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Boundary Value Problems

Volume 2009

Hindawi Publishing Corporationhttp://www.hindawi.com

Editor-in-Chief: Ravi P. Agarwal

Special IssueSingular Boundary Value Problems for Ordinary Differential Equations

Guest EditorsJuan J. Nieto and Donal O’Regan

Singular Boundary Value Problems forOrdinary Differential Equations

Boundary Value Problems

Singular Boundary Value Problems forOrdinary Differential Equations

Guest Editors: Juan J. Nieto and Donal O’Regan

Copyright © 2009 Hindawi Publishing Corporation. All rights reserved.

This is a special issue published in volume 2009 of “Boundary Value Problems.” All articles are open access articles distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided theoriginal work is properly cited.

Editor-in-ChiefRavi P. Agarwal, Florida Institute of Technology, USA

Associate Editors

Ugur G. Abdulla, USAPeter Bates, USAMichel C. Chipot, SwitzerlandEmmanuele Dibenedetto, USAPavel Drabek, Czech RepublicLawrence C. Evans, USARobert Finn, USAAvner Friedman, USARobert Bob Gilbert, USANobuyuki Kenmochi, Japan

Ivan T. Kiguradze, GeorgiaV. Lakshmikantham, USAGary M. Lieberman, USARaul F. Manasevich, ChileJean Mawhin, BelgiumPatrick Joseph McKenna, USASalim A. Messaoudi, Saudi ArabiaDonal O’Regan, IrelandKanishka Perera, USAIrena Rachunkova, Czech Republic

Vicentiu D. Radulescu, RomaniaColin Rogers, AustraliaSandro Salsa, ItalyMartin D. Schechter, USAVeli B. Shakhmurov, TurkeyRoger M. Temam, USARoberto Triggiani, USAZhitao Zhang, ChinaWenming Zou, China

Contents

Singular Boundary Value Problems for Ordinary Differential Equations, Juan J. Nieto and Donal O’ReganVolume 2009, Article ID 895290, 2 pages

A Viral Infection Model with a Nonlinear Infection Rate, Yumei Yu, Juan J. Nieto, Angela Torres,and Kaifa WangVolume 2009, Article ID 958016, 19 pages

An Existence Result for Nonlinear Fractional Differential Equations on Banach Spaces,Mouffak Benchohra, Alberto Cabada, and Djamila SebaVolume 2009, Article ID 628916, 11 pages

Antiperiodic Boundary Value Problems for Finite Dimensional Differential Systems, Y. Q. Chen,D. O’Regan, F. L. Wang, and S. L. ZhouVolume 2009, Article ID 541435, 11 pages

Constant Sign and Nodal Solutions for Problems with the p-Laplacian and a Nonsmooth Potential UsingVariational Techniques, Ravi P. Agarwal, Michael E. Filippakis, Donal O’Regan,and Nikolaos S. PapageorgiouVolume 2009, Article ID 820237, 32 pages

Electroelastic Wave Scattering in a Cracked Dielectric Polymer under a Uniform Electric Field,Yasuhide Shindo and Fumio NaritaVolume 2009, Article ID 949124, 27 pages

Existence and Exponential Stability of Positive Almost Periodic Solutions for a Model of Hematopoiesis,J. O. Alzabut, J. J. Nieto, and G. Tr. StamovVolume 2009, Article ID 127510, 10 pages

Existence and Uniqueness of Positive and Nondecreasing Solutions for a Class of Singular FractionalBoundary Value Problems, J. Caballero Mena, J. Harjani, and K. SadaranganiVolume 2009, Article ID 421310, 10 pages

Existence and Uniqueness of Smooth Positive Solutions to a Class of Singular m-Point Boundary ValueProblems, Xinsheng Du and Zengqin ZhaoVolume 2009, Article ID 191627, 13 pages

Existence of Periodic Solution for a Nonlinear Fractional Differential Equation, Mohammed Belmekki,Juan J. Nieto, and Rosana Rodrıguez-LopezVolume 2009, Article ID 324561, 18 pages

Existence of Positive Solutions for Multipoint Boundary Value Problem on the Half-Line with Impulses,Jianli Li and Juan J. NietoVolume 2009, Article ID 834158, 12 pages

Existence of Solutions for Fractional Differential Inclusions with Antiperiodic Boundary Conditions,Bashir Ahmad and Victoria Otero-EspinarVolume 2009, Article ID 625347, 11 pages

Existence Results for Nonlinear Boundary Value Problems of Fractional Integrodifferential Equationswith Integral Boundary Conditions, Bashir Ahmad and Juan J. NietoVolume 2009, Article ID 708576, 11 pages

First-Order Singular and Discontinuous Differential Equations, Daniel C. Biles and Rodrigo Lopez PousoVolume 2009, Article ID 507671, 25 pages

Homoclinic Solutions of Singular Nonautonomous Second-Order Differential Equations,Irena Rachunkova and Jan TomecekVolume 2009, Article ID 959636, 21 pages

Limit Properties of Solutions of Singular Second-Order Differential Equations, Irena Rachunkova,Svatoslav Stanek, Ewa Weinmuller, and Michael ZenzVolume 2009, Article ID 905769, 28 pages

Multiplicity Results Using Bifurcation Techniques for a Class of Fourth-Order m-Point Boundary ValueProblems, Yansheng Liu and Donal O’ReganVolume 2009, Article ID 970135, 20 pages

Multipoint Singular Boundary-Value Problem for Systems of Nonlinear Differential Equations,Jaromır Bastinec, Josef Diblık, and Zdenek SmardaVolume 2009, Article ID 137451, 20 pages

New Results on Multiple Solutions for Nth-Order Fuzzy Differential Equations under GeneralizedDifferentiability, A. Khastan, F. Bahrami, and K. IvazVolume 2009, Article ID 395714, 13 pages

On Some Generalizations Bellman-Bihari Result for Integro-Functional Inequalities for DiscontinuousFunctions and Their Applications, Angela Gallo and Anna Maria PiccirilloVolume 2009, Article ID 808124, 14 pages

On Step-Like Contrast Structure of Singularly Perturbed Systems, Mingkang Ni and Zhiming WangVolume 2009, Article ID 634324, 17 pages

Positive Solutions for a Class of Coupled System of Singular Three-Point Boundary Value Problems,Naseer Ahmad Asif and Rahmat Ali KhanVolume 2009, Article ID 273063, 18 pages

Positive Solutions of Singular Multipoint Boundary Value Problems for Systems of NonlinearSecond-Order Differential Equations on Infinite Intervals in Banach Spaces, Xingqiu ZhangVolume 2009, Article ID 978605, 22 pages

Positive Solutions to Singular and Delay Higher-Order Differential Equations on Time Scales,Liang-Gen Hu, Ti-Jun Xiao, and Jin LiangVolume 2009, Article ID 937064, 19 pages

Contents

Recent Existence Results for Second-Order Singular Periodic Differential Equations, Jifeng Chu andJuan J. NietoVolume 2009, Article ID 540863, 20 pages

Robust Monotone Iterates for Nonlinear Singularly Perturbed Boundary Value Problems, Igor BoglaevVolume 2009, Article ID 320606, 17 pages

Hindawi Publishing CorporationBoundary Value ProblemsVolume 2009, Article ID 895290, 2 pagesdoi:10.1155/2009/895290

EditorialSingular Boundary Value Problems forOrdinary Differential Equations

Juan J. Nieto1 and Donal O’Regan2

1 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Santiago de Compostela,15782 Santiago de Compostela, Spain

2 Department of Mathematics, National University of Ireland, Galway, Ireland

Correspondence should be addressed to Juan J. Nieto, [email protected]

Received 31 December 2009; Accepted 31 December 2009

Copyright q 2009 J. J. Nieto and D. O’Regan. 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.

Singular boundary value problems for ordinary differential equations model many realworld phenomena ranging from different physics equations to biological, physiological, andmedical processes [1–3].

This special issue places its emphasis on the study, theory, and applications ofboundary value problems involving singularities. It includes some review articles such as [4],equations with discontinuous nonlinearities [5], boundary value problems with uncertainty[6], fractional differential equations [7], periodic or antiperiodic solutions [8], and biological[9] or medical applications [10]. Different methods and techniques are used ranging fromvariational methods [11] to bifurcation techniques [12].

The editors aimed at a volume that may serve as a reference in the topic of the specialissue and collect twenty five original and cutting-edge research articles by some of the topresearchers in boundary value problems for ordinary differential equations worldwide andfrom many different countries (Algeria, Austria, Bulgaria, China, Czech Republic, Greece,Iran, Ireland, Italy, Japan, New Zealand, Pakistan, Saudi Arabia, South Korea, Spain, USA).

We would like to thank the authors for their contributions, the Editor-in-Chief of thejournal, Professor Ravi P. Agarwal, and the Editorial Office of the journal for their support.

References

[1] R. P. Agarwal and D. O’Regan, Singular Differential and Integral Equations with Applications, KluwerAcademic Publishers, Dordrecht, The Netherlands, 2003.

[2] A. Cabada, E. Liz, and J. J. Nieto, Mathematical Models in Engineering, Biology and Medicine, AmericanInstitute of Physics, Melville, NY, USA, 2009.

[3] D. O’Regan, Theory of Singular Boundary Value Problems, World Scientific Publishing, River Edge, NJ,USA, 1994.

2 Boundary Value Problems

[4] J. Chu and J. J. Nieto, “Recent existence results for second-order singular periodic differentialequations,” Boundary Value Problems, vol. 2009, Article ID 540863, 20 pages, 2009.

[5] D. C. Biles and R. Lopez-Pouso, “First-order singular and discontinuous differential equations,”Boundary Value Problems, vol. 2009, Article ID 507671, 25 pages, 2009.

[6] A. Khastan, F. Bahrami, and K. Ivaz, “New results on multiple solutions for Nth-order fuzzydifferential equations under generalized differentiability,” Boundary Value Problems, vol. 2009, ArticleID 395714, 13 pages, 2009.

[7] M. Belmekki, J. J. Nieto, and R. Rodrıguez-Lopez, “Existence of periodic solution for a nonlinearfractional differential equation,” Boundary Value Problems, vol. 2009, Article ID 324561, 18 pages, 2009.

[8] Y. Q. Chen, D. O’Regan, F. L. Wang, and S. L. Zhou, “Antiperiodic boundary value problems for finitedimensional differential systems,” Boundary Value Problems, vol. 2009, Article ID 541435, 11 pages,2009.

[9] Y. Yu, J. J. Nieto, A. Torres, and K. Wang, “A viral infection model with a nonlinear infection rate,”Boundary Value Problems, vol. 2009, Article ID 958016, 19 pages, 2009.

[10] J. O. Alzabut, J. J. Nieto, and G. Tr. Stamov, “Existence and exponential stability of positive almostperiodic solutions for a model of hematopoiesis,” Boundary Value Problems, vol. 2009, Article ID127510, 10 pages, 2009.

[11] R. P. Agarwal, M. E. Filippakis, D. O’Regan, and N. S. Papageorgiou, “Constant sign and nodalsolutions for problems with the p-Laplacian and a nonsmooth potential using variational techniques,”Boundary Value Problems, vol. 2009, Article ID 820237, 32 pages, 2009.

[12] Y. Liu and D. O’Regan, “Multiplicity results using bifurcation techniques for a class of fourth-orderm-point boundary value problems,” Boundary Value Problems, vol. 2009, Article ID 970135, 20 pages,2009.


Research ArticleA Viral Infection Model with a NonlinearInfection Rate

Yumei Yu,1 Juan J. Nieto,2 Angela Torres,3 and Kaifa Wang4

1 School of Science, Dalian Jiaotong University, Dalian 116028, China2 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Santiago de Compostela,15782 Santioga de compostela, Spain

3 Departamento de Psiquiatrıa, Radiologıa y Salud Publica, Facultad de Medicina,Universidad de Santiago de Compostela, 15782 Santioga de compostela, Spain

4 Department of Computers Science, Third Military Medical University, Chongqing 400038, China

Correspondence should be addressed to Kaifa Wang, [email protected]

Received 28 February 2009; Revised 23 April 2009; Accepted 27 May 2009

Recommended by Donal O’Regan

A viral infection model with a nonlinear infection rate is constructed based on empirical evidences.Qualitative analysis shows that there is a degenerate singular infection equilibrium. Furthermore,bifurcation of cusp-type with codimension two (i.e., Bogdanov-Takens bifurcation) is confirmedunder appropriate conditions. As a result, the rich dynamical behaviors indicate that the modelcan display an Allee effect and fluctuation effect, which are important for making strategies forcontrolling the invasion of virus.

Copyright q 2009 Yumei Yu 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.

1. Introduction

Mathematical models can provide insights into the dynamics of viral load in vivo. A basicviral infection model [1] has been widely used for studying the dynamics of infectious agentssuch as hepatitis B virus (HBV), hepatitis C virus (HCV), and human immunodeficiencyvirus (HIV), which has the following forms:

dxdt

= λ − dx − βxv,

dydt

= βxv − ay,

dvdt

= ky − uv,

(1.1)


where susceptible cells (x(t)) are produced at a constant rate λ, die at a density-dependentrate dx, and become infected with a rate βuv; infected cells (y(t)) are produced at rate βuvand die at a density-dependent rate ay; free virus particles (v(t)) are released from infectedcells at the rate ky and die at a rate uv. Recently, there have been many papers on virusdynamics within-host in different aspects based on the (1.1). For example, the influences ofspatial structures on virus dynamics have been considered, and the existence of travelingwaves is established via the geometric singular perturbation method [2]. For more literature,we list [3, 4] and references cited therein.

Usually, there is a plausible assumption that the amount of free virus is simplyproportional to the number of infected cells because the dynamics of the virus is substantiallyfaster than that of the infected cells, u � a, k � λ. Thus, the number of infected cells y(t)can also be considered as a measure of virus load v(t) (e.g., see [5–7]). As a result, the model(1.1) is reduced to

dxdt

= λ − dx − βxy,

dydt

= βxy − ay.(1.2)

As for this model, it is easy to see that the basic reproduction number of virus is given byR0 = βλ/ad, which describes the average number of newly infected cells generated fromone infected cell at the beginning of the infectious process. Furthermore, we know that theinfection-free equilibrium E0 = (λ/d, 0) is globally asymptotically stable if R0 < 1, and so isthe infection equilibrium E1 = (a/β, (βλ − ad)/aβ) if R0 > 1.

Note that both infection terms in (1.1) and (1.2) are based on the mass-action principle(Perelson and Nelson [8]); that is, the infection rate per susceptible cell and per virus is aconstant β. However, infection experiments of Ebert et al. [9] and McLean and Bostock [10]suggest that the infection rate of microparasitic infections is an increasing function of theparasite dose and is usually sigmoidal in shape. Thus, as Regoes et al. [11], we take thenonlinear infection rate into account by relaxing the mass-action assumption that is madein (1.2) and obtain

dxdt

= λ − dx − β(y)x,

dydt

= β(y)x − ay,

(1.3)

where the infection rate per susceptible cell, β(y), is a sigmoidal function of the virus(parasite) concentration because the number of infected cells y(t) can also be considered as ameasure of virus load (e.g., see [5–7]), which is represented in the following form:

β(y)=

(y/ID50)κ

1 + (y/ID50)κ , κ > 1. (1.4)

Here, ID50 denotes the infectious dose at which 50% of the susceptible cells are infected, κmeasures the slope of the sigmoidal curve at ID50 and approximates the average number

Boundary Value Problems 3

of virus that enters a single host cell at the begin stage of invasion, (y/ID50)κ measures

the infection force of the virus, and 1/(1 + (y/ID50)κ) measures the inhibition effect from

the behavioral change of the susceptible cells when their number increases or from theproduction of immune response which depends on the infected cells.

In fact, many investigators have introduced different functional responses into relatedequations for epidemiological modeling, of which we list [12–17] and references cited therein.However, a few studies have considered the influences of nonlinear infection rate on virusdynamics. When the parameter κ = 1, [18, 19] considered a viral mathematical model withthe nonlinear infection rate and time delay. Furthermore, some different types of nonlinearfunctional responses, in particular of the form βxqy or Holling-type functional response, wereinvestigated in [20–23].

Note that κ > 1 in (1.4). To simplify the study, we fix the slope κ = 2 in the presentpaper, and system (1.3) becomes

dxdt

= λ − dx −y2

ID250 + y2

x,

dydt

=y2

ID250 + y2

x − ay.

(1.5)

To be concise in notations, rescale (1.5) by X = x/ID50, Y = y/ID50. For simplicity, we stilluse variables x, y instead of X, Y and obtain

dxdt

= m − dx −y2

1 + y2x,

dydt

=y2

1 + y2x − ay,

(1.6)

where m = λ/ID50. Note that 1/d is the average life time of susceptible cells and 1/a isthe average life-time of infected cells. Thus, a ≥ d is always valid by means of biologicaldetection. If a = d, the virus does not kill infected cells. Therefore, the virus is non cytopathicin vivo. However, when a > d, which means that the virus kills infected cells before itsaverage life time, the virus is cytopathic in vivo.

The main purpose of this paper is to study the effect of the nonlinear infection rateon the dynamics of (1.6). We will perform a qualitative analysis and derive the Allee-typedynamics which result from the appearance of bistable states or saddle-node state in (1.6).The bifurcation analysis indicates that (1.6) undergoes a Bogdanov-Takens bifurcation at thedegenerate singular infection equilibrium which includes a saddle-node bifurcation, a Hopfbifurcation, and a homoclinic bifurcation. Thus, the nonlinear infection rate can induce thecomplex dynamic behaviors in the viral infection model.

The organization of the paper is as follows. In Section 2, the qualitative analysis ofsystem (1.6) is performed, and the stability of the equilibria is obtained. The results indicatethat (1.6) can display an Allee effect. Section 3 gives the bifurcation analysis, which indicatesthat the dynamics of (1.6) is more complex than that of (1.1) and (1.2). Finally, a briefdiscussion on the direct biological implications of the results is given in Section 4.


2. Qualitative Analysis

Since we are interested in virus pathogenesis and not initial processes of infection, we assumethat the initial data for the system (1.6) are such that

x(0) > 0, y(0) > 0. (2.1)

The objective of this section is to perform a qualitative analysis of system (1.6) and derivethe Allee-type dynamics. Clearly, the solutions of system (1.6) with positive initial valuesare positive and bounded. Let g(y) = y/(1 + y2), and note that (1.6) has one and only oneinfection-free equilibrium E0 = (m/d, 0). Then by using the formula of a basic reproductionnumber for the compartmental models in van den Driessche and Watmough [24], we knowthat the basic reproduction number of virus of (1.6) is

R0 =1a· md· g(0) = 0, (2.2)

which describes the average number of newly infected cells generated from one infected cellat the beginning of the infectious process as zero. Although it is zero, we will show that thevirus can still persist in host.

We start by studying the equilibria of (1.6). Obviously, the infection-free equilibriumE0 = (m/d, 0) always exists and is a stable hyperbolic node because the correspondingcharacteristic equation is (ω + d)(ω + a) = 0.

In order to find the positive (infection) equilibria, set

m − dx −y2

1 + y2x = 0,

y

1 + y2x − a = 0,

(2.3)

then we have the equation

a(1 + d)y2 −my + ad = 0. (2.4)

Based on (2.4), we can obtain that

(i) there is no infection equilibria if m2 < 4a2d(1 + d);

(ii) there is a unique infection equilibrium E1 = (x∗, y∗) if m2 = 4a2d(1 + d);

(iii) there are two infection equilibria E11 = (x1, y1) and E12 = (x2, y2) if m2 > 4a2d(1+d).


Here,

y∗ =m

2a(1 + d), x∗ =

a(1 + y∗2

)

y∗,

y1 =m −

√m2 − 4a2d(1 + d)

2a(1 + d), x1 =

a(

1 + y21

)

y1,

y2 =m +

√m2 − 4a2d(1 + d)

2a(1 + d), x2 =

a(

1 + y22

)

y2.

(2.5)

Thus, the surface

SN ={(m,d, a) : m2 = 4a2d(1 + d)

}(2.6)

is a Saddle-Node bifurcation surface, that is, on one side of the surface SN system (1.6) has notany positive equilibria; on the surface SN system (1.6) has only one positive equilibrium; onthe other side of the surface SN system (1.6) has two positive equilibria. The detailed resultswill follow.

Next, we determine the stability of E11 and E12. The Jacobian matrix at E11 is

JE11=

⎡

⎢⎢⎢⎢⎢⎢⎢⎣

−d −y2

1

1 + y21

−2x1y1(

1 + y21

)2

y21

1 + y21

−a +2x1y1(

1 + y21

)2

⎤

⎥⎥⎥⎥⎥⎥⎥⎦

. (2.7)

After some calculations, we have

det(JE11

)= −

a(1 + d)(

4a2d(1 + d) +m

(√m2 − 4a2d(1 + d) −m

))

2a2(1 + d) +m

(m −

√m2 − 4a2d(1 + d)

) . (2.8)

Since m2 > 4a2d(1 + d) in this case, 4a2d(1 + d) + m(√m2 − 4a2d(1 + d) − m)) > 0 is valid.

Thus, det(JE11) < 0 and the equilibrium E11 is a saddle.

The Jacobian matrix at E12 is

JE12=

⎡

⎢⎢⎢⎢⎢⎢⎢⎣

−d −y2

2

1 + y22

−2x2y2(

1 + y22

)2

y22

1 + y22

−a +2x2y2(

1 + y22

)2

⎤

⎥⎥⎥⎥⎥⎥⎥⎦

. (2.9)


By a similar argument as above, we can obtain that det(JE12) > 0. Thus, the equilibrium E12 is

a node, or a focus, or a center.For the sake of simplicity, we denote

mε = 2a√d(1 + d),

m0 =a2(1 + 2d)

√(a − d)(1 + a + d)

, if a > 2d(1 + d).(2.10)

We have the following results on the stability of E12.

Theorem 2.1. Suppose that equilibrium E12 exists; that is, m > mε. Then E12 is always stable ifd ≤ a ≤ 2d(1 + d). When a > 2d(1 + d), we have

(i) E12 is stable ifm > m0;

(ii) E12 is unstable ifm < m0;

(iii) E12 is a linear center ifm = m0.

Proof. After some calculations, the matrix trace of JE12is

tr(JE12

)=

2a3(1 + d)(1 + 2d) −m(1 + a + d)(m +

√m2 − 4a2d(1 + d)

)

2a2(1 + d) +m(m +

√m2 − 4a2d(1 + d)

) , (2.11)

and its sign is determined by

F(m) � 2a3(1 + d)(1 + 2d) −m(1 + a + d)(m +

√m2 − 4a2d(1 + d)

). (2.12)

Note that

F ′(m) = −(1 + a + d)

(

2m +√m2 − 4a2d(1 + d) +

m2

√m2 − 4a2d(1 + d)

)

< 0, (2.13)

which means that F(m) is a monotone decreasing function of variable m.Clearly,

F(mε) = 2a2(1 + d)(a − 2d(1 + d))

⎧⎨

⎩

> 0, if a > 2d(1 + d),

≤ 0, if a ≤ 2d(1 + d).(2.14)

Note that F(m) = 0 implies that

2a3(1 + d)(1 + 2d)m(1 + a + d)

−m =√m2 − 4a2d(1 + d). (2.15)


Squaring (2.15) we find that

4a6(1 + d)2(1 + 2d)2

m2(1 + a + d)2− 4a3(1 + d)(1 + 2d)

1 + a + d+m2 = m2 − 4a2d(1 + d). (2.16)

Thus,

a4(1 + d)(1 + 2d)2

m2(1 + a + d)2=

a(1 + 2d)1 + a + d

− d =(a − d)(1 + d)

1 + a + d,

m =a2(1 + 2d)

√(a − d)(1 + a + d)

.

(2.17)

This means that F(m0) = 0. Thus, under the condition of m > mε and the sign of F(m),tr (JE12) < 0 is always valid if a ≤ 2d(1 + d). When a > 2d(1 + d), tr(JE12) < 0 if m > m0,tr(JE12) > 0 if m < m0, and tr(JE12) = 0 if m = m0.

For (1.6), its asymptotic behavior is determined by the stability of E12 if it does nothave a limit cycle. Next, we begin to consider the nonexistence of limit cycle in (1.6).

