,∥∥u′∥∥,∥∥u′′∥∥}. (2.2)

Obviously, PC2(J, R) is a Banach space.


Definition 2.1. A function u ∈ PC2(J, R) with its Caputo derivative of order α existing on J isa solution of (1.1) if it satisfies (1.1).

Theorem 2.2 (see [52]). Let E be a Banach space. Assume that A : E → E is a completely con-tinuous operator, and the set V = {u ∈ E | u = μAu, 0 < μ < 1} is bounded. Then A has a fixedpoint in E.

Theorem 2.3 (see [52]). Let E be a Banach space. Assume that Ω is an open bounded subset of Ewith θ ∈ Ω, and let A : Ω → E be a completely continuous operator such that

‖Au‖ ≤ ‖u‖, ∀u ∈ ∂Ω. (2.3)

Then, A has a fixed point in Ω.

Lemma 2.4. Let 2 < α ≤ 3, 1/= βη2, η ∈ (tm, tm+1), m is a nonnegative integer, 0 ≤ m ≤ p. For agiven y ∈ C[0, 1], a function u is a solution of the following impulsive boundary value problem:

CDαu(t) = y(t), t ∈ J ′,

Δu′(tk) = Ik(u(tk)), k = 1, 2, . . . , p,

Δu′′(tk) = I∗k(u(tk)), k = 1, 2, . . . , p,

u(0) = u′(0) = 0, βu(η)= u(1),

(2.4)

if and only if u is a solution of the impulsive fractional integral equation:

u(t) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

1Γ(α)

∫ t

0(t − s)α−1y(s)ds + Ct2, t ∈ J0;

1Γ(α)

∫ t

tk

(t − s)α−1y(s)ds +1

Γ(α)

k∑

i=1

∫ ti

ti−1

(ti − s)α−1y(s)ds

+k−1∑

i=1

(tk − ti)Γ(α − 1)

∫ ti

ti−1


+k−1∑

i=1

(tk − ti)2

2Γ(α − 2)

∫ ti

ti−1


+k∑

i=1

(t − tk)Γ(α − 1)

∫ ti

ti−1


+k−1∑

i=1

(t − tk)(tk − ti)Γ(α − 2)

∫ ti

ti−1


+k∑

i=1

(t − tk)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds +k−1∑

i=1

(tk − ti)Ii(u(ti))

+k−1∑

i=1

(tk − ti)2

2I∗i (u(ti)) +

k∑

i=1

(t − tk)Ii(u(ti))

+k−1∑

i=1

(t − tk)(tk − ti)I∗i (u(ti))

+k∑

i=1

(t − tk)2

2I∗i (u(ti)) + Ct

2, t ∈ Jk, k = 1, 2, . . . , p,

(2.5)


where

C =1

1 − βη2

×{

β

Γ(α)

∫η

tm

(η − s)α−1

y(s)ds +m∑

i=1

β

Γ(α)

∫ ti

ti−1


+m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds +m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1


+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds +m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1


+m∑

i=1

β(η − tm)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds − 1Γ(α)

p+1∑

i=1

∫ ti

ti−1


−p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds −p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1


−p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds −p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1


−p∑

i=1

(1 − tp)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds +m−1∑

i=1

β(tm − ti)Ii(u(ti)) +m−1∑

i=1

β(tm − ti)2

2I∗i (u(ti))

+m∑

i=1

β(η − tm

)Ii(u(ti)) +

m−1∑

i=1

β(η − tm

)(tm − ti)I∗i (u(ti)) +

m∑

i=1

β(η − tm)2

2I∗i (u(ti))

−p−1∑

i=1

(tp − ti

)Ii(u(ti)) −

p−1∑

i=1

(tp − ti)2

2I∗i (u(ti)) −

p∑

i=1

(1 − tp

)Ii(u(ti))

−p−1∑

i=1

(1 − tp

)(tp − ti

)I∗i (u(ti)) −

p∑

i=1

(1 − tp

)2

2I∗i (u(ti))

}

.

(2.6)

Proof. Let u is a solution of (2.4), it holds

u(t) = Iαy(t) − c1 − c2t − c3t2 =

1Γ(α)

∫ t

0(t − s)α−1y(s)ds − c1 − c2t − c3t

2, t ∈ J0, (2.7)

for some c1, c2, c3 ∈ R. Then, we have

u′(t) =1

Γ(α − 1)

∫ t

0(t − s)α−2y(s)ds − c2 − 2c3t, t ∈ J0,

u′′(t) =1

Γ(α − 2)

∫ t

0(t − s)α−3y(s)ds − 2c3, t ∈ J0.

(2.8)

In view of u(0) = u′(0) = 0, it follows c1 = c2 = 0.


If t ∈ J1, then

u(t) =1

Γ(α)

∫ t

t1

(t − s)α−1y(s)ds − d1 − d2(t − t1) − d3(t − t1)2,

u′(t) =1

Γ(α − 1)

∫ t

t1

(t − s)α−2y(s)ds − d2 − 2d3(t − t1),

u′′(t) =1

Γ(α − 2)

∫ t

t1

(t − s)α−3y(s)ds − 2d3,

(2.9)

for some d1, d2, d3 ∈ R.Thus, we have

u′(t−1)=

1Γ(α − 1)

∫ t1

0(t1 − s)α−2y(s)ds − 2c3t1, u′

(t+1)= −d2,

u′′(t−1)=

1Γ(α − 2)

∫ t1

0(t1 − s)α−3y(s)ds − 2c3, u′′

(t+1)= −2d3.

(2.10)

In view of u(t−1 ) = u(t+1 ), Δu′(t1) = u′(t+1 ) − u′(t−1 ) = I1(u(t1)) and Δu′′(t1) = u′′(t+1 ) −

u′′(t−1 ) = I∗1(u(t1)), we have

−d1 =1

Γ(α)

∫ t1

0(t1 − s)α−1y(s)ds − c3t

21,

−d2 =1

Γ(α − 1)

∫ t1

0(t1 − s)α−2y(s)ds − 2c3t1 + I1(u(t1)),

−2d3 =1

Γ(α − 2)

∫ t1

0(t1 − s)α−3y(s)ds − 2c3 + I∗1(u(t1)).

(2.11)

Consequently,

u(t) =1

Γ(α)

∫ t

t1

(t − s)α−1y(s)ds +1

Γ(α)

∫ t1

0(t1 − s)α−1y(s)ds

+t − t1

Γ(α − 1)

∫ t1

0(t1 − s)α−2y(s)ds +

(t − t1)2

2Γ(α − 2)

∫ t1

0(t1 − s)α−3y(s)ds

+ (t − t1)I1(u(t1)) +12(t − t1)2I∗1(u(t1)) − c3t

2, t ∈ J1.

(2.12)


Similarly, we can get

u(t) =1

Γ(α)

∫ t

tk

(t − s)α−1y(s)ds +1

Γ(α)

k∑

i=1

∫ ti

ti−1


+k−1∑

i=1

(tk − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds +k−1∑

i=1

(tk − ti)2

2Γ(α − 2)

∫ ti

ti−1


+k∑

i=1

(t − tk)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds +k−1∑

i=1

(t − tk)(tk − ti)Γ(α − 2)

∫ ti

ti−1


+k∑

i=1

(t − tk)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds +k−1∑

i=1

(tk − ti)Ii(u(ti))

+k−1∑

i=1

(tk − ti)2

2I∗i (u(ti)) +

k∑

i=1

(t − tk)Ii(u(ti)) +k−1∑

i=1

(t − tk)(tk − ti)I∗i (u(ti))

+k∑

i=1

(t − tk)2

2I∗i (u(ti)) − c3t

2, t ∈ Jk, k = 1, 2, . . . , p.

(2.13)

By (2.13), it follows

u(1) =1

Γ(α)

p+1∑

i=1

∫ ti

ti−1

(ti − s)α−1y(s)ds +p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1


+p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds +p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1


+p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds +p∑

i=1

(1 − tp)2

2Γ(α − 2)

∫ ti

ti−1


+p−1∑

i=1

(tp − ti

)Ii(u(ti)) +

p−1∑

i=1

(tp − ti)2

2I∗i (u(ti)) +

p∑

i=1

(1 − tp

)Ii(u(ti))

+p−1∑

i=1

(1 − tp

)(tp − ti

)I∗i (u(ti)) +

p∑

i=1

(1 − tp

)2

2I∗i (u(ti)) − c3,

u(η)=

1Γ(α)

∫η

tm

(η − s)α−1y(s)ds +1

Γ(α)

m∑

i=1

∫ ti

ti−1


+m−1∑

i=1

(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds +m−1∑

i=1

(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1


+m∑

i=1

(η − tm

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds +m−1∑

i=1

(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1



+m∑

i=1

(η − tm)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds +m−1∑

i=1

(tm − ti)Ii(u(ti)) +m−1∑

i=1

(tm − ti)2

2I∗i (u(ti))

+m∑

i=1

(η − tm

)Ii(u(ti)) +

m−1∑

i=1

(η − tm

)(tm − ti)I∗i (u(ti)) +

m∑

i=1

(η − tm)2

2I∗i (u(ti)) − c3η

2.

(2.14)

In view of the condition βu(η) = u(1), we have

c3 = − 11 − βη2

{β

Γ(α)

∫η

tm

(η − s)α−1y(s)ds +m∑

i=1

β

Γ(α)

∫ ti

ti−1


+m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds +m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1


+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1


+m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1


+m∑

i=1

β(η − tm)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds − 1Γ(α)

p+1∑

i=1

∫ ti

ti−1


−p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds −p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1


−p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2y(s)ds −p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1


−p∑

i=1

(1 − tp

)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3y(s)ds +m−1∑

i=1

β(tm − ti)Ii(u(ti))

+m−1∑

i=1

β(tm − ti)2

2I∗i (u(ti))

+m∑

i=1

β(η − tm

)Ii(u(ti))+

m−1∑

i=1

β(η − tm

)(tm − ti)I∗i (u(ti)) +

m∑

i=1

β(η − tm)2

2I∗i(u(ti))

−p−1∑

i=1

(tp − ti

)Ii(u(ti)) −

p−1∑

i=1

(tp − ti)2

2I∗i (u(ti)) −

p∑

i=1

(1 − tp

)Ii(u(ti))

−p−1∑

i=1

(1 − tp

)(tp − ti

)I∗i (u(ti)) −

p∑

i=1

(1 − tp)2

2I∗i (u(ti))

}

.

(2.15)


Substituting the value of ci (i = 1, 2, 3) in (2.7) and (2.13) and letting C = −c3, we canget (2.5). Conversely, assume that u is a solution of the impulsive fractional integral equation(2.5), then by a direct computation, it follows that the solution given by (2.5) satisfies (2.4).

This completes the proof.

3. Main Results

Let 2 < α ≤ 3, 1/= βη2, η ∈ (tm, tm+1), m is a nonnegative integer, and 0 ≤ m ≤ p. Define theoperator A : C(J) → C(J) as follows:

Au(t) =1

Γ(α)

∫ t

tk

(t − s)α−1f(s, u(s))ds +1

Γ(α)

k∑

i=1

∫ ti

ti−1

(ti − s)α−1f(s, u(s))ds

+k−1∑

i=1

(tk − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2f(s, u(s))ds +k−1∑

i=1

(tk − ti)2

2Γ(α − 2)

∫ ti

ti−1


+k∑

i=1

(t − tk)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2f(s, u(s))ds +k−1∑

i=1

(t − tk)(tk − ti)Γ(α − 2)

∫ ti

ti−1


+k∑

i=1

(t − tk)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3f(s, u(s))ds +k−1∑

i=1

(tk − ti)Ii(u(ti)) +k−1∑

i=1

(tk − ti)2

2I∗i (u(ti))

+k∑

i=1

(t − tk)Ii(u(ti)) +k−1∑

i=1

(t − tk)(tk − ti)I∗i (u(ti)) +k∑

i=1

(t − tk)2

2I∗i (u(ti)) +

t2

1 − βη2

×{

β

Γ(α)

∫η

tm

(η − s)α−1

f(s, u(s))ds +m∑

i=1

β

Γ(α)

∫ ti

ti−1


+m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1


+m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1


+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1


+m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1


+m∑

i=1

β(η − tm

)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3f(s, u(s))ds − 1Γ(α)

p+1∑

i=1

∫ ti

ti−1


−p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti−s)α−2f(s, u(s))ds−p−1∑

i=1

(tp − ti

)2

2Γ(α − 2)

∫ ti

ti−1



−p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1


−p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1


−p∑

i=1

(1 − tp

)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3f(s, u(s))ds +m−1∑

i=1

β(tm − ti)Ii(u(ti))

+m−1∑

i=1

β(tm − ti)2

2I∗i (u(ti)) +

m∑

i=1

β(η − tm

)Ii(u(ti)) +

m−1∑

i=1

β(η − tm

)(tm − ti)I∗i (u(ti))

+m∑

i=1

β(η − tm

)2

2I∗i (u(ti)) −

p−1∑

i=1

(tp − ti

)Ii(u(ti)) −

p−1∑

i=1

(tp − ti

)2

2I∗i (u(ti))

−p∑

i=1

(1 − tp

)Ii(u(ti)) −

p−1∑

i=1

(1 − tp

)(tp − ti

)I∗i (u(ti)) −

p∑

i=1

(1 − tp

)2

2I∗i (u(ti))

}

,

(3.1)

then (1.1) has a solution if and only if the operator A has a fixed point.

Lemma 3.1. The operator A : C(J) → C(J) is completely continuous.

Proof. Obviously, A is continuous in view of continuity of f , Ik, and I∗k.

Let Ω ⊂ C(J) be bounded. Then, there exist positive constants Li > 0 (i = 1, 2, 3) suchthat |f(t, u)| ≤ L1, |Ik(u)| ≤ L2 and |I∗k(u)| ≤ L3, for all u ∈ Ω. Thus, for all u ∈ Ω, we have

|(Au)(t)|

≤ 1Γ(α)

∫ t

tk

(t − s)α−1∣∣f(s, u(s))∣∣ds +

1Γ(α)

k∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

(tk − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds +

k−1∑

i=1

(tk − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k∑

i=1

(t − tk)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds +

k−1∑

i=1

(t − tk)(tk − ti)Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k∑

i=1

(t − tk)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

k−1∑

i=1

(tk − ti)|Ii(u(ti))| +k−1∑

i=1

(tk − ti)2

2∣∣I∗i (u(ti))

∣∣

+k∑

i=1

(t − tk)|Ii(u(ti))| +k−1∑

i=1

(t − tk)(tk − ti)∣∣I∗i (u(ti))

∣∣ +k∑

i=1

(t − tk)2

2∣∣I∗i (u(ti))

∣∣ +t2

1 − βη2

×{

β

Γ(α)

∫η

tm

(η − s)α−1∣∣f(s, u(s))∣∣ds +

m∑

i=1

β

Γ(α)

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds


+m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

1Γ(α)

p+1∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds +

p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+p∑

i=1

(1 − tp)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

m−1∑

i=1

β(tm − ti)|Ii(u(ti))|

+m−1∑

i=1

β(tm − ti)2

2∣∣I∗i (u(ti))

∣∣ +m∑

i=1

β(η − tm

)|Ii(u(ti))| +m−1∑

i=1

β(η − tm

)(tm − ti)

∣∣I∗i (u(ti))∣∣

+m∑

i=1

β(η − tm)2

2∣∣I∗i (u(ti))

∣∣ +p−1∑

i=1

(tp − ti

)|Ii(u(ti))| +p−1∑

i=1

(tp − ti)2

2∣∣I∗i (u(ti))

∣∣

+p∑

i=1

(1 − tp

)|Ii(u(ti))| +p−1∑

i=1

(1 − tp

)(tp − ti

)∣∣I∗i (u(ti))∣∣ +

p∑

i=1

(1 − tp)2

2∣∣I∗i (u(ti))

∣∣}

≤ L1

Γ(α)

∫ t

tk

(t − s)α−1ds +L1

Γ(α)

p∑

i=1

∫ ti

ti−1

(ti − s)α−1ds +p−1∑

i=1

L1

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds

+p−1∑

i=1

L1

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds +p∑

i=1

L1

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds

+p−1∑

i=1

L1

Γ(α − 2)

∫ ti

ti−1


i=1

L1

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds +p−1∑

i=1

L2 +p−1∑

i=1

12L3

+p∑

i=1

L2 +p−1∑

i=1

L3 +p∑

i=1

12L3 +

11 − βη2

×{L1β

Γ(α)

∫η

tm

(η − s)α−1ds +p∑

i=1

L1β

Γ(α)

∫ ti

ti−1

(ti − s)α−1ds


+p−1∑

i=1

L1β

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1β

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1β

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1β

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1β

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds +p+1∑

i=1

L1

Γ(α)

∫ ti

ti−1

(ti − s)α−1ds

+p−1∑

i=1

L1

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds +p−1∑

i=1

βL2 +p−1∑

i=1

β

2L3 +

p∑

i=1

βL2 +p−1∑

i=1

βL3

+p∑

i=1

β

2L3 +

p−1∑

i=1

L2 +p−1∑

i=1

12L3 +

p∑

i=1

L2 +p−1∑

i=1

L3 +p∑

i=1

12L3

}

≤ L1

Γ(α + 1)+

pL1

Γ(α + 1)+

(p − 1

)L1

Γ(α)+

(p − 1

)L1

2Γ(α − 1)+pL1

Γ(α)+

(p − 1

)L1

Γ(α − 1)+

pL1

2Γ(α − 1)

+(p − 1

)L2 +

p − 12

L3 + pL2 +(p − 1

)L3 +

p

2L3 +

11 − βη2

×{

βL1

Γ(α + 1)+

βpL1

Γ(α + 1)+β(p − 1

)L1

Γ(α)+β(p − 1

)L1

2Γ(α − 1)+βpL1

Γ(α)+β(p − 1

)L1

Γ(α − 1)+

βpL1

2Γ(α − 1)

+

(p + 1

)L1

Γ(α + 1)+

(p − 1

)L1

Γ(α)+

(p − 1

)L1

2Γ(α − 1)+pL1

Γ(α)

+

(p − 1

)L1

Γ(α − 1)+

pL1

2Γ(α − 1)+ β(p − 1

)L2 +

β(p − 1

)

2L3

+ βpL2 + β(p − 1

)L3 +

βp

2L3 +

(p − 1

)L2 +

p − 12

L3 + pL2 +(p − 1

)L3 +

p

2L3

}

=2 + β

(1 − η2)

1 − βη2

{(p + 1

)L1

Γ(α + 1)+

(2p − 1

)L1

Γ(α)+

(4p − 3

)L1

2Γ(α − 1)+(2p − 1

)L2 +

(2p − 3

2

)L3

}

,

(3.2)

which implies

‖Au‖ ≤ 2 + β(1 − η2)

1 − βη2

×{(

p + 1)L1

Γ(α + 1)+

(2p − 1

)L1

Γ(α)+

(4p − 3

)L1

2Γ(α − 1)+(2p − 1

)L2 +

(2p − 3

2

)L3

}

:= L.

(3.3)


On the other hand, for any t ∈ Jk, 0 ≤ k ≤ p, we have

∣∣(Au)′(t)

∣∣ ≤ 1

Γ(α − 1)

∫ t

tk

(t − s)α−2∣∣f(s, u(s))∣∣ds +

k∑

i=1

1Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

(tk − ti)Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k∑

i=1

(t − tk)Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k∑

i=1

|Ii(u(ti))| +k−1∑

i=1

(tk − ti)∣∣I∗i (u(ti))

∣∣ +

k∑

i=1

(t − tk)∣∣I∗i (u(ti))

∣∣ +

2t1 − βη2

×{

β

Γ(α)

∫η

tm

(η − s)α−1∣∣f(s, u(s))∣∣ds +

m∑

i=1

β

Γ(α)

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm

)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+1

Γ(α)

p+1∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds


+p∑

i=1

(1 − tp)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

m−1∑

i=1


+m−1∑

i=1

β(tm − ti)2

2∣∣I∗i (u(ti))

∣∣ +

m∑

i=1

β(η − tm

)|Ii(u(ti))|

+m−1∑

i=1

β(η − tm

)(tm − ti)

∣∣I∗i (u(ti))∣∣

+m∑

i=1

β(η − tm

)2

2∣∣I∗i (u(ti))

∣∣ +

p−1∑

i=1

(tp − ti

)|Ii(u(ti))| +p−1∑

i=1

(tp − ti)2

2∣∣I∗i (u(ti))

∣∣

+p∑

i=1

(1 − tp

)|Ii(u(ti))| +p−1∑

i=1

(1 − tp

)(tp − ti

)∣∣I∗i (u(ti))∣∣

+p∑

i=1

(1 − tp)2

2∣∣I∗i (u(ti))

∣∣}

≤ L1

Γ(α − 1)

∫ t

tk

(t − s)α−2ds +p∑

i=1

L1

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds

+p−1∑

i=1

L1

Γ(α − 2)

∫ ti

ti−1


i=1

L1

Γ(α − 2)

∫ ti

ti−1


i=1

L2 +p−1∑

i=1

L3

+p∑

i=1

L3 +2

1 − βη2

×{L1β

Γ(α)

∫η

tm

(η − s)α−1ds +p∑

i=1

L1β

Γ(α)

∫ ti

ti−1

(ti − s)α−1ds

+p−1∑

i=1

L1β

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1β

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1β

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1β

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1β

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds +p+1∑

i=1

L1

Γ(α)

∫ ti

ti−1

(ti − s)α−1ds

+p−1∑

i=1

L1

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2ds +p−1∑

i=1

L1

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds

+p∑

i=1

L1

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3ds +p−1∑

i=1

βL2 +p−1∑

i=1

β

2L3 +

p∑

i=1

βL2 +p−1∑

i=1

βL3

+p∑

i=1

β

2L3 +

p−1∑

i=1

L2 +p−1∑

i=1

12L3 +

p∑

i=1

L2 +p−1∑

i=1

L3 +p∑

i=1

12L3

}


≤ L1

Γ(α)+pL1

Γ(α)+

(p − 1

)L1

Γ(α − 1)+

pL1

Γ(α − 1)pL2 +

(p − 1

)L3 + pL3 +

21 − βη2

×{

βL1

Γ(α + 1)+

βpL1

Γ(α + 1)+β(p − 1

)L1

Γ(α)+β(p − 1

)L1

2Γ(α − 1)+βpL1

Γ(α)+β(p − 1

)L1

Γ(α − 1)

+βpL1

2Γ(α − 1)+

(p + 1

)L1

Γ(α + 1)+

(p − 1

)L1

Γ(α)+

(p − 1

)L1

2Γ(α − 1)+pL1

Γ(α)+

(p − 1

)L1

Γ(α − 1)

+pL1

2Γ(α − 1)+ β(p − 1

)L2 +

β(p − 1

)

2L3 + βpL2 + β

(p − 1

)L3 +

βp

2L3

+(p − 1

)L2 +

p − 12

L3 + pL2 +(p − 1

)L3 +

p

2L3

}

=

(p + 1

)L1

Γ(α)+

(2p − 1

)L1

Γ(α − 1)pL2 +

(2p − 1

)L3 +

2(β + 1

)

1 − βη2

×{(

p + 1)L1

Γ(α + 1)+

(2p − 1

)L1

Γ(α)+

(4p − 3

)L1

2Γ(α − 1)+(2p − 1

)L2 +

(2p − 3

2

)L3

}

:= L.

(3.4)

Hence, let t1, t2 ∈ Jk, t1 < t2, 0 ≤ k ≤ p, we have

|(Au)(t2) − (Au)(t1)| ≤∫ t2

t1

∣∣(Au)′(s)∣∣ds ≤ L(t2 − t1). (3.5)

This implies that A is equicontinuous on all Jk, k = 0, 1, 2, . . . , p. The Arzela-Ascoli Theoremimplies that A : C(J) → C(J) is completely continuous.

Theorem 3.2. Assume that the nonlinearity f is bounded and that the impulse functions Ik, I∗k, k =1, 2, . . . , p are bounded. Then, the nonlinear impulsive fractional three-point boundary value problems(1.1) have at least one solution.

Proof. Firstly, by Lemma 3.1, we know that the operator A : C(J) → C(J) is completelycontinuous.

Let Li (i = 1, 2, 3) be nonnegative constants such that

∣∣f(t, u)∣∣ ≤ L1, |Ik(u)| ≤ L2,

∣∣I∗k(u)∣∣ ≤ L3, t ∈ J, u ∈ R. (3.6)

For 0 < μ < 1, consider the equation:

u = μAu. (3.7)


If u is a solution of (3.7), then for every t ∈ J we have that

|u(t)| = μ|Au(t)|

≤ 1Γ(α)

∫ t

tk

(t − s)α−1∣∣f(s, u(s))∣∣ds +

1Γ(α)

k∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

(tk − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

(tk − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k∑

i=1

(t − tk)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

(t − tk)(tk − ti)Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k∑

i=1

(t − tk)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

(tk − ti)|Ii(u(ti))| +k−1∑

i=1

(tk − ti)2

2∣∣I∗i (u(ti))

∣∣

+k∑

i=1

(t − tk)|Ii(u(ti))| +k−1∑

i=1

(t − tk)(tk − ti)∣∣I∗i (u(ti))

∣∣

+k∑

i=1

(t − tk)2

2∣∣I∗i (u(ti))

∣∣ +t2

1 − βη2

×{

β

Γ(α)

∫η

tm

(η − s)α−1∣∣f(s, u(s))∣∣ds +

m∑

i=1

β

Γ(α)

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm

)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds


+1

Γ(α)

p+1∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+p∑

i=1

(1 − tp)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

m−1∑

i=1


+m−1∑

i=1

β(tm − ti)2

2∣∣I∗i (u(ti))

∣∣ +m∑

i=1

β(η − tm

)|Ii(u(ti))|

+m−1∑

i=1

β(η − tm

)(tm − ti)

∣∣I∗i (u(ti))∣∣

+m∑

i=1

β(η − tm

)2

2∣∣I∗i (u(ti))

∣∣ +p−1∑

i=1

(tp − ti

)|Ii(u(ti))|

+p−1∑

i=1

(tp − ti)2

2∣∣I∗i (u(ti))

∣∣

+p∑

i=1

(1 − tp

)|Ii(u(ti))| +p−1∑

i=1

(1 − tp

)(tp − ti

)∣∣I∗i (u(ti))∣∣

+p∑

i=1

(1 − tp)2

2∣∣I∗i (u(ti))

∣∣}

≤ 2 + β(1 − η2)

1 − βη2

{(p + 1

)L1

Γ(α + 1)+

(2p − 1

)L1

Γ(α)+

(4p − 3

)L1

2Γ(α − 1)+(2p − 1

)L2 +

(2p − 3

2

)L3

}

,

(3.8)

which implies for any t ∈ J , it holds that

‖u‖ ≤ 2 + β(1 − η2)

1 − βη2

{(p + 1

)L1

Γ(α + 1)+

(2p − 1

)L1

Γ(α)+

(4p − 3

)L1

2Γ(α − 1)+(2p − 1

)L2 +

(2p − 3

2

)L3

}

.

(3.9)


This shows that all the solutions of (3.7) are bounded independently of 0 < μ < 1.Using Schaeffer’s theorem (see Theorem 2.2), we get thatA has at least one fixed point, whichimplies that (1.1) has at least one solution.

Now, we present some existence results when the nonlinearity and the impulse func-tions have sublinear growth.


(H1) there exist nonnegative constants a and b such that

∣∣f(t, u)

∣∣ ≤ a + b|u|ρ, 0 ≤ ρ < 1, t ∈ J, u ∈ R; (3.10)

(H2) there exist nonnegative constants ak, bk, a∗k, b∗ksuch that

|Ik(u)| ≤ ak + bk|u|ρk ,∣∣I∗k(u)

∣∣ ≤ a∗k + b∗k|u|ρ∗k , 0 ≤ ρk, ρ∗k < 1, u ∈ R, k = 1, 2, . . . , p.

(3.11)

Then, problem (1.1) has at least one solution.

Proof. If u ∈ C(J), then we can write that

∣∣f(s, u(s))∣∣ ≤ a + b|u(s)|ρ, s ∈ J,

|Ik(u(tk))| ≤ ak + bk|u(tk)|ρk ,∣∣I∗k(u(tk))

∣∣ ≤ a∗k + b∗k|u(tk)|ρ∗k , k = 1, 2, . . . , p.

(3.12)

So, if u is a solution of (3.7), then, by the similar process used to obtain (3.2), for every t ∈ Jwe have that

|u(t)| ≤ μ

Γ(α)

∫ t

tk

(t − s)α−1∣∣f(s, u(s))∣∣ds +

μ

Γ(α)

k∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

μ(tk − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

μ(tk − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

k∑

i=1

μ(t − tk)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+k−1∑

i=1

μ(t − tk)(tk − ti)Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+k∑

i=1

μ(t − tk)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

k−1∑

i=1

μ(tk − ti)|Ii(u(ti))|

+k−1∑

i=1

μ(tk − ti)2

2∣∣I∗i (u(ti))

∣∣ +k∑

i=1

μ(t − tk)|Ii(u(ti))| +k−1∑

i=1

μ(t − tk)(tk − ti)∣∣I∗i (u(ti))

∣∣


+k∑

i=1

μ(t − tk)2

2∣∣I∗i (u(ti))

∣∣ +

μt2

1 − βη2

×{

β

Γ(α)

∫η

tm

(η − s)α−1∣∣f(s, u(s))∣∣ds

+m∑

i=1

β

Γ(α)

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds +

m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+m∑

i=1

β(η − tm)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

1Γ(α)

p+1∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s))∣∣ds

+p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds

+p∑

i=1

(1 − tp)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s))∣∣ds +

m−1∑

i=1


+m−1∑

i=1

β(tm − ti)2

2∣∣I∗i (u(ti))

∣∣ +m∑

i=1

β(η − tm

)|Ii(u(ti))|

+m−1∑

i=1

β(η − tm

)(tm − ti)

∣∣I∗i (u(ti))∣∣

+m∑

i=1

β(η − tm)2

2∣∣I∗i (u(ti))

∣∣ +p−1∑

i=1

(tp − ti

)|Ii(u(ti))| +p−1∑

i=1

(tp − ti)2

2∣∣I∗i (u(ti))

∣∣

+p∑

i=1

(1 − tp

)|Ii(u(ti))| +p−1∑

i=1

(1 − tp

)(tp − ti

)∣∣I∗i (u(ti))∣∣ +

p∑

i=1

(1 − tp)2

2∣∣I∗i (u(ti))

∣∣}


≤ M1a +M2

(p∑

k=1

ak +p−1∑

k=1

ak

)

+M2

2

(p∑

k=1

a∗k + 3p−1∑

k=1

a∗k

)

+M1b‖u‖ρ +M2

p∑

k=1

bk‖u‖ρk

+M2

p−1∑

k=1

bk‖u‖ρk + M2

2

p∑

k=1

b∗k‖u‖ρ∗k +

3M2

2

p−1∑

k=1

b∗k‖u‖ρ∗k ,

(3.13)

where

M1 =2 + β

(1 − η2)

1 − βη2

( (p + 1

)

Γ(α + 1)+

(2p − 1

)

Γ(α)+

(4p − 3

)

2Γ(α − 1)

)

, M2 =2 + β

(1 − η2)

1 − βη2. (3.14)

In consequence,

‖u‖ ≤M +M1b‖u‖ρ +M2

p∑

k=1

bk‖u‖ρk +M2

p−1∑

k=1

bk‖u‖ρk + M2

2

p∑

k=1

b∗k‖u‖ρ∗k +

3M2

2

p−1∑

k=1

b∗k‖u‖ρ∗k ,

(3.15)

where M =M1a +M2(∑p

k=1 ak +∑p−1

k=1 ak) + (M2/2)(∑p

k=1 a∗k + 3

∑p−1k=1 a

∗k).

Now, taking into account that 0 ≤ ρ, ρk, ρ∗k< 1, k = 1, 2, . . . , p, we can conclude that

there exists a constant C > 0 such that ‖u‖ ≤ C for any solution of (3.7), 0 < μ < 1. Therefore,by Schaeffer’s theorem (Theorem 2.2), we obtain the existence of a fixed point for A, whichimplies the problem (1.1) has at least one solution.


(H3) there exist nonnegative constants a and b such that

∣∣f(t, u)∣∣ ≤ a + b|u|, t ∈ J, u ∈ R; (3.16)

(H4) there exist nonnegative constants ak, bk, a∗k, b∗k such that

|Ik(u)| ≤ ak + bk|u|,∣∣I∗k(u)

∣∣ ≤ a∗k + b∗k|u|, u ∈ R, k = 1, 2, . . . , p. (3.17)

If

M1b +M2

(p∑

k=1

(bk +

12b∗k

)+

p−1∑

k=1

(bk +

32b∗k

))

< 1, (3.18)

where

M1 =2 + β

(1 − η2)

1 − βη2

( (p + 1

)

Γ(α + 1)+

(2p − 1

)

Γ(α)+

(4p − 3

)

2Γ(α − 1)

)

, M2 =2 + β

(1 − η2)

1 − βη2. (3.19)



Proof. The proof is similar to that of Theorem 3.3, so we omit it.

Theorem 3.5. Let limu→ 0 f(t, u)/u = 0, limu→ 0 Ik(u)/u = 0, and limu→ 0 I∗k(u)/u = 0, then

problem (1.1) has at least one solution.

Proof. Now, in view of limu→ 0 f(t, u)/u = 0, limu→ 0 Ik(u)/u = 0, and limu→ 0 I∗k(u)/u = 0,

there exists a constant r > 0 such that |f(t, u)| ≤ δ1|u|, |Ik(u)| ≤ δ2|u|, and |I∗k(u)| ≤ δ3|u| for

0 < |u| < r, where δi > 0 (i = 1, 2, 3) satisfy

2 + β(1 − η2)

1 − βη2

{(p + 1

)δ1

Γ(α + 1)+

(2p − 1

)δ1

Γ(α)+

(4p − 3

)δ1

2Γ(α − 1)+(2p − 1

)δ2 +

(2p − 3

2

)δ3

}

≤ 1.

(3.20)

Let Ω = {u ∈ C(J) | ‖u‖ < r}. Take u ∈ C(J) such that ‖u‖ = r, that is, u ∈ ∂Ω. ByLemma 3.1, we know that A is completely continuous, and

|(Au)(t)| ≤ 2 + β(1 − η2)

1 − βη2

×{(

p + 1)δ1

Γ(α + 1)+

(2p − 1

)δ1

Γ(α)+

(4p − 3

)δ1

2Γ(α − 1)+(2p − 1

)δ2 +

(2p − 3

2

)δ3

}

‖u‖.

(3.21)

Thus, in view of (3.21), we obtain ‖Au‖ ≤ ‖u‖, u ∈ ∂Ω. Therefore, by Theorem 2.3,the operator A has at least one fixed point, which implies that (1.1) has at least one solutionu ∈ Ω.

For the forthcoming analysis, we need the following assumption:

(H5) there exist positive constants Ki (i = 1, 2, 3) such that

∣∣f(t, u) − f(t, v)∣∣ ≤ K1|u − v|, |Ik(u) − Ik(v)| ≤ K2|u − v|, ∣∣I∗k(u) − I∗k(v)∣∣ ≤ K3|u − v|,

(3.22)

for t ∈ J, u, v ∈ R and k = 1, 2, . . . , p.Define the constants:

Λ =2 + β

(1 − η2)

1 − βη2

{(p + 1

)K1

Γ(α + 1)+

(2p − 1

)K1

Γ(α)+

(4p − 3

)K1

2Γ(α − 1)+(2p − 1

)K2 +

(2p − 3

2

)K3

}

.

(3.23)

Theorem 3.6. If condition (H5) holds. Then problem (1.1) has a unique solution provided Λ < 1,where Λ is given by (3.23).


Proof. Let u, v ∈ C(J), by a simple computation, we can get

|(Au)(t) − (Av)(t)|

≤ 1Γ(α)

∫ t

tk

(t − s)α−1∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+1

Γ(α)

k∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+k−1∑

i=1

(tk − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+k−1∑

i=1

(tk − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+k∑

i=1

(t − tk)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+k−1∑

i=1

(t − tk)(tk − ti)Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+k∑

i=1

(t − tk)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+k−1∑

i=1

(tk − ti)|Ii(u(ti)) − Ii(v(ti))| +k−1∑

i=1

(tk − ti)2

2∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣

+k∑

i=1

(t − tk)|Ii(u(ti)) − Ii(v(ti))| +k−1∑

i=1

(t − tk)(tk − ti)∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣

+k∑

i=1

(t − tk)2

2∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣ +t2

1 − βη2

×{

β

Γ(α)

∫η

tm

(η − s)α−1∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+m∑

i=1

β

Γ(α)

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+m−1∑

i=1

β(tm − ti)Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+m−1∑

i=1

β(tm − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+m∑

i=1

β(η − tm

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s)) − f(s, v(s))∣∣ds


+m−1∑

i=1

β(η − tm

)(tm − ti)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+m∑

i=1

β(η − tm)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+1

Γ(α)

p+1∑

i=1

∫ ti

ti−1

(ti − s)α−1∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+p−1∑

i=1

(tp − ti

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+p−1∑

i=1

(tp − ti)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+p∑

i=1

(1 − tp

)

Γ(α − 1)

∫ ti

ti−1

(ti − s)α−2∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+p−1∑

i=1

(1 − tp

)(tp − ti

)

Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+p∑

i=1

(1 − tp)2

2Γ(α − 2)

∫ ti

ti−1

(ti − s)α−3∣∣f(s, u(s)) − f(s, v(s))∣∣ds

+m−1∑

i=1

β(tm − ti)|Ii(u(ti)) − Ii(v(ti))| +m−1∑

i=1

β(tm − ti)2

2∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣

+m∑

i=1

β(η − tm

)|Ii(u(ti)) − Ii(v(ti))| +m−1∑

i=1

β(η − tm

)(tm − ti)

∣∣I∗i (u(ti)) − I∗i (v(ti))∣∣

+m∑

i=1

β(η − tm)2

2∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣ +p−1∑

i=1

(tp − ti

)|Ii(u(ti)) − Ii(v(ti))|

+p−1∑

i=1

(tp − ti)2

2∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣ +p∑

i=1

(1 − tp

)|Ii(u(ti)) − Ii(v(ti))|

+p−1∑

i=1

(1 − tp

)(tp − ti

)∣∣I∗i (u(ti)) − I∗i (v(ti))∣∣ +

p∑

i=1

(1 − tp)2

2∣∣I∗i (u(ti)) − I∗i (v(ti))

∣∣}

≤ 2 + β(1 − η2)

1 − βη2

{(p + 1

)K1

Γ(α + 1)+

(2p − 1

)K1

Γ(α)+

(4p − 3

)K1

2Γ(α − 1)+(2p − 1

)K2 +

(2p − 3

2

)K3

}

× ‖u − v‖

≤ Λ‖u − v‖.(3.24)


Thus, ‖Au −Av‖ ≤ Λ‖u − v‖.As Λ < 1, therefore, A is a contraction. Thus, the conclusion of the theorem follows by

the contraction mapping principle.

4. Examples

Example 4.1. Consider the following nonlinear impulsive fractional differential equations:

CDαu(t) =

etcos3u(t)1 + u2(t)

, 0 < t < 1, t /=12,

Δu′(

12

)=

5 + 3u4(1/2)1 + u4(1/2)

, Δu′′(

12

)= 3 sin u

(12

),

u(0) = u′(0) = 0, βu(η)= u(1),

(4.1)

where 2 < α ≤ 3, 0 < η < 1, η /= 1/2, 0 < β < 1/η2, and p = 1.

Obviously, for L1 = e, L2 = 5, and L3 = 3, it is easy to verify condition of Theorem 3.2holds. Hence, by Theorem 3.2, we can get that the above equation (4.1) has at least one solu-tion.

Example 4.2. Consider the following fractional impulsive the three-point boundary value pro-blem:

CDαu(t) =

e5tcos5[3u(t) + e8u(t)]

1 + u4(t)+

2 sin(2t + 5)√

3 + u2(t)|u(t)|ρ, 0 < t < 1, t /=

14,

Δu′(

14

)= 3 + 2cos2

[ln(

1 + 8u2(

14

))]+ 3 cos u

(14

)∣∣∣∣u(

14

)∣∣∣∣

ρ1

,

Δu′′(

14

)=

12+

5 + 2u2(1/4)2 + u2(1/4)

+ 3e−u2(1/4)

∣∣∣∣u(

14

)∣∣∣∣

ρ2

,

u(0) = u′(0) = 0, βu(η)= u(1),

(4.2)

where 2 < α ≤ 3, 0 < η < 1, η /= 1/4, 0 < β < 1/η2 , and p = 1.

Obviously a = e5, b = 2, a1 = 5, b1 = a∗1 = b∗1 = 3. It is easy to verify that for 0 <ρ, ρk, ρ

∗k < 1, the conditions of Theorem 3.3 hold. Therefore, by Theorem 3.3, the impulsive

the three-point fractional boundary value problem (4.2) has at least one solution.


Example 4.3. Consider the following nonlinear impulsive fractional differential equations:

CDαu(t) = t2u2(t) + u(t) arctan u(t), 0 < t < 1, t /=

13,

Δu′(

13

)= eu(t)u3

(13

), Δu′′

(13

)=

3u2(1/3)

1 + 2u4(1/3),

u(0) = u′(0) = 0, βu(η)= u(1),

(4.3)

where 2 < α ≤ 3, 0 < η < 1, η /= 1/3, 0 < β < 1/η2, and p = 1.

Clearly, all the assumptions of Theorem 3.5 hold. Thus, by the conclusion of Theo-rem 3.5 we can get that (4.3) has at least one solution.

5. Conclusion

The existence and uniqueness of solutions to a three-point boundary value problem forfractional nonlinear differential equation with impulses have been discussed. We apply theconcepts of fractional calculus together with fixed point theorems to establish the existenceresults. First of all, we investigate a linear three-point boundary value problem involvingfractional derivatives and impulses, which in fact provides the platform to prove the existenceof solutions for the associated nonlinear fractional equation with impulses. It is worthmentioning that the conditions of our theorems are easily to verify and that our approachis simple, so they are applicable to a variety of real world problems.

Acknowledgments

The authors would like to express their gratitude to the anonymous reviewers and editorsfor their valuable comments and suggestions which improve the quality of present paper.They also would like to thank Professor Bashir Ahmad for his significant suggestions on theimprovement of this paper. This paper was supported by the Natural Science Foundation forYoung Scientists of Shanxi Province, China (Grant no. 2012021002-3) and Science Foundationof Shanxi Normal University, China (Grant no. ZR1113).

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Research ArticleExistence of Solutions forNonhomogeneous A-Harmonic Equationswith Variable Growth

Yongqiang Fu and Lifeng Guo

Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China

Correspondence should be addressed to Lifeng Guo, [email protected]

Received 29 February 2012; Accepted 18 June 2012

Academic Editor: Shaoyong Lai

Copyright q 2012 Y. Fu and L. Guo. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We study the following nonhomogeneous A-harmonic equations: d∗A(x, du(x)) + B(x, u(x)) =0, x ∈ Ω, u(x) = 0, x ∈ ∂Ω, where Ω ⊂ R

n is a bounded and convex Lipschitz domain,A(x, du(x)) and B(x, u(x)) satisfy some p(x)-growth conditions, respectively. We obtain theexistence of weak solutions for the above equations in subspace K

1,p(x)0 (Ω,Λl−1) of W1,p(x)

0 (Ω,Λl−1).

1. Introduction

Spaces of differential forms have been discussed in great details (see [1, 2] and the referencestherein). The theory of differential forms is an approach to multivariable calculus that isindependent of coordinates and provides a better definition for integrals. Differential formshave played an important role in physical laws of thermodynamics, analytical mechanics,and physical theories, in particular Maxwell’s theory, and the Yang-Mills theory, the theoryof relativity, see for example [3–6].

In recent years, the study of A-harmonic equations for differential forms has developedrapidly. Many interesting results concerning A-harmonic equation have been establishedrecently (see [7–11] and the references therein). In [12], spaces Lp(x)(Ω) and Wk,p(x)(Ω)are first introduced, and they used them to study the solutions of nonlinear Dirichletboundary value problems with p(x)-growth conditions. In [13], spaces Lp(x)(Ω,Λl, ω) andW1,p(x)(Ω,Λl, ω) are first introduced and used to study the weak solutions of obstacleproblems of A-harmonic equations with variable growth for differential forms.


Let Ω ⊂ Rn be a bounded and convex Lipschitz domain. It is our purpose to study the

following systems:

d∗A(x, du(x)) + B(x, u(x)) = 0, x ∈ Ω,

u(x) = 0, x ∈ ∂Ω,(1.1)

where u ∈ Λl−1(Ω), l = 1, 2, . . . , n, and A : Ω × Λl(Ω) → Λl(Ω), B : Ω × Λl−1(Ω) → Λl−1(Ω)satisfy the following conditions.

(H1) A(x, ξ) and B(x, ς) are measurable with respect to x for all ξ, ς and continuous withrespect to ξ, ς, respectively, for a.e. x ∈ Ω.

(H2) |A(x, ξ)|+|B(x, ς)| ≤ C1|ξ|p(x)−1+C2|ς|p(x)−1+G(x), whereG ∈ Lp′(x)(Ω) andC1, C2 ≥ 0are constants.

(H3) 〈A(x, ξ), ξ〉 ≥ a|ξ|p(x) − |h(x)|, where a > 0 is a constant and h ∈ L1(Ω).

(H4) 〈B(x, ς), ς〉 ≥ a|ς|p(x) − |h(x)|, where a ≥ 0 is a constant and h ∈ L1(Ω).

(H5) For a.e. x0 ∈ Ω, the mapping ξ → A(x0, ξ) satisfies

∫

D

〈A(x0, ξ0 + dv(x)), dv(x)〉dx ≥ γ∫

D

|dv(x)|p(x)dx, (1.2)

for each ξ0 ∈ Λl(Ω), D ⊂ Ω and v ∈ C10(Ω,Λ

l−1), where γ > 0 is a constant. Here p′

is the conjugate function of p. Throughout this paper we suppose (unless declarespecially)

p ∈ Plog(Ω), 1 < p∗ = essinfΩ p(x) ≤ p(x) ≤ esssupΩ p(x) = p∗ <∞. (1.3)

2. Preliminaries

Let e1, e2, . . . , en be the standard orthogonal basis of Rn. The space of all l-forms in R

n isdenoted by Λl(Rn). The dual basis to e1, e2, . . . , en is denoted by e1, e2, . . . , en and referred toas the standard basis for 1-form Λ1(Rn). The Grassman algebra Λ(Rn) = ⊕Λl(Rn) is a gradedalgebra with respect to the exterior products. The standard ordered basis for Λ(Rn) consistsof the forms

1, e1, e2, . . . , en, e1 ∧ e2, . . . , en−1 ∧ en, . . . , e1 ∧ e2 . . . ∧ en. (2.1)

For α(x) =∑αI(x)eI ∈ Λl(Rn) and β(x) =

∑βI(x)eI ∈ Λl(Rn), the inner product

is obtained by 〈α, β〉 =∑αI(x)βI(x) with summation over all l–tuples I = (i1, . . . il) and all

integers l = 0, 1, . . . , n. The Hodge star operator (see [14]) : Λ(Rn) → Λ(Rn) is defined bythe formulas

1 = e1 ∧ e2 · · · ∧ en, α ∧ β = β ∧ α =⟨α, β⟩e1 ∧ e2 · · · ∧ en. (2.2)


Hence, the norm of α is given by the formula |α|2 = 〈α, α〉 = (α ∧ α) =∑αI(x)αI(x) ∈

Λ0(Rn) = R. Notice, the Hodge star operator is an isometric isomorphism operator on Λ(Rn).Moreover,

: Λl(Rn) −→ Λn−l(Rn), = (−I)l(n−l) : Λl(Rn) −→ Λl(Rn), (2.3)

where I is the identity map.Let Ω ⊂ R

n be a bounded domain. The coordinate functions x1, x2, . . . , xn in Ω areconsidered to be differential forms of degree 0. The 1-forms dx1, dx2, . . . , dxn are constantfunctions from Ω into Λl(Rn). The value of dxi is simply ei, i = 1, 2, . . . , n. Therefore, everyl-form u : Ω → Λl(Rn) may be written uniquely as

u(x) =∑

I

uI(x)dxI =∑

1≤i1<···<il≤nui1,...il(x)dxi1 ∧ · · · ∧ dxil , (2.4)

where the coefficients ui1,...il(x) are distributions from D′(Ω), dual to the space of smoothfunctions with compact support on Ω.

We use D′(Ω,Λl) to denote the space of all differential l-forms. For each form u(x) ∈D′(Ω,Λl), the exterior differential d : D′(Ω,Λl) → D′(Ω,Λl+1) is expressed by

du(x) =n∑

k=1

∑

1≤i1<···<il≤n

∂ui1,...il(x)∂xk

dxk ∧ dxi1 ∧ · · · ∧ dxil . (2.5)

For u ∈ D′(Ω,Λl), the vector-valued differential form

∇u =(∂u

∂x1,∂u

∂x2, . . . ,

∂u

∂xn

)(2.6)

consists of differential forms ∂u/∂xi ∈ D′(Ω,Λl), where the partial differentiation is appliedto the coefficients of u.

The formal adjoint operator, called the Hodge codifferential, is given by

d = (−1)nl−1 d : D′(Ω,Λl+1

)−→ D′

(Ω,Λl

). (2.7)

By C∞(Ω,Λl) denote the space of infinitely differentiable l-forms on Ω and by C∞0 (Ω,Λl)

denote the subspace of C∞(Ω,Λl) with compact support on Ω.Let P(Ω) be the set of all Lebesgue measurable functions p : Ω → (1,∞). For

p ∈ P(Ω), we put p∗ = essinfΩ p(x) and p∗ = esssupΩ p(x). Given p ∈ P(Ω) we definethe conjugate function p′ ∈ P(Ω) by

p′(x) =p(x)

p(x) − 1, ∀x ∈ Ω. (2.8)


Definition 2.1 (see [15]). A Lebesgue measurable function p : Ω → R is called globally log-Holder continuous in Ω if there exist p∞ ∈ R and a constant C > 0 such that

∣∣p(x) − p(y)∣∣ ≤ C

log(e + 1/

∣∣x − y∣∣) ,

∣∣p(x) − p∞

∣∣ ≤ C

log(e + |x|) (2.9)

hold for all x, y ∈ Ω. P log(Ω) is defined by

Plog(Ω) ={p ∈ P(Ω) :

1p

is globally log-Holder continuous}. (2.10)

For a differential l-form u(x) on Ω, l = 0, 1, . . . , n, define the functional ρp(x) by

ρp(x),Λl(u) =∫

Ω|u(x)|p(x)dx. (2.11)

The space Lp(x)(Ω,Λl) = {u ∈ Λl(Ω) : ∃λ > 0, ρp(x),Λl(λu) < ∞} is a reflexive Banachspace endowed with the norm

‖u‖Lp(x)(Ω,Λl) = inf{λ > 0 : ρp(x),Λl

( u

λ

)≤ 1}. (2.12)

The space W1,p(x)(Ω,Λl) = {u ∈ Λl(Ω) : u ∈ Lp(x)(Ω,Λl) and du ∈ Lp(x)(Ω,Λl+1)} is areflexive Banach space endowed with the norm

‖u‖W1,p(x)(Ω,Λl) = ‖u‖Lp(x)(Ω,Λl) + ‖du‖Lp(x)(Ω,Λl+1). (2.13)

Note that Lp(m)(Ω,Λ0) and W1,p(m)(Ω,Λ0) are spaces of functions on Ω. In this paper,we denote them by Lp(m)(Ω) and W1,p(m)(Ω).

Iwaniec and Lutoborski proved the following results in [2].Let Ω ⊂ R

n be a bounded and convex domain. If u(x) ∈ Λl(Rn) is defined for somex ∈ Ω, then the value of u(x) at the vectors ξ1, . . . , ξl ∈ R

n is denoted by u(x)(ξ1, . . . , ξl). Thento each y ∈ Ω, there corresponds a linear operator Ky : L1

loc(Ω,∧l) → L1loc(Ω,∧l−1) defined by

Kyu(x)(ξ1, ξ2, . . . , ξl−1) =∫1

0tl−1u

(tx + y − ty)(x − y, ξ1, ξ2, . . . , ξl−1

)dt. (2.14)

The homotopy operator T : L1loc(Ω,∧l) → L1

loc(Ω,∧l−1) is defined by averaging Ky over allpoints y ∈ Ω

Tu(x) =∫

Ωϕ(y)Kyu(x)dy, (2.15)


where ϕ ∈ C∞0 (Ω) is normalized so that

∫Ω ϕ(y)dy = 1. Then we have a pointwise estimate

|Tu(x)| ≤ 2nμ(Ω)∫

Ω

∣∣u(y)∣∣

∣∣x − y∣∣n−1

dy, ∀x ∈ Ω, (2.16)

where

μ(Ω) = (diamΩ)n+1 inf

{∥∥∇ϕ∥∥L∞(Ω)∥∥ϕ∥∥L1(Ω)

: ϕ ∈ C∞0 (Ω)

}

, (2.17)

further infimum is attained at ϕ(x) = diam(x, ∂Ω), and the decomposition

u = dTu + Tdu (2.18)

holds for u ∈ L1loc(Ω,Λ

l).

Definition 2.2. For u ∈ L1loc(Ω,Λ

l), define the l-form uΩ ∈ D′(Ω,Λl) by

uΩ =

⎧⎨

⎩

1meas(Ω)

∫

Ωu(x)dx, for l = 0,

dTu, for l = 1, 2, · · · , n,(2.19)

and the Maximal operator is defined by

(Mu)(x) = supr>0

1meas(Br(x))

∫

Br(x)

∣∣u(y)∣∣dy, (2.20)

where Br(x) = {y ∈ Rn : |y − x| < r}.

Lemma 2.3 (see [15]). Let p(x) satisfies (1.3). Then the inequality

‖(Mu)(x)‖Lp(x)(Rn) ≤ C(n, p)‖u(x)‖Lp(x)(Rn) (2.21)

holds for every u ∈ Lp(x)(Rn).

Lemma 2.4 (see [15]). Let Ω ⊂ Rn be a bounded convex domain, x ∈ Ω and u ∈ L1

loc(Rn). Then

∫

Ω

∣∣u(y)∣∣

∣∣x − y∣∣n−1dy ≤ C(n)(diamΩ)(Mu)(x). (2.22)


Lemma 2.5 (see [15]). Let Ψ be a Calderon-Zygmund operator with Calderon-Zygmund kernel Kon R

n × Rn. Then Ψ is bounded on Lp(x)(Rn). Further there exists a constant C = C(n,p) such that

‖Ψu(x)‖Lp(x)(Rn) ≤ C(n, p)‖u(x)‖Lp(x)(Rn) (2.23)

holds for every u ∈ Lp(x)(Rn).

Lemma 2.6. If u ∈ Lp(x)(Ω,Λl), then

‖Tu‖Lp(x)(Ω,Λl−1) ≤ C(n, p)μ(Ω)(diamΩ)‖u‖Lp(x)(Ω,Λl). (2.24)

Moreover, if u ∈W1,p(x)(Ω,Λl), then

‖uΩ‖Lp(x)(Ω,Λl) ≤ C(p)‖u‖Lp(x)(Ω,Λl) + C

(n, p)μ(Ω)(diamΩ)‖du‖Lp(x)(Ω,Λl+1). (2.25)

Proof. First define u(x) = 0 if x ∈ Rn \Ω. From pointwise estimate (2.16) and Lemma 2.4,

|Tu(x)| ≤ C(n)μ(Ω)(diamΩ)M(|u|)(x), ∀x ∈ Ω. (2.26)

In view of Lemma 2.3, we have

‖|Tu|‖Lp(x)(Ω) ≤ C(n, p)μ(Ω)(diamΩ)‖|u|‖Lp(x)(Ω), (2.27)

that is to say, (2.24) holds.From the definition of uΩ and (2.18), we have uΩ = u − Tdu. Therefore,

‖uΩ‖Lp(x)(Ω,Λl) ≤ C(p)‖u‖Lp(x)(Ω,Λl) + C

(n, p)‖Tdu‖Lp(x)(Ω,Λl). (2.28)

Now in (2.24) replace u with du, we obtain (2.25).

Lemma 2.7. Let p(x) satisfies (1.3).

(1) C∞0 (Ω,Λl) is dense in Lp(x)(Ω,Λl),

(2) Lp(x)(Ω,Λl) is separable.

Proof. (1) For any u(x) =∑

I uI(x)dxI ∈ Lp(x)(Ω,Λl), since C∞0 (Ω) is dense in Lp(x)(Ω) and

uI(x) ∈ Lp(x)(Ω) for all I, we can find a sequence {uIk}∞k=1 ⊂ C∞0 (Ω) which converges to uI(x)


in Lp(x)(Ω) for each I. Now let uk(x) =∑

I uIkdxI , then the sequence {uk(x)} ⊂ C∞0 (Ω,Λl)

converges to u(x) in Lp(x)(Ω,Λl), since

∫

Ω|u(x) − uk(x)|p(x)dx =

∫

Ω

⎛

⎝(∑

I

|uI(x) − uIk(x)|2)1/2

⎞

⎠

p(x)

dx

≤∫

Ω

(∑

I

|uI(x) − uIk(x)|)p(x)

dx

≤ 2p∗∑

I

∫

Ω|uI(x) − uIk(x)|p(x)dx.

(2.29)

That is to say, C∞0 (Ω,Λl) is dense in Lp(x)(Ω,Λl).

(2) Let u(x) =∑

I uI(x)dxI ∈ Lp(x)(Ω,Λl). Since Lp(x)(Ω) is separable, there exists acountable dense subset K of Lp(x)(Ω). Then for any uI(x) above we can extract a sequence{uIk(x)} in K which converges to uI(x) in Lp(x)(Ω). Similar to (1), the sequence {uk : uk(x) =∑

I uIk(x)dxI} converges to u(x) in Lp(x)(Ω,Λl). That is to say, Lp(x)(Ω,Λl) is separable.

Let K1,p(x)(Ω,Λl) = {u(x) = ϑ(x) − ϑΩ(x) : ϑ ∈ W1,p(x)(Ω,Λl)}. Note that u ∈K1,p(x)(Ω,Λl) if and only if uΩ = 0.

Lemma 2.8. Let p(x) satisfies (1.3). Then K1,p(x)(Ω,Λl) is a closed subspace of W1,p(x)(Ω,Λl). Inparticular, it is a reflexive Banach space.

Proof. Set a sequence {uk(x)} ⊂ K1,p(x)(Ω,Λl) convergent to u(x) in W1,p(x)(Ω,Λl), then(uk)Ω = 0. By Lemma 2.6, the operator T is continuous on K1,p(x)(Ω,Λl). Therefore, uΩ =0, we have u(x) ∈ K1,p(x)(Ω,Λl). That is to say, K1,p(x)(Ω,Λl) is a closed subspace ofW1,p(x)(Ω,Λl).

In [2], Iwaniec and Lutoborski obtained

∂

∂xi(Tu) = Aiu + Siu, (2.30)

where

|Aiu(x)| ≤2nμ(Ω)

diam(Ω)

∫

Ω

|u(z)||x − z|n−1

dz, (2.31)

Siu(x)(ξ) =∫

Ωu(z)(Ki(z, x − z), ξ)dz, (2.32)


where ξ = (ξ1, ξ2, . . . , ξl−1) and

Ki(z, x − z) = ei|x − z|n

∫∞

0sn−1ϕ

(z − s x − z

|x − z|)ds

− x − z|x − z|n+1

∫∞

0snϕi

(z − s x − z

|x − z|)ds.

(2.33)

Further for each z ∈ Ω and h ∈ Rn − {0}, Ki(z, h) satisfies the following properties:

(i) Ki(z, h) ≤ μ(Ω)|h|−n,

(ii) Ki(z, sh) = s−nKi(z, h), s > 0,

(iii)∫|h|=1 Ki(z, h) = 0 for all z ∈ Ω.

Let Ki(z, h) = (Ki1, Ki2, . . . , Kin). Then Kiα satisfies the conditions of Calderon-Zygmund kernel on R

n × Rn for each α = 1, 2, . . . , n.

Lemma 2.9. Let u ∈ Lp(x)(Ω,Λl). Then

‖|∇Tu|‖Lp(x)(Ω) ≤ C(n, p,Ω

)‖u‖Lp(x)(Ω,Λl). (2.34)

Proof. By Lemmas 2.3 and 2.4, and (2.31),

‖Aiu‖Lp(x)(Ω,Λl) ≤ C(n, p)μ(Ω)‖u‖Lp(x)(Ω,Λl). (2.35)

Let

Siu(x) =∑

1≤j1<j2<···<jl−1≤nωj1,j2,...,jl−1dxj1 ∧ dxj2 ∧ · · · ∧ dxjl−1 , (2.36)

we can write u(x) as

u(x) =∑

1≤α≤n,α /= j1,j2,...,jl−1

∑

1≤j1<j2<···<jl−1≤nuα,j1,j2,...,jl−1dxα ∧ dxj1 ∧ · · · ∧ dxjl−1 . (2.37)

Hence,

ωj1,j2,...,jl−1(x) = Siu(x)(ej1 , ej2 , . . . , ejl−1

). (2.38)

Taking ξ = (ej1 , ej2 , . . . , ejl−1) in (2.32), we obtain

ωj1,j2,...,jl−1(x) =∫

Ω

∑

1≤α≤n,α /= j1,...,jl−1

Kiα(z, x − z)uα,j1,j2,...,jl−1(z)dz. (2.39)


Now define u(x) = 0 if x ∈ Rn \ Ω. Since Kiα satisfies the conditions of Calderon-

Zygmund kernel on Rn × R

n for each α, in view of Lemma 2.5,

∥∥ωj1,j2,...,jl−1

∥∥Lp(x)(Ω) ≤ C

(n, p) ∑

1≤α≤n,α /= j1,...,jl−1

∥∥uα,j1,j2,...,jl−1

∥∥Lp(x)(Ω). (2.40)

So that

‖Siu‖Lp(x)(Ω,Λl) ≤ C(n, p)‖u‖Lp(x)(Ω,Λl). (2.41)

By (2.30), (2.35), and (2.41), we have

‖|∇Tu|‖Lp(x)(Ω) ≤ C(n, p,Ω

)‖u‖Lp(x)(Ω,Λl). (2.42)

Now define another norm

‖|ω|‖K1,p(x)(Ω,Λl) = ‖ω‖Lp(x)(Ω,Λl) + ‖|∇ω|‖Lp(x)(Ω). (2.43)

Remark 2.10. Replacing u with du in (2.34), we get by the definition of uΩ

‖|∇(u − uΩ)|‖Lp(x)(Ω) = ‖|∇Tdu|‖Lp(x)(Ω)

≤ C(n, p)μ(Ω)‖du‖Lp(x)(Ω,Λl)

= C(n, p)μ(Ω)‖d(u − uΩ)‖Lp(x)(Ω,Λl).

(2.44)

Therefore ‖| · |‖K1,p(x)(Ω,Λl) is equivalent to ‖ · ‖K1,p(x)(Ω,Λl).

In this paper we also need the following two lemmas.

Lemma 2.11 (see[15]). Let p(x) satisfies (1.3). Then the embedding W1,p(x)0 (Ω) ↪→ Lp(x)(Ω) is

compact.

Lemma 2.12 (see[15]). Suppose that p ∈ L∞(Ω). Let {uk}∞k=1 be bounded in Lp(x)(Ω). If uk → ua.e. on Ω, then uk ⇀ u weakly in Lp(x)(Ω).

Remark 2.13. Let K1,p(x)0 (Ω,Λl) be the completion of C∞

0 (Ω,∧l) in K1,p(x)(Ω,Λl). Then from

Remark 2.10 and Lemma 2.11, the embedding K1,p(x)0 (Ω,Λl) ↪→ Lp(x)(Ω,Λl) is compact.

Remark 2.14. Suppose p(x) satisfies (1.3), Lemma 2.12 also holds on space Lp(x)(Ω,∧l).


3. Weak Solutions of Dirichlet Problems for theA-Harmonic Equations with Variable Growth

Theorem 3.1. Under conditions (H1)–(H5), the Dirichlet problem (1.1) has at least one weak solutionin K

1,p(x)0 (Ω,Λl), that is to say, there exists at least one u = ϑ − ϑΩ ∈ K

1,p(x)0 (Ω,Λl) satisfying

∫

Ω

⟨A(x, du(x)), dϕ(x)

⟩+⟨B(x, u(x)), ϕ(x)

⟩dx = 0, (3.1)

for all ϕ ∈W1,p(x)0 (Ω,Λl−1). Here, ϑ ∈W1,p(x)(Ω,Λl−1) and p(x) satisfies (1.3).

Let V = W1,p(x)0 (Ω,Λl−1) and K0 = K

1,p(x)0 (Ω,Λl). For u ∈ V , define A : V → V ∗ in the

following way: for each ϕ ∈ V

(Au, ϕ

)=∫

Ω

⟨A(x, du(x)), dϕ(x)

⟩+⟨B(x, u(x)), ϕ(x)

⟩dx. (3.2)

Now we need only to show that there exists u ∈ K0 such that (Au, ϕ) = 0 for all ϕ =∑ϕI(x)dxI ∈ V .

Lemma 3.2. A is strong-weakly continuous on V .

Proof. Let {uk : uk(x) =∑

I ukI(x)dxI} ⊂ V be a sequence strongly convergent to an elementu(x) =

∑uI(x)dxI ∈ V in V . Let duk(x) =

∑J ωkJ(x)dxJ and du(x) =

∑J ωJ(x)dxJ . Then

(h1) ‖uk‖V ≤ C for some constant C,

(h2) {ωkJ(x)} is a sequence strongly convergent to ωJ(x) in Lp(x)(Ω) for each J .

In view of (H2) and (h1), we know that A(x, duk) =∑AkJ(x)dxJ and B(x, uk) =∑

BkI(x)dxI are uniformly bounded in Lp′(x)(Ω,Λl) and Lp

′(x)(Ω,Λl−1), respectively. Hence,AkJ(x) and BkI(x) are uniformly bounded in Lp

′(x)(Ω). On the other hand, by (h2), there existsa subsequence of {ωkJ(x)} (still denoted by {ωkJ(x)}) such that

limk→∞

ωkJ(x) = ωJ(x), a.e. x ∈ Ω, for each J. (3.3)

Then there exists a subsequence of {uk(x)} (still denoted by {uk(x)}) such that

limk→∞

uk(x) = u(x), limk→∞

duk(x) = du(x), a.e. x ∈ Ω. (3.4)

In view of (H1), we obtain

limk→∞

A(x, duk) = A(x, du), a.e. x ∈ Ω,

limk→∞

B(x, uk) = B(x, u), a.e. x ∈ Ω.(3.5)


Let dϕ(x) =∑ψJ(x)dxJ , A(x, du) =

∑AJ(x)dxJ , and B(x, u) =

∑BI(x)dxI , then ψJ(x) ∈

Lp(x)(Ω), in the meantime

limk→∞

AkJ(x) = AJ(x), a.e. x ∈ Ω, (3.6)

limk→∞

BkI(x) = BI(x), a.e. x ∈ Ω, (3.7)

for each J and I.Now by Lemma 2.12, we can show that

∫ΩAkJ(x)ψJ(x)dx → ∫

ΩAJ(x)ψJ(x)dx and∫Ω BkI(x)ϕI(x)dx → ∫

Ω BI(x)ϕI(x)dx as k → ∞. Therefore,

(Auk, ϕ

)=∫

Ω〈A(x, duk), dϕ〉 + 〈B(x, uk), ϕ〉dx

−→∫

Ω〈A(x, du), dϕ〉 + 〈B(x, u), ϕ〉dx

=(Au, ϕ

),

(3.8)

that is to say, A is strong-weakly continuous on V .

Lemma 3.3. A is coercive on K0, that is,

lim‖u‖K →∞

(Au, u)‖u‖K

= +∞, ∀u ∈ K0. (3.9)

Proof. By (H3) and (H4),

(Au, u) =∫

Ω〈A(x, du), du〉 + 〈B(x, u), u〉dx

≥∫

Ω

(a|du|p(x) − |h(x)| + a|u|p(x) −

∣∣∣h(x)∣∣∣)dx

≥∫

Ωa|du|p(x)dx − C

(h, h).

(3.10)

By dϑΩ = 0 and Lemma 2.6, we have

‖u‖Lp(x)(Ω,Λl−1) = ‖Tdϑ‖Lp(x)(Ω,Λl−1) ≤ 2nC(n, p)μ(Ω)(diamΩ)‖du‖Lp(x)(Ω,Λl), (3.11)

for all u = ϑ − ϑΩ ∈ K0. Then ‖du‖Lp(x)(Ω,Λl) → ∞, as ‖u‖K → ∞. Taking

δ =12‖du‖Lp(x)(Ω,Λl) > 1, (3.12)


we have

∫Ω |du|p(x)dx‖du‖Lp(x)(Ω,Λl)

=∫

Ω

(|du|

‖du‖Lp(x)(Ω,Λl) − δ

)p(x)(‖du‖Lp(x)(Ω,Λl) − δ

)p(x)

‖du‖Lp(x)(Ω,Λl)dx

≥

(‖du‖Lp(x)(Ω,Λl) − δ

)p∗

‖du‖Lp(x)(Ω,Λl)=(

12

)p∗‖du‖p∗−1

Lp(x)(Ω,Λl).

(3.13)

Therefore,

∫Ω |du|p(x)dx‖du‖Lp(x)(Ω,Λl)

−→ ∞ as ‖du‖Lp(x)(Ω,Λl) −→ ∞. (3.14)

Then it is immediate to obtain that

(Au, u)‖u‖K

−→ ∞ as ‖u‖K −→ ∞. (3.15)

That is to say, A is coercive on K0.

Lemma 3.4 (see[16]). Suppose g = A(x) is a mapping from Rm into itself such that

lim|x|→∞

A(x) · x|x| = ∞. (3.16)

Then the range of A is the whole of Rm.

Lemma 3.5. There exists a sequence {uk} ⊂ K0 and u0 ∈ K0, such that

(Auk, uk − u0) −→ 0 as k −→ ∞. (3.17)

Proof . By Lemmas 2.7 and 2.8, we can choose a Schauder basis {ωs} of K0 such that the unionof subspace finitely generated from ωs is dense in K0. Let Kk

0 be the subspace of K0 generatedbyω1, ω2, . . . , ωk. Since Kk

0 is topologically isomorphic to Rk. By By Lemmas 3.3, and 3.4, there

exists uk ∈ Kk0 such that

(Auk,ω) = 0 ∀ω ∈ Kk0 . (3.18)

By Lemma 3.3 again, we know that ‖uk‖K ≤ C, where C is independent of k. Since K0 isreflexive, by Remark 2.14 and (H1), we can extract a subsequence of {uk} (still denoted by{uk}) such that

uk ⇀ u0 weakly in K0, Auk ⇀ ξ weakly∗ in K∗0, (ξ, ω) = 0, (3.19)


where ω is in a dense subset of K0. For fixed ξ, by the continuity of (ξ, ·), we get (ξ, ω) = 0 forall ω ∈ K0. For (Auk, uk − u0), we have

(Auk, uk − u0) = (Auk, uk) − (Auk, u0) = −(Auk, u0) −→ 0 as k −→ ∞. (3.20)

This completes the proof of Lemma 3.5.

Set vk = uk − u0 =∑vkIdxI . Then

vk ⇀ 0 weakly in K0 as k −→ ∞. (3.21)

Consider (Auk, uk − u0) once more, then

(Auk, uk − u0) =∫

Ω〈A(x, du0 + dvk), dvk〉 + 〈B(x, u0 + vk), vk〉dx −→ 0, (3.22)

as k → ∞. By Remark 2.13, we get

vk −→ 0 strongly in Lp(x)(Ω,Λl−1

). (3.23)

In view of (3.23) and (H2), it is immediate that

∫

Ω〈B(x, u0 + vk), vk〉dx −→ 0 as k −→ ∞, (3.24)

that is to say,

∫

Ω〈A(x, du0 + dvk), dvk〉dx −→ 0 as k −→ ∞. (3.25)

Now if we can prove that there exists a subsequence of {vk} which is stronglyconvergent in K0, then from the strong-weakly continuity of A, we get Auk ⇀ Au0 = ξ weaklyin K0 as k → ∞ and u0 will be a weak solution of (1.1). We need the following lemmas.

Definition 3.6. Let Ω be an open subset of Rn provided with the Lebesgue measure. The

mapping f : Ω × RN → R is said Caratheodory function if for almost all x ∈ Ω, f(x, ·) is

continuous on RN , for all ξ ∈ R

N is measurable on Ω.

Lemma 3.7 (see[17]). A mapping f : Ω × RN → R is a Caratheodory function if and only if for all

compact setsK ⊂ Ω and all ε > 0, there exists a compact subsetKε ⊂ K such that meas(K −Kε) < εfor with the restriction of f to Kε × R

N is continuous.


Lemma 3.8 (see[15]). Let {fk} be a sequence of bounded function in L1(Rn). For each ε > 0 thereexists (Aε, δ,N) (whereAε is measurable and meas(Aε) < ε, δ > 0,N is an infinite subset of naturalnumbers set N) such that for each k ∈N,

∫

B

∣∣fk(x)

∣∣dx < ε, (3.26)

where B and Aε are disjoint and meas(B) < δ.

Definition 3.9. For u ∈ C10(R

n), define

(M∗u)(x) = (Mu)(x) +n∑

α=1

(M

∂u

∂xα

)(x). (3.27)

Lemma 3.10 (see[18]). If u ∈ C∞0 (Rn), thenM∗u ∈ C0(Rn) and for all x ∈ R

n,

|u(x)| +n∑

α=1

∣∣∣∣∂u

∂xα(x)∣∣∣∣ ≤ (M∗u)(x). (3.28)

Furthermore, if p > 1, then

‖M∗u‖Lp(Rn) ≤ C(n, p)‖u‖W1,p(Rn), (3.29)

and if p = 1, then

meas({x ∈ Rn : (M∗u)(x) > λ}) ≤ C(n)

λ‖u‖W1,1(Rn), (3.30)

for all λ > 0.

Lemma 3.11 (see[19]). Let u ∈ C∞0 (Rn) and λ > 0. Set

Hλ = {x ∈ Rn : (M∗u)(x) < λ}. (3.31)

Then for all x, y ∈ Hλ, we have

∣∣u(y) − u(x)∣∣ ≤ C(n)λ∣∣y − x∣∣. (3.32)

Lemma 3.12 (see[16]). Let X be a metric space, E be a subspace of X, and k be a positive number.Then any k-Lipchitz mapping from E into R can be extended to a k-Lipchitz mapping from X into R.

Proof of Theorem 3.1. We need only to show that there exists subsequence of {vk} which isstrongly convergent in K0.


For each measurable set S ⊂ Ω, define

F(v, S) =∫

S

〈A(x, du0 + dv), dv〉dx, (3.33)

where v ∈ K0. Similar to the proof of Lemma 3.2, F(·, S) is strongly continuous on K0. SinceC∞

0 (Ω,Λl−1) is dense in K0, there exists hk ⊂ C∞0 (Ω,Λl−1) such that

‖hk − vk‖K <1k, |F(hk,Ω) − F(vk,Ω)| < 1

k. (3.34)

So we can suppose that {vk} ⊂ C∞0 (Ω,Λl−1) is bounded in K0.

Next define

vk(x) = 0 when x ∈ Rn \Ω. (3.35)

In this way, we extend the domain of vk to Rn and supp vk ⊂ Ω.

Let β : R+ → R

+ be a continuous increasing function satisfying β(0) = 0 and for eachmeasurable set D ⊂ Ω,

∫

D

(|G(x)|p′(x) + |h(x)| + (C1 + 1)|du0|p(x)

)dx ≤ β(meas(D)), (3.36)

where C1 is the constant in (H2).Let {εj} be a positive decreasing sequence with εj → 0 as j → ∞. For ε1, by

Lemma 3.8, we get a subsequence {k1} of {k}, a set Aε1 ⊂ Ω satisfying meas (Aε1) < ε1,and a real number δ1 > 0 such that

∫

B

(M∗vk1I)p(x)dx < ε1, (3.37)

for each k1, I and B ⊂ Ω \Aε1 satisfying meas(B) < δ1. By Lemma 3.10, we can choose λ > 1so large that for all I and k1,

meas({x ∈ Rn : (M∗vk1I)(x) ≥ λ}) ≤ min{ε1, δ1}. (3.38)

For each I and k1, define

Hλk1I

= {x ∈ Rn : (M∗vk1I)(x) < λ}, Hλ

k1=⋂

I

Hλk1I. (3.39)

In view of Lemma 3.11, we have

∣∣vk1I

(y) − vk1I(x)

∣∣∣∣y − x∣∣ ≤ C(n)λ ∀x, y ∈ Hλ

k1and I. (3.40)


Form Lemma 3.12, there exists a Lipschitz function gk1I which extends vk1I outside Hλk1

andLipschitz constant of gk1I is no more than C(n)λ. As Hλ

k1is an open set, we have gk1I = vk1I

and ∇gk1I(x) = ∇vk1I(x) for all x ∈ Hλk1

, and ‖∇gk1I‖L∞(Rn) ≤ C(n)λ. We can further supposethat

∥∥gk1I

∥∥L∞(Rn) ≤ ‖vk1I‖L∞(Hλ

k1) ≤ λ,

∥∥gk1I

∥∥W1,∞(Ω) ≤ C(n)λ. (3.41)

By the uniformly boundedness of {‖gk1I‖W1,∞(Ω)}, there exists a subsequence of {gk1I} (stilldenoted by {gk1I}) such that

gk1I ⇀ ωI weakly∗ in W1,∞(Ω) as k1 −→ ∞ ∀I. (3.42)

Set ω =∑

I ωIdxI and gk1 =∑

I gk1IdxI . We have

F(vk1 ,Ω) = F(gk1 ,Ω \Aε1

) − F(gk1 , (Ω \Aε1) \Hλ

k1

)+ F(vk1 , Aε1 ∪

(Ω \Hλ

k1

)). (3.43)

Next we estimate F(vk1 ,Ω) in four steps.(1) The estimate of F(gk1 , (Ω \Aε1) \Hλ

k1) and F(vk1 , Aε1 ∪ (Ω \Hλ

k1)). Since

meas((Ω \Aε1) \Hλ

k1

)≤∑

I

meas((Ω \Aε1) \Hλ

k1I

)≤ Cl−1

n min{ε1, δ1}, (3.44)

where Cl−1n = n(n − 1) · · · (n − l + 2)/(l − 1)(l − 2) · · · 1, from (H2), (H3), and the choose of Aε1 ,

we have

∣∣∣F(gk1 , (Ω \Aε1) \Hλ

k1

)∣∣∣

≤∫

(Ω\Aε1 )\Hλk1

(C1∣∣du0 + dgk1

∣∣p(x) −1∣∣dgk1

∣∣ + |G(x)|∣∣dgk1

∣∣)dx

≤∫

(Ω\Aε1 )\Hλk1

(C12p

∗−1(|du0|p(x) +

∣∣dgk1

∣∣p(x))+ C1

∣∣dgk1

∣∣p(x) + |G(x)|p′(x) + ∣∣dgk1

∣∣p(x))dx

≤ 2p∗−1β

(meas

((Ω \Aε1) \Hλ

k1

))+2p

∗(C1+1)

∫

(Ω\Aε1 )\Hλk1

∣∣dgk1

∣∣p(x)dx

≤ 2p∗−1β

(meas

((Ω \Aε1) \Hλ

k1

))+2p

∗(C1+1)

∫

(Ω\Aε1 )\Hλk1

(∑

I

∣∣∇gk1I

∣∣)p(x)

dx


≤ 2p∗−1β

(Cl−1n ε1

)+ 2p

∗C(C1, n, l)

∫

(Ω\Aε1 )\Hλk1

λp(x)dx

≤ 2p∗−1β

(Cl−1n ε1

)+ 2p

∗C(C1, n, l)

∑

I

∫

(Ω\Aε1 )\Hλk1I

(M∗vk1I)p(x)dx

≤ 2p∗−1β

(Cl−1n ε1

)+ 2p

∗C(C1, n, l)ε1 ≤ O(ε1),

(3.45)

F(vk1 , Aε1 ∪

(Ω \Hλ

k1

))

=∫

Aε1∪(Ω\Hλk1)〈A(x, du0 + dvk1), du0 + dvk1〉 − 〈A(x, du0 + dvk1), du0〉dx

≥∫

Aε1∪(Ω\Hλk1)

(a|du0 + dvk1 |p(x) − h(x)

)−(C1|du0 + dvk1 |p(x)−1|du0| + |G(x)||du0|

)dx

≥∫

Aε1∪(Ω\Hλk1)

((a2−(p

∗−1) − C1μ2p∗−1)|dvk1 |p(x) − |h(x)| − |G(x)|p′(x)

)

−(−a2−(p

∗−1) + C1μ2p∗−1 + C1C

(μ)+ 1)|du0|p(x)dx

≥(a2−(p

∗−1)−C1μ2p∗−1)∫

Aε1∪(Ω\Hλk1)|dvk1 |p(x)dx−C

(a, p, C1, μ

)β(

meas(Aε1 ∪

(Ω\Hλ

k1

)))

≥ a2−p∗∫

Aε1∪(Ω\Hλk1)|dvk1 |p(x)dx −O(ε1),

(3.46)

where μ > 0 is small enough.From (3.43)–(3.46), we get

F(vk1 ,Ω) ≥ F(gk1 ,Ω \Aε1

)+ a2−p

∗∫

Aε1∪(Ω\Hλk1)|dvk1 |p(x)dx −O(ε1). (3.47)

(2) The estimate of F(gk1 ,Ω \ Aε1). Set fk1I = gk1I − ωI , where ωI is defined by (3.42).Then

fk1I ⇀ 0 weakly∗ in W1,∞(Ω) as k1 −→ ∞ ∀I,∥∥fk1I

∥∥L∞(Ω) ≤ 2λ,

∥∥dfk1I

∥∥L∞(Ω,Λl) ≤ 2C(n)λ.