Note that E11 is a saddle and E12 is a node, a focus, or a center. A limit cycle of (1.6)must include E12 and does not include E11. Since the flow of (1.6) moves toward down on theline where y = y1 and x < x1 and moves towards up on the line where y = y1 and x > x1,it is easy to see that any potential limit cycle of (1.6) must lie in the region where y > y1.Take a Dulac function D = (1 + y2)/y2, and denote the right-hand sides of (1.6) by P1 and P2,respectively. We have

∂(DP1)∂x

+∂(DP2)

∂y= −

(1 + a + d)y2 − (a − d)y2

, (2.18)

which is negative if y2 > (a − d)/(1 + a + d). Hence , we can obtain the following result.

Theorem 2.2. There is no limit cycle in (1.6) if

y21 >

(a − d)(1 + a + d)

. (2.19)

Note that y1 > 0 as long as it exists. Thus, inequality (2.19) is always valid if a =d. When a > d, using the expression of y1 in (2.5), we have that inequality (2.19) that isequivalent to

2a3(1 + d)(1 + 2d)1 + a + d

< m2 <a4(1 + 2d)2

(a − d)(1 + a + d). (2.20)


Indeed, since

y21 =

m2

2a2(1 + d)2− d

1 + d− m√m2 − 4a2d(1 + d)

2a2(1 + d)2,

m2

2a2(1 + d)2− d

1 + d− a − d

1 + a + d=

m2

2a2(1 + d)2− a(1 + 2d)(1 + d)(1 + a + d)

,

(2.21)

we have (2.19) that is equivalent to

m2

2a2(1 + d)2− a(1 + 2d)(1 + d)(1 + a + d)

>m√m2 − 4a2d(1 + d)

2a2(1 + d)2, (2.22)

that is,

m2 − 2a3(1 + d)2(1 + 2d)(1 + d)(1 + a + d)

> m√m2 − 4a2d(1 + d). (2.23)

Thus,

m2 >2a3(1 + d)2(1 + 2d)(1 + d)(1 + a + d)

. (2.24)

On the other hand, squaring (2.23) we find that

m4 − 4a3(1 + d)2(1 + 2d)(1 + d)(1 + a + d)

m2 +4a6(1 + d)4(1 + 2d)2

(1 + d)2(1 + a + d)2> m4 − 4a2d(1 + d)m2, (2.25)

which is equivalent to

m2 <a4(1 + 2d)2

(a − d)(1 + a + d). (2.26)

The combination of (2.24) and (2.26) yields (2.20).Furthermore,

4a2d(1 + d) <a4(1 + 2d)2

(a − d)(1 + a + d)(2.27)


is equivalent to a/= 2d(1 + d), both

2a3(1 + d)(1 + 2d)1 + a + d

<a4(1 + 2d)2

(a − d)(1 + a + d),

2a3(1 + d)(1 + 2d)1 + a + d

< 4a2d(1 + d)

(2.28)

are equivalent to a < 2d(1 + d). Consequently, we have the following.

Corollary 2.3. There is no limit cycle in (1.6) if either of the following conditions hold:

(i) a = d and m2 > 4a2d(1 + d);

(ii) d < a < 2d(1 + d) and 4a2d(1 + d) < m2 < a4(1 + 2d)2/(a − d)(1 + a + d).

When m2 = 4a2d(1 + d), system (1.6) has a unique infection equilibrium E1. TheJacobian matrix at E1 is

JE1=

⎡

⎢⎢⎢⎢⎢⎢⎣

−d −y∗2

1 + y∗2−

2x∗y∗(1 + y∗2

)2

y∗2

1 + y∗2−a +

2x∗y∗(1 + y∗2

)2

⎤

⎥⎥⎥⎥⎥⎥⎦

. (2.29)

The determinant of JE1is

det(JE1

)= −

a(1 + d)(4a2d(1 + d) −m2)

m2 + 4a2(1 + d)2= 0, (2.30)

and the trace of JE1is

tr(JE1

)=

4a2(1 + d)(a − 2d(1 + d))

m2 + 4a2(1 + d)2. (2.31)

Thus, E1 is a degenerate singular point. Since its singularity, complex dynamic behaviors mayoccur, which will be studied in the next section.

3. Bifurcation Analysis

In this section, the Bogdanov-Takens bifurcation (for short, BT bifurcation) of system (1.6) isstudied when there is a unique degenerate infection equilibrium E1.


For simplicity of computation, we introduce the new time τ by dt = (1+y2)dτ , rewriteτ as t, and obtain

dxdt

= m − dx +my2 − (1 + d)xy2,

dydt

= −ay + xy2 − ay3.

(3.1)

Note that (3.1) and (1.6) are C∞-equivalent; both systems have the same dynamics (only thetime changes).

As the above mentioned, assume that

(H1) m2 = 4a2d(1 + d).

Then (3.1) admits a unique positive equilibrium E1 = (x∗, y∗), where

x∗ =2a2(1 + 2d)

m, y∗ =

m

2a(1 + d). (3.2)

In order to translate the positive equilibrium E1 to origin, we set X = x−x∗, Y = y−y∗and obtain

dXdt

= −2dX − 2aY − 2a2(1 + d)m

Y 2 − m

aXY − (1 + d)XY 2,

dYdt

=d

1 + dX + 2dY +

m

a(1 + d)XY +

2a2(1 − d)m

Y 2 +XY 2 − aY 3.

(3.3)

Since we are interested in codimension 2 bifurcation, we assume further that

(H2) a = 2d(1 + d).

Then, after some transformations, we have the following result.

Theorem 3.1. The equilibrium E1 of (1.6) is a cusp of codimension 2 if (H1) and (H2) hold; that is,it is a Bogdanov-Takens singularity.

Proof. Under assumptions (H1) and (H2), it is clear that the linearized matrix of (3.3)

M =

⎡

⎢⎣−2d −2a

d

1 + d2d

⎤

⎥⎦ (3.4)

has two zero eigenvalues. Let x = X, y = −2dX − 2aY . Since the parameters m, a, d satisfythe assumptions (H1) and (H2), after some algebraic calculations, (3.3) is transformed into

dxdt

= y +md

2a2x2 − 1 + d

2my2 + f1

(x, y),

dydt

=md2(2d + 1)

a2x2 +

2md2

a2xy +

m(2d − 1)4a2

y2 + f2(x, y),

(3.5)


where fi(x, y), i = 1, 2, are smooth functions in variables (x, y) at least of the third order.Using an affine translation u = x + y/2d, v = y to (3.5), we obtain

dudt

= v +m

2au2 − m

a2uv + f1(u, v),

dvdt

=md2(2d + 1)

a2u2 − md

a2uv + f2(u, v),

(3.6)

where fi(u, v), i = 1, 2, are smooth functions in variables (u, v) at least of order three. Toobtain the canonical normal forms, we perform the transformation of variables by

x = u +m

2a2u2, y = v +

m

2au2. (3.7)

Then, (3.6) becomes

dxdt

= y + F1(x, y),

dydt

=md2(2d + 1)

a2x2 +

md(2d + 1)a2

xy + F2(x, y),

(3.8)

where Fi(x, y), i = 1, 2, are smooth functions in (x, y) at least of the third order.Obviously,

md2(2d + 1)a2

> 0,

md(2d + 1)a2

> 0.

(3.9)

This implies that the origin of (3.3), that is, E1 of (1.6), is a cusp of codimension 2 by in [25,Theorem 3, Section 2.11].

In the following we will investigate the approximating BT bifurcation curves. Theparameters m and a are chosen as bifurcation parameters. Consider the following perturbedsystem:

dxdt

= m0 + λ1 − dx −xy2

1 + y2,

dydt

=xy2

1 + y2− (a0 + λ2)y,

(3.10)

where m0, a0 and d are positive constants while (H1) and (H2) are satisfied. That is to say,

m20 = 4a2

0d(1 + d), a0 = 2d(1 + d). (3.11)


λ1 and λ2 are in the small neighborhood of (0, 0); x and y are in the small neighborhood of(x∗, y∗), where

x∗ =2a2

0(1 + 2d)m0

, y∗ =m0

2a0(1 + d). (3.12)

Clearly, if λ1 = λ2 = 0, (x∗, y∗) is the degenerate equilibrium E1 of (1.6). Substituting X =x − x∗, Y = y − y∗ into (3.10) and using Taylor expansion, we obtain

dXdt

=(

1 + y∗2)λ1 −

(d + (1 + d)y∗2

)X − 2

(a0(1 + 2d) − (m0 + λ1)y∗

)Y

+ (m0 − (d + 1)x∗ + λ1)Y 2 − m0

a0XY + f1(X,Y, λ),

dYdt

= −y∗(

1 + y∗2)λ2 + y∗2X +

(2x∗y∗ − a0

(1 + 3y∗2

)−(

1 + 3y∗2)λ2

)Y

+ 2y∗XY +(x∗ − 3a0y

∗ − 3y∗λ2)Y 2 + f2(X,Y, λ),

(3.13)

where λ = (λ1, λ2), fi(X,Y, λ), i = 1, 2, are smooth functions of X, Y and λ at least of orderthree in variables (X,Y ). Making the change of variables x = X, y = −2dX − 2(a0 − y∗λ1)Y to(3.13) and noting the conditions in (3.11) and expressions in (3.12), we have

dxdt

=(

1 + y∗2)λ1 + y +

(m0d

2a22

− d2

a22

λ1

)

x2 +1

4a22

(λ1 −

m0

2d

)y2 + f1

(x, y, λ

),

dydt

= β0 + β1x + β2y + β3x2 + β4xy + β5y

2 + f2(x, y, λ

),

(3.14)

where

a2 = a0 − y∗λ1,

β0 = −2d(

1 + y∗2)λ1 + 2a2y

∗(

1 + y∗2)λ2,

β1 =2d

1 + dy∗λ1 − 2d

(1 + 3y∗2

)λ2,

β2 = −(

1 + 3y∗2)λ2,

β3 =m0d

2(2d + 1)a0a2

− 4m0d2

a2(1 + d)λ1 +

6d2y∗

a2λ2,

β4 =2m0d

2

a0a2− 2m0d

a2(1 + d)λ1 +

6dy∗

a2λ2,

β5 =m0(2d − 1)

4a2a0+

3y∗

2a2λ2.

(3.15)


fi(u, v, λ), i = 1, 2, are smooth functions in variables (u, v) at least of the third order, and thecoefficients depend smoothly on λ1 and λ2.

Let X = x+y/2d, Y = y. Using (3.11) and (3.12), after some algebraic calculations, weobtain

dXdt

= c0 + c1X + c2Y + c3X2 + c4XY + F1(X,Y, λ),

dYdt

= e0 + e1X + e2Y + e3X2 + e4XY + F2(X,Y, λ),

(3.16)

where Fi(X,Y, λ), i = 1, 2, are smooth functions of X, Y and λ at least of the third order invariables (X,Y ),

c0 =1da2y

∗(

1 + y∗2)λ2,

c1 =y∗

1 + dλ1 −

(1 + 3y∗2

)λ2,

c2 = 1 −y∗

a0λ1,

c3 =m0

a0a2

(d(1 + d) +

3d2(1 + d)

λ2 −2a0d

1 + dλ1

),

c4 =m0

a0a2(−1 + 2dλ1),

e0 = −2d(

1 + y∗2)λ1 + 2a2y

∗(

1 + y∗2)λ2,

e1 = 2dc1,

e2 = −y∗

1 + dλ1,

e3 =m0d

2

a0a2

(2d + 1 +

31 + d

λ2 −4a0

1 + dλ1

),

e4 =m0d

a0a2

(−1 +

2a0

1 + dλ1

).

(3.17)

Let x = X, y = c0 + c1X + c2Y + c3X2 + c4XY + F1(X,Y, λ). Then (3.16) becomes

dxdt

= y,

dydt

= b0 + b1x + b2y + b3x2 + b4xy + b5y

2 +G(x, y, λ

),

(3.18)


where

b0 = c2e0 − c0e2,

b1 = c2e1 + c4e0 − c1e2 − c0e4,

b2 = c1 − c0c4

c2+ e2,

b3 = c2e3 + c4e1 − c3e2 − c1e4,

b4 = 2c3 − c1c4

c2+ c0

c24

c22

+ e4,

b5 =c4

c2.

(3.19)

G(x, y, λ) is smooth function in variables (x, y) at least of order three, and all the coefficientsdepend smoothly on λ1 and λ2.

By setting X = x + b2/b4, Y = y to (3.18), we obtain

dXdt

= Y,

dYdt

= r0 + r1X + b3X2 + b4XY + b5Y

2 +G1(X,Y, λ),

(3.20)

where G1(X,Y, λ) is smooth function in variables (X,Y ) at least of the third order and

r0 =b0b

24 − b1b2b4 + b3b

22

b24

,

r1 =b1b4 − 2b2b3

b4.

(3.21)

Now, introducing a new time variable τ to (3.20), which satisfies dt = (1 − b5X)dτ , and stillwriting τ as t, we have

dXdt

= Y (1 − b5X),

dYdt

=(r0 + r1X + b3X

2 + b4XY + b5Y2)(1 − b5X) +G2(X,Y, λ),

(3.22)


where G2(X,Y, λ) is smooth function of X, Y and λ at least of three order in variables (X,Y ).Setting x = X, y = Y (1 − b5X) to (3.22), we obtain

dxdt

= y,

dydt

= r0 + q1x + q2x2 + b4xy +G3

(x, y, λ

),

(3.23)

where G3(x, y, λ) is smooth function of x, y and λ at least of order three in variables (x, y)and

q1 = r1 − 2r0b5,

q2 = r0b25 − 2r1b5 + b3.

(3.24)

If λ1 → 0 and λ2 → 0, it is easy to obtain the following results:

r0 −→ 0,

q1 −→ 0,

q2 −→m0d

2(2d + 1)a2

0

> 0

b4 −→m0d(2d + 1)

a20

> 0.

(3.25)

By setting X = (b24/q2)x + q1b

24/2q2

2, Y = b34/q

22 and τ = (q2/b4)t, and rewriting (X,Y, τ) as

(x, y, t), we obtain

dxdt

= y,

dydt

= μ1 + μ2y + x2 + xy +G4(x, y, λ

),

(3.26)

where

μ1 =r0b

44

q32

−q2

1b44

4q42

,

μ2 = −q1b

24

2q22

,

(3.27)

and G4(x, y, λ) is smooth function of x, y and λ at least of order three in variables (x, y).By the theorem of Bogdanov in [26, 27] and the result of Perko in [25], we obtain the

following local representations of bifurcation curves in a small neighborhood Δ of the origin(i.e., E1 of (1.6)).


H III SN+

II

SN+

0I

IIIII

IV

HL

H

SN−

μ1

μ2

0 IV

HL I SN−

Figure 1: The bifurcation set and the corresponding phase portraits of system (3.26) at origin.

Theorem 3.2. Let the assumptions (H1) and (H2) hold. Then (1.6) admits the following bifurcationbehaviors:

(i) there is a saddle-node bifurcation curve SN± = {(λ1, λ2) : μ1 = 0, μ2 > 0 or μ2 < 0};(ii) there is a Hopf bifurcation curve H = {(λ1, λ2) : μ1 = −μ2

2 + o(‖λ‖2), q1 < 0};(iii) there is a homoclinic-loop bifurcation curve HL= {(λ1, λ2) : μ1 = −(49/25)μ2

2 + o(‖λ‖2)}.

Concretely, as the statement in [28, Chapter 3], when (μ1, μ2) ∈ Δ, the orbital topicalstructure of the system (3.26) at origin (corresponding system (1.6) at E1) is shown inFigure 1.

4. Discussion

Note that most infection experiments suggest that the infection rate of microparasiticinfections is an increasing function of the parasite dose, usually sigmoidal in shape. In thispaper, we study a viral infection model with a type of nonlinear infection rate, which wasintroduced by Regoes et al. [11].

Qualitative analysis (Theorem 2.1) implies that infection equilibrium E12 is alwaysstable if the virus is noncytopathic, a = d, or cytopathic in vivo but its cytopathic effectis less than or equal to an appropriate value, a ≤ 2d(1 + d). When the cytopathic effectof virus is greater than the threshold value, a > 2d(1 + d), the stability of the infectionequilibrium E12 depends on the value of parameter m, which is proportional to the birth rateof susceptible cells λ and is in inverse proportion to the infectious dose ID50. The infectionequilibrium is stable if m > m0 and becomes unstable if m < m0. When m gets to the criticalvalue, m = m0, the infection equilibrium is a linear center, so the oscillation behaviors mayoccur.

If our model (1.6) does not have a limit cycle (see Theorem 2.2 and Corollary 2.3),its asymptotic behavior is determined by the stability of E12. When E12 is stable, thereis a region outside which positive semiorbits tend to E0 as t tends to infinity and inside


0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

y

0 5 10 15 20

x

E0

ExtinctionExtinction

UM

SM

E11

UM

SM

E12 Persistence

Figure 2: Illustrations of the Allee effect for (1.5). Here, λ = 17.06, d = 1.0, a = 3.0, ID50 = 2. E0 = (17.06, 0)is stable, E11 = (13.2311, 1.2763) is a saddle point, E12 = (12.3589, 1.567) is stable. Note that SM is the stablemanifolds of E11 (solid line), UM is the unstable manifolds of E11 (dash line), and the phase portrait of(1.6) is divided into two domains of extinction and persistence of the virus by SM.

which positive semi-orbits tend to E12 as t tends to infinity; that is, the virus will persistif the initial position lies in the region and disappear if the initial position lies outside thisregion. Thus, besides the value of parameters, the initial concentration of the virus canalso affect the result of invasion. An invasion threshold may exist in these cases, whichis typical for the so-called Allee effect that occurs when the abundance or frequency of aspecies is positively correlated with its growth rate (see [11]). Consequently, the unrescaledmodel (1.5) can display an Allee effect (see Figure 2), which is an infrequent phenomenonin current viral infection models though it is reasonable and important in viral infectionprocess.

Furthermore, when infection equilibrium becomes a degenerate singular point, wehave shown that the dynamics of this model are very rich inside this region (see Theorems 3.1and 3.2 and Figure 1). Static and dynamical bifurcations, including saddle-node bifurcation,Hopf bifurcation, homoclinic bifurcation, and bifurcation of cusp-type with codimensiontwo (i.e., Bogdanov-Takens bifurcation), have been exhibited. Thus, besides the Allee effect,our model (1.6) shows that the viral oscillation behaviors can occur in the host based onthe appropriate conditions, which was observed in chronic HBV or HCV carriers (see [29–31]). These results inform that the viral infection is very complex in the development ofa better understanding of diseases. According to the analysis, we find that the cytopathiceffect of virus and the birth rate of susceptible cells are both significant to induce the complexand interesting phenomena, which is helpful in the development of various drug therapystrategies against viral infection.

Acknowledgments

This work is supported by the National Natural Science Fund of China (nos. 30770555 and10571143), the Natural Science Foundation Project of CQ CSTC (2007BB5012), and the ScienceFund of Third Military Medical University (06XG001).


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Research ArticleAn Existence Result for Nonlinear FractionalDifferential Equations on Banach Spaces

Mouffak Benchohra,1 Alberto Cabada,2 and Djamila Seba3

1 Laboratoire de Mathematiques, Universite de Sidi Bel-Abbes, BP 89, 22000 Sidi Bel-Abbes, Algeria2 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Santiago de Compostela,15782, Santiago de Compostela, Spain

3 Departement de Mathematiques, Universite de Boumerdes, Avenue de l’Independance,35000 Boumerdes, Algeria

Correspondence should be addressed to Mouffak Benchohra, [email protected]

Received 30 January 2009; Revised 23 March 2009; Accepted 15 May 2009

Recommended by Juan J. Nieto

The aim of this paper is to investigate a class of boundary value problem for fractional differentialequations involving nonlinear integral conditions. The main tool used in our considerations is thetechnique associated with measures of noncompactness.

Copyright q 2009 Mouffak Benchohra 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.

1. Introduction

The theory of fractional differential equations has been emerging as an important areaof investigation in recent years. Let us mention that this theory has many applicationsin describing numerous events and problems of the real world. For example, fractionaldifferential equations are often applicable in engineering, physics, chemistry, and biology.See Hilfer [1], Glockle and Nonnenmacher [2], Metzler et al. [3], Podlubny [4], Gaul et al.[5], among others. Fractional differential equations are also often an object of mathematicalinvestigations; see the papers of Agarwal et al. [6], Ahmad and Nieto [7], Ahmad and Otero-Espinar [8], Belarbi et al. [9], Belmekki et al [10], Benchohra et al. [11–13], Chang and Nieto[14], Daftardar-Gejji and Bhalekar [15], Figueiredo Camargo et al. [16], and the monographsof Kilbas et al. [17] and Podlubny [4].

Applied problems require definitions of fractional derivatives allowing the utilizationof physically interpretable initial conditions, which contain y(0), y

′(0), and so forth. the same

requirements of boundary conditions. Caputo’s fractional derivative satisfies these demands.For more details on the geometric and physical interpretation for fractional derivatives ofboth the Riemann-Liouville and Caputo types, see [18, 19].


In this paper we investigate the existence of solutions for boundary value problemswith fractional order differential equations and nonlinear integral conditions of the form

cDry (t) = f(t, y (t)

), for each t ∈ J = [0, T] ,

y (0) − y′ (0) =∫T

0g(s, y (s)

)ds,

y (T) + y′(T) =

∫T

0h(s, y (s)

)ds,

(1.1)

where cDr, 1 < r ≤ 2 is the Caputo fractional derivative, f , g, and h : J × E → E are givenfunctions satisfying some assumptions that will be specified later, and E is a Banach spacewith norm ‖ · ‖.

Boundary value problems with integral boundary conditions constitute a veryinteresting and important class of problems. They include two, three, multipoint, andnonlocal boundary value problems as special cases. Integral boundary conditions are oftenencountered in various applications; it is worthwhile mentioning the applications of thoseconditions in the study of population dynamics [20] and cellular systems [21].

Moreover, boundary value problems with integral boundary conditions have beenstudied by a number of authors such as, for instance, Arara and Benchohra [22], Benchohraet al. [23, 24], Infante [25], Peciulyte et al. [26], and the references therein.

In our investigation we apply the method associated with the technique of measuresof noncompactness and the fixed point theorem of Monch type. This technique was mainlyinitiated in the monograph of Bana and Goebel [27] and subsequently developed and used inmany papers; see, for example, Bana and Sadarangoni [28], Guo et al. [29], Lakshmikanthamand Leela [30], Monch [31], and Szufla [32].

2. Preliminaries

In this section, we present some definitions and auxiliary results which will be needed in thesequel.

Denote by C(J, E) the Banach space of continuous functions J → E, with the usualsupremum norm

∥∥y∥∥∞ = sup

{‖y (t) ‖, t ∈ J

}. (2.1)

Let L1(J, E) be the Banach space of measurable functions y : J → E which are Bochnerintegrable, equipped with the norm

∥∥y∥∥L1 =

∫T

0‖y (s) ‖ds. (2.2)


Let L∞(J, E) be the Banach space of measurable functions y : J → E which are bounded,equipped with the norm

∥∥y∥∥L∞ = inf

{c > 0 : ‖y (t) ‖ ≤ c, a.e. t ∈ J

}. (2.3)

Let AC1(J, E) be the space of functions y : J → E, whose first derivative is absolutelycontinuous.

Moreover, for a given set V of functions v : J → E let us denote by

V (t) = {v (t) , v ∈ V } , t ∈ J,

V (J) = {v (t) : v ∈ V } , t ∈ J.(2.4)

Now let us recall some fundamental facts of the notion of Kuratowski measure ofnoncompactness.

Definition 2.1 (see [27]). Let E be a Banach space and ΩE the bounded subsets of E. TheKuratowski measure of noncompactness is the map α : ΩE → [0,∞] defined by

α (B) = inf{ε > 0 : B ⊆

⋃n

i=1Bi and diam (Bi) ≤ ε

}; here B ∈ ΩE. (2.5)

Properties

The Kuratowski measure of noncompactness satisfies some properties (for more details see[27]).

(a) α(B) = 0⇔ B is compact (B is relatively compact).

(b) α(B) = α(B).

(c) A ⊆ B ⇒ α(A) ≤ α(B).

(d) α(A + B) ≤ α(A) + α(B).

(e) α(cB) = |c|α(B); c ∈ R.

(f) α(coB) = α(B).

Here B and coB denote the closure and the convex hull of the bounded set B, respectively.For completeness we recall the definition of Caputo derivative of fractional order.

Definition 2.2 (see [17]). The fractional order integral of the function h ∈ L1([a, b]) of orderr ∈ R+; is defined by

Irah (t) =1

Γ (r)

∫ t

a

h (s)

(t − s)1−r dt, (2.6)


where Γ is the gamma function. When a = 0, we write Irh(t) = [h ∗ ϕr](t), where

ϕr (t) =tr−1

Γ (r)for t > 0, (2.7)

ϕr(t) = 0 for t ≤ 0, and ϕr → δ(t) as r → 0.

Here δ is the delta function.

Definition 2.3 (see [17]). For a function h given on the interval [a, b], the Caputo fractional-order derivative of h, of order r > 0, is defined by

cDra+h (t) =

1Γ (n − r)

∫ t

a

h(n) (s)ds

(t − s)1−n+r . (2.8)

Here n = [r] + 1 and [r] denotes the integer part of r.

Definition 2.4. A map f : J × E → E is said to be Caratheodory if

(i) t �→ f(t, u) is measurable for each u ∈ E;

(ii) u �→ f(t, u) is continuous for almost each t ∈ J.

For our purpose we will only need the following fixed point theorem and the importantLemma.

Theorem 2.5 (see [31, 33]). Let D be a bounded, closed and convex subset of a Banach space suchthat 0 ∈ D, and letN be a continuous mapping of D into itself. If the implication

V = coN (V ) or V = N (V ) ∪ {0} =⇒ α (V ) = 0 (2.9)

holds for every subset V of D, thenN has a fixed point.

Lemma 2.6 (see [32]). Let D be a bounded, closed, and convex subset of the Banach space C(J, E),G a continuous function on J × J, and a function f : J ×E → E satisfies the Caratheodory conditions,and there exists p ∈ L1(J,R+) such that for each t ∈ J and each bounded set B ⊂ E one has

limk→ 0+

α(f (Jt,k × B)

)≤ p (t)α (B) ; where Jt,k = [t − k, t] ∩ J. (2.10)

If V is an equicontinuous subset of D, then

α

({∫

J

G (s, t) f(s, y (s)

)ds : y ∈ V

})

≤∫

J

‖G (t, s)‖ p (s)α (V (s))ds. (2.11)


3. Existence of Solutions

Let us start by defining what we mean by a solution of the problem (1.1).

Definition 3.1. A function y ∈ AC1(J, E) is said to be a solution of (1.1) if it satisfies (1.1).

Let σ, ρ1, ρ2 : J → E be continuous functions and consider the linear boundary valueproblem

cDry (t) = σ (t) , t ∈ J,

y (0) − y′ (0) =∫T

0ρ1 (s)ds,

y (T) + y′(T) =

∫T

0ρ2 (s)ds.

(3.1)

Lemma 3.2 (see [11]). Let 1 < r ≤ 2 and let σ, ρ1, ρ2 : J → E be continuous. A function y is asolution of the fractional integral equation

y (t) = P (t) +∫T

0G (t, s)σ (s)ds (3.2)

with

P (t) =(T + 1 − t)

T + 2

∫T

0ρ1 (s)ds +

(t + 1)T + 2

∫T

0ρ2 (s)ds, (3.3)

G (t, s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(t − s)r−1

Γ (r)− (1 + t) (T − s)r−1

(T + 2) Γ (r)− (1 + t) (T − s)r−2

(T + 2) Γ (r − 1), 0 ≤ s ≤ t,

− (1 + t) (T − s)r−1

(T + 2) Γ (r)− (1 + t) (T − s)r−2

(T + 2) Γ (r − 1), t ≤ s ≤ T,

(3.4)

if and only if y is a solution of the fractional boundary value problem (3.1).

Remark 3.3. It is clear that the function t �→∫T

0 |G(t, s)|ds is continuous on J , and hence isbounded. Let

G := sup

{∫T

0|G (t, s)|ds, t ∈ J

}

. (3.5)


For the forthcoming analysis, we introduce the following assumptions

(H1) The functions f, g, h : J × E → E satisfy the Caratheodory conditions.

(H2) There exist pf , pg, ph ∈ L∞(J,R+), such that

‖f(t, y)‖ ≤ pf (t) ‖y‖ for a.e. t ∈ J and each y ∈ E,

‖g(t, y)‖ ≤ pg (t) ‖y‖, for a.e. t ∈ J and each y ∈ E,

‖h(t, y)‖ ≤ ph (t) ‖y‖, for a.e. t ∈ J and each y ∈ E.

(3.6)

(H3) For almost each t ∈ J and each bounded set B ⊂ E we have

limk→ 0+

α(f (Jt,k × B)

)≤ pf (t)α (B) ,

limk→ 0+

α(g (Jt,k × B)

)≤ pg (t)α (B) ,

limk→ 0+

α (h (Jt,k × B)) ≤ ph (t)α (B) .

(3.7)

Theorem 3.4. Assume that assumptions (H1)–(H3) hold. If

T (T + 1)T + 2

[∥∥pg∥∥L∞

+∥∥ph∥∥L∞

]+ G∥∥pf∥∥L∞

< 1, (3.8)

then the boundary value problem (1.1) has at least one solution.

Proof. We transform the problem (1.1) into a fixed point problem by defining an operatorN : C(J, E) → C(J, E) as

(Ny)(t) = Py (t) +

∫T

0G (t, s) f

(s, y (s)

)ds, (3.9)

where

Py (t) =(T + 1 − t)

T + 2

∫T

0g(s, y (s)

)ds +

(t + 1)T + 2

∫T

0h(s, y (s)

)ds, (3.10)

and the function G(t, s) is given by (3.4). Clearly, the fixed points of the operator N aresolution of the problem (1.1). Let R > 0 and consider the set

DR ={y ∈ C (J, E) :

∥∥y∥∥∞ ≤ R

}. (3.11)

Clearly, the subset DR is closed, bounded, and convex. We will show that N satisfies theassumptions of Theorem 2.5. The proof will be given in three steps.


Step 1. N is continuous.Let {yn} be a sequence such that yn → y in C(J, E). Then, for each t ∈ J ,

∣∣(Nyn

)(t) −

(Ny)(t)∣∣ ≤ T + 1

T + 2

∫T

0

∥∥g(s, yn (s)

)− g(s, y (s)

)∥∥ds

+T + 1T + 2

∫T

0

∥∥h(s, yn (s)

)− h(s, y (s)

)∥∥ds

+∫T

0|G (t, s)|

∥∥f(s, yn (s)

)− f(s, y (s)

)∥∥ds.

(3.12)

Let ρ > 0 be such that

∥∥yn

∥∥∞ ≤ ρ,

∥∥y∥∥∞ ≤ ρ. (3.13)

By (H2) we have

∥∥g(s, yn (s)

)− g(s, y (s)

)∥∥ ≤ 2ρpg (s) := σ1 (s) ; σ1 ∈ L1 (J,R+) ,∥∥h(s, yn (s)

)− h(s, y (s)

)∥∥ ≤ 2ρph (s) := σ2 (s) ; σ2 ∈ L1 (J,R+) ,

|G (·, s)|∥∥f(s, yn (s)

)− f(s, y (s)

)∥∥ ≤ 2ρ |G (·, s)| pf (s) := σ3 (s) ; σ3 ∈ L1 (J,R+) .

(3.14)

Since f, g, and h are Caratheodory functions, the Lebesgue dominated convergencetheorem implies that

∥∥N(yn) −N(y)∥∥∞ −→ 0 as n −→ ∞. (3.15)

Step 2. N maps DR into itself.For each y ∈ DR, by (H2) and (3.8) we have for each t ∈ J

‖N(y)(t) ‖ ≤ T + 1

T + 2

∫T

0

∥∥g(s, y (s)

)∥∥ds +T + 1T + 2

∫T

0

∥∥h(s, y (s)

)∥∥ds

+∫T

0|G (t, s)|

∥∥f(s, y (s)

)∥∥ds

≤ R

[T (T + 1)T + 2

∥∥pg∥∥L∞

+T (T + 1)T + 2

∥∥ph∥∥L∞ + G

∥∥pf∥∥L∞

]

< R.

(3.16)

Step 3. N(DR) is bounded and equicontinuous.By Step 2, it is obvious that N(DR) ⊂ C(J, E) is bounded.


For the equicontinuity of N(DR). Let t1, t2 ∈ J , t1 < t2 and y ∈ DR. Then

‖(Ny)(t2) −

(Ny)(t1) ‖ =

∥∥∥∥∥− t2 − t1T + 2

∫T

0g(s, y (s)

)ds +

t2 − t1T + 2

∫T

0h(s, y (s)

)ds

+∫T

0[G (t2, s) −G (t1, s)] f

(s, y (s)

)ds

∥∥∥∥∥

≤ t2 − t1T + 2

TR[∥∥pg∥∥L∞

+∥∥ph∥∥L∞

]

+ R∥∥pf∥∥L∞

∫T

0|G (t2, s) −G (t1, s)|ds.

(3.17)

As t1 → t2, the right-hand side of the above inequality tends to zero.Now let V be a subset of DR such that V ⊂ co(N(V ) ∪ {0}).

V is bounded and equicontinuous, and therefore the function v → v(t) = α(V (t)) iscontinuous on J . By (H3), Lemma 2.6, and the properties of the measure α we have for eacht ∈ J

v (t) ≤ α (N (V ) (t) ∪ {0})

≤ α (N (V ) (t))

≤∫T

0

∣∣∣∣T + 1 − tT + 2

∣∣∣∣ pg (s)α (V (s))ds +∫T

0

∣∣∣∣t + 1T + 2

∣∣∣∣ ph (s)α (V (s))ds

+∫T

0|G (t, s)| pf (s)α (V (s))ds

≤ T (T + 1)T + 2

∥∥pg∥∥L∞

v (s) +T (T + 1)T + 2

∥∥ph∥∥L∞v (s) + G

∥∥pf∥∥L∞

v (s)

≤ ‖v‖∞[T (T + 1)T + 2

[∥∥pg∥∥L∞

+∥∥ph∥∥L∞

]+ G∥∥pf∥∥L∞

].

(3.18)

This means that

‖v‖∞(

1 −[T (T + 1)T + 2

[∥∥pg∥∥L∞

+∥∥ph∥∥L∞

]+ G∥∥pf∥∥L∞

])≤ 0. (3.19)

By (3.8) it follows that ‖v‖∞ = 0, that is, v(t) = 0 for each t ∈ J , and then V (t) is relativelycompact in E. In view of the Ascoli-Arzela theorem, V is relatively compact in DR. Applyingnow Theorem 2.5 we conclude that N has a fixed point which is a solution of the problem(1.1).


4. An Example

In this section we give an example to illustrate the usefulness of our main results. Let usconsider the following fractional boundary value problem:

cDry (t) =2

19 + et∣∣y (t)

∣∣ , t ∈ J := [0, 1] , 1 < r ≤ 2,

y (0) − y′ (0) =∫1

0

15 + e5s

∣∣y (s)∣∣ds,

y (1) + y′(1) =

∫1

0

13 + e3s

∣∣y (s)

∣∣ds.

(4.1)

Set

f (t, x) =2

19 + etx, (t, x) ∈ J × [0,∞) ,

g (t, x) =1

5 + e5tx, (t, x) ∈ [0, 1] × [0,∞) ,

h (t, x) =1

3 + e3tx, (t, x) ∈ [0, 1] × [0,∞) .

(4.2)

Clearly, conditions (H1),(H2) hold with

pf (t) =2

19 + et, pg (t) =

15 + e5t

, ph (t) =1

3 + e3t. (4.3)

From (3.4) the function G is given by

G (t, s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(t − s)r−1

Γ (r)− (1 + t) (1 − s)r−1

3Γ (r)− (1 + t) (1 − s)r−2

3Γ (r − 1), 0 ≤ s ≤ t,

− (1 + t) (1 − s)r−1

3Γ (r)− (1 + t) (1 − s)r−2

3Γ (r − 1), t ≤ s ≤ 1.

(4.4)

From (4.4), we have

∫1

0G (t, s)ds =

∫ t

0G (t, s)ds +

∫1

t

G (t, s)ds

=tr

Γ (r + 1)+(1 + t) (1 − t)r

3Γ (r + 1)− (1 + t)

3Γ (r + 1)

+(1 + t) (1 − t)r−1

3Γ (r)− (1 + t)

3Γ (r)

− (1 + t) (1 − t)r

3Γ (r + 1)− (1 + t) (1 − t)r−1

3Γ (r).

(4.5)


A simple computation gives

G∗ <3

Γ (r + 1)+

2Γ (r)

. (4.6)

Condition (3.8) is satisfied with T = 1. Indeed

T (T + 1)T + 2

[∥∥pg∥∥L∞

+∥∥ph∥∥L∞

]+ G∥∥pf∥∥L∞

<23

[16+

14

]+

310Γ (r + 1)

+2

10Γ (r)

=5

18+

310Γ (r + 1)

+1

5Γ (r)< 1,

(4.7)

which is satisfied for each r ∈ (1, 2]. Then by Theorem 3.4 the problem (4.1) has a solution on[0, 1].

Acknowledgments

The authors thank the referees for their remarks. The research of A. Cabada has been partiallysupported by Ministerio de Educacion y Ciencia and FEDER, project MTM2007-61724, andby Xunta de Galicia and FEDER, project PGIDIT05PXIC20702PN.

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Research ArticleAntiperiodic Boundary Value Problems for FiniteDimensional Differential Systems

Y. Q. Chen,1 D. O’Regan,2 F. L. Wang,1 and S. L. Zhou1

1 Faculty of Applied Mathematics, Guangdong University of Technology, Guangzhou,Guangdong 510006, China

2 Department of Mathematics, National University of Ireland, Galway, Ireland

Correspondence should be addressed to D. O’Regan, [email protected]

Received 16 March 2009; Accepted 28 May 2009


We study antiperiodic boundary value problems for semilinear differential and impulsivedifferential equations in finite dimensional spaces. Several new existence results are obtained.

Copyright q 2009 Y. Q. Chen 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.

1. IntroductionThe study of antiperiodic solutions for nonlinear evolution equations is closely related tothe study of periodic solutions, and it was initiated by Okochi [1]. During the past twentyyears, antiperiodic problems have been extensively studied by many authors, see [1–31] andthe references therein. For example, antiperiodic trigonometric polynomials are important inthe study of interpolation problems [32, 33], and antiperiodic wavelets are discussed in [34].Moreover, antiperiodic boundary conditions appear in physics in a variety of situations, see[35–40]. In Section 2 we consider the antiperiodic problem

u′(t) = Au(t) + f(t, u(t)), t ∈ R,

u(t) = −u(t + T), t ∈ R,(E 1.1)

where A is an n × n matrix, f : R × Rn → Rn is continuous, and f(t + T, x) = −f(t, x) for all(t, x) ∈ R×Rn. Under certain conditions on the nondiagonal elements of A and f we prove anexistence result for (E 1.1). In Section 3 we consider the antiperiodic boundary value problem

u′(t) = Gu(t) + f(t, u(t)), a.e. t ∈ J = [0, T], t /= tk,

u(0) = −u(T),

Δu(tk) = Ik(u(tk)), k = 1, 2, . . . , p,

(E 1.2)


where G : Rn → Rn is a function satisfying G0 = 0, and f : J × Rn → Rn is a Caratheodoryfunction, Δu(tk) = u(t+

k) − u(t−

k), and Ik ∈ C(Rn, Rn). Under certain conditions on G, f , and

Ik(u) for k = 1, 2, . . . , p, we prove an existence result for (E 1.2).

2. Antiperiodic Problem for Differential Equations in Rn

Let | · | be the norm in Rn. In this section we study

u′(t) = Au(t) + f(t, u(t)), t ∈ R,

u(t) = −u(t + T).(E 2.1)

First, we have the following result.

Theorem 2.1. Let A = (aij) be an n × n matrix, where aij is the element of A in the ith row and jthcolumn, f : R → Rn is continuous and f(t+T) = −f(t) for t ∈ R. Suppose (T/2)Σ1≤i<j≤n|aij −aji| <1. Then the equation

u′(t) = Au(t) + f(t), t ∈ R,

u(t) = −u(t + T), t ∈ R(E 2.2)

has a unique solution.

Proof. Put Wa = {v(·) ∈ C(R;Rn) : v(t) = −v(t + T)}. Then Wa is a Banach space under thenorm |v(·)|∞ = maxt∈[0,T]|v(t)|. For each v(·) ∈Wa, consider the following equation:

u′(t) = Av(t) + f(t), t ∈ R,

u(t) = −u(t + T), t ∈ R.(E 2.3)

It is easy to see that u(t) = −(1/2)∫T

0 [Av(s)+f(s)]ds+∫ t

0[Av(s)+f(s)]ds is the unique solutionof (E 2.3).

We define a mapping K : Wa → Wa as follows:

for any v(·) ∈Wa, Kv(·) = u(·), u(·) is the solution of (E 2.3). (2.1)

First we prove that K is a continuous compact mapping. Now assume vn(·) ∈Wa, n = 1, 2, . . .,and vn(·) → v(·) ∈ Wa. Then |Avn(·) − Av(·)|∞ → 0 as n → ∞. This immediately impliesthat

∫T0 |(Kvn(t))

′ − (Kv(t))′|2dt → 0 as n → ∞.We have Kvn(t) − Kv(t) = (1/2){

∫ t0[(Kvn(s))

′ − (Kv(s))′]ds −∫Tt [(Kvn(s))

′ −(Kv(s))′]ds]}, and so Kvn(·) → Kv(·) in Wa.

Now since (Kv(t))′ = Av(t) + f(t), t ∈ R, it is easy to see that

(∫T

0

∣∣(Kv(t))′∣∣2dt

)1/2

≤√T |Av(·)|∞ +

(∫T

0

∣∣f(t)∣∣2dt

)1/2

. (2.2)


Thus K maps a bounded subset of Wa to a bounded equicontinuous subset in Wa, thereforeK is completely continuous.

Next take r0 > (1 − (T/2)Σ1≤i<j≤n|aij − aji|)−1(√T/2)(

∫T0 |f(t)|

2dt)1/2. We show thatKv(·)/=λv(·) for all λ ≥ 1, and |v(·)|∞ = r0. If this is not true, there exist λ0 ≥ 1, w(·) ∈Wa with|w(·)|∞ = r0 such that Kw(·) = λ0w(·), that is, w(t) = −w(t + T), t ∈ R and

λ0w′(t) = Aw(t) + f(t), t ∈ R. (2.3)

Multiply (2.3) by w′(t) (i.e., take inner product) and integrate over [0, T], and notice that∫T

0wi(t)w′j(t)dt = −∫T

0w′i(t)wj(t)dt to get

λ0

∫T

0|w′(t)|2dt ≤ Σ1≤i<j≤n

∣∣aij − aji

∣∣∫T

0

∣∣∣wi(t)w′j(t)

∣∣∣dt +

(∫T

0

∣∣f(t)

∣∣2dt

)1/2(∫T

0

∣∣w′(t)

∣∣2dt

)1/2

,

(2.4)

where w(t) = (wi(t)), i = 1, 2, . . . , n. Notice that w(t) = (1/2)[∫ t

0w′(s)ds −

∫Tt w

′(s)]ds, so wehave

|w(·)|∞ ≤√T

2

(∫T

0

∣∣w′(t)∣∣2dt

)1/2

. (2.5)

From (2.4), (2.5), we have

λ0

(∫T

0

∣∣w′(t)∣∣2dt

)1/2

≤√TΣ1≤i<j≤n

∣∣aij − aji

∣∣|w(·)|∞ +

(∫T

0

∣∣f(t)∣∣2dt

)1/2

. (2.6)

This with (2.5) gives

λ0|w(·)|∞ ≤T

2Σ1≤i<j≤n

∣∣aij − aji

∣∣|w(·)|∞ +

√T

2

(∫T

0

∣∣f(t)∣∣2dt

)1/2

. (2.7)

As a result

|w(·)|∞ ≤(

1 − T

2Σ1≤i<j≤n

∣∣aij − aji

∣∣)−1√T

2

(∫T

0

∣∣f(t)∣∣2dt

)1/2

, (2.8)

which contradicts |w(·)|∞ = r0.Thus the Leray-Schauder degree deg(I −K,B(0, r0), 0) = 1, where B(0, r0) is the open

ball centered at 0 with radius r0 in Ca. Consequently, K has a fixed point in B(0, r0), that is,(E 2.2) has a solution. For the uniqueness, if u(·), v(·) are two solutions of (E 2.2), set w(t) =u(t) − v(t), then w′(t) = Aw(t), and w(t) = −w(t + T), for t ∈ R. Following the obvious


strategy above (see the clear adjustment of (2.8)) gives |w(·)|∞ = 0. Thus the solution of(E 2.2) is unique.

From Theorem 2.1 we have immediately the following result.

Corollary 2.2. Let A = (aij) be an n × n symmetric matrix, f : R → Rn is continuous andf(t + T) = −f(t) for t ∈ R. Then

u′(t) = Au(t) + f(t), t ∈ R,

u(t) = −u(t + T), t ∈ R,(E 2.4)

has a unique solution.

Using a proof similar to Theorem 2.1, we have the following result.

Theorem 2.3. LetA = (aij) be an n × n matrix, G : Rn → Rn is an even continuously differentiablefunction, and f(t, u) : R × Rn → Rn is continuous and f(t + T, u) = −f(t, u) for (t, u) ∈ R × Rn.Suppose the following conditions are satisfied:

(1) |f(t, x)| ≤ M|x| + g(t), for a.e. (t, x) ∈ R × Rn, where M > 0 is a constant, and g(·) ∈L2(0, T);

(2) (T/2)[Σ1≤i<j≤n|aij − aji| +M] < 1.

Then

u′(t) = Au(t) + ∂Gu(t) + f(t, u(t)), t ∈ R,

u(t) = −u(t + T), t ∈ R(E 2.5)

has a solution.

Remark 2.4. Equation (E 2.5) was studied by Haraux [18] and Chen et al. [14] in the caseA = 0, and also by Chen [12] with different assumptions on f and A.

3. Antiperiodic Boundary Value Problem for Impulsive ODE

In this section, we prove an existence result for the equation


u(0) = −u(T),

Δu(tk) = Ik(u(tk)), k = 1, 2, . . . , p,

(E 3.1)

where G : Rn → Rn is a Lipschitz function. We first introduce some notations. LetJ = [0, T], and 0 = t0 < t1 < · · · < tp < tp+1 = T . PC(J) = {u : J → Rn, u(tk ,tk+1] ∈C((tk, tk+1], Rn), k = 0, 1, . . . , p, u(t−k) exist for k = 1, 2, . . . , p, and u(0+) = u(0)}, andPW1,2(J) = {u ∈ PC(J) : u(tk ,tk+1) ∈ W1,2((tk, tk+1), Rn), k = 1, . . . , p}. It is clear that PC(J)


and PW1,2(J) are Banach spaces with the respective norm ‖u‖PC(J) = sup{|u(t)|, t ∈ J}, and‖u‖PW1,2(J) =

∑p

k=0 ‖uk‖W1,2(tk ,tk+1), where uk : (tk, tk+1] → R is defined by uk(t) = u(t) fort ∈ (tk, tk+1], k = 0, 1, . . . , p.

We say a function u is a solution of (E 3.1) if u ∈ PW1,2(J) and u satisfies (E 3.1).We first prove the following result.