(3.48)


Let G =⋃I GI with GI = {x ∈ Ω : ωI(x)/= 0}. According to Acerbi and Fusco [19], we

have meas(G) ≤ (Cl−1n + 1)ε1 where Cl−1

n = n(n − 1) · · · (n − l + 2)/(l − 1)(l − 2) · · · 1, and setfk1 =

∑I fk1IdxI , then

F(gk1 ,Ω \Aε1

)= F

(fk1 , (Ω \Aε1) \G

)

+ F(vk1 , (Ω \Aε1) ∩Hλ

k1∩G)

+ F(gk1 , (Ω \Aε1) ∩

(G \Hλ

k1

)).

(3.49)

Define

Ωε1,k11 = Aε1 ∪

(Ω \Hλ

k1

), Ωε1

2 = (Ω \Aε1) \G,

Ωε1,k13 = (Ω \Aε1) ∩Hλ

k1∩G, Ωε1,k1

4 = (Ω \Aε1) ∩(G \Hλ

k1

).

(3.50)

Similar to the proof of (3.46), we get

F(vk1 ,Ω

ε1,k13

)≥ a2−p

∗∫

Ωε1 ,k13

|dvk1 |p(x)dx −O(ε1). (3.51)

Since on Ωε1,k14 we have

∫

Ωε1 ,k14

∣∣dgk1

∣∣p(x)dx ≤ C(n, p)(

Cl−1n + 1

)ε1, (3.52)

then similar to the proof of (3.45), we get

∣∣∣F(gk1 ,Ω

ε1,k14

)∣∣∣ ≤ O(ε1). (3.53)

By (3.49)–(3.53), we have

F(gk1 ,Ω \Aε1

) ≥ F(fk1 ,Ωε12

)+ a2−p

∗∫

Ωε1 ,k13

|dvk1 |p(x)dx −O(ε1). (3.54)

Thus, we have

F(vk1 ,Ω) ≥ F(fk1 ,Ωε12

)+ a2−p

∗∫

Ωε1 ,k15

|dvk1 |p(x)dx −O(ε1), (3.55)

where Ωε1,k15 = Ωε1,k1

1 ∪Ωε1,k13 .


Choose an open set Ω′ ⊂ Ω which contains Ωε12 such that

∣∣F(fk1 ,Ω

′) − F(fk1 ,Ωε12

)∣∣ < ε1. (3.56)

From (3.55), we get

F(vk1 ,Ω) ≥ F(fk1 ,Ω′) + a2−p

∗∫

Ωε1 ,k15

|dvk1 |p(x)dx −O(ε1). (3.57)

Approximate Ω′ by hypercubes with edges parallel to coordinate axes, that is, con-struct

Hj ⊂ Ω′,

meas(Ω′ \Hj

) −→ 0 as j −→ ∞,

Hj =hj⋃

s=1

Dj,s,

meas(Dj,s

)= 1/2nj , 1 ≤ s ≤ hj .

(3.58)

Let j > 0 be large enough such that for all k1 > 0, we have

∣∣F(fk1 ,Ω

′) − F(fk1 ,Hj

)∣∣ < ε1,

∫

Ω′\Hj

∣∣dfk1

∣∣p(x)dx < ε1,

meas(Ω′ \Hj

)< min{ε1, δ1}.

(3.59)

Thus,

F(vk1 ,Ω) ≥ F(fk1 ,Hj

)+ a2−p

∗∫

Ωε1 ,k15

|dvk1 |p(x)dx −O(ε1) − 2ε1. (3.60)

(3) The estimate of F(fk1 ,Hj). Let α > 0 be large enough such that for E = {x ∈ Ω′ :η(x) ≤ α}. Then

meas(Ω′ \ E) < ε1

N,

∫

Ω′\Eη(x)dx < ε1, (3.61)

where ‖dfk1‖L∞(Ω,Λl) ≤ 2Cl−1n C(n)λ =N and η(x) = |G(x)|p′(x) + 2p

∗−1(C1 + 1)|du0|p(x).For x ∈ Ω, ξ ∈ Λl(Ω), define

ψ(x, ξ) = 〈A(x, du0(x) + ξ), ξ〉. (3.62)


By Lemma 3.7 and (H1), there exists a compact subset K ⊂ Hj such that ψ(x, ξ) is continuouson K × Λl(Ω) and meas(Hj \ K) < ε1/(α + N). Hence, ψ(x, ξ) is uniformly continuous onbounded subsets of K ×Λl(Ω).

Divide each Dj,s into 2nm hypercubes Qmt,j,s with edge length 2−jm, 1 ≤ t ≤ 2nm. For all

j, s,m, t, take xmt,j,s ∈ Qmt,j,s ∩K ∩ E (if this set is empty, take xmt,j,s ∈ Qm

t,j,s) such that

η(xmt,j,s

)meas

(Qmt,j,s

)≤∫

Qmt,j,s

η(x)dx. (3.63)

Then

F(fk1 ,Hj

)

= F(fk1 ,Hj ∩K ∩ E) + F(fk1 ,Hj \ E

)+ F(fk1 ,

(Hj ∩ E

) \K)

≥ F(fk1 ,Hj ∩K ∩ E) −∫

Hj\Eη(x)dx −

∫

(Hj∩E)\Kη(x)dx

− 2p∗(C1 + 1)

(∫

Hj\E

∣∣dfk1

∣∣p(x)dx +∫

(Hj∩E)\K

∣∣dfk1

∣∣p(x)dx

)

= F(fk1 ,Hj ∩K ∩ E) −O(ε1)

= bm,jk1

+ cm,jk1

+ dm,jk1

−O(ε1),

(3.64)

where

bm,j

k1=∑

t,s

∫

Qmt,j,s∩K∩E

(ψ(x, dfk1(x)

) − ψ(xmt,j,s, dfk1(x)

))dx,

cm,j

k1=∑

t,s

∫

Qmt,j,s

ψ(xmt,j,s, dfk1(x)

)dx,

dm,j

k1= −

∑

t,s

∫

Qmt,j,s\(K∩E)

ψ(xmt,j,s, dfk1(x)

)dx.

(3.65)

By (3.25), we have

limk1 →∞

F(vk1 ,Ω) = 0. (3.66)

Note that if Qmt,j,s ∩K ∩ E is an empty set, then

∫

Qmt,j,s∩K∩E

[ψ(x, dfk1(x)


)]dx = 0. (3.67)


Now we only consider Qmt,j,s which satisfies Qm

t,j,s ∩ K ∩ E/=φ. Since du0(x) is uniformlycontinuous on Hj , then by the uniform continuity of ψ on bounded subsets of K × Λl(Ω),we obtain that for x ∈ Qm

t,j,s, there exists a constant L > 0 such that

∣∣∣ψ(x, dfk1(x)


)∣∣∣

=∣∣∣⟨A(x, du0(x) + dfk1(x)

) −A(xmt,j,s, du0

(xmt,j,s

)+ dfk1(x)

), dfk1(x)

⟩∣∣∣

<1

meas(Hj

)ε1

(3.68)

holds for all m > L and each k1. Therefore, |bm,jk1| < ε1 for all k1.

∣∣∣dm,j

k1

∣∣∣ ≤∑

t,s

∫

Qmt,j,s\(K∩E)

∣∣∣ψ(xmt,j,s, dfk1(x)

)∣∣∣dx

=∑

t,s

∫

Qmt,j,s\(K∩E)

⟨A(xmt,j,s, du0

(xmt,j,s

)+ dfk1(x)

), dfk1(x)

⟩dx

≤∑

t,s

∫

Qmt,j,s\(K∩E)

C1

∣∣∣du0

(xmt,j,s

)+ dfk1(x)

∣∣∣p(x)−1∣∣dfk(x)

∣∣ +∣∣∣G(xmt,j,s

)∣∣∣∣∣dfk1(x)

∣∣dx

≤∑

t,s

∫

Qmt,j,s\(K∩E)

(η(xmt,j,s

)+ 2p

∗(C1 + 1)N

)dx

≤∫

(Hj∩E)\K

(η(xmt,j,s

)+ 2p

∗(C1 + 1)N

)dx + C

(C1, p

)∑

t,s

∫

Qmt,j,s\E

(η(xmt,j,s

)+N

)dx

≤ C(α,N,C1, p

)meas

((Hj ∩ E

) \K) + C(C1, p)∫

Hj\E

[η(x) +N

]dx

≤ C(α,N,C1, p

)ε1 ≤ O(ε1).

(3.69)

Now we suppose that m is large enough that |bm,jk1| < ε1 for each k1 > 0 and there exists

k1 > 0 such that F(vk1Ω) < ε1 for k1 > k1. Therefore, from (3.25), (3.60), and (3.64), we have

ε1 ≥ F(vk1 ,Ω)

≥ cm,j

k1+ a2−p

∗∫

Ωε1 ,k15

|dvk1 |p(x)dx −O(ε1) − 3ε1 − C(C1, p

)ε1

= cm,j

k1+ a2−p

∗∫

Ωε1 ,k15

|dvk1 |p(x)dx −O(ε1).

(3.70)


(4) The estimate of cm,jk1. By fk1I ⇀ 0 weakly∗ in W1,∞(Ω) as k1 → ∞, we obtain

‖fk1I‖L∞(Ω) → 0 ask1 → ∞ for each I. Then

Rk1,mt,s,j =

∥∥|fk1 |

∥∥L∞(Qm

t,s,j )−→ 0 as k1 −→ ∞ for fixed m. (3.71)

Define a hypercube Ek1,mt,s,j contained in Qm

t,s,j with edge length 1/2jm − 2Rk1,mt,s,j such that

dist(∂Qmt,s,j , E

k1,mt,s,j ) = R

k1,mt,s,j .

Next define

ϕk1(x) = 0, x ∈ ∂Qmt,s,j ,

ϕk1(x) = fk1(x), x ∈ Emt,s,j .(3.72)

Since ϕk1I is a Lipschitz mapping on set Emt,s,j ∪ ∂Qmt,s,j and its Lipschitz constant is no more

than 2C(n)λ, by Lemma 3.12, ϕk1I can be extended to the whole Qmt,s,j , where it is also a

Lipschitz mapping with the same Lipchistz constant. We still denote the extension by ϕk1I

and suppose that it is defined on the whole Hj . Then by [20]

∇ϕk1I − ∇fk1I −→ 0 a.e. on Hj. (3.73)

Thus, there exists a k1 > k1 such that for all k1 > k1, we have

∫

Hj

∣∣dϕk1 − dfk1

∣∣p(x)dx ≤ ε1

2,

∑

t,s

∣∣∣∣∣

∫

Qmt,j,s

ψ(xmt,j,s, dfk1(x)

)− ψ(xmt,j,s, dϕk1(x)

)dx

∣∣∣∣∣≤ ε1

2.

(3.74)

In view of (H5), we obtain that

cm,j

k1=∑

t,s

∫

Qmt,j,s

ψ(xmt,j,s, dfk1(x)

)dx

≥∑

t,s

∫

Qmt,j,s

ψ(xmt,j,s, dϕk1(x)

)dx − ε1

2

=∑

t,s

∫

Qmt,j,s

⟨A(xmt,j,s, du0

(xmt,j,s

)+ dϕk1(x)

), dϕk1(x)

⟩dx − ε1

2

≥ γ∑

t,s

∫

Qmt,j,s

∣∣dϕk1

∣∣p(x)dx − ε1

2

≥ γ

2p∗−1

∫

Hj

∣∣dfk1

∣∣p(x)dx −(γ + 1

)ε1

2.

(3.75)


Thus in (3.70) for k1 > k1, we obtain the estimate of F(vk1 ,Ω) from the four steps above

ε1 ≥ F(vk1 ,Ω)

≥ a2−p∗∫

Ωε1 ,k15

|dvk1 |p(x)dx +γ

2p∗−1

∫

Hj

∣∣dfk1

∣∣p(x)dx −

(γ + 1

)ε1

2−O(ε1).

(3.76)

Let K(ε1) = (γ + 1)ε1/(2 + o(ε1))/min{a2−p∗, γ/2p

∗−1}. Then

∫

Ωε1 ,k15

|dvk1 |p(x)dx +∫

Hj

∣∣dfk1

∣∣p(x)dx ≤ K(ε1), for k1 > k1. (3.77)

Form (3.59) and (3.77), we deduce that

∫

Ωε1 ,k15

|dvk1 |p(x)dx ≤ K(ε1),∫

Ω′

∣∣dfk1

∣∣p(x)dx ≤ K(ε1) + ε1. (3.78)

According to the definition of Ωε12 , we have

∫

Ω2ε1

∣∣dgk1

∣∣p(x)dx ≤ K(ε1) + ε1. (3.79)

Since dgk1(x) = dvk1(x) for each x ∈ Hλk1

, we get

∫

Ω2ε1∩Hλ

k1

|dvk1 |p(x)dx ≤ K(ε1) + ε1. (3.80)

By the definitions of Ωε12 and Ωε1,k1

5 , it is immediate that

(Ωε1

2 ∩Hλk1

)∪Ωε1,k1

5 = Ω, (3.81)

which implies that

∫

Ω|dvk1 |p(x)dx ≤ 2K(ε1) + ε1 ≤ O(ε1). (3.82)

For ε2 > 0 and the sequence {vk1}, repeating the above arguments we can extract asubsequence {vk2} of {vk1} such that

∫

Ω|dvk2 |p(x)dx ≤ O(ε2), (3.83)


whenever k2 > k2 for some k2. If {vkn} has been obtained, repeating the above process, wecan extract a subsequence {kn+1} of {kn} such that

∫

Ω|dvkn+1 |p(x)dx ≤ O(εn+1), (3.84)

whenever kn+1 > kn+1 for some kn+1. Finally, by a diagonal argument we get a subsequence{vki}∞i=1 which satisfies

∫

Ω|dvki |p(x)dx −→ 0 as i −→ ∞. (3.85)

Therefore,

‖dvki‖Lp(x)(Ω,Λl) −→ 0 as i −→ ∞, (3.86)

and by (3.23), {vki}∞i=1 strongly converges to zero in K0 as i → ∞. This completes the proof ofTheorem 3.1.

4. Applications

In this section, we explore applications of our results developed in this paper.Let Ω ⊂ R

n be a bounded and convex Lipschitz domain. Suppose that maps A : Ω ×Λl(Ω) → Λl(Ω) and B : Ω ×Λl−1(Ω) → Λl−1(Ω), where l = 1, 2, . . . , n.

Example 4.1. If p(x) satisfies (1.3), let l = 1, A(x, ξ) = ξ|ξ|p(x)−2 and B(x, ς) = ς|ς|p(x)−2 − f(x),where f(x) ∈ Lp

′(x)(Ω). Then A,B satisfy the required conditions, and (1.1) reduce to thefollowing p(x)-Laplacian equations:

−div(|∇u|p(x)−2∇u

)+ |u|p(x)−2u = f(x), x ∈ Ω,

u(x) = 0, x ∈ ∂Ω.(4.1)

Now by Theorem 3.1, we deduce that the p(x)-Laplacian equations (4.1) have at least oneweak solution in K1,p(x)(Ω) with u = 0 on ∂Ω.

Example 4.2. If l = 1, A(x, ξ) =∑

i,j Aij(x)ξjdxi, B(x, ς) = B(x)ς − f(x), where f(x) ∈ L2(Ω),and Aij(x), B(x) satisfy the following conditions:

Aij(x) = Aji(x), ∧|ξ|2 ≥n∑

i,j=1

Aij(x)ξiξj ≥ λ|ξ|2, λ ≤ B(x) ≤ ∧, (4.2)


for some constants λ,∧ > 0. Then A,B satisfy the required conditions, and (1.1) reduce to thefollowing Divergence form equations:

n∑

i,j=1

∇j

(Aij(x)∇iu(x)

)+ B(x)u(x) = f(x), x ∈ Ω, (4.3)

u(x) = 0, x ∈ ∂Ω, (4.4)

where ∇i = (∂/∂xi). Now by Theorem 3.1, we deduce that the divergence form (4.3) haveat least one weak solution u(x) in K1,2(Ω) with u = 0 on ∂Ω. The comparison principles, themaximum principles, and the existence of weak solutions for divergence form equation (4.3)can be found in [21].

Example 4.3. If p(x) satisfies (1.3), let A(x, ξ) = ξ|ξ|p(x)−2 and B(x, ς) = ς|ς|p(x)−2 − f(x), wheref(x) ∈ Lp

′(x)(Ω,∧l−1). Then A,B satisfy the required conditions, and (1.1) reduce to thefollowing p(x)-harmonic equations for differential forms:

d∗(du|du|p(x)−2

)+ u|u|p(x)−2 = f(x), x ∈ Ω, (4.5)

u(x) = 0, x ∈ ∂Ω. (4.6)

Now by Theorem 3.1, we deduce that (4.5) have at least one weak solution u(x) inK

1,p(x)0 (Ω,∧l−1). If p(x) is a constant q and 1 < q < ∞, the equation (4.5) is called

nonhomogeneous q-harmonic equation. In [2], Iwaniec and Lutoborski studied the Lq theoryof weak solution for homogeneous q-harmonic equations.

References

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[2] T. Iwaniec and A. Lutoborski, “Integral estimates for null Lagrangians,” Archive for Rational Mechanicsand Analysis, vol. 125, no. 1, pp. 25–79, 1993.

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[4] F. L. Teixeira, “Differential form approach to the analysis of electromagnetic cloaking and masking,”Microwave and Optical Technology Letters, vol. 49, no. 8, pp. 2051–2053, 2007.

[5] P. Morando, “Liouville condition, Nambu mechanics, and differential forms,” Journal of Physics A, vol.29, no. 13, pp. L329–L331, 1996.

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[7] S. Ding and C. A. Nolder, “Weighted Poincare inequalities for solutions to A-harmonic equations,”Illinois Journal of Mathematics, vol. 46, no. 1, pp. 199–205, 2002.

[8] R. P. Agarwal and S. Ding, “Advances in differential forms and the A-harmonic equation,”Mathematical and Computer Modelling, vol. 37, no. 12-13, pp. 1393–1426, 2003.

[9] Y. Xing and S. Ding, “Caccioppoli inequalities with Orlicz norms for solutions of harmonic equationsand applications,” Nonlinearity, vol. 23, no. 5, pp. 1109–1119, 2010.

[10] Y. Xing and S. Ding, “Norms of the composition of the maximal and projection operators,” NonlinearAnalysis: Theory, Methods & Applications, vol. 72, no. 12, pp. 4614–4624, 2010.

[11] S. Ding and B. Liu, “Global estimates for singular integrals of the composite operator,” Illinois Journalof Mathematics, vol. 53, no. 4, pp. 1173–1185, 2009.


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[13] Y. Fu and L. Guo, “Weighted variable exponent spaces of differential forms and their applications”.[14] G. de Rham, Differentiable Manifolds, vol. 266 of Grundlehren der Mathematischen Wissenschaften,

Springer, Berlin, Germany, 1984.[15] L. Diening, P. Harjulehto, P. Hasto, and M. Ruzicka, Lebesgue and Sobolev Spaces with Variable Exponents,

vol. 2017 of Lecture Notes in Mathematics, Springer, Heidelberg, Germany, 2011.[16] C. B. Morrey, Multiple Integrals in the Calculus of Variations, Springer, 1966.[17] I. Ekeland and R. T’emam, Convex Analysis and Variational Problems, North-Holland, 1976.[18] F. C. Liu, “A Luzin type property of Sobolev functions,” Indiana University Mathematics Journal, vol.

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Mechanics and Analysis, vol. 86, no. 2, pp. 125–145, 1984.[20] B. Dacorogna, Weak Continuity and Weak Lower Semicontinuity of Nonlinear Functionals, vol. 922 of

Lecture Notes in Mathematics, Springer, Berlin, Germany, 1982.[21] D. Gilbarg and N. S. Trudinger, Elliptic Partial Differential Equations of Second Order, Springer, 1998.


Research ArticleForward-Backward Splitting Methods for AccretiveOperators in Banach Spaces

Genaro Lopez,1 Victoria Martın-Marquez,1 Fenghui Wang,2and Hong-Kun Xu3

1 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Sevilla, Apartado. 1160,41080-Sevilla, Spain

2 Department of Mathematics, Luoyang Normal University, Luoyang 471022, China3 Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan

Correspondence should be addressed to Hong-Kun Xu, [email protected]

Received 31 March 2012; Accepted 29 May 2012


Copyright q 2012 Genaro Lopez et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

Splitting methods have recently received much attention due to the fact that many nonlinearproblems arising in applied areas such as image recovery, signal processing, and machine learningare mathematically modeled as a nonlinear operator equation and this operator is decomposedas the sum of two (possibly simpler) nonlinear operators. Most of the investigation on splittingmethods is however carried out in the framework of Hilbert spaces. In this paper, we considerthese methods in the setting of Banach spaces. We shall introduce two iterative forward-backwardsplitting methods with relaxations and errors to find zeros of the sum of two accretive operatorsin the Banach spaces. We shall prove the weak and strong convergence of these methods undermild conditions. We also discuss applications of these methods to variational inequalities, the splitfeasibility problem, and a constrained convex minimization problem.

1. Introduction

Splitting methods have recently received much attention due to the fact that many nonlinearproblems arising in applied areas such as image recovery, signal processing, and machinelearning are mathematically modeled as a nonlinear operator equation and this operator isdecomposed as the sum of two (possibly simpler) nonlinear operators. Splitting methods forlinear equations were introduced by Peaceman and Rachford [1] and Douglas and Rachford[2]. Extensions to nonlinear equations in Hilbert spaces were carried out by Kellogg [3] andLions and Mercier [4] (see also [5–7]). The central problem is to iteratively find a zero of


the sum of two monotone operators A and B in a Hilbert space H, namely, a solution to theinclusion problem

0 ∈ (A + B)x. (1.1)

Many problems can be formulated as a problem of form (1.1). For instance, a stationarysolution to the initial value problem of the evolution equation

∂u

∂t+ Fu � 0, u(0) = u0 (1.2)

can be recast as (1.1) when the governing maximal monotone F is of the form F = A + B [4].In optimization, it often needs [8] to solve a minimization problem of the form

minx∈H

f(x) + g(Tx), (1.3)

where f, g are proper lower semicontinuous convex functions from H to the extended realline R := (−∞,∞], and T is a bounded linear operator on H. As a matter of fact, (1.3) isequivalent to (1.1) (assuming that f and g ◦ T have a common point of continuity) withA := ∂f and B := T ∗ ◦ ∂g ◦ T . Here T ∗ is the adjoint of T and ∂f is the subdifferential operatorof f in the sense of convex analysis. It is known [8, 9] that the minimization problem (1.3) iswidely used in image recovery, signal processing, and machine learning.

A splitting method for (1.1) means an iterative method for which each iterationinvolves only with the individual operators A and B, but not the sum A + B. To solve (1.1),Lions and Mercier [4] introduced the nonlinear Peaceman-Rachford and Douglas-Rachfordsplitting iterative algorithms which generate a sequence {υn} by the recursion

υn+1 =(

2JAλ − I)(

2JBλ − I)υn (1.4)

and respectively, a sequence {υn} by the recursion

υn+1 = JAλ(

2JBλ − I)υn +

(I − JBλ

)υn. (1.5)

Here we use JTλ

to denote the resolvent of a monotone operator T ; that is, JTλ= (I + λT)−1.

The nonlinear Peaceman-Rachford algorithm (1.4) fails, in general, to converge (evenin the weak topology in the infinite-dimensional setting). This is due to the fact that thegenerating operator (2JA

λ− I)(2JB

λ− I) for the algorithm (1.4) is merely nonexpansive.

However, the mean averages of {un} can be weakly convergent [5]. The nonlinear Douglas-Rachford algorithm (1.5) always converges in the weak topology to a point u and u = JBλ υ isa solution to (1.1), since the generating operator JAλ (2J

Bλ − I) + (I − JBλ ) for this algorithm is

firmly nonexpansive, namely, the operator is of the form (I + T)/2, where T is nonexpansive.There is, however, little work in the existing literature on splitting methods for

nonlinear operator equations in the setting of Banach spaces (though there was some workon finding a common zero of a finite family of accretive operators [10–12]).


The main difficulties are due to the fact that the inner product structure of a Hilbertspace fails to be true in a Banach space. We shall in this paper use the technique of dualitymaps to carry out certain initiative investigations on splitting methods for accretive operatorsin Banach spaces. Namely, we will study splitting iterative methods for solving the inclusionproblem (1.1), where A and B are accretive operators in a Banach space X.

We will consider the case where A is single-valued accretive and B is possiblymultivalued m-accretive operators in a Banach space X and assume that the inclusion (1.1)has a solution. We introduce the following two iterative methods which we call Mann-typeand respectively, Halpern-type forward-backward methods with errors and which generatea sequence {xn} by the recursions

xn+1 = (1 − αn)xn + αn(JBrn(xn − rn(Axn + an)) + bn

), (1.6)

xn+1 = αnu + (1 − αn)(JBrn(xn − rn(Axn + αn)) + bn

), (1.7)

where JBr is the resolvent of the operator B of order r (i.e., JBr = (I + rB)−1), and {αn} is asequence in (0, 1]. We will prove weak convergence of (1.6) and strong convergence of (1.7)to a solution to (1.1) in some class of Banach spaces which will be made clear in Section 3.

The paper is organized as follows. In the next section we introduce the class ofBanach spaces in which we shall study our splitting methods for solving (1.1). We alsointroduce the concept of accretive and m-accretive operators in a Banach space. In Section 3,we discuss the splitting algorithms (1.6) and (1.7) and prove their weak and strongconvergence, respectively. In Section 4, we discuss applications of both algorithms (1.6) and(1.7) to variational inequalities, fixed points of pseudocontractions, convexly constrainedminimization problems, the split feasibility problem, and linear inverse problems.

2. Preliminaries

Throughout the paper, X is a real Banach space with norm ‖ ·‖, distance d, and dual space X∗.The symbol 〈x∗, x〉 denotes the pairing between X∗ and X, that is, 〈x∗, x〉 = x∗(x), the valueof x∗ at x. C will denote a nonempty closed convex subset of X, unless otherwise stated, andBr the closed ball with center zero and radius r. The expressions xn → x and xn ⇀ x denotethe strong and weak convergence of the sequence {xn}, respectively, and ωw(xn) stands forthe set of weak limit points of the sequence {xn}.

The modulus of convexity of X is the function δX(ε) : (0, 2] → [0, 1] defined by

δX(ε) = inf

{

1 −∥∥x + y

∥∥

2: ‖x‖ =

∥∥y∥∥ = 1,

∥∥x − y∥∥ ≥ ε}

. (2.1)

Recall thatX is said to be uniformly convex if δX(ε) > 0 for any ε ∈ (0, 2]. Let p > 1. We say thatX is p-uniformly convex if there exists a constant cp > 0 so that δX(ε) ≥ cpεp for any ε ∈ (0, 2].

The modulus of smoothness of X is the function ρX(τ) : R+ → R+ defined by

ρX(τ) = sup

{∥∥x + τy∥∥ +

∥∥x − τy∥∥2

− 1 : ‖x‖ =∥∥y

∥∥ = 1

}

. (2.2)


Recall that X is called uniformly smooth if limτ→ 0ρX(τ)/τ = 0. Let 1 < q ≤ 2. We say that Xis q-uniformly smooth if there is a cq > 0 so that ρX(τ) ≤ cqτ

q for τ > 0. It is known that Xis p-uniformly convex if and only if X∗ is q-uniformly smooth, where (1/p + 1/q = 1). Forinstance, Lp spaces are 2-uniformly convex and p-uniformly smooth if 1 < p ≤ 2, whereasp-uniformly convex and 2-uniformly smooth if p ≥ 2.

The norm of X is said to be the Frechet differentiable if, for each x ∈ X,

limλ→ 0

∥∥x + λy

∥∥ − ‖x‖λ

(2.3)

exists and is attained uniformly for all y such that ‖y‖ = 1. It can be proved thatX is uniformlysmooth if the limit (2.3) exists and is attained uniformly for all (x, y) such that ‖x‖ = ‖y‖ = 1.So it is trivial that a uniformly smooth Banach space has a Frechet differentiable norm.

The subdifferential of a proper convex function f : X → (−∞,+∞] is the set-valuedoperator ∂f : X → 2X defined as

∂f(x) ={x∗ ∈ X∗ :

⟨x∗, y − x⟩ + f(x) ≤ f(y)}. (2.4)

If f is proper, convex, and lower semicontinuous, the subdifferential ∂f(x)/= ∅ for any x ∈intD(f), the interior of the domain of f . The generalized duality mapping Jp : X → 2X

∗is

defined by

Jp(x) ={j(x) ∈ X∗ :

⟨j(x), x

⟩= ‖x‖p, ∥∥j(x)

∥∥ = ‖x‖p−1}. (2.5)

If p = 2, the corresponding duality mapping is called the normalized duality mapping anddenoted by J. It can be proved that, for any x ∈ X,

Jp(x) = ∂(

1p‖x‖p

). (2.6)

Thus we have the following subdifferential inequality, for any x, y ∈ X:

∥∥x + y∥∥p ≤ ‖x‖p + p⟨y, j(x + y

)⟩, j

(x + y

) ∈ Jp

(x + y

). (2.7)

In particular, we have, for x, y ∈ X,

∥∥x + y∥∥2 ≤ ‖x‖2 + 2

⟨y, j

(x + y

)⟩, j

(x + y

) ∈ J(x + y

). (2.8)

Some properties of the duality mappings are collected as follows.

Proposition 2.1 (see Cioranescu [13]). Let 1 < p <∞.

(i) The Banach space X is smooth if and only if the duality mapping Jp is single valued.

(ii) The Banach space X is uniformly smooth if and only if the duality mapping Jp is single-valued and norm-to-norm uniformly continuous on bounded sets of X.


Among the estimates satisfied by p-uniformly convex and p-uniformly smooth spaces,the following ones will come in handy.

Lemma 2.2 (see Xu [14]). Let 1 < p <∞, q ∈ (1, 2], r > 0 be given.

(i) If X is uniformly convex, then there exists a continuous, strictly increasing and convexfunction ϕ : R

+ → R+ with ϕ(0) = 0 such that

∥∥λx + (1 − λ)y∥∥p ≤ λ‖x‖p + λ∥∥y∥∥p −Wp(λ)ϕ

(∥∥x − y∥∥), x, y ∈ Br , 0 ≤ λ ≤ 1, (2.9)

whereWp(λ) = λp(1 − λ) + (1 − λ)λp.(ii) If X is q-uniformly smooth, then there exists a constant κq > 0 such that

∥∥x + y

∥∥q ≤ ‖x‖q + q⟨Jq(x), y

⟩+ κq

∥∥y

∥∥q, x, y ∈ X. (2.10)

The best constant κq satisfying (2.10) will be called the q-uniform smoothness coefficient of X.For instance [14], for 2 ≤ p < ∞, Lp is 2-uniformly smooth with κ2 = p − 1, and for 1 < p ≤ 2, Lp isp-uniformly smooth with κp = (1 + tp−1

p )(1 + tp)1−p, where tp is the unique solution to the equation

(p − 2

)tp−1 +

(p − 1

)tp−2 − 1 = 0, 0 < t < 1. (2.11)

In a Banach space X with the Frechet differentiable norm, there exists a function h :[0,∞) → [0,∞) such that limt→ 0h(t)/t = 0 and for all x, u ∈ X

12‖x‖2 + 〈u,J(x)〉 ≤ 1

2‖x + u‖2 ≤ 1

2‖x‖2 + 〈u,J(x)〉 + h(‖u‖). (2.12)

Recall that T : C → C is a nonexpansive mapping if ‖Tx − Ty‖ ≤ ‖x − y‖, for allx, y ∈ C. From now on, Fix(T) denotes the fixed point set of T . The following lemma claimsthat the demiclosedness principle for nonexpansive mappings holds in uniformly convexBanach spaces.

Lemma 2.3 (see Browder [15]). Let C be a nonempty closed convex subset of a uniformly convexspace X and T a nonexpansive mapping with Fix(T)/= ∅. If {xn} is a sequence in C such that xn ⇀ xand (I − T)xn → y, then (I − T)x = y. In particular, if y = 0, then x ∈ Fix(T).

A set-valued operator A : X → 2X , with domain D(A) and range R(A), is said to beaccretive if, for all t > 0 and every x, y ∈ D(A),

∥∥x − y∥∥ ≤ ∥∥x − y + t(u − υ)∥∥, u ∈ Ax, υ ∈ Ay. (2.13)

It follows from Lemma 1.1 of Kato [16] that A is accretive if and only if, for each x, y ∈ D(A),there exists j(x − y) ∈ J(x − y) such that

⟨u − υ, j(x − y)⟩ ≥ 0, u ∈ Ax, υ ∈ Ay. (2.14)


An accretive operator A is said to be m-accretive if the range R(I + λA) = X for some λ > 0. Itcan be shown that an accretive operator A is m-accretive if and only if R(I + λA) = X for allλ > 0.

Given α > 0 and q ∈ (1,∞), we say that an accretive operator A is α-inverse strong-ly accretive (α-isa) of order q if, for each x, y ∈ D(A), there exists jq(x − y) ∈ Jq(x − y) suchthat

⟨u − υ, jq

(x − y)⟩ ≥ α‖u − υ‖q, u ∈ Ax, υ ∈ Ay. (2.15)

When q = 2, we simply say α-isa, instead of α-isa of order 2; that is, T is α-isa if, for eachx, y ∈ D(A), there exists j(x − y) ∈ J(x − y) such that

⟨u − υ, j(x − y)⟩ ≥ α‖u − υ‖2, u ∈ Ax, υ ∈ Ay. (2.16)

Given a subset K of C and a mapping T : C → K, recall that T is a retraction of C ontoK if Tx = x for all x ∈ K. We say that T is sunny if, for each x ∈ C and t ≥ 0, we have

T(tx + (1 − t)Tx) = Tx, (2.17)

whenever tx + (1 − t)Tx ∈ C.The first result regarding the existence of sunny nonexpansive retractions onto the

fixed point set of a nonexpansive mapping is due to Bruck.

Theorem 2.4 (see Bruck [17]). If X is strictly convex and uniformly smooth and if T : C →C is a nonexpansive mapping having a nonempty fixed point set Fix(T), then there exists a sunnynonexpansive retraction of C onto Fix(T).

The following technical lemma regarding convergence of real sequences will be usedwhen we discuss convergence of algorithms (1.6) and (1.7) in the next section.

Lemma 2.5 (see [18, 19]). Let {an}, {cn} ⊂ R+, {αn} ⊂ (0, 1), and {bn} ⊂ R be sequences such that

an+1 ≤ (1 − αn)an + bn + cn, ∀n ≥ 0. (2.18)

Assume∑∞

n=0 cn <∞. Then the following results hold:(i) If bn ≤ αnM whereM ≥ 0, then {an} is a bounded sequence.(ii) If

∑∞n=0 αn = ∞ and lim supn→∞bn/αn ≤ 0, then liman = 0.

3. Splitting Methods for Accretive Operators

In this section we assume that X is a real Banach space and C is a nonempty closed subsetof X. We also assume that A is a single-valued and α-isa operator for some α > 0 and B isan m-accretive operator in X, with D(A) ⊃ C and D(B) ⊃ C. Moreover, we always use Jr todenote the resolvent of B of order r > 0; that is,

Jr ≡ JBr = (I + rB)−1. (3.1)


It is known that the m-accretiveness of B implies that Jr is single valued, defined onthe entire X, and firmly nonexpansive; that is,

∥∥Jrx − Jry

∥∥ ≤ ∥

∥s(x − y) + (1 − s)(Jrx − Jry

)∥∥, x, y ∈ X, 0 ≤ s ≤ 1. (3.2)

Below we fix the following notation:

Tr := Jr(I − rA) = (I + rB)−1(I − rA). (3.3)

Lemma 3.1. For r > 0, Fix(Tr) = (A + B)−1(0).

Proof. From the definition of Tr , it follows that

x = Trx ⇐⇒ x = (I + rB)−1(x − rAx)⇐⇒ x − rAx ∈ x + rBx

⇐⇒ 0 ∈ Ax + Bx.

(3.4)

This lemma alludes to the fact that in order to solve the inclusion problem (1.1), itsuffices to find a fixed point of Tr . Since Tr is already “split,” an iterative algorithm for Trcorresponds to a splitting algorithm for (1.1). However, to guarantee convergence (weakor strong) of an iterative algorithm for Tr , we need good metric properties of Tr such asnonexpansivity. To this end, we need geometric conditions on the underlying space X (seeLemma 3.3).

Lemma 3.2. Given 0 < s ≤ r and x ∈ X, there holds the relation

‖x − Tsx‖ ≤ 2‖x − Trx‖. (3.5)

Proof. Note that ((x−Trx)/r)−Ax ∈ B(Trx). By the accretivity of B, we have js,r ∈ J(Tsx−Trx)such that

⟨x − Tsx

s− x − Trx

r, js,r

⟩≥ 0. (3.6)

It turns out that

‖Tsx − Trx‖2 ≤ r − sr

⟨x − Trx, js,r

⟩

≤∣∣∣∣1 − s

r

∣∣∣∣‖x − Trx‖‖Tsx − Trx‖.(3.7)


This along with the triangle inequality yields that

‖x − Tsx‖ ≤ ‖x − Trx‖ + ‖Trx − Tsx‖

≤ ‖x − Trx‖ +∣∣∣∣1 − s

r

∣∣∣∣‖x − Trx‖

≤ 2‖x − Trx‖.

(3.8)

We notice that though the resolvent of an accretive operator is always firmlynonexpansive in a general Banach space, firm nonexpansiveness is however insufficient toestimate useful bounds which are required to prove convergence of iterative algorithms forsolving nonlinear equations governed by accretive operations. To overcome this difficulty, weneed to impose additional properties on the underlying Banach space X. Lemma 3.3 belowestablishes a sharper estimate than nonexpansiveness of the mapping Tr , which is useful forus to prove the weak and strong convergence of algorithms (1.6) and (1.7).

Lemma 3.3. Let X be a uniformly convex and q-uniformly smooth Banach space for some q ∈ (1, 2].Assume that A is a single-valued α-isa of order q in X. Then, given s > 0, there exists a continuous,strictly increasing and convex function φq : R

+ → R+ with φq(0) = 0 such that, for all x, y ∈ Bs,

∥∥Trx − Try∥∥q ≤ ∥∥x − y∥∥q − r

(αq − rq−1κq

)∥∥Ax −Ay∥∥q

− φq(∥∥(I − Jr)(I − rA)x − (I − Jr)(I − rA)y

∥∥),(3.9)

where κq is the q-uniform smoothness coefficient of X (see Lemma 2.2).

Proof. Put x = x − rAx and y = y − rAy. Since (x − Jrx)/r ∈ B(Jrx), it follows from theaccretiveness of B that

∥∥Jrx − Jry∥∥ ≤

∥∥∥∥(Jrx − Jry

)+r

2

(x − Jrx

r− y − Jry

r

)∥∥∥∥

=∥∥∥∥

12(x − y) + 1

2(Jrx − Jry

)∥∥∥∥.

(3.10)

Since x, y ∈ Bs, by the accretivity of A it is easy to show that there exists t > 0 such thatx − y ∈ Bt; hence, Jrx − Jry ∈ Bt for Jr is nonexpansive. Now since X is uniformly convex, wecan use Lemma 2.2 to find a continuous, strictly increasing and convex function ϕ : R

+ → R+,

with ϕ(0) = 0, satisfying

∥∥∥∥12(x − y) + 1

2(Jrx − Jry

)∥∥∥∥

q

≤ 12∥∥x − y∥∥q + 1

2∥∥Jrx − Jry

∥∥q

−Wq

(12

)ϕ(∥∥(I − Jr)x − (I − Jr)y

∥∥)

≤ ∥∥x − y∥∥q − 12qϕ(∥∥(I − Jr)x − (I − Jr)y

∥∥),

(3.11)


where the last inequality follows from the nonexpansivity of the resolvent Jr . Letting φq =ϕ/2q and combining (3.10) and (3.11) yield

∥∥Trx − Try

∥∥q ≤ ∥

∥x − y∥∥q − φq(∥∥(I − Jr)x − (I − Jr)y

∥∥). (3.12)

On the other hand, since X is also q-uniformly smooth and A is α-isa of order q, wederive that

∥∥x − y∥∥q = ∥

∥(x − y) − r(Ax −Ay)∥∥q

≤ ∥∥x − y∥∥q + κqrq

∥∥Ax −Ay∥∥q − rq⟨Ax −Ay,Jq

(x − y)⟩

≤ ∥∥x − y∥∥q − r

(αq − rq−1κq

)∥∥Ax −Ay∥∥q.

(3.13)

Finally the required inequality (3.9) follows from (3.12) and (3.13).

Remark 3.4. Note that from Lemma 3.3 one deduces that, under the same conditions, if r ≤(αq/κq)

1/(q−1), then the mapping Tr is nonexpansive.

3.1. Weak Convergence

Mann’s iterative method [20] is a widely used method for finding a fixed point ofnonexpansive mappings [21]. We have proved that a splitting method for solving (1.1) can,under certain conditions, be reduced to a method for finding a fixed point of a nonexpansivemapping. It is therefore the purpose of this subsection to introduce and prove its weakconvergence of a Mann-type forward-backward method with errors in a uniformly convexand q-uniformly smooth Banach space. (See [22] for a similar treatment of the proximal pointalgorithm [23, 24] for finding zeros of monotone operators in the Hilbert space setting.) Tothis end we need a lemma about the uniqueness of weak cluster points of a sequence, whoseproof, included here, follows the idea presented in [21, 25].

Lemma 3.5. Let C be a closed convex subset of a uniformly convex Banach space X with a Frechetdifferentiable norm, and let {Tn} be a sequence of nonexpansive self-mappings on C with a nonemptycommon fixed point set F. If x0 ∈ C and xn+1 := Tnxn + en, where

∑∞n=1 ‖en‖ < ∞, then 〈z1 −

z2, J(y1 − y2)〉 = 0 for all y1, y2 ∈ F and all z1, z2 weak limit points of {xn}.

Proof. We first claim that the sequence {xn} is bounded. As a matter of fact, for each fixedp ∈ F and any n ∈ N,

∥∥xn+1 − p∥∥ =

∥∥Tnxn − Tnp + en∥∥

≤ ∥∥xn − p∥∥ + ‖en‖.

(3.14)

As∑∞

n=1 ‖en‖ < ∞, we can apply Lemma 2.5 to find that limn→∞‖xn − p‖ exists. In particular,{xn} is bounded.

Let us next prove that, for every y1, y2 ∈ F and 0 < t < 1, the limit

limn→∞

∥∥txn + (1 − t)y1 − y2∥∥ (3.15)


exists. To see this, we set Sn,m = Tn+m−1Tn+m−2 · · · Tn which is nonexpansive. It is to see that wecan rewrite {xn} in the manner

xn+m = Sn,mxn + cn,m, n,m ≥ 1, (3.16)

where

cn,m = Tn+m−1(Tn+m−2(· · · Tn−1(Tnxn + en) + en−1 · · · ) + en+m−2)

+ en+m−1 − Sn,mxn.(3.17)

By nonexpansivity, we have that

‖cn,m‖ ≤n+m−1∑

k=n

‖ek‖, (3.18)

and the summability of {en} implies that

limn,m→∞

‖cn,m‖ = 0. (3.19)

Set

an =∥∥txn + (1 − t)y1 − y2

∥∥,

dn,m =∥∥Sn,m

(txn + (1 − t)y1

) − (tSn,mxn + (1 − t)y1

)∥∥.(3.20)

Let K be a closed bounded convex subset of X containing {xn} and {y1, y2}. A resultof Bruck [26] assures the existence of a strictly increasing continuous function g : [0,∞) →[0,∞) with g(0) = 0 such that

g(∥∥U

(tx + (1 − t)y) − (

tUx + (1 − t)Uy)∥∥) ≤ ∥∥x − y∥∥ − ∥∥Ux −Uy∥∥ (3.21)

for all U : K → X nonexpansive, x, y ∈ K and 0 ≤ t ≤ 1. Applying (3.21) to each Sn,m, weobtain

g(dn,m) ≤∥∥xn − y1

∥∥ − ∥∥Sn,mxn − Sn,my1∥∥

=∥∥xn − y1

∥∥ − ∥∥xn+m − y1 − cn,m∥∥

≤ ∥∥xn − y1∥∥ − ∥∥xn+m − y1

∥∥ + ‖cn,m‖.(3.22)

Now since limn→∞‖xn − y1‖ exists, (3.19) and (3.22) together imply that

limn,m→∞

dn,m = 0. (3.23)


Furthermore, we have

an+m ≤ an + dn,m + ‖cn,m‖. (3.24)

After taking first lim supm→∞ and then lim infn→∞ in (3.24) and using (3.19) and(3.23), we get

lim supm→∞

am ≤ lim infn→∞

an + limn,m→∞

(dn,m + ‖cn,m‖) = lim infn→∞

an. (3.25)

Hence the limit (3.15) exists.If we replace now x and u in (2.12) with y1 − y2 and t(xn − y1), respectively, we arrive

at

12∥∥y1 − y2

∥∥2 + t⟨xn − y1,J

(y1 − y2

)⟩

≤ 12∥∥txn + (1 − t)y1 − y2

∥∥2

≤ 12∥∥y1 − y2

∥∥2 + t⟨xn − y1,J

(y1 − y2

)⟩+ h

(t∥∥xn − y1

∥∥).

(3.26)

Since the limn→∞‖xn − y1‖ exists, we deduce that

12∥∥y1 − y2

∥∥2 + t lim supn→∞

⟨xn − y1,J

(y1 − y2

)⟩

≤ limn→∞

12∥∥txn + (1 − t)y1 − y2

∥∥2

≤ 12∥∥y1 − y2

∥∥2 + t lim infn→∞

⟨xn − y1,J

(y1 − y2

)⟩+ o(t),

(3.27)

where limt→ 0 o(t)/t = 0. Consequently, we deduce that

lim supn→∞

⟨xn − y1,J

(y1 − y2

)⟩ ≤ lim infn→∞

⟨xn − y1,J

(y1 − y2

)⟩+o(t)t. (3.28)

Setting t tend to 0, we conclude that limn→∞〈xn − y1,J(y1 − y2)〉 exists. Therefore, for anytwo weak limit points z1 and z2 of {xn}, 〈z1 − y1,J(y1 − y2)〉 = 〈z2 − y1,J(y1 − y2)〉; that is,〈z1 − z2,J(y1 − y2)〉 = 0.

Theorem 3.6. Let X be a uniformly convex and q-uniformly smooth Banach space. Let A : X → Xbe an α-isa of order q and B : X → 2X an m-accretive operator. Assume that S = (A + B)−1(0)/= ∅.We define a sequence {xn} by the perturbed iterative scheme

xn+1 = (1 − αn)xn + αn(Jrn(xn − rn(Axn + an)) + bn), (3.29)


where Jrn = (I + rnB)−1, {an}, {bn} ⊂ X, {αn} ⊂ (0, 1], and {rn} ⊂ (0,+∞). Assume that

(i)∑∞

n=1 ‖an‖ <∞ and∑∞

n=1 ‖bn‖ <∞;

(ii) 0 < lim infn→∞αn;

(iii) 0 < lim infn→∞rn ≤ lim supn→∞rn < (αq/κq)1/(q−1).

Then {xn} converges weakly to some x ∈ S.

Proof. Write Tn = (I + rnB)−1(I − rnA). Notice that we can write

Jrn(xn − rn(Axn + an)) + bn = Tnxn + en, (3.30)

where en = Jrn(xn − rn(Axn + an)) + bn − Tnxn. Then the iterative formula (3.29) turns into theform

xn+1 = (1 − αn)xn + αn(Tnxn + en). (3.31)

Thus, by nonexpansivity of Jrn ,

‖en‖ ≤ ‖Jrn(xn − rn(Axn + an)) − Tnxn‖‖bn‖≤ rn‖an‖ + ‖bn‖.

(3.32)

Therefore, condition (i) implies

∞∑

n=1

‖en‖ <∞. (3.33)

Take z ∈ S to deduce that, as S = Fix(Tn) and Tn is nonexpansive,

‖xn+1 − z‖ ≤ (1 − αn)‖xn − z‖ + αn‖Tnxn − Tnz + en‖≤ ‖xn − z‖ + αn‖en‖.

(3.34)

Due to (3.33), Lemma 2.5 is applicable and we get that limn→∞‖xn − z‖ exists; inparticular, {xn} is bounded. Let M > 0 be such that ‖xn‖ < M, for all n ∈ N, and lets = q(M + ‖z‖)q−1. By (2.7) and Lemma 3.3, we have

‖xn+1 − z‖q ≤ ‖(1 − αn)(xn − z) + αn(Tnxn − z) + αnen‖q

≤ ‖(1 − αn)(xn − z) + αn(Tnxn − z)‖q + αn〈en,J(xn+1 − z)〉

≤ (1 − αn)‖xn − z‖q + αn‖Tnxn − Tnz‖q + αnq‖en‖‖xn+1 − z‖q−1

≤ ‖xn − z‖q − αnrn(qα − rq−1

n κq)‖Axn −Az‖q

− φq(‖xn − rnAxn − Tnxn + rnAz‖) + s‖en‖.

(3.35)


From (3.35), assumptions (ii) and (iii), and (3.33), it turns out that

limn→∞

‖Axn −Az‖q + ‖xn − rnAxn − Tnxn + rnAz‖ = 0. (3.36)

Consequently,

limn→∞

‖Tnxn − xn‖ = 0. (3.37)

Since lim infn→∞rn > 0, there exists ε > 0 such that rn ≥ ε for all n ≥ 0. Then, by Lemma 3.2,

limn→∞

‖Tεxn − xn‖ ≤ 2 limn→∞

‖Tnxn − xn‖ = 0. (3.38)

By Lemmas 3.3 and 3.1, Tε is nonexpansive and Fix(Tε) = S/= ∅. We can therefore makeuse of Lemma 2.3 to assure that

ωw(xn) ⊂ S. (3.39)

Finally we set Un = (1 − αn)I + αnTn and rewrite scheme (3.31) as

xn+1 = Unxn + e′n, (3.40)

where the sequence {e′n} satisfies∑∞

n=1 ‖e′n‖ < ∞. Since {Un} is a sequence of nonexpansivemappings with S as its nonempty common fixed point set, and since the space X is uniformlyconvex with a Frechet differentiable norm, we can apply Lemma 3.5 together with (3.39)to assert that the sequence {xn} has exactly one weak limit point; it is therefore weaklyconvergent.

3.2. Strong Convergence

Halpern’s method [27] is another iterative method for finding a fixed point of nonexpansivemappings. This method has been extensively studied in the literature [28–30] (see also therecent survey [31]). In this section we aim to introduce and prove the strong convergence ofa Halpern-type forward-backward method with errors in uniformly convex and q-uniformlysmooth Banach spaces. This result turns out to be new even in the setting of Hilbert spaces.

Theorem 3.7. Let X be a uniformly convex and q-uniformly smooth Banach space. Let A : X → Xbe an α-isa of order q and B : X → 2X an m-accretive operator. Assume that S = (A + B)−1(0)/= ∅.We define a sequence {xn} by the iterative scheme

xn+1 = αnu + (1 − αn)(Jrn(xn − rn(Axn + an)) + bn), (3.41)


where u ∈ X, Jrn = (I + rnB)−1, {an}, {bn} ⊂ X, {αn} ⊂ (0, 1], and {rn} ⊂ (0,+∞). Assume the

following conditions are satisfied:

(i)∑∞

n=1 ‖an‖ <∞and∑∞

n=1 ‖bn‖ <∞, or limn→∞‖an‖/αn = limn→∞‖bn‖/αn = 0;

(ii) limn→∞αn = 0,∑∞

n=1 αn = ∞;

(iii) 0 < lim infn→∞rn ≤ lim supn→∞rn < (αq/κq)1/(q−1).

Then {xn} converges in norm to z = Q(u), where Q is the sunny nonexpansive retraction ofX onto S.

Proof. Let z = Q(u), where Q is the sunny nonexpansive retraction of X onto S whoseexistence is ensured by Theorem 2.4. Let (yn) be a sequence generated by

yn+1 = αnu + (1 − αn)Tnyn, (3.42)

where we abbreviate Tn := Jrn(I − rnA). Hence to show the desired result, it suffices to provethat yn → z. Indeed, since Jrn and I − rnA are both nonexpansive, it follows that

∥∥xn+1 − yn+1∥∥ ≤ (1 − αn)

∥∥Jrn(xn − rn(Axn + an)) + bn − Jrn(yn − rnAyn

)∥∥

≤ (1 − αn)∥∥(I − rnA)xn − (I − rnA)yn − rnan

∥∥ + ‖bn‖= (1 − αn)

∥∥xn − yn∥∥ + L(‖an‖ + ‖bn‖),

(3.43)

where L := max(1, (αq/κq)1/(q−1)). According to condition (i), we can apply Lemma 2.5(ii) to

conclude that ‖xn − yn‖ → 0 as n → ∞.

We next show yn → z. Indeed, since S = Fix(Tn) and Tn is nonexpansive, we have

∥∥yn+1 − z∥∥ ≤ αn‖u − z‖ + (1 − αn)

∥∥Tnyn − Tnz∥∥

≤ αn‖u − z‖ + (1 − αn)∥∥yn − z

∥∥.(3.44)

Hence, we can apply Lemma 2.5(i) to claim that {yn} is bounded.Using the inequality (2.7) with p = q, we derive that

∥∥yn+1 − z∥∥q =

∥∥αn(u − z) + (1 − αn)Tnyn − z∥∥q

≤ (1 − αn)q∥∥Tnyn − z

∥∥q + qαn⟨u − z,Jq

(yn+1 − z

)⟩.

(3.45)

By condition (iii), we have some δ > 0 such that

1 − αn ≥ δ, (1 − αn)rn(αq − rq−1

n κq)≥ δ, (3.46)


for all n ∈ N. Hence, by Lemma 3.3 we get from (3.45) that

∥∥yn+1 − z

∥∥q ≤ (1 − αn)

∥∥yn − z

∥∥q − δφq

(∥∥yn − rnAyn − Tnyn + rnAz∥∥)

− δ∥∥Ayn −Az∥∥q + qαn

⟨u − z,Jq

(yn+1 − z

)⟩.

(3.47)

Let us define sn = ‖yn − z‖q for all n ≥ 0. Depending on the asymptotic behavior of thesequence {sn} we distinguish two cases.

Case 1. Suppose that there exists N ∈ N such that the sequence {sn}n≥N is nonincreasing;thus, limn→∞sn exists. Since αn → 0 and ‖en‖ → 0, it follows immediately from (3.47) that

limn→∞

∥∥Ayn −Az

∥∥q +

∥∥yn − rnAyn − Tnyn + rnAz

∥∥ = 0. (3.48)

Consequently,

limn→∞

∥∥Tnyn − yn∥∥ = 0. (3.49)

By condition (iii), there exists ε > 0 such that rn ≥ ε for all n ≥ 0. Then, by Lemma 3.2,we get

limn→∞

∥∥Tεyn − yn∥∥ ≤ lim

n→∞∥∥Tnyn − yn

∥∥ = 0. (3.50)

The demiclosedness principle (i.e., Lemma 2.3) implies that

ωw

(yn

) ⊂ S. (3.51)

Note that from inequality (3.47) we deduce that

sn+1 ≤ (1 − αn)sn + qαn⟨u − z,Jq

(yn+1 − z

)⟩. (3.52)

Next we prove that

lim supn→∞

⟨u − z,Jq

(yn − z

)⟩ ≤ 0. (3.53)

Equivalently (should ‖yn − z‖ � 0), we need to prove that

lim supn→∞

⟨u − z,J(

yn − z)⟩ ≤ 0. (3.54)


To this end, let zt satisfy zt = tu + Tεzt. By Reich’s theorem [32], we get zt → QSu = zas t → 0. Using subdifferential inequality, we deduce that

∥∥zt − yn

∥∥2 =

∥∥t(u − yn

)+ (1 − t)(Tεzt − yn

)∥∥2

≤ (1 − t)2∥∥Tεzt − yn∥∥2 + 2t

⟨u − yn,J

(zt − yn

)⟩

≤ (1 − t)2(∥∥Tεzt − Tεyn∥∥ +

∥∥Tεyn − yn

∥∥)2 + 2t

∥∥zt − yn

∥∥2 + 2t

⟨u − zt,J

(zt − yn

)⟩

≤(

1 + t2)∥∥zt − yn

∥∥2 +M

∥∥Tεyn − yn

∥∥ + 2t

⟨u − zt,J

(zt − yn

)⟩,

(3.55)

where M > 0 is a constant such that

M > max{∥∥zt − yn

∥∥2, 2∥∥zt − yn

∥∥ +∥∥Tεyn − yn

∥∥}, t ∈ (0, 1), n ∈ N. (3.56)

Then it follows from (3.55) that

⟨u − zt,J

(yn − zt

)⟩ ≤ M

2t +

M

2t∥∥Tεyn − yn

∥∥. (3.57)

Taking lim supn→∞ yields

lim supn→∞

⟨u − zt,J

(yn − zt

)⟩ ≤ M

2t. (3.58)

Then, letting t → 0 and noting the fact that the duality map J is norm-to-normuniformly continuous on bounded sets, we get (3.54) as desired. Due to (3.53), we can applyLemma 2.5(ii) to (3.52) to conclude that sn → 0; that is, yn → z.

Case 2. Suppose that there exists n1 ∈ N such that sn1 ≤ sn1+1. Let us define

In := {n1 ≤ k ≤ n : sk ≤ sk+1}, n ≥ n1. (3.59)

Obviously In /= ∅ since n1 ∈ In for any n ≥ n1. Set

τ(n) = max In. (3.60)

Note that the sequence {τ(n)} is nonincreasing and limn→∞τ(n) = ∞. Moreover, τ(n) ≤ nand

sτ(n) ≤ sτ(n)+1, (3.61)

sn ≤ sτ(n)+1, (3.62)


for any n ≥ n1 (see Lemma 3.1 of Mainge [33] for more details). From inequality (3.47) weget

sτ(n)+1 ≤ (1 − ατ(n)

)sτ(n) − δφ

(∥∥yτ(n) − rτ(n)Ayτ(n) − Tτ(n)yτ(n) + rτ(n)Az∥∥)

− δ∥∥Ayτ(n) −Az∥∥q + qατ(n)

⟨u − z,Jq

(yτ(n)+1 − z

)⟩.

(3.63)

It turns out that

limn→∞

∥∥Ayτ(n) −Az

∥∥ = 0, lim

n→∞∥∥yτ(n) − rτ(n)Ayτ(n) − Tτ(n)yτ(n) + rτ(n)Az

∥∥ = 0. (3.64)

Consequently,

limn→∞

∥∥Tτ(n)yτ(n) − yτ(n)

∥∥ = 0. (3.65)

Now repeating the argument of the proof of (3.53) in Case 1, we can get

lim supn→∞

⟨u − z,Jq

(yτ(n) − z

)⟩ ≤ 0. (3.66)

By the asymptotic regularity of {yτ(n)} and (3.65), we deduce that

limn→∞

∥∥yτ(n)+1 − yτ(n)∥∥ = 0. (3.67)

This implies that

lim supn→∞

⟨u − z,J(

yτ(n)+1 − z)⟩ ≤ 0. (3.68)

On the other hand, it follows from (3.64) that

sτ(n)+1 − sτ(n) + ατ(n)sτ(n) ≤ qατ(n)⟨u − z,Jq

(yτ(n)+1 − z

)⟩. (3.69)

Taking the lim supn→∞ in (3.69) and using condition (i) we deduce thatlim supn→∞sτ(n) ≤ 0; hence limn→∞sτ(n) = 0. That is, ‖yτ(n) − z‖ → 0. Using the triangleinequality,

∥∥yτ(n)+1 − z∥∥ ≤ ∥∥yτ(n)+1 − yτ(n)

∥∥ +∥∥yτ(n) − z

∥∥, (3.70)

we also get that limn→∞sτ(n)+1 = 0 which together with (3.42) guarantees that ‖yn − z‖ → 0.

4. Applications

The two forward-backward methods previously studied, (3.29) and (3.41), find applicationsin other related problems such as variational inequalities, the convex feasibility problem,fixed point problems, and optimization problems.


Throughout this section, let C be a nonempty closed and convex subset of a Hilbertspace H. Note that in this case the concept of monotonicity coincides with the concept ofaccretivity.

Regarding the problem we concern, of finding a zero of the sum of two accretiveoperators in a Hilbert space H, as a direct consequence of Theorem 3.7, we first obtain thefollowing result due to Combettes [34].

Corollary 4.1. Let A : H → H be monotone and B : H → H maximal monotone. Assume thatκA is firmly nonexpansive for some κ > 0 and that

(i) limn→∞αn > 0,

(ii) 0 < lim infn→∞λn ≤ lim supn→∞λn < 2κ,

(iii)∑∞

n=1 ‖an‖ <∞ and∑∞

n=1 ‖bn‖ <∞,

(iv) S := (A + B)−1(0)/= ∅.

Then the sequence {xn} generated by the algorithm

xn+1 = (1 − αn)xn + αnJλn((xn − λn(Axn + an)) + bn) (4.1)

converges weakly to a point in S.

Proof. It suffices to show that κA is firmly nonexpansive if and only if A is κ-inverse stronglymonotone. This however follows from the following straightforward observation:

⟨κAx − κAy, x − y⟩ ≥ ∥∥κAx − κAy∥∥2 ⇐⇒ ⟨

Ax −Ay, x − y⟩ ≥ κ∥∥Ax −Ay∥∥2, (4.2)

for all x, y ∈ H.

4.1. Variational Inequality Problems

A monotone variational inequality problem (VIP) is formulated as the problem of finding apoint x ∈ C with the property:

〈Ax, z − x〉 ≥ 0, ∀z ∈ C, (4.3)

where A : C → H is a nonlinear monotone operator. We shall denote by S the solution set of(4.3) and assume S/= ∅.

One method for solving VIP (4.3) is the projection algorithm which generates, startingwith an arbitrary initial point x0 ∈ H, a sequence {xn} satisfying

xn+1 = PC(xn − rAxn), (4.4)

where r is properly chosen as a stepsize. If in addition A is κ-inverse strongly monotone(ism), then the iteration (4.4) with 0 < r < 2κ converges weakly to a point in S wheneversuch a point exists.


By [35, Theorem 3], VIP (4.3) is equivalent to finding a point x so that

0 ∈ (A + B)x, (4.5)

where B is the normal cone operator of C. In other words, VIPs are a special case of theproblem of finding zeros of the sum of two monotone operators. Note that the resolvent ofthe normal cone is nothing but the projection operator and that if A is κ-ism, then the set Ωis closed and convex [36]. As an application of the previous sections, we get the followingresults.

Corollary 4.2. LetA : C → H be κ-ism for some κ > 0, and let the following conditions be satisfied:


(ii) 0 < lim infn→∞λn ≤ lim supn→∞λn < 2κ.

Then the sequence {xn} generated by the relaxed projection algorithm

xn+1 = (1 − αn)xn + αnPC(xn − λnAxn) (4.6)

converges weakly to a point in S.

Corollary 4.3. Let A : C → H be κ-ism and let the following conditions be satisfied:

(i) limn→∞αn = 0,∑∞

n=1 αn = ∞;

(ii) 0 < limn→∞λn ≤ lim supn→∞λn < 2κ.

Then, for any given u ∈ C, the sequence {xn} generated by

xn+1 = αnu + (1 − αn)PC(xn − λnAxn), (4.7)

converges strongly to PSu.

Remark 4.4. Corollary 4.3 improves Iiduka-Takahashi’s result [37, Corollary 3.2], where apartfrom hypotheses (i)-(ii), the conditions

∑∞n=1 |αn − αn+1| < ∞ and

∑∞n=1 |λn − λn+1| < ∞ are

required.

4.2. Fixed Points of Strict Pseudocontractions

An operator T : C → C is said to be a strict κ-pseudocontraction if there exists a constantκ ∈ [0, 1) such that

∥∥Tx − Ty∥∥2 ≤ ∥∥x − y∥∥2 + κ∥∥(I − T)x − (I − T)y∥∥2 (4.8)

for all x, y ∈ C. It is known that if T is strictly κ-pseudocontractive, then A = I −T is (1−κ)/2-ism (see [38]). To solve the problem of approximating fixed points for such operators, aniterative scheme is provided in the following result.


Corollary 4.5. Let T : C → C be strictly κ-pseudocontractive with a nonempty fixed point setFix(T). Suppose that

(i) limn→∞αn = 0 and∑∞

n=1 αn = ∞,

(ii) 0 < limn→∞λn ≤ lim supn→∞λn < 1 − κ.Then, for any given u ∈ C, the sequence {xn} generated by the algorithm

xn+1 = αnu + (1 − αn)((1 − λn)xn + λnTxn) (4.9)

converges strongly to the point PFix(T)u.

Proof. Set A = I − T . Hence A is (1 − κ)/2-ism. Moreover we rewrite the above iteration as

xn+1 = αnu + (1 − αn)(xn − λnAxn). (4.10)

Then, by setting B the operator constantly zero, Corollary 4.3 yields the result asdesired.

4.3. Convexly Constrained Minimization Problem

Consider the optimization problem

minx∈C

f(x), (4.11)

where f : H → R is a convex and differentiable function. Assume (4.11) is consistent, andlet Ω denote its set of solutions.

The gradient projection algorithm (GPA) generates a sequence {xn} via the iterativeprocedure:

xn+1 = PC(xn − r∇f(xn)

), (4.12)

where ∇f stands for the gradient of f . If in addition ∇f is (1/κ)-Lipschitz continuous; thatis, for any x, y ∈ H,

∥∥∇f(x) − ∇f(y)∥∥ ≤(

1κ

)∥∥x − y∥∥, (4.13)

then the GPA with 0 < r < 2κ converges weakly to a minimizer of f in C (see, e.g, [39,Corollary 4.1]).

The minimization problem (4.11) is equivalent to VIP [40, Lemma 5.13]:

⟨∇f(x), z − x⟩ ≥ 0, z ∈ C. (4.14)

It is also known [41, Corollary 10] that if ∇f is (1/κ)-Lipschitz continuous, then it is alsoκ-ism. Thus, we can apply the previous results to (4.11) by taking A = ∇f .


Corollary 4.6. Assume that f : H → R is convex and differentiable with (1/κ)-Lipschitzcontinuous gradient ∇f . Assume also that


(ii) 0 ≤ lim infn→∞λn ≤ lim supn→∞λn ≤ 2κ.

Then the sequence {xn} generated by the algorithm

xn+1 = (1 − αn)xn + αnPC(xn − λn∇f(xn)

)(4.15)

converges weakly to x ∈ Ω.

Corollary 4.7. Assume that f : H → R is convex and differentiable with (1/κ)-Lipschitzcontinuous gradient ∇f . Assume also that


n=1 αn = ∞;

(ii) 0 < limn→∞λn ≤ lim supn→∞λn < 2κ.

Then for any given u ∈ C, the sequence {xn} generated by the algorithm

xn+1 = αnu + (1 − αn)PC(xn − λn∇f(xn)

)(4.16)

converges strongly to PΩu whenever such point exists.

4.4. Split Feasibility Problem

The split feasibility problem (SFP) [42] consists of finding a point x satisfying the property:

x ∈ C, Ax ∈ Q, (4.17)

where C and Q are, respectively, closed convex subsets of Hilbert spaces H and K andA : H → K is a bounded linear operator. The SFP (4.17) has attracted much attentiondue to its applications in signal processing [42]. Various algorithms have, therefore, beenderived to solve the SFP (4.17) (see [39, 43, 44] and reference therein). In particular, Byrne[43] introduced the so-called CQ algorithm:

xn+1 = PC(xn − λA∗(I − PQ

)Axn

), (4.18)

where 0 < λ < 2ν with ν = 1/‖A‖2.To solve the SFP (4.17), it is very useful to investigate the following convexly

constrained minimization problem (CCMP):

minx∈C

f(x), (4.19)

where

f(x) :=12∥∥(I − PQ

)Ax

∥∥2. (4.20)


Generally speaking, the SFP (4.17) and CCMP (4.19) are not fully equivalent: everysolution to the SFP (4.17) is evidently a minimizer of the CCMP (4.19); however a solution tothe CCMP (4.19) does not necessarily satisfy the SFP (4.17). Further, if the solution set of theSFP (4.17) is nonempty, then it follows from [45, Lemma 4.2] that

C ∩ (∇f)−1(0)/= ∅, (4.21)

where f is defined by (4.20). As shown by Xu [46], the CQ algorithm need not convergestrongly in infinite-dimensional spaces. We now consider an iteration process with strongconvergence for solving the SFP (4.17).

Corollary 4.8. Assume that the SFP (4.17) is consistent, and let S be its nonempty solution set.Assume also that


n=1 αn = ∞;

(ii) 0 < limn→∞λn ≤ lim supn→∞λn < 2ν.

Then for any given u ∈ C, the sequence (xn) generated by the algorithm

xn+1 = αnu + (1 − αn)PC[xn − λnA∗(I − PQ

)Axn

](4.22)

converges strongly to the solution PSu of the SFP (4.17).

Proof. Let f be defined by (4.19). According to [39, page 113], we have

∇f = A∗(I − PQ)A, (4.23)

which is (1/ν)-Lipschitz continuous with ν = 1/‖A‖2. Thus Corollary 4.7 applies, and theresult follows immediately.Remark 4.9. Corollary 4.8 improves and recovers the result of [44, Corollary 3.7], which usesthe additional condition

∑∞n=1 |αn+1 − αn| < ∞, condition (i), and the special case of condition

(ii) where λn ≡ λ for all n ∈ N.

4.5. Convexly Constrained Linear Inverse Problem

The constrained linear system

Ax = b

x ∈ C,(4.24)

where A : H → K is a bounded linear operator and b ∈ K, is called convexly constrainedlinear inverse problem (cf. [47]). A classical way to deal with this problem is the well-knownprojected Landweber method (see [40]):

xn+1 = PC[xn − λA∗(Axn − b)], (4.25)


where 0 < λ < 2ν with ν = 1/‖A‖2. A counterexample in [8, Remark 5.12] shows that theprojected Landweber iteration converges weakly in infinite-dimensional spaces, in general.To get strong convergence, Eicke introduced the so-called damped projection method (see[47]). In what follows, we present another algorithm with strong convergence, for solving(4.24).

Corollary 4.10. Assume that (4.24) is consistent. Assume also that


n=1 αn = ∞;

(ii) 0 < limn→∞λn ≤ lim supn→∞λn < 2ν.

Then, for any given u ∈ H, the sequence {xn} generated by the algorithm

xn+1 = αnu + (1 − αn)PC[xn − λnA∗(Axn − b)] (4.26)

converges strongly to a solution to problem (4.24) whenever it exists.

Proof. This is an immediate consequence of Corollary 4.8 by taking Q = {b}.

Acknowledgments

The work of G. Lopez, V. Martın-Marquez, and H.-K. Xu was supported by Grant MTM2009-10696-C02-01. This work was carried out while F. Wang was visiting Universidad de Sevillaunder the support of this grant. He was also supported by the Basic and Frontier Projectof Henan 122300410268 and the Peiyu Project of Luoyang Normal University 2011-PYJJ-002.The work of G. Lopez and V. Martın-Marquez was also supported by the Plan Andaluz deInvestigacin de la Junta de Andaluca FQM-127 and Grant P08-FQM-03543. The work of H.-K.Xu was also supported in part by NSC 100-2115-M-110-003-MY2 (Taiwan). He extended hisappreciation to the Deanship of Scientific Research at King Saud University for funding thework through a visiting professor-ship program (VPP).

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Research ArticleBlow-Up Analysis for a Quasilinear DegenerateParabolic Equation with Strongly Nonlinear Source

Pan Zheng, Chunlai Mu, Dengming Liu,Xianzhong Yao, and Shouming Zhou

School of Mathematics and Statistics, Chongqing University, Chongqing 401331, China

Correspondence should be addressed to Pan Zheng, [email protected]

Received 8 January 2012; Accepted 26 April 2012


Copyright q 2012 Pan Zheng et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We investigate the blow-up properties of the positive solution of the Cauchy problem for aquasilinear degenerate parabolic equation with strongly nonlinear source ut = div(|∇um|p−2∇ul) +uq, (x, t) ∈ RN × (0, T), where N ≥ 1, p > 2 , and m, l, q > 1, and give a secondary critical exponenton the decay asymptotic behavior of an initial value at infinity for the existence and nonexistenceof global solutions of the Cauchy problem. Moreover, under some suitable conditions we provesingle-point blow-up for a large class of radial decreasing solutions.

1. Introduction

In this paper, we consider the following Cauchy problem to a quasilinear degenerateparabolic equation with strongly nonlinear source

ut = div(|∇um|p−2∇ul

)+ uq, (x, t) ∈ RN × (0, T),

u(x, 0) = u0(x), x ∈ RN,

(1.1)

where N ≥ 1, p > 2, m , l , q > 1, and the initial data u0(x) is nonnegative bounded andcontinuous.

Equation (1.1) has been suggested as a mathematical model for a variety of physicalproblems (see [1]). For instance, it appears in the non-Newtonian fluids and is a nonlinearform of heat equation. Moreover, it can also be used to model the nonlinear heat propagationin a reaction medium (see [2]).


One of the particular features of problem (1.1) is that the equation is degenerate atpoints where u = 0 or ∇u = 0. Hence, there is no classical solution in general and we introducethe following definition of weak solution (see [3, 4]).

Definition 1.1. A nonnegative measurable function u(x, t) defined in RN × (0, T) is called aweak solution of the Cauchy problem (1.1) if for every bounded open set Ω with smoothboundary ∂Ω, u ∈ Cloc(Ω × (0, T)), um, ul ∈ Lploc(0, T ;W1,p(Ω)), and

∫

Ωuϕdx +

∫ t

t0

∫

Ω

(−uϕt + |∇um|p−2∇ul · ∇ϕ

)dx dt =

∫ t

t0

∫

Ωuqϕdx dt +

∫

Ωu(x, t0)ϕ(x, t0)dx

(1.2)

for all 0 ≤ t0 ≤ t ≤ T and all test functions ϕ ∈ C10(Ω × (0, T)). Moreover,

limt→ t0

∫

Ωu(x, t)η(x)dx =

∫

Ωu(x, t0)η(x)dx (1.3)

for any η(x) ∈ C10(Ω).

Under some suitable assumptions, the existence, uniqueness and regularity of a weaksolution to the Cauchy problem (1.1) and their variants have been extensively investigatedby many authors (see [5–7] and the references therein).

The first goal of this paper is to study the blow-up behavior of solution u(x, t) of (1.1)when the initial data u0(x) has slow decay near x = ∞. For instance, in the following case

u0(x) ∼=M|x|−a with M > 0, a ≥ 0, (1.4)

we investigate the existence of global and nonglobal solutions for the Cauchy problem (1.1)in terms of M and a. In recent years, many authors have studied the properties of solutionsto the Cauchy problem (1.1) and their variants (see [8–17] and the references therein). Inparticular, J.-S. Guo and Y. Y. Guo [18] obtained the secondary critical exponent for thefollowing porous medium type equation in high dimensions:

ut = Δum + up, (x, t) ∈ RN × (0, T),

u(x, 0) = u0(x), x ∈ RN,(1.5)

where p > 1, m > 1 or max{0, 1 − (2/N)} < m < 1, u0(x) is nonnegative bounded andcontinuous, and proved that for p > p∗m = m+(2/N), there exists a secondary critical exponenta∗ = 2/(p −m) such that the solution u(x, t) of (1.5) blows up in finite time for the initial datau0(x), which behaves like |x|−a at x = ∞ if a ∈ (0, a∗), and there exists a global solution forthe initial data u0(x), which behaves like |x|−a at x = ∞ if a ∈ (a∗,N). Here, we say that thesolution blows up in finite time; it means that there exists T ∈ (0,+∞) such that ‖u(·, t)‖L∞ <∞for all t ∈ [0, T), but limt→ T− sup ‖u(., t)‖L∞ = ∞.


Mu et al. [19] studied the secondary critical exponent for the following p-Laplacianequation with slow decay initial values:

ut = div(|∇u|p−2∇u

)+ uq, (x, t) ∈ RN × (0, T),

u(x, 0) = u0(x), x ∈ RN,(1.6)

where p > 2, q > 1, and showed that, for q > q∗c = p − 1 + (p/N), there exists a secondarycritical exponent a∗c = (p/(q + 1 − p)) such that the solution u(x, t) of (1.6) blows up in finitetime for the initial data u0(x) which behaves like |x|−a at x = ∞ if a ∈ (0, a∗c), and there existsa global solution for the initial data u0(x), which behaves like |x|−a at x = ∞ if a ∈ (a∗c,N).

Recently, Mu et al. [20] also investigated the secondary critical exponent for the doublydegenerate parabolic equation with slow decay initial values and obtained similar results.

On the other hand, in this paper, we will also consider single-point blow-up for theCauchy problem (1.1). It is interesting to study the set of blow-up points and the behavior ofthe solution u(x, t) at the blow-up point.

In order to investigate single-point blow-up for the Cauchy problem (1.1), weintroduce the concept of the blow-up point.

Definition 1.2. A point x ∈ Ω is called a blow-up point if there exists a sequence (xn, tn) suchthat xn → x, tn → T− and u(xn, tn) → ∞ as n → ∞, where T is blow-up time.

In recent years, some authors also studied single-point blow-up for the Cauchyproblem to nonlinear parabolic equations (see [21, 22] and the references therein) by differentmethods. In particular, when p = 2, l = 1 and N = 1, the Cauchy problem (1.1) has beeninvestigated by Weissler in [23], and the author obtained that the solution blows up only at asingle point. Galaktionov and Posashkov [24] studied the single-point blow-up and gave theupper and lower bound near the blow-up point for the Cauchy problem (1.1) when p > 2 andm = l = 1. Recently, when p > 2 and m = l, Mu and Zeng [25] extended Galaktionov’s resultsto the doubly degenerate parabolic equation. For more works about single-point blow-up, werefer to [26, 27], where the parabolic systems have been considered.

Motivated by the above works, based on a modification of the energy methods,comparison principle, and regularization methods used in [15, 19, 21, 24], we investigate thesecondary critical exponent and single-point blow-up for the Cauchy problem (1.1). Beforestating the results of the secondary critical exponent, we start with some notations as follows.