Lemma 3.1. Let Ii : Rn → Rn be continuous functions for i = 1, 2, . . . , p, and Σp

k=1|Ik(xk)| ≤α{max1≤k≤p|xk|} + δ for all xk ∈ Rn, k = 1, 2, . . . , p, where α, δ > 0 are constants, and α < 2.Suppose u ∈ PW1,2(J) with u(0) = −u(T), and Δu(ti) = Ii(u(ti)), for i = 1, 2, . . . , p. Then

‖u‖PC(J) ≤(

1 − 12α

)−1⎡

⎣12δ +

√T

2

(∫T

0

∣∣u′(s)

∣∣2ds

)1/2⎤

⎦. (3.1)

Proof. By assumption, we have u(t) = u(0) +∫ t

0u′(s)ds for t ∈ [0, t1), and

u(t) = u(0) + Σki=1 Ii(u(ti)) +

∫ t

0u′(s)ds (3.2)

for t ∈ [tk, tk+1), k = 1, 2, . . . , p. Since u(0) = −u(T), it follows that u(t) = −(1/2)[Σp

i=1Ii(u(ti)) +∫T0u′(s)ds] +

∫ t0u′(s)ds for t ∈ [0, t1), and

u(t) = −12

[

Σp

i=1 Ii(u(ti)) +∫T

0u′(s)ds

]

+ Σki=1 Ii(u(ti)) +

∫ t

0u′(s)ds (3.3)

for t ∈ [tk, tk+1), k = 1, 2, . . . , p. Hence we have

‖u‖PC(J) ≤12

[α‖u‖PC(J) + δ

]+

√T

2

(∫T

0

∣∣u′(s)∣∣2ds

)1/2

. (3.4)

Thus

‖u‖PC(J) ≤(

1 − 12α

)−1⎡

⎣12δ +

√T

2

(∫T

0

∣∣u′(s)∣∣2ds

)1/2⎤

⎦. (3.5)

Theorem 3.2. Let G : Rn → Rn be a function satisfying G0 = 0, and f : [0, T] → Rn such thatf(·) ∈ L2([0, T]), and let Ik : Rn → Rn be continuous functions for k = 1, 2, . . . , p. Suppose thefollowing conditions are satisfied:

(1) |Gu −Gv| ≤ L|u − v| for all u, v ∈ Rn, and L > 0 is a constant;

(2) Σp

k=1|Ik(xk)| ≤ γ{max1≤k≤p|xk|} + δ for all xk ∈ Rn, k = 1, 2, . . . , p, where γ, δ > 0 areconstants;

(3) γ + TL < 2.


Then the problem

u′(t) = Gu(t) + f(t), a.e. t ∈ J = [0, T], t /= tk,

u(0) = −u(T),

Δu(tk) = Ik(u(tk)), k = 1, 2, . . . , p

(E 3.2)

has a solution.

Proof. For each v ∈ PC(J), consider the problem

u′(t) = Gv(t) + f(t) a.e. t ∈ J = [0, T], t /= tk,

u(0) = −u(T),

Δu(tk) = Ik(v(tk)), k = 1, 2, . . . , p.

(E 3.3)

One can easily show that the solution u of (E 3.3) is given by the following:

u(t) = −12

[

Σp

i=1 Ii(v(ti)) +∫T

0

(Gv(s) + f(s)

)ds

]

+∫ t

0

(G(v(s)) + f(s)

)ds, for t ∈ [0, t1),

u(t) = −12

[

Σp

i=1 Ii(v(ti)) +∫T

0

(Gv(s) + f(s)

)ds

]

+ Σki=1 Ii(v(ti))

+∫ t

0

(Gv(s) + f(s)

)ds,

(3.6)

for t ∈ [tk, tk+1), k = 1, . . . , p.Obviously, the solution of (E 3.3) is unique. Now we define K : PC(J) → PW1,2(J) ⊂

PC(J) by u = Kv. We prove that K is continuous. Let vn ∈ PC(J) and vn → v in PC(J). It iseasy to see that

∫T

0

∣∣(Kvn(t) −Kv(t))′∣∣2dt =

∫T

0|Gvn(t) −Gv(t)|2dt ≤ L2

∫T

0|vn(t) − v(t)|2dt. (3.7)

Therefore (∫T

0 |(Kvn(t) −Kv(t))′|2dt)1/2 ≤√TL‖vn − v‖PC(J) → 0 as n → ∞.


Note that Δ(Kvn −Kv)(tk) = Ik(vn(tk)) − Ik(v(tk)), and we have

Kvn(t) −Kv(t) = −12

[

Σp

i=1(Ii(vn(ti)) − Ii(v(ti))) +∫T

0(Kvn −Kv)′(s)ds

]

+∫ t

0(Kvn −Kv)′(s)ds, for t ∈ [0, t1),

Kvn(t) −Kv(t) = −12

[

Σp

i=1(Ii(vn(ti)) − Ii(v(ti))) +∫T


]

+ Σki=1(Ii(vn(ti)) − Ii(v(ti))) +

∫ t


(3.8)

for t ∈ [tk, tk+1), k = 1, 2, . . . , p. From the continuity of Ii, i = 1, 2, . . . , p, and∫T

0 |(Kvn(t) −Kv(t))′|2dt → 0 as n → ∞, we deduce that K is continuous.

For each v ∈ PC(J), notice that 0 = G0, so we have

(∫T

0|Kv|2dt

)1/2

≤√TL‖v‖PC(J) +

(∫T

0

∣∣f(s)∣∣2ds

)1/2

. (3.9)

From (3.9) and Lemma 3.1, we know that K maps bounded subsets of PC(J) to relativelycompact subsets of PC(J).

Finally, for ∀λ ∈ (0, 1], we prove that the set of solutions of u = λKu is bounded. Ifu = λKu for some λ ∈ (0, 1), then

u′(t) = λGu(t) + λf(t) a.e. t ∈ J = [0, T], t /= tk,

u(0) = −u(T),

Δu(tk) = λIk(u(tk)), k = 1, 2, . . . , p.

(3.10)

Therefore we have

u(t) = −12λ

[

Σp

i=1 Ii(ui(ti)) +∫T

0

(Gu(s) + f(s)

)ds

]

+ λ

∫ t

0

(G(u(s)) + f(s)

)ds (3.11)

for t ∈ [0, t1), and

u(t) = −12λ

[

Σp

i=1 Ii(ui(ti)) +∫T

0

(Gu(s) + f(s)

)ds

]

+ λΣki=1Ii(ui(ti))

+ λ

∫ t

0

(G(u(s)) + f(s)

)ds

(3.12)


for t ∈ (tk, tk+1], k = 1, . . . , p. This implies that

‖u‖PC(J) ≤12

[

γ‖u‖PC(J) + δ +∫T

0

(|Gu(s)| +

∣∣f(s)

∣∣)ds

]

. (3.13)

Since 0 = G0, and |Gu| ≤ L|u|, so we have

‖u‖PC(J) ≤12

[1 − 1

2(γ + TL

)]−1(

δ +∫T

0

∣∣f(s)

∣∣ds

)

. (3.14)

The Leray-Schauder principle guarantees a fixed point of K, which is easily seen to be asolution of (E 3.2).

By using a similar method to Theorem 3.2, one can deduce the following result.

Theorem 3.3. Let G : Rn → Rn be a function satisfying G0 = 0, and f(t, x) : [0, T] × Rn → Rn

a Caratheodory function, that is, f is measurable in t for each x ∈ Rn, and f is continuous in x foreach t ∈ [0, T], such that |f(t, x)| ≤ g(t) for (t, x) ∈ [0, T] × Rn, where g(·) ∈ L2([0, T]), andlet Ik : Rn → Rn be continuous functions for k = 1, 2, . . . , p. Suppose the following conditions aresatisfied:

(1) |Gu −Gv| ≤ L|u − v| for all u, v ∈ Rn, and L > 0 is a constant;

(2) Σp

k=1|Ik(xk)| ≤ γ{max1≤k≤p|xk|} + δ for all xk ∈ Rn, k = 1, 2, . . . , p, where γ, δ > 0 areconstants;

(3) γ + TL < 2.

Then the equation


u(0) = −u(T),

Δu(tk) = Ik(u(tk)), k = 1, 2, . . . , p

(E 3.4)

has a solution.

4. Examples

In this section, we give examples to show the application of our results to differential andimpulsive differential equations.


Example 4.1. Consider the antiperiodic problem

u′1(t) = λ1u1(t) + 5u2(t) + sinπt, t ∈ R,

u′2(t) =72u1(t) + λ2u2(t) + cosπt, t ∈ R,

u1(t) = −u1(t + 1), u2(t) = −u2(t + 1), t ∈ R.

(E 4.1)

Set

u =

(u1

u2

)

, f(t) =

(sinπt

cosπt

)

, A =

⎛

⎝λ1 5

72

λ2

⎞

⎠. (4.1)

Now (E 4.1) is equivalent to

u′(t) = Au(t) + f(t), t ∈ R,

u(t) = −u(t + 1), t ∈ R.(E 4.2)

Also f(t) = −f(t+1), for t ∈ R and (1/2)|a12−a21| = 3/4. By Theorem 2.1, (E 4.2) has a uniquesolution, so (E 4.1) has a unique solution.

Example 4.2. Consider the antiperiodic boundary value problem

u′1(t) =1

2 + u21(t) + u2

2(t)[3u1(t) − 2u2(t)] + sinπt, t ∈ (0, 1), t /=

14,

u′2(t) =1

2 + u21(t) + u2

2(t)[2u1(t) + 3u2(t)] − cosπt, t ∈ (0, 1), t /=

14,

Δu1

(14

)=

15(1 + |u2(1/4)|) , Δu2

(14

)=

18(1 + |u1(1/4)|) ,

u1(0) = −u1(1), u2(0) = −u2(1).

(E 4.3)

Set

u =

(u1

u2

)

, f(t) =

(sinπt

− cosπt

)

, Gu =

⎛

⎜⎜⎝

3u1 − 2u2

2 + u21 + u2

22u1 + 3u2

2 + u21 + u2

2

⎞

⎟⎟⎠, I(u)

⎛

⎜⎜⎝

15(1 + |u2|)

18(1 + |u1|)

⎞

⎟⎟⎠.

(4.2)


It is easy to check that |Gu − Gv| ≤ (√

13/2)|u − v| for u, v ∈ R2, |I(u)| < 2/5 for u ∈ R2, and√13/2 < 2. Now (E 4.3) is equivalent to the equation

u′(t) = Gu(t) + f(t), t ∈ (0, 1), t /=14,

Δu

(14

)= I

(u

(14

)), u(0) = −u(1).

(E 4.4)

By Theorem 3.2, we know that (E 4.4) has a solution, so (E 4.3) has a solution.

Acknowledgment

The first author is supported by an NSFC Grant, Grant no. 10871052.

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Research ArticleConstant Sign and Nodal Solutions forProblems with the p-Laplacian and a NonsmoothPotential Using Variational Techniques

Ravi P. Agarwal,1 Michael E. Filippakis,2 Donal O’Regan,3and Nikolaos S. Papageorgiou4

1 Department of Mathematical Sciences, Florida Institute of Technology, Melbourne, FL 32901-6975, USA2 Department of Mathematics, Hellenic Army Academy, Vari, 16673 Athens, Greece3 Department of Mathematics, National University of Ireland, Galway, Ireland4 Department of Mathematics, National Technical University, Zografou Campus, 15780 Athens, Greece

Correspondence should be addressed to Ravi P. Agarwal, [email protected]

Received 10 December 2008; Revised 21 January 2009; Accepted 23 January 2009


We consider a nonlinear elliptic equation driven by the p-Laplacian with a nonsmooth potential(hemivariational inequality) and Dirichlet boundary condition. Using a variational approachbased on nonsmooth critical point theory together with the method of upper and lower solutions,we prove the existence of at least three nontrivial smooth solutions: one positive, the secondnegative, and the third sign changing (nodal solution). Our hypotheses on the nonsmoothpotential incorporate in our framework of analysis the so-called asymptotically p-linear problems.

Copyright q 2009 Ravi P. Agarwal 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.

1. Introduction

The aim of this work is to prove the existence of multiple solutions of constant sign and ofnodal solutions (sign changing solutions) for nonlinear elliptic equations driven by the p-Laplacian and having a nonsmooth potential (hemivariational inequalities). So let Z ⊆ R

N

be a bounded domain with a C2-boundary ∂Z. The problem under consideration is thefollowing:

−div(‖Dx(z)‖p−2Dx(z)

)∈ ∂j(z, x(z)) a.e. on Z,

x|∂Z = 0 1 < p <∞.(1.1)

Here j(z, x) is measurable function on Z × R, which in the x ∈ R variable is locallyLipschitz and ∂j(z, x) stands for the generalized subdifferential of x → j(z, x) in the sense of


Clarke [1]. Problem (1.1) is a hemivariational inequality. Hemivariational inequalities area new type of variational expressions, which arise in applications if one considers morerealistic mechanical laws of multivalued and nonmonotone nature. Then the correspondingenergy (Euler) functional is nonsmooth and nonconvex. Various engineering applications ofhemivariational inequalities can be found in the book of Naniewicz-Panagiotopoulos [2].

Multiple solutions of constant sign for problems monitored by the p-Laplacian andwith a C1-potential were obtained by Ambrosetti et al. [3], Garcıa Azorero-Peral Alonso [4],and Garcıa Azorero et al. [5]. In all these works, the authors consider nonlinear eigenvalueproblems and prove the existence of positive and negative solutions for certain values of theparameter λ ∈ R. The question of existence of nodal solutions was first addressed withinthe framework of semilinear problems (i.e., p = 2). We mention the works of Dancer-Du[6] and Zhang-Li [7], which contain two different approaches to the problem. In Dancer-Du[6], the authors use a combination of the variational method (critical point theory) with themethod of upper and lower solutions. In contrast Zhang-Li [7] use invariance properties ofthe negative gradient flow of the corresponding equation in C1

0(Z). Recently these methodswere extended to “smooth” problems driven by the p-Laplacian differential operator. Carl-Perera [8] extended the work of Dancer-Du [6], by assuming the existence of upper and lowersolutions for the problem. Zhang-Li [9] and Zhang et al. [10] extended the semilinear workof [7], by carefully constructing a pseudogradient vector field whose descent flow exhibitsthe necessary invariance properties. These works were extended recently by Filippakis-Papageorgiou [11]. Recently the approach based on the invariance properties of descentflow was used by Zhang-Perera [12] to produce nodal solutions for certain Kirchhoff typeequations. Other recent works dealing with p-Laplacian equations are those by Ahmad-Nieto[13] (monotone iterative technique), Kim et al. [14] (radial solutions), Lin et al. (singularodes) [15], and Vath [16] (degree theoretic approach).

In this paper using techniques from nonsmooth critical point theory in conjunctionwith the method of upper and lower solutions, we are able to extend the works of Dancer-Du [6] and Carl-Perera [8] to hemivariational inequalities. Helpful in this respect is thenonsmooth second deformation lemma of Corvellec [17]. Recently, sign-changing solutionsfor problems with discontinuous nonlinearities were obtained by Averna et al. [18], but incontrast to our work they deal with p-superlinear problems.

2. Mathematical Background

In our analysis of problem (1.1), we use the nonsmooth critical point theory which is basedon the subdifferential theory for locally Lipschitz functions and some basic facts about thespectrum of the negative p-Laplacian with Dirichlet boundary conditions. For easy reference,we recall some definitions and results from these areas, which will be used in the sequel.

We start with the subdifferential theory for locally Lipschitz functions and thecorresponding nonsmooth critical point theory. Details can be found in the books of Gasinski-Papageorgiou [19] and Motreanu-Panagiotopoulos [20]. So let X be a Banach space and letX∗ be its topological dual. By 〈·, ·〉 we denote the duality brackets for the pair (X,X∗). Givena locally Lipschitz function ϕ : X → R, the generalized directional derivative ϕ0(x;h) of ϕ atx ∈ X in the direction h ∈ X is defined as follows:

ϕ0(x;h) = lim supx′ →xλ↓0

ϕ(x′ + λh) − ϕ(x′)λ

. (2.1)


The function h → ϕ0(x;h) is sublinear continuous and so it is the support function ofa nonempty, convex, and w∗-compact set ∂ϕ(x) ⊆ X∗ defined by

∂ϕ(x) ={x∗ ∈ X∗ :

⟨x∗, h

⟩≤ ϕ0(x;h) ∀h ∈ X

}. (2.2)

The multifunction x → ∂ϕ(x) is called the “generalized gradient” (or generalizedsubdifferential) of ϕ. If ϕ : X → R is also convex, then ∂ϕ(x) coincides with thesubdifferential in the sense of convex analysis ∂cϕ(x), defined by

∂cϕ(x) ={x∗ ∈ X∗ :

⟨x∗, y − x

⟩≤ ϕ(y) − ϕ(x) ∀y ∈ X

}. (2.3)

Moreover if ϕ ∈ C1(X), then ϕ is locally Lipschitz and ∂ϕ(x) = {ϕ′(x)}.We say that x ∈ X is a critical point of the locally Lipschitz function ϕ : X → R, if

0 ∈ ∂ϕ(x). It is easy to see that if x ∈ X is a local extremum of ϕ (i.e., a local minimum or alocal maximum of ϕ), then x ∈ X is a critical point of ϕ.

A locally Lipschitz function ϕ : X → R satisfies the Palais-Smale condition at levelc ∈ R (PSc-condition for short), if every sequence {xn}n≥1 ⊆ X such that ϕ(xn) → c andm(xn) = inf{‖x∗‖ : x∗ ∈ ∂ϕ(xn)} → 0 as n → ∞ has a strongly convergent subsequence. Wesay that ϕ satisfies the PS-condition, if it satisfies the PSc-condition for every c ∈ R.

The following topological notion is crucial in the minimax characterization of thecritical values of a locally Lipschitz functional ϕ : X → R.

Definition 2.1. Let Y be a Hausdorff topological space and E0, E, and D are nonempty closedsubsets of Y with E0 ⊆ E. We say that the pair {E0, E} is linking with D in Y if and only if

(a) E0 ∩D = ∅;

(b) for any γ ∈ C(E, Y ) such that γ |E0 = id|E0 , we have γ(E) ∩D/=∅.

Using this notion, we have the following general minimax principle for the criticalvalues of a locally Lipschitz function ϕ : X → R.

Theorem 2.2. If X is a reflexive Banach space, E0, E, and D are nonempty closed subsets of X suchthat {E0, E} is linking with D in X, ϕ : X → R is locally Lipschitz, supE0

ϕ < infD ϕ, Γ ={γ ∈ C(E,X) : γ |E0 = id|E0}, c = infγ∈Γ supv∈E ϕ(γ(v)), and ϕ satisfies the PSc-condition, thenc ≥ infD ϕ and c is a critical value of ϕ.

Remark 2.3. From this general minimax principle, by appropriate choices of the linking sets,one can produce nonsmooth versions of the mountain pass theorem, of the saddle pointtheorem, and of the generalized mountain pass theorem.

Definition 2.4. If Y is a subset of the Banach space X, a “deformation of Y” is a continuousmap h : [0, 1] × Y → Y such that h(0, ·) = idY . If V ⊆ Y , then we can say that V is a “weakdeformation retract of Y”, if there exists a deformation h : [0, 1]×Y → Y such that h(1, Y ) ⊆ Vand h(t, ·) ⊆ V for all t ∈ [0, 1].


Given a locally Lipschitz function ϕ : X → R and c ∈ R, we define

0ϕc = {x ∈ X : ϕ < c},

Kc = {x ∈ x : 0 ∈ ∂ϕ(x), ϕ(x) = c}.(2.4)

The next theorem is a partial extension to a nonsmooth setting of the so-called “seconddeformation theorem” (see, e.g., Gasinski -Papageorgiou [21, page 628]) and it is due toCorvellec [17]. In fact the result of Corvellec is formulated in the more general context ofmetric spaces, for continuous functions using the so-called weak slope. For our purposes, itsuffices to use a particular form of the result which we state next.

Theorem 2.5. If X is a Banach space, ϕ : X → R is locally Lipschitz and satisfies the PS-condition,a ∈ R, b ∈ R ∪ {+∞}, ϕ has no critical points in ϕ−1(a, b), and Ka is discrete nonempty, then there

exists a deformation h : [0, 1]×0

ϕb→0

ϕb such that

(a) h(t, ·)|Ka= id for all t ∈ [0, 1];

(b) h(1,0

ϕb) ⊆0ϕa ∪Ka;

(c) ϕ(h(t, x)) ≤ ϕ(x) for all t ∈ [0, 1] and all x ∈0

ϕb .

In particular the set0ϕa ∪Ka is a weak deformation retract of

0

ϕb.Next let us recall some basic facts about the spectrum of the negative p-Laplacian with

Dirichlet boundary conditions. So let Z ⊆ RN be a bounded domain with a C2-boundary

∂Z and m ∈ L∞(Z)+, m/= 0. We consider the following nonlinear weighted (with weight m)eigenvalue problem:


)= λm(z)|x(z)|p−2x(z) a.e. on Z,

x|∂Z = 0 1 < p <∞.(2.5)

The least number λ ∈ R for which problem (2.5) has a nontrivial solution is thefirst eigenvalue of (−Δp,W

1,p0 (Z), m) and it is denoted by λ1(m). The first eigenvalue λ1(m)

is strictly positive (i.e., λ1(m) > 0); it is isolated and it is simple (i.e., the associatedeigenspace is one dimensional). Moreover, using the Rayleigh quotient we have a variationalcharacterization of λ1(m), namely,

λ1(m) = min

[‖Dx‖pp∫Zm|x|pdz

: x ∈W1,p0 (Z), x /= 0

]

, (2.6)

(see also Cuccu et al. [22]).The minimum in (2.6) is attained on the corresponding one-dimensional eigenspace.

In what follows by u1 ∈W1,p0 (Z) we denote the normalized eigenfunction. Note that |u1| also

realizes the minimum in (2.6). Hence we may assume that u1(z) ≥ 0 a.e. on Z. Moreover,from nonlinear regularity theory (see, e.g., Gasinski -Papageorgiou [21, page 738]), we have


u1 ∈ C10(Z) = {x ∈ C1(Z) : x(z) = 0 for all z ∈ ∂Z}. The Banach space C1

0(Z) is an orderedBanach space with order cone given by

C10(Z)+ =

{x ∈ C1

0(Z) : x(z) ≥ 0 ∀z ∈ Z}. (2.7)

We know that intC10(Z)+ /=∅ and in fact

intC10(Z)+ =

{x ∈ C1

0(Z)+ : x(z) > 0 ∀z ∈ Z and∂x

∂n(z) < 0 ∀z ∈ ∂Z

}. (2.8)

By virtue of the strong maximum principle of Vazquez [23], we have u1 ∈ intC10(Z)+.

Using the Lusternik-Schnirelmann theory, in addition to λ1(m) > 0, we obtain a wholestrictly increasing sequence {λk(m)}k≥1 ⊆ R+ of eigenvalues of (2.5), such that λk(m) → +∞as k → ∞. These are the so-called “variational eigenvalues” of ( −Δp,W

1,p0 (Z), m). When

p = 2 (linear case), then these are all the eigenvalues. For p /= 2 (nonlinear case), we do notknow if this is true. Nevertheless exploiting the fact that λ1(m) > 0 is isolated, we can define

λ∗2(m) = inf{λ : λ is an eigenvalue of (2.5), λ /= λ1(m)

}> λ1(m). (2.9)

Because the set of eigenvalues of (2.5) is closed, we see that λ∗2(m) is an eigenvalueof ( −Δp,W

1,p0 (Z), m). In fact we have λ∗2(m) = λ2(m); that is, the second eigenvalue and

the second variational eigenvalue of (−Δp,W1,p0 (Z), m) coincide. Then for λ2(m) we have

a variational expression provided by the Lusternik-Schnirelmann theory. The eigenvaluesλ1(m) and λ2(m) exhibit some monotonicity properties with respect to the weight functionm ∈ L∞(Z)+. More precisely, we have the following.

(a) If m(z) ≤ m′(z) a.e. on Z with strict inequality on a set of positive measure, thenλ1(m′) < λ1(m) (this is immediate from (2.6)).

(b) If m(z) < m′(z) a.e. on Z, then λ2(m′) < λ2(m) (see Anane-Tsouli [24]).

If m ≡ 1, then we write λ1(m) = λ1 and λ2(m) = λ2. For λ2 > 0, there is an alternativevariational characterization, due to Cuesta et al. [25]; namely, if ∂BLp(Z)

1 = {x ∈ Lp(Z) : ‖x‖p =1}, S = W

1,p0 (Z) ∩ ∂BLp(Z)

1 , and Γ0 = {γ0 ∈ C([−1, 1], S) : γ0(−1) = −u1, γ0(1) = u1}, then

λ2 = infγ0∈Γ0

supx∈γ0([−1,1])

‖Dx‖pp. (2.10)

Finally we recall the notions of upper and of lower solutions for problem (1.1).

(a) A function x ∈W1,p(Z) with x|∂Z ≥ 0 is an “upper solution” for problem (1.1), if

∫

Z

‖Dx(z)‖p−2(Dx(z), Dψ(z))RN dz ≥

∫

Z

u(z)ψ(z)dz (2.11)

for all ψ ∈ W1,p0 (Z), ψ(z) ≥ 0 a.e. on Z and for some u ∈ Lη(Z), u(z) ∈ ∂j(z, x(z))

a.e. on Z for some 1 < η < p∗ = Np/(N − p) if N > p, +∞ if N ≤ p.