Let Cb(RN) be the space of all bounded continuous functions in RN . For a ≥ 0, wedefine

Φa ={φ(x) ∈ Cb

(RN)| φ(x) ≥ 0, lim

|x|→∞sup |x|aφ(x) <∞

},

Φa ={φ(x) ∈ Cb

(RN)| φ(x) ≥ 0, lim

|x|→∞inf |x|aφ(x) > 0

}.

(1.7)

Moreover, we denote

q∗c = l +m(p − 2

)+p

N, a∗c =

p

q − l −m(p − 2) . (1.8)

Our main results of this paper are stated as follows.


Theorem 1.3. For N ≥ 2, p > 2, m > 1, l > 1, and q > q∗c = l +m(p − 2) + (p/N), suppose thatu0(x) ∈ Φa for some a ∈ (0, a∗c); then the solution u(x, t) of the Cauchy problem (1.1) blows up infinite time.

Theorem 1.4. For N ≥ 2, p > 2, m > 1, l > 1, and q > q∗c = l +m(p − 2) + (p/N), suppose thatu0(x) = λϕ(x) for some λ > 0 and ϕ(x) ∈ Φa for some a ∈ (a∗c,N); then there is λ0 = λ0(ϕ) > 0such that the solution u(x, t) of the Cauchy problem (1.1) exists globally for all t > 0, and if λ < λ0,one has

||u(x, t)||∞ ≤ Ct−aβ, ∀t > 0, (1.9)

where β = 1/(a(l +m(p − 2) − 1) + p), C > 0.

Remark 1.5. When p > 2, N ≥ 2 and q > q∗c , we have q∗c > 1 and 0 < a∗c < N.

Remark 1.6. It follows from Theorems 1.3 and 1.4 that the number a∗c = p/(q − l −m(p − 2))gives another cut-off between the blow-up case and the global existence case. Therefore, thenumber a∗c is a new secondary critical exponent of the Cauchy problem (1.1). Unfortunately,in the critical case a = a∗c, we do not know whether the solution of (1.1) exists globally orblows up in finite time.

Remark 1.7. When m = l = 1 or m = l > 1, the results of Theorems 1.3 and 1.4 are consistentwith those in [19, 20], respectively.

Remark 1.8. In [28], Afanas’eva and Tedeev also established the Fujita type results for (1.1)with m = l. In particular, if u0(x) ∼ |x|−a, 0 < a < N, they obtained that if q < m(p− 1) + (p/a),then every nontrivial solution blows up in finite time, and if q > m(p − 1) + (p/a), then thesolution exists globally for a small initial data u0(x). We note that when m = l in (1.1), ifq > m(p − 1) + (p/N) and 0 < a < p/(q −m(p − 1)), then 0 < a < N and q < m(p − 1) + (p/a),while if q > m(p − 1) + (p/N) and p/(q − m(p − 1)) < a < N, then q > m(p − 1) + (p/a).Therefore, the results of Theorems 1.3 and 1.4 coincide with those in [28].

Finally, we also consider single-point blow-up for a large number of radial decreasingsolutions of the Cauchy problem (1.1) and give upper bound of the radial solution u(r, t) ina small neighborhood of the point (x, t), where x = 0, t = T . We assume that the initial datau0(x) = u0(r) satisfies the following condition:

(H) u0(x) = u0(r) ≥ 0 for r > 0, u0(0) > 0, and u0(r) ∈ C1(R1+), u

′0(0) = 0, and u′0(r) ≤ 0

for r > 0, M0 = sup u0(r) < +∞, K0 = sup |u′0(r)| < +∞.

Theorem 1.9. LetN ≥ 1, p > 2, m > 1, l > 1, and q > l +m(p − 2), and let condition (H) hold. Inaddition, assume that the initial function u0(x) = u0(r) satisfies

uq

0(r) · rN = o(1) as r −→ ∞, (1.10)

λ0 = infr>0, u0(r)>0

⎧⎪⎨

⎪⎩−

∣∣∣(um0)′∣∣∣

p−2(ul0

)′

ruq

0

⎫⎪⎬

⎪⎭∈(

0,p − 2

(p − 2

)(N + 1) + 2

]

. (1.11)


Let T be the blow-up time; then one has

u(r, t) ≤ Cr−p/(q−l−m(p−2)), (r, t) ∈ R1+ × (0, T), (1.12)

where

C =

[q − l −m(p − 2

)

(lmp−2)1/(p−1)pλ

1/(p−1)0

]−((p−1)/(q−l−m(p−2)))

> 0, (1.13)

that is, there is single-point blow-up at point x = 0.

Remark 1.10. By (1.11), the best upper estimate (1.12) obtained by our method has thefollowing form:

u(r, t) ≤⎡

⎣q − l −m(p − 2

)

p(lmp−2

)1/(p−1)

(p − 2

(p − 2

)(N + 1) + 2

)1/(p−1)⎤

⎦

−((p−1)/(q−l−m(p−2)))

· r−(p/(q−l−m(p−2)))

(1.14)

in R1+ × (0, T). But, we do not give the lower bound estimate of the radial solution u(r, t) in a

small neighborhood of the point (x, t), where x = 0, t = T .

Remark 1.11. When m = l = 1 or m = l > 1, the results of Theorem 1.9 are consistent with thosein [24, 25], respectively. For 1 < q < l +m(p − 2), in [29], the authors obtained the results ofglobal blow-up to arbitrary compactly supported initial data.

Remark 1.12. From Theorem 1.9, we obtain the same decay exponent as that of Theorem 1.2in [29] by different methods. Moreover, it is interesting to see that the decay exponent of theupper estimate of Theorem 1.9 is also the same as the secondary critical exponent of Theorems1.3 and 1.4.

This paper is organized as follows. In Section 2, by using the energy method, wewill obtain a blow-up condition and prove Theorem 1.3. In Section 3, using the comparisonprinciple, we can construct a global supersolution to prove Theorem 1.4. Finally, we considerthe single-point blow-up under some suitable conditions and prove Theorem 1.9 in Section 4.

2. Blow-Up Case

By using the energy method, we will obtain a blow-up condition corresponding to (1.1).Therefore, we need to select a suitable test function as follows:

ψε(x) = Aεe−ε|x|, Aε =

1∫RN e

−ε|x|dx=

εN∫ωN

∫∞0 e−rrN−1dr ds

, ∇ψε(x) = −εψε x|x| . (2.1)


Proof of Theorem 1.3. Suppose that u(x, t) is the solution of the Cauchy problem (1.1) and T isthe blow-up time. Let

E(t) =1s

∫

RNus(x, t)ψε(x)dx, t ∈ [0, T), (2.2)

where 0 < s < 1/p, p > 2, Then, E(t) ∈ C[0, T)⋂C1(0, T) and

E′(t) =∫

RNus−1ψεutdx =

∫

RNus−1ψε div

(|∇um|p−2∇ul

)dx +

∫

RNuq+s−1ψεdx

= − l(s − 1)∫

RNus+l−3ψε|∇um|p−2|∇u|2dx + ε

∫

RNus−1ψε|∇um|p−2∇ul x|x|dx

+∫

RNuq+s−1ψεdx

≥ lmp−2(1 − s)∫

RNul+m(p−2)+s−p−1ψε|∇u|pdx

− lmp−2ε

∫

RNul+m(p−2)+s−pψε|∇u|p−1dx +

∫

RNuq+s−1ψεdx.

(2.3)

Using Young’s inequality, we have

ε

∫

RNul+m(p−2)+s−pψε|∇u|p−1dx ≤ p − 1

p

∫

RNul+m(p−2)+s−p−1ψε|∇u|pdx

+εp

p

∫

RNul+m(p−2)+s−1ψεdx.

(2.4)

Since 0 < s < (1/p), it follows from (2.3) and (2.4) that

E′(t) ≥∫

RNuq+s−1ψεdx − lmp−2εp

p

∫

RNul+m(p−2)+s−1ψεdx. (2.5)

By q > q∗c = l +m(p − 2) + (p/N) > l +m(p − 2) > 1,∫RN ψε(x)dx = 1, and Holder’s inequality,

we obtain

∫

RNul+m(p−2)+s−1ψεdx =

∫

RNul+m(p−2)+s−1ψ

(l+m(p−2)+s−1)/(q+s−1)ε ψ

q−[l+m(p−2)]/(q+s−1)ε dx

≤[∫

RNuq+s−1ψεdx

](l+m(p−2)+s−1)/(q+s−1)

.

(2.6)


Therefore, by (2.5) and (2.6), we have

dE

dt≥[∫

RNuq+s−1ψεdx

](l+m(p−2)+s−1)/(q+s−1)

×[(∫

RNuq+s−1ψεdx

)(q−l−m(p−2))/(q+s−1)

− lmp−2εp

p

]

.

(2.7)

Applying Jensen’s inequality, we obtain

(∫

RNuq+s−1ψεdx

)(q−l−m(p−2))/(q+s−1)

≥(∫

RNusψεdx

)(q−l−m(p−2))/s

. (2.8)

Thus, it follows from (2.7) and (2.8) that

dE

dt≥ 1

2

(∫

RNusψεdx

)(q+s−1)/s

=12s(q+s−1)/sE(q+s−1)/s(t) (2.9)

as long as

E(t) ≥ 1s

(2lmp−2εp

p

)s/(q−l−m(p−2))

∀t ∈ [0, T). (2.10)

Hence, if E(0) satisfies

E(0) ≥ 1s

(2lmp−2εp

p

)s/(q−l−m(p−2))

= C0, (2.11)

then E(t) increases and remains below C0 for all t ∈ [0, T).And by (2.9) we have

E(t) ≥(Eq−1/s(0) − C1t

)−s/(q−1), where C1 =

q − 12

s(q−1)/s > 0. (2.12)

Therefore, from (2.11) and (2.12), we obtain that u(x, t) blows up in finite time T =(1/C1)Eq−1/s(0): and get an estimate on the blow-up time T of the solution u(x, t) as follows:

T ≤ 2q − 1

(2lmp−2εp

p

)(1−q)/(q−l−m(p−2))

. (2.13)


Finally, it remains to verify the blow-up condition (2.11). Since u0(x) ∈ Φa for some a ∈(0, a∗c), there exist two positive constants M and R0 > 1 such that u0(x) ≥ M|x|−a for all|x| ≥ R0, and we have

E(0) =1s

∫

RNus0(x)ψε(x)dx

>MsAε

s

∫

|x|≥R0

|x|−ase−ε|x|dx

=MsAε

sε−N+as

∫

|y|≥εR0

∣∣y∣∣ase−|y|dy.

(2.14)

By the definition of Aε, 0 < a < a∗c, we can choose 0 < ε ≤ (1/R0) so small such that (2.11)holds. The proof of Theorem 1.3 is complete.

3. Global Existence

In this section, we shall prove Theorem 1.4 by constructing a global supersolution. To do this,we introduce the radially symmetric self-similar solution UM,a(x, t) to the following Cauchyproblem:

ut = div(|∇um|p−2∇ul

), (x, t) ∈ RN × (0,+∞), (3.1)

u(x, 0) = u0(x) =M|x|−a, x ∈ RN. (3.2)

It is well known that the existence and uniqueness of the solution of (3.1) have been wellestablished (see [7]). By symmetric properties of (3.1), the solution UM,a(x, t) is given by thefollowing form

UM,a(x, t) = t−aβfM(r), with r =|x|tβ, β =

1a(l +m

(p − 2

) − 1)+ p

, (3.3)

where the positive function fM is the solution of the problem

(∣∣∣(fmM)′∣∣∣

p−2(flM

)′)′+N − 1r

∣∣∣(fmM)′∣∣∣

p−2(flM

)′(r) + βrf ′

M(r) + aβfM(r) = 0, r > 0,

fM(r) ≥ 0, r ≥ 0, f ′M(0) = 0, lim

r→+∞rafM(r) =M.

(3.4)

We shall prove the existence of solution fM(r) to (3.4) by the following ordinary differentialequation, and furthermore we obtain the nonincreasing property of the solution fM(r).


Firstly, given a fixed η > 0, we consider the following Cauchy problem:

(∣∣∣(gm)′∣∣∣p−2(

gl)′)′

+N − 1r

∣∣∣(gm)′∣∣∣p−2(

gl)′(r) + βrg ′(r) + aβg(r) = 0, r > 0,

g(0) = η, g ′(0) = 0.

(3.5)

According to the standard of the Cauchy problem for ODE and the methods used in [7, 30],we can obtain that the solution g(r) of the Cauchy problem (3.5) is positive, and g(r) → 0 asr → ∞; moreover,

limr→+∞

rag(r) = M (3.6)

for some M =M(η) > 0.Secondly, we shall prove that there exists a one-to-one correspondence between M ∈

(0,+∞) and η ∈ (0,+∞). Indeed, this can be seen from the following relation:

gη(r) = ηg1(ησr

), σ =

1 − l −m(p − 2)

p, (3.7)

where g1(r) is the solution of (3.5) for η = 1. Then,

M(η)= η1−aσM(1), with M(1) = lim

r→+∞rag1(r). (3.8)

Therefore, we can deduce that, for each M > 0, there exists a positive, bounded, and globalsolution fM(r) satisfying (3.4).

Finally, we shall prove that the solution g(r) is non-increasing, that is, fM(r) is alsonon-increasing. To do this, we need the following lemmas.

Lemma 3.1. Let g(r) be the solution of (3.5); then

limr→ 0

N∣∣∣(gm)′(r)

∣∣∣p−2(

gl)′(r)

r= − aβg(0). (3.9)

Proof. Integrating the (3.5) over (0, ε) with ε > 0, we have

(∣∣∣(gm)′∣∣∣

p−2(gl)′)

(ε) +∫ ε

0

N − 1r

∣∣∣(gm)′∣∣∣

p−2(gl)′dr +

∫ ε

0βrg ′dr +

∫ ε

0aβg dr = 0. (3.10)

Dividing by ε and taking ε → 0 in (3.10), we obtain

limε→ 0

⎡

⎢⎣

∣∣∣(gm)′(ε)

∣∣∣p−2(

gl)′(ε)

ε+N − 1ε

∣∣∣(gm)′(ε)

∣∣∣p−2(

gl)′(ε)

⎤

⎥⎦ = − lim

ε→ 0aβg(ε), (3.11)

which implies that (3.9) holds. The proof of Lemma 3.1 is complete.


Lemma 3.2. If there exists r0 ∈ [0,+∞) such that g(r0) = 0, then g(r) = 0 for all r ≥ r0.

Proof. We shall prove by contradiction. Assuming that Lemma 3.2 does not hold, it is easy tosee that there exists ε > 0 such that

g(r) > 0, g ′(r) > 0 in (r0, r0 + ε). (3.12)

Multiplying (3.5) by rN−1 and integrating over (r0, r) with r ∈ (r0, r0 + ε), we obtain

rN−1∣∣∣(gm)′∣∣∣

p−2(gl)′

+ βrNg(r) =∫ r

r0

NβrN−1g(r)dr −∫ r

r0

aβrN−1g(r)dr. (3.13)

It follows from (3.12) and (3.13) that

βrNg(r) ≤∫ r

r0

NβrN−1g(r)dr −∫ r

r0

aβrN−1g(r)dr ≤ (N − a)βN

g(r)(rN − rN0

), (3.14)

equivalently,

1 ≤ N − aN

(

1 −(r0

r

)N)

. (3.15)

Letting r → r0 in (3.15), we obtain the inequality 1 ≤ 0, which is a contradiction. The proofof Lemma 3.2 is complete.

Lemma 3.3. The solution g(r) of (3.5) is monotone nonincreasing in [0,+∞).

Proof. Our method is based on the contradiction argument. Suppose that, for some r0 > 0,g ′(r0) > 0, by Lemma 3.1, there exists r1 ∈ (0, r0) such that

g ′(r1) = 0 ,(∣∣∣(gm)′∣∣∣

p−2(gl)′)′

(r1) ≥ 0. (3.16)

By Lemma 3.2, we have g(r1) > 0. Using the similar argument in Lemma 3.1, we obtain

limr→ r1

N∣∣∣(gm)′(r)

∣∣∣p−2(

gl)′(r)

r − r1= −aβg(r1) < 0, (3.17)

which is a contradiction with (3.14). The proof of Lemma 3.3 is complete.

Next, we apply the monotone properties of fM(r) to obtain the condition on the globalexistence of the solution to (1.1).

Proof of Theorem 1.4. We prove Theorem 1.4 by the following steps.


Step 1. Since ϕ(x) ∈ Φa, there exists a constant K > 0 such that

ϕ(x) ≤ K(1 + |x|)−a ∀x ∈ RN. (3.18)

Taking M > K and the self-similar solution UM,a(x, t) of (3.1) defined as (3.3), sincelimr→+∞rafM(r) =M > K, there exists a positive constant R0 such that

rafM(r) > K for r ≥ R0. (3.19)

Setting γ = fM(R0) = min{fM(r) | r ∈ [0, R0]} > 0, it is easy to verify that ϕ(x) ≤ UM,a(x, t)for all x ∈ RN , where t0 ∈ (0, 1) and t

−aβ0 γ > ‖ϕ‖∞.

Let λ > 0; then w(x, t) = λUM,a(x, λl+m(p−2)−1t + t0) is the solution of the followingproblem

wt = div(|∇wm|p−2∇wl

), (x, t) ∈ RN × (0,+∞),

w(x, 0) = λUM,a(x, t0) ≥ λϕ(x), x ∈ RN.

(3.20)

Taking η = fM(0) and noting that fM(r) is non-increasing, we have

‖w(x, t)‖∞ = ηλ(λl+m(p−2)−1t + t0

)−aβ. (3.21)

Step 2. Set v(x, t) = A(t)w(x, B(t)), where A(t) and B(t) are solutions of the followingproblem:

A′(t) = ηq−1λq−1[λl+m(p−2)−1B(t) + t0

]−(a(q−1)/(a(l+m(p−2)−1)+p))Aq(t), t ∈ (0,+∞),

B′(t) = Al+m(p−2)−1(t) t ∈ (0,+∞),

A(0) = 1, B(0) = 0.

(3.22)

By a direct calculation, we obtain that v(x, t) satisfies

vt ≥ div(|∇vm|p−2∇vl

)+ vq, (x, t) ∈ RN × (0,+∞),

v(x, 0) = w(x, 0) = λUM,a(x, t0) ≥ λϕ(x), x ∈ RN.

(3.23)

Step 3. We shall prove that there exists a positive constant λ0 = λ0(ϕ) such that theproblem (3.22) has a global solution (A(t), B(t)) with A(t) bounded in [0, T) if λ ∈ [0, λ0).According to the standard theory of ODE, the local existence and uniqueness of solution(A(t), B(t)) of (3.22) hold. By (3.22), we have A′(t) > 0, A(t) > 1 for t > 0; furthermore, thesolution is continuous as long as the solution exists and A(t) is finite.


From (3.22), when A(t) exists in [0, t], then B(t) is uniquely defined by

B(t) =∫ t

0Al+m(p−2)−1(s)ds. (3.24)

Since p > 2 and A(t) is increasing, we obtain

B(s) =∫ s

0Al+m(p−2)−1(τ)dτ ≥ Al+m(p−2)−1(0)s = s ∀s ∈ [0, t]. (3.25)

By (3.22), (3.25), and a > a∗c = p/(q − l −m(p − 2)), it follows that

1 −A1−q(t) =(q − 1

)ηq−1λq−1

∫ t

0

[λl+m(p−2)−1B(s) + t0

]−(a(q−1)/(a(l+m(p−2)−1)+p))ds

≤ (q − 1)ηq−1λq−1

∫ t

0

(λl+m(p−2)−1s + t0

)−(a(q−1)/(a(l+m(p−2)−1)+p))ds

≤(q − 1

)ηq−1λq−l−m(p−2)

((a(q − 1

))/(a(l +m

(p − 2

) − 1)+ p)) − 1

t1−(a(q−1)/(a(l+m(p−2)−1)+p))0 .

(3.26)

Let λ0 = λ0(ϕ) be a positive constant defined by

(q − 1

)[a(l +m

(p − 2

) − 1)+ p]

a(q − l −m(p − 2

)) − p ηq−1λq−l−m(p−2)0 t

1−(a(q−1)/(a(l+m(p−2)−1)+p))0 =

12. (3.27)

Then from (3.26), q > q∗c > l + m(p − 2) > 1 and a > a∗c = p/(q − l − m(p − 2)), we have1 ≤ A(t) ≤ 21/q−1 for any λ ∈ (0, λ0], as long as A(t) exists globally.

On the other hand, by (3.22) and (3.25), we have

t ≤ B(t) ≤ 2(p−2)/(q−1)t ∀t ≥ 0. (3.28)

Therefore, B(t) is also global.Step 4. For any λ ∈ (0, λ0], where λ0 = λ0(ϕ) is defined as (3.27), the solution u(x, t)

of (1.1) with initial value u0(x) = λϕ(x) exists globally, and u(x, t) ≤ v(x, t) in RN × (0,+∞).Moreover, there exists a positive constant C such that

‖u(·, t)‖∞ ≤ ‖v(., t)‖∞ ≤ 21/(q−1)ηλ(λl+m(p−2)−1B(t) + t0

)−aβ ≤ Ct−aβ ∀t > 0. (3.29)

The proof of Theorem 1.4 is complete.


4. Single Point Blow-Up

In this section, under some suitable assumptions, we shall prove that the blow-up set consistsof the single point x = 0. Moreover, we also give the upper estimate of the solution u(x, t) ina small neighborhood of the point (x, t), where x = 0, t = T .

First, we suppose that the solution is radially symmetric, that is, depending only onr = |x| at a given time t > 0. Therefore, we study the following problem:

ut = r−(N−1)[rN−1|(um)r |p−2

(ul)

r

]

r+ uq, (r, t) ∈ (0,+∞) × (0,+∞),

u(r, 0) = u0(r), r ∈ (0,+∞),

rN−1|(um)r |p−2(ul)

r= 0, (r, t) ∈ {0} × [0,+∞).

(4.1)

Proof of Theorem 1.9. It is based on the method in [21]. The main idea is to apply the maximumprinciple to the auxiliary function ωi(r, t), which is defined in (4.7), and to show that ωi issmall enough in (0, i)× (0, T). Then by integrating the obtained inequality and taking limit asi → ∞, one can get upper bound of the solution u(r, t). Therefore, we divide the proof intothe following steps.

Step 1. Since problem (1.1) has no classical solution, we will construct the weaksolution by means of regularization of the degenerate equation.

Now define a strictly monotone sequence {εi}, εi > 0 for all i = 1, 2, 3, . . ., such that

εi −→ 0, as i −→ +∞. (4.2)

Then, the weak solution u(r, t) is the limit function of the solution of the following regularizedproblem (see [31]):

(ui)t = r−(N−1)

[rN−1

(((umi)r

)2 + ε2i

)(p−2)/2((uli

)

r+ εi

)]

r

+ uqi , (r, t) ∈ (0, i) × (0, T),

ui(r, 0) = u0(r), r ∈ (0, i),

ui(i, t) = u0(i), t ∈ (0, T),

rN−1(((

umi)r

)2 + ε2i

)(p−2)/2((uli

)

r+ εi

)= 0, (r, t) ∈ {0} × (0, T).

(4.3)

By the standard methods used in [1, 32], the uniform estimates for the passage to thelimit which do not depend on εi are established. Therefore, for any fixed εi > 0, we mayassume that, for all sufficiently large i, the function ui(r, t) satisfies the following conditions:

|ui| ≤M1, |(ui)r | ≤M2 in Qi,T = (0, i) × (0, T), (4.4)

where M1, M2 do not depend on i, and

(ui(r, t))r |r=0 = 0 ∀t ∈ (0, T). (4.5)


Moreover, by using condition (H) and the maximum principle in [33], we have

(ui(r, t))r ≤ 0 in Qi,T . (4.6)

Step 2. Set

wi(r, t) = rN−1(((umi)r)

2 + ε2i

)(p−2)/2(uli

)

r+ λ0r

Nuq

i , (4.7)

where λ0 is given in (1.11).By a direct calculation, we find that wi(r, t) satisfies the following parabolic equation:

(wi)t = ai(wi)rr + bi(wi)r + ciwi + di + ei, (4.8)

where

ai = lul−1i

(((umi)r

)2 + ε2i

)(p−2)/2+m

(p − 2

)um−1i

(((umi)r

)2 + ε2i

)(p−4)/2(uli

)

r

(umi)r , (4.9)

bi =(((

umi)r

)2 + ε2i

)(p−2)/2[l(1 −N)ul−1

i

r+ l(l − 1)ul−2

i (ui)r

]

+(p − 2

)(uli

)

r

(umi)r

·(((

umi)r

)2 + ε2i

)(p−4)/2[m(1 −N)um−1

i

r+m(m − 1)um−2

i (ui)r

]

,

(4.10)

ci = quq−1i

[(p − 1

) − λ0((p − 2

)(N + 1) + 2

)], (4.11)

di = AqrN−1[1 − λ0(N + 1)]uq−1i , with A ≡ −(p − 2

)ε2i

(uli

)

r

[((umi)r

)2 + ε2i

](p−4)/2, (4.12)

ei =(p − 2

)rN−1

(((umi)r

)2 + ε2i

)(p−4)/2(uli

)

r

(umi)r

×[−(m + q − 2

)mqrλ0u

q+m−3i ((ui)r)

2+((m−1)(1−λ0N)+q(1−λ0(N+1))

)muq+m−2(ui)r

]

+ rN−1(((

umi)r

)2 + ε2i

)(p−2)/2[−(l + q − 2)lqrλ0u

q+l−3i ((ui)r)

2

+((l − 1)(1−λ0N)+q(1−λ0(N+1))

)luq+l−2(ui)r

]

+ λ0q[λ0((p − 2

)(N + 1) + 2

) − (p − 2)]rNu

2q−1i .

(4.13)

Since N ≥ 1, p > 2, m > 1, l > 1, and q > l + m(p − 2), it follows from (1.11) and (4.6) thatei ≤ 0.

Now we consider the coefficients ci and di in Qi,T . By (4.6) and p > 2, we have

A = − (p − 2)ε2i

(uli

)

r

[((umi)r

)2 + ε2i

](p−4)/2 ≥ 0. (4.14)


It follows from (4.4) that |A| = O(ενi ) as i → ∞, where ν = min{2, p − 1}, and we obtain thefollowing estimate:

sup ci ≤ M3, supdi ≤M4iN−1ενi in Qi,T , (4.15)

where M3, M4 are positive constants, which are independent of i.Therefore, we have the parabolic differential inequality

(wi)t ≤ ai(wi)rr + bi(wi)r + ciwi + di, in Qi,T , (4.16)

where ci, di satisfy (4.15).Next, we consider the function wi(r, t) on the parabolic boundary of Qi,T . At first, it is

easy to see that wi(0, t) = 0 for all t ∈ (0, T). By (1.10), we have wi(i, t) ≤ λ0iNu

q

0(i) = o(1), asi → ∞ for all t ∈ (0, T). Finally, it follows from (1.11) that

wi(r, 0) = rN−1(((

um0)r

)2 + ε2i

)(p−2)/2(ul0

)

r+ λ0r

Nuq

0

≤ rN−1[∣∣(um0

)r

∣∣p−2(ul0

)

r+ λ0ru

q

0

]≤ 0 ∀r ∈ [0, i].

(4.17)

Hence, for all sufficiently large i, there exists γi = supwi > 0 on the parabolic boundaryof Qi,T and γi = o(1) as i → ∞.

In order to estimate wi(r, t) in Qi,T , we study the following ODE:

dwi

dt=M3wi +M4i

N−1ενi , t > 0,

wi(0) = γi,(4.18)

which has the solution

wi(t) =

(M4i

N−1ενiM3

+ γi

)

eM3t − M4iN−1

M3ενi , t > 0. (4.19)

Taking the sequence εi such that

iN−1ενi −→ 0 as i −→ ∞, (4.20)

it is obvious that

wi(t) ≤ wi(T) = αi ∀t ∈ (0, T), (4.21)

where

αi −→ 0 as i −→ ∞. (4.22)


Setting

z(r, t) = wi(r, t) −wi(t), (4.23)

we have z(r, t) ≤ 0 on the parabolic boundary, and z(r, t) satisfies the following parabolicinequality:

zt ≤ aizrr + bizr + ciz − (M3 − ci)wi +[di −M4i

N−1ενi

], in Qi,T . (4.24)

It follows from (4.15) that

zt ≤ aizrr + bizr + ciz, in Qi,T . (4.25)

By the maximum principle (Chapter II, [33]), we obtain that z(r, t) ≤ 0 in Qi,T , that is,

wi(r, t) ≤ wi(t) ≤ wi(T) = αi in Qi,T . (4.26)

Step 3. For large i and (ui(r, t))r ∈ [−M2, 0], we have the following estimate

(((umi)r

)2 + ε2i

)(p−2)/2(uli

)

r≥ ∣∣(umi

)r

∣∣p−2(uli

)

r− δi, (4.27)

where

0 < δi = O(ενi) −→ 0 as i −→ ∞. (4.28)

By (4.7) and (4.26), we have

∣∣∣(umi)r

∣∣∣p−2(

uli

)

r

uq

i

≤ r1−Nαi + δiuq

i

− λ0r,(4.29)

namely,

lmp−2u(l−1)+(m−1)(p−2)−qi (ui)r |(ui)r |p−2 ≤ r1−Nαi + δi

uq

i

− λ0r. (4.30)

Letting F(b) = b|b|−((p−2)/(p−1)), we deduce that

(lmp−2

)1/(p−1)u((l+m(p−2)−q−p+1)/(p−1))i (ui)r ≤ F

[r1−Nαi + δi

uq

i

− λ0r

]

. (4.31)


For arbitrary t0 ∈ (0, T), r0 ∈ supp u(r, t0) ≡ {r > 0 | u(r, t0) > 0}, we denote μ0 = u(x0, t0) > 0.By (4.6) and the uniform convergence ui(r, t0) → u(r, t0) as i → ∞, r ∈ [0, r0], we haveui(r, t0) ≥ (μ0/2), r ∈ [0, r0] for all sufficiently large i ≥ i0. Therefore, from (4.31) we obtainlyfor t = t0, i > i0,

(lmp−2

)1/(p−1)u((l+m(p−2)−q−p+1)/(p−1))i (ui)r ≤ F

[2q(r1−Nαi + δi

)

μq

0

− λ0r

]

. (4.32)

Integrating the above inequality over interval [r1, r0], where r1 > 0, we obtain

−(lmp−2)1/(p−1)(

p − 1)

q − l −m(p − 2) u

−((q−l−m(p−2))/(p−1))i |r0

r1 ≤∫ r0

r1

F

[2q(r1−Nαi + δi

)

μq

0

− λ0r

]

dr. (4.33)

Letting i → ∞, by (4.22), (4.28), and the uniform convergence ui(r, t0) → u(r, t0) as i → ∞,r ∈ [0, r0], we have the following estimate:

−(lmp−2)1/(p−1)(

p − 1)

q − l −m(p − 2)

[u−((q−l−m(p−2))/p−1)(r0, t0) − u−((q−l−m(p−2))/p−1)(r1, t0)

]

≤ limi→∞

∫ r0

r1

F

[2q(r1−Nαi + δi

)

μq

0

− λ0r

]

dr

= − λ1/(p−1)0

p − 1p

(rp/(p−1)0 − rp/(p−1)

1

).

(4.34)

Setting r1 → 0, from (4.34), we obtain the upper estimate

u(r0, t0) ≤[q − l −m(p − 2

)

p(lmp−2

)1/p−1λ

1/p−10

]−((p−1)/(q−l−m(p−2)))

r−(p/(q−l−m(p−2)))0 , (4.35)

where (r0, t0) ∈ R1+ × (0, T). The proof of Theorem 1.9 is complete.

Acknowledgments

The authors would like to thank the anonymous referees for their valuable comments on thiswork. The second author is supported in part by NSF of China (11071266) and in part by theNatural Science Foundation Project of CQ CSTC(2010BB9218). This work is supported by theFundamental Research Funds for the Central University, Project No CDJXS 11 10 00 40.

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Research ArticlePositive Solutions for Sturm-Liouville BoundaryValue Problems in a Banach Space

Hua Su,1 Lishan Liu,2 and Yonghong Wu3

1 School of Mathematics and Quantitative Economics, Shandong University of Finance and Economics,Jinan 250014, China

2 School of Mathematical Sciences, Qufu Normal University, Qufu, 273165, China3 Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia

Correspondence should be addressed to Hua Su, [email protected]

Received 20 February 2012; Accepted 13 June 2012


Copyright q 2012 Hua Su et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We consider the existence of single and multiple positive solutions for a second-order Sturm-Liouville boundary value problem in a Banach space. The sufficient condition for the existenceof positive solution is obtained by the fixed point theorem of strict set contraction operators in theframe of the ODE technique. Our results significantly extend and improve many known resultsincluding singular and nonsingular cases.

1. Introduction

Boundary value problems for ordinary differential equations play a very important role inboth theoretical study and practical application in many fields. They are used to describea large number of physical, biological, and chemical phenomena. In this paper, we studythe existence of positive solutions for the following second-order nonlinear Sturm-Liouvilleboundary value problem (BVP) in a Banach Space E

1p(t)

(p(t)u′(t)

)′ + f(u(t)) = 0, 0 < t < 1,

αu(0) − β limt→ 0+

p(t)u′(t) = 0,

γu(1) + δ limt→ 1−

p(t)u′(t) = 0,

(1.1)

where α, β, δ, γ ≥ 0 are constants such that B(t, s) =∫st dτ/p(τ), ω = βγ + αγB(0, 1) + αδ > 0,

and p ∈ C1((0, 1), (0,+∞)). Moreover, p may be singular at t = 0 and/or 1.


The BVP (1.1) is often referred to as a model for the deformation of an elastic beamunder a variety of boundary conditions [1–12]. We notice that previous work is limited to usethe completely continuous operators and the function f is required to satisfy some growthcondition or assumptions of monotonicity.

The aim of this paper is to consider the existence of positive solutions for the moregeneral Sturm-Liouville boundary value problem (1.1) by using the fixed point theorem ofstrict set contraction operators. Here we allow p to have singularity at t = 0, 1. The resultsobtained in this paper improve and generalize many well-known results.

The rest of the paper is organized as follows. In Section 2, we first present someproperties of Green’s functions to be used to define a positive operator. Then we approximatethe singular second-order boundary value problem by constructing an integral operator. InSection 3, the sufficient condition for the existence of single and multiple positive solutionsfor the BVP (1.1) is established. In Section 4, we give a example to demonstrate the applicationof our results.


In this paper, we denote by (E, ‖ · ‖1) a real Banach space. A nonempty closed convex subsetP in E is said to be a cone if λP ∈ P for λ ≥ 0 and P

⋂{−P} = {θ}, where θ denotes the zeroelement of E. The cone P defines a partial ordering in E by x ≤ y if and only if y−x ∈ P . Recallthat the cone P is said to be normal if there exists a positive constant λ such that 0 ≤ x ≤ yimplies ‖x‖1 ≤ λ‖y‖1.

In this paper, we assume P ⊆ E is normal, and without loss of generality, we mayassume that the normal constant of P is 1. Let J = [0, 1], and

C(J, E) = {u : J → E | u(t) continuous},

Ci(J, E) ={u : J → E | u(t) is i-order continuously differentiable

}, i = 1, 2, . . . .

(2.1)

For u = u(t) ∈ C(J, E), let ‖u‖ = maxt∈J‖u(t)‖1, then C(J, E) is a Banach space with the norm‖ · ‖.

Definition 2.1. A function u(t) is said to be a positive solution of the boundary value problem(1.1) if u ∈ C([0, 1], E)

⋂C1((0, 1), E) satisfies u(t) > 0, t ∈ (0, 1], pu′ ∈ C1((0, 1), E) and the

BVP (1.1).

We notice that if u(t) is a positive solution of the BVP (1.1) and p ∈ C1(0, 1), thenu ∈ C2(0, 1).

Now we denote by G(t, s) the Green’s functions for the following boundary valueproblem:

1p(t)

(p(t)u′(t)

)′ = 0, 0 < t < 1,

u(0) − limt→ 0+

βp(t)u′(t) = 0,

γu(1) + limt→ 1−

δp(t)u′(t) = 0.

(2.2)


It is well known that G(t, s) can be written by

G(t, s) =1ω

{(β + αB(0, s)

)(δ + γB(t, 1)

), 0 ≤ s ≤ t ≤ 1,

(β + αB(0, t)

)(δ + γB(s, 1)

), 0 ≤ t ≤ s ≤ 1,

(2.3)

where B(t, s) =∫st dτ/p(τ), ω = αδ + αγB(0, 1) + βγ > 0.

It is easy to verify the following properties of G(t, s):

(I) G(t, s) ≤ G(s, s) ≤ (1/ω)(β + αB(0, 1))(δ + γB(0, 1)) < +∞;

(II) G(t, s) ≥ ρG(s, s), for any t ∈ [a, b] ⊂ (0, 1), s ∈ [0, 1], where

ρ = min{δ + γB(b, 1)δ + γB(0, 1)

,β + αB(0, a)β + αB(0, 1)

}. (2.4)

Throughout this paper, we adopt the following assumptions:

(H1) p ∈ C1((0, 1), (0,+∞)) and satisfies

0 <∫1

0

ds

p(s)< +∞, 0 < e =

∫1

0G(s, s)p(s)ds < +∞. (2.5)

(H2) f : P → P is a uniformly continuous function and there exists M > 0 such that forany bounded set B ⊂ C(J, E), we have

α(f(B(t))

) ≤Mα(B(t)), t ∈ J, 2Meρ < 1, (2.6)

where α(·) denotes the Kuratowski measure of noncompactness in E.The following Lemmas play an important role in this paper (see [13]).

Lemma 2.2. Let B ⊂ C(J, E) be bounded and equicontinuous on J , then αc(B) = supt∈Jα(B(t)).

Lemma 2.3. Let B ⊂ C(J, E) be bounded and equicontinuous on J , then α(B(t)) is continuous on Jand

α

({∫

J

u(t)dt : u ∈ B})

≤∫

J

α(B(t))dt. (2.7)

Lemma 2.4. Let B ⊂ C(J, E) be a bounded set on J . Then α(B(t)) ≤ 2αc(B).

Now, for the given [a, b] ⊂ (0, 1) and the ρ as in (II), we introduce

K ={u ∈ C(J, P) : u(t) ≥ ρu(s), t ∈ [a, b], s ∈ [0, 1]

}. (2.8)

It is easy to check that K is a cone in C(J, E) and for u(t) ∈ K, t ∈ [a, b], we have ‖u(t)‖1 ≥ρ‖u‖.


Next, we define an operator T : K → C(J, P) given by

Tu(t) =∫1

0G(t, s)p(s)f(u(s))ds, ∀u ∈ K, t ∈ [0, 1]. (2.9)

Clearly, u is a solution of the BVP (1.1) if and only if u is a fixed point of the operator T .Through direct calculation, by (II) and for v ∈ K, t ∈ [a, b], s ∈ J , we have

Tu(t) =∫1

0G(t, s)p(s)f(u(s))ds ≥ ρ

∫1

0G(s, s)p(s)f(u(s))ds = ρTu(s). (2.10)

So, this implies that TK ⊂ K.

Lemma 2.5. Assume that, hold. Then T : K → K is a strict set contraction operator.

Proof. Firstly, The continuity of T is easily obtained. In fact, if vn, v ∈ K and vn → v in thenorm in C(J, E), then for any t ∈ J , we get

‖Tvn(t) − Tv(t)‖1 ≤∫1

0G(s, s)p(s)

∥∥f(vn(s)) − f(v(s))∥∥

1ds, (2.11)

so, by the uniformly continuity of f , we have

‖Tvn − Tv‖ = supt∈J

‖Tvn(t) − Tv(t)‖1 −→ 0. (2.12)

This implies that Tvn → Tv in C(J, E), that is, T is continuous.Now, let B ⊂ K be a bounded set. It follows from that there exists a positive number L

such that ‖f(v)‖ ≤ L for any v ∈ B. Then, we can get

‖Tv(t)‖1 ≤ Le <∞, ∀t ∈ J, v ∈ B, (2.13)

where e is as defined in. So, T(B) ⊂ K is a bounded set in K.For any ε > 0, by (H1), there exists a δ′ > 0 such that

∫δ′

0G(s, s)p(s) ≤ ε

6L,

∫1

1−δ′G(s, s)p(s) ≤ ε

6L. (2.14)

Let P = maxt∈[δ′,1−δ′]p(t). It follows from the uniform continuity of G(t, s) on [0, 1]× [0, 1] thatthere exists δ > 0 such that

∣∣G(t, s) −G(t′, s)∣∣ ≤ ε

3PL,

∣∣t − t′∣∣ < δ, t, t′ ∈ [0, 1], s ∈ [0, 1]. (2.15)


Consequently, when |t − t′| < δ, t, t′ ∈ [0, 1], v ∈ B, we have

∥∥Tv(t) − Tv(t′)∥∥1 =

∥∥∥∥∥

∫1

0

(G(t, s) −G(t′, s))p(s)f(v(s))ds

∥∥∥∥∥

1

≤∫δ′

0

∣∣G(t, s) −G(t′, s)∣∣p(s)∥∥f(v(s))∥∥1ds

+∫1−δ′

δ′

∣∣G(t, s) −G(t′, s)∣∣p(s)∥∥f(v(s))∥∥1ds

+∫1

1−δ′

∣∣G(t, s) −G(t′, s)∣∣p(s)∥∥f(v(s))∥∥1ds

≤ 2L∫δ′

0G(s, s)p(s)ds + 2L

∫1

1−δ′G(s, s)p(s)ds

+ PL∫1

0

∣∣G(t, s) −G(t′, s)∣∣ds

≤ ε.

(2.16)

This implies that T(B) is an equicontinuous set on J . Therefore, by Lemma 2.2, we have

α(T(B)) = supt∈J

α(T(B)(t)). (2.17)

Without loss of generality, by condition, we may assume that p(t) is singular at t = 0, 1.So, there exist {ani}, {bni} ⊂ (0, 1), {ni} ⊂N with {ni} being a strictly increasing sequence andlimi→+∞ni = +∞ such that

0 < · · · < ani < · · · < an1 < bn1 < · · · < bni < · · · < 1,

p(t) ≥ ni, t ∈ (0, ani] ∪ [bni , 1), p(ani) = p(bni) = ni,(2.18)

limi→+∞

ani = 0, limi→+∞

bni = 1. (2.19)

Next, we let

pni(t) =

{ni, t ∈ (0, ani] ∪ [bni , 1),p(t), t ∈ [ani , bni].

(2.20)

Then, from the above discussion we know that pni is continuous on J for every i ∈N and

pni(t) ≥ p(t), pni(t) −→ p(t), ∀t ∈ (0, 1), as i −→ +∞. (2.21)


For any ε > 0, by (2.19) and, there exists an i0 such that for any i > i0, we have

2L∫ani

0G(s, s)p(s)ds <

ε

2, 2L

∫1

bni

G(s, s)p(s)ds <ε

2. (2.22)

Therefore, for any bounded set B ⊂ C(J, E), by Lemmas 2.3 and 2.4,, the above discussion andnoting that pni(t) ≤ p(t), t ∈ (0, 1), as t ∈ J, i > i0, we have that

α(T(B)(t)) = α

({∫1

0G(t, s)p(s)f(v(s))ds | v ∈ B

})

≤ α

({∫1

0G(t, s)

[p(s) − pni(s)

]f(v(s))ds | v ∈ B

})

+ α

({∫1

0G(t, s)pni(s)f(v(s))ds | v ∈ B

})

≤ 2L∫ani

0G(s, s)p(s)ds + 2L

∫1

bni

G(s, s)p(s)ds

+∫1

0α(G(t, s)pni(s)f(v(s)) | v ∈ B)ds

≤ ε + ρ∫1

0G(s, s)p(s)α

(f(v(s)) | v ∈ B)ds

≤ ε + 2Meρα(B).

(2.23)

As ε is arbitrarily, we get

α(T(B)(t)) ≤ 2Meραc(B), t ∈ J. (2.24)

So, it follows from (2.17) and (2.24) that for any bounded set B ⊂ C(J, E), we have

αc(T(B)) ≤ 2Meραc(B). (2.25)

And note that 2Meρ < 1, we have T : K → K is a strict set contraction operator. The proof iscompleted.

Remark 2.6. When E = R, (2.6) naturally holds. In this case, we may takeM as 0, consequently,T : K → K is a completely continuous operator. Furthermore, if p(t) ≡ 1, t ∈ J , Dalmasso[1] used the following condition:

0 < e =∫1

0sα(1 − s)βp(s)ds < +∞, α, β ∈ [0, 1). (2.26)

Clearly, our condition is weaker than (2.26).


Our main tool used in this paper is the following fixed point index theorem of cone.

Theorem 2.7 (see [13]). Suppose that E is a Banach space, K ⊂ E is a cone, and let the Ω1, Ω2 betwo bounded open sets of E such that θ ∈ Ω1, Ω1 ⊂ Ω2. Let operator T : K ∩ (Ω2 \ Ω1) → K bestrict set contraction. Suppose that one of the following two conditions holds:

(i) ‖Tx‖ ≤ ‖x‖, for all x ∈ K ∩ ∂Ω1, ‖Tx‖ ≥ ‖x‖, for all x ∈ K ∩ ∂Ω2;

(ii) ‖Tx‖ ≥ ‖x‖, for all x ∈ K ∩ ∂Ω1, ‖Tx‖ ≤ ‖x‖, for all x ∈ K ∩ ∂Ω2.

Then T has at least one fixed point in K ∩ (Ω2 \Ω1).

Theorem 2.8 (see [13]). Suppose E is a real Banach space, K ⊂ E is a cone. Let Ωr = {u ∈ K :‖u‖ ≤ r}, and let the operator T : Ωr → K be completely continuous and satisfy Tx /=x, for allx ∈ ∂Ωr . Then

(i) If ‖Tx‖ ≤ ‖x‖, for all x ∈ ∂Ωr , then i(T,Ωr , K) = 1;

(ii) If ‖Tx‖ ≥ ‖x‖, for all x ∈ ∂Ωr , then i(T,Ωr , K) = 0.

3. The Main Results

Denote

f0 = lim‖u‖1 → 0+

∥∥f(u)∥∥

1

‖u‖1, f∞ = lim

‖u‖1 →∞

∥∥f(u)∥∥

1

‖u‖1. (3.1)

We now present our main results by Theorems 3.1 and 3.2.

Theorem 3.1. Suppose that conditions, hold. Assume that f also satisfies

(A1) f(u) ≥ ru∗, ρr ≤ ‖u‖1 ≤ r;(A2) f(u) ≤ Ru∗, 0 ≤ ‖u‖1 ≤ R,

where u∗ and u∗ satisfy

ρ

∥∥∥∥∥

∫b

a

G(s, s)p(s)u∗(s)ds

∥∥∥∥∥1

≥ 1, ‖u∗‖∫1

0G(s, s)p(s)ds ≤ 1. (3.2)

Then, the boundary value problem (1.1) has a positive solution u such that ‖u‖ is between r and R.

Proof of Theorem 3.1. Without loss of generality, we suppose that r < R. For any u ∈ K, wehave

‖u(t)‖1 ≥ ρ‖u‖, t ∈ [a, b]. (3.3)

We now define two open subset Ω1 and Ω2 of C(J, E)

Ω1 = {u ∈ C(J, E) : ‖u‖ < r}, Ω2 = {u ∈ C(J, E) : ‖u‖ < R}. (3.4)


For u ∈ K⋂∂Ω1, by (3.3), we have

r = ‖u‖ ≥ ‖u(t)‖1 ≥ ρ‖u‖ = ρr, t ∈ [a, b]. (3.5)

For u ∈ K ∩ ∂Ω1, if holds, we have

‖Tu(t)‖1 =

∥∥∥∥∥

∫1

0G(t, s)p(s)f(u(s))ds

∥∥∥∥∥

1

≥∥∥∥∥∥

∫b

a

G(t, s)p(s)u∗(s)rds

∥∥∥∥∥

1

≥ ρr∥∥∥∥∥

∫b

a

G(s, s)p(s)u∗(s)ds

∥∥∥∥∥

1

≥ r = ‖u‖, t ∈ J.(3.6)

Therefore, we have

‖Tu‖ = maxt∈[0,1]

‖Tu(t)‖1 ≥ ‖u‖, ∀u ∈ K ∩ ∂Ω1. (3.7)

On the other hand, as u ∈ K ∩ ∂Ω2, we have u(t) ≤ ‖u‖ = R and by, we know

‖Tu(t)‖1 =

∥∥∥∥∥

∫1


∥∥∥∥∥1

≤ R∥∥∥∥∥

∫1

0G(t, s)p(s)u∗(s)ds

∥∥∥∥∥1

≤ R‖u∗‖∫1

0G(s, s)p(s)ds ≤ R = ‖u‖.

(3.8)

Thus

‖T(u)‖ = maxt∈[0,1]

‖Tu(t)‖1 ≤ ‖u‖, ∀u ∈ K ∩ ∂Ω2. (3.9)

Therefore, by (3.7), (3.9), Theorem 2.7 and r < R, we have that T has a fixed point u ∈K ∩ (Ω2 \ Ω1). Obviously, u is a positive solution of the problem (1.1) and r < ‖u‖ < R.The proof of Theorem 3.1 is complete.

Theorem 3.2. Suppose that conditions,, and in Theorem 3.1 hold. Assume that f also satisfies

(A3) f0 = 0;

(A4) f∞ = 0.

Then, the boundary value problem (1.1) has at least two solutions.

Proof of Theorem 3.2. Firstly, by condition, we can have lim‖u‖1 → 0+(‖f(u)‖1/‖u‖1) = 0. Then,there exists an adequately small positive number m > 0 such that m

∫ba G(s, s)p(s)ds ≤ 1 and

there exists a constant ρ∗ ∈ (0, r) such that

∥∥f(u)∥∥

1 ≤ m‖u‖1, 0 < ‖u‖1 ≤ ρ∗, u /= 0. (3.10)


Set Ωρ∗ = {u ∈ K : ‖u‖ < ρ∗}, for any u ∈ ∂Ωρ∗ , by (3.10), we have

∥∥f(u)

∥∥

1 ≤ m‖u‖1 ≤ mρ∗. (3.11)

For u ∈ ∂Ωρ∗ , we have

‖Tu(t)‖1 =

∥∥∥∥∥

∫1


∥∥∥∥∥

1

≤∫1

0G(t, s)p(s)

∥∥f(u(s))

∥∥

1

≤∫b

a

G(t, s)p(s)mρ∗ds ≤ ρ∗m∫b

a

G(s, s)p(s)ds ≤ ρ∗ = ‖u‖.(3.12)

Therefore, we have

‖Tu‖ ≤ ‖u‖, ∀u ∈ ∂Ωρ∗ . (3.13)

Then by Theorem 2.8, we have

i(T,Ωρ∗ , K

)= 1. (3.14)

Next, by condition (A4), we have lim‖u‖1 →∞(‖f(u)‖1/‖u‖1) = 0. Then, there exists anadequately small positive number m > 0 such that m

∫ba G(s, s)p(s)ds ≤ 1, and there exists a

constant ρ0 > 0 such that

∥∥f(u)∥∥

1 ≤ m‖u‖1, ‖u‖1 > ρ0. (3.15)

We choose a constant ρ∗ > r, obviously, ρ∗ < r < ρ∗. Set Ωρ∗ = {u ∈ K : ‖u‖ < ρ∗}, then for anyu ∈ ∂Ωρ∗ , by (3.15), we have

∥∥f(u)∥∥

1 ≤ m‖u‖1 ≤ mρ∗. (3.16)

For u ∈ ∂Ωρ∗ , we have

‖Tu(t)‖1 =

∥∥∥∥∥

∫1


∥∥∥∥∥1

≤∫1

0G(t, s)p(s)

∥∥f(u(s))∥∥

1

≤∫b

a

G(t, s)p(s)mρ∗ds ≤ ρ∗m∫b

a

G(s, s)p(s)ds ≤ ρ∗ = ‖u‖.(3.17)

Therefore, we have

‖Tu‖ ≤ ‖u‖, ∀u ∈ ∂Ωρ∗ . (3.18)



i(T,Ωρ∗ , K

)= 1. (3.19)

Finally, set Ωr = {u ∈ K : ‖u‖ < r}, For any u ∈ ∂Ωr , by, Lemma 2.3 and proceeding asfor the proof of Theorem 3.1, we have

‖Tu‖ ≥ ‖u‖, ∀u ∈ ∂Ωr . (3.20)


i(T,Ωr , K) = 0. (3.21)

Therefore, by (3.14), (3.19), (3.21) and ρ∗ < R < ρ∗, we have

i(T,Ωr \Ωρ∗ , k

)= −1, i

(T,Ωρ∗ \Ωr , k

)= 1. (3.22)

Then T have fixed points u1 ∈ Ωr \Ωρ∗ and u2 ∈ Ωρ∗ \Ωr . Obviously, u1 and u2 are all positivesolutions of the BVP (1.1). The proof of Theorem 3.2 is complete.

4. Application

In order to illustrate the application of our results, we give an example in this section.

Example 4.1. Consider the following singular boundary value problem (SBVP):

33√t

(13

3√tu′(t)

)+ 160

[u1/2 + u1/3

]= θ, 0 < t < 1,

u(0) − limt→ 0+

13

3√tu′(t) = θ, u(1) + lim

t→ 1−

13

3√tu′(t) = θ,

(4.1)

where

β = γ = δ = 1, α = 3, p(t) =13

3√t, f(u) = 160

(u1/2 + u1/3

). (4.2)

Then obviously,

∫1

0

1p(t)

dt =32, f∞ = 0, f0 = 0. (4.3)


By computing, we know that the Green’s function is

G(t, s) =17

{(1 + 3s)(2 − t), 0 ≤ s ≤ t ≤ 1,(1 + 3t)(2 − s), 0 ≤ t ≤ s ≤ 1.

(4.4)

It is easy to note that 0 ≤ G(s, s) ≤ 1 and conditions,,, hold.Next, by computing, we know that ρ = 0.44. We choose r = 3, u∗ = 104, as 1.32 = ρr ≤

‖u‖ = max{u(t), t ∈ J} ≤ 3 and ρ‖ ∫ba G(s, s)p(s)u∗(s)ds‖ = 1.3 > 1, because of the monotoneincreasing of f(u) on [0,∞), then

f(u) ≥ f(1.32) = 359.3 ≥ 312 = ru∗, 1.32 ≤ ‖u‖ ≤ 3. (4.5)

Thus condition (A1) holds. Hence by Theorem 3.2, SBVP (4.1) has at least two positivesolutions u1, u2 and 0 < ‖u1‖ < 3 < ‖u2‖.

Acknowledgment

The authors were supported financially by Shandong Province Natural Science Foundation(ZR2009AQ004, ZR2010GL013), National Natural Science Foundation of China (11071141).

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Research ArticlePositive Solutions for Second-OrderSingular Semipositone Differential EquationsInvolving Stieltjes Integral Conditions

Jiqiang Jiang,1 Lishan Liu,1 and Yonghong Wu2

1 School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, China2 Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia

Correspondence should be addressed to Lishan Liu, [email protected]

Received 18 March 2012; Accepted 3 May 2012


Copyright q 2012 Jiqiang Jiang et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

By means of the fixed point theory in cones, we investigate the existence of positive solutionsfor the following second-order singular differential equations with a negatively perturbed term:−u′′(t) = λ[f(t, u(t)) − q(t)], 0 < t < 1, αu(0) − βu′(0) = ∫1

0 u(s)dξ(s), γu(1) + δu′(1) =

∫10 u(s)dη(s),

where λ > 0 is a parameter; f : (0, 1) × (0,∞) → [0,∞) is continuous; f(t, x) may be singular att = 0, t = 1, and x = 0, and the perturbed term q : (0, 1) → [0,+∞) is Lebesgue integrable and mayhave finitely many singularities in (0, 1), which implies that the nonlinear term may change sign.

1. Introduction

In this paper, we are concerned with positive solutions of the following second-order singularsemipositone boundary value problem (BVP):

−u′′(t) = λ[f(t, u(t)) − q(t)], 0 < t < 1,

αu(0) − βu′(0) =∫1

0u(s)dξ(s),

γu(1) + δu′(1) =∫1

0u(s)dη(s),

(1.1)

where λ > 0 is a parameter, α, γ ≥ 0, β, δ > 0 are constants such that ρ = αγ + αδ + βγ > 0,and the integrals in (1.1) are given by Stieltjes integral with a signed measure, that is, ξ, η are


suitable functions of bounded variation, q : (0, 1) → [0,+∞) is a Lebesgue integral and mayhave finitely many singularities in [0, 1], f : (0, 1) × (0,∞) → [0,∞) is continuous, f(t, x)may be singular at t = 0, t = 1, and x = 0.

Semipositone BVPs occur in models for steady-state diffusion with reactions [1], andinterest in obtaining conditions for the existence of positive solutions of such problems hasbeen ongoing for many years. For a small sample of such work, we refer the reader to thepapers of Agarwal et al. [2, 3], Kosmatov [4], Lan [5–7], Liu [8], Ma et al. [9, 10], and Yao[11]. In [12], the second-order m-point BVP,

−u′′(t) = λf(t, u(t)), t ∈ (0, 1),

u′(0) =m−2∑

i=1

aiu′(ξi), u(1) =

m−2∑

i=1

biu(ξi),(1.2)

is studied, where ai, bi > 0 (i = 1, 2, . . . , m − 2), 0 < ξ1 < ξ2 < · · · < ξm−2 < 1, λ is a positiveparameter. By using the Krasnosel’skii fixed point theorem in cones, the authors establishedthe conditions for the existence of at least one positive solution to (1.2), assuming that 0 <∑m−2

i=1 ai < 1, 0 <∑m−2

i=1 bi < 1, f : [0, 1] × [0,+∞) → (−∞,+∞) is continuous, and there existsA > 0 such that f(t, u) ≥ −A for (t, u) ∈ [0, 1] × [0,+∞). If the constant A is replaced by anycontinuous function A(t) on [0, 1], f also has a lower bound and the existence results are stilltrue.

Recently, Webb and Infante [13] studied arbitrary-order semipositone boundary valueproblems. The existence of multiple positive solutions is established via a Hammerstein inte-gral equation of the form:

u(t) =∫1

0k(t, s)g(s)f(s, u(s))ds, (1.3)

where k is the corresponding Green function, g ∈ L1[0, 1] is nonnegative and may havepointwise singularities, f : [0, 1]×[0,+∞) → (−∞,+∞) satisfies the Caratheodory conditionsand f(t, u) ≥ −A for some A > 0. Although A is a constant, because of the term g, [13]includes nonlinearities that are bounded below by an integral function. It is worth mentioningthat the boundary conditions cover both local and nonlocal types. Nonlocal boundary con-ditions are quite general, involving positive linear functionals on the space C[0, 1], given byStieltjes integrals.

For the cases where the nonlinear term takes only nonnegative values, the existence ofpositive solutions of nonlinear boundary value problems with nonlocal boundary conditions,including multipoint and integral boundary conditions, has been extensively studied bymany researchers in recent years [14–25]. Kong [17] studied the second-order singular BVP:

u′′(t) + λf(t, u(t)) = 0, t ∈ (0, 1),

u(0) =∫1

0u(s)dξ(s),

u(1) =∫1

0u(s)dη(s),

(1.4)


where λ is a positive parameter, f : (0, 1)×(0,+∞) → [0,+∞) is continuous, ξ(s) and η(s) arenondecreasing, and the integrals in (1.4) are Riemann-Stieltjes integrals. Sufficient conditionsare obtained for the existence and uniqueness of a positive solution by using the mixedmonotone operator theory.

Inspired by the above work, the purpose of this paper is to establish the existence ofpositive solutions to BVP (1.1). By using the fixed point theorem on a cone, some new exis-tence results are obtained for the case where the nonlinearity is allowed to be sign changing.We will address here that the problem tackled has several new features. Firstly, as q ∈ L1[0, 1],the perturbed effect of q on f may be so large that the nonlinearity may tend to negativeinfinity at some singular points. Secondly, the BVP (1.1) possesses singularity, that is, theperturbed term q may has finitely many singularities in [0, 1], and f(t, x) is allowed to besingular at t = 0, t = 1, and x = 0. Obviously, the problem in question is different from thosein [2–13]. Thirdly,

∫10 u(s)dξ(s) and

∫10 u(s)dη(s) denote the Stieltjes integrals where ξ, η are of

bounded variation, that is, dξ and dη can change sign. This includes the multipoint problemsand integral problems as special cases.

The rest of this paper is organized as follows. In Section 2, we present some lemmasand preliminaries, and we transform the singularly perturbed problem (1.1) to an equivalentapproximate problem by constructing a modified function. Section 3 gives the main resultsand their proofs. In Section 4, two examples are given to demonstrate the validity of our mainresults.

Let K be a cone in a Banach space E. For 0 < r < R < +∞, let Kr = {x ∈ K : ‖x‖ < r},∂Kr = {x ∈ K : ‖x‖ = r}, and Kr,R = {x ∈ K : r ≤ ‖x‖ ≤ R}. The proof of the main theorem ofthis paper is based on the fixed point theory in cone. We list here one lemma [26, 27] whichis needed in our following argument.

Lemma 1.1. LetK be a positive cone in real Banach space E, T : Kr,R → K is a completely continu-ous operator. If the following conditions hold:

(i) ‖Tx‖ ≤ ‖x‖ for x ∈ ∂KR,

(ii) there exists e ∈ ∂K1 such that x /= Tx +me for any x ∈ ∂Kr andm > 0,

then, T has a fixed point in Kr,R.

Remark 1.2. If (i) and (ii) are satisfied for x ∈ ∂Kr and x ∈ ∂KR, respectively, then Lemma 1.1is still true.


Denote

φ1(t) =1ρ

(δ + γ(1 − t)), φ2(t) =

1ρ

(β + αt

), e(t) = G(t, t), t ∈ [0, 1],

k1 = 1 −∫1

0φ1(t)dξ(t), k2 =

∫1

0φ2(t)dξ(t), k3 =

∫1

0φ1(t)dη(t),

k4 = 1 −∫1

0φ2(t)dη(t), k = k1k4 − k2k3, σ =

ρ(α + β

)(γ + δ

) ,

(2.1)


where

G(t, s) =1ρ

⎧⎨

⎩

(β + αs

)(δ + γ(1 − t)), 0 ≤ s ≤ t ≤ 1,

(β + αt

)(δ + γ(1 − s)), 0 ≤ t ≤ s ≤ 1.

(2.2)

Obviously,

e(t) = ρφ1(t)φ2(t) =1ρ

(β + αt

)(δ + γ(1 − t)), t ∈ [0, 1], (2.3)

σe(t)e(s) ≤ G(t, s) ≤ e(s) (or e(t)) ≤ σ−1, t, s ∈ [0, 1]. (2.4)

Throughout this paper, we adopt the following assumptions.

(H1) k1, k4 ∈ (0, 1], k2 ≥ 0, k3 ≥ 0, k > 0, and

Gξ(s) =∫1

0G(t, s)dξ(t) ≥ 0, Gη(s) =

∫1

0G(t, s)dη(t) ≥ 0, s ∈ [0, 1]. (2.5)

(H2) q : (0, 1) → [0,+∞) is a Lebesgue integral and∫1

0 q(t)dt > 0.

(H3) For any (t, x) ∈ (0, 1) × (0,+∞),

0 ≤ f(t, x) ≤ p(t)(g(x) + h(x)), (2.6)

where p ∈ C(0, 1) with p > 0 on (0, 1) and∫1

0 p(t)dt < +∞, g > 0 is continuousand nonincreasing on (0,+∞), h ≥ 0 is continuous on [0,+∞), and for any constantr > 0,

0 <∫1

0p(s)g(re(s))ds < +∞. (2.7)

Remark 2.1. If dξ and dη are two positive measures, then the assumption (H1) can be replacedby a weaker assumption:

(H ′1) k1 > 0, k4 > 0, k > 0.

Remark 2.2. It follows from (2.4) and (H3) that

∫1

0e(s)p(s)g(re(s))ds ≤ σ−1

∫1

0p(s)g(re(s))ds < +∞. (2.8)


For convenience, in the rest of this paper, we define several constants as follows:

L1 = 1 +k4(γ + δ

)+ k3

(α + β

)

ρk

∫1

0dξ(τ) +

k2(γ + δ

)+ k1

(α + β

)

ρk

∫1

0dη(τ),

L2 = σ

[

1 +k4(γ + δ

)+ k3

(α + β

)

ρk

∫1

0e(τ)dξ(τ) +

k2(γ + δ

)+ k1

(α + β

)

ρk

∫1

0e(τ)dη(τ)

]

,

L3 = 1 +k4δ + k3β

kβδ

∫1

0e(τ)dξ(τ) +

k2δ + k1β

kβδ

∫1

0e(τ)dη(τ).

(2.9)

Remark 2.3. If x ∈ C[0, 1] ∩ C2(0, 1) satisfies (1.1), and x(t) > 0 for any t ∈ (0, 1), then we saythat x is a C[0, 1] ∩ C2(0, 1) positive solution of BVP (1.1).

Lemma 2.4. Assume that (H1) holds. Then, for any y ∈ L1[0, 1], the problem,

−u′′(t) = y(t), t ∈ (0, 1),

αu(0) − βu′(0) =∫1

0u(s)dξ(s),

γu(1) + δu′(1) =∫1

0u(s)dη(s),

(2.10)

has a unique solution

u(t) =∫1

0H(t, s)y(s)ds, (2.11)

where

H(t, s) = G(t, s) +k4φ1(t) + k3φ2(t)

k

∫1

0G(τ, s)dξ(τ) +

k2φ1(t) + k1φ2(t)k

∫1

0G(τ, s)dη(τ).

(2.12)

Proof. The proof is similar to Lemma 2.2 of [28], so we omit it.

Lemma 2.5. Suppose that (H1) holds, then Green’s functionH(t, s) defined by (2.12) possesses thefollowing properties:

(i) H(t, s) ≤ L1e(s), t, s ∈ [0, 1];

(ii) L2e(t)e(s) ≤ H(t, s) ≤ L3e(t), t, s ∈ [0, 1],

where L1, L2, and L3 are defined by (2.9).


Proof. (i) It follows from (2.4) that

H(t, s) ≤ e(s) + k4(γ + δ

)+ k3

(α + β

)

ρk

∫1

0e(s)dξ(τ)

+k2(γ + δ

)+ k1

(α + β

)

ρk

∫1

0e(s)dη(τ)

= L1e(s), t, s ∈ [0, 1].

(2.13)

(ii) By the monotonicity of φ1, φ2 and the definition of G(t, s), we have

e(t)α + β

≤ φ1(t) =e(t)ρφ2(t)

=e(t)αt + β

≤ e(t)β, t ∈ [0, 1],

e(t)γ + δ

≤ φ2(t) =e(t)ρφ1(t)

=e(t)

γ(1 − t) + δ ≤ e(t)δ, t ∈ [0, 1].

(2.14)

By (2.4) and the left-hand side of inequalities (2.14), we have

H(t, s) ≥ σe(t)e(s) + σe(t)e(s)k4/(α + β

)+ k3/

(γ + δ

)

k

∫1

0e(τ)dξ(τ)

+ σe(t)e(s)k2/

(α + β

)+ k1/

(γ + δ

)

k

∫1

0e(τ)dη(τ)

= σe(t)e(s)

[

1 +k4(γ + δ

)+ k3

(α + β

)

ρk

∫1

0e(τ)dξ(τ)

+k2(γ + δ

)+ k1

(α + β

)

ρk

∫1

0e(τ)dη(τ)

]

= L2e(t)e(s), t, s ∈ [0, 1].

(2.15)

Similarly, by (2.4) and the right-hand side of inequalities (2.14), we have

H(t, s) = G(t, s) +k4φ1(t) + k3φ2(t)

k

∫1

0G(τ, s)dξ(τ) +

k2φ1(t) + k1φ2(t)k

∫1

0G(τ, s)dη(τ)

≤ e(t) + e(t)[k4/β + k3/δ

k

∫1

0G(τ, s)dξ(τ) +

k2/β + k1/δ

k

∫1

0G(τ, s)dη(τ)

]

≤ e(t) + e(t)[k4δ + k3β

kβδ

∫1

0e(τ)dξ(τ) +

k2δ + k1β

kβδ

∫1

0e(τ)dη(τ)

]


= e(t)

[

1 +k4δ + k3β

kβδ

∫1

0e(τ)dξ(τ) +

k2δ + k1β

kβδ

∫1

0e(τ)dη(τ)

]

= L3e(t), t, s ∈ [0, 1].

(2.16)

The proof of Lemma 2.5 is completed.

Lemma 2.6. Suppose that (H1) and (H2) hold. Then, the boundary value problem,

−w′′(t) = 2λp(t), t ∈ (0, 1),

αw(0) − βw′(0) =∫1

0w(s)dξ(s),

γw(1) + δw′(1) =∫1

0w(s)dη(s),

(2.17)

has unique solution

w(t) = 2λ∫1

0H(t, s)p(s)ds, (2.18)

which satisfies

w(t) ≤ 2λL3e(t)∫1

0p(s)ds, t ∈ [0, 1]. (2.19)

Proof. It follows from (2.11), Lemma 2.5, (H1) and (H2) that (2.18) and (2.19) hold.

Let X = C[0, 1] be a real Banach space with the norm ‖x‖ = maxt∈[0,1]|x(t)| for x ∈ X.We let

K = {x ∈ X : x is concave on [0, 1], x(t) ≥ Λe(t)‖x‖ for t ∈ [0, 1]}, (2.20)

where Λ = L2/L1. Clearly, K is a cone of X.For any u ∈ X, let us define a function [·]+:

[u(t)]+ =

⎧⎨

⎩

u(t), u(t) ≥ 0,

0, u(t) < 0.(2.21)


Next, we consider the following approximate problem of (1.1):

−x′′(t) = λ[f(t, [x(t) −w(t)]+

)+ q(t)

], t ∈ (0, 1),

αx(0) − βx′(0) =∫1

0x(s)dξ(s),

γx(1) + δx′(1) =∫1

0x(s)dη(s).

(2.22)

Lemma 2.7. If x ∈ C[0, 1] ∩ C2(0, 1) is a positive solution of problem (2.22) with x(t) ≥ w(t) forany t ∈ [0, 1], then x−w is a positive solution of the singular semipositone differential equation (1.1).

Proof. If x is a positive solution of (2.22) such that x(t) ≥ w(t) for any t ∈ [0, 1], then from(2.22) and the definition of [u(t)]+, we have

−x′′(t) = λ[f(t, x(t) −w(t)) + q(t)

], t ∈ (0, 1),

αx(0) − βx′(0) =∫1

0x(s)dξ(s),

γx(1) + δx′(1) =∫1

0x(s)dη(s).

(2.23)

Let u = x −w, then u′′ = x′′ −w′′, which implies that

−x′′ = −u′′ −w′′ = −u′′ + 2λq(t). (2.24)

Thus, (2.23) becomes

−u′′(t) = λ[f(t, u(t)) − q(t)], t ∈ (0, 1),

αu(0) − βu′(0) =∫1

0u(s)dξ(s),

γu(1) + δu′(1) =∫1

0u(s)dη(s),

(2.25)

that is, x −w is a positive solution of (1.1). The proof is complete.

To overcome singularity, we consider the following approximate problem of (2.22):

−x′′(t) = λ[f(t, [x(t) −w(t)]+ + n−1

)+ q(t)

], t ∈ (0, 1),

αx(0) − βx′(0) =∫1

0x(s)dξ(s),

γx(1) + δx′(1) =∫1

0x(s)dη(s),

(2.26)


where n is a positive integer. For any n ∈ N, let us define a nonlinear integral operator Tλn :K → X as follows:

Tλnx(t) = λ∫1

0H(t, s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds. (2.27)

It is obvious that solving (2.26) in C[0, 1] ∩ C2(0, 1) is equivalent to solving the fixed pointequation Tλnx = x in the Banach space C[0, 1].

Lemma 2.8. Assume that (H1)–(H3) hold, then for each n ∈ N, λ > 0, R > r ≥ 4λL3Λ−1∫1

0 q(s)ds,Tλn : Kr,R → K is a completely continuous operator.

Proof. Let n ∈ N be fixed. For any x ∈ K, by (2.27) we have

(Tλnx

)′′(t) = −λ

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]≤ 0,

Tλnx(0) = λ∫1

0H(0, s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds ≥ 0,

Tλnx(1) = λ∫1

0H(1, s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds ≥ 0,

(2.28)

which implies that Tλn is nonnegative and concave on [0, 1]. For any x ∈ K and t ∈ [0, 1], itfollows from Lemma 2.5 that

Tλnx(t) = λ∫1

0H(t, s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

≤ λL1

∫1

0e(s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds.

(2.29)

Thus,

∥∥∥Tλnx∥∥∥ ≤ λL1

∫1

0e(s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds. (2.30)

On the other hand, from Lemma 2.5, we also obtain

Tλnx(t) = λ∫1

0H(t, s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

≥ λL2e(t)∫1

0e(s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds.

(2.31)

So,

Tλnx(t) ≥ Λe(t)∥∥∥Tλnx

∥∥∥, t ∈ [0, 1]. (2.32)

This yields that Tλn (K) ⊂ K.


Next, we prove that Tλn : Kr,R → K is completely continuous. Suppose xm ∈ Kr,R andx0 ∈ Kr,R with ‖xm − x0‖ → 0 (m → ∞). Notice that

∣∣[xm(s) −w(s)]+ − [x0(s) −w(s)]+

∣∣

=∣∣∣∣|xm(s) −w(s)| + xm(s) −w(s)

2− |x0(s) −w(s)| + x0(s) −w(s)

2

∣∣∣∣

=∣∣∣∣|xm(s) −w(s)| − |x0(s) −w(s)|

2+xm(s) − x0(s)

2

∣∣∣∣

≤ |xm(s) − x0(s)|.

(2.33)

This, together with the continuity of f , implies

∣∣∣f(s, [xm(s) −w(s)]+ + n−1

)+ q(s) −

[f(s, [x0(s) −w(s)]+ + n−1

)+ q(s)

]∣∣∣

=∣∣∣f(s, [xm(s) −w(s)]+ + n−1

)− f(s, [x0(s) −w(s)]+ + n−1

)∣∣∣ −→ 0, m −→ ∞.

(2.34)

Using the Lebesgue dominated convergence theorem, we have

∥∥∥Tλnxm − Tλnx0

∥∥∥

≤λL1

∫1

0e(s)

∣∣∣f(s, [xm(s) −w(s)]++n−1

)−f(s, [x0(s) −w(s)]++n−1

)∣∣∣ ds −→ 0, m −→ ∞.

(2.35)

So, Tλn : Kr,R → K is continuous.Let B ⊂ Kr,R be any bounded set, then for any x ∈ B, we have x ∈ K, r ≤ ‖x‖ ≤ R.

Therefore, we have

[x(t) −w(t)]+ ≤ x(t) ≤ ‖x‖ ≤ R, t ∈ [0, 1],

x(t) −w(t) ≥ x(t) − 2λL3e(t)∫1

0q(s)ds ≥ x(t) − 2λL3

x(t)Λr

∫1

0q(s)ds

≥ 12x(t) ≥ 1

2rΛe(t) > 0, t ∈ [0, 1].

(2.36)

By (H3), we have

Lr :=∫1

0p(s)g

(12Λre(s)

)ds < +∞. (2.37)

It is easy to show that Tλn (B) is uniformly bounded. In order to show that Tλn is a compactoperator, we only need to show that Tλn (B) is equicontinuous. By the continuity of H(t, s) on


[0, 1]× [0, 1], for any ε > 0, there exists δ1 > 0 such that for any t1, t2, s ∈ [0, 1] and |t1 − t2| < δ1,we have

|H(t1, s) −H(t2, s)| < ε. (2.38)

By (2.36)–(2.37), and (2.27), we have

∣∣∣Tλnx(t1) − Tλnx(t2)

∣∣∣ ≤ λ

∫1

0|H(t1, s) −H(t2, s)|

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

< ελ

∫1

0

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

≤ ελ∫1

0

[p(s)

(g([x(s) −w(s)]+ + n−1

)+ h([x(s) −w(s)]+ + n−1

))

+q(s)]ds

≤ ελ[

Lr + (1 + h∗(R))∫1

0

[p(s) + q(s)

]ds

]

,

(2.39)

where

h∗(r) = maxy∈[0,1+r]

h(y). (2.40)

This means that Tλn (B) is equicontinuous. By the Arzela-Ascoli theorem, Tλn (B) is a relativelycompact set. Now since λ and n are given arbitrarily, the conclusion of this lemma is valid.

3. Main Results

Theorem 3.1. Assume that conditions (H1)–(H3) are satisfied. Further assume that the followingcondition holds.

(H4) There exists an interval [a, b] ⊂ (0, 1) such that

lim infu→+∞

mint∈[a,b]

f(t, u)u

= +∞. (3.1)

Then, there exists λ∗ > 0 such that the BVP (1.1) has at least one positive solution u(t) ∈ C[0, 1] ∩C2(0, 1) provided λ ∈ (0, λ∗). Furthermore, the solution also satisfies u(t) ≥ le(t) for some positiveconstant l.


Proof. Take r > 4L3Λ−1∫1

0 q(s)ds. Let

λ∗ = min

⎧⎨

⎩1,

r

2L1∫1

0 e(s)p(s)g((1/2)rΛe(s))ds,

r

2L1(h∗(r) + 1)∫1

0 e(s)[p(s) + q(s)

]ds

⎫⎬

⎭,

(3.2)

where h∗ is defined by (2.40). For any λ ∈ (0, λ∗), x ∈ ∂Kr , noticing that λ∗ ≤ 1, we have

[x(t) −w(t)]+ ≤ x(t) ≤ ‖x‖ ≤ r, t ∈ [0, 1],

x(t) −w(t) ≥ x(t) − 2λL3e(t)∫1

0q(s)ds ≥ x(t) − 2L3e(t)

∫1

0q(s)ds

≥ x(t) − 2L3x(t)Λr

∫1

0q(s)ds ≥ 1

2x(t) ≥ 1

2rΛe(t) > 0, t ∈ [0, 1].

(3.3)

For any λ ∈ (0, λ∗), by (3.3), we have

∣∣∣Tλnx(t)∣∣∣ = λ

∫1

0H(t, s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

≤ λL1

∫1

0e(s)

[p(s)

(g([x(s) −w(s)]+ + n−1

)+ h([x(s) −w(s)]+ + n−1

))+ q(s)

]ds

≤ λL1

∫1

0e(s)

[p(s)

(g

(12rΛe(s)

)+ h∗(r)

)+ q(s)

]ds

≤ λL1

∫1

0e(s)p(s)g

(12rΛe(s)

)ds + λL1(h∗(r) + 1)

∫1

0e(s)

[p(s) + q(s)

]ds

≤ r

2+r

2= r,

(3.4)

which means that

∥∥∥Tλnx∥∥∥ ≤ ‖x‖, x ∈ ∂Kr. (3.5)

On the other hand, choose a real number M > 0 such that λML2l21Λ∫ba e(s)ds > 2,

where l1 = min0≤t≤1e(t) > 0, L2 is defined by (2.9). By (H4), there exists N > 0 such that forany t ∈ [a, b], we have

f(t, u) ≥Mu, u ≥N. (3.6)


Take R > max{r, 2N/l1Λ}. Next, we take ϕ1 ≡ 1 ∈ ∂K1 = {x ∈ K : ‖x‖ = 1}, and for anyx ∈ ∂KR, m > 0, n ∈ N, we will show

x /= Tλnx +mϕ1. (3.7)

Otherwise, there exist x0 ∈ ∂KR and m0 > 0 such that

x0 = Tλnx0 +m0ϕ1. (3.8)

From x0 ∈ ∂KR, we know that ‖x0‖ = R. Then, for t ∈ [a, b], we have

x0(t) −w(t) ≥ x0(t) − 2λL3e(t)∫1

0q(s)ds ≥ x0(t) − 2L3

x0(t)ΛR

∫1

0q(s)ds

≥ 12x0(t) ≥ 1

2RΛe(t) ≥ l1RΛ

2≥N > 0.