(b) A function x ∈W1,p(Z) with x|∂Z ≤ 0 is a “lower solution” for problem (1.1), if

∫

Z

‖Dx(z)‖p−2(Dx(z), Dψ(z))RN dz ≤

∫

Z

u(z)ψ(z)dz (2.12)

for all ψ ∈ W1,p0 (Z), ψ(z) ≥ 0 a.e. on Z and for some u ∈ Lη(Z), u(z) ∈ ∂j(z, x(z))

a.e. on Z for some 1 < η < p∗.

3. Solutions of Constant Sign

In this section, we produce two nontrivial solutions of (1.1) which have constant sign. Thefirst is positive and the second is negative. To do this, we will need the following hypotheseson the nonsmooth potential j(z, x).

H(j)1: j : Z × R → R is a function such that j(z, 0) = 0 a.e. on Z, ∂j(z, 0) = {0} a.e. on Z,and

(i) for every x ∈ R, z → j(z, x) is measurable;(ii) for almost all z ∈ Z, x → j(z, x) is locally Lipschitz;(iii) for almost all z ∈ Z, all x ∈ R, and all u ∈ ∂j(z, x), we have

|u| ≤ a(z) + c|x|p−1 with a ∈ L∞(Z)+, c > 0; (3.1)

(iv) there exists θ ∈ L∞(Z)+ satisfying θ(z) ≤ λ1 a.e. on Z with strict inequality ona set of positive measure, such that

lim sup|x|→∞

u

|x|p−2x≤ θ(z) (3.2)

uniformly for almost all z ∈ Z and all u ∈ ∂j(z, x);(v) there exist η, η ∈ L∞(Z)+ satisfying λ1 ≤ η(z) ≤ η(z) a.e. on Z, where the first

inequality is strict on a set of positive measure, such that

η(z) ≤ lim infx→ 0

u

|x|p−2x≤ lim sup

x→ 0

u

|x|p−2x≤ η(z) (3.3)

uniformly for almost all z ∈ Z and all u ∈ ∂j(z, x);(vi) for almost all z ∈ Z, all x ∈ R, and all u ∈ ∂j(z, x), we have ux ≥ 0 (sign

condition).

Remark 3.1. Hypotheses H(j)1(iv) and (v) are nonuniform nonresonance conditions at zeroand at ±∞, respectively. Moreover, as we move from 0 to ±∞, the “slopes” u/|x|p−2x. u ∈∂j(z, x) cross the first eigenvalue λ1 > 0. So our framework incorporates the so-calledasymptotically p-linear equations. For p = 2, since the appearance of the pioneering workof Amann-Zehnder [26], these problems have attracted a lot of interest.


The next lemma is an easy consequence of the strict positivity of u1 ∈ C10(Z) and of the

hypotheses on θ ∈ L∞(Z)+ (see H(j)(iv)). We omit the proof.

Lemma 3.2. If θ ∈ L∞(Z)+ satisfies θ(z) ≤ λ1 a.e. on Z with strict inequality on a set of positivemeasure, then there exists ξ0 > 0 such that

‖Dx‖pp −∫

Z

θ(z)|x(z)|p dz ≥ ξ0‖Dx‖pp ∀x ∈W1,p0 (Z). (3.4)

Given ε > 0 and γε ∈ L∞(Z)+, γε /= 0, we consider the following nonlinear Dirichletproblem:


)= (θ(z) + ε)|x(z)|p−2x(z) + γε(z) a.e. on Z,

x|∂Z = 0.(3.5)

In the next proposition, we establish the solvability of (3.5).

Proposition 3.3. If θ ∈ L∞(Z)+ satisfies θ ≤ λ1 a.e. on Z with strict inequality on a set of positivemeasure, then for all ε > 0 small problem (3.5) admits a solution x ∈ intC1

0(Z)+.

Proof. In what follows by 〈·, ·〉 we denote the duality brackets for the pair(W−1,p′(Z),W1,p

0 (Z))(1/p + 1/p′ = 1). We introduce the nonlinear operator A : W1,p0 (Z) →

W−1,p′(Z) defined by

〈A(x), y〉 =∫

Z

‖Dx(z)‖p−2(Dx(z), Dy(z))RN dz ∀x, y ∈W

1,p0 (Z). (3.6)

It is straightforward to check that A is strictly monotone and demicontinuous, hencemaximal monotone too. Also let Nε : Lp(Z) → Lp′(Z) be the nonlinear, bounded, continuousmap defined by

Nε(x)(·) = (θ(·) + ε)|x(·)|p−2x(·). (3.7)

Because of the compact embedding of W1,p0 (Z) into Lp(Z), Nε viewed as a map from

W1,p0 (Z) into Lp′(Z) is completely continuous. Therefore x → Gε(x) = A(x) − Nε(x) is

pseudomonotone from W1,p0 (Z) into W−1,p′(Z). Also for every x ∈W

1,p0 (Z), we have

⟨Gε(x), x

⟩= ‖Dx‖pp −

∫

Z

θ(z)|x(z)|pdz − ε‖x‖pp

≥(ξ0 −

ε

λ1

)‖Dx‖pp (see Lemma 3.2 and (2.6)).

(3.8)


Therefore, if ε < λ1ξ0, then by virtue of Poincare’s inequality Gε(·) is coercive. But apseudomonotone coercive operator is surjective. Hence we can find that x ∈ W

1,p0 (Z) such

that

Gε(x) = A(x) −Nε(x) = γε, (3.9)

=⇒{−div

(‖Dx(z)‖p−2Dx(z)

)= (θ(z) + ε)|x(z)|p−2x(z) + γε(z) a.e. on Z,

x|∂Z = 0

}

. (3.10)

Thus x ∈ W1,p0 (Z) is a solution of (3.5). We take duality brackets of (3.9) with the test

function −x− = −max{−x, 0} ∈W1,p0 (Z). We obtain

‖Dx−‖pp −∫

Z

θ(z)|x−(z)|p dz ≤ ε‖x−‖pp(since γε ≥ 0

),

=⇒ ξ0‖Dx−‖pp ≤ε

λ1‖Dx−‖pp (see Lemma 3.2 and (2.6)).

(3.11)

But recall that ε < λ1ξ0. So it follows that ‖Dx−‖p = 0, hence x − = 0; that is, x ≥ 0.Since γε /= 0, from (3.10) it follows that x /= 0 and x ∈ C1

0(Z) (nonlinear regularity theory). Inaddition, from (3.10), we see that

div(‖Dx(z)‖p−2Dx(z)

)≤ 0 a.e. on Z,

=⇒ x ∈ intC10(Z)+ (see Vazquez [23]).

(3.12)

In fact the solution x ∈ intC10(Z)+ of (3.5) is an upper solution for problem (1.1).

Proposition 3.4. If hypotheses H(j)1(i)→ (iv) hold and ε > 0 is small, then the solution x ∈intC1

0(Z)+ of problem (3.5) obtained in Proposition 3.3 is a strict upper solution of problem (1.1)(strict means that x is an upper solution of (1.1) which is not a solution).

Proof. Because of hypotheses H(j)1(iv), given ε > 0, we can find M1 = M1(ε) > 0 such thatfor almost all z ∈ Z, all x ≥M1, and all u ∈ ∂j(z, x), we have

u ≤ (θ(z) + ε)xp−1. (3.13)

Also due to hypothesis H(j)1(iii), we can find γε ∈ L∞(Z)+, γε /= 0 such that for almostall z ∈ Z, all x ∈ [0,M1], and all u ∈ ∂j(z, x), we have

u < γε(z). (3.14)

Therefore it follows that for almost all z ∈ Z, all x ≥ 0, and all u ∈ ∂j(z, x), we have

u < (θ(z) + ε)xp−1 + γε(z). (3.15)


So for 0 < ε < λ1ξ0 and γε as above, we consider problem (3.5). From Proposition 3.3,we have a solution x ∈ intC1

0(Z)+. Then due to (3.15), for all u ∈ Lp′(Z)+ with u(z) ∈∂j(z, x(z)) a.e. on Z, we have

u(z) < (θ(z) + ε)x(z)p−1 + γε(z) a.e. on Z,

=⇒ x ∈ intC10(Z)+ is a strict upper solution for problem (1.1).

(3.16)

Since ∂j(z, 0) = {0} a.e. on Z, x ≡ 0 is a lower solution for problem (1.1).We introduce the set

C = {x ∈W1,p0 (Z) : 0 ≤ x(z) ≤ x(z) a.e. on Z} (3.17)

and the truncation function τ+ : R → R+ defined by

τ+(x) =

{0 if x ≤ 0,x if x > 0.

(3.18)

Then we set j+(·, x) = j(z, τ+(x)) and we consider the locally Lipschitz functional ϕ+ :W

1,p0 (Z) → R defined by

ϕ+(x) =1p‖Dx‖pp −

∫

Z

j+(z, x(z))dz ∀x ∈W1,p0 (Z). (3.19)

We will show that we can find a nontrivial solution of (1.1) in C, which is a localminimizer of ϕ+ and of ϕ. To do this we will need the following simple result about orderedBanach spaces.

Lemma 3.5. IfX is an ordered Banach space,K is the order cone ofX, intK/=∅, and x0 ∈ intK, thenfor every y ∈ X, we can find t = t(y) > 0 such that tx0 − y ∈ intK.

Proof. Since x0 ∈ intK, we can find δ > 0 such that

Bδ(x0) ={x ∈ X : ‖x − x0‖ ≤ δ

}⊆ intK. (3.20)

Let y ∈ X, y /= 0 (if y = 0, then clearly the lemma holds for all t > 0). We have thefollowing:

x0 − δy

‖y‖ ∈ intK,

=⇒‖y‖δ

x0 − y ∈ intK.

(3.21)

So, if t = ‖y‖/δ, then tx0 − y ∈ intK.


Using this lemma, we can prove the following result.

Proposition 3.6. If hypotheses H(j)1 hold, then there exists x0 ∈ C which is a local minimizer of ϕ+

and of ϕ.

Proof. From (3.15), we know that given ε > 0, we can find γε ∈ L∞(Z)+, γε /= 0 such that

u < (θ(z) + ε)xp−1 + γε(z) for a.a. z ∈ Z, all x ≥ 0,

and all u ∈ ∂j+(z, x) = ∂j(z, x).(3.22)

Because of hypotheses H(j)1(i), (ii), for almost all z ∈ Z, x → j+(z, x) isalmost everywhere differentiable on R (Rademacher’s theorem) and at every point ofdifferentiability we have

d

dxj+(z, x) ∈ ∂j+(z, x),

=⇒ d

dxj+(z, x) < (θ(z) + ε)xp−1 + γε(z) for a.a. z ∈ Z, all x ≥ 0 (see (3.22)).

(3.23)

Integrating this inequality and since j+(z, x)|R− = 0 for almost all z ∈ Z, we obtain

j+(z, x) <1p(θ(z) + ε)|x|p + γε(z)|x| for a.a. z ∈ Z, all x ∈ R. (3.24)

Then for every x ∈W1,p0 (Z), we have

ϕ+(x) =1p‖Dx‖pp −

∫

Z

j+(z, x(z))dz

>1p‖Dx‖pp −

1p

∫

Z

θ(z)|x(z)|pdz − ε

p‖x‖pp − c1‖Dx‖p

for some c1 > 0 (see (3.24))

≥ 1p

(ξ0 −

ε

λ1

)‖Dx‖pp − c1‖Dx‖p (see Lemma 3.2).

(3.25)

Choosing ε < λ1ξ0, because p > 1, from (3.25) and Poincare’s inequality, we infer thatϕ+ is coercive. Also it is easy to see that ϕ+ is weakly lower semicontinuous on W

1,p0 (Z). Hence

by virtue of the theorem of Weierstrass, we can find x0 ∈ C such that

ϕ+(x0) = infCϕ+. (3.26)

First we show that x0 /= 0. To this end, note that hypothesis H(j)1(v) implies that givenε > 0, we can find δ = δ(ε) > 0 such that

u ≥ (η(z) − ε)xp−1 for a.a. z ∈ Z, all x ∈ [0, δ] and all u ∈ ∂j+(z, x) = ∂j(z, x). (3.27)


As before, integrating (3.27), we obtain

j+(z, x) ≥1p(η(z) − ε)xp for a.a. z ∈ Z, allx ∈ [0, δ]. (3.28)

We know that x ∈ intC10(Z)+ (see Proposition 3.3). So using Lemma 3.5, we can find

μ > 0 small such that

μu1(z) ≤ min{x(z), δ} ∀z ∈ Z. (3.29)

Then, because of (3.28), we have

ϕ+(μu1)=

μp

p

∥∥Du1

∥∥pp −∫

Z

j+(z, μu1(z)

)dz

≤μp

p

∥∥Du1∥∥pp −

μp

p

∫

Z

η(z)u1(z)p dz +

μpε

p

∥∥u1∥∥pp

=μp

p

[∫

Z

(λ1 − η(z)

)u1(z)

p dz + ε∥∥u1∥∥pp

]

.

(3.30)

Let σ =∫Z(λ1 − η(z))u1(z)

pdz. Using the hypothesis on η (see H(j)1(v)) and the factthat u1(z) > 0 for all z ∈ Z, we see that σ < 0. So, if we choose ε < −σ/‖u1‖

pp, we have

ϕ+(μu1)< 0 ∀μ > 0 small. (3.31)

Note that for μ > 0 small, μu1 ∈ C. Hence

ϕ+(x0)= inf

Cϕ+ ≤ ϕ+

(μu1)< 0 = ϕ+(0) (see (3.31)),

=⇒ x0 /= 0, x0 ∈ C.(3.32)

Given any y ∈ C, we define k0(t) = ϕ+(ty + (1 − t)x0), t ∈ [0, 1]. Then k is Lipschitzcontinuous, hence differentiable almost everywhere and k0(0) ≤ k0(t) for all t ∈ [0, 1].From Chang [27, page 106], we know that we can find u ∈ Lp′(Z), u(z) ∈ ∂j+(z, x0(z)) =∂j(z, x0(z)) a.e. on Z, such that

0 ≤⟨A(x0), y − x0

⟩−∫

Z

u(z)(y − x0

)(z)dz, ∀y ∈ C. (3.33)

For any v ∈W1,p0 (Z) and ε > 0, we define

y(z) =

⎧⎪⎪⎨

⎪⎪⎩

0 if z ∈{x0 + εv ≤ 0

},

x0(z) + εv(z) if z ∈{

0 < x0 + εv < x},

x(z) if z ∈{x ≤ x0 + εv

}.

(3.34)


Clearly y ∈ C. We use this y ∈ C in (3.33). Hence we obtain

0 ≤ ε

∫

{0<x0+εv<x}‖Dx0‖p−2(Dx0, Dv)

RN dz −∫

{0<x0+εv<x}u(εv)dz

−∫

{x0+εv≤0}‖Dx0‖p dz +

∫

{x0+εv≤0}ux0 dz

+∫

{x0+εv≥x}‖Dx0‖p−2(Dx0, D(x − x0))RN dz −

∫

{x0+εv≥x}u(x − x0)dz

= ε

∫

Z

‖Dx0‖p−2(Dx0, Dv)RN dz − ε

∫

Z

uv dz

−∫

{x0+εv≥x}‖Dx‖p−2(Dx,D(x0 + εv − x)

RN dz +∫

{x0+εv≥x}u(x0 + εv − x)dz

(u ∈ Lp′(Z), u(z) ∈ ∂j+(z, x(z)) a.e. and clearly from the definition of u,

we can always assume u = u a.e. on{x = x0

})

+∫

{x0+εv≤0}u(x0 + εv

)dz +

∫

{x0+εv≥x}(u − u)

(x − x0 − εv

)dz

−∫

{x0+εv≤0}

∥∥Dx0∥∥pdz − ε

∫

{x0+εv≤0}

∥∥Dx0∥∥p−2(

Dx0, Dv)

RN dz

+∫

{x0+εv≥x}

(∥∥Dx∥∥p−2

Dx −∥∥Dx0

∥∥p−2Dx0, D

(x0 − x

))RN dz

+ ε

∫

{x0+εv≥x}

(∥∥Dx∥∥p−2

Dx −∥∥Dx0

∥∥p−2Dx0, Dv

)RN dz.

(3.35)

Using h = (x0 + εv − x)+ ∈W1,p0 (Z)+ as a test function, from the definition of an upper

solution for problem (1.1), we have

−∫

{x0+εv≥x}‖Dx‖p−2(Dx,D

(x0 + εv − x

))RN dz +

∫

{x0+εv≥x}u(x0 + εv − x

)dz ≤ 0. (3.36)

Also from the (strict) monotonicity of the operator A, we have

∫

{x0+εv≥x}

(∥∥Dx∥∥p−2

Dx −∥∥Dx0

∥∥p−2Dx0, D

(x0 − x

))RN dz ≤ 0. (3.37)

From hypothesis H(j)1(vi), it follows that

∫

{x0+εv≤0}u(x0 + εv

)dz ≤ 0. (3.38)


Since x0 ∈ C10(Z), we have

∫

{x0+εv≥x}(u − u)

(x − x0 − εv

)dz

=∫

{x0+εv≥x>x0}(u − u)

(x − x0 − εv

)dz

(recall that u = u a.e. on

{x = x0

}, x0 ≤ x since x0 ∈ C

)

≤ c2

∫

{x0+εv≥x>x0}

(x0 + εv − x

)dz for some c2 > 0

(see hypothesis H(j)1(iii)

)

≤ εc2

∫

{x0+εv≥x>x0}v dz

(since x0 ≤ x

).

(3.39)

Returning to (3.35) and using (3.36)→ (3.39), we obtain

0 ≤ ε

∫

Z

∥∥Dx0∥∥p−2(

Dx0, Dv)

RN dz − ε∫

Z

uv dz

+ εc2

∫

{x0+εv≥x>x0}v dz − ε

∫

{x0+εv≤0}

∥∥Dx0∥∥p−2(

Dx0, Dv)

RN dz

+ ε

∫

{x0+εv≥x}

(∥∥Dx∥∥p−2

Dx −∥∥Dx0

∥∥p−2Dx0, Dv

)RN dz.

(3.40)

We denote by | · |N the Lebesgue measure on RN. Then

∣∣{x0 + εv ≥ x > x0}∣∣

N ↓ 0 as ε ↓ 0. (3.41)

Moreover, from Stampacchia’s theorem, we know that

Dx0(z) = 0 a.e. on{x0 = 0

}, Dx0(z) = Dx(z) a .e. on

{x0 = x

}. (3.42)

If we divide (3.40) by ε > 0 and then we pass to the limit as ε ↓ 0, because of (3.41) and(3.42), we obtain

0 ≤⟨A(x0), v⟩−∫

Z

uv dz =⟨A(x0)− u, v

⟩. (3.43)

Recall that v ∈W1,p0 (Z) was arbitrary. So from (3.43), it follows that

A(x0) = u,

=⇒ x0 ∈W1,p0 (Z) is a solution of problem (1.1).

(3.44)


The nonlinear regularity theory implies that x0 ∈ C10(Z) and then since x0 /= 0, x0 ≥ 0

from the nonlinear strong maximum principle of Vazquez [23], we have x0 ∈ intC10(Z)+.

From (3.22), we know that

u(z) < (θ(z) + ε)x(z)p−1 + γε(z) a.e. on Z(recall that x0 ≤ x

). (3.45)

Then Proposition 2.2 of Guedda-Veron [28] implies that

x0(z) < x(z) ∀z ∈ Z,∂x

∂n(z) <

∂x0

∂n(z) ∀z ∈ ∂Z,

=⇒ x − x0 ∈ intC10(Z)+.

(3.46)

Recall also that x0 ∈ intC10(Z)+. Thus we can find δ > 0 such that

BC1

0(Z)δ

(x − x0

)={y ∈ C1

0(Z) :∥∥y −

(x − x0

)∥∥C1

0(Z) < δ}⊆ intC1

0(Z)+,

BC1

0(Z)δ

(x0)={y ∈ C1

0(Z) :∥∥y − x0

∥∥C1

0(Z) < δ}⊆ intC1

0(Z)+.(3.47)

These inclusions imply that

x −(x0 + B

C10(Z)

δ

)⊆ intC1

0(Z)+, x0 + BC1

0(Z)δ

⊆ intC10(Z)+. (3.48)

The solution x0 was obtained as a minimizer of ϕ+ on C. Then (3.48) implies that x0

is also a local minimizer of ϕ+ and of ϕ on C10(Z). But then from Motreanu-Papageorgiou

[29] (see also Gasinski -Papageorgiou [21, pages 655-656]), it follows that x0 is also a localW

1,p0 (Z)-minimizer of ϕ+ and of ϕ too.

So far we have worked on the positive semiaxis. Next we repeat the same analysis onthe negative semiaxis. More precisely, given ε > 0 and γε ∈ L∞(Z)+, γε /= 0, we consider thefollowing auxiliary problem:

−div(∥∥Dv(z)

∥∥p−2Dv(z)

)=(θ(z) + ε

)∣∣v(z)∣∣p−2

v(z) − γε(z) a.e. on Z,

v|∂Z = 0.(3.49)

Then as in the proof of Proposition 3.3, through the surjectivity of the pseudomonotonecoercive operator v → Gε(v) = A(v) −Nε(v), we obtain a solution u ∈ −intC1

0(Z)+ of (3.49).We can check that v is a strict lower solution of (1.1), while clearly v ≡ 0 is an upper solution(in fact a solution) of (1.1). This time we consider the set

D ={v ∈W

1,p0 (Z) : v(z) ≤ v(z) ≤ 0 a.e. on Z

}(3.50)


and the truncation function τ− : R → R− defined by

τ−(x) =

{x if x < 0,0 if x ≥ 0.

(3.51)

We set j−(z, x) = j(z, τ−(x)) and then introduce the locally Lipschitz functional ϕ− :W

1,p0 (Z) → R defined by

ϕ−(x) =1p‖Dx‖pp −

∫

Z

j−(z, x(z)

)dz ∀x ∈W

1,p0 (Z). (3.52)

We consider the minimization problem

infDϕ−. (3.53)

Arguing as in Proposition 3.3, we obtain the following.

Proposition 3.7. If hypothesesH(j)1 hold, then there exists v0 ∈ D which is a local minimizer of ϕ−and of ϕ.

Now combining Propositions 3.6 and 3.7, we obtain a multiplicity result for problem(1.1) with solutions of constant sign.

Theorem 3.8. If hypotheses H(j)1 hold, then the problem (1.1) has at least two solutions x0 ∈intC1

0(Z)+ and v0 ∈ −intC10(Z)+.

Remark 3.9. From Propositions 3.6 and 3.7, we know that both x0 and v0 are local minimizersof ϕ. So we must have a third critical point of ϕ, distinct from x0, v0. However, at this pointwe cannot guarantee that it is nontrivial. In the next section by strengthening our hypothesison j(z, ·) near the origin (see H(j)1(v)), we will be able to show that this third critical pointis nontrivial and in fact is a nodal solution.

4. Existence of Nodal Solution

Recall that every eigenfunction of (2.5) corresponding to an eigenvalue λ /= λ1 must changesign. So we expect that in general the sign changing solutions of (1.1) must be more than thesolutions of constant sign. Nevertheless to produce a sign-changing solution (also known asnodal solution) for (1.1) is a rather involved process.

Here we follow an approach first employed by Dancer-Du [6] for semilinear problems(i.e., p = 2) and recently extended to problems with the p-Laplacian and a smooth potential byCarl-Perera [8]. Roughly speaking, the strategy is as follows. Continuing with the argumentused in Section 3, we produce the smallest positive solution y+ and the largest negativesolution y−. Then we form order interval [y−, y+]. Using variational techniques (in particularTheorem 2.2) we produce a solution y0 of (1.1) in [y−, y+] different from y− and y+. Evidentlyif y0 /= 0, then y0 must be sign changing. To show that y0 is nontrivial, we employ Theorem 2.5and (2.10). In addition to the works of Dancer-Du [6] and Carl-Perera [8], variants of this


method can also be found in the works of Ambrosetti-Garcia Azorero-Peral Alonso [3] andJin [30]. A different approach based on the construction of a pseudogradient vector field withappropriate invariance properties can be found in Zhang-Li [7, 9], Zhang et al. [10] (see alsoLi-Wang [31]).

We start executing the solution strategy outlined above by proving first a lemma whichestablishes that the set of upper solutions for problem (1.1) is downward directed.

Lemma 4.1. If y1, y2 ∈ W1,p(Z) are two upper solutions for problem (1.1) and y = min{y1, y2} ∈W1,p(Z), then y is also an upper solution for problem (1.1).

Proof. Given ε > 0, we consider the truncation function ξε : R → R defined by

ξε(s) =

⎧⎪⎪⎨

⎪⎪⎩

ε if s ≥ ε,

s if s ∈ [−ε, ε],−ε if s ≤ −ε.

(4.1)

Clearly ξε is Lipschitz continuous. So from Marcus-Mizel [32], we have

ξε((y1 − y2

)−) ∈W1,p(Z),

Dξε((y1 − y2

)−) = ξ′ε((y1 − y2

)−)D(y1 − y2

)−.

(4.2)

Consider a test function ψ ∈ C1c(Z) with ψ ≥ 0. Then

ξε((y1 − y2

)−)ψ ∈W1,p(Z) ∩ L∞(Z),

D(ξε((y1 − y2

)−)ψ)= ψDξε

((y1 − y2

)−) + ξε((y1 − y2

)−)Dψ.

(4.3)

Because y1, y2 ∈W1,p(Z) are upper solutions for problem (1.1), we have

⟨A(y1), ξε((y1 − y2

)−)ψ⟩≥⟨u1, ξε

((y1 − y2

)−)ψ⟩,

⟨A(y2),(ε − ξε

((y1 − y2)

−))ψ⟩≥⟨u2,(ε − ξε

((y1 − y2

)−))ψ⟩ (4.4)

for some uk ∈ Lp′(Z) with uk(z) ∈ ∂j(z, yk(z)) a.e. on Z, k = 1, 2. Adding these inequalities,we obtain

⟨A(y1), ξε((y1 − y2

)−)ψ〉 + 〈A

(y2),(ε − ξε

((y1 − y2

)−)ψ⟩

≥⟨u1, ξε

((y1 − y2

)−)ψ〉 +

⟨u2,(ε − ξε

((y1 − y2

)−)ψ〉.

(4.5)


Note that

⟨A(y1), ξε((y1 − y2

)−)ψ⟩

=∫

Z

∥∥Dy1

∥∥p−2(

Dy1, D(y1 − y2

)−)RNξ

′ε

((y1 − y2

)−)ψ dz

+∫

Z

∥∥Dy1

∥∥p−2(

Dy1, Dψ)

RNξε((y1 − y2

)−)dz

= −∫

{−ε≤y1−y2≤0}

∥∥Dy1

∥∥p−2(

Dy1, D(y1 − y2

))RNψ dz

+∫

Z

∥∥Dy1

∥∥p−2(

Dy1, Dψ)

RNξε((y1 − y2

)−)dz,

⟨A(y2),(ε − ξε

((y1 − y2

)−))ψ⟩

=∫

{−ε≤y1−y2≤0}

∥∥Dy2∥∥p−2(

Dy2, D(y1 − y2

))RNψ dz

+∫

Z

∥∥Dy2∥∥p−2(

Dy2, Dψ)

RN

(ε − ξε

((y1 − y2

)−))dz.

(4.6)

Adding (4.6) and recalling that ψ ≥ 0, we obtain

⟨A(y1), ξε((y1 − y2

)−)ψ⟩+⟨A(y2),(ε − ξε

((y1 − y2

)−))ψ⟩

=∫

{−ε≤y1−y2≤0}

(∥∥Dy2∥∥p−2

Dy2 −∥∥Dy1

∥∥p−2Dy1, D

(y1 − y2

))RNψ dz

+∫

Z

∥∥Dy1∥∥p−2(

Dy1, Dψ)

RNξε((y1 − y2

)−)dz

+∫

Z

∥∥Dy2∥∥p−2(

Dy2, Dψ)

RN(ε − ξε((y1 − y2

)−))dz

≤∫

Z

∥∥Dy1∥∥p−2(

Dy1, Dψ)

RNξε((y1 − y2

)−)dz

+∫

Z

∥∥Dy2∥∥p−2(

Dy2, Dψ)

RN

(ε − ξε

((y1 − y2

)−))dz.

(4.7)

Returning to (4.5), using (4.7), and dividing by ε > 0, we get

∫

Z

∥∥Dy1∥∥p−2(

Dy1, Dψ)

RN

1εξε((y1 − y2

)−)dz

+∫

Z

∥∥Dy2∥∥p−2(

Dy2, Dψ)(

1 − 1εξε((y1 − y2

)−))dz

≥⟨u,

1εξε((y1 − y2

)−)ψ

⟩+⟨u2,

(1 − 1

εξε((y1 − y2

)−))ψ

⟩.

(4.8)


We observe that

1εξε((y1 − y2

)−(z))−→ χ{y1<y2}(z) a.e. on Z as ε ↓ 0,

χ{y1≥y2} = 1 − χ{y1<y2}.

(4.9)

Therefore, if we pass to the limit as ε ↓ 0 in (4.8), we obtain

∫

{y1<y2}

∥∥Dy1

∥∥p−2(

Dy1, Dψ)

RN dz +∫

{y1≥y1}

∥∥Dy2

∥∥p−2(

Dy2, Dψ)

RN dz

≥∫

{y1<y2}u1ψ dz +

∫

{y1≥y2}u2ψ dz.

(4.10)

Since y = min{y1, y2} ∈W1,p(Z), we have

Dy(z) =

⎧⎨

⎩Dy1(z) for a.a. z ∈

{y1 < y2

},

Dy2(z) for a.a. z ∈{y1 ≥ y2

}.

(4.11)

Also if u = χ{y1<y2}u1 + χ{y1≥y2}u2, then u ∈ Lp′(Z) and u(z) ∈ ∂j(z, y(z)) a.e. on Z.Therefore

∫

Z

‖Dy‖p−2(Dy,Dψ)RNdz ≥

∫

Z

uψdz (4.12)

for some u ∈ Lp′(Z) with u(z) ∈ ∂j(z, y(z)) a.e. on Z. Since ψ ∈ C1c(Z)+ was arbitrary and

C1c(Z)+ is dense in W

1,p0 (Z)+, from (4.12) we conclude that y = min{y1, y2} ∈ W1,p(Z) is an

upper solution for problem (1.1).

Arguing similarly, we can also show that the set of lower solutions for problem (1.1)is upward directed. Namely, we have the following.

Lemma 4.2. If v1, v2 ∈ W1,p(Z) are lower solutions for problem (1.1) and v = max{v1, v2} ∈W1,p(Z), then v is also a lower solution for problem (1.1).

Now that we have established that the sets of upper solutions and of lower solutionsare directed, we will show that problem (1.1) admits the smallest positive solution and thelargest negative solution. To this end, we need to strengthen the hypotheses on j(z, x).

H(j)2: j : Z × R → R is a function such that j(z, 0) = 0 a.e. on Z, ∂j(z, 0) = {0} a.e. on Z;it satisfies hypotheses H(j)1(i)→ (iv) and (vi) and

(v) there exists η ∈ L∞(Z)+ such that

λ1 < lim infx→ 0

u

|x|p−2x≤ lim sup

x→ 0

u

|x|p−2x≤ η(z) (4.13)

uniformly for almost all z ∈ Z and all u ∈ ∂j(z, x).


Remark 4.3. Note that in the new hypotheses, we have strengthened the condition concerningthe behavior of ∂j(z, ·) near the origin. This has as a consequence that we can replace theorigin as a lower solution in the positive axis and as an upper solution in the negative axis(see Section 3), by functions which are strictly positive and strictly negative, respectively. Thisis done in the next lemma.

Lemma 4.4. If hypotheses H(j)2 hold, then problem (1.1) has a strict lower solution x ∈ intC10(Z)+

and a strict upper solution v ∈ −intC10(Z)+.

Proof. By virtue of hypothesis H(j)2(v), we can find c > λ1 and δ > 0 such that for almost allz ∈ Z, all x ∈ [0, δ], and all u ∈ ∂j(z, x), we have

cxp−1 ≤ u. (4.14)

We know that for u1 the principal eigenfunction of (−Δp,W1,p0 (Z)) (i.e., m ≡ 1), we

have that u1 ∈ intC10(Z)+. Thus we can find μ ∈ (0, 1) small enough such that 0 < μu1(z) ≤ δ

for all z ∈ Z. Let x ∈ intC10(Z)+ be the strict upper solution for problem (1.1) obtained in

Proposition 3.4. Invoking Lemma 3.5, we can find t > 1 such that tx−μu1 ∈ intC10(Z)+. We set

x =μ

tu1 ∈ intC1

0(Z)+. (4.15)

Note that since t > 1, x(z) ∈ (0, δ] for all z ∈ Z. Hence

−div(∥∥Dx(z)

∥∥p−2)Dx(z) = λ1|x(z)|p−2x(z)

< c|x(z)|p−2x(z)

≤ u(z) a.e. on Z

(4.16)

for every u ∈ Lp′(Z) with u(z) ∈ ∂j(z, x(z)) a.e. on Z (see (4.14)). Hence for all ψ ∈W1,p(Z)+,we have

∫

Z

∥∥Dx∥∥p−2(Dx,Dψ)

RN dz <

∫

Z

uψ dz,

=⇒ x ∈ intC10(Z)+ is a strict lower solution for problem (1.1).

(4.17)

Note that from the definition of x, we have x − x ∈ intC10(Z)+.

A similar reasoning applied on the negative semiaxis produces an upper solution v =(μ/t)(−u1) for some 0 ≤ μ < 1 < t. Then v ∈ −intC1

0(Z)+ and as for x, we will have v − v ∈intC1

0(Z)+.

Using {x, x} and {v, v}, we introduce the following order intervals in W1,p0 (Z):

[x, x] ={x ∈W

1,p0 (Z) : x(z) ≤ x(z) ≤ x(z) a.e. on Z

},

[v, v] ={v ∈W

1,p0 (Z) : v(z) ≤ v(z) ≤ v(z) a.e. onZ

}.

(4.18)


In the next proposition, we establish the existence of the smallest solution of (1.1) in[x, x] and of the greatest solution of (1.1) in [v, v].

Proposition 4.5. If hypotheses H(j)2 hold, then problem (1.1) admits the smallest solution in theorder interval [x, x] and the greatest solution in the order interval [v, v].

Proof. We will show that the existence of the smallest solution in [x, x] and the proof of thegreatest solution in [v, v]is similar.

Let S+ be the set of solutions of (1.1) belonging in the order interval E+ = [x, x]. Wewill show that S+ is downward directed. So let x1, x2 ∈ S+. In particular both x1, x2 are uppersolutions for problem (1.1). Then Lemma 4.1 implies that x = min{x1, x2} ∈ W

1,p0 (Z) is also

an upper solution for problem (1.1). We set

E+ = [x, x] ={x ∈W

1,p0 (Z) : x(z) ≤ x(z) ≤ x(z)

}. (4.19)

Using standard truncation and penalization techniques, we can obtain x0 ∈ E+

a solution of (1.1) (see Carl-Heikkila [33] and Gasinski -Papageorgiou [19]). Nonlinearregularity theory implies that x0 ∈ C1

0(Z)+ and we have

x ≤ x0 ≤ min{x1, x2

},

=⇒ S+ is downward directed.(4.20)

Consider a chain Γ ⊆ S+ (i.e., a totally ordered subset of S+). By virtue of [34, Corollary7, page 336] by Dunford-Schwartz, we can find {xn}n≥1 ⊆ Γ such that

infn≥1

xn = inf Γ. (4.21)

Because of (4.20), we may assume that {xn}n≥1 is decreasing. Also since the xn’s aresolutions of (1.1) in E+, we see that there exists c3 > 0 such that

∥∥Dxn

∥∥p ≤ c3 ∀n ≥ 1. (4.22)

Therefore {xn}n≥1 ⊆W1,p0 (Z) is bounded and so we may assume that

xnw−→ y in W

1,p0 (Z), xn −→ y in Lp(Z) as n −→ ∞. (4.23)

We can find un ∈ Lp′(Z) with un(z) ∈ ∂j(z, xn(z)) a.e. on Z such that

A(xn

)= un ∀n ≥ 1. (4.24)

Because of hypothesis H(j)2(iii), {un}n≥1 ⊆ Lp′(Z) is bounded. So we may assume thatun

w→ u in Lp′(Z) as n → ∞. By virtue of Hu-Papageorgiou [35, Proposition 3.10, page 694],


we have u(z) ∈ ∂j(z, y(z)) a.e. on Z (recall that the multifunction x → ∂j(z, x) has closedgraph; see Clarke [1, page 29]). From (4.24), we have

⟨A(xn

), xn − y

⟩=∫

Z

un

(xn − y

)dz −→ 0 as n −→ ∞. (4.25)

Since A is maximal monotone, we have (see Gasinski -Papageorgiou [19, page 84])

⟨A(xn

), xn

⟩−→⟨A(y), y

⟩,

=⇒∥∥Dxn

∥∥p −→

∥∥Dy

∥∥p.

(4.26)

Recalling that Dxnw→ Dy in Lp(Z,RN), we infer that Dxn → Dy in Lp(Z,RN) (Kadec-

Klee property) and so we conclude that xn → y in W1,p0 (Z) as n → ∞. So, if we pass to the

limit as n → ∞ in (4.24) and since A is demicontinuous, it follows that

A(y) = u (4.27)

with u ∈ Lp′(Z), u(z) ∈ ∂j(z, y(z)) a.e. on Z. Therefore y ∈ S+ and y = inf Γ. Because Γ wasan arbitrary chain, invoking Zorn’s lemma, we obtain x∗ ∈ S+ a minimal element. Then from(4.20), we conclude that x∗ is the smallest solution of (1.1) in E+.

We will use this proposition to produce the smallest positive and the greatest negativesolutions for problem (1.1).

Proposition 4.6. If hypotheses H(j)2 hold, then problem (1.1) has the smallest nontrivial solutiony+ ∈ intC1

0(Z)+ and the greatest nontrivial negative solution y− ∈ −intC10(Z)+.

Proof. Let xn = εnu1 with εn ↓ 0 and let En+ = [xn, x]. From Proposition 4.5, we know that

problem (1.1) admits the smallest solution xn∗ in the order interval En

+. We know that {xn∗ }n≥1 ⊆

W1,p0 (Z) is bounded and so we may assume that

xn∗

w−→ y+ in W1,p0 (Z), xn

∗ −→ y+ in Lp(Z) as n −→ ∞. (4.28)

We know that

A(xn∗)= un

∗ ∀n ≥ 1, (4.29)

with un∗ ∈ Lp′(Z), un

∗ (z) ∈ ∂j(z, xn∗ (z)) a.e. on Z. Taking duality brackets with xn

∗ − y+ andarguing as before , we can check that

xn∗ −→ y+ in W

1,p0 (Z) as n −→ ∞. (4.30)


Suppose that y+ = 0. Then ‖xn∗ ‖ → 0 as n → ∞. We set wn = xn

∗/‖xn∗ ‖, n ≥ 1. Then by

passing to a suitable subsequence if necessary, we may assume that

wnw−→ w in W

1,p0 (Z), wn −→ w in Lp(Z) n −→ ∞. (4.31)

From (4.29), we have

A(wn

)=

un∗

‖xn∗ ‖p−1

∀n ≥ 1 (4.32)

and so taking duality brackets with wn −w, again we obtain

limn→∞

⟨A(wn

), wn −w

⟩= 0 (4.33)

from which it follows that

wn −→ w in W1,p0 (Z) (4.34)

and so we have ‖w‖ = 1, w /= 0. By virtue of hypothesis H(j)2(v), we can find δ > 0 such thatfor almost all z ∈ Z, all 0 < |x| ≤ δ, and all u ∈ ∂j(z, x), we have

β ≤ u

|x|p−2x≤ η(z) + 1 with β > λ1. (4.35)

In addition, due to hypothesis H(j)2(iii), for almost all z ∈ Z, all |x| ≥ δ, and allu ∈ ∂j(z, x), we have

|u| ≤ α(z) + c|x|p−1 ≤(α(z)δp−1

+ c

)|x|p−1. (4.36)

So finally from (4.35) and (4.36), we infer that

|u| ≤ c3|x|p−1 for a.a. z ∈ Z, all x ∈ R, and all u ∈ ∂j(z, x). (4.37)

From (4.37) it follows that

{un∗

‖xn∗ ‖p−1

}

n≥1⊆ Lp′(Z) is bounded. (4.38)

Therefore we may assume that

hn =un∗

‖xn∗ ‖p−1

w−→ h in Lp′(Z). (4.39)


For ε > 0 and n ≥ 1 we introduce the sets

Σn+ ={z ∈ Z : xn(z) > 0, β − ε ≤ un

∗ (z)

xn∗ (z)

p−1≤ η(z) + ε

},

Σn− ={z ∈ Z : xn(z) < 0, β − ε ≤ un

∗ (z)

xn∗ (z)

p−1≤ η(z) + ε

}.

(4.40)

Since xn∗ → 0 in W

1,p0 (Z), we may assume (at least for a subsequence) that xn

∗ (z) → 0a.e. on Z. Then xn

∗ (z) → 0+ a.e. on {w > 0} and xn∗ (z) → 0− a.e. on {w < 0} and so because

of hypothesis H(j)2(v), we have

χΣn+(z) −→ 1 a.e. on {w > 0}, χΣn

−(z) −→ 1 a.e. on {w < 0}. (4.41)

Then we have

χΣn+

un∗

∥∥xn∗∥∥p−1

w−→ h in Lp′({w > 0}), χΣn−

un∗

∥∥xn∗∥∥p−1

w−→ h in Lp′({w < 0}). (4.42)

From the definition of the set Σn+, we have

χΣn+(z)(β − ε)wn(z)

p−1 ≤ χΣn+(z)

un∗ (z)

xn∗ (z)

p−1wn(z)

p−1 = χΣn+(z)hn(z)

≤ χΣn+(z)(η(z) + ε)wn(z)

p−1 a.e. on Z.

(4.43)

Taking weak limits in Lp′({w > 0}), via Mazur’s lemma, and since ε > 0 was arbitrary,we obtain

βw(z)p−1 ≤ h(z) ≤ η(z)w(z)p−1 a.e. on {w > 0}. (4.44)

Similarly working on Σn−, we obtain

η(z)|w(z)|p−2w(z) ≤ h(z) ≤ β|w(z)|p−2w(z) a.e. on {w < 0}. (4.45)

Moreover, from (4.36) we see that

h(z) = 0 a.e. on {w = 0}. (4.46)

So from (4.44), (4.45), and (4.46) it follows that

h(z) = ξ1(z)|w(z)|p−2w(z) a.e. on Z, (4.47)


with ξ1 ∈ L∞(Z)+ and λ1 < ξ1(z) ≤ η(z) a.e. on Z. Therefore, if we pass to the limit as n → ∞in (4.32), we obtain

A(w) = ξ1|w|p−2w,

=⇒{−div

(∥∥Dw(z)∥∥p−2

Dw(z))= ξ1(z)|w(z)|p−2w(z) a.e. on Z,

w|∂Z = 0 w/= 0

}

.(4.48)

Note that λ1(ξ1) < λ(λ1) = 1. So from (4.48), it follows that w must change sign. Butwn = xn

∗/‖xn∗ ‖ ≥ 0 for all n ≥ 1 and so w ≥ 0, a contradiction. This proves that we cannot have

y+ = 0, hence y+ /= 0 and of course y+ ≥ 0. Moreover, as before we can check that

xn∗ −→ y+ in W

1,p0 (Z), A(y+) = u+ (see (4.29)), (4.49)

with u+ ∈ Lp′(Z), u+(z) ∈ ∂j(z, y+(z)) a.e. on Z. It follows that

−div(∥∥Dy+(z)

∥∥p−2Dy+(z)

)= u+(z) a.e. on Z,

y+|∂Z = 0.(4.50)

From nonlinear regularity theory we have y+ ∈ C10(Z)+, y+ /= 0. Moreover, from

hypothesis H(j)2(v) (the sign condition), we have u+(z) ≥ 0 a.e. on Z. So via the nonlinearstrong maximum principle of Vazquez [23], we obtain that y+ ∈ intC1

0(Z)+.We claim that y+ is the smallest nontrivial positive solution of (1.1). Indeed let y be

another nontrivial positive solution of (1.1) and assume that y ≤ x. As above we can verifythat y ∈ intC1

0(Z)+. Using Lemma 3.5, we can find ε > 0 such that εu1 ≤ y. Then for n ≥ 1 largewe have εnu1 ≤ εu1 ≤ y ≤ x. So for n ≥ 1 large, working on the order interval [εnu1, y], we canobtain a solution y0 of (1.1) in the interval. Then xn

∗ ≤ y0 for n ≥ 1 large and so y+ ≤ y0 ≤ y.This proves the claim.

In a similar fashion working on En− = [v, vn] with vn = εn(−u1), we obtain y− ∈

intC10(Z)+, the largest nontrivial negative solution of (1.1).

Now we are ready to conclude our plan and produce a nontrivial nodal solution. Moreprecisely, using variational methods and Theorem 2.5, we will obtain a nontrivial solution y0

of (1.1) in the order interval [y−, y+] distinct from y− and y+. Evidently y0 must be a signchanging (nodal) solution. We will achieve this by exploiting the variational characterizationof λ2 provided in (2.10). This requires a further strengthening of hypotheses H(j)1. Supposef : Z × R → R is a measurable function with the following property, for every r > 0 thereexists αr ∈ L∞(Z)+ such that |f(z, x)| ≤ αr(z) for a.a. z ∈ Z and all |x| ≤ r. Then we define

f1(z, x) = lim infx′ →x

f(z, x), f2(z, x) = lim infx′ →x

f(z, x), (4.51)


which for almost all z ∈ Z are finite. We assume that f1 and f2 are supmeasurable, namely,for every x : Z → R measurable function the functions z → f1(z, x(z)) and z → f2(z, x(z))are both measurable. We set

j(z, x) =∫x

0f(z, r)dr ∀(z, x) ∈ Z × R. (4.52)

Evidently (z, x) → j(z, x) is measurable and for almost all z ∈ Z, x → j(z, x) islocally Lipschitz and its generalized subdifferential satisfies

∂j(z, x) ⊆[f1(z, x), f2(z, x)

](see Chang [27]). (4.53)

Clearly j(z, 0) = 0 a.e. on Z and if for almost all z ∈ Z, f(z, ·) is continuous at x = 0,then ∂j(z, 0) = {0} a.e. on Z. Then the new hypotheses on the nonsmooth potential j(z, x) arethe following:

H(j)3: j : Z × R → R is a function such that j(z, x) =∫x

0f(z, r)dr with f : Z × R → R

satisfying

(i) (z, x) → f(z, x) is measurable with f1 and f2 supmeasurable;

(ii) for almost all z ∈ Z, f(z, ·) is continuous at x = 0;

(iii) |f(z, x)| ≤ α(z) + c|x|p−1 for a.a. z ∈ Z, all x ∈ R, with α ∈ L∞(Z)+, c > 0;

(iv) there exists θ ∈ L∞(Z)+ satisfying θ(z) ≤ λ1 a.e. on Z with strict inequality ona set of positive measure, such that

lim sup|x|→∞

f2(z, x)|x|p−2x

≤ θ(z) uniformly for a.a. z ∈ Z; (4.54)

(v) there exists η ∈ L∞(Z)+ such that

λ2 < lim infx→ 0

f1(z, x)|x|p−2x

≤ lim supx→ 0

f2(z, x)|x|p−2x

≤ η(z) uniformly for a.a. z ∈ Z; (4.55)

(vi) for almost all z ∈ Z and all x ∈ R, we have f1(z, x)x ≥ 0 (sign condition).

Theorem 4.7. If hypotheses H(j)3 hold, then problem (1.1) has at least three nontrivial solutionsx0, v0, y0 such that x0 ∈ intC1

0(Z)+, v0 ∈ −intC10(Z)+, and y0 ∈ C1

0(Z) a (nodal) solution.