(3.9)

So, by (3.6), (3.9), we have

x0(t) = λ∫1

0H(t, s)

[f(s, [x0(s) −w(s)]+ + n−1

)+ q(s)

]ds +m0

≥ λL2e(t)∫1

0e(s)f

(s, [x0(s) −w(s)]+ + n−1

)ds +m0

≥ λL2e(t)∫b

a

e(s)f(s, [x0(s) −w(s)]+ + n−1

)ds +m0

≥ λML2e(t)∫b

a

e(s)([x0(s) −w(s)]+ + n−1

)ds +m0

≥ 12λML2l

21ΛR

∫b

a

e(s)ds +m0

≥ R +m0 > R.

(3.10)

This implies that R > R, which is a contradiction. This yields that (3.7) holds. By (3.5), (3.7),and Lemma 1.1, for any n ∈ N and λ ∈ (0, λ∗), we obtain that Tλn has a fixed point xn in Kr,R.

Let {xn}∞n=1 be the sequence of solutions of the boundary value problems (2.26). It iseasy to see that they are uniformly bounded. Next, we show that {xn}∞n=1 are equicontinuouson [0, 1]. From xn ∈ Kr,R, we know that

[xn(t) −w(t)]+ ≤ xn(t) ≤ ‖xn‖ ≤ R, t ∈ [0, 1],

xn(t) −w(t) ≥ xn(t) − 2λL3e(t)∫1

0q(s)ds ≥ xn(t) − 2λL3

xn(t)Λ‖xn‖

∫1

0q(s)ds

≥ 12xn(t) ≥ 1

2Λe(t)‖xn‖ ≥ 1

2Λre(t) > 0, t ∈ [0, 1].

(3.11)


For any ε > 0, by the continuity of H(t, s) in [0, 1]× [0, 1], there exists δ2 > 0 such that for anyt1, t2, s ∈ [0, 1] and |t1 − t2| < δ2, we have

|H(t1, s) −H(t2, s)| < ε. (3.12)

This, combined with (2.11) and (2.37), implies that for any t1, t2 ∈ [0, 1] and |t1 − t2| < δ2, wehave

|xn(t1) − xn(t2)| ≤∫1

0|H(t1, s) −H(t2, s)|

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

< ε

∫1

0

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

≤ ε∫1

0

[p(s)

(g([x(s) −w(s)]+ + n−1

)+ h([x(s) −w(s)]+ + n−1

))+ q(s)

]ds

≤ ε[

Lr + (1 + h∗(R))∫1

0

(p(s) + q(s)

)ds

]

.

(3.13)

By the Ascoli-Arzela theorem, the sequence {xn}∞n=1 has a subsequence being uniformlyconvergent on [0, 1]. Without loss of generality, we still assume that {xn}∞n=1 itself uniformlyconverges to x on [0, 1]. Since {xn}∞n=1 ∈ Kr,R ⊂ K, we have xn ≥ 0. By (2.26), we have

xn(t) = xn(

12

)+(t − 1

2

)x′n

(12

)

− λ∫ t

1/2ds

∫ s

1/2

[f(τ, xn(τ) −w(τ) + n−1

)+ q(τ)

]dτ, t ∈ (0, 1).

(3.14)

From (3.14), we know that {x′n(1/2)}∞n=1 is bounded sets. Without loss of generality, we may

assume x′n(1/2) → c1 as n → ∞. Then, by (3.14) and the Lebesgue dominated convergence

theorem, we have

x(t) = x(

12

)+ c1

(t − 1

2

)− λ

∫ t

1/2ds

∫s

1/2

[f(τ, x(τ) −w(τ)) + q(τ)

]dτ, t ∈ (0, 1). (3.15)

By (3.15), direct computation shows that

−x′′(t) = λ[f(t, x(t) −w(t)) + q(t)

], 0 < t < 1. (3.16)


On the other hand, letting n → ∞ in the following boundary conditions:

αxn(0) − βx′n(0) =

∫1

0xn(s)dξ(s),

γxn(1) + δx′n(1) =

∫1

0xn(s)dη(s),

(3.17)

we deduce that x is a positive solution of BVP (2.22).Let u(t) = x(t) − w(t) and l = (1/2)Λr. By (3.11) and the convergence of sequence

{xn}∞n=1, we have u(t) ≥ le(t) > 0, t ∈ [0, 1]. It then follows from Lemma 2.7 that BVP (1.1) hasat least one positive solution u satisfying u ≥ le(t) for any t ∈ [0, 1]. The proof is completed.

Theorem 3.2. Assume that conditions (H1)–(H3) are satisfied. In addition, assume that the fol-lowing condition holds.

(H5) There exists an interval [c, d] ⊂ (0, 1) such that

lim infu→+∞

mint∈[c,d]

f(t, u) >4L3

∫10 q(s)ds

ΛL2l2∫dc e(s)ds

, (3.18)

where l2 = minc≤t≤de(t) and

limu→+∞

h(u)u

= 0. (3.19)

Then there exists λ∗ > 0 such that the BVP (1.1) has at least one positive solution u(t) ∈C[0, 1] ∩ C2(0, 1) provided λ ∈ (λ∗,+∞). Furthermore, the solution also satisfies u(t) ≥le(t) for some positive constant l.

Proof. By (3.18), there exists N0 > 0 such that, for any t ∈ [c, d], u ≥N0, we have

f(t, u) ≥ 4L3∫1

0 q(s)ds

ΛL2l2∫dc e(s)ds

. (3.20)

Choose λ∗ = N0/2l2L3∫1

0 q(s)ds. Let r = 4λL3Λ−1∫1

0 q(s)ds as λ > λ∗. Next, we take ϕ1 ≡ 1 ∈∂K1 = {x ∈ K : ‖x‖ = 1}, and for any x ∈ ∂Kr , m > 0, n ∈ N, we will show that

x /= Tλnx +mϕ1. (3.21)

Otherwise, there exist x0 ∈ ∂Kr and m0 > 0 such that

x0 = Tλnx0 +m0ϕ1. (3.22)


From x0 ∈ ∂Kr , we know that ‖x0‖ = r, and

x0(t) −w(t) ≥ Λre(t) − 2λL3e(t)∫1

0q(s)ds = 2λL3e(t)

∫1

0q(s)ds ≥ N0

l2e(t). (3.23)

So, we have x0(t) −w(t) ≥N0 on t ∈ [c, d], x0 ∈ ∂Kr . Then, by (3.20) we have

x0(t) = λ∫1

0H(t, s)

[f(s, [x0(s) −w(s)]+ + n−1

)+ q(s)

]ds +m0

≥ λL2e(t)∫1

0e(s)f

(s, [x0(s) −w(s)]+ + n−1

)ds +m0

≥ λL2e(t)∫d

c

e(s)f(s, [x0(s) −w(s)]+ + n−1

)ds +m0

≥ λL2e(t)4L3

∫10 q(s)ds

ΛL2l2∫dc e(s)ds

∫d

c

e(s)ds +m0

≥ 4λΛ−1L3

∫1

0q(s)ds +m0

= r +m0 > r.

(3.24)

This implies that r > r, which is a contradiction. This yields that (3.21) holds.On the other hand, by (3.19) and the continuity of h(u) on [0,+∞), we have

limu→+∞

h∗(u)u

= 0, (3.25)

where h∗(u) is defined by (2.40). For

ε =

[

4λL1

∫1

0e(s)[p(s) + q(s)]ds

]−1

, (3.26)

there exists M0 > 0 such that when x > M0, for any 0 ≤ y ≤ x, we have h(y) ≤ εx. Take

R > max

{

2, r,M0, 2λL1

∫1

0e(s)p(s)g(Λe(s))ds

}

. (3.27)


Then, for any x ∈ ∂KR, t ∈ [0, 1], we have

[x(t) −w(t)]+ ≤ x(t) ≤ ‖x‖ ≤ R,

x(t) −w(t) ≥ x(t) − 2λL3e(t)∫1

0q(s)ds ≥ x(t) − 2λL3

x(t)ΛR

∫1

0q(s)ds

≥ 12x(t) ≥ 1

2RΛe(t) > Λe(t) > 0, t ∈ [0, 1].

(3.28)


∣∣∣Tλnx(t)

∣∣∣ = λ

∫1

0H(t, s)

[f(s, [x(s) −w(s)]+ + n−1

)+ q(s)

]ds

≤ λL1

∫1

0e(s)p(s)g

([x(s) −w(s)]+ + n−1

)ds

+ λL1

∫1

0e(s)

[p(s)h

([x(s) −w(s)]+ + n−1

)+ q(s)

]ds

≤ λL1

∫1

0e(s)p(s)g(Λe(s))ds + λL1

∫1

0e(s)

[p(s)ε(R + 1) + q(s)

]ds

≤ λL1

∫1

0e(s)p(s)g(Λe(s))ds + ελL1(R + 2)

∫1

0e(s)

[p(s) + q(s)

]ds

≤ R

2+R

2= R,

(3.29)

which means that

∥∥∥Tλnx∥∥∥ ≤ ‖x‖, x ∈ ∂KR. (3.30)

By (3.21), (3.30), and Lemma 1.1, for any n ∈ N and λ > λ∗, we obtain that Tλn has a fixed pointxn in Kr,R satisfying r ≤ ‖xn‖ ≤ R. The rest of proof is similar to Theorem 3.1. The proof iscomplete.

Remark 3.3. From the proof of Theorem 3.2, we can see that if (H5) is replaced by the followingcondition.

(H ′5) There exists an interval [c, d] ⊂ (0, 1) such that

lim infu→+∞

mint∈[c,d]

f(t, u) = +∞, limu→+∞

h(u)u

= 0, (3.31)

then, the conclusion of Theorem 3.2 is still true.


4. Applications

In this section, we construct two examples to demonstrate the application of our main results.

Example 4.1. Consider the following 4-point boundary value problem:

−u′′(t) = λ[

1√t(1 − t)

(1u2

+ u2 + 1)− q(t)

]

, 0 < t < 1,

u(0) − u′(0) = 14u

(13

)+

19u

(23

),

u(1) + u′(1) =38u

(13

)+ u(

23

),

(4.1)

where λ > 0 is a parameter and

q(t) =1

4 + 3 3√

4

⎡

⎢⎣

1√t+

1√1 − t

+1

3√(t − 1/2)2

⎤

⎥⎦. (4.2)

The BVP (4.1) can be regarded as a boundary value problem of the form of (1.1). In thissituation, α = β = γ = δ = 1 and

ξ(s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

0, s ∈[

0,13

),

14, s ∈

[13,

23

),

1336, s ∈

[23, 1],

η(s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

0, s ∈[

0,13

),

38, s ∈

[13,

23

),

118, s ∈

[23, 1].

(4.3)

Let

f(t, x) =1

√t(1 − t)

(1x2

+ x2 + 1), for (t, x) ∈ (0, 1) × (0,+∞) (4.4)

and let p(t) = 1/√t(1 − t), g(x) = 1/x2, h(x) = x2+1. By direct calculation, we have

∫10 p(t)dt =

π ,∫1

0 q(t)dt = 1, and

ρ = 3, σ =34, φ1(t) =

2 − t3

, φ2(t) =1 + t

3, e(t) =

13(2 − t)(1 + t),

k1 =263324

, k2 =1481, k3 =

4772, k4 =

518, k =

73648

,

L1 = 11, L2 =22736

, L3 =1099.

(4.5)


Clearly, the conditions (H1)–(H3) hold. Taking t ∈ [1/4, 3/4] ⊂ [0, 1], we have

lim infx→+∞

mint∈[1/4,3/4]

f(t, x)x

= lim infx→+∞

mint∈[1/4,3/4]

(1/√t(1 − t)

)(1/x2 + x2 + 1

)

x= +∞. (4.6)

Thus (H4) also holds. Consequently, by Theorem 3.1, we infer that the singular BVP (4.1) hasat least one positive solution provided λ is small enough.

Example 4.2. Consider the following problem:

−u′′(t) = λ[

1√t(1 − t)

(1u2

+√u + 1

)−

√2

4√t3(1 − t)

]

, 0 < t < 1,

u(0) − u′(0) = −∫1

0u(t) cos 2πtdt,

u(1) + u′(1) =15u

(14

)+ u(

34

),

(4.7)

where λ > 0 is a parameter. Let

f(t, x) =1

√t(1 − t)

(1x2

+√x + 1

), (t, x) ∈ (0, 1) × (0,+∞),

q(t) =√

24√t3(1 − t)

, p(t) =1

√t(1 − t)

, g(x) =1x2, h(x) =

√x + 1.

(4.8)

Then,∫1

0 p(t)dt = π ,∫1

0 q(t)dt = 2π . Here, dξ(t) = − cos 2πt dt, so the measure dξ changes signon [0, 1]. By direct calculation, we have

k1 = 1, k2 = 0, k3 =8

15, k4 =

13, k =

13,

Gξ(s) =1

4π2 (1 − cos 2πs) ≥ 0, Gη(s) ≥ 0,(4.9)

where

Gη(s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

15G2

(14, s

)+G2

(34, s

), 0 ≤ s < 1

4,

15G1

(14, s

)+G2

(34, s

),

14≤ s ≤ 3

4,

15G1

(14, s

)+G1

(34, s

),

34< s ≤ 1,

G(t, s) =

⎧⎪⎪⎨

⎪⎪⎩

G1(t, s) =(2 − s)(1 + t)

3, 0 ≤ t ≤ s ≤ 1,

G2(t, s) =(2 − t)(1 + s)

3, 0 ≤ s ≤ t ≤ 1.

(4.10)


Taking t ∈ [1/4, 3/4] ⊂ [0, 1], we have

lim infx→+∞

mint∈[1/4,3/4]

f(t, x) = lim infx→+∞

mint∈[1/4,3/4]

1√t(1 − t)

(1x2

+√x + 1

)= +∞,

limx→+∞

h(x)x

= limx→+∞

√x + 1x

= 0.

(4.11)

So all assumptions of Theorem 3.2 are satisfied. By Theorem 3.2, we know that BVP (4.7) hasat least one positive solution provided λ is large enough.

Acknowledgments

J. Jiang and L. Liu were supported financially by the National Natural Science Foundationof China (11071141, 11126231), the Natural Science Foundation of Shandong Province ofChina (ZR2011AQ008), and a Project of Shandong Province Higher Educational Science andTechnology Program (J11LA06). Y. Wu was supported financially by the Australia ResearchCouncil through an ARC Discovery Project Grant.

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Research ArticleThe Asymptotic Solution ofthe Initial Boundary Value Problem to aGeneralized Boussinesq Equation

Zheng Yin and Feng Zhang

Department of Applied Mathematics, Southwestern University of Finance and Economics,Chengdu 610074, China

Correspondence should be addressed to Zheng Yin, [email protected]

Received 15 April 2012; Accepted 3 May 2012


Copyright q 2012 Z. Yin and F. Zhang. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

The L2 space solution of an initial boundary problem for a generalized damped Boussinesqequation is constructed. Certain assumptions on the coefficients of the equation are found to showthe existence and uniqueness of the global solution to the initial boundary problem. The explicitexpression for the large time asymptotic solution is obtained.

1. Introduction

The classical Boussinesq equation can be expressed by

utt = −αuxxxx + uxx + β(u2)

xx, (1.1)

where u(x, t) is the elevation of the free surface of fluid, the subscripts denote partialderivatives, and the constant coefficients α and β depend on the depth of fluid and thecharacteristic speed of the long waves. The Boussinesq equation (1.1) was first derived byBoussinesq [1], in 1872, to describe propagation of long waves with small amplitude on thesurface of shallow water. Since then, extensive research has been carried out to study theproperty and solutions of the equation and its associated initial boundary value problems.Clarkson [2] proposed a general approach for constructing exact solutions of (1.1). Hirota[3] deduced conservation laws and then examined the numerical solution of a Boussinesqequation. Yajima [4] investigated the nonlinear evolution of a linearly stable solution for(1.1). The exponentially decaying solution of the spherical Boussinesq equation has been


investigated by Nakamura [5]. Galkin et al. [6] developed rational solutions of the one-dimensional Boussinesq equation with both zero and nonzero boundary conditions at theinfinity in the space from the known solutions of the Kadomtzev-Petviashvili equation. Thestructure of their solutions generalizes a family of rational solutions of the Korteweq-deVries equation to the case of two-wave processes. For many other methods to investigatethe Boussinesq system and other shallow water models, the reader is referred to [7–13] andthe references therein.

Various generalizations of the classical Boussinesq equation have been proposed andstudied from many aspects, in particular the well-posedness of the Cauchy problem for thefollowing equation,

utt = −uxxxx + uxx +(f(u)

)xx. (1.2)

Bona and Sachs [14] showed that the special solitary wave solutions of (1.2) are nonlinearlystable for a range of wave speeds. The local and global well-posedness of the problem hasbeen proved by transforming (1.2) into a system of nonlinear Schrodinger equations [15].

Varlamov [16] considered the initial-boundary condition for the following dampedBoussinesq equation:

utt − 2butxx = −αuxxxx + uxx + β(u2)

xx, (1.3)

where the second term on the left-hand side is responsible for dissipation, β ∈ R1, and α andb are assumed to be positive constants and satisfy the assumption that α > b2. An initial-boundary value problem for (1.3) with small initial data is considered for the case of onespace dimension. The classical solution of the problem is constructed, and the long-timeasymptotics is obtained in explicit form. The asymptotics show the presence of both timeand space oscillations and the exponential decay of the solution in time due to dissipations.Varlamov [17] has also used the eigenfunction expansion method to consider the long-timeasymptotics for a damped Boussinesq equation in a ball which is similar to (1.3).

The aim of this paper is to study the initial-boundary value problem for the followinggeneralized damped Boussinesq equation:

utt − auttxx − 2butxx +2baut = −cuxxxx + uxx − p2u + β

(u2)

xx, (1.4)

where a, b and c are positive constants and constants p ≥ 0 and β ∈ R1. We assume thecases a + c > b2 and p ≥ √

2b/a. The classical solutions of the initial-boundary value problemfor (1.4) will be constructed in the form of a Fourier series with coefficients in their ownterm represented as series in terms of a small parameter involved in the initial conditions.It is shown that the new solution of the initial-boundary value problem for (1.4) is well-posedness. In addition, the long time behavior of our solution also shows the presence ofdamped oscillations decaying exponentially in time as t → ∞.


2. Theorem of Existence, Uniqueness, and Asymptotics

In this paper, we consider the following initial-boundary value problem for the dampedBoussinesq equation:

utt − auttxx − 2butxx +2baut = −cuxxxx + uxx − p2u + β

(u2)

xx,

u(0, t) = u(π, t) = 0, t > 0,

uxx(0, t) = uxx(π, t) = 0, t > 0,

u(x, 0) = ε2ϕ(x), ut(x, 0) = ε2ψ(x), x ∈ (0, π),

(2.1)

where a, b, c, and p are positive constants, β ∈ R, and ε is a small parameter.

Definition 2.1. The function u(x) ∈ C2n(0, π), n ≥ 1, if u(0) = u(π) = u′′ (0) = u′′ (π) = · · · =u(2n−2)(0) = u(2n−2)(π) = 0 and u(2n)(x) ∈ L2(0, π).

Definition 2.2. The function u(x, t) defined on [0, π]×[0,+∞) is said to be the classical solutionof the problem defined by (2.1), if it has two continuous derivatives on [0, π] × [0,+∞) andsatisfies system (2.1).

The main results obtained are summarized in the following theorem.

Theorem 2.3. If a > 0, a + c > b2, p ≥ √2b/a, ϕ(x) ∈ C8(0, π), ψ(x) ∈ C6(0, π), then there is a

ε0 > 0 such that, for 0 < ε ≤ ε0, problem (2.1) has a unique classical solution of which is representedin the form of

u(x, t) =∞∑

N=0

εN+1u(N)(x, t), (2.2)

where the function u(N)(x, t) is as that defined by expressions (3.24) in the proof below. This seriesand the series for the derivatives of u(x, t) involved in (2.1) converge absolutely and uniformly withrespect to x ∈ [0, π], t ≥ 0, ε ∈ [0, ε0]. In addition, the solution of system (2.1) has the following longtime behavior as t → +∞ :

u(x, t) = e−(b/a)t[(A cos σt + B sin σt) sinx +O

(e[(−ηb/a)t]

)], (2.3)

where σ =√ac + (a + c − b2) + (1 + ap2 − (b2/a2)) + p2 − (b2/a2)/1 + a is a positive constant,

0 < η < 1, and the coefficients A and B are defined by (3.44).

3. Proof of Theorem 2.3

The proof of Theorem 2.3 includes three parts, namely, existence of a solution, uniquenessof the solution, and the asymptotics of the solution. The main techniques used in this paperare based on those presented in [16]. However, it should be emphasized that the method forproving uniqueness of the solution is different from that of [16].


3.1. Existence of a Solution

We make an odd extension of u(x, t) in x to the interval [−π, 0] and represent u in the form ofa complex Fourier series; namely,

u(x, t) =∞∑

n=−∞n/= 0

un(t)einx, un(t) =1

2π

∫π

−πu(x, t)e−inxdx, (3.1)

where u−n(t) = −un(t) for n ≥ 1. For the above equalities, we shall use the fact that u(x, t)belongs to the space L2(−π,π) for each fixed t > 0, and we shall denote the correspondingnorm by

‖u(t)‖ = ‖u(t)‖L2(−π,π) =(∫π

−π|u(x, t)|2dx

)1/2

. (3.2)

From (3.1), we obtain, through a simple calculation, that

u(x, t) = 2i∞∑

n=1

un(t) sinnx, un(t) =1iπ

∫π

0u(x, t) sinnx dx. (3.3)

Noting the initial functions, we can write on [−π,π] with

ϕ−n = ϕn, ψ−n = ψn, (3.4)

for n ≥ 1 that

ϕ(x) =∞∑

n=−∞n/= 0

ϕneinx, ψ(x) =

∞∑

n=−∞n/= 0

ψneinx. (3.5)

Then on the interval [0, π], we obtain

ϕ(x) = 2i∞∑

n=1

ϕn sinnx, ψ(x) = 2i∞∑

n=1

ψn sinnx,

ϕn =1iπ

∫π

0ϕ(x) sinnx dx, ψn =

1iπ

∫π

0ψ(x) sinnx dx.

(3.6)


Integrating (3.6) by parts and using the smoothness assumption of the initial data, we get thefollowing inequalities:

∣∣ϕn

∣∣ ≤ C1n

−8,∣∣ψn

∣∣ ≤ C1n

−6, n ≥ 1, (3.7)

where C1 is a positive constant. Substituting (3.1) and (3.5) into (2.1), we have the followingCauchy problem for the function un(t), n ∈ Z:

(1 + an2

)u′′n(t) + 2

(bn2 +

b

a

)u′n(t) +

(cn4 + n2 + p2

)un(t) = −βn2P(un(t)),

un(0) = ε2ϕn, u′n(0) = ε2ψn,

(3.8)

where

P(un(t)) =∞∑

g=−∞g /= 0,n

un−g(t)ug(t), u−n(t) = −un(t) for n ≥ 1. (3.9)

For n = 1, we have

P(u1(t)) = 2∞∑

g=1

u−g(t)u1+g(t) = −2∞∑

g=1

ug(t)u1+g(t). (3.10)

For n ≥ 2, it has

P(un(t)) =n−1∑

g=1

un−g(t)ug(t) − 2∞∑

g=1

ug(t)un+g(t). (3.11)

Letting Φn = εϕn, Ψn = εψn and solving the initial value problem (3.8), we get

un(t) = εe−(b/a)t{[

cos(σnt) +b

a· sin(σnt)

σn

]Φn +

sin(σnt)σn

Ψn

}

− βn2

σn(1 + an2)

∫ t

0exp

[−ba(t − τ)

]sin [σn(t − τ)]P(un(τ))dτ,

(3.12)

where

σn =

√acn6 + (a + c − b2)n4 +

(1 + ap2 − (2b/a)

)n2 + p2 − (b2/a2)

1 + an2,

n ≥ 1, a + c > b2, p ≥√

2ba

.

(3.13)


Now, in order to solve the integral equation, we use the perturbation theory. Firstly, weexpress un(t), n ≥ 1, as a formal series in ε:

un(t) =∞∑

N=0

εN+1ξ(N)n (t). (3.14)

Then, by substituting (3.14) into (3.12) and equating the coefficients of the like powers of ε,we get the following formulas, respectively, for N = 0 and N ≥ 1 :

ξ(0)n (t) = εe−(b/a)t

{[cos(σnt) +

b

a· sin(σnt)

σn

]Φn +

sin(σnt)σn

Ψn

}, (3.15)

ξ(N)n (t) = − βn2

σn(1 + an2)

∫ t

0exp

[−ba(t − τ)

]× sin [σn(t − τ)]QN

(ξ(j)n (τ)

)dτ, N ≥ 1, (3.16)

where n ≥ 1 and

QN

(ξ(j)n (τ)

)= εn

n−1∑

g=1

N∑

j=1

ξ(j−1)n−g (τ)ξ(N−j)

g (τ) − 2∞∑

g=1

N∑

j=1

ξ(j−1)n+g (τ)ξ(N−j)

g (τ), (3.17)

in which ε1 = 0 and εn = 1 for n ≥ 2.Now we must prove that the formally constructed function (3.3), together with (3.12)–

(3.14), does represent a solution of problem (2.1). To do this, we shall show that the series

u(x, t) =∞∑

n=−∞n/= 0

einx∞∑

N=0

εN+1ξ(N)n (t)

= 2i∞∑

n=1

sin nx∞∑

N=0

εN+1ξ(N)n (τ)

(3.18)

converges absolutely and uniformly. For this purpose, we firstly establish the following timeestimates for n ≥ 1, t > 0, N ≥ 0:

∣∣∣ξ(N)n (t)

∣∣∣ ≤ CN(N + 1)−2n−6e−(b/a)t. (3.19)

Here and in the sequel we denote by C any positive constant independent of N, n, ε, and t,but possibly depending on the coefficients of the equation and the initial functions.

We shall use the induction on the number N. For N = 0, we have from (3.7) and (3.15)that

∣∣∣ξ(0)n (t)∣∣∣ ≤ εe−(b/a)t

[(1 +

b

aσn

)∣∣∣Φn

∣∣∣ +1σn

|Ψn|]

≤ εn−6e−(b/a)t.

(3.20)


Assuming that (3.19) is valid for all ξ(s)n (t) with 0 ≤ s ≤ N − 1, we shall prove that (3.19) alsoholds for s =N. According to [16], for any integer n ≥ 1, g ≥ 1, and g /=n, we have

∣∣n − g∣∣−6

g−6 ≤ 26n−6[g−6 +

∣∣n − g∣∣−6

],

j−2(N + 1 − j)−2 ≤ 22(N + 1)−2[j−2 +

(N + 1 − j)−2

].

(3.21)

From (3.16) and noting that

bn2

1 + an2≥ b

afor n/= 0, (3.22)

we have

∣∣∣ξ(N)n (t)

∣∣∣ ≤ C∣∣β∣∣(N + 1)−2n−6

∞∑

g=1

(g−6 +

∣∣n − g∣∣−6)

×∞∑

j=1

Cj−1CN−j[(N + 1 − j)−2 + j−2

]|SN(n, t)|, |SN(n, t)|

≤ Ce−(b/a)t∫ t

0e((b/a)−(2b/a))τdτ

≤ Ce−(b/a)t ×∣∣∣∣∣e((b/a)−(2b/a))t − 1(b/a) − (2b/a)

∣∣∣∣∣

≤ Ce−(b/a)t.

(3.23)

Therefore, we know that inequality (3.19) holds.For the derivation of (3.15), we recall (3.3) and (3.14) with ξ

(N)n defined by (3.15) and

(3.16) and interchange the order of summations in the series; namely,

u(x, t) = 2i∞∑

n=1

un(t) sinnx = 2i∞∑

n=1

sinnx∞∑

N=0

εN+1ξ(N)n (t) =

∞∑

N=0

εN+1u(N)(x, t), (3.24)

where

u(N)(x, t) = 2i∞∑

n=1

ξ(N)n sinnx. (3.25)


The interchange of the order of summations is allowable as the series is absolutely anduniformly convergent for x ∈ [0, π], t ≥ 0 and ε ∈ [0, ε0]. Differentiating (3.15) and (3.16), weget, for k = 1, 2, that

∂tkξ(0)n (t) =

k∑

l=0

clk(−1)l(b

a

)l

e−(b/a)t∂k−lk

{[cos(σnt) +

b

a

sin(σnt)σn

]Φn +

sin(σnt)σn

Ψn

},

∂kt ξ(N)n (t) = − βn2

(1 + an2)σn

∫ t

0gk(n, t − τ)QN

(ξ(j)n (τ)

)dτ + Rk(n, t),

(3.26)

where

gk(n, t) =k∑

l=0

clk(−1)l(b

a

)l

e−(b/a)tσk−ln sin[σnt +

(k − l)π2

], (3.27)

QN(ξ(j)n (τ)) is defined by (3.16), clk(l = 0, 1, . . . , k) are binomial coefficients, and Rk(n, t) (k =

1, 2) are obtained by differentiating the integral in (3.16) and are as follows:

R1(n, t) = 0, R2(n, t) = −βn2QN

(ξ(j)n (t)

). (3.28)

Hence, using (3.19), we deduce that for, n ≥ 1, N ≥ 0, t > 0, and k ≥ 0, 1, 2,

∣∣∣∂kt ξ(N)n (t)

∣∣∣ ≤ CN(N + 1)−2n2k−6e−(b/a)t,

∣∣∣∂kt un(t)∣∣∣ ≤ Cn2k−6e−(b/a)t.

(3.29)

Using these estimates and calculating the necessary derivatives of (3.29), we can provestraightforwardly that (3.29) represents the classical solution of the initial-boundary valueproblem defined by system (2.1).

3.2. Uniqueness of the Solution

For proving the uniqueness of the constructed solution, we shall assume that there exist twoclassical solutions u(1)(x, t) and u(2)(x, t) to system (2.1) and then deduce that u(1)(x, t) mustbe equal to u(2)(x, t).


Making an odd extension of the two solutions to the segment (−π, 0], we notice thatboth of them belong to the L2(−π,π) space, and according to Definition 2.2, we have for eachfixed time t > 0 that

maxx∈[−π,π]

∣∣∣u(1)(x, t)

∣∣∣ ≤ Ct, max

x∈[−π,π]

∣∣∣u(2)(x, t)

∣∣∣ < Ct, (3.30)

where Ct is a constant depending on t. Let w(x, t) = u(1)(x, t) − u(2)(x, t), and make an evenextension of w(x, t) to . . . (−3π,−2π), (−2π,−π), (π, 2π), (2π, 3π), . . . Then, we have from(2.1) that

wtt − awttxx − 2bwtxx +2bawt = −cwxxxx +wxx − p2w + β

[w(x, t)

(u(1)(x, t) + u(2)(x, t)

)]

xx,

w(x, 0) = wt(x, t) = 0.(3.31)

Taking the Fourier transform of w in the interval (−∞,+∞); namely,

w(ξ, t) =∫+∞

−∞e−ixξw(x, t)dx, (3.32)

we have

(1 + aξ2

)w′′(ξ, t) + 2b

(ξ2 +

b

a

)w(ξ, t) +

(cξ4 + ξ2 + p2

)w(ξ, t) = −βξ2f(ξ, t), (3.33)

where

f(x, t) = w(u(1) + u(2)

)(x, t). (3.34)

From (3.33), we have

w(ξ, t) = − βξ2

σξ(1 + aξ2)

∫ t

0exp

[−ba(t − τ)

]sin

(σξ(t − τ)

)f(ξ, τ)dτ, (3.35)

where

σξ =

√acξ6 + (a + c − b)ξ4 +

(1 + ap2 − (2b2/a)

)ξ2 + p2 − (b2/a2)

1 + aξ2. (3.36)


Thus, it has

|w(ξ, t)| ≤ C

∫ t

0

∣∣∣∣exp

[−ba(t − τ)

]f(ξ, τ)

∣∣∣∣dτ

≤ C

[∫ t

0exp

[−ba(t − τ)

]dτ

]1/2[∫ t

0

∣∣∣f(ξ, τ)

∣∣∣

2dτ

]1/2

≤ C

[∫ t

0

∣∣∣f(ξ, τ)

∣∣∣

2dτ

]1/2

.

(3.37)

It follows from (3.37) and the Parseval inequality that

∫+∞

−∞|w(ξ, t)|2dξ ≤ C

∫+∞

−∞

∫ t

0

∣∣∣f(ξ, τ)∣∣∣

2dτ dξ

≤ C

∫ t

0

∥∥∥f(ξ, t)∥∥∥

2

L2dτ

≤ C

∫ t

0

∥∥∥w(x, τ)(u(1)(x, τ) + u(2)(x, τ)

)∥∥∥2

L2dτ

≤ C

∫ t

0Cτ‖w(x, τ)‖2

L2dτ.

(3.38)

Using the Gronwall inequality, we obtain

w(x, t) = 0(

in L2), (3.39)

which implies that u(1)(x, t) ≡ u(2)(x, t) and thus the solution is unique.

3.3. Long Time Asymptotics

To find the long-time behavior of the constructed solution, we firstly determine a subtleasymptotic estimate of u1(t) which will contribute to the major term and then estimate theremaining terms

∑∞n=2 un(t) sinnx.

Since

u1(t) =∞∑

N=0

εN+1ξ1(N)

(t), (3.40)


we substitute it into the integral equations (3.15) and (3.16) to obtain

ξ1(0)(t) = e−(b/a)t

[A(0) cos(σt) + B(0) sin(σt)

],

ξ1(N)

(t) = e−(b/a)t{[A(N) + R(N)

A (t)]

cos(σt) +[B(N) + R(N)

B (t)]

sin(σt)},

(3.41)

where

A(0) = εϕ1,

B(0) =ε

σ

(b

1 + aϕ1 + ψ1

),

σ =

√ac + 1 + a + c − b2

1 + a,

A(N) =β

σ(1 + a)

∫ t

0e(b/a)τ sin(στ)Sun(τ)dτ,

B(N) = − β

σ(1 + a)

∫ t

0e(b/a)τ cos(στ)Sun(τ)dτ,

R(N)A =

β

σ(1 + a)

∫+∞

0e(b/a)τ sin(στ)Sun(τ)dτ,

R(N)B = − β

σ(1 + a)

∫+∞

0e(b/a)τ cos(στ)Sun(τ)dτ,

Sun(t) =∞∑

g=1

N∑

j=1

ξ(j−1)1+g (t)ξ(N−j)

g (t), N ≥ 1,

(3.42)

and the function ξ(j)(t), j = 0, 1, . . . ,N − 1 are defined by (3.15)-(3.16). Taking n ≥ 2, N ≥ 1and using a method similar to that used in [16], it follows from (3.19) and (3.24) that thereexists a positive number 0 < η < 1 such that

∣∣∣R(N)A (t)

∣∣∣ ≤ ce−(ηbt/a),∣∣∣R(N)

B (t)∣∣∣ ≤ ce−(ηbt/a). (3.43)

Hence, we have proved that, as t → +∞,

2iu1(t) = e−(b/a)t[A cos(σt) + B sin(σt)] +O(e−((1+η)b/a)t

),

A = 2i∞∑

N=0

εN+1A(N), B = 2i∞∑

N=0

εN+1B(N),(3.44)

where 0 < η < 1,A(N) andB(N) are defined by (3.41), and the series above converge absolutelyand uniformly for ε ∈ [0, ε0].


Now, we can represent the solution by

u(x, t) = 2iu1(t) + Ru(x, t),

Ru(x, t) = 2i∞∑

n=2

sinnx∞∑

N=0

ξn(N)

(t).(3.45)

Using (3.24), we deduce that

|Ru(x, t)| ≤ exp[−ηba

t

] ∞∑

n=0

cNεN+1(N + 1)−2∞∑

N=2

n−6 ≤ c exp[−ηba

t

], (3.46)

u(x, t) = e−(b/a)t[(A cosσt + B sinσt) sinx +O

(e−[(ηb/a)t]

)], 0 < η < 1. (3.47)

It follows from (3.47) that the inequality (2.3) holds.

Acknowledgments

This work is supported by the Fundamental Research Funds for Central Universities (SWUFE2012).

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Research ArticleMultiple Solutions for Degenerate Elliptic SystemsNear Resonance at Higher Eigenvalues

Yu-Cheng An1, 2 and Hong-Min Suo1

1 School of Sciences, Guizhou Minzu University, Guiyang 550025, China2 School of Mathematics and Computer Science, Bijie University, Bijie 551700, China

Correspondence should be addressed to Hong-Min Suo, [email protected]

Received 24 February 2012; Accepted 29 April 2012


Copyright q 2012 Y.-C. An and H.-M. Suo. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

We study the degenerate semilinear elliptic systems of the form −div(h1(x)∇u) = λ(a(x)u +b(x)v) + Fu(x, u, v), x ∈ Ω,−div(h2(x)∇v) = λ(d(x)v + b(x)u) + Fv(x, u, v), x ∈ Ω, u|∂Ω = v|∂Ω = 0,where Ω ⊂ RN(N ≥ 2) is an open bounded domain with smooth boundary ∂Ω, the measurable,nonnegative diffusion coefficients h1, h2 are allowed to vanish in Ω (as well as at the boundary∂Ω) and/or to blow up inΩ. Some multiplicity results of solutions are obtained for the degenerateelliptic systems which are near resonance at higher eigenvalues by the classical saddle pointtheorem and a local saddle point theorem in critical point theory.

1. Introduction

In this paper, we study a class of degenerate elliptic systems:

−div(h1(x)∇u) = λ(a(x)u + b(x)v) + Fu(x, u, v), x ∈ Ω,

−div(h2(x)∇v) = λ(d(x)v + b(x)u) + Fv(x, u, v), x ∈ Ω

u|∂Ω = v|∂Ω = 0,

, (1.1)

where Ω ⊂ RN(N ≥ 2) is an open bounded domain with smooth boundary ∂Ω, F ∈ C1(Ω ×R2, R) satisfies the following sublinear growth condition:

lim|s|→∞

|∇F(x, s)||s| = 0 (1.2)


uniformly in x ∈ Ω, where∇F = (Fu, Fv) denotes the gradient of F with respect to (u, v) ∈ R2.The degeneracy of this system is considered in the sense that the measurable, nonnegativediffusion coefficients h1, h2 are allowed to vanish inΩ (as well as at the boundary ∂Ω) and/orto blow up in Ω. The consideration of suitable assumptions on the diffusion coefficients willbe based on the work [1], where the degenerate scalar equation was studied. We introducethe function space (N)h, which consists of functions h : Ω ⊂ RN → R, such that h ∈ L1(Ω),h−1 ∈ L1(Ω), and h−s ∈ L1(Ω), for some s > N/2.

Then for the weight functions h1, h2 we assume the following hypothesis. (N) thereexist functions μ satisfying condition (N)μ, for some sμ, and ν satisfying condition (N)ν, forsome sν, such that

μ(x)k1

≤ h1(x) ≤ k1μ(x), ν(x)k2

≤ h2(x) ≤ k2ν(x), (1.3)

a.e. in Ω, for some constants k1 > 1 and k2 > 1.The mathematical modeling of various physical processes, ranging from physics to

biology, where spatial heterogeneity plays a primary role, is reduced to nonlinear evolutionequations with variable diffusion or dispersion. Also note that problem (1.1) is closely related(see [1]) to the following system:

−div(h1(x, u, v)∇u) = f(λ, x, u, v,∇u,∇v), x ∈ Ω,

−div(h2(x, u, v)∇v) = g(λ, x, u, v,∇u,∇v), x ∈ Ω,

u|∂Ω = v|∂Ω = 0.

(1.4)

Problems of such a type have been successfully applied to the heat propagation inheterogeneous materials, to the study of transport of electron temperature in a confinedplasma, to the propagation of varying amplitude waves in a nonlinear medium, to the studyof electromagnetic phenomena in nonhomogeneous superconductors and the dynamics ofJosephson junctions, to electrochemistry, to nuclear reaction kinetics, to image segmentation,to the spread ofmicroorganisms, to the growth and control of brain tumors, and to populationdynamics (see [2–4] and the references therein).

An example of the physical motivation of the assumptions (N), (N)h may be foundin [3]. These assumptions are related to the modeling of reaction diffusion processes incomposite materials occupying a bounded domain Ω, which at some point they behave asperfect insulators. When at some point the medium is perfectly insulated, it is natural toassume that h1(x) and/or h2(x) vanish in Ω. For more information we refer the reader to [4]and the references therein.

For the perturbed problem,Mawhin and Schmitt [5] first considered the following twopoint boundary value problem:

−u′′ − λu = f(x, u) + h(x), u(0) = u(π) = 0. (1.5)

Under the assumption that f is bounded and satisfies a sign condition, if the parameter λis sufficiently close to λ1 from left, problem (1.5) has at least three solutions, if λ1 ≤ λ < λ2,problem (1.5) has at least one solution, where λ1, λ2 are the first and the second eigenvaluesof the corresponding linear problem. Ma et al. [6] considered the boundary value problem


Δu + λu + f(x, u) = h(x) defined on a bounded open set Ω ⊂ RN , no matter whether theboundary conditions are Dirichlet or Neumann condition, as the parameter λ approaches λ1from left, there exist three solutions. Moreover, existence of three solutions was obtained forthe quasilinear problem in bounded domains as the parameter λ approaches λ1 from left. In[7, 8], these results were extended to the perturbed p-Laplacian equation in RN . In [9], Ouand Tang extended above some results to some elliptic systems with the Dirichlet boundaryconditions. Especially, de Paiva and Massa in [10] studied the semilinear elliptic boundaryvalue problem in any spatial dimension and by using variational techniques, they showedthat a suitable perturbation will turn the almost resonant situation (λ near to λk, i.e., nearresonance with a nonprincipal eigenvalue) in a situation where the solutions are at leasttwo. In [11], those results were extended to the cooperative elliptic systems in the boundeddomain. Motivated by the idea above, we have the goal in this paper of extending theseresults in [10, 11] to some degenerate elliptic systems with the Dirichlet boundary conditions.

2. Preliminaries and Main Results

Let h(x) be a nonnegative weight function in Ω which satisfies condition (N)h. We considerthe weighted Sobolev space D1,2

0 (Ω, h) to be defined as the closure of C∞0 (Ω) with respect to

the following norm:

‖u‖h =(∫

Ωh(x)|∇u|2dx

)1/2

, (2.1)

and the following scalar product:

〈u, v〉h =∫

Ωh(x)∇u∇v dx, (2.2)

for all u, v ∈ D1,20 (Ω, h). The space D1,2

0 (Ω, h) is a Hilbert space. For a discussion about thespace setting we refer to [1] and the references therein. Let

2∗s =2Ns

N(s + 1) − 2s. (2.3)

Lemma 2.1. Assume that Ω is a bounded domain in RN and the weight h satisfies (N)h. Then thefollowing embeddings hold:

(i) D1,20 (Ω, h) ↪→ L2∗s(Ω) continuously,

(ii) D1,20 (Ω, h) ↪→ Lr(Ω) compactly for any r ∈ [1, 2∗s).

In the sequel one denotes by 2∗μ and 2∗ν the quantities 2∗sμ and 2∗sν , respectively, where sμ andsν are induced by condition (N), recall that h1, h2 satisfy (N). The assumptions concerning thecoefficient functions of systems (1.1) are as follows.

(AD) The functions a, d ∈ C(Ω, R) and there exists x0 ∈ Ω, such that a(x0) > 0, d(x0) > 0.

(B) The function b ∈ C(Ω, (0,+∞)).


The space setting for our problem is the product space H = D1,20 (Ω, h1) × D1,2

0 (Ω, h2)equipped with the following norm:

‖z‖ =(‖u‖2h1 + ‖v‖2h2

)1/2, z = (u, v) ∈ H, (2.4)

and the following scalar product:

⟨z, φ

⟩= 〈u, ξ〉h1 + 〈v, τ〉h2 , (2.5)

for all z = (u, v), φ = (ξ, τ). Observe that inequalities (1.3) in condition (N) imply thatthe functional spaces D1,2

0 (Ω, h1) × D1,20 (Ω, h2) and D1,2

0 (Ω, μ) × D1,20 (Ω, ν) are equivalent.

Especially, by Lemma 2.1 we know that for any 1 ≤ δ ≤ min{2∗μ, 2∗ν}, the embeddingH ↪→ Lδ(Ω) × Lδ(Ω) is continuous and there is a positive constant S = S(δ,N,Ω) such that∫Ω |z|δdx ≤ Sδ‖z‖δ for all z ∈ H. Moreover, if δ < min{2∗μ, 2∗ν}, the embedding above is alsocompact. Let

H(x) = diag(h1(x), h2(x)), A(x) =(a(x) b(x)b(x) d(x)

). (2.6)

Assume that hypothesis (N) is satisfied and the coefficient functions a, d and b satisfyconditions (AD) and (B), respectively.

We consider the eigenvalue problem with weight A(x),

−div(h1(x)∇u) = λ(a(x)u + b(x)v), x ∈ Ω,

−div(h2(x)∇v) = λ(d(x)v + b(x)u), x ∈ Ω,

u|∂Ω = v|∂Ω = 0.

(2.7)

A simple calculation shows that λ is an eigenvalue of (2.7) if and only if

TAz = λ−1z, (2.8)

where TA : H → H is the symmetric bounded linear operator defined by

⟨TAz, φ

⟩=∫

Ω

(A(x)z, φ

)dx, ∀z, φ ∈ H. (2.9)

Since the coefficient of A are continuous functions and the embedding H ↪→ L2(Ω) × L2(Ω)is compact, we can check that the operator TA is also compact. Thus, we may invoke thespectral theory for compact operators to conclude thatH possesses a Hilbertian basis formedby eigenfunctions of (2.7).

Let us denote z = (u, v) and

λ−11 = μ1 = sup{〈TAz, z〉 : ‖z‖ = 1}. (2.10)


Recalling that A satisfies (AD) and (B), we can use [2, Theorem 1.1] (p = q = 2, α = β = 0)to conclude that the eigenvalue μ1 is positive, simple, and isolated in the spectrum of TA.Moreover, if we denote by φ1 the normalized eigenfunction associated to λ1, we can supposethat the φ1 is positive on Ω. By using induction, if we suppose that μ1 > μ2 ≥ . . . ≥ μk−1 arethe k − 1 first eigenvalues of TA and {φi}k−1i=1 are the associated normalized eigenfunctions, wecan define

λ−1k = μk = sup{〈TAz, z〉 := ‖z‖ = 1, z ∈ (

span{φ1, . . . , φk−1

})⊥}. (2.11)

It is proved in [12, Proposition 1.3], that, if μk > 0, then it is an eigenvalue of TA withassociated normalized eigenfunction φk. In view of the condition (AD), we can argue as inthe proof of [12, Proposition 1.11(c)], and conclude that μk > 0. Thus, we obtain a sequenceof eigenvalues for (2.7).

0 < λ1 < λ2 ≤ · · · ≤ λk ≤ · · · (2.12)

such that λk → ∞ as k → ∞. We denote by Ek the eigenfunction space corresponding toλk. Moreover, if we set Vk = span{φ1, . . . , φk}, we can decomposeH = Vk⊕V ⊥

k. Moreover, the

following inequalities hold:

∫

Ω(H(x)∇z,∇z)dx ≤ λk

∫

Ω(A(x)z, z)dx, ∀z ∈ Vk, (2.13)

∫

Ω(H(x)∇z,∇z)dx ≥ λk+1

∫

Ω(A(x)z, z)dx, ∀z ∈ V ⊥

k . (2.14)

Lemma 2.2 (from Lemma 4.6 of [10]). Let X be a Hilbert space with orthonormal direct sumsplitting X = V ⊕ Z ⊕W . Moreover, let dim(V ⊕ Z) <∞. For ρ > R > 0, set

A = {u ∈W : ‖u‖ ≥ R} ∪ {u ∈ Z ⊕W : ‖u‖ = R},B =

{u ∈ V ⊕ Z : ‖u‖ = ρ

}.

(2.15)

Then A links with B.

Lemma 2.3 (from Theorem 8.1 of [13]). Let H = X1 ⊕ X2 be a Hilbert space where X1 has finitedimension, J ∈ C1(H,R) satisfying the (P.S.) condition and such that, for given ρ1, ρ2 > 0,

supz∈ρ1S1

J(z) < a = infz∈ρ2B2

J(z) ≤ b = supz∈ρ1B1

J(z) < infz∈ρ2S2

J(z), (2.16)

where Bi and Si represent the unit ball and the unit sphere in Xi : i = 1, 2.Then there exists a critical point z0 such that J(z0) ∈ [a, b].


Next, in order to state our main results, we introduce the following assumptions onthe nonlinear term:

(F1) lim|s|→∞(∇F(x, s), s)/|s| = +∞ uniformly with respect to x ∈ Ω.

(F2) lim|s|→∞F(x, s) = +∞ uniformly with respect to x ∈ Ω.

(F3) lim|s|→∞(∇F(x, s), s)/|s| = −∞ uniformly with respect to x ∈ Ω.

(F4) lim|s|→∞F(x, s) = −∞ uniformly with respect to x ∈ Ω.

Our main results are given by the following theorems.

Theorem 2.4. Let λk(k ≥ 2) be an eigenvalue of multiplicitym. Suppose that condition (N) and thecoefficient functions a, d and b satisfy conditions (AD) and (B), respectively. Assume, in addition,that F satisfies (1.2) and ∇F(x, 0) = 0 for all x ∈ Ω, and one of the sets of hypotheses (F1) or (F2).Then there exists δ0 > 0 such that for λ ∈ (λk − δ0, λk) problem (1.1) has at least two solutions.

Theorem 2.5. Let λk(k ≥ 2) be an eigenvalue of multiplicitym. Suppose that conditions (N) and thecoefficient functions a, d and b satisfy conditions (AD) and (B), respectively. Assume, in addition,that F satisfies (1.2) and ∇F(x, 0) = 0 for all x ∈ Ω, and one of the sets of hypotheses (F3) or (F4).Then there exists δ1 > 0 such that for λ ∈ (λk, λk + δ1) problem (1.1) has at least two solutions.

3. Proof of Theorems

Consider the C1 functional J : H → R, ∀z ∈ H,

J(z) =12

∫

Ω(H(x)∇z,∇z)dx − λ

2

∫

Ω(A(x)z, z)dx −

∫

ΩF(x, z)dx. (3.1)

Since the problems in Theorems 2.4 and 2.5 are not resonant, J satisfies the Palais-Smalecondition of compactness (see, e.g., in [14]). In addition, z ∈ H is a weak solution of problem(1.1) if and only if z is a critical point of J .

We set

V = span{φ1, . . . , φk−1

}, Z = span

{φk, . . . , φk+m−1

}= Ek, W = (V ⊕ Z)⊥, (3.2)

and we define

BV = {z ∈ V : ‖z‖ ≤ 1}, BVZ = {z ∈ V ⊕ Z : ‖z‖ ≤ 1}, BZW = {z ∈ Z ⊕W : ‖z‖ ≤ 1},(3.3)

and SV , SVZ, SZW , respectively, their relative boundaries.Theorems 2.4 and 2.5 will be a consequence of the geometry in Propositions 3.1 and

3.2, whose proofs will be postponed to Sections 4 and 5.


Proposition 3.1. If λ ∈ (λk−1, λk) and hypothesis (1.2) is satisfied, then there exist constantsDλ andρλ such that

J(z) ≥ Dλ, for z ∈ Z ⊕W, (3.4)

J(z) < Dλ, for z ∈ ρλSV . (3.5)

Moreover, if one of the sets of hypotheses (F1) or (F2) is satisfied, then there exists δ0 such that forλ ∈ (λk − δ0, λk) there exist DW,Dλ ∈ R, ρλ > R1 > 0 such that, in addition to (3.4) and (3.5),

J(z) ≥ DW, for z ∈W, (3.6)

J(z) < DW, for z ∈ R1SVZ, (3.7)

J(z) < DW, for z ∈ V, ‖z‖ ≥ R1. (3.8)

(The values with index λ depend on λ, the others may be fixed uniformly.)Based on this geometry one gives the following proof.

Proof of Theorem 2.4. Since the functional J satisfies the (P.S.) condition, we can apply thesaddle point theorem (see, e.g., in [15]) for two times, let

Γk−1 ={γ ∈ C0(ρλBV ;H

)s.t. γ |ρλSV = id

},

Γk ={γ ∈ C0(R1BVZ;H)s.t. γ |R1SVZ = id

}.

(3.9)

The first solution, that we denote by zk−1 and may be obtained for any λ ∈ (λk−1, λk)with just hypothesis (1.2), corresponds to a critical point at the level

ck−1 = infγ∈Γk−1

supw∈ρλBV

J(γ(w)

), (3.10)

the criticality of this level is guaranteed by the estimates (3.4) and (3.5), since ρλSV and Z⊕Wlink, that is, the image of any map in Γk−1 intersects Z ⊕W .

The second solution, that we denote by zk, corresponds to a critical point at the criticallevel

ck = infγ∈Γk

supw∈R1BVZ

J(γ(w)

), (3.11)

actually, this is a critical level because of the estimates (3.6) and (3.7), since R1SVZ and Wlink.

To conclude the proof, we need to show that these two solutions are distinct.We observe first that by estimate (3.6) we have that ck ≥ DW , then we observe that

we may build a map γ0 ∈ Γk−1 in such a way that its image is the union between the annulus{z ∈ V : ‖z‖ ∈ [R1, ρλ]} and the image of a (k−1)-dimensional ball in R1SVZ whose boundaryis R1SV . By the estimates (3.7) and (3.8), we deduce that supw∈ρλBV J(γ0(w)) < DW , and as a


consequence ck−1 < DW , proving that the two solutions are distinct, for being at differentcritical levels.

Proposition 3.2. If λ ∈ (λk, λk+m) and hypothesis (1.2) is satisfied, then there exist constants Kλ

and βλ such that

J(z) ≥ Kλ, for z ∈W, (3.12)

J(z) < Kλ, for z ∈ βλSVZ. (3.13)

Moreover, if one of the sets of hypotheses (F3) or (F4) is satisfied, then there exists δ1 such that forλ ∈ (λk, λk + δ1) there exist Kλ,KV , E ∈ R, βλ > R2 > 0, ξ > 0 such that, in addition to (3.12) and(3.13),

J(z) < KV , for z ∈ V, (3.14)

J(z) > KV , for z ∈ R2SZW, (3.15)

J(z) > KV , for z ∈W, ‖z‖ ≥ R2, (3.16)

J(z) > E, for z ∈ R2BZW, (3.17)

J(z) < E, for z ∈ ξSV . (3.18)

The values with index λ depend on λ, the others may be fixed uniformly.This geometry, along with Lemma 2.2, allows one to give the following.

Proof of Theorem 2.5. Since the functional J satisfies the (P.S.) condition, we can apply thesaddle point theorem and Lemma 2.3.

The first solution that we denote by wk and may be obtained for any λ ∈(λk, λk+m) with just hypothesis (1.2) is again obtained through the saddle point theorem andcorresponds to a critical point at the critical level

dk = infγ∈Γk

supw∈βλBVZ

J(γ(w)

), (3.19)

where now

Γk ={γ ∈ C0(βλBVZ;H

)s.t. γ |βλSVZ = id

}, (3.20)

the criticality is guaranteed by estimates (3.12) and (3.13), since βλSVZ andW link.The second solution that we denote by wk−1 comes from Lemma 2.3, where we set

X1 = V and X2 = Z ⊕W , actually we have the following structure:

supξSV

J(z) < E = infR2BZW

J(z) ≤ supξBV

J(z) < KV < infR2SZW

J(z), (3.21)

and then we have a critical point wk−1 at the level dk−1 ≤ KV .


Finally, in order to prove that these two solutions are distinct, we need a sharperestimate for dk than that given by (3.13). For this we use Lemma 2.2 to guarantee that forany map γ ∈ Γk, since βλ > R2, one has that the image of γ either intersects R2SZW or has apoint z ∈ W with ‖z‖ ≥ R2. This implies that supw∈βλBVZJ(γ(w)) > KV , by estimates (3.15)and (3.16), and then dk > KV proving that the two solutions are distinct, for being at differentcritical levels.

4. Proof of Estimates

In this section we will prove all the estimates in Propositions 3.1 and 3.2.From (1.2) and the continuity of the potential F, for any ε > 0, there exists a positive

constantMε =M(ε) such that

|∇F(x, s)| ≤ ε|s| +Mε, (4.1)

for all (x, s) ∈ Ω × R2. By (4.1), Holder’s inequality, we have

∣∣∣∣

∫

ΩF(x, z)dx

∣∣∣∣ ≤∫

Ω

∣∣∣∣∣

∫1

0(∇F(x, tz), z)dt

∣∣∣∣∣dx ≤

∫

Ω

∫1

0|∇F(x, tz)||z|dt dx

≤∫

Ω

(ε|z|2 +Mε|z|

)dx ≤ ε‖z‖2L2 +Mε|Ω|1/2‖z‖L2

≤ εS2‖z‖2 +MεS|Ω|1/2‖z‖,

(4.2)

where S is the best embedding constant.

4.1. Estimates of the Saddle Geometry

Lemma 4.1. Under hypothesis (1.2), one gets the following:

(i) for λ ∈ (λk−1, λk), there exists Dλ satisfying (3.4) and DW ∈ R satisfying (3.6);

(ii) for λ ∈ (λk, λk+m):

(a) there existsKλ ∈ R satisfying (3.12),(b) for a given R2 > 0, there exists E ∈ R satisfying (3.17).

Proof. Let z ∈W : using estimates (4.2) and (2.14)we get

J(z) ≥(λk+m − λ2λk+m

− εS2)‖z‖2 −Mε|Ω|1/2S‖z‖. (4.3)

For λ ∈ (λk−1, λk), letting ε < (λk+m −λk)/2S2λk+m < (λk+m −λ)/2S2λk+m, it follows thatJ is bounded below inW , that is, there exists a DW as in (3.6).

For λ ∈ (λk, λk+m), then the same estimate holds but the constant cannot be madeindependent of λ, giving (3.12).


In the same way, let z ∈ Z ⊕W and set δ = λk − λ > 0, we get

J(z) ≥(λk − λ2λk

− εS2)‖z‖2 −Mε|Ω|1/2S‖z‖

≥(

δ

2λk− εS2

)‖z‖2 −Mε|Ω|1/2S‖z‖.

(4.4)

Letting ε < δ/2S2λk, it follows that J is bounded below in Z ⊕W , that is, there exists a Dλ

such that for all z ∈ Z ⊕W we have (3.4), where again the constant Dλ depends on δ, that is,on λ.

Finally, (4.4)with λ ∈ (λk, λk+m) implies

J(z) ≥(λk − λk+m

2λk− εS2

)‖z‖2 −Mε|Ω|1/2S‖z‖, (4.5)

then, no matter the value of λ, J is bounded from below in any bounded subset of Z ⊕W ,giving (3.17) for a suitable value of E.

Lemma 4.2. Under hypothesis (1.2), one gets the following:

(i) for λ ∈ (λk−1, λk), given the constant Dλ ∈ R, there exists ρλ > 0 satisfying (3.5);

(ii) for λ ∈ (λk, λk+m):

(a) there existsKV ∈ R satisfying (3.14),

(b) for a given Kλ ∈ R, there exists βλ > 0 satisfying (3.13),

(c) for a given E ∈ R, there exists ξ > 0 satisfying (3.18).

Moreover, given the values R1, R2, one may always choose ρλ > R1, βλ > R2 as claimed inPropositions 3.1 and 3.2.

Proof. Let z ∈ V , by estimates (4.2) and (2.13)we get

J(z) ≤(λk−1 − λ2λk−1

+ εS2)‖z‖2 +Mε|Ω|1/2S‖z‖. (4.6)

For λ ∈ (λk−1, λk), letting ε < (λ − λk−1)/2S2λk−1, then one obtains (3.5) for suitablylarge ρλ > R1.

For λ ∈ (λk, λk+m), letting ε < (λk − λk−1)/2S2λk−1, one obtains, for suitable KV andξ > 0, (3.14) and (3.18).

Finally, let z ∈ V ⊕ Z and set δ = λ − λk > 0, we get

J(z) ≤(λk − λ2λk

+ εS2)‖z‖2 +Mε|Ω|1/2S‖z‖

≤(− δ

2λk+ εS2

)‖z‖2 +Mε|Ω|1/2S‖z‖.

(4.7)


Letting ε < δ/2S2λk, it is clear that (once that δ is fixed) this goes to −∞ and then we mayfind the claimed βλ > R2 such that (3.13) holds.

Observe that KV and E can be chosen uniformly for λ ∈ (λk, λk+m), while ρλ, βλ in factdepend on λ.

4.2. Estimating the Effect of the Nontrivial Perturbation

In this section we will prove the remaining inequalities in Propositions 3.1 and 3.2, whichrely on the hypotheses (F1), or (F2), or (F3), or (F4), which, roughly speaking, say that theperturbation F is nontrivial in such a way that a new solution arises when it is sufficientlynear to the eigenvalue λk. The proof is simpler for Theorem 2.4, since we need to estimate thefunctional in the compact set SVZ, while for Theorem 2.5 the same kind of estimate is requiredin the noncompact set SZW .

4.2.1. Estimating J in SVZ

For the next estimates, we will need the following lemma.

Lemma 4.3. Hypotheses (F2) implies that there exists a nondecreasing function D : (0,+∞) → Rsuch that

limR→+∞

D(R) = +∞, infz∈RSVZ

∫

ΩF(x, z)dx > D(R). (4.8)

Proof. First we claim that there exists a constant η > 0 such that the setsΩz = {x ∈ Ω : |z(x)| >η} have measure |Ωz| > η, for all z ∈ SVZ.

Actually, V ⊕Z is a finite-dimensional subspace and the functions z ∈ SVZ are smooth,they are uniformly bounded, that is, there exists M > 0 such that |z(x)| ≤ M for all x ∈ Ω.Suppose that for ηn → 0(ηn < 1) there exists {zn} ⊂ SVZ such that |Ωzn | ≤ ηn.

On one hand, by (2.13), one obtains

1λk

≤∫

Ω(A(x)zn, zn)dx. (4.9)

On the other hand,

∫

Ω(A(x)zn, zn)dx =

∫

Ω

(a(x)u2n + 2b(x)unvn + d(x)v2

n

)dx

≤∫

Ω(a(x) + b(x))u2n +

∫

Ω(b(x) + d(x))v2

ndx

≤M∫

Ω|zn|2dx

=M

(∫

Ωzn

|zn|2dx +∫

Ω−Ωzn

|zn|2dx)

≤M(M2|Ωzn | + η2n|Ω −Ωzn |

)

≤ ηnC−→ 0,

(4.10)


where M = maxx∈Ω{|a(x) + b(x)|, |b(x) + d(x)|}, zn = (un, vn), C = M(M2 + |Ω|). That is acontradiction.

Now for any H > 0, we will show that we can find an R large enough so that∫Ω F(x,Rz)dx ≥ H for any z ∈ SVZ and R ≥ R, which means that

limR→∞

infz∈RSVZ

∫

ΩF(x, z)dx = +∞. (4.11)

Actually, letting M = (H + |Ω|CF)η−1, by (F2) we have that there exists s0 such thatF(x, s) > M for |s| > s0.

For R > s0/η, one has Ωz ⊆ {x ∈ Ω : |Rz(x)| > s0}, and then one gets

∫

|Rz|≥s0F(x,Rz)dx ≥Mη. (4.12)

ForR ≤ s0/η, by (F2) and F ∈ C1(Ω×R2, R), there existsCF > 0 such that F(x, s) ≥ −CF ,for all (x, s) ∈ (Ω, R2).

One finally obtains

∫

ΩF(x,Rz)dx =

∫

|Rz|≥s0F(x,Rz)dx +

∫

|Rz|≤s0F(x,Rz)dx

≥Mη − |Ω|CF

= H,

(4.13)

it is elementary that

D(R) = infρ≥R

infz∈RSVZ

∫

ΩF(x, z)dx (4.14)

is well defined and satisfies the claim.

Now we may prove the following.

Lemma 4.4. Consider Theorem 2.4 with one of the sets of hypotheses (F1) or (F2). Given the constantDW ∈ R, there exist R1, δ0 > 0 such that, for any λ ∈ (λk − δ0, λk), (3.7) and (3.8) hold.

Proof. We consider the two sets of hypotheses separately.(i) In case (F1), assuming (1.2) and (F1) hold, we claim that there exists CM such that

F(x, s) ≥M|s| − CM, ∀M ∈ R, (4.15)

uniformly in x ∈ Ω, in particular we setM = 1.


In fact, by (F1), there exits R0 > 0 such that for |s| ≥ R0, (∇F(x, s), s) ≥ |s| uniformly inx ∈ Ω. For any s ∈ R2(s /= 0), from (4.1) (letting ε = 2) we have

F(x, s) =∫1

0(∇F(x, ts), s)dt

=∫1

R0/|s|(∇F(x, ts), s)dt +

∫R0/|s|

0(∇F(x, ts), s)dt

≥∫1

R0/|s||s|dt +

∫R0/|s|

0(∇F(x, ts), s)dt

≥ |s| − R0 −∫R0/|s|

0

(2t|s|2 + |M2||s|

)dt

= |s| − R0 − R20 −M2R0

= |s| − C1,

(4.16)

where C1 = R0 + R20 +M2R0.

Let z ∈ RSVZ, for being in a finite-dimensional subspace, all the norms are equivalent,so that (set δ = λk − λ > 0 and uses estimates (4.15) and (2.13))

J(z) ≤ λk − λ2λk

‖z‖2 − ‖z‖ + C1|Ω|

≤ δ

2λk‖z‖2 − ‖z‖ + C1|Ω|

≤ δ

2λkR2 − R + C1|Ω|.

(4.17)

(ii) In case (F2), let D(R) be as in Lemma 4.3, for ‖z‖ = R, let z = w + φ with w ∈ Vand φ ∈ Z = Ek,

J(z) =12

∫

Ω(H(x)∇z,∇z)dx − λ

2

∫

Ω(A(x)z, z) −

∫

ΩF(x, z)dx

≤ λk−1 − λ2λk−1

‖w‖2 + λk − λ2λk

∥∥φ∥∥2 −

∫

ΩF(x, z)dx

≤ λk−1 − λk + δ2λk−1

‖w‖2 + δ

2λk

∥∥φ∥∥2 −

∫

ΩF(x, z)dx

≤ δ

2λk

∥∥φ∥∥2 − λk − λk−1

4λk−1‖w‖2 −

∫

ΩF(x, z)dx.

(4.18)

Assume that δ ≤ (λk − λk−1)/2, it is easy to see that (−(λk − λk−1)/4λk−1)‖w‖2 ≤ C for someconstant C, we estimate

J(z) ≤ δ

2λk

∥∥φ∥∥2 + C −D(R)

≤ δ

2λkR2 −D(R) + C.

(4.19)


Considering (4.17) and (4.19), we see that since limR→∞D(R) = +∞ by Lemma 4.3, we mayfix R1 so that C − D(R1) < DW − 1 (or DM − R1 < DW − 1 for the case (F1)) and then for0 < δ < min{2λk/R2

1, (λk − λk−1)/2} one gets (3.7).To obtain (3.8), we observe that (since λ > λk−1) if φ = 0, that is, if z ∈ V , then in

estimates (4.17) and (4.19)we may avoid the term (δ/2λk)R2 so that (remember thatD(R) isnondecreasing) J(z) < DW − 1 for ‖z‖ > R1.

4.2.2. Estimating J in SZW

We consider the corresponding of the previous lemma, for Theorem 2.5.

Lemma 4.5. Considering Theorem 2.5 with one of the sets of hypotheses (F3) or (F4). Given theconstant KV ∈ R, there exists R2, δ1 > 0 such that, for any λ ∈ (λk, λk + δ1), (3.15) and (3.16) hold.

Proof. Letting λ = λk +δ, we see from (4.3), that property (3.16)will be satisfied provided thatR2 is large enough (say R2 > R) and observing that this value can be made independent fromλ once that δ is small enough.

Now we consider the two sets of hypotheses separately.(i) In case (F3), suppose z ∈ Ek ⊕W , we can assume that z = w + φ, with w ∈ W and

φ ∈ Ek. Since Ek is a finite dimension subspace, all the norms are equivalent, so that thereexists C > 0 such that for all φ ∈ Ek we have ‖φ‖ ≤ C‖φ‖L1 . By (F3), from the proof of (4.15),we also have the similar inequality: there exists C2 such that

−F(x, s) ≥ C|s| − C2, (4.20)

uniformly in x ∈ Ω. So by (2.14) and (4.20),

J(w + φ

)=

12

∫

Ω

(H(x)∇(

w + φ), ∇(

w + φ))dx

− λ

2

∫

Ω

(A(x)

(w + φ

), w + φ

)dx −

∫

ΩF(x,w + φ

)dx

≥ λk+m − (λk + δ)2λk+m

‖w‖2 − δ

2λk

∥∥φ∥∥2 + C

∥∥w + φ∥∥L1 − C2|Ω|


‖w‖2 − δ

2λk

∥∥φ∥∥2 + C

∥∥φ∥∥L1 − C‖−w‖L1 − C2|Ω|


‖w‖2 − δ

2λk

∥∥φ∥∥2 +

∥∥φ∥∥ − C3‖w‖ − C4,

(4.21)

where C3 = |Ω|1/2SC, C4 = C2|Ω|. Since

(1 − δ

2λk‖z‖

)‖z‖ ≤ ‖w‖ + ∥∥φ

∥∥ − δ

2λk

(∥∥φ∥∥2 + ‖w‖2

)

≤(1 − δ

2λk

∥∥φ∥∥)∥∥φ

∥∥ + ‖w‖,(4.22)


suppose δ ≤ (λk+m − λk)/2, (4.21) becomes

J(w + φ

) ≥ λk+m − (λk + δ)2λk+m

‖w‖2 − C3‖w‖ − C4 − ‖w‖ +(1 − δ

2λk‖z‖

)‖z‖

≥ λk+m − λk4λk+m

‖w‖2 − (C3 + 1)‖w‖ − C4 +(1 − δ

2λk‖z‖

)‖z‖,

(4.23)

since (λk+1 − λk)/4λk+1 > 0, so ((λk+1 − λk)/4λk+1)‖w‖2 − (C3 + 1)‖w‖ − C4 is bounded belowfor all w ∈W , that is, there exists C5 ∈ R such that

λk+m − λk4λk+m

‖w‖2 − C3‖w‖ − C4 ≥ C5, (4.24)

by (4.23) one gets

J(z) ≥(1 − δ

2λk‖z‖

)‖z‖ + C5

= − δ

2λk‖z‖2 + ‖z‖ + C5.

(4.25)

(ii) In case (F4), first we give some conclusions which are similar to Lemma 3 of [16].Under the property of F, there exists a constant C, and G ∈ C(R2, R) which is subadditive,that is,

G(s + t) ≤ G(s) +G(t), (4.26)

for all s, t ∈ R2, and coercive, that is,

G(s) −→ +∞, (4.27)

as |s| → ∞, and satisfies that

G(s) ≤ |s| + 4, (4.28)

for all s ∈ R2, such that

−F(x, s) ≥ G(s) − C, (4.29)

for all s ∈ R2 and x ∈ Ω.In fact, since −F(x, s) → +∞ as |s| → ∞ uniformly for all x ∈ Ω, there exists a

sequence of positive integers nk with nk+1 > 2nk for all positive integers k such that

−F(x, s) ≥ k, (4.30)


for all |s| ≥ nk and all x ∈ Ω. Let n0 = 0 and define

G(s) = k + 2 +|s| − nk−1nk − nk−1 , (4.31)

for nk−1 ≤ |s| < nk, where k ∈N.By the definition of Gwe have

k + 2 ≤ G(s) ≤ k + 3, (4.32)

for all nk−1 ≤ |s| < nk. By (F4) and F ∈ C1(Ω × R2, R), there exists CF > 0 such that

−F(x, s) ≥ −CF, ∀(x, s) ∈(Ω, R2

). (4.33)

It follows that

−F(x, s) ≥ G(s) − C, (4.34)

where C = CF + 4. In fact, when nk−1 ≤ |s| < nk for some k ≥ 2, one has, by (4.30) and (4.32),

−F(x, s) ≥ k − 1 ≥ G(s) − 4 ≥ G(s) − C, (4.35)

for all x ∈ Ω. When |s| < n1, we have, by (4.32) and (4.33),

−F(x, s) ≥ −CF = 4 − C ≥ G(s) − C, (4.36)

for all x ∈ Ω.It is obvious that G is continuous and coercive. Moreover one has

G(s) ≤ |s| + 4, (4.37)

for all s ∈ R2. In fact, for every s ∈ R2 there exists k ∈N such that

nk−1 ≤ |s| < nk, (4.38)

which implies that

G(s) ≤ (k − 1) + 4 ≤ nk−1 + 4 ≤ |s| + 4, (4.39)

for all s ∈ R2 by (4.32) and the fact that nk ≥ k for all integers k ≥ 0.Now we only need to prove the subadditivity of G. Let

nk−1 ≤ |s| < nk, nj−1 ≤ |t| < nj, (4.40)


andm = max{k, j}. Then we have

|s + t| ≤ |s| + |t| < nk + nj ≤ 2nm < nm+1. (4.41)

Hence we obtain, by (4.32),

G(s + t) ≤ m + 4 ≤ k + 2 + j + 2 ≤ G(s) +G(t), (4.42)

which shows that G is subadditive.For z ∈ = Ek ⊕ W , assuming that z = w + φ, with w ∈ W and φ ∈ Ek, and letting

0 < δ < (λk+m − λk)/2, by (2.13), (4.29), (4.26), and (4.28), one gets

J(w + φ

)=

12

∫

Ω

(H(x)∇(

w + φ),∇(

w + φ))dx

− λ

2

∫

Ω

(A(x)

(w + φ

), w + φ

)dx −

∫

ΩF(x,w + φ

)dx


‖w‖2 − δ

2λk

∥∥φ∥∥2

+∫

ΩG(φ +w

)dx − C|Ω|


‖w‖2 − δ

2λk

∥∥φ∥∥2

+∫

ΩG(φ)dx −

∫

ΩG(−w)dx − C|Ω|


‖w‖2 − δ

2λk

∥∥φ∥∥2

+∫

ΩG(φ)dx −

∫

Ω(|w| + 4)dx − C|Ω|


‖w‖2 − δ

2λk‖z‖2

+∫

ΩG(φ)dx − S|Ω|‖w‖ − C1

= g(w) +∫

ΩG(φ)dx − δ

2λk‖z‖2,

(4.43)

where g(w) = ((λk+m − λk)/4λk+m)‖w‖2 − S|Ω|‖w‖ −C1, C1 = (4 +C)|Ω|. Since φ ∈ Ek, Ek is afinite-dimensional subspace, and G is coercive, from the proof of (4.8), one can get

lim‖φ‖→∞

∫

ΩG(φ)dx = +∞, (4.44)


that is,∫ΩG(φ)dx is coercive on Ek. Since (λk+m − λk)/4λk+m > 0, so g(w) is coercive on W ,

and∫ΩG(φ)dx and g(w) is bounded below, it is obvious that

lim‖z‖→∞

(g(w) +

∫

ΩG(φ)dx

)= +∞, (4.45)

for all z ∈ Z ⊕W .Considering (4.25), (4.43), and (4.45), we can choose R2 large enough such that for all

‖z‖ ≥ R2 one gets

g(w) +∫

ΩG(φ)dx > KV + 1, (4.46)

(or R2 + C5 > KV + 1 for the case (F3)) and property (3.16) holds, then for 0 < δ <min{2λk/R2

2, (λk+m − λk)/2} = δ1 and z ∈ R2SZW one gets J(z) > KV , that is, the property(3.15) holds.

5. Proof of the Geometry in Propositions 3.1 and 3.2

We finally give the proof of Propositions 3.1 and 3.2, which is nothing but a resume ofthe lemmata above, verifying that all the constants can be chosen sequentially withoutcontradictions.

Proof of Proposition 3.1. Under hypothesis (1.2), if we fix a value λ, then we obtain the constantDλ from Lemma 4.1 and with this we get ρλ from Lemma 4.2. If we also consider one of thetwo sets of hypotheses (F1) or (F2), then we proceed as follows: first of all, we determine(once for ever) the constant DW from Lemma 4.1, with this we obtain from Lemma 4.4 thevalues R1 and δ0. Then, for any (now fixed) λ ∈ (λk − δ0, λk), we obtain from Lemma 4.1 thevalue Dλ. Finally, we can get from Lemma 4.2 the corresponding value of ρλ > R1.

Proof of Proposition 3.2. Under hypothesis (1.2), if we fix a value λ ∈ (λk, λk+m), then we obtainthe constantKλ from Lemma 4.1 and with this we get βλ from Lemma 4.2. If we also considerone of the two sets of hypotheses (F3) or (F4), then we proceed as follows: first of all,we determine (once for ever) the constant KV from Lemma 4.2, with this we obtain fromLemma 4.5 the values R2 and δ1. Since we have R2, we can get from Lemma 4.1 the constantE and with this obtain ξ from Lemma 4.2.

Finally, for any (now fixed) λ ∈ (λk, λk + δ1), we obtain from Lemma 4.1 the constantKλ and with this we get from Lemma 4.2 the corresponding value of βλ > R2.

Acknowledgments

This paper was supported by National Commission of Ethnic Affairs, Science Foundationof China (no. [2011]02) and by Natural Science Foundation of the Education Department ofGuizhou Province (no. [2010]305).


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Research ArticleThe Solution of a Class of SingularlyPerturbed Two-Point Boundary Value Problems bythe Iterative Reproducing Kernel Method

Zhiyuan Li,1 YuLan Wang,1 Fugui Tan,2Xiaohui Wan,1 and Tingfang Nie1

1 Department of Mathematics, Inner Mongolia University of Technology, Hohhot 010051, China2 Jining Teachers College, Jining 012000, China

Correspondence should be addressed to YuLan Wang, [email protected]

Received 14 February 2012; Revised 7 April 2012; Accepted 18 April 2012


Copyright q 2012 Zhiyuan Li et al. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

In (Wang et al., 2011), we give an iterative reproducing kernel method (IRKM). The maincontribution of this paper is to use an IRKM (Wang et al., 2011), in singular perturbation problemswith boundary layers. Two numerical examples are studied to demonstrate the accuracy of thepresent method. Results obtained by the method indicate that the method is simple and effective.

1. Introduction

Singularly perturbed problems (SPPs) arise frequently in applications including geophysicalfluid dynamics, oceanic and atmospheric circulation, chemical reactions, and optimal control.In this paper, we consider the following singularly perturbed two-point boundary valueproblem:

εUxx(x) + a(x)Ux(x) + b(x)U(x) = F(x,U(x)), x ∈ (0, 1),

U(0) = α, U(1) = β,(1.1)

where ε is a positive small parameter, a(x), b(x), and F(x, v) are known functions, andU(x) is a unknown function to be determined. In this paper, we assume that (1.1)has a unique solution that belongs to W3

2 [0, 1]. Like in [1–5], we give reproducing


kernel spaces W12 [a, b] and W3

2 [a, b]. (i) We define the inner product space W12 [a, b] =

{u | u is one-variable absolutely continuous function, u′ ∈ L2[a, b]}. The inner product isgiven by 〈u(x), v(x)〉W1

2= u(a)v(a) +

∫ba u

′(x)v′(x)dx. The space W12 [a, b] is a repro-

ducing kernel space, and its reproducing kernel is R{1}x (y). (ii) Space W3

2 [a, b] ={u | u, u′, u′′ is one-variable absolutely continuous function, u(a) = u(b) = 0, u′′′ ∈ L2[a, b]}.

The inner product is given by 〈u(x), v(x)〉W32

= u′(a)v′(a) + u′′(a)v′′(a) +∫ba u

′′′(x)v′′′(x)dx. The space W32 [a, b] is a reproducing kernel space, and its reproducing

kernel is R{2}x (y).

2. Iterative Reproducing Kernel Method (IRKM)

In order to solve (1.1), we first give the analytical and approximate solutions of the followingoperator equation:

(Lu)(x) = F(x, u(x)), (2.1)

where L : H[a, b] → H1[a, b] is a bounded linear operator and L−1 is existent. H1[a, b] is anRKHS with the reproducing kernel K(x, t), H[a, b] is also an RKHS with the reproducingkernel K(x, t).

Theorem 2.1. If L−1 is existent and {xi}∞i=1 are countable dense points in [a, b], Letting ψi(x) =∑i

k=1 βikψk(x), where the βik are the coefficients resulting from the Gram-Schmidt orthonormalization,ψi(x) = (LyK(x, y))(xi), i = 1, 2, . . ., then

u(x) =∞∑

i=1

i∑

k=1

βikf(xk, u(xk))ψi(x) (2.2)

is an analytical solution of (2.1).

Proof. u(x) can be expanded to the Fourier series in terms of normal orthogonal basis{ψi(x)}∞i=1 in H[a, b]:

u(x) =∞∑

i=1

⟨u(x), ψi(x)

⟩ψi(x)

=∞∑

i=1

i∑

k=1

βik⟨u(x), ψk(x)

⟩ψi(x)

=∞∑

i=1

i∑

k=1

βik〈u(x), (LsKx(s))(xk)〉ψi(x)

=∞∑

i=1

i∑

k=1

βik(Ls〈u(x), Kx(s)〉)(xk)ψi(x)


=∞∑

i=1

i∑

k=1

βik(Lsu(s))(xk)ψi(x)

=∞∑

i=1

i∑

k=1

βikf(xk, u(xk))ψi(x).

(2.3)

(i) Linear Problem

Suppose (2.1) is a linear problem, that is, f(x, u) = F(x). We define an approximate solutionun(x) by

un(x) =n∑

i=1

i∑

k=1

βikF(xk)ψi(x). (2.4)

Theorem 2.2 (convergence analysis). Let ε2n = ‖u(x)−un(x)‖2; then the sequence of real numbers

εn is monotonously decreasing and εn → 0 and the sequence un(x) is convergent uniformly to u(x),k = 0, 1, 2.

Proof. We have

ε2n = ‖u(x) − un(x)‖2 =

∞∑

i=n+1

⟨u(x), ψi(x)

⟩ψi(x) =

∞∑

i=n+1

(⟨u(x), ψi(x)

⟩)2, (2.5)

and clearly εn−1 ≥ εn and consequently {εn} is monotone decreasing in the sense of ‖ · ‖. ByTheorem 2.1, we know that

∑∞i=1〈u(x), ψi(x)〉ψi(x) is convergent in the norm of ‖ · ‖, then we

have ε2n = ‖u(x) − un(x)‖2 → 0.For any x ∈ [a, b], k = 0, 1, 2,

∣∣∣u(k)n (x) − u(k)(x)∣∣∣ =

∣∣∣∣∣

⟨

un(t) − u(t), ∂kK(x, t)∂xk

⟩∣∣∣∣∣≤ ‖un(t) − u(t)‖ ·

∥∥∥∥∥∂kK(x, t)∂xk

∥∥∥∥∥, (2.6)

and by the expression of K(x, t), there exists Ck > 0, such that ‖∂kK(x, t)/∂xk‖ < Ck; thus

∣∣∣u(k)n (x) − u(k)(x)∣∣∣ ≤ Ck‖un(t) − u(t)‖ = Ckεn −→ 0. (2.7)

(ii) Nonlinear Problem

Suppose that (1.1) is a nonlinear problem, that is, f(x, u) =N(u)+F(x), whereN : H[a, b] →H1[a, b] is a nonlinear operator, and we give an iterative sequence un(x):

u0,∗(x) is the solution of the linear equation Lu = F(x),

un+1,∗(x) is the solution of the linear equation Lu =N(un,∗) + F(x), n = 0, 1, 2 . . ..


Lemma 2.3. If un,∗(x) → u(x), then u(x), is the solution of (1.1).

Theorem 2.4. Suppose that the nonlinear operator A � (L−1N) : H1[a, b] → H[a, b] satisfies thecontractive mapping principle, that is,

‖A(u) −A(v)‖ ≤ λ‖u − v‖, λ < 1; (2.8)

then un,∗(x) is convergent.

3. Solution of Singularly Perturbed Problems

We notice that a small variation in the parameter ε produces a large variation in the solution.In other words, we are treating an ill-posed problem. In this paper, by dividing the domain[0, 1] into three subdomains [0, d], [d, 1 − d], and [1 − d, 1].

(i) Outer Region

We have

εUxx(x) + a(x)Ux(x) + b(x)U(x) = F(x,U(x)), x ∈ [d, 1 − d],u(d) = p, u(1 − d) = q.

(3.1)

Letting u(x) = U(x) − ((x − d)/(1 − 2d))q + ((x + d − 1)/(1 − 2d)) and (L1u)(x) = εuxx(x) +a(x)ux + b(x)u, (3.1) can further be converted into

(L1u)(x) = f(x, u(x)), (3.2)

where f(x, u(x)) = F(x, u(x))−a(x)((1/(1−2d))q− (1/(1−2d))p)−b(x)(((x−d)/(1−2d))q+((x + d − 1)/(1 − 2d))p). Using IRKM, we can get the solution of the outer region problem.

(ii) Left Layer

We have

εUxx(x) + a(x)Ux(x) + b(x)U(x) = F(x,U(x)), x ∈ [0, d),

U(0) = α, U(d) = p is known.(3.3)

Letting u(x) = U(x) − α − (x/d)(p − α), x/d = x1, then x = dx1, u(x) = u(dx1).= u(x1),

du/dx = (1/d)(du/dx1), and d2u/dx2 = (1/d2)(d2u/dx21). In space W3

2 [0, 1], (3.3) canfurther be converted into following form:

(Lεu)(x1) = f(x1, u(x1)), (3.4)

where f(x1, u(x1)) = d2(F(x,U(x)) − (a(x)/d)(p − α) − b(x)(α + (x/d)(p − α))), (Lεu)(x1) =εux1x1(x1) + a(x1)dux1(x1) + b(x1)d2u(x1). Using IRKM, we can get the solution of the innerregion (left layer near) problem.


Table 1: Comparison of results, x ∈ [0, d], ε = 2−10, and d = 0.1.

x uT (x) u100 (x) |uT − u100| |uT − u200|0 0 0 0 0

0.01 −0.272864 −0.272866 1.8788 ×10−6 7.48551 ×10−7

0.02 −0.468765 −0.46877 5.38727 ×10−6 1.15911 ×10−7

0.03 −0.608251 −0.608257 6.10734 ×10−6 1.17267 ×10−7

0.04 −0.706254 −0.70626 5.54546 ×10−6 1.01462 ×10−7

0.05 −0.773632 −0.773636 4.52354 ×10−6 8.05672 ×10−7

0.06 −0.818281 −0.818285 3.45181 ×10−6 6.04003 ×10−7

0.07 −0.845955 −0.845958 2.49497 ×10−6 4.30798 ×10−7

0.08 −0.860849 −0.86085 1.66887 ×10−6 2.84958 ×10−7

0.09 −0.866029 −0.86603 8.9114 ×10−7 1.50666 ×10−7

0.1 −0.863746 −0.863746 3.33067 ×10−16 2.22045 ×10−7

Table 2: Comparison of results, x ∈ [d, 1 − d], ε = 2−30, and d = 0.001.

x uT (x) u100(x) |uT − u10| |uT − u100|0.001 −0.99999 −0.99999 1.51212 × 10−13 4.71756 × 10−12

0.1008 −0.903026 −0.903045 1.89196 × 10−5 7.69723 × 10−10

0.2006 −0.652715 −0.652709 5.8384 × 10−6 1.58393 × 10−10

0.3004 −0.344297 −0.344299 1.66259 × 10−6 4.10425 × 10−10

0.4002 −0.0951225 −0.0951222 3.12214 × 10−7 2.67338 × 10−10

0.5998 −0.0951225 −0.0951223 1.7619 × 10−7 6.92579 × 10−10

0.6996 −0.344297 −0.344298 1.11205 × 10−6 2.66992 × 10−10

0.7994 −0.652715 −0.652711 3.96364 × 10−6 1.77961 × 10−10

0.8992 −0.903026 −0.903039 1.27772 × 10−5 6.22121 × 10−8

0.999 −0.99999 −0.99999 6.67721 × 10−12 1.91558 × 10−12

(iii) Right Layer

We have

εUxx(x) + a(x)Ux(x) + b(x)U(x) = F(x,U(x)), x ∈ (1 − d, 1],U(1) = β, U(1 − d) = q.

(3.5)

Letting u(x) = U(x) − β + ((x − 1)/d)(q − β), (x/d) − (1/d) + 1 = x1, then x = dx1 + 1 − d,u(x) = u(dx1 + 1 − d) .= u(x1), du/dx = (1/d)(du/dx1), and d2u/dx2 = (1/d2)(d2u/dx2

1). Inspace W3

2 [0, 1], (3.5) can further be converted into following form:

(Lεu)(x1) = f(x1, u(x1)), (3.6)

where f(x, u(x)) = F(x,U(x))+ (a(x)/d)(q−β)−b(x)(β− ((x−1)/d)(q−β)), q is known (theouter solution has been given), f(x1, u(x1)) = d2f(x, u(x)), and (Lεu)(x1) = εux1x1(x1) +a(x1)dux1(x1) + b(x1)d2u(x1). Using IRKM, we can get the solution of the inner region


Table 3: Comparison of results, x ∈ [1 − d, 1], ε = 2−10, and d = 0.01.

x uT (x) u10(x) |uT − u10| u200(x) |uT − u200|0.991 −0.249439 −0.249431 8.11552 × 10−6 −0.249439 2.72812 × 10−9

0.992 −0.225227 −0.225219 7.75753 × 10−6 −0.225227 3.72799 × 10−9

0.993 −0.200201 −0.200194 7.17326 × 10−6 −0.200201 4.18195 × 10−9

0.994 −0.174338 −0.174331 6.40087 × 10−6 −0.174338 4.18298 × 10−9

0.995 −0.147609 −0.147604 5.47897 × 10−6 −0.147609 3.82517 × 10−9

0.996 −0.119989 −0.119984 4.44655 × 10−6 −0.119989 3.20353 × 10−9

0.997 −0.0914472 −0.0914438 3.34288 × 10−6 −0.0914472 2.41376 × 10−9

0.998 −0.0619555 −0.0619533 2.20746 × 10−6 −0.0619555 1.55216 × 10−9

0.999 −0.0314835 −0.0314825 1.07993 × 10−6 −0.0314835 7.1531 × 10−10

1. −2.22045 ×10−16 −3.62743 ×10−17 1.8577 × 10−16 −3.62852 ×10−17 1.85759 × 10−16

Table 4: Comparison of results, x ∈ [0, d], and d = 0.01

x uT (x) u3,10(x) |uT − u3,10| uT (x) u3,10(x) |uT − u3,10|ε = 2−10 ε = 2−10 ε = 2−10 ε = 2−5 ε = 2−5 ε = 2−5

0.001 0.0324619 0.0324564 5.56236 × 10−6 0.00628391 0.0062835 −4.09563 × 10−7

0.002 0.063871 0.0638658 5.18059 × 10−6 0.0125208 0.0125204 −3.72485 × 10−7

0.003 0.0942614 0.0942567 4.67451 × 10−6 0.0187109 0.0187106 −3.57056 × 10−7

0.004 0.123666 0.123662 4.20082 × 10−6 0.0248545 0.0248541 −3.30969 × 10−7

0.005 0.152117 0.152113 3.72089 × 10−6 0.0309517 0.0309514 −3.0024 × 10−7

0.006 0.179645 0.179642 3.23223 × 10−6 0.0370029 0.0370026 −2.61703 × 10−7

0.007 0.20628 0.206277 2.78314 × 10−6 0.0430082 0.043008 −2.14219 × 10−7

0.008 0.232051 0.232049 2.18254 × 10−6 0.048968 0.0489678 −1.56195 × 10−7

0.009 0.256986 0.256984 2.14645 × 10−6 0.0548823 0.0548822 −8.66593 × 10−8

(right layer near) problem. After solving the inner and outer region problems, we combinetheir solutions to obtain an approximate solution to the original problem (1.1) over theinterval 0 ≤ x ≤ 1.


Example 4.1. This example is from [6–8]:

−εuxx + u = f(x), 0 ≤ x ≤ 1,

u(0) = 0, u(1) = 0.(4.1)

We determine f(x) to get the true solution, the true solution uT (x) = (e(x−1)/√ε + e−x/

√ε)/(1 +

e−1/√ε) − cos2(πx). The numerical results are given in Tables 1, 2, and 3.


Example 4.2. Considering the following nonlinear singularly perturbed problem withboundary layers

εuxx +ex

xux + eu = f(x), 0 < x ≤ 1,

u(0) = 0, u(1) = 0,(4.2)

we determine f(x) to get the true solution, the true solution uT (x) = 1 + (x − 1)e−x/√ε −

xe(x−1)/√ε. The numerical results are given in Tables 3 and 4.

5. Conclusions

In this paper, IRKM was employed successfully for solving a class of SPPs with boundarylayers. The numerical results show that the present method is an accurate and reliableanalytical technique for SPP with boundary layers.

Acknowledgments

The authors thank the reviewers for their valuable suggestions, which greatly improved thequality of the paper. This paper is supported by the Natural Science Foundation of InnerMongolia (no. 2009MS0103) and the project of Inner Mongolia University of Technology (no.ZS201036).

References

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[2] Y.-l. Wang and L. Chao, “Using reproducing kernel for solving a class of partial differential equationwith variable-coefficients,” Applied Mathematics and Mechanics, vol. 29, no. 1, pp. 129–137, 2008.

[3] Z. Chen and Z.-j. Chen, “The exact solution of system of linear operator equations in reproducingkernel spaces,” Applied Mathematics and Computation, vol. 203, no. 1, pp. 56–61, 2008.

[4] Y. Wang, T. Chaolu, and Z. Chen, “Using reproducing kernel for solving a class of singular weaklynonlinear boundary value problems,” International Journal of Computer Mathematics, vol. 87, no. 1–3,pp. 367–380, 2010.

[5] Y. Wang, X. Cao, and X. Li, “A new method for solving singular fourth-order boundary value problemswith mixed boundary conditions,” AppliedMathematics and Computation, vol. 217, no. 18, pp. 7385–7390,2011.

[6] R. K. Bawa and S. Natesan, “A computational method for self-adjoint singular perturbation problemsusing quintic spline,” Computers & Mathematics with Applications, vol. 50, no. 8-9, pp. 1371–1382, 2005.

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[8] D. Herceg and D. Herceg, “On a fourth-order finite-difference method for singularly perturbed bound-ary value problems,” Applied Mathematics and Computation, vol. 203, no. 2, pp. 828–837, 2008.


Research ArticleStrong Global Attractors for 3D Wave Equationswith Weakly Damping

Fengjuan Meng1, 2

1 Department of Mathematics, Nanjing University, Nanjing 210093, China2 Department of Mathematics, Taizhou College, Nanjing Normal University, Taizhou 225300, China

Correspondence should be addressed to Fengjuan Meng, [email protected]

Received 30 March 2012; Accepted 13 April 2012


Copyright q 2012 Fengjuan Meng. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We consider the existence of the global attractor A1 for the 3D weakly damped wave equation.We prove that A1 is compact in (H2(Ω) ∩ H1

0 (Ω)) × H10 (Ω) and attracts all bounded subsets of

(H2(Ω) ∩H10 (Ω)) ×H1

0 (Ω) with respect to the norm of (H2(Ω) ∩H10 (Ω)) ×H1

0 (Ω). Furthermore,this attractor coincides with the global attractor in the weak energy space H1

0 (Ω) × L2(Ω).

1. Introduction

Let Ω ⊂ R3 be a bounded domain with smooth boundary ∂Ω. We consider the followingweakly damped wave equation:

utt + αut −Δu + ϕ(u) = f in Ω × R+ (1.1)

with the boundary condition

u|∂Ω = 0, (1.2)

and initial conditions:

u(·, 0) = u0, ut(·, 0) = u1, in Ω, (1.3)

where α > 0, ϕ is the nonlinear term, and f is a given external forcing term.


Nonlinear wave equation of the type (1.1) arises as an evolutionary mathematicalmodel in many branched of physics, for example, (i) modeling a continuous Josephsonjunction with ϕ(u) = β sinu; (ii) modeling a relativistic quantum mechanics with ϕ(u) =|u|γu. A relevant problem is to investigate the asymptotic dynamical behavior of thesemathematical models. The understanding of the asymptotic behavior of dynamical systemsis one of the most important problems of modern mathematical physics. One way to treat thisproblem is to analyse the existence of its global attractor.

The existence of global attractors for the classical wave equations inH10(Ω)×L2(Ω) and

the regularities of the global attractors has been studied extensively in many monographs andlectures, for example, see [1–7] and references therein.

However, to our knowledge, the research about the stronger attraction of globalattractors for the damped wave equations with respect to the norm of (H2(Ω) ∩ H1

0(Ω)) ×H1

0(Ω) is fewer, only has been found in [8–10]. In the above three papers, the global attractorsin strong topological space (H2(Ω) ∩H1

0(Ω)) ×H10(Ω) were established, the attraction with

respect to the norm of (H2(Ω)∩H10(Ω))×H1

0(Ω) was proved by the asymptotic compactnessof the operator semigroup.

Recently, we consider (1.1) in n dimensional space where the nonlinear term ϕwithoutpolynomial growth is in [11].

In this paper, our aim is to prove the existence of a global attractor for (1.1) instrong topological space (H2(Ω) ∩H1

0(Ω)) ×H10(Ω) where the nonlinear term ϕ with some

polynomial growth. For simplicity, we consider the space dimension is 3, as we know,when the space dimension is lagerer than 3, the case is similar as in 3D, when the spacedimension is 1 or 2, the case is more easier. The attraction with respect to the norm of(H2(Ω) ∩H1

0(Ω)) ×H10(Ω) will be proved by a method different from [8–10]. Furthermore,

this attractor coincides with the global attractor in the weak energy space H10(Ω) × L2(Ω).

The basic assumptions about the external forcing term f and the nonlinear term ϕ areas follows. Let f ∈ L2(Ω) be independent of time, and let the nonlinear term ϕ ∈ C1(R,R)satisfy the following assumptions:

lim inf|s|→∞

Φ(s)s2

≥ 0, here Φ(s) =∫s

0ϕ(τ)dτ ; (1.4)

lim sup|s|→∞

∣∣ϕ′(s)

∣∣

s2= 0; (1.5)

moreover, there exists a constant C0 > 0 such that

lim inf|s|→∞

sϕ(s) − C0Φ(s)s2

≥ 0. (1.6)

Throughout this paper, we use the following notations. Let Ω be a bounded subset ofRn with sufficiently smooth boundary, A = −Δ. V = H1

0(Ω), H = L2(Ω), and D(A) = H2(Ω)∩H1

0(Ω) with the corresponding norms ‖u‖ = (∫Ω |∇u(x)|2dx)1/2, |u| = (

∫Ω |u(x)|2dx)1/2 and


|Δu| = (∫Ω |Δu(x)|2dx)1/2, respectively. The norms in Lp(Ω), 1 ≤ p < ∞ are denoted by |u|p =

(∫Ω |u|pdx)1/p, the scalar products of V,H are denoted by

((u, v)) =∫

Ω∇u(x)∇v(x)dx, (u, v) =

∫

Ωu(x)v(x)dx, (1.7)

respectively. We have D(A) ⊂ V ⊂ H = H∗ ⊂ V ∗,H∗ and V ∗ are the dual spaces of H and V ,respectively, and each space is dense in the following one and the injections are continuous.Then, we introduce the product Hilbert spaces H0 = V ×H = H1

0(Ω)×L2(Ω),H1 = D(A)×V =(H2(Ω) ∩H1

0(Ω)) ×H10(Ω), endowed with the standard product norms:

‖{u1, u2}‖2H0

= ‖u1‖2V + ‖u2‖2

H, ‖{u1, u2}‖2H1

= ‖u1‖2D(A) + ‖u2‖2

V . (1.8)

Denote by C any positive constant which may be different from line to line and even inthe same line, we also denote the different positive constants by Ci, i ∈ N, for specialdifferentiation.

The rest of the paper is organized as follows. In the next section, for the convenienceof the reader, we recall some basic concepts about the global attractors and recapitulate someabstract results. In Section 3, we present our main results.

2. Preliminaries

In this section, we first recall some basic concepts and theorems, which are important forgetting our main results. We refer to [2, 5, 6, 12, 13] and the references therein for more details.Then, we outline some known results about (1.1)–(1.3).

Definition 2.1. The mappings S(t), where S : X × [0,+∞) → X, is said to be a C0 semigroupon X, if {S(t)}t≥0 satisfies

(1) S(0)u = u for all u ∈ X;

(2) S(t1)(S(t2)u) = S(t1 + S2)(u) for all u ∈ X and t1, t2 ∈ R+;

(3) the mapping S : X × (0,∞) → X is continuous.

Definition 2.2. Let {S(t)}t≥0 be a semigroup on a metric space (E, d). A subset A of E is calleda global attractor for the semigroup, if A is compact and enjoys the following properties:

(1) A is invariant, that is, S(t)A = A, for all t ≥ 0;

(2) A attracts all bounded sets of E. That is, for any bounded subset B of E,

d(S(t)B,A) −→ 0, as t −→ 0, (2.1)

where d(B,A) is the semidistance of two sets B and A:

d(B,A) = supx∈B

infy∈A

d(x, y

). (2.2)


Definition 2.3. A C0 semigroup {S(t)}t≥0 in a Banach space X is said to satisfy the condition(C) if for any ε > 0 and for any bounded set B of X, there exist t(B) > 0 and a finitedimensional subspace X1 of X such that {‖PS(t)x‖X, x ∈ B, t ≥ t(B)} is bounded and

‖(I − P)S(t)x‖X < ε, t ≥ t(B), x ∈ B, (2.3)

where P : X → X1 is a bounded projector.

Definition 2.4. Let {S(t)}t≥0 be a semigroup on a metric space (E, d). A set B0 ⊂ E is called anabsorbing set for the semigroup {S(t)}t≥0, if and only if for every bounded set B ⊂ E, thereexists a T0 = T0(B) > 0 such that S(t)B ⊂ B0 for all t ≥ T0.

Theorem 2.5. Let X be a Banach space and let {S(t)}t≥0 be a C0 semigroup in X. Then, there is aglobal attractor for {S(t)}t≥0 in X if the following conditions hold true:

(1) {S(t)}t≥0 satisfies the condition (C), and

(2) there is a bounded absorbing set B ⊂ X.

In [12], the authors have discussed the relations between Condition (C) and ω-limitcompact and proved that, in uniformly convex Banach space, Condition (C) is equivalent toω-limit compact, if the semigroup has a bounded absorbing set.

Next, we recall the result about the global attractor in H0 whose proofs are omittedhere, the reader is referred to [6] and the reference therein.

Theorem 2.6. Under the conditions (1.4), (1.5), (1.6), the solution semigroup {S(t)}t≥0 of theproblem (1.1)–(1.3) has a global attractorA0 inH0. A0 is included and bounded inH1.

3. Main Results

According to the standard Fatou-Galerkin method, it is easy to obtain the existence anduniqueness of solutions and the continuous dependence to the initial value of (1.1)–(1.3). Weaddress the reader to [6] and the reference therein. Here, we only state the result as follows.

Lemma 3.1. Let conditions (1.4), (1.5), (1.6) hold, then for any T > 0 and (u0, u1) ∈ H0, thereexists a unique solution of (1.1)–(1.3) such that

{u, ut} ∈ C([0, T];H0). (3.1)

If, furthermore,

(u0, u1) ∈ H1, (3.2)

then u satisfies

{u, ut} ∈ C([0, T];H1). (3.3)


We define the mappings:

S(t) : {u0, u1} −→ {u(t), ut(t)} ∀t ∈ R. (3.4)

By Lemma 3.1, it is easy to see that {S(t)}t≥0 is C0 semigroup in the energy phase spaces H0

and H1.In order to verify the existence of the bounded absorbing set in H1, we need the result

about the existence of the bounded absorbing set in H0. First, we establish the boundedabsorbing set in H0. Its proof is essentially established in [6] and the reference therein, andwe only need to make a few minor changes for our problem. Here, we only give the followinglemma.

Lemma 3.2. Under the conditions (1.4), (1.5), (1.6), {S(t)}t≥0 has a bounded absorbing set B0 �BH0(0, ρ0) inH0, that is, for any ε > 0 and any bounded subset B0 ⊂ H0, there is a positive constantt0 = t(B0, ρ0) such that

S(t)B ⊂ B0 for any t ≥ t0, u0, u1 ∈ B0. (3.5)

Next, let us establish the existence of the bounded absorbing set in H1.

Lemma 3.3. Under the conditions (1.4), (1.5), (1.6), {S(t)}t≥0 has a bounded absorbing set B �BH1(0, ρ1) in H1, that is, for any ε > 0 and any bounded subset B ⊂ H1, there is a positive constantT = T(B, ρ1) such that

S(t)B ⊂ B for any t ≥ T, u0, u1 ∈ B. (3.6)

Proof. Take the scalar product in H of (1.1) with Av = Aut + σAu, we have

12d

dt

(|Au|2 + ‖v‖2

)+ σ|Au|2 + (α − σ)‖v‖2 − σ(α − σ)(Au, v) + (

ϕ(u), Av)=(f,Av

). (3.7)

For 0 < σ ≤ σ0, σ0 = {α/4, λ1/2α}, by Holder inequality, Poincare inequality, and Cauchyinequality we have

σ|Au|2 + (α − σ)‖v‖2 − σ(α − σ)(Au, v)

≥ σ|Au|2 + (α − σ)‖v‖2 − σ(α − σ)√λ1

|Au|‖v‖

≥ σ|Au|2 + 34α‖v‖2 − σα

√λ1

|Au|‖v‖

≥ σ|Au|2 + 34α‖v‖2 −

(σ

2|Au|2 + σα2

2λ1‖v‖2

)

≥ σ

2|Au|2 + α

2‖v‖2.

(3.8)


It follows from (1.5) that, for any ε > 0, there exists a constant C1 > 0, such that

∣∣ϕ′(s)

∣∣ ≤ ε|s|2 + C1, ∀s ∈ R. (3.9)

Hence,

∣∣((ϕ(u), v

))∣∣

=∣∣∣∣

∫

Ωϕ′(u) · ∇u · ∇v

∣∣∣∣dx

≤∫

Ω

∣∣ϕ′(u)

∣∣ · |∇u| · |∇v|dx

≤ ε∫

Ω|u|2 · |∇u| · |∇v|dx + C1

∫

Ω|∇u| · |∇v|dx

≤ ε(∫

Ω|u|4 · |∇u|2dx

)1/2

· ‖v‖ + C1‖u‖ · ‖v‖

≤ ε(∫

Ω|u|6dx

)1/3

·(∫

Ω|∇u|6dx

)1/6

· ‖v‖ + C1‖u‖ · ‖v‖

≤ ε‖u‖2 · C22 · |Au| · ‖v‖ +

α

16‖v‖2 +

4C21

α‖u‖2,

(3.10)

where C2 is the positive constant satisfying

C2‖u‖2 ≥(∫

Ω|u|6dx

)1/3

,

C2|Au| ≥(∫

Ω|∇u|6dx

)1/6

,

(3.11)

(f,Av

)=d

dt

(f,Au

)+ σ

(f,Au

) ≤ d

dt

(f,Au

)+σ

8|Au|2 + 2σ

∣∣f∣∣2. (3.12)

If (u0, v0) belongs to a bounded set B of H1, then B is also bounded in H0, and for t ≥ t0, byLemma 3.2, we have

‖u‖2 + |ut|2 ≤ ρ20; (3.13)

t0, ρ0 are given in Lemma 3.2 Choose

0 < ε2 ≤ ασ

16ρ40C

42

, (3.14)


it follows from (3.10) that

∣∣((ϕ(u), v

))∣∣ ≤ α

8‖v‖2 +

4ε2

α‖u‖4C4

2|Au|2 +4C2

1

α

≤ α

8‖v‖2 +

σ

4|Au|2 + 4C2

1

αρ2

0, t ≥ t0.(3.15)

Combining with (3.8), (3.12), and (3.15), by the Holder inequality and the Young inequality,we deduce from that (3.7):

d

dt

(|Au|2 + ‖v‖2 − 2

(f,Au

))+σ

4|Au|2 + α

4|v|2 ≤ 4σ

∣∣f

∣∣2 +

8C21

αρ2

0. (3.16)

Let y = |Au − f |2 + ‖v‖2, from the above inequality, we can obtain

dy

dt+σ

8y ≤ dy

dt+σ

8

(|Au|2 + ‖v‖2 +

∣∣f∣∣2)+σ

4|Au|∣∣f∣∣

=dy

dt+σ

4

(|Au|2 + ‖v‖2

)− σ

8

(|Au|2 + ‖v‖2

)+σ

4|Au|∣∣f∣∣ + σ

8∣∣f∣∣2

≤ 4σ∣∣f

∣∣2 +8C2

1

αρ2

0 −σ

8

(|Au|2 + ‖v‖2

)+σ

8

(|Au|2 + ∣∣f

∣∣2)+σ

8∣∣f∣∣2

≤ 4σ∣∣f

∣∣2 +8C2

1

αρ2

0 +σ

4∣∣f∣∣2 − σ

8‖v‖2

≤ σ

4∣∣f∣∣2 + 4σ

∣∣f∣∣2 +

8C21

αρ2

0

≤ 92σ∣∣f∣∣2 +

8C21

αρ2

0.

(3.17)

Let C3 = (9/2)σ|f |2 + (8C21/α)ρ

20. By the Gronwall lemma, we have

y(t) ≤ y(t0) exp(−σ

8(t − t0)

)+

8C3

σ. (3.18)

Defining ρ′1 by

ρ′21 =

8C3

σ, (3.19)

we see that

lim supt→∞

y(t) ≤ ρ′21 , (3.20)

and we conclude that


The ball of H1, B � BH1(0, ρ1), centered at (A−1f, 0) of radius ρ1 > ρ′1 + ρ0, is absorbing

in H1 for the semigroup S(t), t ≥ 0.

We now give the property of compactness about the nonlinear operator ϕ which willbe needed in the proof of the condition (C).

Lemma 3.4. Assume that ϕ ∈ C1(R,R) and ϕ : D(A) → V are defined by

((ϕ(u), v

))=∫

Ωϕ′(u)∇u∇vdx, (3.21)

for all u ∈ D(A), v ∈ H10(Ω). Then, ϕ is continuous compact.

Proof. Let {um} be a bounded sequence in D(A). Without loss of generality, we assume that{um} weakly converges to u0 in D(A), since D(A) is reflexive. By the Sobolev embeddingtheorem, we know that H2(Ω) ↪→ L∞ ⋂

H1(Ω) and the embedding H2(Ω) ↪→ H1(Ω) iscompact in R3. Hence, we have that

um −→ u0 in H1. (3.22)

Furthermore, there exists a constant C such that

‖u0‖L∞⋂

H1(Ω)≤ C, ‖um‖L∞

⋂H1(Ω)

≤ C. (3.23)

It is sufficient to prove that {ϕ(um)} converges to {ϕ(u0)} in V :

∥∥ϕ(um)−ϕ(u0)∥∥=

(∫

Ω

∣∣ϕ′(um)∇(um)−ϕ′(u0)∇u0∣∣2dx

)1/2

≤(∫

Ω

∣∣ϕ′(um)∣∣2|∇(um)−∇u0|2dx

)1/2

+(∫

Ω|∇u0|2

∣∣ϕ′(um)−ϕ′(u0)∣∣2dx

)1/2

.

(3.24)

On the one hand, for the first term in (3.24), combining with (3.23) and the continuity of ϕ′(·),we have

∫

Ω

∣∣ϕ′(um)∣∣2|∇(um) − ∇u0|2dx ≤ C‖um‖L∞(Ω)

‖∇(um) − ∇u0‖2. (3.25)

On the other hand, for the second term in (3.24), using the continuity of ϕ′(·),∫

Ω|∇u0|2

∣∣ϕ′(um) − ϕ′(u0)∣∣2dx −→ 0, as m −→ 0, (3.26)

follows immediately by dominated convergence theorem.


Also, considering (3.22), passing to the limit in (3.24), we can obtain

limm→∞

∥∥ϕ(um) − ϕ(u0)

∥∥ = 0. (3.27)


Lemma 3.5. Suppose the conditions (1.4), (1.5), (1.6) hold, the solution semigroup {S(t)}t≥0 of theproblem (1.1)–(1.3) satisfies the condition (C) inH1.

Proof. Let {ωi} be an orthonormal basis of L2(Ω) which consists of eigenvalues of A. Thecorresponding eigenvalues are denoted by {λj}∞j=1:

0 < λ1 ≤ λ2 ≤ λ3 ≤ · · · ≤ λj ≤ · · · , λj −→ ∞ (3.28)

with

Aωi = λiωi, ∀i ∈ N. (3.29)

Let Vm = span{ω1, . . . , ωm} in V and let Pm : V → Vm be an orthogonal projector. We write

u = Pmu + (I − Pm)u � u1 + u2. (3.30)

Taking the scalar product of (1.1) in H with Av2 = Au2t + σAu2, we find

12d

dt

(|Au2|2 + ‖v2‖2

)+ σ|Au2|2 + (α − σ)‖v2‖2 − σ(α − σ)(Au2, v2) +

(ϕ(u), Av2

)=(f,Av2

).

(3.31)

Choose 0 < σ ≤ σ0, similar to (3.8), we have

σ|Au2|2 + (α − σ)‖v2‖2 − σ(α − σ)(Au2, v2) ≥ σ

2|Au2|2 + α

2‖v2‖2. (3.32)

Since f ∈ L2(Ω), ϕ : D(A) → V is compact by Lemma 3.4, for any ε > 0, there exists some msuch that

∣∣(I − Pm)f∣∣H ≤ ε, (3.33)

∥∥(I − Pm)ϕ(u)∥∥V ≤ ε, ∀u ∈ B1

(0, ρ1

), (3.34)

where ρ1 is given by Lemma 3.2.


By exploiting the Holder inequality and Cauchy inequality, we have

(ϕ(u), Av2

) ≤ ∥∥(I − Pm)ϕ

∥∥‖v2‖ ≤ ε‖v2‖ ≤ α

4‖v2‖2 +

ε2

α, (3.35)

(f,Av

)=d

dt

(f,Au

)+ σ

(f,Au

)

≤ d

dt

(f,Au

)+σ

4|Au|2 + σ∣∣f∣∣2

.

(3.36)

By (3.33) and (3.36), we have

(f,Av2

)=d

dt

(f2, Au2

)+ σ

(f,Au2

)

≤ d

dt

(f2, Au2

)+σ

4|Au2|2 + σ

∣∣f2∣∣2

≤ d

dt

(f2, Au2

)+σ

4|Au2|2 + σε2,

(3.37)

where f2 � (I − Pm)f .Hence, combining with (3.32), (3.35), and (3.37), we obtain from (3.31) that

d

dt

(|Au2|2 + ‖v2‖2 − 2

(f2, Au2

))+σ

2

(|Au2|2 + ‖v2‖2

)≤(

2σ +2α

)ε2. (3.38)

Let y = |Au2 − f2|2 + ‖v2‖2, from the above inequality, similar to (3.17), we can obtain

dy

dt+σ

4y ≤ dy

dt+σ

4

(|Au2|2 + ‖v2‖2 +

∣∣f2∣∣2)+σ

2|Au2|

∣∣f2∣∣

≤(

5σ2

+2α

)ε2.

(3.39)

Let C4 = 5σ/2 + 2/α. By the Gronwall lemma, we have

y(t) ≤ y(t1) exp(−σ

4(t − t1)

)+

4C4ε2

σ. (3.40)

Choosing t2 = t1 + (4/σ) ln(ρ21/ε

2), it follows that

y(t) ≤(

1 +4C4

σ

)ε2, t ≥ t2. (3.41)


That is,

∣∣Au2 − f2

∣∣2 + ‖v2‖2 ≤ Cε2, (3.42)

for all t ≥ t2, where C = (1 + 4C4/σ).Thus we complete the proof.

We are now in a position to state our main results as follows.

Theorem 3.6. Under the conditions (1.4), (1.5), (1.6), problem (1.1)–(1.3) has a global attractorA1

inH1; it attracts all bounded subsets ofH1 with respect to the norm ofH1.

Proof. By Lemmas 3.1, 3.3, and 3.5, the conditions of Theorem 2.5 are satisfied. The proof iscomplete.

Corollary 3.7. The global attractorA0 inH0 is coincides withA1 inH1, that is, A0 = A1.

Proof. By Theorem 2.6, A0 is a bounded set of H1, combining with Theorem 3.6, we can easilyget A0 = A1.

Acknowledgment

The author of this paper would like to express her gratitude to Professor ChengkuiZhong for his guidance and encouragement, and to the reviewer, for valuable commentsand suggestions. This work was supported in part by the Scientific Research Foundationof Graduate School of Nanjing University (2012CL21) and Taizhou Natural Science andTechnology Development Project 2011.

References

[1] J. M. Ball, “Global attractors for damped semilinear wave equations,” Discrete and ContinuousDynamical Systems. Series A, vol. 10, no. 1-2, pp. 31–52, 2004.

[2] A. V. Babin and M. I. Vishik, Attractors of Evolution Equations, vol. 25 of Studies in Mathematics and ItsApplications, North-Holland, Amsterdam, The Netherlands, 1992.

[3] V. V. Chepyzhov and M. I. Vishik, Attractors for Equations of Mathematical Physics, vol. 49 of AmericanMathematical Society Colloquium Publications, American Mathematical Society, Providence, RI, USA,2002.

[4] I. Chueshov and I. Lasiecka, “Long-time behavior of second order evolution equations with nonlineardamping,” Memoirs of the American Mathematical Society, vol. 195, no. 912, 183 pages, 2008.

[5] J. K. Hale, Asymptotic Behavior of Dissipative Systems, vol. 25 of Mathematical Surveys and Monographs,American Mathematical Society, Providence, RI, USA, 1988.

[6] R. Temam, Infinite-Dimensional Dynamical Systems in Mechanics and Physics, vol. 68 of AppliedMathematical Sciences, Springer, New York, NY, USA, 2nd edition, 1997.

[7] V. Pata and S. Zelik, “A remark on the damped wave equation,” Communications on Pure and AppliedAnalysis, vol. 5, no. 3, pp. 609–614, 2006.

[8] A. Kh. Khanmamedov, “On a global attractor for the strong solutions of the 2-D wave equation withnonlinear damping,” Applicable Analysis, vol. 88, no. 9, pp. 1283–1301, 2009.

[9] A. Kh. Khanmamedov, “Global attractors for 2-D wave equations with displacement-dependentdamping,” Mathematical Methods in the Applied Sciences, vol. 33, no. 2, pp. 177–187, 2010.


[10] A. Kh. Khanmamedov, “Global attractors for strongly damped wave equations with displacementdependent damping and nonlinear source term of critical exponent,” Discrete and ContinuousDynamical Systems. Series A, vol. 31, no. 1, pp. 119–138, 2011.

[11] F. J. Meng and C. K. Zhong, “Strong global attractor for weakly damped wave equations,” submitted.[12] Q. Ma, S. Wang, and C. Zhong, “Necessary and sufficient conditions for the existence of global

attractors for semigroups and applications,” Indiana University Mathematics Journal, vol. 51, no. 6, pp.1541–1559, 2002.

[13] J. C. Robinson, Infinite-Dimensional Dynamical Systems, Cambridge Texts in Applied Mathematics,Cambridge University Press, Cambridge, UK, 2001.


Research ArticleA Regularity Criterion for the Navier-StokesEquations in the Multiplier Spaces

Xiang’ou Zhu1, 2

1 College of Physics and Electronic Information Engineering, Wenzhou University, Zhejiang,Wenzhou 325035, China

2 The Key Laboratory of Low-voltage Apparatus Intellectual Technology of Zhejiang,Wenzhou 325035, China

Correspondence should be addressed to Xiang’ou Zhu, [email protected]

Received 16 February 2012; Accepted 23 April 2012


Copyright q 2012 Xiang’ou Zhu. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

We exhibit a regularity condition concerning the pressure gradient for the Navier-Stokes equationsin a special class. It is shown that if the pressure gradient belongs to L2/(2−r)((0, T);M(Hr(R3) →H−r(R3))), where M(Hr(R3) → H−r(R3)) is the multipliers between Sobolev spaces whosedefinition is given later for 0 < r < 1, then the Leray-Hopf weak solution to the Navier-Stokesequations is actually regular.

1. Introduction

Consider the Navier-Stokes equations in R3:

∂tu + u · ∇u −Δu +∇p = 0, (x, t) ∈ R3 × (0,∞),

divu = 0, (x, t) ∈ R3 × (0,∞),

u(x, 0) = u0(x), x ∈ R3,

(1.1)

where u = u(x, t) is the velocity field, p = p(x, t) is the scalar pressure, and u0(x) with divu0 = 0 in the sense of distribution is the initial velocity field. For simplicity, we assume thatthe external force has a scalar potential and is included into the pressure gradient.

In the famous paper, Leray [1] and Hopf [2] constructed a weak solution u of (1.1) forarbitrary u0 ∈ L2(R3) with divu0 = 0. The solution is called the Leray-Hopf weak solution.Regularity of such Leray-Hopf weak solutions is one of the most significant open problems


in mathematical fluid mechanics. We note here that there are partial regularity results fromScheffer and from Caffarelli et al., see [3, 4] and references therein. Besides, more work waspioneered by Serrin [5] and extended and improved by Giga [6], Struwe [7, 8], and Zhou [9].Further results can be found in [10–16] and references therein.

Introducing the class Lγ((0, T);Lα(R3)), Serrin [5] showed that if we have a Leray-Hopf weak solution u belonging to Lγ((0, T);Lα(R3)) with the exponents α and γ satisfying2/γ + 3/α < 1, 2 < γ < ∞, 3 < α < ∞, then the solution u(x, t) ∈ C∞((0, T) × R

3), while thelimit case 2/γ + 3/α = 1 was shown much later by Sohr [17] (see also [18]).

Regularity results including assumptions on the pressure gradient have been given byZhou [15], and it was extended later by Struwe [8] to any dimension n ≥ 3. It is shown that ifthe gradient of pressure ∇p ∈ Lα((0, T);Lq(R3)) with 2/α + 3/q ≤ 3, then the correspondingweak solution is actually strong. For the recent work on the regularity problem containing thepressure, velocity field, and the quotient of pressure-velocity, we refer to [19–21] for details.

The purpose of this short paper is to establish a regularity criterion in termsof the pressure gradient for weak solutions to the Navier-Stokes equations in the classL2/(2−r)((0, T);M(Hr → H−r)). This work is motivated by the recent results [22, 23] onthe Navier-Stokes equations. It is an unusual, larger space considered in the current paperthan L3/2r (the following Lemma 2.3) and possesses more information. Obviously, the presentresult extends some previous ones. For more facts concerning regularity of weak solutions,we refer the readers to the celebrated papers [24–30].

2. Preliminaries

We recall the definition of the multiplier space, which was introduced in [31] (see also [32,33]). The space M(Hr(R3) → H−r(R3)) of pointwise multipliers, which map Hr into H−r , isdefined in the following way.

Definition 2.1. For 0 ≤ r < 3/2, M(Hr(R3) → H−r(R3)) is a Banach space of all distributionsf on R

3 such that there exists a constant C such that for all u ∈ D(R3) we have fu ∈ H−r(R3)and

∥∥fu∥∥H−r ≤ C‖u‖Hr , (2.1)

where we denote by Hr(R3) the completion of the space C∞0 (R3) with respect to the norm

‖u‖Hr = ‖(−Δ)r/2u‖L2 and denote by D(R3) the Schwarz class.

The norm of M(Hr(R3) → H−r(R3)) is given by the operator norm of pointwisemultiplication

∥∥f∥∥M(Hr(R3)→ H−r(R3)) = sup

{∥∥fu∥∥H−r : ‖u‖Hr ≤ 1, u ∈ D

(R

3)}. (2.2)

Remark 2.2. Equivalently, we will say that f ∈ M(Hr(R3) → H−r(R3)) if and only if theinequality

∣∣⟨fu, v⟩∣∣ ≤ C‖u‖Hr‖v‖Hr (2.3)

holds for all u, v ∈ D(R3).


Lemma 2.3. Let 0 ≤ r < 3/2. Then the following embedding:

L3/2r(R

3)⊂ M

(Hr(R

3)−→ H−r

(R

3))

(2.4)

holds.

Proof. Indeed, let f ∈ L3/2r(R3). By using the following well-known Sobolev embedding:

Lq(R

3)⊂ H−r

(R

3)

(2.5)

with 3/q = 3/2 + r, we have by Holder’s inequality

∥∥fg∥∥H−r ≤ C

∥∥fg∥∥Lq ≤ C

∥∥f∥∥L3/2r

∥∥g∥∥Lσ

≤ C∥∥f∥∥L(3/2r)

∥∥g∥∥Hr

(Hr(R

3)⊂ Lσ

(R

3)),

(2.6)

where 1/σ = 1/2 − r/3. Then, it follows that

∥∥f∥∥M(Hr → H−r) = sup

‖g‖Hr ≤1

∥∥fg∥∥H−r ≤ C

∥∥f∥∥L3/2r . (2.7)


Example 2.4. Due to the well-known inequality

∥∥∥∥u

|x|∥∥∥∥L2

≤ 2‖∇u‖L2 , (2.8)

we see that |x|−2 ∈ M(H1(R3) → H−1(R3)).

Indeed, since the functions of class C∞0 (R3) are dense in H1(R3) in the norm ‖ · ‖Hr(R3),

suppose u, v ∈ C∞0 (R3). Then by virtue of the Cauchy-Schwarz inequality we obtain

∣∣∣⟨|x|−2u, v

⟩∣∣∣ ≤∥∥∥∥u

|x|∥∥∥∥L2

∥∥∥∥v

|x|∥∥∥∥L2

≤ 4‖∇u‖L2‖∇v‖L2 ,

(2.9)

and thus for u, v ∈ D, ‖u‖Hr ≤ 1, and ‖v‖Hr ≤ 1

∥∥∥|x|−2∥∥∥M(Hr → H−r)

= supu∈D,‖u‖Hr ≤1

∥∥∥|x|−2u∥∥∥H−r

= supu,v∈D

∣∣∣⟨|x|−2u, v

⟩∣∣∣ ≤ 4 <∞. (2.10)


3. Regularity Theorem

Now we state our result as following.

Theorem 3.1. Let u0 ∈ L2(R3) ∩ Lq(R3) for some q ≥ 3 and ∇ · u0 = 0 in the sense of distributions.Suppose that u(t, x) is a Leray-Hopf solution of (1.1) in [0, T). If the pressure gradient satisfies

∇p ∈ L2/(2−r)((0, T);M(Hr −→ H−r)) with 0 < r < 1, (3.1)

then u(t, x) is a regular solution in the sense that

u ∈ C∞([0, T] × R

3). (3.2)

Proof. In order to prove this result, we have to do a priori estimates for the Navier-Stokes equations and then show that the solution satisfies the well-known Serrin regularitycondition. Multiply both sides of the first equation of (1.1) by 4u|u|2 and integrate by parts toobtain (see, e.g., [30])

d

dt‖u(·, t)‖4

L4 − 4∫

R3(Δu) · u|u|2dx = −4

∫

R3∇p · u|u|2dx, (3.3)

for t ∈ (0, T). Then we have

d

dt‖u(·, t)‖4

L4 + 2∥∥∥∇|u|2

∥∥∥2

L2≤ −4

∫

R3∇p · u|u|2dx, (3.4)

where we have used

4∫

R3(Δu) · u|u|2dx ≤ −2

∫

R3

∣∣∣∇|u|2∣∣∣

2dx. (3.5)

Let us estimate the integral

I =∫

R3∇p · u|u|2dx (3.6)

on the right-hand side of (3.4). By the Holder inequality and the Young inequality, we have

I ≤ ∥∥∇p · u∥∥H−r

∥∥∥|u|2∥∥∥Hr

≤ ∥∥∇p∥∥M(Hr → H−r)‖u‖Hr

∥∥∥|u|2∥∥∥Hr

≤ C∥∥∇p∥∥M(Hr → H−r)‖u‖1−rL2 ‖∇u‖rL2

∥∥∥|u|2∥∥∥

1−r

L2

∥∥∥∇|u|2∥∥∥r

L2

≤ C∥∥∇p∥∥M(Hr → H−r)

∥∥∥|u|2∥∥∥

1−r

L2

∥∥∥∇|u|2∥∥∥r

L2


≤ C(ε)∥∥∇p∥∥2/(2−r)M(Hr → H−r)

∥∥∥|u|2

∥∥∥

2(1−r)/(2−r)

L2+ ε∥∥∥∇|u|2

∥∥∥

2

L2

= C(ε)∥∥∇p∥∥2/(2−r)

M(Hr → H−r)‖u‖4ξL4 + ε

∥∥∥∇|u|2

∥∥∥

2

L2,

(3.7)

where ξ = (1 − r)/(2 − r) < 1; we have used the inequality

‖w‖Hr ≤ C‖w‖1−rL2 ‖∇w‖rL2 (3.8)

and the Young inequality with ε:

ab ≤ εap + C(ε)bq (a, b > 0, ε > 0),(

1p+

1q= 1), (3.9)

for C(ε) = (εp)−q/pq−1. Hence by (3.4) and the above inequality, we derive

d

dt‖u(·, t)‖4

L4 + (2 − ε)∥∥∥∇|u|2

∥∥∥2

L2≤ C(ε)∥∥∇p∥∥2/(2−r)

M(Hr → H−r)‖u‖4ξL4 . (3.10)

Now by Gronwall’s lemma (see for instance in [28, Lemma 2]), we have

‖u(·, t)‖4L4 ≤ C

⎡

⎣‖u(0)‖4L4 +

(∫ t

0

∥∥∇p∥∥2/(2−r)M(Hr → H−r)dτ

)1/(1−ξ)⎤

⎦

= C

[

‖u(0)‖4L4 +

(∥∥∇p∥∥2/(2−r)L2/(2−r)((0,T);M(Hr → H−r))

)1/(1−ξ)]

.

(3.11)

Due to the integrability of the pressure gradient, it follows that

u ∈ L∞(

0, T ;L4(R

3)). (3.12)

Consequently u falls into the well-known Serrin’s regularity framework. Therefore, thesmoothness of u follows immediately. This completes the proof of Theorem 3.1.

Remark 3.2. By a strong solution we mean a weak solution of the Navier-Stokes equation suchthat

u ∈ L∞((0, T);H1

)∩ L2

((0, T);H2

). (3.13)

It is wellknown that strong solutions are regular (we say classical) and unique in the class ofweak solutions.


Acknowledgment

The author thanks the anonymous referee for his/her comments on this paper.

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Research ArticleThe Well-Posedness of Solutions fora Generalized Shallow Water Wave Equation

Shaoyong Lai and Aiyin Wang


Correspondence should be addressed to Shaoyong Lai, [email protected]

Received 9 January 2012; Accepted 28 March 2012


Copyright q 2012 S. Lai and A. Wang. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

A nonlinear partial differential equation containing the famous Camassa-Holm and Degasperis-Procesi equations as special cases is investigated. The Kato theorem for abstract differential equa-tions is applied to establish the local well-posedness of solutions for the equation in the Sobolevspace Hs(R) with s > 3/2. Although the H1-norm of the solutions to the nonlinear model does notremain constant, the existence of its weak solutions in the lower-order Sobolev space Hs with1 ≤ s ≤ 3/2 is proved under the assumptions u0 ∈ Hs and ‖u0x‖L∞ <∞.

1. Introduction

Constantin and Lannes [1] derived the shallow water equation

ut + ux +32ρuux + μ

(αuxxx + βutxx

)= ρμ

(γuxuxx + δuuxxx

), (1.1)

where the constants α, β, γ, δ, ρ, and μ satisfy certain conditions. Under several restrictions onthe coefficients of model (1.1), the large time well-posedness was established on a time scaleO(|ρ|−1) provided that the initial value u0 belongs toHs(R) with s > 5/2, and the wave-break-ing phenomena were also discussed in [1]. As stated in [1], using suitable mathematical trans-formations, one can turn (1.1) into the form

ut − utxx + 2kux +muux = auxuxx + buuxxx, (1.2)


where a, b, k, and m are constants. Obviously, (1.2) is a generalization of both the Camassa-Holm equation [2]

ut − utxx + 2kux + 3uux = 2uxuxx + uuxxx, k is a constant (1.3)

and the Degasperis-Procesi model [3]

ut − uxxt + 2kux + 4uux = 3uxuxx + uuxxx. (1.4)

Equations (1.3) and (1.4) are bi-Hamiltonian and arise in the modeling of shallow waterwaves. These two equations pertain to waves of medium amplitude (cf. the discussionsin [1, 4]) and accommodate wave-breaking phenomena. Moreover, the Camassa-Holm andDegasperis-Procesi models admit peaked (periodic as well as solitary) traveling waves cap-turing the main feature of the exact traveling wave solutions of the greatest height of thegoverning equations for water waves (cf. [5, 6]). For other dynamic properties about (1.3)and (1.4), the reader is referred to [7–20].

Recently, Lai and Wu [21] investigate (1.2) in the case where k = 0,m = a+b, a > 0, andb > 0. The well-posedness of global solutions is established in [21] in Sobolev space Hs(R)with s > 3/2 under certain assumptions on the initial value. The local strong and weak solu-tions for (1.2) are discussed in [22] in the case where b > 0, a, k, andm are arbitrary constants.

Motivated by the desire to extend the work in [22], we investigate the following gene-ralized model of (1.2):

ut − utxx + 2kux +muux = auxuxx + buuxxx + βunx, (1.5)

where b > 0, a, k,m, and β are arbitrary constants, and n is a positive integer.The aim of this paper is to investigate (1.5). Since a, b,m, and β are arbitrary constants,

we do not have the result that the H1 norm of the solution of (1.5) remains constant. We willapply the Kato theorem [23] to prove the existence and uniqueness of local solutions for (1.5)in the space C([0, T),Hs(R))

⋂C1([0, T),Hs−1(R))(s > 3/2) provided that the initial value

u0(x) belongs to Hs(R)(s > 3/2). Moreover, it is shown that there exists a weak solution of(1.5) in lower-order Sobolev space Hs(R) with 1 ≤ s ≤ 3/2.

The structure of this paper is as follows. The main results are given in Section 2. Theexistence and uniqueness of the local strong solution for the Cauchy problem (1.5) are provedin Section 3. The existence of weak solutions is established in Section 4.

2. Main Results

Firstly, we give some notations.The space of all infinitely differentiable functions φ(t, x) with compact support in

[0,+∞) × R is denoted by C∞0 . We let Lp = Lp(R)(1 ≤ p < +∞) be the space of all measurable

functions h such that ‖h‖pLp =∫R |h(t, x)|pdx < ∞. We define L∞ = L∞(R) with the standard

norm ‖h‖L∞ = infm(e)=0 supx∈R\e|h(t, x)|. For any real number s, we let Hs = Hs(R) denote theSobolev space with the norm defined by

‖h‖Hs =(∫

R

(1 + |ξ|2

)s∣∣∣h(t, ξ)∣∣∣

2dξ

)1/2

<∞, (2.1)



−ixξh(t, x)dx. Here, we note that the norms ‖ · ‖pLp , ‖ · ‖L∞ , and ‖ · ‖Hs dependon variable t.

For T > 0 and nonnegative number s, letC([0, T);Hs(R)) denote the space of functionsu : [0, T)×R → R with the properties that u(t, ·) ∈ Hs(R) for each t ∈ [0, T], and the mappingu : [0, T) → Hs(R) is continuous and bounded.

For simplicity, throughout this paper, we let c denote any positive constant which isindependent of parameter ε and set Λ = (1 − ∂2

x)1/2.

In order to study the existence of solutions for (1.5), we consider its Cauchy problemin the form

ut − utxx = −∂x(

2ku +m

2u2)+ auxuxx + buuxxx + βunx,

u(0, x) = u0(x),(2.2)

where b > 0, a, k,m, β, and n are arbitrary constants. Now, we give the theorem to describe thelocal well-posedness of solutions for problem (2.2).

Theorem 2.1. Let u0(x) ∈ Hs(R)with s > 3/2, then the Cauchy problem (2.2) has a unique solutionu(t, x) ∈ C([0, T);Hs(R))

⋂C1([0, T);Hs−1(R)) where T > 0 depends on ‖u0‖Hs(R).

For a real number s with s > 0, suppose that the function u0(x) is in Hs(R), and let uε0

be the convolution uε0 = φε � u0 of the function φε(x) = ε−1/4φ(ε−1/4x) and u0 such that theFourier transform φ of φ satisfies φ ∈ C∞

0 , φ(ξ) ≥ 0, and φ(ξ) = 1 for any ξ ∈ (−1, 1). Thus onehas uε0(x) ∈ C∞. It follows from Theorem 2.1 that for each ε satisfying 0 < ε < 1/4, theCauchy problem

ut − utxx = −∂x(

2ku +m

2u2)+ auxuxx + buuxxx + βunx,

u(0, x) = uε0(x), x ∈ R(2.3)

has a unique solution uε(t, x) ∈ C∞([0, Tε);H∞), in which Tε may depend on ε. However, onewill show that under certain assumptions, there exist two constants c and T > 0, bothindependent of ε, such that the solution of problem (2.3) satisfies ‖uεx‖L∞ ≤ c for any t ∈[0, T), and there exists a weak solution u(t, x) ∈ L2([0, T],Hs) for problem (2.2). These resultsare summarized in the following two theorems.

Theorem 2.2. If u0(x) ∈ Hs(R) with s ∈ [1, 3/2] such that ‖u0x‖L∞ < ∞, let uε0 be defined as insystem (2.3), then there exist two constants c and T > 0, which are independent of ε, such that thesolution uε of problem (2.3) satisfies ‖uεx‖L∞ ≤ c for any t ∈ [0, T).

Theorem 2.3. Suppose that u0(x) ∈ Hs with 1 ≤ s ≤ 3/2 and ‖u0x‖L∞ <∞, then there exists a T >0 such that problem (2.2) has a weak solution u(t, x) ∈ L2([0, T],Hs(R)) in the sense of distributionand ux ∈ L∞([0, T] × R).


3. Proof of Theorem 2.1

Consider the abstract quasilinear evolution equation

dv

dt+A(v)v = f(v), t ≥ 0, v(0) = v0. (3.1)

Let X and Y be Hilbert spaces such that Y is continuously and densely embedded in X, andlet Q : Y → X be a topological isomorphism. Let L(Y,X) be the space of all bounded linearoperators from Y to X. If X = Y , we denote this space by L(X). We state the following con-ditions in which ρ1, ρ2, ρ3, and ρ4 are constants depending only on max{‖y‖Y , ‖z‖Y}:

(I) A(y) ∈ L(Y,X) for y ∈ X with

∥∥(A(y) −A(z)

)w∥∥X ≤ ρ1

∥∥y − z∥∥X‖w‖Y , y, z, w ∈ Y, (3.2)

and A(y) ∈ G(X, 1, β) (i.e., A(y) is quasi-m-accretive), uniformly on bounded setsin Y .

(II) QA(y)Q−1 = A(y) + B(y), where B(y) ∈ L(X) is bounded, uniformly on boundedsets in Y . Moreover,

∥∥(B(y) − B(z))w∥∥X ≤ ρ2

∥∥y − z∥∥Y‖w‖X, y, z ∈ Y, w ∈ X. (3.3)

(III) f : Y → Y extends to a map from X into X, is bounded on bounded sets in Y , andsatisfies

∥∥f(y) − f(z)∥∥Y ≤ ρ3

∥∥y − z∥∥Y , y, z ∈ Y,∥∥f(y) − f(z)∥∥X ≤ ρ4

∥∥y − z∥∥X, y, z ∈ Y.(3.4)

Kato Theorem (see [23])

Assume that (I), (II), and (III) hold. If v0 ∈ Y , there is a maximal T > 0 depending only on‖v0‖Y and a unique solution v to problem (3.1) such that

v = v(·, v0) ∈ C([0, T);Y )⋂C1([0, T);X). (3.5)

Moreover, the map v0 → v(·, v0) is a continuous map from Y to the space

C([0, T);Y )⋂C1([0, T);X). (3.6)

In fact, problem (2.2) can be written as

ut − utxx = −[2ku +

m

2u2]

x+b

2∂3x u

2 − 3b − a2

∂x(u2x

)+ βunx,

u(0, x) = u0(x),(3.7)


which is equivalent to

ut + buux = −Λ−2[(

2ku +m

2u2)

x+ buux − 3b − a

2∂x(u2x

)+ βunx

],

u(0, x) = u0(x).(3.8)

We set A(u) = bu∂x with constant b > 0, Y = Hs(R), X = Hs−1(R), Λ = (1 − ∂2x)

1/2,f(u) = −Λ−2(2ku + (m/2)u2)x + bΛ−2(uux) − ((3b − a)/2)Λ−2∂x(u2

x) + βΛ−2unx, and Q = Λs.

We know that Q is an isomorphism of Hs onto Hs−1. In order to prove Theorem 2.1, we onlyneed to check that A(u) and f(u) satisfy assumptions (I)–(III).

Lemma 3.1. The operator A(u) = u∂x with u ∈ Hs(R), s > 3/2 belongs to G(Hs−1, 1, β).

Lemma 3.2. Let A(u) = bu∂x with u ∈ Hs and s > 3/2, then A(u) ∈ L(Hs,Hs−1) for all u ∈ Hs.Moreover,

‖(A(u) −A(z))w‖Hs−1 ≤ ρ1‖u − z‖Hs−1‖w‖Hs , u, z,w ∈ Hs(R). (3.9)

Lemma 3.3. For s > 3/2, u, z ∈ Hs, and w ∈ Hs−1, it holds that B(u) = [Λ, u∂x]Λ−1 ∈ L(Hs−1)for u ∈ Hs and

‖(B(u) − B(z))w‖Hs−1 ≤ ρ2‖u − z‖Hs‖w‖Hs−1 . (3.10)

Proofs of the above Lemmas 3.1–3.3 can be found in [24] or [25].

Lemma 3.4 (see [23]). Let r and q be real numbers such that −r < q ≤ r, then


‖uv‖Hr+q−1/2 ≤ c‖u‖Hr‖v‖Hq , if r <12.

(3.11)

Lemma 3.5. Let u, z ∈ Hs with s > 3/2 and f(u) = −Λ−2(2ku+(m/2)u2)x + bΛ−2(uux) − ((3b −

a)/2)Λ−2∂x(u2x) + βΛ

−2(unx), then f is bounded inHs and satisfies

∥∥f(u) − f(z)∥∥Hs ≤ ρ3‖u − z‖Hs,∥∥f(u) − f(z)∥∥Hs−1 ≤ ρ4‖u − z‖Hs−1 .

(3.12)


Proof. Using the algebra property of the space Hs0 with s0 > 1/2 and s − 1 > 1/2, we have

∥∥f(u) − f(z)∥∥Hs

≤∥∥∥∥Λ

−2((

2ku +m

2u2)

x−(

2kz +m

2z2)

x

)∥∥∥∥Hs

+∥∥∥bΛ−2(uux − zzx)

∥∥∥Hs

+∥∥∥∥

3b − a2

Λ−2∂x(u2x − z2

x

)∥∥∥∥Hs

+∥∥∥βΛ−2(unx − znx)

∥∥∥Hs

≤ c⎛

⎝‖u − z‖Hs−1(1 + ‖u‖Hs−1 + ‖z‖Hs−1) +∥∥∥Λ−2

(u2 − z2

)

x

∥∥∥Hs

+∥∥∥u2

x − z2x

∥∥∥Hs−1

+ ‖ux − zx‖Hs−1

n−1∑

j=0‖ux‖n−jHs−1‖zx‖jHs−1

⎞

⎠

≤ c⎛

⎝‖u − z‖Hs(1 + ‖u‖Hs + ‖z‖Hs) + ‖(u − z)(u + z)‖Hs−1

+∥∥∥u2

x − z2x

∥∥∥Hs−1

+ ‖ux − zx‖Hs−1

n−1∑

j=0‖u‖n−jHs ‖z‖jHs

⎞

⎠

≤ c‖u − z‖Hs

⎛

⎝1 + ‖u‖Hs + ‖z‖Hs +n−1∑


⎞

⎠

≤ cρ3‖u − z‖Hs,

(3.13)

from which we obtain (3.12).Applying Lemma 3.4, uux = 1/2(u2)x, s > 3/2, we get

∥∥f(u) − f(z)∥∥Hs−1

≤ c(∥∥∥2ku +

m

2u2 −

(2kz +

m

2z2)∥∥∥

Hs−2+∥∥∥u2 − z2

∥∥∥Hs−2

+‖(ux − zx)(ux + zx)‖Hs−2 + ‖unx − znx‖Hs−1

)

≤ c‖u − z‖Hs−2(1 + ‖u‖Hs−1 + ‖z‖Hs−1)

+ c‖ux − zx‖Hs−2(‖ux‖Hs−1 + ‖zx‖Hs−1)

+ c‖ux − zx‖Hs−2

n−1∑

j=0‖ux‖n−jHs−1‖zx‖jHs−1

≤ c‖u − z‖Hs−1

⎛

⎝1 + ‖u‖Hs + ‖z‖Hs +n−1∑


⎞

⎠,

(3.14)

which completes the proof of (3.12).


Proof of Theorem 2.1. Using the Kato theorem, Lemmas 3.1, 3.2, 3.3, and 3.5, we know that sys-tem (3.11) or problem (2.2) has a unique solution

u(t, x) ∈ C([0, T);Hs(R))⋂C1([0, T);Hs−1

). (3.15)

4. Proofs of Theorems 2.2 and 2.3

Before establishing the proofs of Theorems 2.2 and 2.3, we give several lemmas.

Lemma 4.1 (Kato and Ponce [26]). If r ≥ 0, thenHr⋂L∞ is an algebra. Moreover,

‖uv‖Hr ≤ c(‖u‖L∞‖v‖Hr + ‖u‖Hr‖v‖L∞), (4.1)

where c is a constant depending only on r.

Lemma 4.2 (Kato and Ponce [26]). Let r > 0. If u ∈ Hr⋂W1,∞ and v ∈ Hr−1⋂L∞, then

‖[Λr , u]v‖L2 ≤ c(‖∂xu‖L∞

∥∥∥Λr−1v∥∥∥L2

+ ‖Λru‖L2‖v‖L∞

), (4.2)

where [Λr , u]v = Λr(uv) − uΛrv.Using the first equation of problem (2.2) gives rise to

d

dt

[∫

R

(u2 + u2

x

)dx

]+ (a − 2b)

∫

R

(ux)3dx = 2β∫

R

uunx dx, (4.3)

from which one has

∫

R

(u2 + u2

x

)dx +

∫ t

0

[∫

R

((a − 2b)u3

x − 2βuunx)dx

]dτ =

∫

R

(u2

0 + u20x

)dx. (4.4)

Lemma 4.3. Let s ≥ 3/2, and the function u(t, x) is a solution of the problem (2.2) and the initial datau0(x) ∈ Hs, then it holds that

‖u‖L∞ ≤ c‖u‖H1 ≤ c‖u0‖H1ec0∫ t

0(‖ux‖L∞+‖ux‖n−1L∞ )dτ , (4.5)

where c0 = 1/2 max(|a − 2b|, |β|).For q ∈ (0, s − 1], there is a constant c depending only on q such that

∫

R

(Λq+1u

)2dx ≤

∫

R

(Λq+1u0

)2dx + c

∫ t

0

(‖ux‖L∞ + ‖ux‖n−1

L∞

)‖u‖2

Hq+1 dτ. (4.6)

If q ∈ [0, s − 1], there is a constant c depending only on q such that

‖ut‖Hq ≤ c‖u‖Hq+1

(1 + ‖u‖H1 + ‖ux‖n−1

L∞

). (4.7)


Proof. Using ‖u‖2H1 ≤ c

∫R(u

2 + u2x)dx and (4.4) derives (4.5).

We write (1.5) in the equivalent form

ut − utxx = −[2ku +

m

2u2]

x+b

2∂3xu

2 − 3b − a2

∂x(u2x

)+ βunx. (4.8)

Applying ∂2x = −Λ2 + 1 and the Parseval’s equality gives rise to

∫

R

ΛquΛq∂3x

(u2)dx = −2

∫

R

(Λq+1u

)Λq+1(uux)dx + 2

∫

R

(Λqu)Λq(uux)dx. (4.9)

For q ∈ (0, s − 1], applying (Λqu)Λq on both sides of (4.8), noting the above equality,and integrating the new equation with respect to x by parts, we obtain the equation

12d

dt

[∫

R

((Λqu)2 + (Λqux)

2)dx

]

= −∫

R

(Λqu)Λq[2ku +

m

2u2]

xdx − b

∫

R

(Λq+1u

)Λq+1(uux)dx

+3b − a

2

∫

R

(Λqux)Λq(u2x

)dx + b

∫

R

(Λqu)Λq(uux)dx + β∫

R

ΛquΛqunx dx.

(4.10)

We will estimate each of the terms on the right-hand side of (4.10). For the first and thefourth terms, using integration by parts, the Cauchy-Schwartz inequality, and Lemmas 4.1-4.2, we have

∫R(Λ

qu)Λq(uux)dx =∫

R

(Λqu)[Λq(uux) − uΛqux]dx +∫

R

(Λqu)uΛqux dx

≤ c‖u‖Hq(‖ux‖L∞‖u‖Hq + ‖ux‖L∞‖u‖Hq)

+12‖ux‖L∞‖Λqu‖2

L2

≤ c‖u‖2Hq‖ux‖L∞ ,

(4.11)

where c only depends on q. Using the above estimate to the second term yields

∫

R

(Λq+1u

)Λq+1(uux)dx ≤ c‖u‖2

Hq+1‖ux‖L∞ . (4.12)


For the third term, using Lemma 4.1 gives rise to

∫

R

(Λqux)Λq(u2x

)dx ≤ ‖Λqux‖L2

∥∥∥Λqu2

x

∥∥∥L2

≤ c‖u‖Hq+1‖ux‖L∞‖ux‖Hq

≤ c‖u‖2Hq+1‖ux‖L∞ .

(4.13)

For the last term, using Lemma 4.1 repeatedly, we get

∣∣∣∣

∫

R

(Λqu)Λq(ux)ndx∣∣∣∣ ≤ ‖u‖Hq‖unx‖Hq ≤ c‖u‖2

Hq+1‖ux‖n−1L∞ . (4.14)

It follows from (4.10)–(4.14) that

12

∫

R

[(Λqu)2 + (Λqux)

2]dx − 1

2

∫

R

[(Λqu0)

2 + (Λqu0x)2]dx

≤ c∫ t

0

(‖ux‖L∞ + ‖ux‖n−1

L∞

)‖u‖2

Hq+1dτ,

(4.15)

which results in (4.6). Applying the operator (1 − ∂2x)

−1 on both sides of (4.8) yields the equa-tion

ut =(

1 − ∂2x

)−1[−[2ku +

m

2u2]

x+b

2∂3xu

2 − 3b − a2

∂x(u2x

)+ βunx

].

(4.16)

Multiplying both sides of (4.16) by (Λqut)Λq for q ∈ [0, s − 1] and integrating the resultantequation by parts give rise to

∫

R

(Λqut)2dx

=∫

R

(Λqut)(

1 − ∂2x

)−1Λq

[−[2ku +

m

2u2]

x+b

2∂3xu

2 − 3b − a2

∂x(u2x

)+ βunx

]dx.

(4.17)

On the right-hand side of (4.17), we have

∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq(−2kux)dx

∣∣∣∣ ≤ |2k|‖ut‖Hq‖u‖Hq , (4.18)

10 Abstract and Applied Analysis∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq∂x

(−m

2u2 − 3b − a

2u2x

)dx

∣∣∣∣

≤‖ut‖Hq ×(∫

R

(1 + ξ2

)q−1dξ

(∫

R

[m

2u(ξ − η)u(η) + 3b − a

2ux(ξ − η)ux

(η)]dη

)2)1/2

≤ c‖ut‖Hq

(∫

R

c(‖u‖Hq‖u‖L2 + ‖ux‖L2‖ux‖Hq)1 + ξ2

dξ

)1/2

≤ c‖ut‖Hq‖u‖H1‖u‖Hq+1 ,

(4.19)

in which we have used Lemma 4.1. As

∫

R

(Λqut)(

1 − ∂2x

)−1Λq∂2

x(uux)dx = −∫

R

(Λqut)Λq(uux)dx

+∫

R

(Λqut)(

1 − ∂2x

)−1Λq(uux)dx,

(4.20)

by using ‖uux‖Hq ≤ c‖u2‖Hq+1 ≤ c‖u‖L∞‖u‖Hq+1 ≤ c‖u‖H1‖u‖Hq+1 , we have

∣∣∣∣

∫

R

(Λqut)Λq(uux)dx∣∣∣∣ ≤ c‖ut‖Hq‖uux‖Hq ≤ c‖ut‖Hq‖u‖H1‖u‖Hq+1 , (4.21)

∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq(uux)dx

∣∣∣∣ ≤ c‖ut‖Hq‖u‖H1‖u‖Hq+1 . (4.22)

Using the Cauchy-Schwartz inequality and Lemma 4.1 yields

∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq(unx)dx

∣∣∣∣ ≤ c‖ut‖Hq‖ux‖n−1L∞ ‖u‖Hq+1 . (4.23)

Applying (4.18)–(4.23) to (4.17) yields the inequality


(1 + ‖u‖H1 + ‖ux‖n−1

), (4.24)

for a constant c > 0.


Lemma 4.4. For s > 0, u0 ∈ Hs(R), and uε0 = φε � u0, the following estimates hold for any ε withε ∈ (0, 1/4) :

‖uε0x‖L∞ ≤ c‖u0x‖L∞ , ‖uε0‖Hq ≤ c if q ≤ s,

‖uε0‖Hq ≤ cε(s−q)/4 if q > s,

‖uε0 − u0‖Hq ≤ cε(s−q)/4 if q ≤ s,‖uε0 − u0‖Hs = o(1),

(4.25)

where c is a constant independent of ε.

The proof of this lemma can be found in [21].Applying Lemmas 4.3 and 4.4, we can now state the following lemma, which plays an

important role in proving existence of weak solutions.

Lemma 4.5. For s ≥ 1 and u0 ∈ Hs(R), there exists a constant c independent of ε, such that thesolution uε of problem (2.3) satisfies

‖uε‖H1 ≤ cec0∫ t

0(‖uεx‖L∞+‖uεx‖n−1L∞ )dτ for t ∈ [0, Tε), (4.26)

where c0 = 1/2 max(|a − 2b|, |β|).

Proof. The proof can be directly obtained from Lemma 4.4 and inequality (4.5).

Proof of Theorem 2.2. Using notation u = uε and differentiating (4.16) with respect to x giverise to

uxt + buuxx +a − b

2u2x = 2ku +

m − b2

u2

−Λ−2[

2ku +m − b

2u2 +

3b − a2

u2x − β(unx)x

].

(4.27)

Using

∫

R

uuxx(ux)2p+1dx =∫

R

u(ux)2p+1dux

= −∫

R

ux[(ux)2p+2 +

(2p + 1

)u(ux)2puxx

]dx,

(4.28)

we get

∫

R

uuxx(ux)2p+1dx = − 12p + 2

∫

R

(ux)2p+3dx. (4.29)


Letting p > 0 be an integer and multiplying (4.27) by (ux)2p+1 and then integrating the

resulting equation with respect to x yield the equality

12p + 2

d

dt

∫

R

(ux)2p+2dx +(a − b)p + a − 2b

2p + 2

∫

R

(ux)2p+3dx

=∫

R

(ux)2p+1(

2ku +m − b

2u2)dx

−∫

R

(ux)2p+1Λ−2[

2ku +m − b

2u2 +

3b − a2

u2x − β(unx)x

]dx.

(4.30)

Applying the Holder’s inequality, we get

12p + 2

d

dt

∫

R

(ux)2p+2dx ≤⎧⎨

⎩

(∫

R

∣∣∣∣2ku +m − b

2u2∣∣∣∣

2p+2

dx

)1/(2p+2)

+(∫

R

|G|2p+2dx

)1/(2p+2)}(∫

R

|ux|2p+2dx

) (2p+1)/(2p+2)

+∣∣∣∣(a − b)p + a − 2b

2p + 2

∣∣∣∣‖ux‖L∞

∫

R

|ux|2p+2dx,

(4.31)

or

d

dt

(∫

R

(ux)2p+2dx

) 1/(2p+2)

≤⎧⎨

⎩

(∫

R

∣∣∣∣2ku +m − b

2u2∣∣∣∣

2p+2

dx

) 1/(2p+2)

+(∫

R

|G|2p+2dx

)1/(2p+2)⎫⎬

⎭

+∣∣∣∣(a − b)p + a − 2b

2p + 2

∣∣∣∣‖ux‖L∞

(∫

R

|ux|2p+2dx

)1/(2p+2)

,

(4.32)

where

G = Λ−2[

2ku +m − b

2u2 +

3b − a2

u2x − β(unx)x

]. (4.33)

Since ‖f‖Lp → ‖f‖L∞ as p → ∞ for any f ∈ L∞⋂L2, integrating (4.32) with respect to t andtaking the limit as p → ∞ result in the estimate

‖ux‖L∞ ≤ ‖u0x‖L∞ +∫ t

0

[c

(∥∥∥∥2ku +m − b

2u2∥∥∥∥L∞

+ ‖G‖L∞

)+a − b

2‖ux‖2

L∞

]dτ. (4.34)


Using the algebraic property of Hs0(R) with s0 > 1/2 and Lemma 4.5 leads to

∥∥∥∥2ku +

m − b2

u2∥∥∥∥L∞

≤ c∥∥∥∥2ku +

m − b2

u2∥∥∥∥H1/2+

≤ c(‖u‖H1 + ‖u‖2

H1

)

≤ ce2c0∫ t

0(‖ux‖L∞+‖ux‖n−1L∞ )dτ ,

(4.35)

‖G‖L∞ =∥∥∥∥Λ

−2[

2ku +m − b

2u2 +

3b − a2

u2x − β(unx)x

]∥∥∥∥L∞

≤ c(∥∥∥∥Λ

−2(

2ku +m − b

2u2)∥∥∥∥H1/2+

+∥∥∥Λ−2u2

x

∥∥∥H1/2+

+∥∥∥Λ−2(unx)x

∥∥∥H1/2+

)

≤ c(∥∥∥∥2ku +

m − b2

u2∥∥∥∥H1/2+

+ ‖u‖H1‖ux‖L∞ + ‖unx‖H0

)

≤ c(‖u‖H1 + ‖u‖2

H1 + ‖u‖H1‖ux‖L∞ + ‖u‖H1‖ux‖n−1L∞

)

≤ c(

1 + ‖ux‖L∞ + ‖ux‖n−1L∞

)e2c0

∫ t0(‖ux‖L∞+‖ux‖n−1

L∞ )dτ ,

(4.36)

where c is independent of ε, c0 = 1/2 max(|a − 2b|, |β|) and H1/2+ means that there exists asufficiently small δ > 0 such that ‖uε‖H1/2+ = ‖uε‖H1/2+δ . From Lemma 4.5, we have

∫ t

0‖G‖L∞dτ ≤ c

∫ t

0

(1 + ‖ux‖L∞ + ‖ux‖n−1

L∞

)× e2c0

∫τ0 (‖ux‖L∞+‖ux‖n−1

L∞ )dξ dτ. (4.37)

Applying (4.25), (4.34), (4.35), and (4.37) and writing out the subscript ε of u, weobtain

‖uεx‖L∞ ≤ c‖u0x‖L∞

+ c∫ t

0

[(1 + ‖uεx‖L∞ + ‖uεx‖n−1

L∞

)× e2c0

∫τ0 (‖uεx‖L∞+‖uεx‖n−1

L∞ )dξ + ‖uεx‖2L∞

]dτ.

(4.38)

It follows from the contraction mapping principle that there is a T > 0 such that theequation

‖W‖L∞ = c‖u0x‖L∞

+ c∫ t

0

[(1 + ‖W‖L∞ + ‖W‖n−1

L∞

)× e2c0

∫τ0 (‖W‖L∞+‖W‖n−1

L∞ )dξ + ‖W‖2L∞

]dτ

(4.39)

has a unique solution W ∈ C[0, T]. From (4.39), we know that the variable T only dependson c and ‖u0x‖L∞ . Using the theorem presented on page 51 in [16] or Theorem 2 in Section 1.1


in [27] derives that there are constants T > 0 and c > 0 independent of ε such that ‖uεx‖L∞ ≤W(t) for arbitrary t ∈ [0, T], which leads to the conclusion of Theorem 2.