Proof. From Theorem 3.8, we have the two constant sign solutions x0 ∈ intC10(Z)+ and

v0 ∈ −intC10(Z)+. Let y+ ∈ intC1

0(Z)+ and y− ∈ −intC10(Z)+ be the two extremal constant

sign solutions from Proposition 4.6. We have

A(y±) = u± with u± ∈ Lp′(Z), u±(z) ∈ ∂j(z, u±(z)) a.e. on Z. (4.56)


We introduce the following functions

f+(z, x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

0 if x < 0,

f(z, x) if 0 ≤ x ≤ y+(z),

u+(z) if y+(z) < x,

f−(z, x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

u−(z) if x < y−(z),

f(z, x) if y−(z) ≤ x ≤ 0,

0 if 0 < x,

f(z, x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

u−(z) if x < y−(z),

f(z, x) if y−(z) ≤ x ≤ y+(z),

u+(z) if y+(z) < x.

(4.57)

We consider the corresponding potential functions defined by

j+(z, x) =∫x

0f+(z, r)dr, j−(z, x) =

∫x

0f−(z, r)dr , j(z, x) =

∫x

0f(z, r)dr. (4.58)

Then we define the following locally Lipschitz Euler functionals on W1,p0 (Z):

ϕ0,+(x) =1p‖Dx‖pp −

∫

Z

j+(z, x(z))dz,

ϕ0,−(x) =1p‖Dx‖pp −

∫

Z

j−(z, x(z))dz,

ϕ0(x) =1p‖Dx‖pp −

∫

Z

j(z, x(z))dz.

(4.59)

We will use the following order intervals in W1,p0 (Z):

I+ = [0, y+], I− = [y−, 0], I = [y−, y+]. (4.60)

The critical points of ϕ0,+ are in I+, the critical points of ϕ0,− are in I−, and the criticalpoints of ϕ0 are in I. We show this for ϕ0,+; the proof for the rest is similar.


So let x ∈W1,p0 (Z) be a critical point of ϕ0,+. Then we have 0 ∈ ∂ϕ0,+(x) and so A(x) = u

with u ∈ Lp′(Z), u(z) ∈ ∂j+(z, x(z)) a.e. on Z. If we act with the test function (x − y+)+ ∈

W1,p0 (Z) we obtain

⟨A(x),

(x − y+

)+⟩ =∫

Z

u(x − y+

)+dz

=∫

{x>y+}u(x − y+

)dz

=∫

Z

u+(x − y+

)+dz

=⟨A(y+),(x − y+

)+⟩,

=⇒⟨A(x) −A

(y+),(x − y+

)+⟩

=∫

{x>y+}

(‖Dx‖p−2Dx −

∥∥Dy+∥∥p−2

Dy+, Dx −Dy+)

RN dz = 0,

=⇒∣∣{x>y+

}∣∣N =0

(|·|N being the Lebesgue measure on R

N), i.e., x≤y+.

(4.61)

Similarly we show that 0 ≤ x. Hence x ∈ I+.Since the critical points of ϕ0,+ are in I+, it follows that {0, y+} are the only critical points

of ϕ0,+. From hypothesis H(j)3(v), we can find δ > 0 small such that

λ2xp−1 < u for a.a. z ∈ Z and all x ∈ [0, δ] and all u ∈ ∂j(z, x). (4.62)

We choose ε > 0 small such that

εu1(z) ≤ min{y+(z), δ} ∀z ∈ Z. (4.63)

We know that

j(z, x) =∫x

0f(z, r)dr for a.a. z ∈ Z, all x ∈ R,

f(z, r) ∈ ∂j(z, r) for a.a. z ∈ Z, a.a. r ∈ R.

(4.64)

Then from (4.62), (4.63), and (4.64), we have

j+(z, εu1(z)) = j(z, εu1(z)) =∫ εu1(z)

0f(z, r)dr >

λ2

pεpu1(z)

p a.e. on Z. (4.65)


Hence

ϕ0,+(εu1)=

εp

p

∥∥Du1

∥∥pp −∫

Z

j+(z, εu1(z)

)dz

<εp

p

(∥∥Du1∥∥pp − λ2

∥∥u1∥∥pp

)(see (4.65))

<εp

p

(∥∥Du1∥∥pp − λ1

∥∥u1∥∥pp

)= 0 (see (2.6) with m ≡ 1),

=⇒ infW

1,p0 (Z)

ϕ0,+ < 0 = ϕ0,+(0).

(4.66)

By hypothesis we have j(z, x) =∫x

0f(z, r)dr for almost all z ∈ Z and all x ∈ R. So usinghypothesis H(j)3(iii), it follows that

∣∣j+(z, x)∣∣ ≤ α(z) for a.a. z ∈ Z, all x ∈ R, with α ∈ L∞(Z)+. (4.67)

Therefore from this and Poincare’s inequality, we infer that ϕ0,+ is coercive. It is easy tosee that ϕ0,+ is weakly lower semicontinuous on W

1,p0 (Z). So by the Weierstrass theorem, we

can find y0 a minimizer of ϕ0,+ and ϕ0,+(y0) < 0 = ϕ0,+(0), that is, y0 /= 0 (see (4.66)). Since y0 isa nonzero critical point of ϕ0,+, we must have y0 = y+. Clearly y+ is a local C1

0(Z)-minimizerof ϕ0 and so y+ is a local W1,p

0 (Z) minimizer of ϕ0. We can assume that y+ is an isolated criticalpoint of ϕ0. If this is not the case, we can find a sequence {xn}n≥1 ⊆ W

1,p0 (Z) of critical points

of ϕ0 such that

xn −→ y+ in W1,p0 (Z) as n −→ ∞, xn /= 0, y+, y− ∀n ≥ 1. (4.68)

Since xn is a critical point of ϕ0, we must have xn ∈ I. Thus we have produced a wholesequence of distinct nontrivial nodal solutions for problem (1.1).

Similarly working with ϕ0,− on I−, we have that y− is a global minimizer ofϕ0,−, ϕ0,−(y−) = ϕ0(y−) < 0 = ϕ0(0) and we can assume that it is an isolated critical pointof ϕ0. As in Motreanu et al. [36], we can find δ > 0 small such that

ϕ0(y−)< inf

[ϕ0(x) : x ∈ ∂Bδ

(y−)]≤ 0,

ϕ0(y+)< inf

[ϕ0(x) : x ∈ ∂Bδ

(y+)]≤ 0,

(4.69)

where ∂Bδ(y±) = {x ∈W1,p0 (Z) : ‖x0 − y±‖ = δ}.

If we set S = ∂Bδ(y+) ∪ ∂Bδ(y−), I = [y−, y+], and I0 = {y−, y+}, then we can easily seethat the pair {I0, I} is linking with S in W

1,p0 (Z). Moreover, as for ϕ0,+ we can check that ϕ0 is

coercive and so we can easily verify the PS-condition. Therefore, we can apply Theorem 2.2and produce y0 ∈W

1,p0 (Z), a critical point of ϕ0, such that

ϕ0(y±)< ϕ0

(y0)= inf

γ∈Γmaxt∈[−1,1]

ϕ0(γ(t)), (4.70)


where Γ = {γ ∈ C([−1, 1]) : γ(−1) = y−, γ(1) = y+}. Note that from (4.70) we have thaty0 /=y±.

We claim that ϕ0(y0) < 0 = ϕ0(0) and so y0 /= 0. To show this, it is enough to produce apath γ0 ∈ Γ such that

ϕ0(γ0(t)

)< 0 ∀t ∈ [−1, 1]. (4.71)

So in what follows we construct such a path γ0.

Recall that ∂BLp(Z)1 = {x ∈ Lp(Z) : ‖x‖p = 1} and S = W

1,p0 (Z) ∩ ∂BLp(Z)

1 endowed withthe W

1,p0 (Z)-topology. We also set

Sc = W1,p0 (Z) ∩ C1

0(Z) ∩ ∂BLp(Z)1 (4.72)

equipped with the C10(Z)-topology.

Then Sc is dense in S in the W1,p0 (Z)-topology. Because of (2.10), given δ > 0, we can

find γ0 ∈ Γ0 = {γ0 ∈ C([−1, 1], S) : γ0(−1) = −u1, γ0(1) = u1} satisfying γ0([−1, 1]) ⊆ Sc and

max[‖Dx‖pp : x ∈ γ0([−1, 1])

]≤ λ2 + δ (4.73)

(since C([−1, 1], Sc) is dense in C([−1, 1], S).We can always choose δ > 0 small such that

λ2 + 2δ < lim infx→ 0

u

|x|p−2xuniformly for a.a. z ∈ Z, allu ∈ ∂j(z, x) (4.74)

(see hypothesis H(j)3(v)). Then we can find δ0 > 0 such that

λ2 + δ <u

|x|p−2xfor a.a. z ∈ Z, all 0 < |x| ≤ δ0, and all u ∈ ∂j(z, x). (4.75)

As before exploiting the fact that (d/dx)f(z, x) ∈ ∂j(z, x) for a.a. z ∈ Z and almost allx ∈ R, from (4.75) we obtain

1p

(λ2 + δ

)|x|p < j(z, x) for a.a. z ∈ Z and all 0 < |x| ≤ δ0. (4.76)

Since γ0([−1, 1]) ⊆ Sc and −y−, y+ ∈ intC10(Z)+, we can find ε > 0 small such that

|εx(z)| ≤ δ0 ∀z ∈ Z, all x ∈ γ0([−1, 1]),

εx ∈ [−y−, y+] and all x ∈ γ0([−1, 1]).(4.77)


If x ∈ γ0([−1, 1]), we have

ϕ0(εx) = ϕ(εx) =εp

p‖Dx‖pp −

∫

Z

j(z, εx(z))dz

<εp

p‖Dx‖pp −

εp

p

(λ2 + δ

)‖x‖pp (see (4.76))

≤ 0 (see (4.73) and recall ‖x‖p = 1).

(4.78)

So, if we consider the continuous path γ0 = εγ0 which joins −εu1 and εu1, we have

ϕ0|γ0 < 0. (4.79)

Next with the help of Theorem 2.5, we will produce a continuous path joining εu1

and y+ along which ϕ is strictly negative. We know that {0, y+} are the only critical pointsof ϕ0,+. Let α+ = ϕ0,+(y+) = infϕ0,+ < 0 and let b+ = 0. Recall that ϕ0,+ is coercive and soit satisfies the PS-condition. Therefore according to Theorem 2.5, we can find a deformation

h : [0, 1] × 0ϕ0

b+→ 0

ϕ0

b+such that

h(t, ·)|Kα+= id ∀t ∈ [0, 1],

h(

1,0

ϕ0,+

b+)⊆ 0ϕ0,+

a+∪Ka+ ,

ϕ0,+(h(t, z)) ⊆ ϕ0,+(x) ∀(t, x) ∈ [0, 1] × 0ϕ0,+

b+.

(4.80)

We consider the path γ+ : [0, 1] → 0ϕ0,+

b+defined by

γ+(t) = h(t, εu1

)∀t ∈ [0, 1]. (4.81)

Clearly this is a continuous path and we have

γ+(0) = h(0, εu1

)= εu1 (since h is a deformation),

γ+(1) = h(1, εu1

)= y+

( 0ϕ0,+

a+= ∅, Ka+ = {y+}

),

ϕ0,+(γ+(t)) = ϕ0,+(h(t, εu1

)) ≤ ϕ0,+

(εu1)< 0 ∀t ∈ [0, 1] (see (4.79)).

(4.82)

Therefore we have produced a continuous path γ+ joining εu1 and y+ such that

ϕ0,+|γ+ < 0. (4.83)

But note that ϕ0,+ ≥ ϕ0 (see H(j)3(vi)). Hence

ϕ0|γ+ < 0. (4.84)


In a similar fashion, we produce a continuous path γ− joining −εu1 and y− such that

ϕ0|γ− < 0. (4.85)

If we join the paths γ−, γ0, γ+, we produce a continuous path γ0 ∈ Γ such that

ϕ0|γ0< 0, (see (4.79), (4.84), (4.85)). (4.86)

From (4.70) it follows that ϕ0(y0) < 0 = ϕ(0) and so y0 /= 0.Therefore y0 is the third nontrivial solution of (1.1), which is (nodal) and from the

nonlinear regularity theory we have y0 ∈ C10(Z).

Acknowledgments

The authors wish to thank the referees for their remarks. Researcher supported by a grant ofthe National Scholarship Foundation of Greece (I.K.Y.).

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Research ArticleElectroelastic Wave Scattering in a CrackedDielectric Polymer under a Uniform Electric Field

Yasuhide Shindo and Fumio Narita

Department of Materials Processing, Graduate School of Engineering, Tohoku University,Aoba-yama 6-6-02, Sendai 980-8579, Japan

Correspondence should be addressed to Yasuhide Shindo, [email protected]

Received 25 April 2009; Revised 2 May 2009; Accepted 18 May 2009


We investigate the scattering of plane harmonic compression and shear waves by a Griffith crack inan infinite isotropic dielectric polymer. The dielectric polymer is permeated by a uniform electricfield normal to the crack face, and the incoming wave is applied in an arbitrary direction. Bythe use of Fourier transforms, we reduce the problem to that of solving two simultaneous dualintegral equations. The solution of the dual integral equations is then expressed in terms of a pairof coupled Fredholm integral equations of the second kind having the kernel that is a finite integral.The dynamic stress intensity factor and energy release rate for mode I and mode II are computedfor different wave frequencies and angles of incidence, and the influence of the electric field on thenormalized values is displayed graphically.

Copyright q 2009 Y. Shindo and F. Narita. 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.

1. Introduction

Elastic dielectrics such as insulating materials have been reported to have poor mechanicalproperties. Mechanical failure of insulators is also a well-known phenomenon. Therefore,understanding the fracture behavior of the elastic dielectrics will provide useful informationto the insulation designers. Toupin [1] considered the isotropic elastic dielectric material andobtained the form of the constitutive relations for the stress and electric fields. Kurlandzka[2] investigated a crack problem of an elastic dielectric material subjected to an electrostaticfield. Pak and Herrmann [3, 4] also derived a material force in the form of a path-independentintegral for the elastic dielectric medium, which is related to the energy release rate. Recently,Shindo and Narita [5] considered the planar problem for an infinite dielectric polymercontaining a crack under a uniform electric field, and discussed the stress intensity factorand energy release rate under mode I and mode II loadings.

This paper investigates the scattering of in-plane compressional (P) and shear (SV)waves by a Griffith crack in an infinite dielectric polymer permeated by a uniform electric


field. The electric field is normal to the crack surface. Fourier transforms are used to reducethe problem to the solution of two simultaneous dual integral equations. The solution of theintegral equations is then expressed in terms of a pair of coupled Fredholm integral equationsof the second kind. In literature, there are two derivations of dual integral equations. One isthe one mentioned in this paper. The other one is for the dual boundary element methods(BEM) [6, 7]. Numerical calculations are carried out for the dynamic stress intensity factorand energy release rate under mode I and mode II, and the results are shown graphically todemonstrate the effect of the electric field.

2. Basic Equations

Consider the rectangular Cartesian coordinate system with axes x1, x2, and x3. We decomposethe electric field intensity vector Ei, the polarization vector Pi, and the electric displacementvector Di into those representing the rigid body state, indicated by overbars, and those forthe deformed state, denoted by lower case letters:

Ei = Ei + ei, Pi = Pi + pi, Di = Di + di. (2.1)

We assume that the deformation will be small even with large electric fields, and the secondterms will have only a minor influence on the total fields. The formulations will then belinearized with respect to these unknown deformed state quantities.

The linearized field equations are obtained as

σLji,j + Ei,jpj + Pjei,j = ρui,tt,

Di,i = 0,

di,i = 0,

(2.2)

where ui is the displacement vector, σLij is the local stress tensor, ρ is the mass density, a comma

followed by an index denotes partial differentiation with respect to the space coordinate xi orthe time t, and the summation convention for repeated indices is applied.

The linearized constitutive equations can be written as

σLij = λuk,kδij + μ

(ui,j + uj,i

)+A1

(EkEk + 2Ekek

)δij +A2

(EiEj + Eiej + Ejei

),

σMij = ε0εr

(EiEj + Eiej + Ejei

)− 1

2ε0

(EkEk + 2Ekek

)δij ,

Di = ε0Ei + Pi = ε0εrEi, di = ε0ei + pi = ε0εrei,

Ei =1ε0η

Pi, ei =1ε0η

pi,

(2.3)

where σMij is the Maxwell stress tensor, λ and μ are the Lame constants, A1 and A2 are

the electrostrictive coefficients, ε0 is the permittivity of free space, εr= 1 +η is the specificpermittivity, η is the electric susceptibility, and δij is the Kronecker delta.


The linearized boundary conditions are found as

[∣∣∣σL

ji

∣∣∣]nj +

12ε0

[(Pknk

)2+ 2Pkplnknl

]ni = 0,

[∣∣∣Di

∣∣∣]ni = 0,

eijknj

[∣∣∣Ei

∣∣∣]= 0,

[|di|]ni −[∣∣∣Di

∣∣∣]ui,jnj = 0,

eijk{nj[|ei|] − nlnl,j

[∣∣∣Ei

∣∣∣]}

= 0,

(2.4)

where ni is an outer unit vector normal to an undeformed body, eijk is the permutationsymbol, and [|fi|] means the jump in any field quantity fi across the discontinuity surface.

3. Problem Statement

Let a Griffith crack be located in the interior of an infinite elastic dielectric. We consider arectangular Cartesian coordinate system (x, y, z) such that the crack is placed on the x-axisfrom −a to a as shown in Figure 1, and assume that plane strain is perpendicular to the z-axis.A uniform electric field E0 is applied perpendicular to the crack surface. For convenience, allelectric quantities outside the solid will be denoted by the superscript +. The solution for therigid body state is

E+y = εrE0, D

+y = ε0εrE0, P

+y = 0,

Ey = E0, Dy = ε0εrE0, Py = ε0ηE0.(3.1)

The equations of motion are given by

∇21ux +

11 − 2ν

(ux,x + uy,y

),x+

2A1E0

μey,x +

A3E0

μex,y =

1c2

2

ux,tt,

∇21uy +

11 − 2ν

(ux,x + uy,y

),y+A2E0

μex,x +

E0

μ(2A1 +A2 +A3)ey,y =

1c2

2

uy,tt,

(3.2)

where ∇21 = ∂2/∂x2 + ∂2/∂y2 is the two-dimensional Laplace operator in the variables x, y, ν

is the Poisson’s ratio, c2 = (μ/ρ)1/2 is the shear wave velocity, and A3 = A2 + ε0η. The electricfield equations for the perturbed state are

ex,x + ey,y = 0, e+x,x + e+y,y = 0. (3.3)


Incident waves

γ

y

xO

E0

−a a

Figure 1: Scattering of waves in a dielectric medium with a Griffith crack.

The electric field equations (3.3) are satisfied by introducing an electric potential φ(x, y, t)such that

ei = −φ,i, ∇21φ = 0,

e+i = −φ+,i , ∇2

1φ+ = 0.

(3.4)

The displacement components can be written in terms of two scalar potentials ϕe(x, y, t) andψe(x, y, t) as

ux = ϕe,x + ψe,y, uy = ϕe,y − ψe,x. (3.5)

The equations of motion become

∇21ϕe −

E0

μ(2A1 +A2 +A3)

(c2

c1

)2

φ,y =1c2

1

ϕe,tt,

∇21ψe +

E0

μA2φ,x =

1c2

2

ψe,tt,

(3.6)

where c1 = {(λ + 2μ)/ρ}1/2 is the compression wave velocity.Let an incident plane harmonic compression wave (P-wave) be directed at an angle γ

with the x-axis so that

ϕie = ϕe0 exp

[−iω

{t +

x cos γ + y sin γ

c1

}], ψi

e = 0 (P-wave), (3.7)


where ϕe0 is the amplitude of the incident P-wave, and ω is the circular frequency. Thesuperscript i stands for the incident component. Similarly, if an incident plane harmonic shearwave (SV-wave) impinges on the crack at an angle γ with x-axis, then

ϕie = 0, ψi

e = ψe0 exp[−iω

{t +

x cos γ + y sin γ

c2

}](SV-wave), (3.8)

where ψe0 is the amplitude of the incident SV-wave. In view of the harmonic time variation ofthe incident waves given by (3.7) and (3.8), the field quantities will all contain the time factorexp(−iωt) which will henceforth be dropped.

The problem may be split into two parts: one symmetric (opening mode, Mode I) andthe other skew-symmetric (sliding mode, Mode II). Hence, the boundary conditions for thescattered fields are

Mode I:

σLyx(x, 0) = 0 (0 ≤ |x| <∞),

φ,x(x, 0) = −ηE0uy,x(x, 0) + φ+,x(x, 0) (0 ≤ |x| < a),

φ(x, 0) = 0 (a ≤ |x| <∞),

σLyy(x, 0) = −ε0η

2E0φ,y − pj exp(−iαjx cos γ

) (j = 1, 2

)(0 ≤ |x| < a),

uy(x, 0) = 0 (a ≤ |x| <∞),

(3.9)

Mode II:

σLyy(x, 0) = 0 (0 ≤ |x| <∞),

φ,x(x, 0) = −ηE0uy,x(x, 0) + φ+,x(x, 0) (0 ≤ |x| < a),

φ,y(x, 0) = 0 (a ≤ |x| <∞),

σLxy(x, 0) = −qj exp

(−iαjx cos γ

) (j = 1, 2

)(0 ≤ |x| < a),

ux(x, 0) = 0, (a ≤ |x| <∞),

(3.10)

where the subscript j = 1 and 2 correspond to the incident P- and SV-waves, p1 = μα22ϕe0(1 −

2σ2cos2γ), p2 = μα22ψe0 sin 2γ , q1 = μα2

2ϕe0σ2 sin 2γ , q2 = μα2

2ψe0 cos 2γ , α1 = p/c1 and, α2 =p/c2 are the compression and shear wave numbers, respectively, and σ = c2/c1.

4. Method of Solution

The desired solution of the original problem can be obtained by superposition of the solutionsfor the two cases: mode I and mode II. The problem will further be divided into two parts:(1) symmetric with respect to x and (2) antisymmetric with respect to x.


4.1. Mode I Problem

4.1.1. Symmetric Solution for Mode I Crack

The boundary conditions for symmetric scattered fields can be written as

σLyxs(x, 0) = 0 (0 ≤ x <∞), (4.1)

φs,x(x, 0) = −ηE0uys,x(x, 0) + φ+s,x(x, 0) (0 ≤ x < a),

φs(x, 0) = 0 (a ≤ x <∞),(4.2)

σLyys(x, 0) = −ε0η

2E0φs,y − pj cos(αjx cos γ

) (j = 1, 2

)(0 ≤ x < a),

uys(x, 0) = 0 (a ≤ x <∞),(4.3)

where the subscript s stands for the symmetric part. It can be shown that solutions φs, ϕes,ψes, and φ+

s of (3.4) and (3.6) for y ≥ 0 are

φs = −2π

∫∞

0as(α)e−αy cos(αx)dα,

ϕes =2π

∫∞

0

{

A1s(α)e−γ1(α)y +(c2

p

)2 E0

μ(2A1 +A2 +A3)αas(α)e−αy

}

cos(αx)dα,

(4.4)

ψes =2π

∫∞

0

{

A2s(α)e−γ2(α)y −(c2

p

)2 E0

μA2αas(α)e−αy

}

sin(αx)dα, (4.5)

φ+s = − 2

π

∫∞

0a+s (α) sinh

(αy

)cos(αx)dα, (4.6)

where as(α), A1s(α), A2s(α), and a+s (α) are unknown functions, and γ1(α) and γ2(α) are

γ1(α) =

{

α2 −(

pc1

)2}1/2

, γ2(α) =

{

α2 −(

pc2

)2}1/2

. (4.7)

The functions γ1(α) and γ2(α) should be restricted as

Re γk(α) > 0, Imγk(α) < 0 (k = 1, 2) (4.8)

in the upper half-space y ≥ 0, because of a radiation condition at infinity and an edgecondition near the crack tip. A simple calculation leads to the displacement and stress


expressions:

uxs = −2π

∫∞

0

[

αA1s(α)e−γ1(α)y + γ2(α)A2s(α)e−γ2(α)y

+(c2

p

)2 E0

μ(2A1 +A3)α2as(α)e−αy

]

sin(αx)dα,

uys = −2π

∫∞

0

[

γ1(α)A1s(α)e−γ1(α)y + αA2s(α)e−γ2(α)y

+(c2

p

)2 E0

μ(2A1 +A3)α2as(α)e−αy

]

cos(αx)dα,

σLxxs = −

4πμ

∫∞

0

[{λ

2μ

(p

c1

)2

+ α2

}

A1s(α)e−γ1(α)y + αγ2(α)A2s(α)e−γ2(α)y

+E0

μ

{(c2

p

)2

(2A1 +A3)α2 +A1

}

αas(α)e−αy]

cos(αx)dα +A1E20,

σLxys =

2πμ

∫∞

0

[

2αγ1(α)A1s(α)e−γ1(α)y +

{

2α2 −(

p

c2

)2}

A2s(α)e−γ2(α)y

+E0

μ

{

2(c2

p

)2

(2A1 +A3)α2 −A2

}

αas(α)e−αy]

sin(αx)dα,

σLyys =

4πμ

∫∞

0

[{

−12

(p

c2

)2

+ α2

}

A1s(α)e−γ1(α)y + αγ2(α)A2s(α)e−γ2(α)y

+E0

μ

{(c2

p

)2

(2A1+A3)α2−(A1+A2)

}

αas(α)e−αy]

cos(αx)dα + (A1 +A2)E20,

σMxxs =

2πε0E0

∫∞

0αas(α)e−αy cos(αx)dα −

ε0E20

2,

σMxys = −

2πε0εrE0

∫∞

0αas(α)e−αy sin(αx)dα,

σMyys = −

2π

(1 + 2η

)ε0E0

∫∞

0αas(α)e−αy cos(αx)dα +

ε0E20

(1 + 2η

)

2.