Remark 4.6. Under the assumptions of Theorem 2.2, there exist two constants T and c, bothindependent of ε, such that the solution uε of problem (2.3) satisfies ‖uεx‖L∞ ≤ c for any t ∈[0, T]. This states that in Lemma 4.5, there exists a T independent of ε such that (4.26) holds.

Using Theorem 2.2, Lemma 4.5, (4.6), (4.7), notation uε = u, and Gronwall’s inequalityresults in the inequalities

‖uε‖Hq ≤ c exp

[

c

∫ t

0

(‖uεx‖L∞ + ‖uεx‖n−1

L∞

)dτ

]

≤ c,

‖uεt‖Hr ≤ ‖uε‖Hr+1

(1 + ‖uε‖H1 + ‖uεx‖n−1

L∞

)≤ c,

(4.40)

where q ∈ (0, s], r ∈ (0, s − 1], and t ∈ [0, T). It follows from Aubin’s compactness theo-remthat there is a subsequence of {uε}, denoted by {uεn1

}, such that {uεn1} and their temporal

derivatives {uεn1 t} are weakly convergent to a function u(t, x) and its derivative ut in L2([0, T],

Hs) and L2([0, T],Hs−1), respectively. Moreover, for any real number R1 > 0, {uεn1} is con-

vergent to the function u strongly in the space L2([0, T],Hq(−R1, R1)) for q ∈ (0, s], and {uεn1 t}

converges to ut strongly in the space L2([0, T],Hr(−R1, R1)) for r ∈ [0, s − 1].

Proof of Theorem 2.3. From Theorem 2.2, we know that {uεn1x}(εn1 → 0) is bounded in the

space L∞. Thus, the sequences{uεn1}, {uεn1x

}, {u2εn1x

}, and {unεn1x} are weakly convergent to u,

ux, u2x, and unx in L2([0, T],Hr(−R1, R1)) for any r ∈ [0, s−1), separately. Hence, u satisfies the

equation

−∫T

0

∫

R

u(gt − gxxt

)dx dt =

∫T

0

∫

R

[(2ku +

m

2u2 +

3b − a2

u2x

)gx − b

2u2gxxx − βunxg

]dxdt,

(4.41)

with u(0, x) = u0(x) and g ∈ C∞0 . Since X = L1([0, T] × R) is a separable Banach space and

{uεn1x} is a bounded sequence in the dual space X∗ = L∞([0, T] × R) of X, there exists a

subsequence of {uεn1x}, still denoted by {uεn1x

}, weakly star convergent to a function v inL∞([0, T]×R). As {uεn1x

} weakly converges to ux in L2([0, T]×R), it results that ux = v almosteverywhere. Thus, we obtain ux ∈ L∞([0, T] × R).

Acknowledgment

This work is supported by the Applied and Basic Project of Sichuan Province (2012JY0020).

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Research ArticlePositive Solutions of a Nonlinear Fourth-OrderDynamic Eigenvalue Problem on Time Scales

Hua Luo1 and Chenghua Gao2

1 School of Mathematics and Quantitative Economics, Dongbei University of Finance and Economics,Dalian 116025, China

2 Department of Mathematics, Northwest Normal University, Lanzhou 730070, China

Correspondence should be addressed to Hua Luo, [email protected]

Received 25 December 2011; Accepted 27 March 2012


Copyright q 2012 H. Luo and C. Gao. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

Let T be a time scale and a, b ∈ T, a < ρ2(b). We study the nonlinear fourth-order eigenvalueproblem on T, uΔ4(t) = λh(t)f(u(t), uΔ2(t)), t ∈ [a, ρ2(b)]

T, u(a) = uΔ(σ(b)) = uΔ

2(a) =uΔ

3(ρ(b)) = 0 and obtain the existence and nonexistence of positive solutions when 0 < λ ≤ λ∗

and λ > λ∗, respectively, for some λ∗. The main tools to prove the existence results are the Schauderfixed point theorem and the upper and lower solution method.

1. Introduction

The deformation of an elastic beam with one end fixed and the other end free can be describedby the nonlinear fourth-order dynamic eigenvalue problem on T

uΔ4(t) = λh(t)f

(u(t), uΔ

2(t)

), t ∈ [

a, ρ2(b)]

T,

u(a) = uΔ(σ(b)) = uΔ2(a) = uΔ

3(ρ(b)

)= 0,

(1.1)

where T is a time scale, λ > 0 is a parameter, a, b ∈ T, and a < ρ2(b).Nonlinear dynamic eigenvalue problems of the above type have been studied by

some authors, but most of them study only second-order dynamic equations. In 2000,Chyan and Henderson [1] obtained the existence of at least one positive solution forsome λ to second-order case of the dynamic equation in problem (1.1) under conjugate


boundary value condition and right focal boundary value condition, respectively. Anderson[2] discussed the same second-order dynamic equation under the Sturm-Liouville boundaryvalue condition and directly generalized the result of [1]. Erbe et al. [3] then studied thegeneral second-order Sturm-Liouville dynamic boundary value problem and obtained theexistence, nonexistence, and multiplicity results of positive solutions. In 2005, Li and Liu[4] further studied the dependence of positive solutions on the parameter λ > 0 for thesecond-order dynamic equation under the right focal boundary value condition. Luo andMa [5] in 2006 were concerned with the existence and multiplicity of nodal solutions andobtained eigenvalue intervals of the nonlinear second-order dynamic eigenvalue problemunder conjugate boundary value condition by using bifurcation methods. In 2007, Sun etal. [6] obtained some sufficient conditions for the nonexistence and existence of at least oneor two positive solutions for the p-Laplacian three-point dynamic eigenvalue problem withmixed derivatives by using the Krasnosel’skii’s fixed point theorem in a cone. In 2009, Luo [7]derived the eigenvalue intervals in which there exist positive solutions of a singular second-order multipoint dynamic eigenvalue problem with mixed derivatives by making use of thefixed point index theory.

As for the nonlinear higher-order dynamic eigenvalue problems, few papers canbe found in the literature to the best of our knowledge. L. Kong and Q. Kong [8], andBoey and Wong [9] discussed the even-order dynamic eigenvalue problem and the rightfocal eigenvalue problem, respectively, but their problems do not contain (1.1). Particularlyfor fourth-order problems and special case λ ≡ 1, Wang and Sun [10] studied theexistence of positive solutions for dynamic equations under nonhomogeneous boundary-value conditions describing an elastic beam that is simply supported at its two ends. Andboth Karaca [11] and Pang and Bai [12] obtained the existence of a solution for two classes offourth-order four-point problems on time scales by the Leray-Schauder fixed point theoremand the upper and lower solution method, respectively, but the problems they studied aredifferent to (1.1).

This paper studies the relationship between the existence and nonexistence of positivesolutions and the value of parameter λ > 0. We find the existence of a λ∗ such that problem(1.1) has positive solutions for 0 < λ ≤ λ∗ and no positive solutions for λ > λ∗.

The rest of this paper is organized as follows: in Section 2, we firstly introduce thetime scales concepts and notations and present some basic properties on time scales whichare needed later. Next, Section 3 gives some preliminary results relevant to our discussion,and Section 4 is devoted to establish our main theorems.

2. Introduction to Time Scales

The calculus theory on time scales, which unifies continuous and discrete analysis, is nowstill an active area of research. We refer the reader to [13–16] and the references thereinfor introduction on this theory. For the convenience of readers, we present some necessarydefinitions and results here.

A time scale T is a nonempty closed subset of R, assuming that T has the topology thatit inherits from the standard topology on R. Define the forward and backward jump operatorsσ, ρ : T → T by

σ(t) = inf{τ > t | τ ∈ T}, ρ(t) = sup{τ < t | τ ∈ T}. (2.1)


Here we put inf ∅ = sup T, sup ∅ = inf T. Let μ(t) = σ(t) − t, t ∈ T be the graininess function.And T

k which are derived from the time scale T is

Tk :=

{t ∈ T : t is nonmaximal or ρ(t) = t

}, (2.2)

and Tkn := (Tkn−1

)k, n > 1, n ∈ N. Define interval I on T by IT = I ∩ T.

Definition 2.1. If u : T → R is a function and t ∈ Tk, then the Δ-derivative of u at the point t

is defined to be the number uΔ(t) (provided it exists) with the property that for each ε > 0,there is a neighborhood U of t such that

∣∣∣u(σ(t)) − u(s) − uΔ(t)(σ(t) − s)

∣∣∣ � ε|σ(t) − s| (2.3)

for all s ∈ U. The function u is called Δ-differentiable on T if uΔ(t) exists for all t ∈ Tk.

The second Δ-derivative of u at t ∈ Tk2

, if it exists, is defined to be uΔ2(t) = uΔΔ(t) :=

(uΔ)Δ(t). Similarly,

uΔi

(t) :=(uΔ

i−1)Δ

(t), i > 2, i ∈ N (2.4)

is called the ith Δ-derivative of u at t ∈ Tki . We also define the function uσ := u ◦ σ.

Definition 2.2. If UΔ = u holds on Tk, we define the Cauchy Δ-integral by

∫ t

s

u(τ)Δτ = U(t) −U(s), s, t ∈ Tk. (2.5)

Lemma 2.3. If a, b ∈ T, a < ρ(b), h : [a, b]T

→ [0,+∞) is continuous and h(t)/≡ 0 on [a, b)T,

then

∫b

a

h(t)Δt > 0. (2.6)

Proof. From [15, Theorems 1.28 (Viii) and 1.29], it is clear.

Lemma 2.4 (See [14, Theorem 1.16]). If the Δ-derivative of u exists at t ∈ Tk, then

u(σ(t)) = u(t) + μ(t)uΔ(t). (2.7)

Define the Banach space C(T) to be the set of continuous functions u : T → R withthe norm

‖u‖∞ = max{|u(t)| | t ∈ T}. (2.8)


For i ∈ N, we define the Banach space CiΔ(T) to be the set of the ith Δ-differential functions

u : T → R for which uΔi ∈ C(Tki) with the norm

‖u‖i = max{‖u‖∞,

∥∥∥uΔ

∥∥∥∞, . . . ,

∥∥∥uΔ

i∥∥∥∞

}, (2.9)

where

∥∥∥uΔ

j∥∥∥∞= max

{∣∣∣uΔ

j

(t)∣∣∣ | t ∈ T

kj}, j = 0, 1, . . . , i. (2.10)

3. Preliminaries

Throughout this paper, we assume that both

ξ = min

{

t ∈ T | t ≥ σ2(b) + 3a4

}

, ω = max

{

t ∈ T | t ≤ 3σ2(b) + a4

}

(3.1)

exist and a < ξ < ρ(ω) ≤ ω ≤ ρ(b). So there exists a number m > 0 such that

ω − ξ ≥ m. (3.2)

We also make the following assumptions:

(H1) h : [a, σ(b)]T→ [0,+∞) is continuous and h(t)/≡ 0 on [ξ, ω)

T;

(H2) f : [0,+∞) × (−∞, 0] → [0,+∞) is continuous. f(u,w) is nondecreasing in u,nonincreasing in w and f(0, 0) > 0.

Set v(t) = uΔ2(t), t ∈ [a, b]

T. Then problem (1.1) is equivalent to the system

uΔ2(t) = v(t), t ∈ [a, b]

T,

vΔ2(t) = λh(t)f(u(t), v(t)), t ∈

[a, ρ2(b)

]

T

,

u(a) = uΔ(σ(b)) = 0,

v(a) = vΔ(ρ(b))= 0.

(3.3)

According to [14, Corollary 4.84 and Theorem 4.70], the Green’s function of problems

uΔ2(t) = 0, t ∈ [a, b]

T,

u(a) = uΔ(σ(b)) = 0,

vΔ2(t) = 0, t ∈

[a, ρ2(b)

]

T

,

v(a) = vΔ(ρ(b))= 0,

(3.4)


is of the same form

G(t, s) =

{t − a, t ≤ s,σ(s) − a, σ(s) ≤ t, (3.5)

and the solution of system (3.3) is

u(t) = −∫σ(b)

a

G(t, s)v(s)Δs, t ∈[a, σ2(b)

]

T

,

v(t) = −λ∫ρ(b)

a

G(t, s)h(s)f(u(s), v(s))Δs, t ∈ [a, b]T.

(3.6)

Therefore, the solution of problem (1.1) is

u(t) = λ∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs, t ∈[a, σ2(b)

]

T

. (3.7)

Lemma 3.1. Green’s function (3.5) is of the following properties:

0 ≤ G(t, s) ≤ min{t − a, σ(s) − a}, (t, s) ∈[a, σ2(b)

]

T

× [a, σ(b)]T, (3.8)

G(t, s) ≥ 14(σ(s) − a), (t, s) ∈ [ξ, ω]

T× [a, σ(b)]

T, (3.9)

G(t, s) ≥ t − aσ2(b) − aG(x, s), (t, s) ∈

[a, σ2(b)

]

T

× [a, σ(b)]T, x ∈

[a, σ2(b)

]

T

, (3.10)

G(t, s) ≥ 14G(x, s), (t, s) ∈ [ξ, ω]

T× [a, σ(b)]

T, x ∈

[a, σ2(b)

]

T

, (3.11)

G(t, s) ≥ σ2(b) − a4

, (t, s) ∈ [ξ, ω]T× [ξ, ω]

T. (3.12)

Proof. We here only give the proof of (3.10), and the others can be obtained easily. We dividethe proof into the following four cases.

Case 1 (t ≤ s, x ≤ s). We have

G(t, s) = t − a ≥ x − aσ2(b) − a (t − a) =

t − aσ2(b) − a(x − a) = t − a

σ2(b) − aG(x, s). (3.13)

Case 2 (t ≤ s, σ(s) ≤ x). We have

G(t, s) = t − a ≥ σ(s) − aσ2(b) − a (t − a) =

t − aσ2(b) − a (σ(s) − a) =

t − aσ2(b) − aG(x, s). (3.14)


Case 3 (x ≤ s, σ(s) ≤ t). We have

G(t, s) = σ(s) − a ≥ x − a ≥ t − aσ2(b) − a (x − a) = t − a

σ2(b) − aG(x, s). (3.15)

Case 4 (σ(s) ≤ t, σ(s) ≤ x). We have

G(t, s) = σ(s) − a ≥ t − aσ2(b) − a(σ(s) − a) =

t − aσ2(b) − aG(x, s). (3.16)

Define

E ={u ∈ C2

Δ

([a, σ2(b)

])| u(a) = uΔ(σ(b)) = uΔ2

(a) = uΔ3(ρ(b)

)= 0

}. (3.17)

Lemma 3.2. For u ∈ E, one has

‖u‖∞ ≤ 2[σ2(b) − a

]∥∥∥uΔ∥∥∥∞, (3.18)

∥∥∥uΔ∥∥∥∞≤ 2

[σ2(b) − a

]∥∥∥uΔ2∥∥∥∞, (3.19)

where ‖u‖∞ = max{|u(t)| | t ∈ [a, σ2(b)]T}, ‖uΔ‖∞ = max{|uΔ(t)| | t ∈ [a, σ(b)]

T}, ‖uΔ2‖∞ =

max{|uΔ2(t)| | t ∈ [a, b]

T}.

Proof. Firstly, we show that (3.18) holds.For all t ∈ [a, σ(b)], we have from u(a) = 0 that

|u(t)| =∣∣∣∣∣u(a) +

∫ t

a

uΔ(s)Δs

∣∣∣∣∣

≤∫ t

a

∣∣∣uΔ(s)∣∣∣Δs

≤∥∥∥uΔ

∥∥∥∞(t − a)

≤[σ2(b) − a

]∥∥∥uΔ∥∥∥∞.

(3.20)

Combining this with

∣∣∣u(σ2(b)

)∣∣∣ =∣∣∣u(σ(b)) + μ(σ(b))uΔ(σ(b))

∣∣∣

≤[σ2(b) − a

]∥∥∥uΔ∥∥∥∞+[σ2(b) − a

]∥∥∥uΔ∥∥∥∞

= 2[σ2(b) − a

]∥∥∥uΔ∥∥∥∞,

(3.21)

we have ‖u‖∞ ≤ 2[σ2(b) − a]‖uΔ‖∞.


Secondly, we show that (3.19) holds.For all t ∈ [a, b], we have from uΔ(σ(b)) = 0 that

∣∣∣uΔ(t)

∣∣∣ =

∣∣∣∣∣−∫b

t

uΔ2(s)Δs + uΔ(b)

∣∣∣∣∣

=

∣∣∣∣∣−∫b

t

uΔ2(s)Δs + uΔ(σ(b)) − μ(b)uΔ2

(b)

∣∣∣∣∣

≤∥∥∥uΔ

2∥∥∥∞(b − t) +

∥∥∥uΔ

2∥∥∥∞μ(b)

≤ 2[σ2(b) − a

]∥∥∥uΔ

2∥∥∥∞.

(3.22)

Combining this with

∣∣uΔ(σ(b))∣∣ = 0 ≤ 2

[σ2(b) − a]

∥∥∥uΔ2∥∥∥∞, (3.23)

we have ‖uΔ‖∞ ≤ 2[σ2(b) − a]‖uΔ2‖∞.

Define

‖u‖E = max{

‖u‖∞, 2[σ2(b) − a

]∥∥∥uΔ∥∥∥∞, 4

[σ2(b) − a

]2∥∥∥uΔ2∥∥∥∞

}

= 4[σ2(b) − a

]2∥∥∥uΔ2∥∥∥∞.

(3.24)

Then E is a Banach space under the norm ‖ · ‖E. Set

(Aλu)(t) = λ∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs, t ∈[a, σ2(b)

]

T

. (3.25)

Then Aλ : E → E. Since Aλ : C2Δ([a, σ

2(b)]) → C4Δ([a, σ

2(b)]) ↪→ C2Δ([a, σ

2(b)]), we havethat Aλ is completely continuous.

Lemma 3.3. Suppose (H1) and (H2) hold, and u(t) is a solution of problem (1.1), then

u(t) ≥ 0, t ∈[a, σ2(b)

]

T

, uΔ2(t) ≤ 0, t ∈ [a, b]

T, (3.26)

u(t) > 0, uΔ2(t) < 0, t ∈ [ξ, ω]

T, (3.27)

mint∈[ξ,ω]

T

(−uΔ2

(t))≥ 1

4

∥∥∥uΔ2∥∥∥∞=

1

16[σ2(b) − a]2 ‖u‖E, (3.28)

mint∈[ξ,ω]

T

u(t) ≥ 14‖u‖∞, (3.29)

mint∈[ξ,ω]

T

u(t) ≥ m

64[σ2(b) − a]‖u‖E. (3.30)


Proof. By λ > 0, (3.7), (3.8) and the fact that h, f are nonnegative functions, (3.26) holds. Fort ∈ [ξ, ω]

T, from (3.7), (3.26), (3.2) and (3.12), we have

u(t) = λ∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ λf(0, 0)∫ω

ξ

[∫ω

ξ

G(t, s)G(s, j

)h(j)Δj

]

Δs

≥ λm[σ2(b) − a]2

16f(0, 0)

∫ω

ξ

h(j)Δj

> 0,

−uΔ2(t) = λ

∫ρ(b)

a

G(t, s)h(s)f(u(s), uΔ

2(s)

)Δs

≥ λf(0, 0)∫ω

ξ

G(t, s)h(s)Δs

≥ λσ2(b) − a

4f(0, 0)

∫ω

ξ

h(s)Δs

> 0

(3.31)

Therefor (3.27) holds.For all t ∈ [ξ, ω]

T, for all x ∈ [a, σ2(b)]

T, from (3.7), (3.11), and (3.12), we have

−uΔ2(t) = λ

∫ρ(b)

a


2(s)

)Δs

≥ 14λ

∫ρ(b)

a

G(x, s)h(s)f(u(s), uΔ

2(s)

)Δs

=14

[−uΔ2

(x)],

u(t) = λ∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ 14λ

∫σ(b)

a

[∫ρ(b)

a

G(x, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

=14u(x),

u(t) = λ∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ σ2(b) − a4

λ

∫ω

ξ

[∫ρ(b)

a

G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ σ2(b) − a16

λ

∫ω

ξ

[∫ρ(b)

a

G(x, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs


≥ m[σ2(b) − a]

16

[−uΔ2

(x)]

=m

64[σ2(b) − a]4[σ2(b) − a

]2[−uΔ2(x)

].

(3.32)

Thus (3.28), (3.29), and (3.30) hold.

At the end of this section, we state a lemma of the upper and lower solution method,which is needed for some proofs in next section.

Lemma 3.4 (See [17, Theorem 3.3.8]). Let P be a cone with nonempty interior in Banach space E,and A : P → P a completely continuous and increasing operator. Suppose the following conditionshold:

(i) there exist x0, y0 ∈ P , such that x0 ≤ Ax0, Ay0 ≤ y0, x0 � y0;

(ii) there exist φ ∈ P and a constant α > 0, such that Ax ≥ α‖Ax‖φ, for all x ∈ P ;(iii) y0 is in the interior of P , and there exists β > 0, such that φ ≥ βx0.

Then A has at least one fixed point x /= 0 in P .

4. The Main Result

Our main result is the following existence theorem.

Theorem 4.1. Suppose (H1) and (H2) hold, and either (H3) or (H4) holds. Here

(H3) lim(u,w)→ (+∞,−∞)(f(u,w)/u) = +∞;

(H4) lim(u,w)→ (+∞,−∞)(f(u,w)/(−w)) = +∞.

Then there exists λ∗ > 0, such that problem (1.1) has at least one positive solution for λ ∈ (0, λ∗], andhas no positive solution for λ ∈ (λ∗,+∞).

To prove Theorem 4.1, first, we show that problem (1.1) has positive solutions for someλ small enough.

Theorem 4.2. Suppose (H1) and (H2) hold. Then there exists λ∗ > 0, such that problem (1.1) hasat least one positive solution for λ ∈ (0, λ∗].

Proof. The fixed point of Aλ defined in (3.25) is the solution of problem (1.1), so it will beenough to find the fixed point of Aλ.

Set

M = maxt∈[a,b]

T

∫ρ(b)

a

G(t, s)h(s)Δs > 0. (4.1)


Let

λ∗ =1

Mf(

4[σ2(b) − a]2,−1) . (4.2)

For λ ∈ (0, λ∗], we have

f

(4[σ2(b) − a

]2,−1

)≤ 1

Mλ. (4.3)

Set

D1 ={u ∈ E | ‖u‖E ≤ 4

[σ2(b) − a

]2}. (4.4)

If u ∈ D1, then

‖u‖∞ ≤ ‖u‖E ≤ 4[σ2(b) − a

]2,

∥∥∥uΔ2∥∥∥∞=

‖u‖E4[σ2(b) − a]2

≤ 1. (4.5)

Consequently, for t ∈ [a, b]T

,

∣∣∣ (Aλu)Δ2(t)

∣∣∣ =

∣∣∣∣∣−λ

∫ρ(b)

a


2(s)

)Δs

∣∣∣∣∣

≤ Mλf(‖u‖∞,−

∥∥∥uΔ2∥∥∥∞

)

≤ Mλf

(4[σ2(b) − a

]2,−1

)

≤ 1.

(4.6)

Then

‖Aλu‖E = 4[σ2(b) − a

]2∥∥∥(Aλu)Δ2∥∥∥∞≤ 4

[σ2(b) − a

]2, (4.7)

that is, Aλ : D1 → D1. By the Schauder fixed point theorem, Aλ has at least one fixed pointuλ in E satisfying ‖uλ‖E ≤ 4[σ2(b)−a]2. From Lemma 3.3, uλ is a positive solution of problem(1.1).

Next, we show that there exist no positive solution for some λ large enough.

Theorem 4.3. Suppose that (H1) and (H2), hold, and either (H3) or (H4) holds. Then problem(1.1) has no positive solution for λ 1.

Proof. Suppose uλ ∈ E is a solution to problem (1.1) for some λ > 0. We divide our discussionsinto two cases.


Case 1 ((H1), (H2) and (H3) hold). By (H3), for a fixed M1 > 0, there is H > 0 such that

f(u,w) ≥M1u, u ≥ mH

64[σ2(b) − a] , w ≤ −H16[σ2(b) − a]2

. (4.8)

If u ∈ E with ‖uλ‖E ≥ H, then from (3.30) and (3.28), we have

mint∈[ξ,ω]

T

uλ(t) ≥ m

64[σ2(b) − a]‖un‖E ≥ mH

64[σ2(b) − a] ,

mint∈[ξ,ω]

T

(−uΔ2

λ (t))≥ 1

4

∥∥∥uΔ

2

λ

∥∥∥∞=

‖uλ‖E16[σ2(b) − a]2

≥ H

16[σ2(b) − a]2.

(4.9)

Further by (4.8),

f(uλ(t), uΔ

2

λ (t))≥M1uλ(t), t ∈ [ξ, ω]

T. (4.10)

If u ∈ E with ‖uλ‖E < H, then from (3.27), there is

f(uλ(t), uΔ

2

λ (t))≥M2uλ(t), t ∈ [ξ, ω]

T(4.11)

with

M2 = mint∈[ξ,ω]

T

f(uλ(t), uΔ

2

λ (t))

uλ(t)> 0. (4.12)

Set M = min{M1,M2}, we have from (4.10) and (4.11) that

f(uλ(t), uΔ

2

λ (t))≥ Muλ(t), t ∈ [ξ, ω]

T. (4.13)

Combining this with (3.7), (4.13), (3.29), (3.12), and (3.2), we have for t ∈ [ξ, ω]T

,

uλ(t) = λ∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(uλ

(j), uΔ

2

λ

(j))

Δj

]

Δs

≥ λM∫σ(b)

a

[∫ω

ξ

G(t, s)G(s, j

)h(j)uλ

(j)Δj

]

Δs

≥ λM‖uλ‖∞4

∫ω

ξ

[∫ω

ξ

G(t, s)G(s, j

)h(j)Δj

]

Δs

≥ λMm[σ2(b) − a]2

64‖uλ‖∞

∫ω

ξ

h(j)Δj.

(4.14)


If we choose λ such that

λMm

[σ2(b) − a]2

64

∫ω

ξ

h(j)Δj > 1, (4.15)

then

uλ(t) > ‖uλ‖∞, t ∈ [ξ, ω]T, (4.16)

which is a contradiction.

Case 2 ((H1), (H2), and (H4) hold). Similar to Case 1, by (H4), there is also an M, such that

f(uλ(t), uΔ

2

λ (t))≥ M

(−uΔ2

λ (t)), t ∈ [ξ, ω]

T. (4.17)

Thus for t ∈ [ξ, ω]T

, we have from (3.7), (4.17), (3.28), and (3.12) that

−uΔ2

λ (t) = λ∫ρ(b)

a

G(t, s)h(s)f(uλ(s), uΔ

2

λ (s))Δs

≥ λM∫ω

ξ

G(t, s)h(s)(−uΔ2

λ (s))Δs

≥ λM[σ2(b) − a]

16

∥∥∥uΔ2

λ

∥∥∥∞

∫ω

ξ

h(s)Δs.

(4.18)

If we take λ such that

λM

[σ2(b) − a]

16

∫ω

ξ

h(s)Δs > 1, (4.19)

then

−uΔ2

λ (t) >∥∥∥uΔ

2

λ

∥∥∥∞, t ∈ [ξ, ω]

T, (4.20)

which is a contradiction.

Therefore, problem (1.1) has no positive solution for λ 1.

Define the set B = {λ > 0 : problem (1.1) has at least one positive solution in E}.

Theorem 4.4. One has

λ ∈ B =⇒ λ ∈ B, ∀λ ∈(

0, λ]. (4.21)


Proof. Let λ∗ be defined as Theorem 4.2. For λ ≤ λ∗, the result holds from Theorem 4.2. So, wediscuss the case that λ > λ∗.

For λ ∈ (0, λ], three cases will be discussed.

(1) λ ∈ (0, λ∗];

(2) λ ∈ (λ∗, λ);

(3) λ = λ.

Cases (1) and (3) are clear from Theorem 4.2 and the assumption λ ∈ B, respectively. Now,we deal with Case (2).

Define

D2 ={u ∈ E | u(t) ≥ 0, t ∈

[a, σ2(b)

], uΔ

2(t) ≤ 0, t ∈ [a, b]

}. (4.22)

Then D2 is a cone with nonempty interior in E. For λ ∈ (λ∗, λ), let Aλ be defined as (3.25).Then Aλ : D2 → D2 and Aλ is an increasing operator from (H2). Set u∗ and u as positivesolutions of problem (1.1) at λ∗ and λ, respectively. Then

(Aλu∗)(t) > (Aλ∗u∗)(t) = u∗(t),

(Aλu)(t) <(Aλu

)(t) = u(t).

(4.23)

So u is an upper solution of the operator Aλ and u∗ is a lower solution. If u∗ ≤ u, then thereexists a positive solution u satisfying u∗(t) ≤ u(t) ≤ u(t) for t ∈ [a, σ2(b)]

Tby the upper and

lower solution method. If u∗ � u, we verify the conditions of Lemma 3.4.Clearly, the condition (i) in Lemma 3.4 holds for u∗ = x0, u = y0.For all u ∈ D2, for all t, z ∈ [a, σ2(b)]

T, x ∈ [ξ, ω]

T, we have from (3.10), (3.12), (3.11),

and (3.2) that

(Aλu)(t) = λ∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ λ t − aσ2(b) − a

∫σ(b)

a

[∫ρ(b)

a

G(x, s)G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ λt − a4

∫ω

ξ

[∫ρ(b)

a

G(s, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ λt − a4

∫ω

ξ

[∫ρ(b)

a

14G(z, j

)h(j)f(u(j), uΔ

2(j))

Δj

]

Δs

≥ m

16(t − a)

[−(Aλu)Δ

2(z)

].

(4.24)


Then

Aλu ≥ m

16

∥∥∥(Aλu)Δ

2∥∥∥∞(t − a) = m

64[σ2(b) − a]2 ‖Aλu‖E(t − a), (4.25)

and the condition (ii) of Lemma 3.4 is satisfied for φ = t − a, α = m/64[σ2(b) − a]2> 0.

From (3.26) and (3.27), we have that u is in the interior of D2. From (3.8), there is

u∗(t) = λ∗

∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(u∗(j), uΔ

2

∗(j))

Δj

]

Δs

≤ λ∗(t − a)∫σ(b)

a

[∫ρ(b)

aG(s, j

)h(j)f(u∗(j), uΔ

2

∗(j))

Δj

]

Δs

≤ λ∗(t − a)[∫σ(b)

a

(s − a)Δs][∫ρ(b)

a

h(j)f(u∗(j), uΔ

2

∗(j))

Δj

]

.

(4.26)

This implies φ(t) = t − a ≥ βu∗(t), t ∈ [a, σ2(b)]T

for

β =

{

λ∗

[∫σ(b)

a

(s − a)Δs][∫ρ(b)

a

h(j)f(u∗(j), uΔ

2

∗(j))

Δj

]}−1

> 0, (4.27)

and the condition (iii) of Lemma 3.4 is satisfied.By Lemmas 3.4 and 3.3, we get a positive solution uλ in D2 ∈ E. That is, λ ∈ B.

Now, we give the proof of Theorem 4.1.

Proof of Theorem 4.1. From Theorems 4.2 and 4.3, B /= ∅ is bounded. Thus, we can define λ∗ =supB. Firstly, we show that λ∗ ∈ B.

Choose a sequence {λn}∞n=1, λn ∈ B (n = 1, 2, . . .) which belongs to a compactsubinterval in (0,∞), and λn → λ∗(n → ∞). Then there exists N1 > 0, for n ≥N1,

λn >λ∗

2. (4.28)

Let un ∈ E satisfy

Aλnun = un. (4.29)

If {un}∞n=1 is uniformly bounded, then there exists u∗ such that un → u∗ (n → ∞) and

un = Aλnun −→ Aλ∗u∗ (n −→ ∞). (4.30)


Consequently,

u∗ = Aλ∗u∗. (4.31)

By Lemma 3.3, u∗ is a positive solution of problem (1.1), and λ∗ ∈ B.Next, we will prove that {un}∞n=1 is uniformly bounded and the discussions will be

divided into two cases.

Case 1 ((H1), (H2), and (H3) hold). From (H3), there exists H1 > 0,

f(u,w) ≥M1u, for u ≥ mH1

64[σ2(b) − a] , −w ≥ H1

16[σ2(b) − a]2, (4.32)

where M1 > 0 satisfies

M1[σ2(b) − a]2

mλ∗

128

∫ω

ξ

h(j)Δj > 1. (4.33)

Suppose on the contrary that {un}∞n=1 is unbounded. Then

‖un‖E −→ ∞, n −→ ∞, (4.34)

which implies that there exists N2 > 0, ‖un‖E ≥ H1 for n ≥ N2. From (3.30) and (3.28), wehave

mint∈[ξ,ω]

T

un(t) ≥ m

64[σ2(b) − a]‖un‖E ≥ mH1

64[σ2(b) − a] ,

mint∈[ξ,ω]

T

(−uΔ2

n (t))≥ 1

16[σ2(b) − a]2 ‖un‖E ≥ H1

16[σ2(b) − a]2.

(4.35)

Subsequently, for t ∈ [ξ, ω]T

, n ≥ max{N1,N2},

un(t) = λn

∫σ(b)

a

[∫ρ(b)

a

G(t, s)G(s, j

)h(j)f(un

(j), uΔ

2

n

(j))

Δj

]

Δs

≥ λnM1

∫σ(b)

a

[∫ω

ξ

G(t, s)G(s, j

)h(j)un

(j)Δj

]

Δs

≥ λnM1

4‖un‖∞

∫ω

ξ

[∫ω

ξ

G(t, s)G(s, j

)h(j)Δj

]

Δs

≥ M1[σ2(b) − a]2

mλ∗

128‖un‖∞

∫ω

ξ

h(j)Δj

> ‖un‖∞

(4.36)

by (4.32), (3.29), (3.12), (4.28), (3.2), and (4.33). This is a contradiction.


Case 2 ((H1), (H2), and (H4) hold). From (H4), there exists H2 > 0,

f(u,w) ≥M2(−w), for u ≥ mH2

64[σ2(b) − a] , −w ≥ H2

16[σ2(b) − a]2, (4.37)

where M2 > 0 satisfies

M2[σ2(b) − a]λ∗

32

∫ω

ξ

h(s)Δs > 1. (4.38)

Suppose on the contrary that {un}∞n=1 is unbounded. Then similar to Case 1, there existsN3 > 0such that ‖un‖E ≥ H2 for n ≥N3. Thus for t ∈ [ξ, ω]

T, n ≥ max{N1,N3}, we have

−uΔ2

n (t) = λn

∫ρ(b)

a

G(t, s)h(s)f(un(s), uΔ

2

n (s))Δs

≥ λnM2

∫ω

ξ

G(t, s)h(s)(−uΔ2

n (s))Δs

>M2

4λ∗

2

∥∥∥uΔ2

n

∥∥∥∞σ2(b) − a

4

∫ω

ξ

h(s)Δs

>∥∥∥uΔ

2

n

∥∥∥∞

(4.39)

by (4.37), (3.28), (3.12), (4.28) and (4.38). This is also a contradiction.

According to the definition of λ∗ and B, and Theorem 4.4, we complete the proof.

Acknowledgments

The authors would like to thank editor and the referees for carefully reading this paper andsuggesting many valuable comments. This paper is supported by China Postdoctoral ScienceFoundation Funded Projects (no. 201104602 and no. 20100481239) and General Project forScientific Research of Liaoning Educational Committee (no. L2011200).

References

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[3] L. Erbe, A. Peterson, and R. Mathsen, “Existence, multiplicity, and nonexistence of positive solutionsto a differential equation on a measure chain,” Journal of Computational and Applied Mathematics, vol.113, no. 1-2, pp. 365–380, 2000.

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Research ArticleThe Local Strong and Weak Solutions fora Generalized Pseudoparabolic Equation

Nan Li


Correspondence should be addressed to Nan Li, [email protected]

Received 7 February 2012; Accepted 21 March 2012


Copyright q 2012 Nan Li. This is an open access article distributed under the Creative CommonsAttribution License, which permits unrestricted use, distribution, and reproduction in anymedium, provided the original work is properly cited.

The Cauchy problem for a nonlinear generalized pseudoparabolic equation is investigated. Thewell-posedness of local strong solutions for the problem is established in the Sobolev spaceC([0, T);Hs(R))

⋂C1([0, T);Hs−1(R)) with s > 3/2, while the existence of local weak solutions

is proved in the space Hs(R) with 1 ≤ s ≤ 3/2. Further, under certain assumptions of the nonlinearterms in the equation, it is shown that there exists a unique global strong solution to the problemin the space C([0,∞);Hs(R))

⋂C1([0,∞);Hs−1(R)) with s ≥ 2.

1. Introduction

Davis [1] investigated the pseudoparabolic equation

ut(t, x) =∂

∂xϕ(ux) + αutxx, (1.1)

where the constant α ≥ 0, the function ϕ ∈ C2(−∞,∞), ϕ(0) = 0 and ϕ′(ξ) > 0, and thesubscripts x and t indicate partial derivatives. Equation (1.1) arises from the study ofshearing flows of incompressible simple fluids. The quantity ϕ(ux) + αutx is viewed as anapproximation to the stress functional during such a flow. Much attention has been given tothis approximation when the function ϕ is linear (see [2, 3]). The existence and uniquenessof the global weak solution of the initial value problem for (1.1) were established in [1].


Recently, Chen and Xue [4] investigated the Cauchy problem for the nonlinear gener-alized pseudoparabolic equation

ut − αutxx − λuxx + γux + f(u)x =∂

∂xϕ(ux) + g(u) − αg(u)xx, x ∈ R, t > 0, (1.2)

where u(t, x) is an unknown function, α > 0, λ ≥ 0, γ is a real number, f(s), ϕ(s), and g(s)denote given nonlinear functions. The well-posedness of global strong solution in a Sobolevspace, the global classical solution and its asymptotic behavior are studied in [4] in whichseveral key assumptions are imposed on the functions ϕ(s) and g(s). In fact, various dynamicproperties for many special cases of (1.2) have been established in [5–7]. For example, whenϕ(s) = g(s) = 0, (1.2) becomes the generalized regularized long wave Burger equation.

Motivated by the works in [1, 4], we study the problem

ut − αutxx =∂

∂xϕ(ux) + βu2muxx, x ∈ R, t > 0,

u(0, x) = u0(x), x ∈ R,(1.3)

where α > 0 and β ≥ 0, m is a nature number, ϕ(s) is a given function, and u0(x) is a giveninitial value function. Here we should address that (1.2) does not include the first equationof problem (1.3) due to the term βu2muxx. Letting β = 0, the first equation of problem (1.3)reduces to (1.1).

The objectives of this work are threefold. The first objective is to establish the localwell-posedness of system (1.3) in the space C([0, T);Hs(R))

⋂C1([0, T);Hs−1(R)) with s >

3/2. We should address that the Sobolev index s ≥ 2 is required to guarantee the local well-posedness of (1.1) and (1.2) in the works of Davis [1] and Chen and Xue [4]. The second aimis to study the existence of local weak solutions for system (1.3). The third aim is to discuss thewell-posedness of the global strong solution for problem (1.3). Under the assumptions of thefunction ϕ(s) and the initial value u0(x) similar to those presented in [1, 4], problem (1.3) isshown to have a unique global solution in the space C([0,∞);Hs(R))

⋂C1([0,∞);Hs−1(R)).

The organization of this paper is as follows. The well-posedness of local strong solu-tions for problem (1.3) is investigated in Section 2, and the existence of local weak solutions isestablished in Section 3. Section 4 deals with the well-posedness of the global strong solution.


Let Lp = Lp(R) (1 ≤ p < +∞) be the space of all measurable functions h such that‖h‖pLp =

∫R |h(t, x)|pdx < ∞. We define L∞ = L∞(R) with the standard norm ‖h‖L∞ =

infm(e)=0 supx∈R\e|h(t, x)|. For any real number s, Hs = Hs(R) denotes the Sobolev space withthe norm defined by

‖h‖Hs =(∫

R

(1 + |ξ|2)s|h(t, ξ)|2dξ)1/2

<∞, (2.1)


−ixξh(t, x)dx.


For T > 0 and nonnegative number s, C([0, T);Hs(R)) denotes the Frechet space of allcontinuous Hs-valued functions on [0, T). We set Λ = (1 − ∂2

x)1/2. For simplicity, throughout

this paper, we let c denote any positive constants.The local well-posedness theorem is stated as follows.

Theorem 2.1. Provided that s ≥ 3/2, u0 ∈ Hs(R), ϕ is a polynomial of order N with ϕ(0) = 0.Then problem (1.3) admits a unique local solution:

u(t, x) ∈ C([0, T);Hs(R))⋂C1

([0, T);Hs−1(R)

). (2.2)

Proof. In fact, the first equation of problem (1.3) is equivalent to the equation

ut = Λ−2(∂

∂xϕ(ux) + βu2muxx

), (2.3)

which leads to

u = u0 +∫ t

0Λ−2

(∂


)dτ. (2.4)

Suppose that both u and v are in the closed ball BM0(0) of radius M0 > 1 about the zerofunction in C([0, T];Hs(R)) and A is the operator in the right-hand side of (2.4), for fixedt ∈ [0, T], we get

∥∥∥∥∥

∫ t

0Λ−2

(ϕ(ux)x + βu

2muxx)dt −

∫ t

0Λ−2

(ϕ(vx)x + βv

2mvxx)dt

∥∥∥∥∥Hs

≤ T(

sup0≤t≤T

∥∥ϕ(ux) − ϕ(vx)∥∥Hs−1 + sup

0≤t≤T

∥∥∥u2muxx − v2mvxx∥∥∥Hs−2

)

.

(2.5)

The algebraic property of Hs0(R) with s0 > 1/2 (see [8–10]) and s > 3/2 derives that

∥∥∥ujx − vjx

∥∥∥Hs−1

=∥∥∥(ux − vx)

(uj−1x + uj−2

x vx + · · · + uxvj−2x + vj−1

x

)∥∥∥Hs−1

≤∥∥∥ux − vx‖Hs−1‖

(uj−1x + uj−2

x vx + · · · + uxvj−2x + vj−1

x

)∥∥∥Hs−1

≤ c‖ux − vx‖Hs−1

j−1∑

i=0‖ux‖j−1−i

Hs−1 ‖vx‖iHs−1 ≤ cMj−10 ‖u − v‖Hs,

‖ϕ(ux) − ϕ(vx)‖Hs−1 ≤ cN∑

j=0

‖(ux)j − (vx)j‖Hs−1 ≤ cMN−10 ‖u − v‖Hs.

(2.6)


Using u2muxx = ∂x[u2mux] − 2mu2m−1(ux)2 and v2mvxx = ∂x[v2mvx] − 2mv2m−1(vx)

2, we get

∥∥∥u2muxx − v2mvxx

∥∥∥Hs−2

≤ ‖∂x[u2mux − v2mvx

]‖Hs−2 + c‖u2m−1u2

x − v2m−1v2x‖Hs−2

≤ c∥∥∥(u2m − v2m

)vx + u2m(ux − vx)

∥∥∥Hs−1

+ c∥∥∥u2m−1

(u2x − v2

x

)+(u2m−1 − v2m−1

)v2x

∥∥∥Hs−1

≤ c(‖(u2m − v2m

)vx‖Hs−1 +

∥∥∥u2m(ux − vx)

∥∥∥Hs−1

+∥∥∥u2m−1

(u2x − v2

x

)∥∥∥Hs−1

+∥∥∥(u2m−1 − v2m−1

)v2x

∥∥∥Hs−1

)

≤ cM2m0 ‖u − v‖Hs,

(2.7)

in which s > 3/2 is used.From (2.5)–(2.7), we obtain

‖Au −Av‖Hs ≤ ‖u − v‖Hs, (2.8)

where θ = max(cTMN−10 , cTM2m

0 ) and c is independent of T . Choosing T sufficiently smallsuch that θ < 1, we know that A is a contractive mapping. Applying the above inequality and(2.4) yields

‖Au‖Hs ≤ ‖u0‖Hs + θ‖u‖Hs. (2.9)

Choosing T sufficiently small such that θM0 + ‖u0‖Hs < M0, we know that A maps BM0(0)to itself. It follows from the contractive mapping principle that the mapping A has a uniquefixed point u in BM0(0). This completes the proof of Theorem 2.1.

3. Existence of Local Weak Solutions

In this section, we assume that ϕ(η) = η2N+1 whereN is a nature number. In order to establishthe existence of local weak solution, we need the following lemmas.

Lemma 3.1 (see Kato and Ponce [8]). If r ≥ 0, thenHr⋂L∞ is an algebra. Moreover,

‖uv‖r ≤ c(‖u‖L∞‖v‖r + ‖u‖r‖v‖L∞), (3.1)


Lemma 3.2 (see Kato and Ponce [8]). Let r > 0. If u ∈ Hr⋂W1,∞ and v ∈ Hr−1 ⋂L∞, then

‖[Λr , u]v‖L2 ≤ c(‖∂xu‖L∞‖Λr−1v‖L2 + ‖Λru‖L2‖v‖L∞

). (3.2)


Lemma 3.3. Let s ≥ 3/2, ϕ(ux) = u2N+1x , and the function u(t, x) is a solution of problem (1.3) and

the initial data u0(x) ∈ Hs. Then the following results hold.For q ∈ (0, s − 1], there is a constant c such that

∫

R

(Λq+1u

)2dx ≤

∫

R

[(Λq+1u0

)2]dx

+ c∫ t

0‖u‖2

Hq+1

(‖ux‖2N

L∞ + ‖ux‖2L∞‖u‖2m−2

L∞

+‖ux‖L∞‖u‖2m−1L∞ + ‖u‖2m

L∞

)dτ.

(3.3)

For q ∈ [0, s − 1], there is a constant c such that


(‖u‖2m−1

L∞ + ‖ux‖2N−1L∞

). (3.4)

Proof. For q ∈ (0, s − 1], applying (Λqu)Λq to both sides of the first equation of system (1.3)and integrating with respect to x by parts, we have the identity

12d

dt

∫

R

[(Λqu)2 + α(Λqux)

2]dx =

∫

R

(Λqu)Λq(ϕ(ux)x)dx + β

∫

R

ΛquΛq[u2muxx

]dx. (3.5)

We will estimate the two terms on the right-hand side of (3.5), respectively. For the first term,by using the Cauchy-Schwartz inequality and Lemmas 3.1 and 3.2, we have

∣∣∣∣

∫

R

(Λqu)Λq(ϕ(ux)x)dx

∣∣∣∣ =∣∣∣∣

∫

R

(Λqux)Λq(ϕ(ux))dx

∣∣∣∣

≤ c‖Λqux‖L2

∥∥∥Λq(ux)2N+1∥∥∥L2

≤ c‖u‖2Hq+1‖ux‖2N

L∞ .

(3.6)

For the second term, we have

∫

R

ΛquΛq[u2muxx

]dx =

∫

R

ΛquΛq[(u2mux

)

x− 2mu2m−1u2

x

]dx

=∫

R

ΛquxΛq(u2mux

)dx − 2m

∫

R

ΛquΛq[u2m−1u2

x

]dx = K1 +K2.

(3.7)

For K1, applying Lemma 3.1 derives

|K1| ≤ c‖u‖2Hq+1

(‖u‖2m

L∞ + ‖ux‖L∞‖u‖2m−1L∞

). (3.8)


For K2, we get

|K2| ≤ c‖u‖Hq‖u2m−1u2x‖Hq

≤ c‖u‖Hq

(‖u2m−1ux‖L∞‖ux‖Hq + ‖u2m−1ux‖Hq‖ux‖L∞

)

≤ c‖u‖2Hq+1

(‖ux‖L∞‖u‖2m−1

L∞ + ‖ux‖2L∞‖u‖2m−2

L∞

).

(3.9)

It follows from (3.5)–(3.9) that there exists a constant c such that

12d

dt

∫

R

[(Λqu)2 + (Λqux)

2]dx

≤ c‖u‖2Hq+1

(‖ux‖2N

L∞ + ‖ux‖2L∞‖u‖2m−2

L∞ + ‖ux‖L∞‖u‖2m−1L∞ + ‖u‖2m

L∞

).

(3.10)

Integrating both sides of the above inequality with respect to t results in inequality (3.3).To estimate the norm of ut, we apply the operator (1 − ∂2

x)−1 to both sides of the first

equation of system (1.3) to obtain the equation

ut = Λ−2(∂


). (3.11)

Applying (Λqut)Λq to both sides of (3.11) for q ∈ [0, s − 1] gives rise to

∫

R

(Λqut)2dx =

∫

R

(Λqut)Λq−2[∂xϕ(ux) + u2muxx

]dτ. (3.12)

For the right-hand of (3.12), we have

∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq∂xϕ(ux)dx

∣∣∣∣

≤ c‖ut‖Hq

(∫

R

(1 + ξ2

)q−1[∫

R

[u2Nx

(ξ − η)ux

(η)]dη

]2)1/2

≤ c‖ut‖Hq‖u‖H1‖u‖Hq+1‖ux‖2N−1L∞ ,

∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq∂x

(u2mux

)dx

∣∣∣∣

≤ c‖ut‖Hq

(∫

R

(1 + ξ2

)q−1[∫

R

[u2m

(ξ − η)ux

(η)]dη

]2)1/2

≤ c‖ut‖Hq‖u‖H1‖u‖Hq+1‖u‖2m−1L∞ ,

∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq

(u2m−1u2

x

)dx

∣∣∣∣

≤ c‖ut‖Hq

(∫

R

(1 + ξ2

)q−1[∫

R

[u2m−1ux

(ξ − η)ux

(η)]dη

]2)1/2

≤ c‖ut‖Hq‖u‖H1‖u‖Hq+1‖u‖2m−1L∞ .

(3.13)


Applying (3.13) into (3.12) yields the inequality

‖ut‖Hq ≤ c‖u‖H1‖u‖Hq+1

(‖u‖2m−1

L∞ + ‖ux‖2N−1L∞

)(3.14)

for a constant c > 0. This completes the proof of Lemma 3.3.

Lemma 3.4. If u(t, x) is a solution of problem (1.3), α > 0, ϕ(η) = η2N+1, then

‖u‖L∞ ≤ c‖u‖H1(R) ≤ c‖u0‖H1(R), (3.15)

where c is a constant.

Proof. Multiplying both sides of the first equation of (1.3) by u(t, x) and integrating withrespect to x over R, we have

12d

dt

∫

R

[u(t, x)2 + αux(t, x)2

]dx =

∫

R

ϕ(ux)xu(t, x)dx + β∫

R

u2m+1uxxdx. (3.16)

Since

∫

R

ϕ(ux)xu(t, x)dx + β∫

R

u2m+1uxxdx = −∫

R

u2N+2x dx − β(2m + 1)

∫

R

u2mu2xdx < 0, (3.17)

we derive that

12d

dt

∫

R

[u(t, x)2 + αux(t, x)2

]dx < 0, (3.18)

which results in

∫

R

[u(t, x)2 + αux(t, x)2

]<

∫

R

[u(0, x)2 + αux(0, x)2

]≤ c‖u0‖2

H1 . (3.19)

From (3.19), we know that (3.15) holds. This completes the proof.

Defining

φ(x) =

{e1/(x2−1), |x| < 1,0, |x| ≥ 1

(3.20)

and setting φε(x) = ε−1/4φ(ε−1/4x) with 0 < ε < 1/4 and uε0 = φε � u0, we know that uε0 ∈ C∞

for any u0 ∈ Hs(R) and s > 0.


It follows from Theorem 2.1 that for each ε the Cauchy problem

ut − utxx =∂

∂xϕ(ux) + βu2muxx,

u(0, x) = uε0(x), x ∈ R(3.21)

has a unique solution uε(t, x) ∈ C∞([0, T);H∞).

Lemma 3.5. Under the assumptions of problem (3.21), the following estimates hold for any ε with0 < ε < 1/4, u0 ∈ Hs(R) and s > 0 :

‖uε0x‖L∞ ≤ c1‖u0x‖L∞ ,

‖uε0‖Hq ≤ c1, if q ≤ s,(3.22)

where c1 is a constant independent of ε.

Proof. Using the definition of uε0 and uε0x results in the conclusion of the lemma.

Lemma 3.6. Suppose that u0(x) ∈ Hs(R) with s ∈ [1, 3/2] such that ‖u0x‖L∞ < ∞. Let uε0 bedefined as in system (3.21) and let ϕ(η) = η2N+1. Then there exist two positive constants T and c,independent of ε, such that the solution uε of problem (3.21) satisfies ‖uεx‖L∞ ≤ c for any t ∈ [0, T).

Proof. Using notation u = uε and differentiating both sides of the first equation of problem(3.11) with respect to x give rise to

utx = −ϕ(ux) − βu2mux + Λ−2[ϕ(ux) + βu2mux − 2mβ

(u2m−1ux

)

x

]. (3.23)

Letting p > 0 be an integer and multiplying the above equation by (ux)2p+1 and then

integrating the resulting equation with respect to x yield the equality

12p + 2

d

dt

∫

R

(ux)2p+2dx = −∫

R

ϕ(ux)u2p+1x dx − β

∫

R

u2mu2p+2x dx +

∫

R

Ju2p+1x dx, (3.24)

where

J = Λ−2[ϕ(ux) + βu2mux − 2mβ

(u2m−1ux

)

x

]. (3.25)

Applying the Holder’s inequality to (3.24) and noting Lemmas 3.4 and 3.5, we obtain

12p + 2

d

dt

∫

R

(ux)2p+2dx ≤ c‖ux‖2NL∞

∫

R

|ux|2p+2dx + c∫

R

u2p+2x dx

+(∫

R

|J |2p+2dx

)1/(2p+2)(u

2p+2x dx

)2(p+1)/2(p+2)(3.26)


or

d

dt

(∫

R

(ux)2(p+2)dx

)1/(2p+2)

≤ c‖ux‖2NL∞

(∫

R

u2p+2x dx

)1/(2p+2)

+ c(∫

R

u2p+2x dx

)1/(2p+2)

+(∫

R

|J |2p+2dx

)1/(2p+2)

.

(3.27)

Since ‖f‖Lp → ‖f‖L∞ as p → ∞ for any f ∈ L∞ ⋂L2, integrating both sides of the

inequality (3.27) with respect to t and taking the limit as p → ∞ result in the estimate

‖ux‖L∞ ≤ ‖u0x‖L∞ +∫ t

0c(‖ux‖L∞ + ‖ux‖2N+1

L∞ + ‖J‖L∞)dτ. (3.28)

Using the algebra property of Hs0(R) with s0 > 1/2 yields (‖uε‖H1/2+ means that there existsa sufficiently small δ > 0 such that ‖uε‖1/2+ = ‖uε‖H1/2+δ)

‖J‖L∞ ≤ c‖J‖H1/2+ ≤ c‖Λ−2[ϕ(ux) + βu2mux − 2mβ

(u2m−1ux

)

x

]‖H1/2+

≤ c(‖ϕ(ux)‖H0 + ‖u‖H1 + ‖u2m−1ux‖H0

)

≤ c(‖ux‖2N‖u‖H1 + ‖u‖H1 + ‖u‖2m−1

L∞ ‖u‖H1

)

≤ c(‖ux‖2N + 1

),

(3.29)

in which Lemmas 3.4 and 3.5 are used. From (3.28) and (3.29), one has

‖ux‖L∞ ≤ ‖u0x‖L∞ + c∫ t

0

[1 + ‖ux‖L∞ + ‖ux‖2N

L∞ + ‖ux‖2N+1L∞

]dτ. (3.30)

From Lemma 3.5, it follows from the contraction mapping principle that there is aT > 0 such that the equation

‖W‖L∞ = ‖u0x‖L∞ + c∫ t

0

[1 + ‖W‖L∞ + ‖W‖2N

L∞ + ‖W‖2N+1L∞

]dτ (3.31)

has a unique solution W ∈ C[0, T]. Using the result presented on page 51 in [11] yields thatthere are constants T > 0 and c > 0 independent of ε such that ‖ux‖L∞ ≤ ‖W(t)‖L∞ ≤ c forarbitrary t ∈ [0, T], which leads to the conclusion of Lemma 3.6.

Using Lemmas 3.3–3.6, notation uε = u and Gronwall’s inequality result in theinequalities

‖uε‖Hq ≤ CTeCT ,

‖uεt‖Hr ≤ CTeCT ,

(3.32)


where q ∈ (0, s], r ∈ (0, s − 1] (1 ≤ s ≤ 3/2) and CT depends on T . It follows fromthe Aubin’s compactness theorem that there is a subsequence of {uε}, denoted by {uεn},such that {uεn} and their temporal derivatives {uεnt} are weakly convergent to a functionu(t, x) and its derivative ut in L2([0, T],Hs) and L2([0, T],Hs−1), respectively. Moreover,for any real number R1 > 0, {uεn} is convergent to the function u strongly in the spaceL2([0, T],Hq(−R1, R1)) and {uεnt} converges to ut strongly in the space L2([0, T],Hr(−R1, R1))for r ∈ [0, s − 1]. Thus, we can prove the existence of a weak solution to (1.3).

Theorem 3.7. Suppose that u0(x) ∈ Hs with 1 ≤ s ≤ 3/2, ‖u0x‖L∞ < ∞ and ϕ(η) = η2N+1.Then there exists a T > 0 such that (1.3) subject to initial value u0(x) has a weak solution u(t, x) ∈L2([0, T],Hs) in the sense of distribution and ux ∈ L∞([0, T] × R).

Proof. From Lemma 3.6, we know that {uεnx} (εn → 0) is bounded in the space L∞. Thus, thesequences {uεn}, {uεnx}, {u2

εnx}, and {u2N+1εnx } are weakly convergent to u, ux, u2

x, and u2N+1x in

L2[0, T],Hr(−R,R) for any r ∈ [0, s − 1), separately. Therefore, u satisfies the equation

∫T

0

∫

R

u(gt − gxxt

)dx dt =

∫T

0

∫

R

[u2N+1x gx + βu2muxgx − 2mβu2m−1u2

xg]dx dt, (3.33)

with u(0, x) = u0(x) and g ∈ C∞0 . Since X = L1([0, T] × R) is a separable Banach space

and {uεnx} is a bounded sequence in the dual space X∗ = L∞([0, T] × R) of X, there existsa subsequence of {uεnx}, still denoted by {uεnx}, weakly star convergent to a function v inL∞([0, T] × R). It derives from the weakly convergence of {uεnx} to ux in L2([0, T] × R) thatux = v almost everywhere. Thus, we obtain ux ∈ L∞([0, T] × R).

4. Well-Posedness of Global Solutions

Lemma 4.1. If u(t, x) is a solution of problem (1.3), α > 0, ϕ(η) = η2N+1, then

‖ux‖L∞ ≤ A1/2, (4.1)

where

A =∫

R

[1 + αα

(u′0(x)

)2 + (1 + α)(u′′0(x)

)2]dx. (4.2)

Proof. Multiplying each side of the first equation of problem (1.3) by uxx and integrating over[0, t] × R yields

∫ t

0

∫

R

(u2xxϕ

′(ux) + u2mu2xx +

α

2∂

∂t

(u2xx

))dx dt =

∫ t

0

∫

R

utuxxdx dt. (4.3)

Integrating the right-hand side of the above identity by parts and using ux(±∞) = 0, we get

2∫ t

0

∫

R

utuxxdxdt =∫

R

[u′0(x)

]2dx −

∫

R

u2x(t, x)dx. (4.4)


From (4.3), (4.4) and the assumption of this lemma, we have

α‖uxx‖2L2 + ‖ux‖L2 ≤

∫

R

[(u′0(x)

)2 + α(u′′0(x)

)2]dx, (4.5)

from which we obtain (4.1).

Theorem 4.2. Suppose that s ≥ 2, u0 ∈ Hs(R), ϕ(ux) = u2N+1x with positive integer N. Then

problem (1.3) has a unique global solution:

u(t, x) ∈ C([0,∞);Hs(R))⋂C1

([0,∞);Hs−1(R)

). (4.6)

Proof. Using the Gronwall inequality and Lemma 3.3 and choosing s = q + 1, we have

‖u‖Hs ≤ c‖u0‖Hse∫ t

0(‖ux‖2NL∞+‖ux‖2

L∞‖u‖2m−2L∞ +‖ux‖L∞‖u‖2m−1

L∞ +‖u‖2mL∞ )dτ . (4.7)

From Lemma 4.1, we have

‖ux‖ ≤ A1/2 =(∫

R

[1 + αα

(u′0(x)

)2 + (1 + α)(u′′0(x)

)2]dx

)1/2

≤ c‖u0‖H2(R). (4.8)

Using (4.7) and (4.8) derives

‖u‖Hs ≤ c‖u0‖Hsect, (4.9)

which completes the proof of Theorem 4.2.

References

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[2] B. D. Coleman and W. Noll, “An approximation theorem for functionals, with applications incontinuum mechanics,” Archive for Rational Mechanics and Analysis, vol. 6, no. 1, pp. 355–370, 1960.

[3] R. E. Showalter and T. W. Ting, “Pseudoparabolic partial differential equations,” SIAM Journal onMathematical Analysis, vol. 1, pp. 1–26, 1970.

[4] G. W. Chen and H. X. Xue, “Global existence of solution of Cauchy problem for nonlinear pseudo-parabolic equation,” Journal of Differential Equations, vol. 245, no. 10, pp. 2705–2722, 2008.

[5] P. G. Kevrekidis, I. G. Kevrekidis, A. R. Bishop, and E. S. Titi, “Continuum approach to discreteness,”Physical Review E, vol. 65, no. 4, Article ID 046613, 13 pages, 2002.

[6] J. L. Bona and L. Luo, “Asymptotic decomposition of nonlinear, dispersive wave equations withdissipation,” Physica D, vol. 152–153, pp. 363–383, 2001.

[7] J. Albert, “Dispersion of low-energy waves for the generalized Benjamin-Bona-Mahony equation,”Journal of Differential Equations, vol. 63, no. 1, pp. 117–134, 1986.

[8] T. Kato and G. Ponce, “Commutator estimates and the Euler and Navier-Stokes equations,” Commu-nications on Pure and Applied Mathematics, vol. 41, no. 7, pp. 891–907, 1988.

[9] S. Y. Lai and Y. H. Wu, “The local well-posedness and existence of weak solutions for a generalizedCamassa-Holm equation,” Journal of Differential Equations, vol. 248, no. 8, pp. 2038–2063, 2010.


[10] S. Y. Lai and Y. H. Wu, “Existence of weak solutions in lower order Sobolev space for a Camassa-Holm-type equation,” Journal of Physics A, vol. 43, no. 9, Article ID 095205, 13 pages, 2010.

[11] Y. A. Li and P. Olver, “Well-posedness and blow-up solutions for an integrable nonlinearly dispersivemodel wave equation,” Journal of Differential Equations, vol. 162, no. 1, pp. 27–63, 2000.


Research ArticleThe Cauchy Problem to a Shallow Water WaveEquation with a Weakly Dissipative Term

Ying Wang and YunXi Guo

College of Science, Sichuan University of Science and Engineering, Zigong 643000, China

Correspondence should be addressed to Ying Wang, [email protected]

Received 21 February 2012; Accepted 11 March 2012


Copyright q 2012 Y. Wang and Y. Guo. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

A shallow water wave equation with a weakly dissipative term, which includes the weaklydissipative Camassa-Holm and the weakly dissipative Degasperis-Procesi equations as specialcases, is investigated. The sufficient conditions about the existence of the global strong solutionare given. Provided that (1 − ∂2

x)u0 ∈ M+(R), u0 ∈ H1(R), and u0 ∈ L1(R), the existence anduniqueness of the global weak solution to the equation are shown to be true.

1. Introduction

The Camassa-Holm equation (C-H equation)

ut − utxx + 3uux − 2uxuxx − uuxxx = 0, t > 0, x ∈ R, (1.1)

as a model for wave motion on shallow water, has a bi-Hamiltonian structure and iscompletely integrable. After the equation was derived by Camassa and Holm [1], a lot ofworks was devoted to its investigation of dynamical properties. The local well posedness ofsolution for initial data u0 ∈ Hs(R) with s > 3/2 was given by several authors [2–4]. Undercertain assumptions, (1.1) has not only global strong solutions and blow-up solutions [2, 5–7] but also global weak solutions in H1(R) (see [8–10]). For other methods to handle theproblems related to the Camassa-Holm equation and functional spaces, the reader is referredto [11–14] and the references therein.

To study the effect of the weakly dissipative term on the C-H equation, Guo [15] andWu and Yin [16] discussed the weakly dissipative C-H equation

ut − utxx + 3uux − 2uxuxx − uuxxx + λ(u − uxx) = 0, t > 0, x ∈ R. (1.2)


The global existence of strong solutions and blow-up in finite time were presented in[16] provided that y0 = (1 − ∂2

x)u0 changes sign. The sufficient conditions on the infinitepropagation speed for (1.2) are offered in [15]. It is found that the local well posednessand the blow-up phenomena of (1.2) are similar to the C-H equation in a finite interval oftime. However, there are differences between (1.2) and the C-H equation in their long timebehaviors. For example, the global strong solutions of (1.2) decay to zero as time tends toinfinite under suitable assumptions, which implies that (1.2) has no traveling wave solutions(see [16]).

Degasperis and Procesi [17] derived the equation (D-P equation)

ut − utxx + 4uux − 3uxuxx − uuxxx = 0, t > 0, x ∈ R, (1.3)

as a model for shallow water dynamics. Although the D-P equation (1.3) has a similar form tothe C-H equation (1.1), it should be addressed that they are truly different (see [18]). In fact,many researchers have paid their attention to the study of solutions to (1.3). Constantin etal. [19] developed an inverse scattering approach for smooth localized solutions to (1.3). Liuand Yin [20] and Yin [21, 22] investigated the global existence of strong solutions and globalweak solutions to (1.3). Henry [23] showed that the smooth solutions to (1.3) have infinitespeed of propagation. Coclite and Karlsen [24] obtained global existence results for entropysolutions in L1(R) ∩ BV (R) and in L2(R) ∩ L4(R).

The weakly dissipative D-P equation

ut − utxx + 4uux − 3uxuxx − uuxxx + λ(u − uxx) = 0, t > 0, x ∈ R (1.4)

is investigated by several authors [25–27] to find out the effect of the weakly dissipative termon the D-P equation. The global existence, persistence properties, unique continuation andthe infinite propagation speed of the strong solutions to (1.4) are studied in [26]. The blow-up solution modeling wave breaking and the decay of solution were discussed in [27]. Theexistence and uniqueness of the global weak solution in spaceW1,∞

loc (R+×R)⋂L∞

loc(R+;H1(R))were proved (see [25]).

In this paper, we will consider the Cauchy problem for the weakly dissipative shallowwater wave equation

ut − utxx + (a + b)uux − auxuxx − buuxxx + λ(u − uxx) = 0, t > 0, x ∈ R,u(0, x) = u0(x), x ∈ R,

(1.5)

where a > 0, b > 0, and λ ≥ 0 are arbitrary constants, u is the fluid velocity in the x direction,λ(u − uxx) represents the weakly dissipative term. For λ = 0, we notice that (1.5) is a specialcase of the shallow water equation derived by Constantin and Lannes [28].

Since (1.5) is a generalization of the Camassa-Holm equation and the Degasperis-Procesi equation, (1.5) loses some important conservation laws that C-H equation and D-P equation possesses. It needs to be pointed out that Lai and Wu [12] studied globalexistence and blow-up criteria for (1.5) with λ = 0. To the best of our knowledge, thedynamical behaviors related to (1.5) with λ = 0, such as the global weak solution in spaceW1,∞

loc (R+ × R)⋂L∞loc(R+;H1(R)), have not been yet investigated. The objective of this paper

is to investigate several dynamical properties of solutions to (1.5). More precisely, we firstly


use the Kato theorem [29] to establish the local well-posedness for (1.5) with initial valueu0 ∈ Hs with s > 3/2. Then, we present a precise blow-up scenario for (1.5). Provided thatu0 ∈ Hs(R)

⋂L1(R) and the potential y0 = (1−∂2

x)u0 does not change sign, the global existenceof the strong solution is shown to be true. Finally, under suitable assumptions, the existenceand uniqueness of global weak solution in W1,∞(R+ × R)⋂L∞

loc(R+;H1(R)) are proved. Ourmain ideas to prove the existence and uniqueness of the global weak solution come fromthose presented in Constantin and Molinet [8] and Yin [22].

2. Notations

The space of all infinitely differentiable functions φ(t, x) with compact support in [0,+∞)×Ris denoted by C∞

0 . Let 1 ≤ p < +∞, and let Lp = Lp(R) be the space of all measurable functionsh(t, x) such that ‖h‖PLP =

∫R |h(t, x)|pdx < ∞. We define L∞ = L∞(R) with the standard norm

‖h‖L∞ = infm(e)=0supx∈R\e|h(t, x)|. For any real number s, let Hs = Hs(R) denote the Sobolev

space with the norm defined by ‖h‖Hs = (∫R (1 + |ξ|2)s|h(t, ξ)|2dξ)1/2 < ∞, where h(t, ξ) =∫

R e−ixξh(t, x)dx.

We denote by ∗ the convolution. Let ‖ · ‖X denote the norm of Banach space X and〈·, ·〉 the H1(R), H−1(R) duality bracket. Let M(R) be the space of the Radon measures on Rwith bounded total variation and M+(R) the subset of positive measures. Finally, we writeBV (R) for the space of functions with bounded variation, V (f) being the total variation off ∈ BV (R).

Note that if G(x) := (1/2)e−|x|, x ∈ R. Then, (1 − ∂2x)

−1f = G ∗ f for all f ∈ L2(R) and

G ∗ (u − uxx) = u. Using this identity, we rewrite problem (1.5) in the form

ut + buux + ∂xG ∗[a

2u2 +

3b − a2

(ux)2]+ λu = 0, t > 0, x ∈ R,

u(0, x) = u0(x), x ∈ R,(2.1)


yt + buyx + ayux + λy = 0, t > 0, x ∈ R,y = u − uxx,

u(0, x) = u0(x).

(2.2)

3. Preliminaries

Throughout this paper, let {ρn}n≥1 denote the mollifiers

ρn(x) :=(∫

R

ρ(ξ)dξ)−1

nρ(nx), x ∈ R, n ≥ 1, (3.1)


where ρ ∈ C∞c (R) is defined by

ρ(x) :=

⎧⎨

⎩

e1/(x2−1) for |x| < 1,

0 for |x| ≥ 1.(3.2)

Thus, we get∫

R

ρn(x)dx = 1, ρn ≥ 0, x ∈ R, n ≥ 1. (3.3)

Next, we give some useful results.

Lemma 3.1 (see [8]). Let f : R → R be uniformly continuous and bounded. If μ ∈M(R), then[ρn ∗

(fμ) − (ρn ∗ f

)(ρn ∗ μ

)]�n −→ ∞ 0 in L1(R). (3.4)

Lemma 3.2 (see [8]). Let f : R → R be uniformly continuous and bounded. If g ∈ L∞(R), then

[ρn ∗

(fg

) − (ρn ∗ f)(ρn ∗ g

)]�n −→ ∞ 0 in L∞(R). (3.5)

Lemma 3.3 (see [30]). Let T > 0. If f, g ∈ L2((0, T);H1(R)) and df/dt, dg/dt ∈L2((0, T);H−1(R)), then f, g are a.e. equal to a function continuous from [0, T] into L2(R) and

⟨f(t), g(t)

⟩ − ⟨f(s), g(s)⟩ =∫ t

s

⟨d(f(τ)

)

dτ, g(τ)

⟩

dτ +∫ t

s

⟨d(g(τ)

)

dτ, f(τ)

⟩

dτ (3.6)

for all s, t ∈ [0, T].

Lemma 3.4 (see [8]). Assume that u(t, ·) ∈W1,1(R) is uniformly bounded inW1,1(R) for all t ∈ R+.Then, for a.e. t ∈ R+

d

dt

∫

R

∣∣ρn ∗ u

∣∣dx =∫

R

(ρn ∗ ut

)sgn

(ρn ∗ u

)dx,

d

dt

∫

R

∣∣ρn ∗ ux∣∣dx =

∫

R

(ρn ∗ uxt

)sgn

(ρn ∗ ux

)dx.

(3.7)

4. Global Strong Solution

We firstly present the existence and uniqueness of the local solutions to the problem (2.1).

Theorem 4.1. Let u0 ∈ Hs(R), s > 3/2. Then, the problem (2.1) has a unique solution u, such that

u = u(t, x) ∈ C([0, T);Hs(R)) ∩ C1([0, T);Hs−1(R)

), (4.1)

where T > 0 depends on ‖u0‖Hs(R).


Proof. The proof of Theorem 4.1 can be finished by using Kato’s semigroup theory (see [29]or [4]). Here, we omit the detailed proof.

Theorem 4.2. Given u0 ∈ Hs, s > 3/2, the solution u = (·, u0) of problem (2.1) blows up in finitetime T < +∞ if and only if

limt→ T

inf{

infx∈R

[ux(t, x)]}

= −∞. (4.2)

Proof. Setting y(t, x) = u(t, x) − uxx(t, x), we get

∥∥y

∥∥2L2 =

∫

R

(u − uxx)2dx =∫

R

(u2 + 2u2

x + u2xx

)dx, (4.3)

which yields

‖u‖2H2 ≤

∥∥y∥∥2L2 ≤ 2‖u‖2

H2 . (4.4)

Using system (2.2), one has

d

dt

∫

R

y2(t, x)dx = 2∫

R

yytdx

= −2b∫

R

uyyxdx − 2a∫

R

uxy2dx − 2λ

∫

R

y2dx

= −(2a − b)∫

R

uxy2dx − 2λ

∫

R

y2dx.

(4.5)

Assume that there is a constant M > 0 such that

−ux(t, x) ≤M on [0, T) × R. (4.6)

From (4.5), we get

d

dt

∫

R

y2(t, x)dx ≤ |2a − b|M∫

R

y2dx − 2λ∫

R

y2dx. (4.7)

Using Gronwall’ inequality, we deduce the ‖u‖H2 is bounded on [0, T). On the other hand,

u(t, x) =(

1 − ∂2x

)−1y(t, x) =

∫

R

G(x − s)y(s)ds. (4.8)

Therefore, using (4.4) leads to

‖ux‖L∞ ≤∣∣∣∣

∫

R

Gx(x − s)y(s)ds∣∣∣∣ ≤ ‖Gx‖L2

∥∥y∥∥L2 =

12∥∥y

∥∥L2 ≤ ‖u‖H2 . (4.9)

It shows that if ‖u‖H2 is bounded, then ‖ux‖L∞ is also bounded. This completes the proof.


We consider the differential equation

qt = bu(t, q

), t ∈ [0, T), x ∈ R,

q(0, x) = x, x ∈ R,(4.10)

where u solves (1.5) and T > 0.

Lemma 4.3. Let u0 ∈ Hs(R) (s > 3); T is the maximal existence time of the corresponding solutionu to (1.5). Then, system (4.10) has a unique solution q ∈ C1([0, T)×R;R). Moreover, the map q(t, ·)is an increasing diffeomorphism of R with

qx(t, x) = exp

(∫ t

0bux

(s, q(s, x)

)ds

)

> 0, ∀(t, x) ∈ [0, T) × R,

y(t, q(t, x)

)q2x(t, x) = y0(x) exp

(∫ t

0

[−(a − 2b)ux(s, q(s, x)

) − λ]ds)

.

(4.11)

Proof. From Theorem 4.1, we have u ∈ C1([0, T);Hs−1(R)) and Hs−1 ∈ C1(R). We concludethat both functions u(t, x) and ux(t, x) are bounded, Lipschitz in space, and C1 in time.Applying the existence and uniqueness theorem of ordinary differential equations impliesthat system (4.10) has a unique solution q ∈ C1([0, T) × R,R).

Differentiating (4.10) with respect to x leads to

d

dtqx = bux

(t, q

)qx, t ∈ [0, T), b > 0,

qx(0, x) = 1, x ∈ R,(4.12)

which yields

qx = exp

(∫ t

0bux

(s, q(s, x)

)ds

)

. (4.13)

For every T ′ < T , using the Sobolev embedding theorem gives rise to

sup(s,x)∈[0,T ′)×R

|ux(s, x)| <∞. (4.14)

It is inferred that there exists a constant K0 > 0 such that qx ≥ e−K0t for (t, x) ∈ [0, T) × R.By computing directly, we derive

d

dt

[y(t, q(t, x)

)q2x(t, x)

]= −(a − 2b)ux(t, x)yq2

x − λyq2x, (4.15)


which results in

y(t, q(t, x)

)q2x(t, x) = y0(x) exp

(∫ t

0

[−(a − 2b)ux(s, q(s, x)

) − λ]ds)

. (4.16)

The proof of Lemma 4.3 is completed.

Theorem 4.4. Let u0 ∈ L1(R) ∩Hs(R), s > 3/2, and (1 − ∂2x)u0 ≥ 0 for all x ∈ R (or equivalently

(1 − ∂2x)u0 ≤ 0 for all x ∈ R). Then, problem (2.1) has a global strong solution

u = u(·, u0) ∈ C([0,∞);Hs(R)) ∩ C1([0,∞);Hs−1(R)

). (4.17)

Moreover, if y(t, ·) = u − uxx, then one has for all t ∈ R+,

(i) y(t, ·) ≥ 0, u(t, ·) ≥ 0, and |ux(t, ·)| ≤ u(t, ·) on R,

(ii) e−λt‖y0‖L1(R) = ‖y(t, ·)‖L1(R) = ‖u(t, ·)‖L1(R) = e−λt‖u0‖L1(R) and ‖ux(t, ·)‖L∞(R) ≤e−λt‖u0‖L1(R),

(iii) ‖u‖2H1 ≤ ‖u0‖2

H1 exp[(|a − 2b|/λ)(1 − e−λt)‖u0‖L1 − 2λt].

Proof. Let u0 ∈ Hs, s > 3/2, and let T > 0 be the maximal existence time of the solution uwith initial date u0 (cf. Theorem 4.1). If y0 ≥ 0, then Theorem 4.2 ensures that y ≥ 0 for allt ∈ [0, T).

Note that u0 = G ∗ y0 and y0 = (u0 − u0,xx) ∈ L1(R). By Young’s inequality, we get

‖u0‖L1(R) =∥∥G ∗ y0(t, ·)

∥∥L1(R) ≤ ‖G‖L1(R) ≤ ‖G‖L1(R)

∥∥y0(t, ·)∥∥L1(R) ≤

∥∥y0∥∥L1(R). (4.18)

Integrating the first equation of problem (2.1) by parts, we get

d

dt

∫

R

udx = −b∫

R

uuxdx −∫

R

∂x

(G ∗

[a

2u2 +

3b − a2

(ux)2])

dx + λ∫

R

udx = 0. (4.19)

It follows that

∫

R

udx = e−λt∫

R

u0dx. (4.20)

Since y = u − uxx, we have

∫

R

y dx =∫

R

udx −∫

R

uxxdx =∫

R

udx

= e−λt∫

R

u0dx = e−λt(∫

R

u0dx −∫

R

u0,xxdx

)= e−λt

∫

R

y0dx.

(4.21)


Given t ∈ [0, T), due to u(t, x) ∈ Hs, s ≥ 3/2, from Theorem 4.1 and (4.21), we obtain

−ux(t, x) +∫x

−∞udx =

∫x

−∞u − uxxdx =

∫x

−∞y dx

≤∫∞

−∞ydx = e−λt

∫

R

y0dx = e−λt∫

R

u0dx.

(4.22)

Note that u = G ∗ y, y ≥ 0 on [0, T) and the positivity of G. Thus, we can infer that u ≥ 0 on[0, T). From (4.22) we have

ux(t, x) ≥ −e−λt∫

R

u0dx, ∀(t, x) ∈ [0, T) × R. (4.23)

From Theorem 4.2 and (4.23), we find T = ∞. This implies that problem (2.1) has a uniquesolution

u = u(·, u0) ∈ C([0,∞);Hs(R))⋂C1([0,∞);Hs−1(R)

), s ≥ 3

2. (4.24)

Due to y(t, x) ≥ 0 and u(t, x) ≥ 0 for all t ≥ 0, it shows that

u(t, x) =e−x

2

∫x

−∞eξy(ξ)dξ +

ex

2

∫∞

x

e−ξy(ξ)dξ,

ux(t, x) = −e−x

2

∫x

−∞eξy(ξ)dξ +

ex

2

∫∞

x

e−ξy(ξ)dξ.

(4.25)

From the two identities above, we infer that (ux)2 ≤ u2 on R for all t ≥ 0. This proves (i).

Due to y ≥ 0, we obtain

ux(t, x) −∫x

−∞udx = −

∫x

−∞(u − uxx)dx = −

∫x

−∞ydx ≤ 0. (4.26)

From u ≥ 0 and the inequality above, we get

ux(t, x) ≤∫x

−∞udx ≤

∫∞

−∞udx = e−λt

∫∞

−∞u0dx = e−λt‖u0‖L1(R). (4.27)

On the other hand, from (4.23), we have that ux(t, x) ≥ −e−λt‖u0‖L1(R). This proves (ii).


Multiplying the first equation of problem (2.1) by u and integrating by parts, we find

12d

dt

∫

R

(u2 + u2

x

)dx = −(a + b)

∫

R

u2uxdx + a∫

R

uuxuxxdx + b∫

R

u2uxxxdx

− λ∫

R

(u2 + u2

x

)dx

= −a − 2b2

∫

R

u3xdx − λ

∫

R

(u2 + u2

x

)dx,

(4.28)

which yields

∫

R

(u2 + u2

x

)dx = −(a − 2b)

∫

R

u3xdx − 2λ

∫

R

(u2 + u2

x

)dx

≤ |a − 2b|∫

R

|ux|3dx − 2λ∫

R

(u2 + u2

x

)dx

≤ |a − 2b|‖ux‖L∞‖ux‖2L2 − 2λ

∫

R

(u2 + u2

x

)dx

≤ |a − 2b|‖ux‖L∞‖u‖2H1 − 2λ

∫

R

(u2 + u2

x

)dx.

(4.29)

From Gronwall’s inequality, one has

‖u‖2H1 ≤ ‖u0‖2

H1 exp[ |a − 2b|

λ

(1 − e−λt

)‖u0‖L1 − 2λt

]. (4.30)

This proves (iii) and completes the proof of the theorem.

5. Global Weak Solution

Theorem 5.1. Let u0 ∈ H1(R)⋂L1(R) and y0 = (u0 − u0xx) ∈ M+(R). Then equation (1.5) has

a unique solution u ∈ W1,∞(R+ × R)⋂L∞loc(R+;H1(R)) with initial data u(0) = u0 and such that

(u − uxx) ∈M+, a.e. ∈ R+ is uniformly bounded on R.

Proof. We split the proof of Theorem 5.1 in two parts.Let u0 ∈ H1(R) and y0 = u0−u0,xx ∈M+(R). Note that u0 = G∗y0. Thus, for ϕ ∈ L∞(R),

we have

‖u0‖L1(R) =∥∥G ∗ y0

∥∥L1(R) = sup

‖ϕ‖L∞(R)≤1

∫

R

ϕ(x)(G ∗ y0

)(x)dx

= sup‖ϕ‖L∞(R)

≤1

∫

R

ϕ(x)∫

R

G(x − ξ)dy0(ξ)dx



≤1

∫

R

(G ∗ ϕ)(ξ)dy0(ξ)


≤1

∥∥G ∗ ϕ∥∥L∞(R)

∥∥y0

∥∥M(R)

≤ sup‖ϕ‖L∞(R)

≤1

‖G‖L1(R)

∥∥ϕ

∥∥L∞(R)

∥∥y0

∥∥M(R) =

∥∥y0

∥∥M(R).

(5.1)

Let us define un0 := ρn ∗ u0 ∈ H∞(R) for n ≥ 1. Obviously, we get

un0 −→ u0 in H1(R) for n −→ ∞,∥∥un0

∥∥H1(R) =

∥∥ρn ∗ u0∥∥H1(R) =

∥∥ρn ∗ u0∥∥L2 +

∥∥ρn ∗ u′0∥∥L2 ≤ ‖u0‖H1(R),

‖un0‖L1(R) = ‖ρn ∗ u0‖L1(R) ≤ ‖u0‖L1(R).

(5.2)

Note that, for all n ≥ 1,

yn0 := un0 − un0,xx = ρn ∗(y0) ≥ 0. (5.3)

Referring to the proof of (5.1), we have

∥∥yn0∥∥L1(R) ≤

∥∥y0∥∥M(R), n ≥ 1. (5.4)

From the Theorem 4.4, we know that there exists a global strong solution

un = un(·, un0

) ∈ C([0,∞);Hs(R)) ∩ C1([0,∞);Hs−1(R)

), s ≥ 3

2, (5.5)

and un(t, x) − unxx(t, x) ≥ 0 for all (t, x) ∈ R+ × R.Note that for all (t, x) ∈ R+ × R

(un)2 =∫x

−∞2ununxdξ ≤

∫

R

[(un)2 + (unx)

2]dξ = ‖un‖2

H1 . (5.6)

From Theorem 4.4 and (5.2), we obtain

‖unx‖2L∞(R) ≤ ‖un‖2

L∞(R) ≤ ‖un‖2H1(R)

≤ ∥∥un0∥∥2H1 exp

[ |a − 2b|λ

(1 − e−λt

)‖un0‖L1 − 2λt

]

≤ ‖u0‖2H1 exp

[ |a − 2b|λ

(1 − e−λt

)‖u0‖L1 − 2λt

].

(5.7)


From the Holder inequality, Theorem 4.4, and (5.2), for all t ≥ 0 and n ≥ 1, we have

‖bun(t)unx(t)‖L2(R) ≤ b‖un(t)‖L∞(R)‖unx(t)‖L2(R) ≤ b‖un‖2H1(R)

≤ b∥∥un0∥∥2H1 exp

[ |a − 2b|λ

(1 − e−λt

)∥∥un0

∥∥L1 − 2λt

]

≤ b‖u0‖2H1 exp

[ |a − 2b|λ

(1 − e−λt

)‖u0‖L1 − 2λt

].

(5.8)

Using Young’s inequality, we get

∥∥∥∥∂xG ∗

[a

2(un)2 +

3b − a2

(unx)2]∥∥∥∥L2(R)

≤ a

2

∥∥∥∂xG ∗ (un)2∥∥∥L2(R)

+3b + a

2

∥∥∥∂xG ∗ (unx)2∥∥∥L2(R)

≤ a

2‖∂xG‖L2(R)

∥∥∥(un)2∥∥∥L1(R)

+3b + a

2‖∂xG‖L2(R)

∥∥∥(unx)2∥∥∥L1(R)

≤ 3b + a2

‖∂xG‖L2(R)‖un‖2H1(R)

≤ 3b + a2

‖∂xG‖L2(R)

∥∥un0∥∥2H1 exp

[ |a − 2b|λ

(1 − e−λt

)∥∥un0∥∥L1 − 2λt

]

≤ 3b + a2

‖∂xG‖L2(R)‖u0‖2H1 exp

[ |a − 2b|λ

(1 − e−λt

)‖u0‖L1 − 2λt

],

(5.9)

where ‖∂xG‖L2(R) is bounded, and

‖λun‖L2 ≤ λ‖un‖H1

≤ λ‖u0‖H1 exp[ |a − 2b|

2λ

(1 − e−λt

)‖u0‖L1 − λt

].

(5.10)

Applying (5.8)–(5.10) and problem (2.1), we have

‖ ddtun‖L2(R) ≤

(b +

3b + a2

‖∂xG‖L2(R)

)‖u0‖2

H1 exp[ |a − 2b|

λ

(1 − e−λt

)‖u0‖L1 − 2λt

]

+ λ‖u0‖H1 exp[ |a − 2b|

2λ

(1 − e−λt

)‖u0‖L1 − λt

].

(5.11)

For fixed T > 0, from (5.7) and (5.11), we deduce

∫T

0

∫

R

([un(t, x)]2 + [unx(t, x)]

2 +[unt (t, x)

]2)dxdt ≤M, (5.12)


where M is a positive constant depending only on ‖Gx‖L2(R), ‖u0‖H1(R), ‖u0‖L1(R), and T . Itfollows that the sequence {un}n≥1 is uniformly bounded in the space H1((0, T)×R). Thus, wecan extract a subsequence such that

unk ⇀ u, weakly in H1((0, T) × R) for nk −→ ∞, (5.13)

unk −→ u, a.e. on (0, T) × R for nk −→ ∞, (5.14)

for some u ∈ H1((0, T)×R). From Theorem 4.4 and (5.2), for fixed t ∈ (0, T), we have that thesequence unkx (t, ·) ∈ BV (R) satisfies

V[unkx (t, x)

]=∥∥unkxx(t, ·)

∥∥L1(R) ≤ ‖unk(t, ·)‖L1(R) +

∥∥ynk(t, ·)∥∥L1(R)

≤ 2e−λt∥∥unk0 (t, ·)∥∥

L1(R) ≤ 2e−λt‖u0(t, ·)‖L1(R) ≤ 2e−λt∥∥y0(t, ·)

∥∥M(R),

∥∥unkx (t, ·)∥∥L∞ ≤ e−λt∥∥unk0

∥∥L1(R) ≤ e−λt‖u0‖L1(R) ≤ e−λt

∥∥y0∥∥M(R).

(5.15)

Applying Helly’s theorem [31], we infer that there exists a subsequence, denoted again by{unkx (t, ·)}, which converges at every point to some function ν(t, ·) of finite variation with

V (ν(t, ·)) ≤ 2e−λt∥∥y0

∥∥M(R). (5.16)

From (5.14), we get that for almost all t ∈ (0, T), unkx (t, ·) → ux(t, ·) in D′(R), it follows thatν(t, ·) = ux(t, ·) for a.e. t ∈ (0, T). Therefore, we have

unkx (t, ·) −→ ux(t, ·) a.e. on (0, T) × R for nk −→ ∞, (5.17)

and, for a.e. t ∈ (0, T),

V [ux(t, ·)] = ‖uxx(t, ·)‖M(R) = 2e−λt‖u0‖L1 ≤ 2e−λt∥∥y0

∥∥M(R). (5.18)

By Theorem 4.4 and (5.7), we have

∥∥∥∥a

2(un)2 +

3b − a2

(unx)2∥∥∥∥L2(R)

≤ a

2

∥∥∥(un)2∥∥∥L2(R)

+3b + a

2

∥∥∥(unx)2∥∥∥L2

≤ a

2‖un‖L∞‖un‖L2(R) +

3b + a2

‖unx‖L∞‖unx‖L2


≤ a

2‖un‖2

H1(R) +3b + a

2‖un‖2

H1(R)

=(a +

3b2

)‖un‖2

H1(R)

≤(a +

3b2

)‖u0‖2

H1 exp[ |a − 2b|

λ

(1 − e−λt

)x‖u0‖L1 − 2λt

].

(5.19)

Note that for fixed t ∈ (0, T), the sequence {(a/2)(un)2 + ((3b − a)/2)(unx)2}n≥1 is uniformly

bounded in L2(R). Therefore, it has a subsequence {(a/2)(unk)2 + ((3b − a)/2)(unkx )2}nk≥1,which converges weakly in L2(R). From (5.14), we infer that the weak L2(R)-limit is{(a/2)(u)2 + ((3b − a)/2)(ux)

2}. It follows from Gx ∈ L2(R) that

∂xG ∗(a

2(unk)2 +

3b − a2

(unkx

)2)

−→ ∂xG ∗(a

2(u)2 +

3b − a2

(ux)2)

for nk −→ ∞. (5.20)

From (5.14), (5.17), and(5.20), we have that u solves (2.1) in D′((0, T) × R).For fixed T > 0, note that unkt is uniformly bounded in L2(R) as t ∈ [0, T) and

‖unk(t)‖H1(R) is uniformly bounded for all t ∈ [0, T) and n ≥ 1, and we infer that the familyt → unk ∈ H1(R) is weakly equicontinous on [0,T]. An application of the Arzela-Ascolitheorem yields that {unk} has a subsequence, denoted again {unk}, which converges weaklyin H1(R), uniformly in t ∈ [0, T). The limit function is u. T being arbitrary, we have that u islocally and weakly continuous from [0,∞) into H1(R), that is, u ∈ Cw,loc(R+;H1(R)).

Since, for a.e. t ∈ R+, unk(t, ·)⇀ u(t, ·) weakly in H1(R), from Theorem 4.4, we get

‖u(t, ·)‖L∞(R) ≤ ‖u(t, ·)‖H1(R) ≤ lim infnk →∞

‖unk(t, ·)‖H1(R)

≤ ‖u0‖H1 exp[ |a − 2b|

2λ

(1 − e−λt

)‖u0‖L1(R) − λt

].

(5.21)

Inequality (5.21) shows that

u ∈ L∞loc(R+ × R) ∩ L∞

loc

(R+;H1(R)

). (5.22)

From Theorem 4.4, (5.1) and (5.2), for t ∈ R+, we obtain

‖unx(t, ·)‖L∞ ≤ e−λt∥∥un0∥∥L1(R) ≤ e−λt‖u0‖L1(R) ≤ e−λt

∥∥y0∥∥M(R). (5.23)

Combining with (5.14), we have

ux ∈ L∞(R+ × R). (5.24)

Next, we will prove that∫R u(t, ·)dx = e−λt

∫R u(0, ·)dx by using a regularization approach.


Since u satisfies (2.1) in distribution sense, convoluting (2.1) with ρn, we have that, fora.e. t ∈ R+,

ρn ∗ ut + ρn ∗ (buux) + ρn ∗ ∂xp ∗[a

2u2 +

3b − a2

(ux)2]+ λρn ∗ u = 0. (5.25)

Integrating the above equation with respect to x on R, we obtain

d

dt

∫

R

ρn ∗ udx +∫

R

ρn ∗ (buux)dx

+∫

R

ρn ∗ ∂xp ∗[a

2u2 +

3b − a2

(ux)2]dx + λ

∫

R

ρn ∗ udx = 0.

(5.26)

Integration by parts gives rise to

d

dt

∫

R

ρn ∗ udx = −λ∫

R

ρn ∗ udx, t ∈ R+, n ≥ 1. (5.27)

Utilizing Lemma 3.3, we obtain that

∫

R

ρn ∗ u(t, ·)dx = e−λt∫

R

ρn ∗ u0dx. (5.28)

Since

limn→∞

∥∥ρn ∗ u(t, ·) − u(t, ·)∥∥L1(R) = lim

n→∞∥∥ρn ∗ u0 − u0

∥∥L1(R) = 0. (5.29)

it follows that, for a.e. t ∈ R+,

∫

R

u(t, ·)dx = limn→∞

∫

R

ρn ∗ u(t, ·)dx = limn→∞

e−λt∫

R

ρn ∗ u0dx = e−λt∫

R

u0dx. (5.30)

Finally, we prove that (u(t, ·) − uxx(t, ·)) ∈ M+ is uniformly bounded on R and u(t, x) ∈W1,∞(R+ × R).

Due to

L1(R) ⊂ (L∞)∗ ⊂ (C0(R))∗ =M(R), (5.31)

from (5.18), we get that, for a.e. t ∈ R+,

‖u(t, ·) − uxx(t, ·)‖M(R) ≤ ‖u(t, ·)‖L1(R) + ‖uxx(t, ·)‖M(R)

≤ e−λt‖u0‖L1(R) + 2e−λt∥∥y0

∥∥M(R) ≤ 3e−λt

∥∥y0∥∥M(R).

(5.32)


The above inequality implies that, for a.e. t ∈ R+, (u(t, ·) − uxx(t, ·)) ∈ M(R) is uniformlybounded on R. For fixed T ≥ 0, applying (5.13) and (5.14), we have

[unk(t, ·) − unkxx(t, ·)

] −→ [u(t, ·) − uxx(t, ·)] in D′(R) for n −→ ∞. (5.33)

Since (unk(t, ·) − unkxx(t, ·)) ≥ 0 for all (t, x) ∈ R+ × R, we obtain that for a.e. t ∈ R+, (u(t, ·) −uxx(t, ·)) ∈M+(R).

Note that u(t, x) = G ∗ (u(t, x) − uxx(t, x)). Then we get

|u(t, x)| = |G ∗ (u(t, x) − uxx(t, x))| ≤ ‖G‖L∞(R)‖u(t, x) − uxx(t, x)‖M(R)

≤ 3e−λt∥∥y0

∥∥M(R).

(5.34)

Combining with (5.24), it implies that u(t, x) ∈W1,∞(R+ × R).This completes the proof of the existence of Theorem 5.1.Next, we present the uniqueness proof of the Theorem 5.1.Let u, v ∈W1,∞(R+×R)∩L∞

loc(R+;H1(R)) be two global weak solutions of problem (2.1)

with the same initial data u0. Assume that (u(t, ·)−uxx(t, ·)) ∈M+(R) and (v(t, ·)− vxx(t, ·)) ∈M+(R) are uniformly bounded on R+ and set

N := supt∈R+

{‖u(t, ·) − uxx(t, ·)‖M(R) + ‖v(t, ·) − vxx(t, ·)‖M(R)

}. (5.35)

From assumption, we know that N <∞. Then, for all (t, x) ∈ R+ × R,

|u(t, x)| = |G ∗ (u(t, x) − uxx(t, x))|

≤ ‖G‖L∞(R)‖u(t, x) − uxx(t, x)‖M(R) ≤N

2,

(5.36)

|ux(t, x)| = |Gx ∗ (u(t, x) − uxx(t, x))|

≤ ‖Gx‖L∞(R)‖u(t, x) − uxx(t, x)‖M(R) ≤N

2.

(5.37)

Similarly,

|v(t, x)| ≤ N

2, |vx(t, x)| ≤ N

2, (t, x) ∈ R+ × R. (5.38)

Following the same procedure as in (5.1), we may also get that

‖u(t, x)‖L1 = ‖G ∗ (u(t, x) − uxx(t, x))‖L1(R)

≤ ‖G‖L1(R)‖u(t, x) − uxx(t, x)‖M(R) ≤N,

‖ux(t, x)‖L1(R) = ‖Gx ∗ (u(t, x) − uxx(t, x))‖L1(R)

≤ ‖Gx‖L1(R)‖u(t, x) − uxx(t, x)‖M(R) ≤N

(5.39)


and, for all (t, x) ∈ R+ × R,

|v(t, x)| ≤N, |vx(t, x)| ≤N, (t, x) ∈ R+ × R. (5.40)

We define

w(t, x) := u(t, x) − v(t, x), (t, x) ∈ R+ × R. (5.41)

Convoluting (2.1) for u and v with ρn, we get that for a.e. t ∈ R+ and all n ≥ 1,

ρn ∗ ut + ρn ∗ (buux) + ρn ∗ ∂xG ∗[a

2u2 +

3b − a2

(ux)2]+ λρn ∗ u = 0, (5.42)

ρn ∗ vt + ρn ∗ (bvvx) + ρn ∗ ∂xG ∗[a

2v2 +

3b − a2

(vx)2]+ λρn ∗ v = 0. (5.43)

Subtracting (5.43) from (5.42) and using Lemma 3.4, integration by parts shows that, for a.e.t ∈ R+ and all n ≥ 1

d

dt

∫

R

∣∣ρn ∗w∣∣dx =

∫

R

(ρn ∗wt

)sgn

(ρn ∗w

)dx

= −b∫

R

(ρn ∗wux

)sgn

(ρn ∗w

)dx

− b∫

R

(ρn ∗ vwx

)sgn

(ρn ∗w

)dx

− a

2

∫

R

(ρn ∗ ∂xG ∗w(u + v)

)sgn

(ρn ∗w

)dx

− 3b − a2

∫

R

(ρn ∗ ∂xG ∗wx(ux + vx)

)sgn

(ρn ∗w

)dx

− λ∫

R

(ρn ∗w

)sgn

(ρn ∗w

)dx.

(5.44)

Using (5.36)–(5.38) and Young’s inequality to the first term on the right-hand side of (5.44)yields,

∣∣∣∣

∫

R

(ρn ∗ (wux)

)sgn

(ρn ∗w

)dx

∣∣∣∣

≤∫

R

∣∣(ρn ∗ (wux))∣∣dx

≤∫

R

∣∣ρn ∗w∣∣∣∣ρn ∗ ux

∣∣dx +∫

R

∣∣ρn ∗ (wux) −(ρn ∗w

)(ρn ∗ ux

)∣∣dx


≤ ∥∥ρn ∗ ux∥∥L∞

∫

R

∣∣ρn ∗w

∣∣dx +

∫

R

∣∣ρn ∗ (wux) −

(ρn ∗w

)(ρn ∗ ux

)∣∣dx

≤ ∥∥ρn∥∥L1‖ux‖L∞

∫

R

∣∣ρn ∗w∣∣dx +

∫

R

∣∣ρn ∗ (wux) −(ρn ∗w

)(ρn ∗ ux

)∣∣dx

≤ N

2

∫

R

∣∣ρn ∗w

∣∣dx +

∫

R

∣∣ρn ∗ (wux) −

(ρn ∗w

)(ρn ∗ ux

)∣∣dx.

(5.45)

Similarly, for the second term and the third term on the right-hand side of (5.44), we have

∣∣∣∣

∫

R

(ρn ∗ (wxv)

)sgn

(ρn ∗w

)dx

∣∣∣∣

≤∫

R

∣∣(ρn ∗ (wxv))∣∣ dx

≤∫

R

∣∣ρn ∗wx

∣∣∣∣ρn ∗ v∣∣dx +

∫

R

∣∣ρn ∗ (wxv) −(ρn ∗wx

)(ρn ∗ v

)∣∣dx

≤ N

2

∫

R

∣∣ρn ∗wx

∣∣dx +∫

R

∣∣ρn ∗ (wxv) −(ρn ∗wx

)(ρn ∗ v

)∣∣dx,

∣∣∣∣

∫

R

(ρn ∗ ∂xG ∗ [w(u + v)]

)sgn

(ρn ∗w

)dx

∣∣∣∣

≤∫

R

∣∣ρn ∗G ∗ [wx(u + v)]∣∣dx

+∫

R

∣∣ρn ∗G ∗ [w(u + v)x]∣∣dx

≤ 12

∫

R

∣∣ρn ∗ [wx(u + v)]∣∣dx +

12

∫

R

∣∣ρn ∗ [w(ux + vx)]∣∣dx

≤ N

2

∫

R

∣∣ρn ∗wx

∣∣dx +N

2

∫

R

∣∣ρn ∗w∣∣dx

+∫

R

∣∣ρn ∗ (wx(u + v)) − (ρn ∗wx

)[ρn ∗ (u + v)

]∣∣dx

+∫

R

∣∣ρn ∗ [w(u + v)x] −(ρn ∗w

)[ρn ∗ (u + v)x

]∣∣dx.

(5.46)

For the last term on the right-hand side of (5.44), we have

∣∣∣∣

∫

R

(ρn ∗ ∂xG ∗ [wx(u + v)x]

)sgn

(ρn ∗w

)dx

∣∣∣∣

≤∫

R

ρn ∗ ∂xG ∗ [|wx|(|ux| + |vx|)]dx


≤N∫

R

∣∣ρn ∗ ∂xG ∗ |wx|

∣∣dx

≤N‖∂xG‖L1(R)

∫

R

ρn ∗ |wx|dx

≤N∫

R

∣∣ρn ∗wx

∣∣dx +N

[∫

R

(ρn ∗ |wx| −

∣∣ρn ∗wx

∣∣)dx

].

(5.47)

From (5.45)–(5.47), for a.e. t ∈ R+ and all n ≥ 1, we find

d

dt

∫

R

∣∣ρn ∗w

∣∣dx ≤

(a + 2b

4N + λ

)∫

R

∣∣ρn ∗w

∣∣dx

+ (a + 2b)N∫

R

∣∣ρn ∗wx

∣∣dx + Rn(t),

(5.48)

where

Rn(t) −→ 0 as t −→ ∞,

|Rn(t)| ≤ K, n ≥ 1, t ∈ R+,(5.49)

where K is a positive constant depending on N and the H1(R)-norms of u(0) and v(0).In the same way, convoluting (2.1) for u and v with ρn,x and using Lemma 3.4, we get

that, for a.e. t ∈ R+ and all n ≥ 1,

d

dt

∫

R

∣∣ρn ∗wx

∣∣dx =∫

R

(ρn ∗wxt

)sgn

(ρn,x ∗w

)dx

= −b∫

R

(ρn ∗wx(ux + vx)

)sgn

(ρn,x ∗w

)dx

− b∫

R

(ρn ∗ vxxw

)sgn

(ρn,x ∗w

)dx

− b∫

R

(ρn ∗ uwxx

)sgn

(ρn,x ∗w

)dx

−∫

R

(ρn ∗ ∂xxG ∗

[a

2(u2 − v2) +

3b − a2

(u2x − v2

x

)]sgn

(ρn,x ∗w

)dx

− λ∫

R

(ρn ∗wx

)sgn

(ρn,x ∗w

)dx.

(5.50)


Using the identity ∂2x(G ∗ g) = G ∗ g − g for g ∈ L2(R) and Young’s inequality, we estimate the

forth term of the right-hand side of (5.50)

∣∣∣∣

∫

R

(ρn ∗ ∂xxG ∗

[a

2

(u2 − v2

)+

3b − a2

(u2x − v2

x

)]dx

∣∣∣∣

≤∫

R

∣∣∣∣

(ρn ∗G ∗

[a

2

(u2 − v2

)+

3b − a2

(u2x − v2

x

)]∣∣∣∣dx

+∫

R

∣∣∣∣

(ρn ∗

[a

2

(u2 − v2

)+

3b − a2

(u2x − v2

x

)]∣∣∣∣dx

≤(‖G‖L1(R) + 1

)∫

R

∣∣∣∣

(ρn ∗

[a

2w(u + v) +

3b − a2

wx(ux + vx)]∣∣∣∣dx

≤ a∫

R

∣∣(ρn ∗ [w(u + v)]∣∣dx + (3b + a)

∫

R

∣∣(ρn ∗ [wx(ux + vx)]∣∣dx

≤ aN∫

R

∣∣ρn ∗w∣∣dx + (3b + a)N

∫

R

∣∣ρn ∗wx

∣∣dx + Rn.

(5.51)

Using (5.36)–(5.38) and Young’s inequality to the first term on the right-hand side of (5.50)gives rise to

− b∫

R

(ρn ∗wx(ux + vx)

)sgn

(ρn,x ∗w

)dx

≤ b∫

R

∣∣ρn ∗wx(ux + vx)∣∣dx

≤ b∫

R

∣∣ρn ∗wx

∣∣∣∣ρn ∗ (ux + vx)∣∣dx

+ b∫

R

∣∣ρn ∗wx(ux + vx) −(ρn ∗wx

)(ρn ∗ (ux + vx)

)∣∣dx

≤ bN∫

R

∣∣ρn ∗wx

∣∣dx + Rn.

(5.52)

To treat the second term of the right-hand side of (5.50), we note that

∣∣∣∣b∫

R

(ρn ∗ vxxw

)sgn

(ρn,x ∗w

)dx

∣∣∣∣

≤ b∫

R

∣∣(ρn ∗w)(ρn ∗ vxx

)∣∣dx

+ b∫

R

∣∣(ρn ∗ vxxw) − (ρn ∗w

)(ρn ∗ vxx

)∣∣dx.

(5.53)


Applying Lemma 3.1, the second expression of the right-hand side of (5.53) can be estimatedby a function Rn(t) belonging to (5.49). Making use of the Holder inequality and (5.1), for a.e.t ∈ R+ and all n ≥ 1, we have

∫

R

∣∣(ρn ∗w

)(ρn ∗ vxx

)∣∣dx ≤ ∥∥ρn ∗w∥∥L∞(R)

∥∥ρn ∗ vxx

∥∥L1(R)

≤ ∥∥ρn ∗w∥∥W1,1(R)‖vxx‖M(R).

(5.54)


∣∣∣∣b∫

R

(ρn ∗ vxxw

)sgn

(ρn,x ∗w

)dx

∣∣∣∣

≤ bN∫

R

∣∣ρn ∗w∣∣dx

+ bN∫

R

∣∣ρn ∗wx

∣∣ + Rn(t).

(5.55)

Now, we deal with the third term on the right-hand side of (5.50)

− b∫

R

(ρn ∗ uwxx

)sgn

(ρn,x ∗w

)dx

= −b∫

R

(ρn ∗ u

)(ρn ∗wxx

)sgn

(ρn,x ∗w

)dx

− b∫

R

[(ρn ∗ uwxx

) − (ρn ∗ u)(ρn ∗wxx

)]sgn

(ρn,x ∗w

)dx

≤ −b∫

R

(ρn ∗ u

) ∂∂x

∣∣ρn ∗wx

∣∣dx

+ b∫

R

∣∣(ρn ∗ uwxx

) − (ρn ∗ u)(ρn ∗wxx

)∣∣dx

= b∫

R

(ρn ∗ ux

)∣∣ρn ∗wx

∣∣dx + Rn.

(5.56)

Therefore, (5.56) implies that, for a.e. t ∈ R+ and all n ≥ 1,

∣∣∣∣−b∫

R

(ρn ∗ uwxx

)sgn

(ρn,x ∗w

)dx

∣∣∣∣ ≤ bN∫

R

∣∣ρn ∗wx

∣∣dx + Rn. (5.57)


From (5.51), (5.52), (5.55), and (5.57), for a.e. t ∈ R+ and all n ≥ 1, we deduce that

d

dt

∫

R

∣∣ρn ∗wx

∣∣dx ≤ (a + b)N

∫

R

∣∣ρn ∗w

∣∣dx

+ [(6b + a)N + λ]∫

R

∣∣ρn ∗wx

∣∣dx + Rn.

(5.58)

Combining with (5.48) and (5.58), we find

d

dt

∫

R

(∣∣ρn ∗w∣∣ +

∣∣ρn ∗wx

∣∣)dx ≤

[5a + 6b

4N + λ

] ∫

R

∣∣ρn ∗w

∣∣dx

+ [(2a + 8b)N + λ]∫

R

∣∣ρn ∗wx

∣∣dx + Rn

≤ [(2a + 8b)N + λ]∫

R

(∣∣ρn ∗w∣∣ +

∣∣ρn ∗wx

∣∣)dx + Rn.

(5.59)

It follows from Gronwall’ inequality that, for a.e. t ∈ R+ and all n ≥ 1,

∫

R

(∣∣ρn ∗w∣∣ +

∣∣ρn ∗wx

∣∣)dx

≤[∫ t

0Rn(s)ds +

∫

R

(∣∣ρn ∗w∣∣ +

∣∣ρn ∗wx

∣∣)(0, x)dx

]

e[(2a+8b)N+λ]t.

(5.60)

Fix t > 0, and let n → ∞ in (5.60). Sincew = u−v ∈W1,1(R) and relation (5.49) holds, makinguse of Lebesgue’s dominated convergence theorem yields

∫

R

(|w| + |wx|)dx ≤[∫

R

(|w| + |wx|)(0, x)dx]e[(2a+8b)N+λ]t. (5.61)

Note that w(0) = wx(0) = 0; therefore, we obtain u(t, x) = v(t, x) for a.e. (t, x) ∈ R+ × R. Thiscompletes the proof of the theorem.

Acknowledgments

This work was supported by RFSUSE (no. 2011KY12), RFSUSE (no. 2011RC10) and theproject (no. 12ZB080). The authors thanks the referee for valuable comments.

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Research ArticlePositive Solutions of EigenvalueProblems for a Class of Fractional DifferentialEquations with Derivatives

Xinguang Zhang,1 Lishan Liu,2Benchawan Wiwatanapataphee,3 and Yonghong Wu4

1 School of Mathematical and Informational Sciences, Yantai University, Yantai, Shandong 264005, China2 School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165, China3 Department of Mathematics, Faculty of Science, Mahidol University, Bangkok 10400, Thailand4 Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia

Correspondence should be addressed to Xinguang Zhang, [email protected] Benchawan Wiwatanapataphee, [email protected]

Received 2 January 2012; Accepted 15 March 2012


Copyright q 2012 Xinguang Zhang et al. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

By establishing a maximal principle and constructing upper and lower solutions, the existenceof positive solutions for the eigenvalue problem of a class of fractional differential equations isdiscussed. Some sufficient conditions for the existence of positive solutions are established.

1. Introduction

In this paper, we discuss the existence of positive solutions for the following eigenvalueproblem of a class fractional differential equation with derivatives

−Dtαx(t) = λf

(t, x(t),Dt

βx(t)), t ∈ (0, 1),

Dtβx(0) = 0, Dt

γx(1) =p−2∑

j=1

ajDtγx(ξj),

(1.1)

where λ is a parameter, 1 < α ≤ 2, α − β > 1, 0 < β ≤ γ < 1, 0 < ξ1 < ξ2 < · · · < ξp−2 < 1,aj ∈ [0,+∞) with c =

∑p−2j=1 ajξ

α−γ−1j < 1, and Dt is the standard Riemann-Liouville derivative.


f : (0, 1) × (0,+∞) × (0,+∞) → [0,+∞) is continuous, and f(t, u, v) may be singular atu = 0, v = 0, and t = 0, 1.

As fractional order derivatives and integrals have been widely used in mathematics,analytical chemistry, neuron modeling, and biological sciences [1–6], fractional differentialequations have attracted great research interest in recent years [7–17]. Recently, ur Rehmanand Khan [8] investigated the fractional order multipoint boundary value problem:

Dtαy(t) = f

(t, y(t),Dt

βy(t)), t ∈ (0, 1),

y(0) = 0, Dtβy(1) −

m−2∑

i=1

ζiDtβy(ξi) = y0,

(1.2)

where 1 < α ≤ 2, 0 < β < 1, 0 < ξi < 1, ζi ∈ [0,+∞) with∑m−2

i=1 ζiξα−β−1i < 1. The Schauder fixed

point theorem and the contraction mapping principle are used to establish the existence anduniqueness of nontrivial solutions for the BVP (1.2) provided that the nonlinear function f :[0, 1]×R×R is continuous and satisfies certain growth conditions. But up to now, multipointboundary value problems for fractional differential equations like the BVP (1.1) have seldombeen considered when f(t, u, v) has singularity at t = 0 and (or) 1 and also at u = 0, v = 0.We will discuss the problem in this paper.

The rest of the paper is organized as follows. In Section 2, we give some definitions andseveral lemmas. Suitable upper and lower solutions of the modified problems for the BVP(1.1) and some sufficient conditions for the existence of positive solutions are established inSection 3.


For the convenience of the reader, we present here some definitions about fractional calculus.

Definition 2.1 (See [1, 6]). Let α > 0 with α ∈ R. Suppose that x : [a,∞) → R. Then the αthRiemann-Liouville fractional integral is defined by

Iαx(t) =1

Γ(α)

∫ t

a

(t − s)α−1x(s)ds (2.1)

whenever the right-hand side is defined. Similarly, for α ∈ R with α > 0, we define the αthRiemann-Liouville fractional derivative by

Dαx(t) =1

Γ(n − α)(d

dt

)(n) ∫ t

a

(t − s)n−α−1x(s)ds, (2.2)

where n ∈ N is the unique positive integer satisfying n − 1 ≤ α < n and t > a.


Dtα(x(t) + y(t)

)= Dt

αx(t) +Dtαy(t). (2.3)


Lemma 2.3 (See [6]). One has the following.

(1) If x ∈ L1(0, 1), ν > σ > 0, then

IνIσx(t) = Iν+σx(t), DtσIνx(t) = Iν−σx(t), Dt

σIσx(t) = x(t). (2.4)

(2) If ν > 0, σ > 0, then

Dtνtσ−1 =

Γ(σ)Γ(σ − ν) t

σ−ν−1. (2.5)

Lemma 2.4 (See [6]). Let α > 0. Assume that x ∈ C(0, 1) ∩ L1(0, 1). Then

IαDtαx(t) = x(t) + c1t

α−1 + c2tα−2 + · · · + cntα−n, (2.6)

where ci ∈ R (i = 1, 2, . . . , n), and n is the smallest integer greater than or equal to α.

Let

k1(t, s) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

tα−β−1(1 − s)α−γ−1 − (t − s)α−β−1

Γ(α − β) , 0 ≤ s ≤ t ≤ 1,

tα−β−1(1 − s)α−γ−1

Γ(α − β) , 0 ≤ t ≤ s ≤ 1,

k2(t, s) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

(t(1 − s))α−γ−1 − (t − s)α−γ−1

Γ(α − β) , 0 ≤ s ≤ t ≤ 1,

(t(1 − s))α−γ−1

Γ(α − β) , 0 ≤ t ≤ s ≤ 1,

(2.7)

and for t, s ∈ [0, 1], we have

ki(t, s) ≤ (1 − s)α−γ−1

Γ(α − β) , i = 1, 2. (2.8)

Lemma 2.5. Let h ∈ C(0, 1); If 1 < α − β ≤ 2, then the unique solution of the linear problem

−Dtα−βy(t) = h(t), t ∈ (0, 1),

y(0) = 0, Dtγ−βy(1) =

p−2∑

j=1

ajDtγ−βy

(ξj) (2.9)

is given by

y(t) =∫1

0K(t, s)h(s)ds, (2.10)


where

K(t, s) = k1(t, s) +tα−β−1

1 −∑p−2j=1 ajξ

α−γ−1j

p−2∑

j=1

ajk2(ξj , s)

(2.11)

is the Green function of the boundary value problem (2.9).

Proof. Applying Lemma 2.4, we reduce (2.9) to an equivalent equation:

y(t) = −Iα−βh(t) + c1tα−β−1 + c2t

α−β−2, c1, c2 ∈ R. (2.12)

From (2.12) and noting that y(0) = 0, we have c2 = 0. Consequently the general solution of(2.9) is

y(t) = −Iα−βh(t) + c1tα−β−1. (2.13)

Using (2.13) and Lemma 2.3, we have

Dtγ−βy(t) = −Dt

γ−βIα−βh(t) + c1Dtγ−βtα−β−1

= −Iα−γh(t) + c1Γ(α − β)

Γ(α − γ) t

α−γ−1

= −∫ t

0

(t − s)α−γ−1

Γ(α − γ) h(s)ds + c1

Γ(α − β)

Γ(α − γ) t

α−γ−1.

(2.14)

Thus,

Dtγ−βy(1) = −

∫1

0

(1 − s)α−γ−1


Γ(α − β)

Γ(α − γ) , (2.15)

and for j = 1, 2, . . . , p − 2,

Dtγ−βw

(ξj)= −∫ ξj

0

(ξj − s

)α−γ−1


Γ(α − β)

Γ(α − γ) ξ

α−γ−1j . (2.16)

Using Dtγ−βy(1) =

∑p−2j=1 ajDt

γ−βy(ξj), (2.15), and (2.16), we obtain

c1 =

∫10 (1 − s)α−γ−1 h(s)ds −∑p−2

j=1 aj∫ ξj

0

(ξj − s

)α−γ−1h(s)ds

Γ(α − β)

(1 −∑p−2

j=1 ajξα−γ−1j

) . (2.17)


So the unique solution of the problem (2.9) is

y(t) = −∫ t

0

(t − s)α−β−1

Γ(α − β) h(s)ds +

tα−β−1

1 −∑p−2j=1 ajξ

α−γ−1j

×⎧⎨

⎩

∫1

0

(1 − s)α−γ−1

Γ(α − β) h(s)ds −

p−2∑

j=1

aj

∫ ξj

0

(ξj − s

)α−γ−1

Γ(α − β) h(s)ds

⎫⎬

⎭

= −∫ t

0

(t − s)α−β−1

Γ(α − β) h(s)ds +

∫1

0

(1 − s)α−γ−1tα−β−1


+tα−β−1

1 −∑p−2j=1 ajξ

α−γ−1j

p−2∑

j=1

aj

∫1

0

(1 − s)α−γ−1ξα−γ−1j


− tα−β−1

1 −∑p−2j=1 ajξ

α−γ−1j

p−2∑

j=1

aj

∫ ξj

0

(ξj − s

)α−γ−1


=∫1

0

⎛

⎝k1(t, s) +tα−β−1

1 −∑p−2j=1 ajξ

α−γ−1j

p−2∑

j=1

ajk2(ξj , s)⎞

⎠h(s)ds

=∫1

0K(t, s)h(s)ds.

(2.18)


Lemma 2.6. The function K(t, s) has the following properties.

(1) K(t, s) > 0, for t, s ∈ (0, 1)

(2) tα−β−1M(s) ≤ K(t, s) ≤ M(1 − s)α−γ−1, for t, s ∈ [0, 1],

where

M(s) =

∑p−2j=1 ajk2

(ξj , s)

1 −∑p−2j=1 ajξ

α−γ−1j

, M =1 +∑p−2

j=1 aj(

1 − ξα−γ−1j

)

Γ(α − β)

(1 −∑p−2


) . (2.19)

Proof. It is obvious that (1) holds.From (2.11), we obtain

K(t, s) ≥ tα−β−1

1 −∑p−2j=1 ajξ

α−γ−1j

p−2∑

j=1

ajk2(ξj , s)= tα−β−1M(s). (2.20)


From (2.8), we have

K(t, s) = k1(t, s) +tα−β−1

1 −∑p−2j=1 ajξ

α−γ−1j

p−2∑

j=1

ajk2(ξj , s)

≤ (1 − s)α−γ−1

Γ(α − β) +

∑p−2j=1 aj

1 −∑p−2j=1 ajξ

α−γ−1j

(1 − s)α−γ−1

Γ(α − β)

≤⎛

⎝1 +

∑p−2j=1 aj

1 −∑p−2j=1 ajξ

α−γ−1j

⎞

⎠ (1 − s)α−γ−1

Γ(α − β) .

(2.21)


Consider the modified problem of the BVP (1.1):

−Dtα−βy(t) = λf

(t, Iβy(t), y(t)

),

y(0) = 0, Dtγ−βy(1) =

p−2∑

j=1

ajDtγ−βy

(ξj).

(2.22)

Lemma 2.7. Let x(t) = Iβy(t) and y(t) ∈ C[0, 1]; then problem (1.1) is turned into (2.22).Moreover, if y ∈ C([0, 1], [0,+∞)) is a solution of problem (2.22), then the function x(t) = Iβy(t) isa positive solution of the problem (1.1).

Proof. Substituting x(t) = Iβy(t) into (1.1) and using Definition 2.1 and Lemmas 2.3 and 2.4,we obtain

Dtαx(t) =

dn

dtnIn−αx(t) =

dn

dtnIn−αIβy(t)

=dn

dtnIn−α+βy(t) = Dt

α−βy(t),

Dtβx(t) = Dt

βIβy(t) = y(t).

(2.23)

Consequently, Dtβx(0) = y(0) = 0. It follows from Dt

γx(t) = dn/dtnIn−γx(t) =(dn/dtn)In−γ Iβy(t) = (dn/dtn)In−γ+βy(t) = Dt

γ−βy(t) that Dtγ−βy(1) =

∑p−2j=1 ajDt

γ−βy(ξj).Using x(t) = Iβy(t), y ∈ C[0, 1], we transform (1.1) into (2.22).

Now, let y ∈ C([0, 1], [0,+∞)) be a solution for problem (2.22). Using Lemma 2.3,(2.22), and (2.23), one has

−Dtαx(t) = − d

n

dtnIn−αx(t) = − d

n

dtnIn−αIβy(t) = − d

n

dtnIn−α+βy(t) = −Dt

α−βy(t)

= λf(t, Iβy(t), y(t)

)= λf

(t, x(t),Dt

βx(t)), 0 < t < 1.

(2.24)


Noting

Dtβx(t) = Dt

βIβy(t) = y(t), Dtγx(t) = Dt

γ−βy(t), (2.25)

we have

Dtβx(0) = 0, Dt

γx(1) =p−2∑

j=1

ajDtγx(ξj). (2.26)

It follows from the monotonicity and property of Iβ that

Iβy ∈ C([0, 1], [0,+∞)). (2.27)

Consequently, x(t) = Iβy(t) is a positive solution of the problem (1.1).

Definition 2.8. A continuous function ψ(t) is called a lower solution of the BVP (2.22), if itsatisfies

−Dtα−βψ(t) ≤ λf

(t, Iβψ(t), ψ(t)

),

ψ(0) ≥ 0, Dtγ−βψ(1) ≥

p−2∑

j=1

ajDtγ−βψ

(ξj).

(2.28)

Definition 2.9. A continuous function φ(t) is called an upper solution of the BVP (2.22), if itsatisfies

−Dtα−βφ(t) ≥ λf

(t, Iβφ(t), φ(t)

),

φ(0) ≤ 0, Dtγ−βφ(1) ≤

p−2∑

j=1

ajDtγ−βφ

(ξj).

(2.29)

By Lemmas 2.5 and 2.6, we have the maximal principle.

Lemma 2.10. If 1 < α − β ≤ 2 and y ∈ C([0, 1], R) satisfies

y(0) = 0, Dtγ−βy(1) =

p−2∑

j=1

ajDtγ−βy

(ξj), (2.30)

and −Dtα−βy(t) ≥ 0 for any t ∈ (0, 1), then y(t) ≥ 0, for t ∈ [0, 1].

Set

G(t) = tα−β−1, L(t) = IβG(t) = 1Γ(β)∫ t

0(t − s)β−1sα−β−1ds =

Γ(α − β)

Γ(α)tα−1. (2.31)


To end this section, we present here two assumptions to be used throughout the restof the paper.

(B1) f ∈ C((0, 1) × (0,∞) × (0,∞), [0,+∞)) is decreasing in u and v, and for any (u, v) ∈(0,∞) × (0,∞),

limσ→+∞

σf(t, σu, σv) = +∞ (2.32)

uniformly on t ∈ (0, 1).

(B2) For any μ, ν > 0, f(t, μ, ν)/≡ 0, and

∫1

0(1 − s)α−γ−1f

(s, μL(s), μG(s))ds < +∞. (2.33)

3. Main Results

The main result is summarized in the following theorem.

Theorem 3.1. Provided that (B1) and (B2) hold, then there is a constant λ∗ > 0 such that for anyλ ∈ (λ∗,+∞), the problem (1.1) has at least one positive solution x(t), which satisfies x(t) ≥ L(t),t ∈ [0, 1].

Proof. Let E = C[0, 1]; we denote a set P and an operator Tλ in E as follows:

P ={y ∈ E : there exists positive number ly such that y(t) ≥ lyG(t), t ∈ [0, 1]

}, (3.1)

(Tλy)(t) = λ

∫1

0K(t, s)f

(s, Iβy(s), y(s)

)ds, for any y ∈ P. (3.2)

Clearly, P is a nonempty set since G(t) ∈ P . We claim that Tλ is well defined andTλ(P) ⊂ P .

In fact, for any ρ ∈ P , by the definition of P , there exists one positive numberlρ such that ρ(t) ≥ lρG(t) for any t ∈ [0, 1]. It follows from Lemma 2.6 and (B2)that

(Tλρ)(t) = λ

∫1

0K(t, s)f

(s, Iβρ(s), ρ(s)

)ds ≤ λM

∫1

0(1 − s)α−γ−1f

(s, Iβρ(s), ρ(s)

)ds

≤ λM∫1

0(1 − s)α−γ−1f

(s, lρL(s), lρG(s)

)ds < +∞.

(3.3)


Setting B = maxt∈[0,1]ρ(t) > 0, from (B2), we have f(t, B/Γ(β + 1), B)/≡ 0. By thecontinuity of f(t, u, v) on (0, 1)× (0,∞)× (0,∞), we have

∫10 M(s)f(s, B/Γ(β+1), B)ds > 0. On

the other hand,

IβB =1

Γ(β)∫ t

0(t − s)β−1Bds =

Btβ

βΓ(β) ≤ B

Γ(β + 1

) ,

M(s) =

∑p−2j=1 ajk2

(ξj , s)

1 −∑p−2j=1 ajξ

α−γ−1j

≤∑p−2

j=1 aj(1 − s)α−γ−1

Γ(α − β)

(1 −∑p−2


) .

(3.4)

From (3.3), one has

0 <∫1

0M(s)f

(

s,B

Γ(β + 1

) , B

)

ds ≤∫1

0M(s)f

(s, IβB, B

)ds ≤

∫1

0M(s)f

(s, Iβρ(s), ρ(s)

)ds

≤∑p−2

j=1 aj

Γ(α − β)

(1 −∑p−2


)∫1

0(1 − s)α−γ−1f

(s, Iβρ(s), ρ(s)

)ds < +∞.

(3.5)

It follows from Lemma 2.6 and (3.3) that

(Tλρ)(t) ≥ λG(t)

∫1

0M(s)f

(s, Iβρ(s), ρ(s)

)ds = l′ρG(t), (3.6)

where

l′ρ = λ∫1

0M(s)f

(s, Iβρ(s), ρ(s)

)ds. (3.7)

Using (3.3) and (3.6), we know that Tλ is well defined and Tλ(P) ⊂ P .Next we will focus on the upper and lower solutions of problem (2.22). From (B1) and

(3.2), we know that the operator Tλ is decreasing in y. Using

∫1

0K(t, s)f(s,L(s),G(s))ds ≥ G(t)

∫1

0M(s)f(s,L(s),G(s))ds, ∀t ∈ [0, 1], (3.8)

and letting

λ1 =1

∫10 M(s)f(s,L(s),G(s))ds

, (3.9)


we have

λ1

∫1

0K(t, s)f(s,L(s),G(s))ds ≥ G(t), ∀t ∈ [0, 1]. (3.10)

On the other hand, letting b(t) =∫1

0 K(t, s)f(s,L(s),G(s))ds, since f(t, u, v) isdecreasing with respect to u and v, for any λ > λ1, we have

∫1

0K(t, s)f

(s, λIβb(s), λb(s)

)ds ≤

∫1

0K(t, s)f

(s, λ1I

βb(s), λ1b(s))ds

≤∫1

0K(t, s)f(s,L(s),G(s))ds ≤M

∫1

0(1 − s)α−γ−1f(s,L(s),G(s))ds

< +∞.

(3.11)

From (3.2), (3.3), and (B1), for all (u, v) ∈ (0,∞) × (0,∞), we have

limμ→+∞

μf(t, μu, μv

)= +∞ (3.12)

uniformly on t ∈ (0, 1). Thus there exists large enough λ∗ > λ1 > 0, such that, for any t ∈ (0, 1),

λ∗f(s, λ∗L(s), λ∗G(s)) ≥ 1∫1

0 M(s)ds. (3.13)

From Lemma 2.6, one has

λ∗∫1

0K(t, s)f(s, λ∗L(s), λ∗G(s))ds ≥

∫10 K(t, s)ds∫1

0 M(s)ds

≥∫1

0 G(t)M(s)ds∫1

0 M(s)ds= G(t), ∀t ∈ [0, 1].

(3.14)

Letting

φ(t) = λ∗∫1

0K(t, s)f(s,L(s),G(s))ds = λ∗b(t),

ψ(t) = λ∗∫1

0K(t, s)f

(s, λ∗Iβb(s), λ∗b(s)

)ds,

(3.15)


and using Lemmas 2.3 and 2.7, we obtain

φ(t) = λ∗∫1

0 K(t, s)f(s,L(s),G(s))ds ≥ G(t), t ∈ [0, 1],

φ(0) = 0, Dtγ−βφ(1) =

p−2∑

j=1ajDt

γ−βφ(ξj),

ψ(t) = λ∗∫1

0 K(t, s)f(s, λ∗Iβb(s), λ∗b(s)

)ds ≥ G(t), t ∈ [0, 1],

ψ(0) = 0, Dtγ−βψ(1) =

p−2∑

j=1ajDt

γ−βψ(ξj).

(3.16)

Obviously, φ(t), ψ(t) ∈ P . By (3.16), we have

G(t) ≤ ψ(t) = (Tλ∗φ)(t), G(t) ≤ φ(t), ∀t ∈ [0, 1], (3.17)

which implies that

ψ(t) =(Tλ∗φ

)(t) = λ∗

∫1

0K(t, s)f

(s, Iβφ(s), φ(s)

)ds

≤ λ∗∫1

0K(t, s)f(s,L(s),G(s))ds = φ(t), ∀t ∈ [0, 1].

(3.18)

Consequently, it follows from (3.17)-(3.18) that

Dtψ(t) + λ∗f(t, Iβψ(t), ψ(t)

)= Dt

(Tλ∗φ

)(t) + λ∗f

(t, Iβ(Tλ∗φ

)(t),(Tλ∗φ

)(t))

≥ Dt(Tλ∗φ

)(t) + λ∗f

(t, Iβφ(t), φ(t)

)

= −λ∗f(t, Iβφ(t), φ(t)

)+ λ∗f

(t, Iβφ(t), φ(t)

)= 0,

(3.19)

Dtφ(t) + λ∗f(t, Iβφ(t), φ(t)

)= −λ∗f(t,L(t),G(t)) + λ∗f

(t, Iβφ(t), φ(t)

)

≤ −λ∗f(t,L(t),G(t)) + λ∗f(t,L(t),G(t)) = 0.(3.20)

From (3.16) and (3.18)–(3.20), we know that ψ(t) and φ(t) are upper and lower solutions ofthe problem (2.22), and ψ(t), φ(t) ∈ P .

Define the function F and the operator Aλ∗ in E by

F(t, y)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f(t, Iβψ(t), ψ(t)

), y < ψ(t),

f(t, Iβy(t), y(t)

), ψ(t) ≤ y ≤ φ(t),

f(t, Iβφ(t), φ(t)

), y > φ(t),

(Aλ∗y

)(t) = λ∗

∫1

0K(t, s)F

(s, y(s)

)ds, ∀y ∈ E.

(3.21)


It follows from (B1) and (3.21) that F : (0, 1)× [0,+∞) → [0,+∞) is continuous. Consider thefollowing boundary value problem:

−Dtα−βy(t) = λ∗F

(t, y), t ∈ [0, 1],

y(0) = 0, Dtγ−βy(1) =

p−2∑

j=1

ajDtγ−βy

(ξj).

(3.22)

Obviously, a fixed point of the operator Aλ∗ is a solution of the BVP (3.22). For all y ∈ E, itfollows from Lemma 2.6, (3.21), and ψ(t) ≥ G(t) that

(Aλ∗y

)(t) ≤ λ∗M

∫1

0(1 − s)α−γ−1F

(s, y(s)

)ds

≤ λ∗M∫1

0(1 − s)α−γ−1f

(s, Iβψ(s), ψ(s)

)ds

≤ λ∗M∫1

0(1 − s)α−γ−1f(s,L(s),G(s))ds

< +∞.

(3.23)

So Aλ∗ is bounded. From the continuity of F(t, y) and K(t, s), it is obviously that Aλ∗ : E → Eis continuous.

From the uniform continuity of K(t, s) and the Lebesgue dominated convergencetheorem, we easily get that Aλ∗(Ω) is equicontinuous. Thus from the Arzela-Ascoli theorem,Aλ∗ : E → E is completely continuous. The Schauder fixed point theorem implies that Aλ∗

has at least one fixed point w such that w = Aλ∗w.Now we prove

ψ(t) ≤ w(t) ≤ φ(t), t ∈ [0, 1]. (3.24)

Let z(t) = φ(t) −w(t), t ∈ [0, 1]. Since φ(t) is the upper solution of problem (2.22) and w is afixed point of Aλ∗ , we have

w(0) = 0, Dtγ−βw(1) =

p−2∑

j=1

ajDtγ−βw

(ξj). (3.25)

From (3.17), (3.18), and the definition of F, we obtain


)≤ F(t, y(t)) ≤ f

(t, Iβψ(t), ψ(t)

), ∀y ∈ E,

f(t,L(t),G(t)) ≥ f(t, Iβψ(t), ψ(t)

), ∀t ∈ [0, 1].

(3.26)


So


)≤ F(t, y(t)) ≤ f(t,L(t),G(t)), ∀y ∈ E. (3.27)

From (3.18) and (3.20), one has

Dtα−βz(t) = Dt

α−βφ(t) −Dtα−βw(t)

= −λ∗f(t,L(t),G(t)) + λ∗F(t,w(t))

≤ 0, ∀t ∈ [0, 1].

(3.28)

By (3.27), (3.28), and Lemma 2.10, we get z(t) ≥ 0 which implies that w(t) ≤ φ(t) on [0, 1]. Inthe same way, we have w(t) ≥ ψ(t) on [0, 1]. Thus we obtain

ψ(t) ≤ w(t) ≤ φ(t), t ∈ [0, 1]. (3.29)

Consequently, F(t,w(t)) = f(t, Iβw(t), w(t)), t ∈ [0, 1]. Then w(t) is a positive solution of theproblem (2.22). It thus follows from Lemma 2.7 that x(t) = Iβw(t) is a positive solution of theproblem (1.1).

Finally, by (3.29), we have

w(t) ≥ ψ(t) ≥ G(t). (3.30)

Thus,

x(t) = Iβw(t) =1

Γ(β)∫ t

0(t − s)β−1w(s)ds ≥ 1

Γ(β)∫ t

0(t − s)β−1G(s)ds = L(t). (3.31)

Corollary 3.2. Suppose that condition (B1) holds, and that for any μ, ν > 0, f(t, μ, ν)/≡ 0, and

∫1

0f(s, μL(s), μG(s))ds < +∞. (3.32)

Then there exists a constant λ∗ > 0 such that for any λ ∈ (λ∗,+∞), the problem (1.1) has at least onepositive solution x(t), which satisfies x(t) ≥ L(t), t ∈ [0, 1].

We consider some special cases in which f(t, u, v) has no singularity at u, v = 0 ort = 0, 1.


We give the following assumption.(B∗1) f ∈ C((0, 1) × [0,∞) × [0,∞), (0,+∞)) is decreasing in u, v.

Then, f(t, u, v) is nonsingular at u = v = 0 and for all u, v ≥ 0, f(t, u, v) > 0, t ∈ (0, 1), whichimplies that f(t, 0, 0) > 0, t ∈ (0, 1). Thus

limμ→+∞

μf(t, 0, 0) = +∞, uniformly for t ∈ (0, 1) (3.33)

naturally holds; we then have the following corollary.

Corollary 3.3. If (B∗1) holds and(B∗2)

∫1

0(1 − s)α−γ−1f(s, 0, 0)ds < +∞, (3.34)

then there exists a constant λ∗ > 0 such that for any λ ∈ (λ∗,+∞), the problem (1.1) has at least onepositive solution x(t), which satisfies x(t) ≥ L(t), t ∈ [0, 1].

Proof. In the proof of Theorem 3.1, we replace the set P by

P1 ={y ∈ E : y(t) ≥ 0, t ∈ [0, 1]

}(3.35)

and the inequalities (3.18)–(3.20) by

0 ≤ ψ(t) = Tλ0, 0 ≤ φ(t) = Tλψ(t) ≤ Tλ0 = ψ(t). (3.36)

Since Tλ0, Tλψ(t) ∈ P , we have

Dtα−βTλ0 + f

(t, IβTλ, Tλ0

)= −f(t, 0, 0) + f

(t, IβTλ0, Tλ0

)≤ 0,

Dtα−β Tλψ(t) + f

(t, IβTλψ(t), Tλψ(t)

)= −f

(t, Iβψ(t), ψ(t)

)+ f(t, IβTλψ(t), Tλψ(t)

)≥ 0

, t ∈ [0, 1].(3.37)

The rest of the proof is similar to that of Theorem 3.1.

If f(t, u, v) is nonsingular at u = 0, v = 0 and t = 0, 1, we have the conclusion.

Corollary 3.4. If f(t, u, v) : [0, 1] × [0,∞) × [0,∞) → (0,+∞) is continuous and decreasingin u and v, the problem (1.1) has at least one positive solution x(t), which satisfies x(t) ≥ L(t),t ∈ [0, 1].


Example 3.5. Consider the existence of positive solutions for the following eigenvalueproblem of fractional differential equation:

−Dt3/2x(t) =

λ

et(1 − t)1/8

(x−1/2(t) +

(Dt

1/8x(t))−1/8

),

Dt1/8x(0) = 0, Dt

3/8x(1) = 2Dt3/8x

(12

)−Dt

3/8x

(34

).

(3.38)

Let

f(t, u, v) =1

et(1 − t)1/8

(u−1/2 + v−1/8

), (t, u, v) ∈ (0, 1) × (0,+∞) × (0,+∞). (3.39)

Then f ∈ C((0, 1) × (0,+∞) × (0,+∞), (0,+∞)) is decreasing in u and v, and for any (u, v) ∈(0,∞) × (0,∞),

limσ→+∞

σf(t, σu, σv) = limσ→+∞

σ1/2u−1/2 + σ7/8v−1/8

et(1 − t)1/8= +∞, (3.40)

uniformly on t ∈ (0, 1). Thus (B1) holds.On the other hand, for any μ, ν > 0 and t ∈ (0, 1),

f(t, μ, ν

)=

1

et(1 − t)1/8

(μ−1/2 + ν−1/8

)/≡ 0,

L(t) =∫ t

0

(t − s)−7/8s3/8

Γ(1/8)ds =

Γ(11/8)Γ(3/2)

t1/2.

(3.41)

thus we have

∫1

0(1 − s)α−γ−1f

(s, μL(s), μG(s))ds =

∫1

0(1 − s)1/8

[1

es(1 − s)1/8

(μ−1/2L−1/2(s)

+μ−1/8G−1/8(s))]

≤∫1

0

[

μ−1/2(Γ(11/8)Γ(3/2)

t1/2)−1/2

+ μ−1/8s−3/64

]

ds

=∫1

0

[μ−1/2

(Γ(11/8)Γ(3/2)

t−1/4)+ μ−1/8s−3/64

]ds < +∞,

(3.42)


which implies that (B2) holds. From Theorem 3.1, there is a constant λ∗ > 0 such that for anyλ ∈ (λ∗,+∞) the problem (3.38) has at least one positive solution x(t) and

x(t) ≥ L(t) =Γ(11/8)Γ(3/2)

t1/2 ≈ 1.003t1/2, t ∈ [0, 1]. (3.43)

Acknowledgments

This work is supported financially by the National Natural Science Foundation of China(11071141, 11126231) and the Natural Science Foundation of Shandong Province of China(ZR2010AM017).

References

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Research ArticlePositive Solutions of a Fractional BoundaryValue Problem with Changing Sign Nonlinearity

Yongqing Wang,1 Lishan Liu,1, 2 and Yonghong Wu2

1 School of Mathematical Sciences, Qufu Normal University, Shandong, Qufu 273165, China2 Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia

Correspondence should be addressed to Lishan Liu, [email protected]

Received 12 December 2011; Revised 8 February 2012; Accepted 22 February 2012


Copyright q 2012 Yongqing Wang et al. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

We discuss the existence of positive solutions of a boundary value problem of nonlinear fractionaldifferential equation with changing sign nonlinearity. We first derive some properties of theassociated Green function and then obtain some results on the existence of positive solutions bymeans of the Krasnoselskii’s fixed point theorem in a cone.

1. Introduction

Recently, much attention has been paid to the existence of solutions for fractional differentialequations due to its wide range of applications in engineering, economics, and many otherfields, and for more details see, for instance, [1–17] and the references therein. In most of theworks in literature, the nonlinearity needs to be nonnegative to get positive solutions [10–17]. In particular, by using the Krasnosel’skii fixed-point theorem and the Leray-Schaudernonlinear alternative, Bai and Qiu [14] consider the positive solution for the followingboundary value problem:

cDα0+u(t) + f(t, u(t)) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(P)

where 2 < α ≤ 3 is a real number, cDα0+ is the Caputo fractional derivative, f : (0, 1]×[0,∞) →

[0,∞) is continuous and singular at t = 0.To the best of our knowledge, there are only very few papers dealing with the

exis-tence of positive solutions of semipositone fractional boundary value problems due


to the difficulties in finding and analyzing the corresponding Green function. The purposeof this paper is to establish the existence of positive solutions to the following nonlinearfractional differential equation boundary value problem:

cDα0+u(t) + λf(t, u(t)) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(1.1)

where 2 < α ≤ 3 is a real number, cDα0+ is the Caputo fractional derivative, λ is a positive

parameter, and f may change sign and may be singular at t = 0, 1. In this paper, by a positivesolution to (1.1), we mean a function u ∈ C[0, 1], which is positive on (0, 1] and satisfies (1.1).

The rest of the paper is organized as follows. In Section 2, we present some preliminar-ies and lemmas that will be used to prove our main results. We also develop some propertiesof the associated Green function. In Section 3, we discuss the existence of positive solutionsof the semipositone BVP (1.1). In Section 4, we give two examples to illustrate the applicationof our main results.

2. Basic Definitions and Preliminaries

In this section, we present some preliminaries and lemmas that are useful to the proof of ourmain results. For the convenience of the reader, we also present here some necessary defini-tions from fractional calculus theory. These definitions can be found in the recent literature.

Definition 2.1. The Riemann-Liouville fractional integral of order α > 0 of a function u :(0,+∞) → R is given by

Iα0+u(t) =1

Γ(α)

∫ t

0(t − s)α−1u(s)ds (2.1)

provided that the right-hand side is pointwise defined on (0,+∞).

Definition 2.2. The Caputo’s fractional derivative of order α > 0 of a function u : (0,+∞) → Ris given by

cDα0+u(t) =

1Γ(n − α)

∫ t

0(t − s)n−α−1u(n)(s)ds, (2.2)

where n − 1 < α ≤ n, provided that the right-hand side is pointwise defined on (0,+∞).

Lemma 2.3 (see [14]). Given y(t) ∈ C(0, 1) ∩ L(0, 1), the unique solution of the problem

cDα0+u(t) + y(t) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0(2.3)


is

u(t) =∫1

0G(t, s)y(s)ds, (2.4)

where

G(t, s) =1

Γ(α)

{(α − 1)t(1 − s)α−2, 0 ≤ t ≤ s ≤ 1,(α − 1)t(1 − s)α−2 − (t − s)α−1, 0 ≤ s ≤ t ≤ 1.

(2.5)

Lemma 2.4. The function G(t, s) has the following properties:

(1) G(t, s) ≤ (1/Γ(α − 1)) t(1 − s)α−2, for t,s ∈ [0, 1],

(2) G(t, s) ≤ (1/Γ(α − 1))(α − 2 + s)(1 − s)α−2, for t, s ∈ [0, 1],

(3) G(t, s) ≥ (1/Γ(α))(α − 2 + s)t(1 − s)α−2, for t, s ∈ [0, 1].

Proof. It is obvious that (1) holds. In the following, we will prove (2) and (3).(i) When 0 ≤ s ≤ t ≤ 1, as 2 < α ≤ 3, we have

∂G(t, s)∂t

=(1 − s)α−2 − (t − s)α−2

Γ(α − 1)≥ 0, (2.6)

therefore

G(t, s) ≤ G(1, s) = (α − 2 + s)Γ(α)

(1 − s)α−2 ≤ 1Γ(α − 1)

(α − 2 + s)(1 − s)α−2. (2.7)

On the other hand, since 0 < α − 2 ≤ 1, we have

G(t, s) =(α − 1)t(1 − s)α−2 − (t − s)(t − s)α−2

Γ(α)

≥ (α − 1)t(1 − s)α−2 − (t − s)(1 − s)α−2

Γ(α)

=(α − 2)t + s

Γ(α)(1 − s)α−2 ≥ (α − 2)t + st

Γ(α)(1 − s)α−2

=1

Γ(α)(α − 2 + s)t(1 − s)α−2.

(2.8)

(ii) When 0 ≤ t ≤ s ≤ 1, we have

G(t, s) =(α − 1)t(1 − s)α−2

Γ(α)≤ (α − 1)s(1 − s)α−2

Γ(α)≤ 1

Γ(α − 1)(α − 2 + s)(1 − s)α−2. (2.9)


On the other hand, as α − 1 ≥ α − 2 + s for 0 ≤ s ≤ 1, we have

G(t, s) =(α − 1)t(1 − s)α−2

Γ(α)≥ 1

Γ(α)(α − 2 + s)t(1 − s)α−2. (2.10)


Remark 2.5. By Lemma 2.4, there exists K > 0 such that the positive solution u in [14] satisfies

u(t) ≥ t

α − 1‖u‖, u(t) ≤ Kt, (2.11)

where ‖u‖ = max0≤t≤1|u(t)|.

Proof. In [14], the positive solution of (P) is equivalent to the fixed point of A in Q, where Q ={u(t) ∈ C[0, 1] : u(t) ≥ 0} and

Au(t) =∫1

0G(t, s)f(s, u(s))ds. (2.12)

For any u ∈ Q, by (1) of Lemma 2.4, we have

Au(t) ≤ t

Γ(α − 1)

∫1

0(1 − s)α−2f(s, u(s))ds. (2.13)

On the other hand, by (2), (3) of Lemma 2.4, we get

Au(t) ≥ t

Γ(α)

∫1

0(α − 2 + s)(1 − s)α−2f(s, u(s))ds,

Au(t) ≤ 1Γ(α − 1)

∫1

0(α − 2 + s)(1 − s)α−2f(s, u(s))ds,

(2.14)

which implies Au(t) ≥ (t/(α − 1))‖Au(t)‖.If u is a positive solution of (P), then u is a fixed point of A in Q, therefore

u(t) ≥ t

α − 1‖u‖, u(t) ≤ Kt, (2.15)

where K = (1/(Γ(α − 1)))∫1

0 (1 − s)α−2f(s, u(s))ds. The proof is completed.

For the convenience of presentation, we list here the hypotheses to be used later.

(H1) f ∈ C((0, 1) × [0,+∞), (−∞,+∞)) and satisfies

−r(t) ≤ f(t, x) ≤ z(t)g(x), (2.16)


where r, z ∈ C((0, 1), [0,+∞)), g ∈ C([0,+∞), [0,+∞)).

(H2) 0 <∫1

0 r(s)ds < +∞, 0 <∫1

0 (α − 2 + s)(1 − s)α−2(z(s) + r(s))ds < +∞.

(H3) There exists [a, b] ⊂ (0, 1) such that

lim infu→+∞

mint∈[a,b]

f(t, u)u

= +∞. (2.17)

(H4) There exists [c, d] ⊂ (0, 1) such that

lim infu→+∞

mint∈[c,d]

f(t, u) >2(α − 1)2 ∫1

0 (1 − s)α−2r(s)ds∫dc (α − 2 + s)(1 − s)α−2ds

,

limu→+∞

g(u)u

= 0.

(2.18)

Lemma 2.6. Assume that (H1) and (H2) hold, then the boundary value problem

cDα0+u(t) + r(t) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(2.19)

has a unique solution ω(t) =∫1

0 G(t, s)r(s)ds with

ω(t) ≤ t∫1

0

1Γ(α − 1)

(1 − s)α−2r(s)ds, t ∈ [0, 1]. (2.20)

Proof. By Lemma 2.3, we have that ω(t) =∫1

0 G(t, s)r(s)ds is the unique solution of (2.19). By(1) of Lemma 2.4, we have

ω(t) =∫1

0G(t, s)r(s)ds ≤ t

∫1

0

1Γ(α − 1)

(1 − s)α−2r(s)ds. (2.21)


Let E = C[0, 1] be endowed with the maximum norm ‖u‖ = max0≤t≤1|u(t)|. Define acone P by

P ={u(t) ∈ E : u(t) ≥ t

α − 1‖u‖}. (2.22)

Set Br = {u(t) ∈ E : ‖u‖ < r}, Pr = P ∩ Br , ∂Pr = P ∩ ∂Br .


Next we consider the following boundary value problem:

cDα0+u(t) + λ

[f(t, [u(t) − λω(t)]+) + r(t)] = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(2.23)

where λ > 0, ω(t) is defined in Lemma 2.6, [u(t) − λω(t)]+ = max{u(t) − λω(t), 0}.Let

Tu(t) = λ∫1

0G(t, s)

[f(s, [u(s) − λω(s)]+) + r(s)]ds. (2.24)

It is easy to check that u is a solution of (2.23) if and only if u is a fixed point of T .

Lemma 2.7. T : P → P is a completely continuous operator.

Proof. For any u ∈ P , Lemma 2.4 implies that

Tu(t) ≥ λt

Γ(α)

∫1

0(α − 2 + s)(1 − s)α−2[f

(s, [u(s) − λω(s)]+) + r(s)]ds. (2.25)

On the other hand

Tu(t) ≤ λ

Γ(α − 1)

∫1

0(α − 2 + s)(1 − s)α−2[f

(s, [u(s) − λω(s)]+) + r(s)]ds. (2.26)

Then Tu(t) ≥ (t/(α − 1))‖Tu(t)‖, which implies T : P → P .According to the Ascoli-Arzela theorem, we can easily get that T : P → P is a com-

pletely continuous operator. The proof is completed.

Lemma 2.8 (see [18]). Let E be a real Banach space, and let P ⊂ E be a cone. Assume thatΩ1 andΩ2

are two bounded open subsets of E with θ ∈ Ω1,Ω1 ⊂ Ω2, and T : P ∩ (Ω2 \Ω1) → P is a completelycontinuous operator such that either

(1) ‖Tu‖ ≤ ‖u‖, u ∈ P ∩ ∂Ω1 and ‖Tu‖ ≥ ‖u‖, u ∈ P ∩ ∂Ω2, or

(2) ‖Tu‖ ≥ ‖u‖, u ∈ P ∩ ∂Ω1 and ‖Tu‖ ≤ ‖u‖, u ∈ P ∩ ∂Ω2.

Then T has a fixed point in P ∩ (Ω2 \Ω1).

3. Existence of Positive Solutions

Theorem 3.1. Suppose that (H1)–(H3) hold. Then there exists λ∗ > 0 such that the boundary valueproblem (1.1) has at least one positive solution for any λ ∈ (0, λ∗).


Proof. Choose r1 > ((α − 1)/Γ(α − 1))∫1

0 (1 − s)α−2r(s)ds. Let

λ∗ = min

⎧⎨

⎩1,

r1Γ(α − 1)[g∗(r1) + 1

] ∫10 (α − 2 + s)(1 − s)α−2(z(s) + r(s))ds

⎫⎬

⎭, (3.1)

where

g∗(r) = maxx∈[0,r]

g(x). (3.2)

In the rest of the proof, we suppose λ ∈ (0, λ∗).For any u ∈ ∂Pr1 , noting that

u(t) ≥ t

α − 1r1, t ∈ [0, 1] (3.3)

and using (2.20), we have

0 ≤ t[

r1

α − 1−∫1

0

1Γ(α − 1)

(1 − s)α−2r(s)ds

]

≤ u(t) − λω(t) ≤ r1. (3.4)

Therefore,

Tu(t) = λ∫1

0G(t, s)

[f(s, [u(s) − λω(s)]+) + r(s)]ds

≤ λ

Γ(α − 1)

∫1

0(α − 2 + s)(1 − s)α−2[z(s)g

([u(s) − λω(s)]+) + r(s)]ds

≤ λ

Γ(α − 1)[g∗(r1) + 1

]∫1

0(α − 2 + s)(1 − s)α−2[z(s) + r(s)]ds

<λ∗

Γ(α − 1)[g∗(r1) + 1

]∫1

0(α − 2 + s)(1 − s)α−2[z(s) + r(s)]ds ≤ r1.

(3.5)

Thus,

‖Tu‖ ≤ ‖u‖, ∀u ∈ ∂Pr1 . (3.6)

Now choose a real number

L >4Γ(α)

λa∫ba(α − 2 + s)(1 − s)α−2ds

. (3.7)

By (H3), there exists a constant N > 0 such that for any t ∈ [a, b], x ≥N, we have

f(t, x) > Lx. (3.8)


Select

r2 > max

{

r1,2(α − 1)Γ(α − 1)

∫1

0(1 − s)α−2r(s)ds,

4Na

}

. (3.9)

Then for any u ∈ ∂Pr2 , we have u(t) − λω(t) ≥ 0, t ∈ [0, 1]. Moreover, by the selection of r2 wehave

1Γ(α − 1)

∫1

0(1 − s)α−2r(s)ds <

r2

2(α − 1). (3.10)

Thus for any t ∈ [a, b], as 1 < α − 1 ≤ 2, we get

u(t) − λω(t) ≥ t[

r2

α − 1− 1Γ(α − 1)

∫1

0(1 − s)α−2r(s)ds

]

>ar2

2(α − 1)≥ ar2

4. (3.11)

Noting that r2 > 4N/a, we have

u(t) − λω(t) > ar2

4> N, t ∈ [a, b]. (3.12)

Hence we get

Tu(1) =λ

Γ(α)

∫1

0(α − 2 + s)(1 − s)α−2[f

(s, [u(s) − λω(s)]+) + r(s)]ds

≥ λ

Γ(α)

∫b

a

(α − 2 + s)(1 − s)α−2f(s, [u(s) − λω(s)]+)ds

≥ λL

Γ(α)

∫b

a

(α − 2 + s)(1 − s)α−2[u(s) − λω(s)]ds

≥ λLar2

4Γ(α)

∫b

a

(α − 2 + s)(1 − s)α−2ds > r2.

(3.13)

Thus,

‖Tu‖ ≥ ‖u‖, ∀u ∈ ∂Pr2 . (3.14)

By Lemma 2.8, T has a fixed point u such that r1 ≤ ‖u‖ ≤ r2. Since ‖u‖ ≥ r1, by (3.4) we haveu(t) − λω(t) > 0, t ∈ (0, 1]. Let u(t) = u(t) − λω(t). As ω(t) is the solution of (2.19) and u(t) isthe solution of (2.23), u(t) is a positive solution of the singular semipositone boundary valueproblem (1.1). The proof is completed.

Theorem 3.2. Suppose that (H1), (H2), and (H4) hold. Then there exists λ∗ > 0 such that theboundary value problem (1.1) has at least one positive solution for any λ ∈ (λ∗,+∞).


Proof. By the first limit of (H4), we have that there exists N > 0 such that, for any t ∈ [c, d]and u ≥N, we have

f(t, u) ≥ 2(α − 1)2 ∫10 (1 − s)α−2r(s)ds

∫dc (α − 2 + s)(1 − s)α−2ds

. (3.15)

Select

λ∗ =NΓ(α − 1)

c∫1

0 (1 − s)α−2r(s)ds. (3.16)

In the rest of the proof, we suppose λ > λ∗.Let

R1 =2λ(α − 1)

∫10 (1 − s)α−2r(s)dsΓ(α − 1)

. (3.17)

Then, for any u ∈ ∂PR1 , we have

u(t) − λω(t) ≥ t[R1

α − 1− λ∫1

0

1Γ(α − 1)

(1 − s)α−2r(s)ds

]

=λt

Γ(α − 1)

∫1

0(1 − s)α−2r(s)ds

≥ λ∗tΓ(α − 1)

∫1

0(1 − s)α−2r(s)ds =

Nt

c,

(3.18)

and therefore u(t) − λω(t) ≥N on t ∈ [c, d], u ∈ ∂PR1 . Then,

Tu(t) = λ∫1

0G(t, s)

[f(s, [u(s) − λω(s)]+) + r(s)]ds

≥ λ∫d

c

G(t, s)f(s, [u(s) − λω(s)]+)ds

≥ 2λ(α − 1)2 ∫10 (1 − s)α−2r(s)ds

∫dc (α − 2 + s)(1 − s)α−2ds

∫d

c

G(t, s)ds

≥ 2tλ(α − 1)2 ∫10 (1 − s)α−2r(s)ds

∫dc (α − 2 + s)(1 − s)α−2ds

∫d

c

(α − 2 + s)Γ(α)

(1 − s)α−2ds

=2λt(α − 1)Γ(α − 1)

∫1

0(1 − s)α−2r(s)ds = tR1,

(3.19)

which implies

‖Tu‖ ≥ ‖u‖, ∀u ∈ ∂PR1 . (3.20)


On the other hand, as g(t) is continuous on [0,+∞), from the second limit of (H4), wehave

limu→+∞

g∗(u)u

= 0, (3.21)

where g∗(u) is defined by (3.2). In fact, by limu→+∞g(u)/u = 0, for any ε > 0, there existsN1 >0 such that for any u > N1 we have 0 ≤ g(u) < εu. LetN = max{N1, g

∗(N1)/ε}, for any u > Nwe have 0 ≤ g∗(u) < εu + g∗(N1) < 2εu. Therefore, limu→+∞g∗(u)/u = 0. For

ε = Γ(α − 1)

[

2λ∫1

0(α − 2 + s)(1 − s)α−2z(s)ds

]−1

, (3.22)

there exists X0 > 0 such that when x ≥ X0, for any 0 ≤ u ≤ x, we have

g(u) ≤ g∗(x) ≤ εx. (3.23)

Select

R2 ≥ max

{

X0, R1,2λ

Γ(α − 1)

∫1

0(α − 2 + s)(1 − s)α−2r(s)ds

}

. (3.24)

Then, for any u ∈ ∂PR2 , we get

‖Tu‖ ≤ λ

Γ(α − 1)

∫1

0(α − 2 + s)(1 − s)α−2[z(s)g

([u(s) − λω(s)]+) + r(s)]ds

≤ λεR2

Γ(α − 1)

∫1

0(α − 2 + s)(1 − s)α−2z(s)ds

+λ

Γ(α − 1)

∫1

0(α − 2 + s)(1 − s)α−2r(s)ds ≤ R2

2+R2

2= R2.

(3.25)

Thus,

‖Tu‖ ≤ ‖u‖, ∀u ∈ ∂PR2 . (3.26)

By Lemma 2.8, T has a fixed point u such that R1 ≤ ‖u‖ ≤ R2. Since R1 ≤ ‖u‖, by (3.18), wehave u(t) − λω(t) > 0, t ∈ (0, 1]. Let u(t) = u(t) − λω(t). As ω(t) is a solution of (2.19) and u(t)is a solution of (2.23), u(t) is a positive solution of the singular semipositone boundary valueproblem (1.1). The proof is completed.


Corollary 3.3. The conclusion of Theorem 3.2 is valid if (H4) is replaced by (H∗4): there exists [c, d] ⊂

(0, 1) such that

lim infu→+∞

mint∈[c,d]

f(t, u) = +∞;

limu→+∞

g(u)u

= 0.(3.27)

4. Examples


cD5/20+ u(t) + λf(t, u) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(4.1)

where f(t, u) = u2/√t(1 − t) + ln t. Let z(t) = 1/

√t(1 − t), r(t) = − ln t, g(u) = u2. By direct

calculation, we have∫1

0 r(t)dt = 1,∫1

0 z(t)dt = π , and

lim infu→+∞

mint∈[1/4,3/4]

f(t, u)u

= +∞. (4.2)

So all conditions of Theorem 3.1 are satisfied. By Theorem 3.1, BVP (4.1) has at least onepositive solution provided λ is sufficiently small.


cD9/40+ u(t) + λf(t, u) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(4.3)

where f(t, u) = ln(1+u)/√t(1 − t)+ln t. Let z(t) = 1/

√t(1 − t) , r(t) = − ln t, g(u) = ln(1+u).

By direct calculation, we have∫1

0 r(t)dt = 1,∫1

0 z(t)dt = π , and

lim infu→+∞

mint∈[1/4,3/4]

f(t, u) = +∞;

limu→+∞

g(u)u

= 0.(4.4)

So all conditions of Theorem 3.2 are satisfied. By Theorem 3.2, BVP (4.3) has at least one pos-itive solution provided λ is sufficiently large.

Acknowledgments

The first and second authors were supported financially by the National Natural ScienceFoundation of China (11071141, 11101237) and the Natural Science Foundation of Shandong


Prov-ince of China (ZR2011AQ008, ZR2011AL018). The third author was supported finan-cially by the Australia Research Council through an ARC Discovery Project Grant.

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Research ArticleThe Local Strong and Weak Solutions fora Generalized Novikov Equation

Meng Wu and Yue Zhong


Correspondence should be addressed to Meng Wu, [email protected]

Received 8 December 2011; Accepted 13 January 2012


Copyright q 2012 M. Wu and Y. Zhong. This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, andreproduction in any medium, provided the original work is properly cited.

The Kato theorem for abstract differential equations is applied to establish the local well-posednessof the strong solution for a nonlinear generalized Novikov equation in space C([0, T),Hs(R)) ∩C1([0, T),Hs−1(R)) with s > (3/2). The existence of weak solutions for the equation in lower-orderSobolev space Hs(R) with 1 ≤ s ≤ (3/2) is acquired.

1. Introduction

The Novikov equation with cubic nonlinearities takes the form

ut − utxx + 4u2ux = 3uuxuxx + u2uxxx, (1.1)

which was derived by Vladimir Novikov in a symmetry classification of nonlocal partialdifferential equations [1]. Using the perturbed symmetry approach, Novikov was able toisolate (1.1) and investigate its symmetries. A scalar Lax pair for it was discovered in [1, 2]and was shown to be related to a negative flow in the Sawada-Kotera hierarchy. Manyconserved quantities were found as well as a bi-Hamiltonian structure. The scattering theorywas employed by Hone et al. [3] to find nonsmooth explicit soliton solutions with multiplepeaks for (1.1). This multiple peak property is common with the Camassa-Holm andDegasperis-Procesi equations (see [4–10]). Ni and Zhou [11] proved that the Novikov equa-tion associated with initial value is locally well-posedness in Sobolev spaceHs with s > (3/2)by using the abstract Kato theorem. Two results about the persistence properties of the strongsolution for (1.1) were established. It is shown in [12] that the local well-posedness for theperiodic Cauchy problem of the Novikov equation in Sobolev space Hs with s > (5/2).


The orbit invariants are employed to get the existence of periodic global strong solution ifthe Sobolev index s ≥ 3 and a sign condition holds. For analytic initial data, the existence anduniqueness of analytic solutions for (1.1) are also obtained in [12].

In this paper, motivated by the work in [7, 13], we study the following generalizedNovikov equation:

ut − utxx + 4u2ux = 3uuxuxx + u2uxxx + β∂x[(ux)N

], (1.2)

where N ≥ 1 is a natural number. In fact, (1.1) has the property

∫

R

(u2 + u2

x

)dx = constant. (1.3)

Due to the term β∂x[(ux)N] appearing in (1.2), the conservation law (1.3) for (1.2) is

not valid. This brings us a difficulty to obtain the bounded estimates for the solution of (1.2).However, we will overcome this difficulty to investigate the local existence and uniquenessof the solution to (1.2) subject to initial value u0(x) ∈ Hs(R) with s > (3/2). Meanwhile, asufficient condition is presented to guarantee the existence of local weak solution for (1.2).

The main tasks of this work are two-fold. Firstly, by using the Kato theorem forabstract differential equations, we establish the local existence and uniqueness of solutionsfor the (1.2) with any β and arbitrary positive integer N in space C([0, T),Hs(R))

⋂C1([0, T),

Hs−1(R)) with s > (3/2). Secondly, it is shown that there exist local weak solutions in lower-order Sobolev space Hs(R) with 1 ≤ s ≤ (3/2). The ideas of proving the second result comefrom those presented in Li and Olver [8].

2. Main Results

Firstly, some notations are presented as follows.The space of all infinitely differentiable functions φ(t, x) with compact support in

[0,+∞)×R is denoted by C∞0 . Lp = Lp(R)(1 ≤ p < +∞) is the space of all measurable functions

h such that ‖h‖pLp =∫R |h(t, x)|pdx < ∞. We define L∞ = L∞(R) with the standard norm

‖h‖L∞ = infm(e)=0supx∈R\e|h(t, x)|. For any real number s, Hs = Hs(R) denotes the Sobolevspace with the norm defined by

‖h‖Hs =(∫

R

(1 + |ξ|2

)s∣∣∣h(t, ξ)∣∣∣

2dξ

)1/2

<∞, (2.1)


−ixξh(t, x)dx.For T > 0 and nonnegative number s, C([0, T);Hs(R)) denotes the Frechet space of all

continuous Hs-valued functions on [0, T). We set Λ = (1 − ∂2x)

1/2.The Cauchy problem for (1.2) is written in the form

ut − utxx = −43

(u3)

x+

13∂3xu

3 − 2∂x(uu2

x

)+ uuxuxx + β∂x

[(ux)N

],

u(0, x) = u0(x),(2.2)



ut + u2ux = Λ−2[−3u2ux − 3

2∂x(uu2

x

)− 1

2u3x + β∂x

[(ux)N

]],

u(0, x) = u0(x).

(2.3)

Now, we state our main results.

Theorem 2.1. Let u0(x) ∈ Hs(R) with s > (3/2). Then problem (2.2) or problem (2.3) has a uniquesolution u(t, x) ∈ C([0, T);Hs(R))

⋂C1([0, T);Hs−1(R)), where T > 0 depends on ‖u0‖Hs(R).

Theorem 2.2. Suppose that u0(x) ∈ Hs with 1 ≤ s ≤ (3/2) and ‖u0x‖L∞ < ∞. Then there exists aT > 0 such that (1.2) subject to initial value u0(x) has a weak solution u(t, x) ∈ L2([0, T],Hs) in thesense of distribution and ux ∈ L∞([0, T] × R).


Consider the abstract quasilinear evolution equation

dv

dt+A(v)v = f(v), t ≥ 0, v(0) = v0. (3.1)

Let X and Y be Hilbert spaces such that Y is continuously and densely embedded in X, andlet Q : Y → X be a topological isomorphism. Let L(Y,X) be the space of all bounded linearoperators from Y to X. If X = Y , we denote this space by L(X). We state the following con-ditions in which ρ1, ρ2, ρ3, and ρ4 are constants depending on max{‖y‖Y , ‖z‖Y}.

(I) A(y) ∈ L(Y,X) for y ∈ X with

∥∥(A(y) −A(z)

)w∥∥X ≤ ρ1‖y − z‖X‖w‖Y , y, z,w ∈ Y, (3.2)

and A(y) ∈ G(X, 1, β) (i.e., A(y) is quasi-m-accretive), uniformly on bounded setsin Y .

(II) QA(y)Q−1 = A(y) + B(y), where B(y) ∈ L(X) is bounded, uniformly on boundedsets in Y . Moreover,

∥∥(B(y) − B(z))w∥∥X ≤ ρ2‖y − z‖Y‖w‖X, y, z ∈ Y, w ∈ X. (3.3)

(III) f : Y → Y extends to a map from X into X, is bounded on bounded sets in Y , andsatisfies

∥∥f(y) − f(z)∥∥Y ≤ ρ3‖y − z‖Y , y, z ∈ Y,

∥∥f(y) − f(z)∥∥X ≤ ρ4‖y − z‖X, y, z ∈ Y.

(3.4)


Kato Theorem (see [14])

Assume that (I), (II), and (III) hold. If v0 ∈ Y , there is a maximal T > 0 depending only on‖v0‖Y and a unique solution v to problem (3.1) such that

v = v(·, v0) ∈ C([0, T);Y )⋂C1([0, T);X). (3.5)

Moreover, the map v0 → v(·, v0) is a continuous map from Y to the space

C([0, T);Y )⋂C1([0, T);X). (3.6)

For problem (2.3), we set A(u) = u2∂x, Y = Hs(R), X = Hs−1(R), Λ = (1 − ∂2x)

1/2,

f(u) = Λ−2[−3u2ux − 3

2∂x(uu2

x

)− 1

2u3x + β∂x

[(ux)N

]], (3.7)

and Q = Λ. In order to prove Theorem 2.1, we only need to check that A(u) and f(u) satisfyassumptions (I)–(III).

Lemma 3.1 (Ni and Zhou [11]). The operator A(u) = u2∂x with u ∈ Hs(R), s > (3/2) belongs toG(Hs−1, 1, β).

Lemma 3.2 (Ni and Zhou [11]). Let A(u) = u2∂x with u ∈ Hs and s > (3/2). Then A(u) ∈L(Hs,Hs−1) for all u ∈ Hs. Moreover,

‖(A(u) −A(z))w‖Hs−1 ≤ ρ1‖u − z‖Hs−1‖w‖Hs, u, z,w ∈ Hs(R). (3.8)

Lemma 3.3 (Ni and Zhou [11]). For s > (3/2), u, z ∈ Hs and w ∈ Hs−1, it holds that B(u) =[Λ, u2∂x]Λ−1 ∈ L(Hs−1) for u ∈ Hs and

‖(B(u) − B(z))w‖Hs−1 ≤ ρ2‖u − z‖Hs‖w‖Hs−1 . (3.9)

Lemma 3.4. Let r and q be real numbers such that −r < q ≤ r. Then


‖uv‖Hr+q−1/2 ≤ c‖u‖Hr‖v‖Hq , if r <12,

‖uv‖Hr1 ≤ c‖u‖L∞‖v‖Hr1 , if r1 ≤ 0.

(3.10)

The above first two inequalities of this lemma can be found in [14, 15], and the thirdinequality can be found in [7].


Lemma 3.5. Letting u, z ∈ Hs with s > (3/2), then f(u) is bounded on bounded sets in Hs andsatisfies

∥∥f(u) − f(z)∥∥Hs ≤ ρ3‖u − z‖Hs, (3.11)∥∥f(u) − f(z)∥∥Hs−1 ≤ ρ4‖u − z‖Hs−1 . (3.12)

Proof. Using the algebra property of the space Hs0 with s0 > (1/2) and s− 1 > (1/2), we have∥∥∥Λ−2

[∂x(uu2

x

)− ∂x

(zz2

x

)]∥∥∥Hs

≤ c∥∥∥uu2

x − zz2x

∥∥∥Hs−1

≤ c‖u − z‖Hs−1

∥∥∥u2

x

∥∥∥Hs−1

+ ‖z‖Hs−1

∥∥∥u2

x − z2x

∥∥∥Hs−1

≤ c‖u − z‖Hs‖u‖2Hs + ‖z‖Hs−1‖u − z‖Hs−1‖u + z‖Hs−1

≤ c‖u − z‖Hs

(‖u‖2

Hs + ‖z‖Hs−1(‖u‖Hs + ‖z‖Hs)),

∥∥∥Λ−2∂x[(ux)N − (zx)N

]∥∥∥Hs

≤ c∥∥∥(ux)N − (zx)N

∥∥∥Hs−1

≤ c‖ux − zx‖Hs−1

N−1∑

j=0

‖ux‖N−jHs−1‖zx‖jHs−1 ≤ c‖u − z‖Hs

N−1∑

j=0

‖u‖N−jHs ‖z‖jHs .

(3.13)

It follows from Lemma 3.4 and s − 1 > (1/2) that∥∥∥Λ−2

[∂x(uu2

x

)− ∂x

(zz2

x

)]∥∥∥Hs−1

≤ c∥∥∥uu2

x − zz2x

∥∥∥Hs−2

≤ c∥∥∥(u − z)u2

x

∥∥∥Hs−2

+∥∥∥z(u2x − z2

x

)∥∥∥Hs−2

≤ c‖u − z‖Hs−2

∥∥∥u2x

∥∥∥Hs−1

+ ‖z‖Hs−1

∥∥∥u2x − z2

x

∥∥∥Hs−2

≤ c‖u − z‖Hs−1‖ux‖2Hs−1 + ‖z‖Hs−1‖ux − zx‖Hs−2‖ux + zx‖Hs−1

≤ c‖u − z‖Hs−1

(‖u‖2

Hs + ‖z‖Hs−1(‖u‖Hs + ‖z‖Hs)),

∥∥∥Λ−2∂x[(ux)N − (zx)N

]∥∥∥Hs−1

≤ c∥∥∥(ux)N − (zx)N

∥∥∥Hs−2

≤ c‖ux − zx‖Hs−2

∥∥∥∥∥∥

N−1∑

j=0

uN−jx z

jx

∥∥∥∥∥∥Hs−1

≤ c‖u − z‖Hs−1

N−1∑

j=0

‖ux‖N−jHs−1‖zx‖jHs−1 ≤ c‖u − z‖Hs−1

N−1∑

j=0

‖u‖N−jHs ‖z‖jHs .

(3.14)


From (3.7), and (3.13), we know that (3.11) is valid, while inequality (3.12) follows from(3.14).

Proof of Theorem 2.1. Using the Kato Theorem, Lemmas 3.1–3.3 and 3.5, we know that system(2.2) or problem (2.3) has a unique solution

u(t, x) ∈ C([0, T);Hs(R))⋂C1([0, T);Hs−1(R)

). (3.15)

4. Existence of Weak Solutions

For s ≥ 2, using the first equation of system (1.3) derives

d

dt

∫

R

(

u2 + u2x + 2β

∫ t

0uN+1x dτ

)

dx = 0, (4.1)

from which we have the conservation law

∫

R

(

u2 + u2x + 2β

∫ t

0uN+1x dτ

)

dx =∫

R

(u2

0 + u20x

)dx. (4.2)

Lemma 4.1 (Kato and Ponce [15]). If r ≥ 0, thenHr⋂L∞ is an algebra. Moreover

‖uv‖r ≤ c(‖u‖L∞‖v‖r + ‖u‖r‖v‖L∞), (4.3)


Lemma 4.2 (Kato and Ponce [15]). Letting r > 0. If u ∈ Hr⋂W1,∞ and v ∈ Hr−1⋂L∞, then

‖[Λr , u]v‖L2 ≤ c(‖∂xu‖L∞

∥∥∥Λr−1v∥∥∥L2

+ ‖Λru‖L2‖v‖L∞

). (4.4)

Lemma 4.3. Let s ≥ (3/2) and the function u(t, x) is a solution of problem (2.2) and the initial datau0(x) ∈ Hs. Then the following results hold:

‖u‖L∞ ≤ ‖u‖H1 ≤ c‖u0‖H1 e|β|∫ t

0 ‖ux‖N−1L∞ dτ . (4.5)

For q ∈ (0, s − 1], there is a constant c such that

∫

R

(Λq+1u

)2dx ≤

∫

R

[(Λq+1u0

)2]dx

+ c∫ t

0‖u‖2

Hq+1

(‖ux‖L∞‖u‖L∞ + ‖ux‖2

L∞ + ‖ux‖N−1L∞

)dτ.

(4.6)


For q ∈ [0, s − 1], there is a constant c such that


(‖u‖L∞‖u‖H1 + ‖u‖L∞‖ux‖L∞ + ‖ux‖2

L∞ + ‖ux‖N−1L∞

). (4.7)

Proof. The identity ‖u‖2H1 =

∫R(u

2 + u2x)dx, (4.2), and the Gronwall inequality result in (4.5).

Using ∂2x = −Λ2 + 1 and the Parseval equality gives rise to

∫

R

ΛquΛq∂3x

(u3)dx = −3

∫

R

(Λq+1u

)Λq+1

(u2ux

)dx + 3

∫

R

(Λqu)Λq(u2ux

)dx. (4.8)

For q ∈ (0, s − 1], applying (Λqu)Λq to both sides of the first equation of system (2.2)and integrating with respect to x by parts, we have the identity

12

∫

R

((Λqu)2 + (Λqux)

2)dx

= −3∫

R

ΛquΛq(u2ux

)dx

−∫

R

(Λq+1u

)Λq+1

(u2ux

)dx + 2

∫

R

(Λqux)Λq(uu2

x

)dx

+∫

R

ΛquΛq(uuxuxx)dx − β∫

R

ΛquxΛq[(ux)N

]dx.

(4.9)

We will estimate the terms on the right-hand side of (4.9) separately. For the first term, byusing the Cauchy-Schwartz inequality and Lemmas 4.1 and 4.2, we have∣∣∣∣

∫

R

(Λqu)Λq(u2ux

)dx

∣∣∣∣

=∣∣∣∣

∫

R

(Λqu)[Λq(u2ux

)− u2Λqux

]dx +

∫

R

(Λqu)u2Λquxdx

∣∣∣∣

≤ c‖u‖Hq(2‖u‖L∞‖ux‖L∞‖u‖Hq + ‖ux‖L∞‖u‖L∞‖u‖Hq) + ‖u‖L∞‖ux‖L∞‖Λqu‖2L2

≤ c‖u‖2Hq‖u‖L∞‖ux‖L∞ .

(4.10)

Using the above estimate to the second term yields∣∣∣∣

∫

R

(Λq+1u

)Λq+1

(u2ux

)dx

∣∣∣∣ ≤ c‖u‖2Hq+1‖u‖L∞‖ux‖L∞ . (4.11)

For the third term, using the Cauchy-Schwartz inequality and Lemma 4.2, we obtain∣∣∣∣

∫

R

(Λqux)Λq(uu2

x

)dx

∣∣∣∣

≤ ‖Λqux‖L2

∥∥∥Λq(uu2

x

)∥∥∥L2

≤ c‖u‖Hq+1(‖uux‖L∞‖ux‖Hq + ‖ux‖L∞‖uux‖Hq)

≤ c‖u‖2Hq+1‖ux‖L∞‖u‖L∞ ,

(4.12)

in which we have used ‖uux‖Hq ≤ c‖(u2)x‖Hq ≤ c‖u‖L∞‖u‖Hq+1 .


For the fouth term in (4.9), using u(u2x)x = (uu2

x)x − uxu2x results in

∣∣∣∣

∫

R

(Λqu)Λq(uuxuxx)dx∣∣∣∣ ≤

12

∣∣∣∣

∫

R

ΛquxΛq(uu2

x

)dx

∣∣∣∣ +

12

∣∣∣∣

∫

R

ΛquΛq[uxu

2x

]dx

∣∣∣∣

= K1 +K2.

(4.13)

For K1, it follows from (4.12) that

K1 ≤ c‖u‖2Hq+1‖ux‖L∞‖u‖L∞ . (4.14)

For K2, applying Lemma 4.2 derives

K2 ≤ c‖u‖Hq

∥∥∥uxu2x

∥∥∥Hq

≤ c‖u‖Hq

(‖ux‖L∞

∥∥∥u2x

∥∥∥Hq

+ ‖ux‖Hq

∥∥∥u2x

∥∥∥L∞

)

≤ c‖u‖2Hq+1‖ux‖2

L∞ .

(4.15)

For the last term in (4.9), using Lemma 4.1 repeatedly results in

∣∣∣∣

∫

R

ΛquxΛq(ux)Ndx∣∣∣∣ ≤ c‖ux‖Hq

∥∥∥uNx∥∥∥Hq

≤ c‖u‖2Hq+1‖ux‖N−1

L∞ . (4.16)

It follows from (4.10)–(4.16) that there exists a constant c such that

12d

dt

∫

R

[(Λqu)2 + (Λqux)

2]dx ≤ c‖u‖2

Hq+1

(‖ux‖L∞‖u‖L∞ + ‖ux‖2

L∞ + ‖ux‖N−1L∞

). (4.17)

Integrating both sides of the above inequality with respect to t results in inequality (4.6).To estimate the norm of ut, we apply the operator (1 − ∂2

x)−1 to both sides of the first

equation of system (2.2) to obtain the equation

ut =(

1 − ∂2x

)−1[−4

3

(u3)

x+

13∂3x

(u3)− 2∂x

(uu2

x

)+ uuxuxx + β∂x

[(ux)N

]]. (4.18)

Applying (Λqut)Λq to both sides of (4.18) for q ∈ [0, s − 1] gives rise to

∫

R

(Λqut)2dx =

∫

R

(Λqut)Λq−2[∂x

(−4

3u3 +

13∂2xu

3 − 2uu2x + β(ux)

N)+ uuxuxx

]dτ. (4.19)


For the right-hand of (4.19), we have

∫

R

(Λqut)(

1 − ∂2x

)−1Λq∂x

(−4

3u3 − 2uu2

x

)dx

≤ c‖ut‖Hq

(∫

R

(1+ξ2

)q−1×[∫

R

[−4

3u2(ξ−η)u(η)−2uux

(ξ−η)ux

(η)]dη

]2)1/2

≤ c‖ut‖Hq‖u‖H1‖u‖Hq+1‖u‖L∞ .

(4.20)

Since∫(Λqut)

(1 − ∂2

x

)−1Λq∂2

x

(u2ux

)dx = −

∫(Λqut)Λq

(u2ux

)dx

+∫(Λqut)

(1 − ∂2

x

)−1Λq(u2ux

)dx,

(4.21)

using Lemma 4.2, ‖u2ux‖Hq ≤ c‖(u3)x‖Hq ≤ c‖u‖2L∞‖u‖Hq+1 , and ‖u‖L∞ ≤ ‖u‖H1 , we have

∫(Λqut)Λq

(u2ux

)dx ≤ c‖ut‖Hq

∥∥∥u2ux∥∥∥Hq

≤ c‖ut‖Hq‖u‖L∞‖u‖H1‖u‖Hq+1 ,

∣∣∣∣

∫(Λqut)

(1 − ∂2

x

)−1Λq(u2ux

)dx

∣∣∣∣ ≤ c‖ut‖Hq‖u‖L∞‖u‖H1‖u‖Hq+1 .

(4.22)

Using the Cauchy-Schwartz inequality and Lemmas 4.1 and 4.2 yields

∣∣∣∣

∫(Λqut)

(1 − ∂2

x

)−1Λq∂x

(uNx

)dx

∣∣∣∣ ≤ c‖ut‖Hq‖ux‖N−1L∞ ‖u‖Hq+1 , (4.23)

∣∣∣∣

∫

R

(Λqut)(

1 − ∂2x

)−1Λq(uuxuxx)dx

∣∣∣∣

≤ c‖ut‖Hq‖uuxuxx‖Hq−2

≤ c‖ut‖Hq

∥∥∥u(u2x)x∥∥∥Hq−2

≤ c‖ut‖Hq

∥∥∥[u(u2x

)]

x− (u)xu

2x

∥∥∥Hq−2

≤ c‖ut‖Hq

(∥∥∥uu2x

∥∥∥Hq−1

+∥∥∥uxu2

x

∥∥∥Hq−2

)≤ c‖ut‖Hq

(∥∥∥uu2x

∥∥∥Hq

+∥∥∥uxu2

x

∥∥∥Hq

)

≤ c‖ut‖Hq‖u‖Hq+1

(‖u‖L∞‖ux‖L∞ + ‖ux‖2

L∞

),

(4.24)

in which we have used inequality (4.15).Applying (4.20)–(4.24) into (4.19) yields the inequality


(‖u‖L∞‖u‖H1 + ‖u‖L∞‖ux‖L∞ + ‖ux‖2

L∞ + ‖ux‖N−1L∞

)(4.25)

for a constant c > 0. This completes the proof of Lemma 4.3.


Defining

φ(x) =

⎧⎨

⎩

e1/(x2−1), |x| < 1,

0, |x| ≥ 1,(4.26)

and setting φε(x) = ε−1/4φ(ε−1/4x) with 0 < ε < (1/4) and uε0 = φε u0, we know that uε0 ∈ C∞

for any u0 ∈ Hs(R) and s > 0.It follows from Theorem 2.1 that for each ε the Cauchy problem

ut − utxx = −43

(u3)

x+

13∂3xu

3 − 2∂x(uu2

x

)+ uuxuxx + β∂x

[(ux)N

]

= −43(u3)x +

13∂3xu

3 − 32∂x(uu2

x

)− 1

2u3x + β∂x

[(ux)N

],

u(0, x) = uε0(x), x ∈ R,

(4.27)

has a unique solution uε(t, x) ∈ C∞([0, T);H∞).

Lemma 4.4. Under the assumptions of problem (4.27), the following estimates hold for any ε with0 < ε < (1/4) and s > 0:

‖uε0x‖L∞ ≤ c1‖u0x‖L∞ ,

‖uε0‖Hq ≤ c1, if q ≤ s,

‖uε0‖Hq ≤ c1ε(s−q)/4, if q > s,

‖uε0 − u0‖Hq ≤ c1ε(s−q)/4, if q ≤ s,

‖uε0 − u0‖Hs = o(1),

(4.28)

where c1 is a constant independent of ε.

The proof of this Lemma can be found in Lai and Wu [7].

Lemma 4.5. If u0(x) ∈ Hs(R) with s ∈ [1, (3/2)] such that ‖u0x‖L∞ < ∞, and uε0 is defined as insystem (4.27). Then there exist two positive constants T and c, which are independent of ε, such thatthe solution uε of problem (4.27) satisfies ‖uεx‖L∞ ≤ c for any t ∈ [0, T).

Proof. Using notation u = uε and differentiating both sides of the first equation of problem(4.27) with respect to x give rise to

utx +12uu2

x + u2uxx = u3 − β(ux)N −Λ−2

[u3 +

32uu2

x +12

(u3x

)

x− β(ux)N

]. (4.29)


Integrating by parts leads to

∫

R

u2uxx(ux)p+1dx = − 1p + 1

∫

R

uu2p+3x dx, integer p > 0, (4.30)

from which we obtain

∫

R

(12uu2

x + u2uxx

)(ux)p+1dx =

p − 12(p + 1

)∫

R

uu2p+3x dx. (4.31)

Multiplying the above equation by (ux)2p+1 and then integrating the resulting equation with

respect to x yield the equality

12p + 2

d

dt

∫

R

(ux)2p+2dx +p − 1

2p + 2

∫

R

u(ux)2p+3dx

=∫

R

(ux)2p+1(u3 − βuNx

)dx

−∫

R

(ux)2p+1Λ−2[u3 +

32uu2

x +12

(u3x

)

x− βuNx

]dx.

(4.32)

Applying the Holder’s inequality yields

12p + 2

d

dt

∫

R

(ux)2p+2dx

≤((∫

R

∣∣∣u3∣∣∣

2p+2dx

)1/(2p+2)

+ β(∫

R

∣∣∣uNx∣∣∣

2p+2dx

)1/(2p+2)

+(∫

R

|G|2p+2dx

)1/(2p+2))

×(∫

R

|ux|2p+2dx

)(2p+1)/(2p+2)

+∣∣∣∣p − 1

2p + 2

∣∣∣∣‖uux‖L∞

∫

R

|ux|2p+2dx

(4.33)

or

d

dt

(∫

R

(ux)2p+2dx

)1/(2p+2)

≤(∫

R

∣∣∣u3∣∣∣

2p+2dx

)1/(2p+2)

+ β(∫

R

∣∣∣uNx∣∣∣

2p+2dx

)1/(2p+2)

+(∫

R

|G|2p+2dx

)1/(2p+2)

+∣∣∣∣p − 1

2p + 2

∣∣∣∣‖uux‖L∞

(∫

R

|ux|2p+2dx

)1/(2p+2)

,

(4.34)


where

G = Λ−2[u3 +

32uu2

x +12

(u3x

)

x− βuNx

]. (4.35)

Since ‖f‖Lp → ‖f‖L∞ as p → ∞ for any f ∈ L∞⋂L2, integrating both sides of the inequality(4.34) with respect to t and taking the limit as p → ∞ result in the estimate

‖ux‖L∞ ≤ ‖u0x‖L∞ +∫ t

0c(‖u‖3

L∞ + β‖ux‖NL∞ + ‖G‖L∞ + ‖u‖L∞‖ux‖2L∞

)dτ. (4.36)

Using the algebra property ofHs0(R) with s0 > (1/2) yields (‖uε‖H(1/2)+ means that there existsa sufficiently small δ > 0 such that ‖uε‖H(1/2)+ = ‖uε‖H(1/2)+δ):

‖G‖L∞ ≤ c‖G‖H(1/2)+

≤ c∥∥∥∥Λ

−2[u3 +

32uu2

x +12

(u3x

)

x− βuNx

]∥∥∥∥H(1/2)+

≤ c(‖u‖3

H1 +∥∥∥Λ−2

(uu2

x

)∥∥∥H(1/2)+

+∥∥∥Λ−2

(u3x

)

x

∥∥∥H(1/2)+

+∥∥∥Λ−2

(uNx

)∥∥∥H(1/2)+

)

≤ c(‖u‖3

H1 +∥∥∥uu2

x

∥∥∥H−1

+∥∥∥u3

x

∥∥∥H−(1/2)+

+∥∥∥uNx

∥∥∥H0

)

≤ c(‖u‖3

H1 + ‖uux‖H−1‖ux‖L∞ +∥∥∥u3

x

∥∥∥H0

+ ‖ux‖N−1L∞ ‖u‖H0

)

≤(‖u‖3

H1 + ‖u‖2H1‖ux‖L∞ + ‖u‖H1‖ux‖2

L∞ + ‖u‖H1‖ux‖N−1L∞

),

(4.37)

in which Lemma 3.4 is used. From Lemma 4.3, we get

∫ t

0‖G‖L∞dτ ≤ c

∫ t

0ec∫τ

0 ‖ux‖N−1L∞ dξ

(1 + ‖ux‖L∞ + ‖ux‖2

L∞ + ‖ux‖N−1L∞

)dτ. (4.38)

Using ‖u‖L∞ ≤ ‖u‖H1 , from (4.36) and (4.38), it has

‖ux‖L∞ ≤ ‖u0x‖L∞ + c∫ t

0ec∫ t

0 ‖ux‖N−1L∞ dτ

[1 + ‖ux‖L∞ + ‖ux‖2

L∞ + ‖ux‖N−1L∞ + ‖ux‖NL∞

]dτ. (4.39)

From Lemma 4.4, it follows from the contraction mapping principle that there is aT > 0 such that the equation

‖W‖L∞ = ‖u0x‖L∞ + c∫ t

0ec∫ t

0 ‖W‖N−1L∞ dτ

[1 + ‖W‖L∞ + ‖W‖2

L∞ + ‖W‖N−1L∞ + ‖W‖NL∞

]dτ (4.40)

has a unique solution W ∈ C[0, T]. Using the theorem presented at page 51 in [8] yields thatthere are constants T > 0 and c > 0 independent of ε such that ‖ux‖L∞ ≤ W(t) for arbitraryt ∈ [0, T], which leads to the conclusion of Lemma 4.5.


Using Lemmas 4.3 and 4.5, notation uε = u, and Gronwall’s inequality results in theinequalities

‖uε‖Hq ≤ CTeCT ,

‖uεt‖Hr ≤ CTeCT ,

(4.41)

where q ∈ (0, s], r ∈ (0, s − 1], and CT depends on T . It follows from Aubin’s compactnesstheorem that there is a subsequence of {uε}, denoted by {uεn}, such that {uεn} and their tem-poral derivatives {uεnt} are weakly convergent to a function u(t, x) and its derivative ut inL2([0, T],Hs) and L2([0, T],Hs−1), respectively. Moreover, for any real number R1 > 0, {uεn}is convergent to the function u strongly in the space L2([0, T],Hq(−R1, R1)) for q ∈ [0, s) and{uεnt} converges to ut strongly in the space L2([0, T],Hr(−R1, R1)) for r ∈ [0, s − 1]. Thus, wecan prove the existence of a weak solution to (1.2).

Proof of Theorem 2.2. From Lemma 4.5, we know that {uεnx} (εn → 0) is bounded in the spaceL∞. Thus, the sequences {uεn} and {uNεnx} are weakly convergent to u and uNx in L2[0, T],Hr(−R,R) for any r ∈ [0, s − 1), respectively. Therefore, u satisfies the equation

−∫T

0

∫

R

u(gt − gxxt

)dx dt =

∫T

0

∫

R

[(43u3 +

32uu2

x

)gx − 1

2u3xg(x) −

13u3gxxx − β(ux)Ngx

]dx dt

(4.42)

with u(0, x) = u0(x) and g ∈ C∞0 . Since X = L1([0, T] × R) is a separable Banach space and

{uεnx} is a bounded sequence in the dual space X∗ = L∞([0, T] × R) of X, there exists asubsequence of {uεnx}, still denoted by {uεnx}, weakly star convergent to a function v inL∞([0, T]×R). It derives from the {uεnx} weakly convergent to ux in L2([0, T]×R) that ux = valmost everywhere. Thus, we obtain ux ∈ L∞([0, T] × R).

Acknowledgment

This work is supported by the Applied and Basic Project of Sichuan Province (2012JY0020).

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Research ArticleRegularity and Exponential Growth ofPullback Attractors for Semilinear ParabolicEquations Involving the Grushin Operator

Nguyen Dinh Binh

Faculty of Applied Mathematics and Informatics, Hanoi University of Technology, 1 Dai Co Viet,Hai Ba Trung, Hanoi, Vietnam

Correspondence should be addressed to Nguyen Dinh Binh, [email protected]

Received 24 November 2011; Accepted 28 December 2011


Copyright q 2012 Nguyen Dinh Binh. This is an open access article distributed under the CreativeCommons Attribution License, which permits unrestricted use, distribution, and reproduction inany medium, provided the original work is properly cited.

Considered here is the first initial boundary value problem for a semilinear degenerate parabolicequation involving the Grushin operator in a bounded domain Ω. We prove the regularity andexponential growth of a pullback attractor in the space S2

0(Ω) ∩ L2p−2(Ω) for the nonautonomousdynamical system associated to the problem. The obtained results seem to be optimal and, inparticular, improve and extend some recent results on pullback attractors for reaction-diffusionequations in bounded domains.

1. Introduction

Let Ω be a bounded domain in RN1 × R

N2 (N1,N2 ≥ 1), with smooth boundary ∂Ω. In thispaper, we consider the following problem:

ut −Gsu + f(u) = g(t, x), (t, x) ∈ Qτ,T = (τ, T] ×Ω,

u(x, t) = 0, x ∈ ∂Ω, t ∈ (τ, T],

u(x, τ) = uτ(x), x ∈ Ω,

(1.1)

where

Gsu = Δx1u + |x1|2sΔx2u, x = (x1, x2) ∈ Ω ⊂ RN1 × R

N2 , s � 0, (1.2)


is the Grushin operator, uτ ∈ L2(Ω) is given, the nonlinearity f and the external force g satisfythe following conditions.

(H1) The nonlinearity f ∈ C1(R,R) satisfies

f(u)u ≥ C1|u|p − C2, p ≥ 2, (1.3)∣∣f ′(u)

∣∣ ≤ C3|u|p−2 + C4, (1.4)

f ′(u) ≥ −�, (1.5)

where �, Ci, (i = 1, 2, 3, 4) are positive constants. Relation (1.3) and (1.4) imply that

α1|u|p − α2 ≤ F(u) ≤ α3|u|p + α4, (1.6)

where F(s) =∫s

0 f(τ)dτ , and αi (i = 1, 2, 3, 4) are positive constants.

(H2) g ∈W1,2loc(R;L2(Ω)) satisfies

∫ t

−∞eλ1s(∥∥g(s)

∥∥2L2(Ω) +

∥∥g ′(s)∥∥2L2(Ω)

)ds < +∞, ∀t ∈ R, (1.7)

where λ1 is the first eigenvalue of the operator −Gs in Ω with the homogeneous Dirichletboundary condition.

The Grushin operator Gs was first introduced in [1]. Noting that if s > 0, then Gs isnot elliptic in domains of R

N1 × RN2 which intersect the hyperplane {x1 = 0}. In the last few

years, the existence and long-time behavior of solutions to parabolic equations involving theGrushin operator have been studied widely in both autonomous and nonautonomous cases(see, e.g., [2–7]). In particular, the existence of a pullback attractor in S1

0(Ω) ∩ Lp(Ω) for theprocess associated to problem (1.1) is considered in [2].

In this paper we continue the study in the paper [2]. First, we will prove the existenceof pullback attractors in S2

0(Ω) (see Section 2 for its definition) and L2p−2(Ω). As we know,if the external force g is only in L2(Ω), then solutions of problem (1.1) are at most inL2p−2(Ω) ∩ S2

0(Ω) and have no higher regularity. Therefore, there are no compact embeddingresults that hold for this case. To overcome the difficulty caused by the lack of embeddingresults, we exploit the asymptotic a priori estimate method which was initiated in [8, 9] forautonomous equations and developed recently for nonautonomous equations in the caseof pullback attractors in [10]. Noting that, to prove the existence of pullback attractors inS1

0(Ω) ∩ Lp(Ω), we only need assumption (H2) of the external force g; however, to prove theexistence of pullback attractors in S2

0(Ω) and L2p−2(Ω), we need an additional assumptionof g, namely, (3.18) in Section 3. Next, following the general lines of the approach in [11],we give exponential growth conditions in S2

0(Ω) ∩ L2p−2(Ω) for the pullback attractors. Itis noticed that, as far as we know, the best known results on the pullback attractors fornonautonomous reaction-diffusion equations are the boundedness and exponential growthinH2(Ω) of the pullback attractors [11, 12]. Therefore, the obtained results seem to be optimaland, in particular when s = 0, improve the recent results on pullback attractors for thenonautonomous reaction-diffusion equations in [11–15].


The content of the paper is as follows. In Section 2, for the convenience of the reader,we recall some concepts and results on function spaces and pullback attractors which wewill use. In Section 3, we prove the existence of pullback attractors in the spaces S2

0(Ω) andL2p−2(Ω) by using the asymptotic a priori estimate method. In Section 4, under additionalassumptions of g, an exponential growth in S2

0(Ω) ∩ L2p−2(Ω) for the pullback attractors isdeduced.

2. Preliminaries

2.1. Operator and Function Spaces

In order to study the boundary value problem for equations involving the Grushin operator,we have usually used the natural energy space S1

0(Ω) defined as the completion of C∞0 (Ω) in

the following norm:

‖u‖S10(Ω) =

(∫

Ω

(∣∣∇x1u∣∣2 + |x1|2s|∇x2u|2

)dx

)1/2

(2.1)

and the scalar product

((u, v)) :=(∫

Ω

(∇x1u∇x1v + |x1|2s∇x2u∇x2v

)dx

)1/2

. (2.2)

The following lemma comes from [16].

Lemma 2.1. Assume that Ω is a bounded domain in RN1 × R

N2 (N1,N2 ≥ 0). Then the followingembeddings hold:

(i) S10(Ω) ↪→ L2∗α(Ω) continuously;

(ii) S10(Ω) ↪→ Lp(Ω) compactly if p ∈ [1, 2∗s),

where 2∗s = 2N(s)/(N(s) − 2),N(s) =N1 + (s + 1)N2.

Now, we introduce the space S20(Ω) defined as the closure of C∞

0 (Ω) with the norm

‖u‖S20(Ω) =

(∫

Ω

(|Δx1u|2 + |x1|2s|Δx2u|2

)dx

)1/2

=(∫

Ω|Gsu|2dx

)1/2

. (2.3)

The following lemma comes directly from the definitions of S10(Ω) and S2

0(Ω).

Lemma 2.2. Assume thatΩ is a bounded domain inRN1 ×R

N2 (N1,N2 ≥ 0), with smooth boundary∂Ω. Then S2

0(Ω) ⊂ S10(Ω) continuously.

It is known that (see, e.g., [3]) for the operator A = −Gs, there exist {ej}j≥1


such that

(ej , ek

)= δjk, Aej = λjej , j, k = 1, 2, . . . ,

0 < λ1 ≤ λ2 ≤ λ3 ≤ · · · , λj −→ +∞ as j −→ ∞,(2.4)

and {ej}j≥1 is a complete orthonormal system in L2(Ω).

2.2. Pullback Attractors

LetX be a Banach space with the norm ‖·‖. B(X) denotes all bounded sets ofX. The Hausdorffsemidistance between A and B is defined by

dist(A,B) = supx∈A

infy∈B

∥∥x − y∥∥. (2.5)

Let {U(t, τ) : t ≥ τ, τ ∈ R} be a process in X, that is, U(t, τ) : X → X such that U(τ, τ) = Idand U(t, s)U(s, τ) = U(t, τ) for all t ≥ s ≥ τ , τ ∈ R. The process {U(t, τ)} is said to be norm-to-weak continuous if U(t, τ)xn ⇀ U(t, τ)x, as xn → x in X, for all t ≥ τ , τ ∈ R. The follow-ing result is useful for proving the norm-to-weak continuity of a process.

Proposition 2.3 (see [9]). Let X,Y be two Banach spaces, and let X∗, Y ∗ be, respectively, their dualspaces. Suppose thatX is dense in Y , the injection i : X → Y is continuous, and its adjoint i∗ : Y ∗ →X∗ is dense, and {U(t, τ)} is a continuous or weak continuous process on Y . Then {U(t, τ)} is norm-to-weak continuous on X if and only if for t ≥ τ , τ ∈ R, U(t, τ) maps compact sets of X into bound-ed sets of X.

Definition 2.4. The process {U(t, τ)} is said to be pullback asymptotically compact if for anyt ∈ R, any D ∈ B(X), any sequence τn → −∞, and any sequence {xn} ⊂ D, the sequence{U(t, τn)xn} is relatively compact in X.

Definition 2.5. A family of bounded sets B = {B(t) : t ∈ R} ⊂ X is called a pullback absorbingset for the process {U(t, τ)} if for any t ∈ R and any D ∈ B(X), there exist τ0 = τ0(D, t) ≤ tand B(t) ∈ B such that

⋃

τ≤τ0

U(t, τ)D ⊂ B(t). (2.6)

Definition 2.6. The family A = {A(t) : t ∈ R} ⊂ B(X) is said to be a pullback attractor for{U(t, τ)} if

(1) A(t) is compact for all t ∈ R,

(2) A is invariant, that is,

U(t, τ)A(τ) = A(t), ∀t ≥ τ, (2.7)


(3) A is pullback attracting, that is,

limτ→−∞

dist(U(t, τ)D,A(t)) = 0, ∀D ∈ B(X), and all t ∈ R, (2.8)

(4) if {C(t) : t ∈ R} is another family of closed pullback attracting sets, thenA(t) ⊂ C(t),for all t ∈ R.

Theorem 2.7 (see [13]). Let {U(t, τ)} be a norm-to-weak continuous process which is pullbackasymptotically compact. If there exists a pullback absorbing set B = {B(t) : t ∈ R}, then {U(t, τ)}has a unique pullback attractor A = {A(t) : t ∈ R} and

A(t) =⋂

s≤t

⋃

τ≤sU(t, τ)B(τ). (2.9)

In the rest of the paper, we denote by | · |2, (·, ·) the norm and inner product in L2(Ω),respectively, and by | · |p the norm in Lp(Ω). By ‖·‖ we denote the norm in S1

0(Ω). For a Banachspace E, ‖ · ‖E will be the norm. We also denote by C an arbitrary constant, which is differentfrom line to line, and even in the same line.

3. Existence of Pullback Attractors in S20(Ω) ∩ L2p−2(Ω)

It is well known (see, e.g., [2] or [14]) that under conditions (H1) − (H2), problem (1.1) de-fines a process

U(t, τ) : L2(Ω) −→ S10(Ω) ∩ Lp(Ω), ∀t ≥ τ, (3.1)

where U(t, τ)uτ is the unique weak solution of (1.1) with initial datum uτ at time τ . Theprocess {U(t, τ)} has a pullback attractor in S1

0(Ω) ∩ Lp(Ω).In this section, we will prove that the pullback attractor is in fact in S2

0(Ω) ∩ L2p−2(Ω).

Lemma 3.1. Assuming that f and g satisfy (H 1)-(H 2), u(t) is a weak solution of (1.1). Then thefollowing inequality holds for t > τ :

‖u‖2 + |u|pp ≤ C(

e−λ1(t−τ)|uτ |22 + 1 + e−λ1t

∫ t

−∞eλ1s∣∣g(s)

∣∣22ds

)

, (3.2)

where C is a positive constant.

Proof. Multiplying (1.1) by u and then integrating over Ω, we get

12d

dt|u|22 + ‖u‖2 +

∫

Ωf(u)udx =

∫

Ωg(t)udx ≤ 1

λ1

∣∣g(t)∣∣2

2 +λ1

4|u|22. (3.3)


Using hypothesis (H1) and the inequality ‖u‖2 ≥ λ1|u|22, we have

d

dt|u|22 + λ1|u|22 + C

(‖u‖2 + |u|pp

)≤ C(

1 +∣∣g(t)

∣∣22

). (3.4)

Letting F(s) =∫s

0 f(τ)dτ , by (H1), we have

α1|u|p − α2 ≤ F(u) ≤ α3|u|p + α4. (3.5)

Now multiplying (3.4) by eλ1t and using (3.5), we get

d

dt

(eλ1t|u(t)|22

)+ Ceλ1t

(‖u‖2 + 2

∫

ΩF(u(t))dx

)≤ Ceλ1t

(1 +∣∣g(t)

∣∣2

2

). (3.6)

Integrating (3.6) from τ to s ∈ [τ, t − 1] and s to s + 1, respectively, we obtain

eλ1s|u(s)|22 ≤ eλ1τ |uτ |22 + Ceλ1s + C∫ s

τ

eλ1r∣∣g(r)

∣∣22dr, ∀s ∈ [τ, t − 1], (3.7)

C

∫s+1

s

eλ1r

(‖u(r)‖2 + 2

∫

ΩF(u(x, r))dx

)dr

≤ eλ1s|u(s)|22 + C∫s+1

s

eλ1r(

1 +∣∣g(r)

∣∣22

)dr

≤ eλ1τ |uτ |22 + Ceλ1s + C∫s

τ

eλ1r∣∣g(r)

∣∣22dr

+Ceλ1(s+1) + C∫s+1

s

eλ1r∣∣g(r)

∣∣22dr

≤ C(

eλ1τ |uτ |22 + eλ1t +∫ t

τ

eλ1r∣∣g(r)

∣∣22dr

)

.

(3.8)

Multiplying (1.1) by ut and integrating over Ω, we have

|ut(s)|22 +12d

ds

(‖u(s)‖2 + 2

∫

ΩF(u(x, s))dx

)=∫

Ωg(s)ut(s) ≤ 1

2∣∣g(s)

∣∣22 +

12|ut(s)|22. (3.9)

Thus

eλ1s|ut(s)|22 +d

ds

[eλ1s(‖u(s)‖2 + 2

∫Ω F(u(x, s))dx

)]

≤ λ1eλ1s

(‖u(s)‖2 + 2

∫

ΩF(u(x, s))dx

)+ eλ1s

∣∣g(s)∣∣2

2.

(3.10)


Combining (3.8) and (3.10), and using the uniform Gronwall inequality, we have

eλ1t

(‖u(t)‖2 + 2

∫

ΩF(u(x, t))dx

)≤ C(

eλ1τ |uτ |2 + eλ1t +∫ t


∣∣2

2ds

)

. (3.11)

Using (H1) once again and thanks to eλ1τ |uτ |22 → 0 as τ → −∞, we get the desired resultfrom (3.11).

Lemma 3.2. Assume that (H 1), (H 2) hold. Then for any t ∈ R and anyD ⊂ L2(Ω) that is bounded,there exists τ0 ≤ t − 1 such that

|ut(t)|22 ≤ C(

1 + e−λ1t

∫ t

−∞eλ1s(∣∣g(s)

∣∣2

2 +∣∣g ′(s)

∣∣2

2

)ds

)

, (3.12)

for any τ ≤ τ0 and any uτ ∈ D, where ut(s) = (d/dt)(U(t, τ)uτ)|t=s.

Proof. Integrating (3.10) from r to r + 1, r ∈ [τ, t − 1] and using (3.8) and (3.11), in particularwe find

∫ r+1

r

eλ1s|ut|22ds ≤ eλ1r

(‖u(r)‖2 + 2

∫

ΩF(u(x, r))dx

)

+ λ1

∫ r+1

r

eλ1s

(‖u(s)‖2 + 2

∫

ΩF(u(x, s))dx

)ds

+∫ r+1

r

eλ1s∣∣g(s)

∣∣22ds ≤ C

(

eλ1τ |uτ |22 + eλ1t +∫ t


∣∣22ds

)

.

(3.13)

On the other hand, differentiating (1.1) and denoting v = ut, we have

vt −Gsv + f ′(u)v = g ′(t). (3.14)

Taking the inner product of (3.14) with v in L2(Ω), we get

12d

dt|v|22 + ‖v‖2 +

(f ′(u)v, v

)=(g ′(t), v

). (3.15)

Using (1.5) and Young’s inequality, after a few computations, we see that

d

dt

(eλ1t|v|22

)+ 2eλ1t‖v‖2 ≤ Ceλ1t|v|22 + Ceλ1t

∣∣g ′(t)∣∣2

2. (3.16)

Combining (3.16) and (3.13) and using the uniform Gronwall inequality, we obtain

eλ1t|v(t)|22 ≤ C(

eλ1τ |uτ |22 + eλ1t +∫ t

−∞eλ1s(∣∣g(s)

∣∣22 +∣∣g ′(s)

∣∣22

)ds

)

. (3.17)

The proof is now complete because eλ1τ |uτ |22 → 0 as τ → −∞.


3.1. Existence of a Pullback Attractor inL2p−2(Ω)

In this section, following the general lines of the method introduced in [9], we prove theexistence of a pullback attractor in L2p−2(Ω). In order to do this, we need an additional con-dition of g

∫ t

−∞eλ1t∥∥g ′(t)

∥∥m′

Lm(Ω)dt < +∞, ∀t ∈ R, (3.18)

where m, m′ are defined as in (3.30).

Lemma 3.3. The process {U(t, τ)} associated to problem (1.1) has a pullback absorbing set inL2p−2(Ω).

Proof. Multiplying (1.1) by |u|p−2u and integrating over Ω, we get

∫

Ωut|u|p−2udx +

(p − 1

)∫

Ω

(|∇x1u|2 + |x1|2s|∇x2u|2

)|u|p−2dx +

∫

Ωf(u)

∣∣∣|u|p−2udx∣∣∣

=∫

Ω

∣∣g(t)u∣∣p−2

udx.

(3.19)

From (1.3) and the fact that Lp(Ω) ⊂ Lp−2(Ω) continuously, we have

∫

Ωf(u)|u|p−2udx ≥

∫

Ω

(C1|u|p − C2

)|u|p−2dx ≥ C1‖u‖2p−2L2p−2(Ω) − C2|u|pp. (3.20)

On the other hand, by Cauchy’s inequality, we see that

∣∣∣∣

∫

Ωut|u|p−2udx

∣∣∣∣ ≤C1

4‖u‖2p−2

L2p−2(Ω) +1C1

|ut|22, (3.21)

∣∣∣∣

∫

Ωg(t)|u|p−2udx

∣∣∣∣ ≤C1

4‖u‖2p−2

L2p−2(Ω) +14∣∣g(t)

∣∣22. (3.22)

Combining (3.19)–(3.22) imply that

‖u‖2p−2L2p−2(Ω) ≤ C

(|ut|22 + |u|pp +

∣∣g(t)∣∣2

2

). (3.23)

Applying (3.2) and Lemma 3.2, we conclude the existence of a pullback absorbing set inL2p−2(Ω) for the process U(t, τ).


Lemma 3.4. For any s ∈ R, any 2 ≤ p < ∞, and any bounded set B ⊂ L2(Ω), there exists τ0 suchthat

∫

Ω|ut(s)|pdx ≤M, ∀τ ≤ τ0, uτ ∈ B, (3.24)

whereM depends on s, p but not on B, and ut(s) = (d/dt)(U(t, τ)uτ)|t=s.

Proof. We will prove the lemma by induction argument. Letting β =N(s)/(N(s)−2) > 1 anddenoting v = ut we prove that for k = 0, 1, 2, . . ., there exist τk and Mk(s) such that

eλ1s

∫

Ω|v(s)|2βkdx ≤Mk(s) for any uτ ∈ B, τ ≤ τk, (Pk)

∫s+1

s

(eλ1r

∫

Ω|v(r)|2βk+1

dx

)1/β

dr ≤Mk(s) for any uτ ∈ B, τ ≤ τk, (Qk)

where τk depends on k and B and Mk depends only on k.For k = 0, we have (P0) from (3.17). Integrating (3.16) and using S1

0(Ω) ↪→ L2β(Ω)continuously, we get (Q0).

Assuming that (Pk), (Qk) hold, we prove so are (Pk+1) and (Qk+1). Multiplying (3.14)by |v|2βk+1−2v and integrating over Ω, we obtain

Cd

dt

∫

Ω|v|2βk+1

dx + C∫

Ω

(|∇x1v|2 + |x1|2s|∇x2v|2

)|v|β2k+1−2dx

≤ �∫

Ω|v|2βk+1

dx +(g ′(t), |v|2βk+1−2v

).

(3.25)

Using the imbedding S10(Ω) ↪→ L2β(Ω) once again, we get

∫

Ω

(|∇x1v|2 + |x1|2s|∇x2v|2

)|v|β2k+1

dx ≥∫

Ω|v|2|v|2β2k+1−2dx

=∥∥∥vβ

k+1∥∥∥

2

L2β(Ω)=(∫

Ω|v|2βk+2

dx

)1/β

.

(3.26)

Combining Holder’s and Young’s inequalities, we see that

∫

Ωg ′(t)|v|2βk+1−2vdx ≤

(∫

Ω

∣∣g ′(t)∣∣mdx

)1/m(∫

Ω|v|(2βk+1−1)ndx

)1/n

≤(∫

Ω

∣∣g ′(t)∣∣mdx

)m′/m

m′ +

(∫Ω |v|(2βk+1−1)ndx

)n′/n

n′,

(3.27)


where 1/m + 1/n = 1/m′ + 1/n′ = 1. Choose n, n′ such that

(2βk+1 − 1

)n = 2βk+2,

n′

n=

1β, (3.28)

thus

n =2βk+2

2βk+1 − 1, n′ =

2βk+1

2βk+1 − 1. (3.29)

Hence

m =n

n − 1=

2βk+1 − 12βk+2 − 2βk+1 + 1

, m′ = 2βk+1. (3.30)

Then from (3.27), we infer that

∫

Ωg ′(t)|v|2βk+1−2vdx ≤ 1

m′∥∥g ′(t)

∥∥m′

Lm(Ω) +1n′

(∫

Ω|v|2βk+2

dx

)1/β

. (3.31)

Applying (3.26) and (3.31) in (3.25), we find that

d

dt

(eλ1t

∫

Ω|v|2βk+1

dx

)+ Ceλ1t

(∫

Ω|v|2βk+2

dx

)1/β

≤ Ceλ1t

∫

Ω|v|2βk+1

dx + Ceλ1t∥∥g ′(t)

∥∥m′

Lm(Ω).

(3.32)

Combining (Qk) and (3.32), using the uniform Gronwall inequality and taking into accountassumption (3.18), we get (Pk+1). On the other hand, integrating (3.32) from t to t+ 1, we find(Qk+1). Now since β > 1, and taking k ≥ logβp/2, we get the desired estimate.

We will use the following lemma.

Lemma 3.5 (see [15]). If there exists σ > 0 such that∫ t−∞ eσs|ϕ(s)|2ds <∞, for all t ∈ R, then

limγ→+∞

∫ t

−∞e−γ(t−s)

∣∣ϕ(s)∣∣2ds = 0, t ∈ R. (3.33)

Let Hm = span{e1, e2, . . . , em} in L2(Ω), and let Pm : L2(Ω) → Hm be the orthogonalprojection, where {ei}∞i=1 are the eigenvectors of operator A = −Gs. For any u ∈ L2(Ω), wewrite

u = Pmu + (I − Pm)u = u1 + u2. (3.34)


Lemma 3.6. For any t ∈ R, any B ⊂ L2(Ω) and any ε, there exist τ0(t, B, ε) andm0 ∈ N such that

|(I − Pm)v|22 < ε, ∀τ ≤ τ0, ∀uτ ∈ B, m ≥ m0. (3.35)

Proof. Multiplying (3.14) by v2 = (I−Pm)v and then integrating over Ω, using |∇v2|22 ≥ λm|v2|22and Cauchy’s inequality we get

d

dt|v2|22 + λm|v2|22 ≤ C

∫

Ω

∣∣f ′(u)v

∣∣2dx + C

∣∣g ′(t)

∣∣2

2. (3.36)

We multiply (3.36) by eλmt and use assumption (1.4). We get

d

dt

(eλmt|v2|22

)≤ Ceλmt

∫

Ω|u|2(p−2)|v|2dx + Ceλmt

∣∣g ′(t)∣∣2

2. (3.37)

Integrating (3.37) from s to t,

eλmt|v2(t)|22 ≤ eλms|v2(s)|22 + C∫ t

s

eλmr∫

Ω|u|2(p−2)|v|2dx dr + C

∫ t

s

eλmr∣∣g ′(r)

∣∣22dr

≤ eλms|v(s)|22 + C∫ t

−∞eλmr

∫

Ω|u|2(p−2)|v|2dx dr + C

∫ t

−∞eλmr∣∣g ′(r)

∣∣22dr.

(3.38)

Now integrating (3.38) with respect to s from τ to t, we infer that

(t − τ)eλmt|v2(t)|22 ≤∫ t

τ

eλmr |v(r)|22dr + C(t − τ)∫ t

−∞eλmr

∫

Ω|u|2(p−2)|v|2dx dr

+C(t − τ)∫ t

−∞eλmr∣∣g ′(r)

∣∣22dr.

(3.39)

Thus

|v2(t)|22 ≤ 1t − τ

∫ t

−∞e−λm(t−r)|v(r)|22dr + C

∫ t

−∞e−λm(t−r)

∫


+ C∫ t


∣∣g ′(r)∣∣2

2dr.

(3.40)

By Lemma 3.5 and since λm → +∞ as m → +∞, there exist τ1 and m1 such that

1t − τ

∫ t

−∞e−λm(t−r)C

∫ t


∣∣g ′(r)∣∣2

2dr <ε

3,

|v(r)|22dr <ε

3,

(3.41)


for all τ ≤ τ1 and m ≥ m1. For the second term of the right-hand side of (3.40), using Holder’sinequality we have

∫ t


∫


≤∫ t

−∞

(∫

Ωe((−p−1)/(p−2)λm )(t−r)|u|2p−2dx

)(p−2)/(p−1)(∫

Ωe−(p−1)λm(t−r)|v|2p−2dx

)1/(p−1)

dr

≤(∫ t

−∞e((−p−1)/(p−2)λm )(t−r)‖u‖2p−2

L2p−2(Ω)dr

)(p−2)/(p−1)(∫ t

−∞e−(p−1)λm(t−r)dr

∫

Ω|v|2p−2dx

)1/(p−1)

.

(3.42)

From Lemmas (3.5)–(3.7), we see that there exist τ2 and m2 ∈ N such that

C

∫ t


∫

Ω|u|2(p−2)|v|2dx dr < ε

3, ∀τ ≤ τ0, m ≥ m2. (3.43)

Let τ0 = min{τ1, τ2} andm0 = max{m1, m2}, from (3.40), taking into account (3.41) and (3.43),we obtain (3.35).

Lemma 3.7 (see [9]). Let B be a bounded subset in Lq(Ω)(q ≥ 1). If B has a finite ε-net in Lq(Ω),then there exists anM =M(B, ε), such that for any u ∈ B, the following estimate is valid:

∫

Ω(|u|≥M)|u|qdx < ε. (3.44)

Using Lemma 3.7 and taking into account Lemmas 3.2 and 3.6 we conclude that theset {ut(s) : s ≤ t, uτ ∈ B} has a finite ε-net in L2(Ω). Therefore, we get the following result.

Lemma 3.8. For any t ∈ R, any B ⊂ L2(Ω) that is bounded, and any ε > 0, there exists τ0 ≤ t andM0 > 0 such that

∫

Ω(|u|≥M)|ut(t)|2dx < ε, ∀τ < τ0, M > M0, uτ ∈ B. (3.45)

Lemma 3.9 (see [9]). For any t ∈ R, any bounded set B ⊂ L2(Ω), and any ε > 0, there exist τ0 andM0 > 0 such that

mes(Ω(u(t) ≥M)) < ε ∀τ ≤ τ0, M ≥M0, uτ ∈ B, (3.46)

where mes is the Lebesgue measure in RN and Ω(u(t) ≥M) = {x ∈ Ω : u(t, x) ≥M}.

Lemma 3.10 (see [2]). Let {U(t, τ)} be a norm-to-weak continuous process in L2(Ω) and Lq(Ω), q ≥2. Then {U(t, τ)} is pullback asymptotically compact in Lq(Ω) if


(i) {U(t, τ)} is pullback asymptotically compact in L2(Ω);

(ii) for any t ∈ R, any bounded set D ⊂ L2(Ω), and any ε > 0, there existM > 0 and τ0 ≤ t

supτ≤τ0

supuτ∈D

(∫

Ω(|U(t,τ)uτ |≥M)|U(t, τ)uτ |qdx

)

≤ Cε, (3.47)

where C is independent ofM, τ , uτ , and ε.

We are now ready to prove the existence of a pullback attractor in L2p−2(Ω).

Theorem 3.11. Assume that assumptions (1.3)–(1.7) and (3.18) hold. Then the process {U(t, τ)}associated to problem (1.1) possesses a pullback attractorA2p−2 = {A2p−2(t)}t∈R

in L2p−2(Ω).

Proof. Because of Lemma 3.10, since {U(t, τ)} has a pullback absorbing set in L2p−2(Ω), weonly have to prove that for any t ∈ R, any B ⊂ L2(Ω), and any ε > 0, there exist τ2 ≤ t andM2 > 0 such that

∫

Ω(|u|≥M)|u|2p−2dx ≤ Cε, ∀τ ≤ τ2, M ≥M2, uτ ∈ B. (3.48)

Taking the inner product of (1.1) with (u −M)p−1+ in L2(Ω), where

(u −M)+ =

{u −M if u ≥M,

0 if u < M,(3.49)

we have∫

Ωut(u −M)p−1

+ dx +(p − 1

)∫

Ω

(|∇x1u|2dx + |x1|2s|∇x2u|2

)(u −M)p−2

+ dx

+∫

Ωf(u)(u −M)p−1

+ dx

≤∫

Ωg(t)(u −M)p−1

+ dx.

(3.50)

Some standard computations give us

∫

Ωf(u)(u −M)p−1

+ dx ≥ C∫

Ω(u≥M)|u|2p−2dx + C

∫

Ω(u≥M)|u|pdx, (3.51)

−∫

Ωut(u −M)p−1

+ dx ≤ C

4

∫

Ω(u≥M)|u|2p−2dx +

1C

∫

Ω(u≥M)|ut|2dx, (3.52)

∫

Ωg(t)(u −M)p−1

+ dx ≤ C

4

∫

Ω(u≥M)|u|2p−2dx +

1C

∫

Ω(u≥M)

∣∣g(t)∣∣2dx. (3.53)


Combining (3.50)–(3.53), we find

∫

Ω(u≥M)|u|2p−2dx ≤ C

(∫

Ω(u≥M)|ut|2dx +

∫

Ω(u≥M)

∣∣g(t)

∣∣2dx + C

∫

Ω(u≥M)|u|pdx

)

. (3.54)

Applying Lemmas 3.7 and 3.8 to (3.54) we find there exist τ0 and M0 such that

∫

Ω(u≥M)|u|2p−2dx < ε ∀τ ≤ τ0, M ≥M0. (3.55)

Repeating the above arguments with |(u +M)−|p−2(u +M)− in place of (u −M)p−1+ , we have

∫

Ω(u≤−M)|u|2p−2dx < ε ∀τ ≤ τ1, M ≥M1, (3.56)

for some τ1 ≤ t and M1 > 0, where

(u +M)− =

⎧⎨

⎩

u +M if u ≤ −M,

0 if u > M.(3.57)

Letting τ2 = min{τ0, τ1} and M2 = max{M0,M1} we have

∫

Ω(|u|≥M2)|u|2p−2dx < Cε, ∀τ ≤ τ2, M ≥M2. (3.58)


3.2. Existence of a Pullback Attractor in S20(Ω)

In this section, we prove the existence of a pullback attractor in S20(Ω).

Lemma 3.12. The process {U(t, τ)} associated to (1.1) has a pullback absorbing set in S20(Ω).

Proof. We multiply (1.1) by −Gsu; then, using f(0) = 0, we have

‖u‖2S2

0(Ω) =∫

ΩutGsudx −

∫

Ωf ′(u)

(|∇x1u|2 + |x1|2s|∇x2u|2

)dx −

∫

Ωg(t)Gsudx. (3.59)

Using f ′(u) ≥ −�, Cauchy’s inequality, and argument as in Lemma 3.3, from (3.59) we have

‖u‖2S2

0(Ω) ≤ 2(|ut|22 + �‖u‖2 +

∣∣g(t)∣∣2

2

). (3.60)

Taking into account (3.11), the proof is complete.


In order to prove the existence of the pullback attractor in S20(Ω), we will verify so-

called “(PDC) condition”, which is defined as follow

Definition 3.13. A process {U(t, τ)} is said to satisfy (PDC) condition in X if for any t ∈ R, anybounded set B ⊂ L2(Ω) and any ε > 0, there exists τ0 ≤ t and a finite dimensional subspaceX1 of X such that

(i) P(⋃τ≤τ0

U(t, τ)B) is bounded in X; and

(ii) ‖(IX−P)U(t, τ)uτ‖X < ε, for all τ ≤ τ0 and uτ ∈ B, where P : X → X1 is a canonicalprojection and IX is the identity.

Lemma 3.14 (see [13]). If a process {U(t, τ)} satisfies (PDC) condition in X then it is pullbackasymptotically compact in X. Moreover, if X is convex then the converse is true.

Lemma 3.15 (see [9]). Assume that f satisfies (1.3) and (1.5). Then for any subset A ⊂ L2p−2(Ω),if κ(A) < ε in L2p−2(Ω), then we have

κ(f(A)

)< Cε in L2(Ω), (3.61)

where the Kuratowski noncompactness measure κ(B) in a Banach space X defined as

κ(B) = inf{δ : B has a finite covering by balls in X with radii δ}. (3.62)

Theorem 3.16. Assume that f satisfies (1.3)–(1.5), g satisfies (1.7) and (3.18). Then the process{U(t, τ)} generated by (1.1) has a pullback attractorA = {A(t) : t ∈ R} in S2

0(Ω).

Proof. We consider a complete trajectory u(t) lies on pullback attractor A2p−2 in L2p−2(Ω) forU(t, τ), that is, u(t) ∈ A2p−2(t) and U(t, τ)uτ = u(t), for all t ≥ τ . Denoting A = −Gs andmultiplying (1.1) by Au2 = A(I − Pm)u = (I − Pm)Au we have

(ut,Au2) + ‖u2‖2S2

0(Ω) +∫

Ωf(u)Au2dx =

(g(t), Au2

). (3.63)

Using Holder’s inequality we get

‖u2‖2S2

0(Ω) ≤ C(|(I − Pm)ut|22 +

∣∣(I − Pm)g(t)∣∣2

2 +∫

Ω

(f(u)

)2dx

). (3.64)

Thanks to Lemmas 3.6 and 3.15 and the fact that g ∈ Cloc(R;L2(Ω)), we see that {U(t, τ)}satisfies condition (PDC) in S2

0(Ω). Now from Lemmas 3.3 and 3.14 we get the desired result.

4. Exponential Growth in S20(Ω) ∩ L2p−2(Ω) of Pullback Attractors

In this section, we will give an exponential growth condition in S20(Ω) ∩ L2p−2(Ω) for the

pullback attractor A(τ).First, we recall a result in [17] which is necessary for the proof of our results.


Lemma 4.1. Let X, Y be Banach spaces such that X is reflexive, and the inclusion X ⊂ Y is con-tinuous. Assume that {un} is bounded sequence in L∞(t0, T ;X) such that un → u weakly inLq(t0, T ;X) for some q ∈ [1,+∞) and u ∈ C0([t0, T];Y ). Then, u(t) ∈ X for all t ∈ [t0, T] and

‖u‖X ≤ supn≥1

‖un‖L∞(t0,T ;X), ∀t ∈ [t0,T]. (4.1)

In the following theorem, instead of evaluating the functions un which are differen-tiable enough and then using Lemma 4.1, we will formally evaluate the function u.

Theorem 4.2. Assume that f satisfies (1.3)-(1.5), g satisfies (H 2), (3.18) and the following con-ditions

limτ→−∞

eλ1τ

∫ τ+1

τ

∣∣g ′(s)∣∣2

2ds = 0,

limτ→−∞

(eλ1τ∣∣g(s)

∣∣22

)= 0.

(4.2)

ThenA(τ) satisfies

limτ→−∞

eλ1τ

{

supw∈A(τ)

‖w‖2p−2L2p−2(Ω) + sup

w∈A(τ)‖w‖2

S20(Ω)

}

= 0. (4.3)

Proof. We differentiate with respect to time in (1.1), then multiply by ut, we get

12d

dr

∣∣u′(r)∣∣2

2 + ‖u(r)‖2 = −∫

Ωf ′(u)u′(r)u′(r)dx +

∫

Ωg ′(r)u′(r)dx

≤ �∣∣u′(r)∣∣22 +12u′(r)2

2 +12∣∣g ′(r)

∣∣22.

(4.4)

Integrating in the last inequality, in particular, we get

∣∣u′(r)∣∣2

2 ≤ ∣∣u′(s)∣∣22 + (2� + 1)∫ t

τ+ε/2

∣∣u′(θ)∣∣2

2dθ +∫ t

τ+ε/2

∣∣g ′(θ)∣∣2

2dθ. (4.5)

for all τ + ε/2 ≤ s ≤ τ ≤ t. Now, integrating with respect to s, between τ + ε/2 and r

(r − τ − ε

2

)∣∣u′(r)∣∣2

2 ≤[(r − τ − ε

2

)(2l + 1) + 1

] ∫ t

τ+ε/2

∣∣u′(θ)∣∣2

2dθ

+(r − τ − ε

2

)∫ t

τ+ε/2

∣∣g ′(θ)∣∣2

2dθ.

(4.6)


for all τ + ε/2 ≤ r ≤ t, in paricular,

∣∣u′(r)

∣∣2

2 ≤ 2ε−1[(r − τ − ε

2

)(2l + 1) + 1

] ∫ t

τ+ε/2

∣∣u′(θ)

∣∣2

2dθ

+∫ t

τ+ε/2

∣∣g ′(θ)

∣∣2

2dθ.

(4.7)

for all r ∈ [τ + ε, t].Multiplying (1.1) by u and then integrating on Ω, we get

12d

dt|u|22 + ‖u‖2 +

∫

Ωf(u)udx =

∫

Ωg(t)udx ≤ 1

2λ1

∣∣g(t)

∣∣2

2 +λ1

2|u|22. (4.8)

Using hypothesis (H1) and the fact that ‖u‖2 ≥ (1/2)‖u‖2 + (λ1/2)|u|22, we have

d

dt|u|22 + λ1|u|22 + 2C1|u|pp − 2C2|Ω| ≤ 1

λ1

∣∣g(t)∣∣2

2. (4.9)

Integrating (4.9) from τ to r ∈ [τ, t], we have

|u(r)|22 + λ1

∫ r

τ

‖u(s)‖2ds + 2C1

∫ r

τ

|u(s)|ppds ≤ |u(τ)|22 +1λ1

∫ r

τ

∣∣g(s)∣∣2

2ds + 2C2|Ω|(t − τ). (4.10)

Thus,

|u(r)|22 +∫ r

τ

‖u(s)‖2ds +∫ r

τ

|u(s)|ppds ≤ C[|u(τ)|22 +

∫ r

τ

∣∣g(s)∣∣2

2ds + (t − τ)]. (4.11)

Multiplying (1.1) by ut then integrating over Ω, we have

∣∣u′(r)

∣∣22 +

12d

dr

(‖u(r)‖2 + 2

∫

ΩF(u(x, r))dx

)≤ 1

2∣∣g(r)

∣∣22 +

12∣∣u′(r)

∣∣22. (4.12)

Integrating now between s ∈ [τ, r] and r ≤ t, we obtain

∫ r

s

∣∣u′(θ)∣∣2

2dθ + ‖u(r)‖2 +∫

ΩF(u(x, r))dx ≤

∫ r

s

∣∣g(θ)∣∣2

2dθ + ‖u(s)‖2 +∫

ΩF(u(x, s))dx. (4.13)

From (4.13) and using hypothesis (H1), we get

∫ r

s

∣∣u′(θ)∣∣2

2dθ + ‖u(r)‖2 + α1

∫

Ω|u(x, r)|pdx − α2|Ω|

≤ ‖u(s)‖2 +∫ r

s

∣∣g(θ)∣∣2

2dθ + α3

∫

Ω|u(x, r)|pdx + α4|Ω|.

(4.14)


Hence

∫ r

s

∣∣u′(θ)

∣∣2

2dθ + ‖u(r)‖2 + α1u(r)pp ≤ ‖u(s)‖2 +

∫ t

s

∣∣g(θ)

∣∣2

2dθ + α3|u(s)|pp + (α2 + α4)|Ω|. (4.15)

Integrating inequality (4.15) with respect to s from τ to r, we obtain

(r − τ)[‖u(r)‖2 + |u(r)|pp

]

≤ C[∫ r

τ

‖u(s)‖2ds +∫ r

τ

|u(s)|ppds]+ C(t − τ)

∫ t

τ

∣∣g(s)

∣∣2

2ds + C|Ω|(t − τ),(4.16)

for all t ≥ τ , r ∈ [τ, t].From (4.16) and (4.11), we obtain that

‖u(r)‖2 + |u(r)|pp ≤ C[

|uτ |22 +∫ τ+2

τ

∣∣g(s)∣∣2

2ds + 1

]

, (4.17)

for all r ∈ [τ + 1, τ + 2].From (4.7), taking t = τ + 3 and ε = 2 we have

∣∣u′(r)∣∣2

2 ≤ (4l + 3)∫ τ+3

τ+1

∣∣u′(θ)∣∣2

2dθ +∫ τ+3

τ+1

∣∣g ′(θ)∣∣2

2dθ (4.18)

for all r ∈ [τ + 2, τ + 3].Analogously, and if we take s = τ + 1 and r = t = τ + 3 in inequlity (4.15), then

∫ τ+3

τ+1

∣∣u′(s)∣∣2

2ds + ‖u(τ + 3)‖2 + α1|u(τ + 3)|pp

≤ ‖u(τ + 1)‖2 +∫ τ+3

τ

∣∣g(s)∣∣2

2ds + α3|u(τ + 1)|pp + (α2 + α4)|Ω|.(4.19)

From (4.18) and (4.19), we obtain

∣∣u′(r)∣∣2

2 ≤ (4l + 3)[‖u(τ + 1)‖2 + α3|u(τ + 1)|pp

]

+ (4l + 3)

[

(α2 + α4)|Ω| +∫ τ+3

τ

∣∣g(s)∣∣2

2ds

]

+∫ τ+3

τ+1

∣∣g ′(θ)∣∣2

2dθ

(4.20)

for all r ∈ [τ + 2, τ + 3].Owing to this inequality and (4.17), we have

∣∣u′(r)∣∣2

2 ≤ C[

|uτ |22 +∫ τ+3

τ

[∣∣g(s)∣∣2

2 +∣∣g ′(s)

∣∣22

]ds + 1

]

, (4.21)

for all r ∈ [τ + 2, τ + 3].


From (3.23), (3.60) and using Young’s inequality, we have

‖u(r)‖2p−2L2p−2(Ω) + ‖u(r)‖2

S20(Ω) ≤ C

(∣∣u′(r)

∣∣2

2 + ‖u(r)‖2 + |u(r)|pp + 1 +∣∣g(r)

∣∣2

2

), (4.22)

for all r ≥ τ .From (4.22) and thank to (4.21), we have

‖u(r)‖2p−2L2p−2(Ω) + ‖u(r)‖2

S20(Ω) ≤ C

[

|uτ |22 +∫ τ+3

τ

[∣∣g(s)

∣∣2

2 +∣∣g ′(s)

∣∣2

2

]ds + 1

]

+ C(‖u(r)‖2 + |u(r)|pp +

∣∣g(r)

∣∣2

2

)(4.23)

for all r ∈ [τ + 2, τ + 3]. From (4.23) and thanks to (4.17), we have

‖u(r)‖2p−2L2p−2(Ω) + ‖u(r)‖2

S20(Ω) ≤ C

[

|uτ | +∫ τ+3

τ

[∣∣g(s)∣∣2

2 +∣∣g ′(s)

∣∣22

]ds + 1 + sup

r∈[τ+2,τ+3]

∣∣g(r)∣∣2

2

]

,

(4.24)

for all r ∈ [τ + 2, τ + 3]. Now, observe that by Cauchy’s inequality

∣∣g(r)∣∣ ≥ ∣∣g(τ + 2)

∣∣ +

(∫ τ+3

τ+2

∣∣g ′(s)∣∣2

2ds

)1/2

, (4.25)

for all r ∈ [τ + 2, τ + 3], τ ∈ R, uτ ∈ L2(Ω).Thus, from (4.23), we have

‖U(τ + 2, τ)uτ‖2p−2L2p−2(Ω) + ‖U(τ + 2, τ)uτ‖2

S20(Ω)

≤ C[

|uτ |22 +∫ τ+3

τ

(∣∣g(s)∣∣2

2 +∣∣g ′(s)

∣∣22

)ds + 1 +

∣∣g(τ + 2)∣∣2

2

]

.

(4.26)

for all τ ∈ R,uτ ∈ L2(Ω). From this inequlity, and the fact that A(τ) = U(τ, τ − 2)A(τ − 2), weobtain

‖v‖2p−2L2p−2(Ω) + ‖v‖2

S20(Ω) ≤ C

[

supw∈A(τ−2)

|w|22 +∫ τ+1

τ−2

(∣∣g(s)∣∣2

2 +∣∣g ′(s)

∣∣22

)ds + 1 +

∣∣g(τ)∣∣2

2

]

. (4.27)

for all v ∈ A(τ), and any τ ∈ R. Now, thank to (4.2), (4.3), we obtain (4.3) from (4.27).

Acknowledgment

The authors thanks the reviewers very much for valuable comments and suggestions.


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