(4.9)

The boundary condition of (4.1) leads to the following relation between unknownfunctions:

2αγ1(α)A1s(α) +

{

2α2 −(

p

c2

)2}

A2s(α) +E0

μ

{

2(c2

p

)2

(2A1 +A3)α2 −A2

}

αas(α) = 0.

(4.10)


The satisfaction of the two mixed boundary conditions (4.2) and (4.3) leads to twosimultaneous dual integral equations of the following form:

∫∞

0α[as(α) + ηE0As(α)

]sin(αx)dα = 0 (0 ≤ x < a),

∫∞

0as(α) cos(αx)dα = 0 (a ≤ x <∞),

(4.11)

∫∞

0α

[fe(α)As(α)+

E0

μfm(α)as(α)

]cos(αx)dα=− π

4μpj cos

(αjx cos γ

)(0≤x<a),

∫∞

0As(α) cos(αx)dα = 0 (a ≤ x <∞),

(4.12)

in which fe(α) and fm(α) are known functions given by

fe(α) =1

γ1(α)(p/c2

)2

⎡

⎣−{

2α2 −(

p

c2

)2}2

+ 4α2γ1(α)γ2(α)

⎤

⎦ 12α

,

fm(α) =1

γ1(α)(p/c2)2

[

−{

2α2 −(

p

c2

)2}(2A1 + ε0η

)+ 2γ1(α)γ2(α)A2

+2αγ1(α)(2A1+A3)−1αγ1(α)

(p

c2

)2(2A1+2A2−ε0η

2)]α

2,

(4.13)

and the original unknowns A1s(α) and A2s(α) are related to the new one As(α) through

A1s(α) = −1

γ1(α)(p/c2

)2

[{2α2 −

(p/c2

)2}As(α) +

E0

μ

(2A1 + ε0η

)α2as(α)

],

A2s(α) =1

(p/c2

)2

[2αAs(α) −

E0

μA2αas(α)

].

(4.14)

The set of two simultaneous dual integral equations (4.11) and (4.12) may be solvedby using a new function Φs(u), and the result is

As(α) =π

4

(pja

2

μc0y0

)∫1

0u1/2Φs(u)J0(αau)du,

as(α) = −π

4

(ηE0pja

2

μc0y0

)∫1

0u1/2Φs(u)J0(αau)du,

(4.15)


where J0() is the zero-order Bessel function of the first kind, and c0 and y0 are

c0 =(c2

c1

)2

− 1,

y0 = 1 +12

[(1 − 2ν)

(2A1 + ε0η

)− 2(1 − ν)

(ε0η

2 + ε0η −A2

)]ημE2

0.

(4.16)

The function Φs(u) is governed by the following Fredholm integral equation of second kind:

Φs(u) −∫1

0Φs(s)(su)1/2Ks(u, s)ds = u1/2J0

(αjau cos γ

), (4.17)

where the kernel Ks(u, s) is given by

Ks(u, s) =∫∞

0

[α +

1c0y0P 2

{f∗e (α) − ηE2

μf∗m(α)

}]J0(αu)J0(αs)dα, (4.18)

f∗e (α) =1

2γ∗1 (α)

[−(

2α2 − P 2)2

+ 4α2γ∗1 (α)γ∗2 (α)

],

f∗m(α) =1

2γ∗1 (α)

[−α2

(2α2 − P 2

)(2Ae1 + η

)− 2α2γ∗1 (α)γ

∗2 (α)Ae2

+ 2α3γ∗1 (α)(2Ae1 +Ae2 + η

)− αγ∗1 (α)P

2(

2Ae1 +Ae2 − η2)]

,

(4.19)

γ∗1 (α) ={α2 − (Pσ)2

}1/2, γ∗2 (α) =

(α2 − P 2

)1/2,

E2μ =

ε0E20

μ, Ae1 =

A1

ε0, Ae2 =

A2

ε0, P =

ap

c2, σ =

c2

c1.

(4.20)

The kernel function Ks(u, s) (4.18) is an infinite integral that has a rather slow ofconvergence. To improve this problem the infinite integral is converted into integrals withfinite limits. Thus, for the calculation of the integral, we consider the contour integrals

Ie1 =∮

Γ1

Le

(k, γ∗1 ,−γ

∗2)J0(ks)H

(1)0 (ku)dk (u > s),

Ie2 =∮

Γ2

Le

(k, γ∗1 , γ

∗2)J0(ks)H

(2)0 (ku)dk (u > s),

(4.21)

where the contours Γ1,Γ2 are defined in Figure 2, H(1)0 (),H(2)

0 () are, respectively, the zero-order Hankel functions of the first and second kinds, and

Le

(k, γ∗1 , γ

∗2)= k +

1c0y0P

{f∗e (k) − ηE2

μf∗m(k)

}. (4.22)


ImK

Branch line

−α1−α2 OReK

Γ1

Γ2

iv′1 v1 v1

iv′2 iv′2 v2

−iv′1 v1 v1

−iv′2 −iv′2 v2

α1 α2

Figure 2: The counters of integration.

The integrands in (4.21) satisfy Jordan’s lemma on the infinite quarter circles, so that,

Ie1 + Ie2 =∫α1

0

{Le

(α, iν′1, iν

′2)H

(1)0 (αu)dα + Le

(α,−iν′1,−iν′2

)H

(2)0 (αu)

}J0(αs)dα

+∫α2

α1

{Le

(α, ν1, iν

′2)H

(1)0 (αu)dα + Le

(α, ν1,−iν′2

)H

(2)0 (αu)

}J0(αs)dα

+ 2∫∞

α2

Le(α, ν1, ν2)J0(αs)J0(αu)dα

+∫0

∞

{Le

(iα, iν′1, iν

′2)+ Le

(−iα,−iν′1,−iν′2

)}J0

(eiπ/2αs

)H

(1)0

(eiπ/2αu

)i dα = 0,

(4.23)

where

ν1 =(α2 − P 2σ2

)1/2, ν2 =

(α2 − P 2

)1/2,

ν′1 =(P 2σ2 − α2

)1/2, ν′2 =

(P 2 − α2

)1/2.

(4.24)

Because of the second of (4.8), the integral in (4.18) must be taken along a path located slightlybelow the real k-axis as in Γ2. Therefore Ks(u, s) for u > s can be finally written as

Ks(u, s) = iP 2∫1

0

{M1(α)J0(αPs)H

(1)0 (αPu) +M2(α)J0(ασPs)H

(1)0 (ασPu)

}dα, (u > s),

(4.25)


where

M1(α) = −1

c0y0

(2 + ηE2

μAe2

)α2(

1 − α2)1/2

,

M2(α) = −1

c0y0

1

2(1 − α2)1/2

{(2α2σ2 − 1

)2− ηE2

μσ2(

2α2σ2 − 1)α2(2Ae1 + η

)}.

(4.26)

The kernel Ks(u, s) is symmetric in u, s, and the value of this kernel for u < s is obtained byinterchanging u and s in (4.25).

4.1.2. Antisymmetric Solution for Mode I Crack

The boundary conditions for anti-symmetric scattered fields can be written as

σLyxa(x, 0) = 0 (0 ≤ x <∞), (4.27)

φa,x(x, 0) = −ηE0uya,x(x, 0) + φ+a,x(x, 0), (0 ≤ x < a),

φa(x, 0) = 0 (a ≤ x <∞),(4.28)

σLyya(x, 0) = −ε0η

2E0φa,y − pj sin(αjx cos γ

) (j = 1, 2

)(0 ≤ x < a),

uya(x, 0) = 0 (a ≤ x <∞),(4.29)

where the subscript a stands for the anti-symmetric part. The solutions φa, ϕea, ψea and φ+a

are

φa = − 2π

∫∞

0aa(α)e−αy sin(αx)dα,

ϕea =2π

∫∞

0

{

A1a(α)e−γ1(α)y +(c2

p

)2 E0

μ(2A1 +A2 +A3)αaa(α)e−αy

}

sin(αx)dα,

(4.30)

ψea =2π

∫∞

0

{

A2a(α)e−γ2(α)y −(c2

p

)2 E0

μA2αaa(α)e−αy

}

cos(αx)dα, (4.31)

φ+a = − 2

π

∫∞

0a+a(α) cosh

(αy

)sin(αx)dα, (4.32)


where aa(α), A1a(α), A2a(α), and a+a(α) are unknown functions. The displacements and

stresses are obtained as

uxa =2π

∫∞

0

{

αA1a(α)e−γ1(α)y − γ2(α)A2a(α)e−γ2(α)y

+(c2

p

)2 E0

μ(2A1 +A3)α2aa(α)e−αy

}

cos(αx)dα,

uya = − 2π

∫∞

0

{

γ1(α)A1a(α)e−γ1(α)y − αA2a(α)e−γ2(α)y

+(c2

p

)2 E0

μ(2A1 +A3)α2aa(α)e−αy

}

sin(αx)dα,

(4.33)

σLxxa = − 4

πμ

∫∞

0

[{λ

2μ

(p

c1

)2

+ α2

}

A1a(α)e−γ1(α)y − αγ2(α)A2a(α)e−γ2(α)y

+E0

μ

{(c2

p

)2

(2A1 +A3)α2 +A1

}

αaa(α)e−αy]

sin(αx)dα,

σLxya = − 2

πμ

∫∞

0

[

2αγ1(α)A1ae−γ1(α)y −

{

2α2 −(

p

c2

)2}

A2a(α)e−γ2(α)y

+E0

μ

{

2(c2

p

)2

(2A1 +A3)α2 −A2

}

αaa(α)e−αy]

cos(αx)dα,

σLyya =

4πμ

∫∞

0

[{

−12

(p

c2

)2

+ α2

}

A1a(α)e−γ1(α)y − αγ2(α)A2a(α)e−γ2(α)y

+E0

μ

{(c2

p

)2

(2A1 +A3)α2 − (A1 +A2)

}

αaa(α)e−αy]

sin(αx)dα,

(4.34)

σMxxa =

2πε0E0

∫∞

0αaa(α)e−αy sin(αx)dα,

σMxya =

2πε0εrE0

∫∞

0αaa(α)e−αy cos(αx)dα,

σMyya = − 2

π

(1 + 2η

)ε0E0

∫∞

0αaa(α)e−αy sin(αx)dα.

(4.35)

The relation between unknown functions can be found by the same procedure as inthe symmetric case. The boundary condition of (4.27) leads to the following relation:

2αγ1(α)A1a(α) −{

2α2 −(

p

c2

)2}

A2a(α) +E0

μ

{

2(c2

p

)2

(2A1 +A3)α2 −A2

}

αaa(α) = 0.

(4.36)


The boundary conditions in (4.28) and (4.29) lead to two simultaneous dual integralequations of the following form:

∫∞

0α{aa(α) + ηE0Aa(α)

}cos(αx)dα = 0 (0 ≤ x < a),

∫∞

0aa(α) sin(αx)dα = 0 (a ≤ x <∞),

(4.37)

∫∞

0α

{fe(α)Aa(α) +

E0

μfm(α)aa(α)

}sin(αx)dα = − π

4μpj sin

(αjx cos γ

)(0 ≤ x < a),

∫∞

0Aa(α) sin(αx)dα = 0 (a ≤ x <∞),

(4.38)

in which the original unknowns A1a(α), A2a(α) are related to the new one Aa(α) through

A1a(α) = −1

γ1(α)(p/c2

)2

[{

2α2 −(

p

c2

)2}

Aa(α) +E0

μ

(2A1 + ε0η

)α2aa(α)

]

,

A2a(α) = −1

(p/c2

)2

{2αAa(α) −

E0

μA2αaa(α)

}.

(4.39)

The unknowns Aa(α) and aa(α) can be found by the same method of approach as inthe symmetric case. The results are

Aa(α) =π

4

(pja

2

μc0y0

)∫1

0u1/2Φa(u)J1(αau)du,

aa(α) = −π

4

(ηE0pja

2

μc0y0

)∫1

0u1/2Φa(u)J1(αau)du,

(4.40)

where J1() is the first-order Bessel function of the first kind, and Φa(u) in (4.40) is the solutionof the following Fredholm integral equation of the second kind:

Φa(u) −∫1

0Φa(s)(su)1/2Ka(u, s)ds = u1/2J1

(αjau cos γ

), (4.41)

where

Ka(u, s) =∫∞

0

[α +

1c0y0P 2

{f∗e (α) − ηE2

μf∗m(α)

}]J1(αu)J1(αs)dα (u > s). (4.42)


By using the contours of integration in Figure 2, the kernel Ka(u, s) for u > s can be rewrittenin the form

Ka(u, s) = iP 2∫1

0

{M1(α)J1(αPs)H


(1)1 (ασPu)

}dα (u > s),

(4.43)

where H(1)1 () is the first-order Hankel function of the first kind. The value of Ka(u, s) for u < s

is obtained by interchanging u and s in (4.43).

4.1.3. Mode I Dynamic Singular Stresses Near the Crack Tip

The mode I dynamic electric stress intensity factor KID is

KID = limx→a+

{2π(x − a)}1/2{σLyys + σL

yya + σMyys + σM

yya

}

y=0

= pj(πa)1/2 z0

y0[Φs(1) − iΦa(1)],

(4.44)

where

z0 = 1 +12{(1 − 2ν)

(2Ae1 + η

)+ 2(1 − ν)

(Ae2 + η + 1

)}ηE2

μ. (4.45)

Next, we examine the static electroelastric crack problem. The boundary conditions may bewritten as

σLyx(x, 0) = 0 (0 ≤ x <∞), (4.46)

φ,x(x, 0) = −ηE0uy,x(x, 0) + φ+,x(x, 0), (0 ≤ x < a),

φ(x, 0) = 0 (a ≤ x <∞),(4.47)

σLyy(x, 0) = ε0η

2

{E2

0

2− E0φ,y

}

(0 ≤ x < a),

uy(x, 0) = 0 (a ≤ x <∞).

(4.48)

The electric stress intensity factor KIS may be obtained as

KIS = μE2μ(πa)

1/2(z0

y0

)2Ae1 + 2Ae2 − η2

2. (4.49)

The dynamic stress intensity factor KI can be found as

KI = |KID| +KIS. (4.50)


The dynamic electroelastic stress is given by

σcij = σ

L(i)ij + σ

L(s)ij + σ

M(i)ij + σ

M(s)ij . (4.51)

The singular parts of the dynamic local stresses and Mexwell stresses near the crack tip canbe expressed as

σLxx ∼

KI

2z0

{[2 +

{2(1 − 2ν)Ae1 + 2(1 − ν)Ae2 − η

}E2μη

]

−[2 +

{(1 − 2ν)

(2Ae1 + η

)+ 2(1 − ν)Ae2

}E2μη

]sin

θ

2sin

3θ2

}cos

θ

21

(2πr)1/2,

σLxy ∼

KI

2z0

[2 +

{2(1 − 2ν)

(2Ae1 + η

)+ 2(1 − ν)Ae2

}E2μη

]sin

θ

2cos

θ

2cos

3θ2

1

(2πr)1/2,

σLyy ∼

KI

2z0

{[2 +

{2(1 − 2ν)Ae1 + 2(1 − ν)Ae2 − η

}E2μη

]

+[2 +

{(1 − 2ν)

(2Ae1 + η

)+ 2(1 − ν)Ae2

}E2μη

]sin

θ

2sin

3θ2

}cos

θ

21

(2πr)1/2,

(4.52)

σMxx ∼ −

KI

z0(1 − ν)ηE2

μ cosθ

21

(2πr)1/2,

σMxy ∼ −

KI

z0(1 − ν)ηεrE2

μ sinθ

21

(2πr)1/2,

σMyy ∼

KI

z0(1 − ν)

(1 + 2η

)ηE2

μ cosθ

21

(2πr)1/2,

(4.53)

where r = {(x − a)2 + y2}1/2and θ = tan−1(y/(x − a)) are the polar coordinates. Also, the

singular parts of the displacements and electric fields near the crack tip are

ux ∼KI

2z0μ

(r

2π

)1/2{2(1 − 2ν) −

{(1 − 2ν)

(Ae1 + η

)− 2(1 − ν)Ae2

}E2μη

+[2 +

{(1 − 2ν)

(Ae1 + η

)+ 2(1 − ν)Ae2

}E2μη

]sin2 θ

2

}cos

θ

2,

uy ∼KI

2z0μ

(r

2π

)1/2{4(1 − ν) +

[2 +

{(1 − 2ν)

(Ae1 + η

)+ 2(1 − ν)Ae2

}E2μη

]cos2 θ

2

}sin

θ

2,

(4.54)

Ex ∼ −KI

z0μ

1

(2πr)1/2(1 − ν)ηE0 sin

θ

2,

Ey ∼KI

z0μ

1

(2πr)1/2(1 − ν)ηE0 cos

θ

2.

(4.55)


4.2. Mode II Problem

Since the mode II problem may also be reduced to the solution of two simultaneous dualintegral equations in the same way as the mode I, many of the details of solution procedurewill be omitted and only the essential steps will be provided.

4.2.1. Symmetric Solution for Mode II Crack

The boundary conditions for symmetric scattered fields are

σLyys(x, 0) = 0 (0 ≤ x <∞), (4.56)

φs,x(x, 0) = −ηE0uys,x(x, 0) + φ+s,x(x, 0) (0 ≤ x < a),

φs,y(x, 0) = 0 (a ≤ x <∞),(4.57)

σLxys(x, 0) = −qj cos

(αjx cos γ

) (j = 1, 2

)(0 ≤ x < a),

uxs(x, 0) = 0 (a ≤ x <∞).(4.58)

Replace the subscript a by s, aa(α), A1a(α), A2a(α), and a+a(α) by bs(α), B1s(α), B2s(α), and

b+s (α), respectively, in (4.30)–(4.35). The boundary condition of (4.56) leads to

{

2α2 −(

p

c2

)2}

B1s(α) − 2αγ2(α)B2s(α) + 2E0

μ

{(c2

p

)2

(2A1 +A3)α2 − (A1 +A2)

}

αbs(α) = 0.

(4.59)

Introducing the abbreviation

Bs(α) = αB1s(α) − γ2(α)B2s(α) +1

(p/c2

)2

E0

μ(2A1 +A3)α2bs(α), (4.60)

and in view of two mixed boundary conditions (4.57) and (4.58), together with (4.59) and(4.60), we have the following two simultaneous dual integral equations for the determinationof the function Bs(α):

∫∞

0α{ηE0f1(α)Bs(α) + f2(α)bs(α)

}cos(αx)dα = 0 (0 ≤ x < a),

∫∞

0αbs(α) sin(αx)dα = 0 (a ≤ x <∞),

(4.61)

Boundary Value Problems 17∫∞

0α

{f3(α)Bs(α) +

E0

μf4(α)bs(α)

}cos(αx)dα = −

πqj

2μcos

(αjx cos γ

)(0 ≤ x < a),

∫∞

0Bs(α) cos(αx)dα = 0 (a ≤ x <∞),

(4.62)

where

f1(α) =α

γ2(α)(p/c2

)2

{(p

c2

)2

+ 2γ1(α)γ2(α) − 2α2

}

,

f2(α) = 1 +α

γ2(α)(p/c2

)2

ηE20

μ

{−2γ1(α)γ2(α)(A1 +A2) + γ2(α)α(2A1 +A3) + (2A2 −A3)α2

}

f3(α) =1

γ2(α)(p/c2

)2

⎡

⎣−{(

p

c2

)2

− 2α2

}2

+ 4γ1(α)γ2(α)α2

⎤

⎦ 1α,

f4(α) =α

γ2(α)(p/c2

)2

{(

2α2 −p

c22

)

(2A2 −A3) − 4γ1(α)γ2(α)(A1 +A2)

+γ2(α)α(2A1 + A3) −1αγ2(α)

(p

c2

)2

A2

}

,

(4.63)

The solution of (4.61) and (4.62) are obtained by using two new functions gs(u) andhs(u), and the results are

Bs(α) =π

2

(qja

2

μ

)∫1

0u1/2gs(u)J0(αau)du,

bs(α) =π

2

(ηE0qja

2

μ

)∫1

0u1/2hs(u)J0(αau)du,

(4.64)

where gs(u) and hs(u) are the solutions of the following Fredholm integral equations of thesecond kind:

F1gs(u) + F2hs(u) −∫1

0(su)1/2{gs(s)K1s(u, s) + hs(s)K2s(u, s)

}ds = 0, (4.65)

F3gs(u) + ηE2μF4hs(u) −

∫1

0(su)1/2{gs(s)K3s(u, s) + hs(s)K4s(u, s)

}ds = −u1/2J0

(αjau cos γ

).

(4.66)


The kernels are given by

K1s(u, s) = iP 2∫1

0

{M11(α)J0(αPs)H


(1)0 (ασPu)

}dα (u > s),

K2s(u, s) = iP 2ηE2μ

∫1

0

{M21(α)J0(αPs)H


(1)0 (ασPu)

}dα (u > s),

K3s(u, s) = iP 2∫1

0

{M31(α)J0(αPs)H


(1)0 (ασPu)

}dα (u > s),

K4s(u, s) = iP 2ηE2μ

∫1

0

{M41(α)J0(αPs)H


(1)0 (ασPu)

}dα (u > s),

(4.67)

where

M11(α) = −α2 − 2α4

(1 − α2)1/2,

M12(α) = 2σ4α2(

1 − α2)1/2

,

M21(α) = −α4

(1 − α2)1/2

(Ae2 − η

),

M22(α) = −2σ4α2(

1 − α2)1/2

(Ae1 +Ae2),

M31(α) =

(1 − 2α2)2

(1 − α2)1/2,

M32(α) = 4σ4α2(

1 − α2)1/2

,

M41(α) = −α2(2α2 − 1

)

(1 − α2)1/2

(Ae2 − η

),

M42(α) = −4σ4α2(

1 − α2)1/2

(Ae1 +Ae2),

(4.68)

and Fi= limα→∞fi(α) (i=1, . . . , 4). The kernels Kis(u, s) (i=1, . . . , 4) are symmetric in u and s.

4.2.2. Antisymmetric Solution for Mode II Crack

The boundary conditions for anti-symmetric scattered fields are

σLyya(x, 0) = 0 (0 ≤ x <∞), (4.69)

φa,x(x, 0) = −ηE0uya,x(x, 0) + φ+a,x(x, 0) (0 ≤ x < a),

φa,y(x, 0) = 0 (a ≤ x <∞),(4.70)


σLxya(x, 0) = −qj sin

(αjx cos γ

) (j = 1, 2

)(0 ≤ x < a),

uxa(x, 0) = 0, (a ≤ x <∞).(4.71)

Let replace the subscript s by a, as(α), A1s(α), A2s(α), and a+s (α) by ba(α), B1a(α), B2a(α) and

b+a(α) in (4.4)–(4.6). The boundary condition of (4.69) leads to

{

2α2 −(

p

c2

)2}

B1a(α) + 2αγ2(α)B2a(α) + 2E0

μ

{(c2

p

)2

(2A1 +A3)α2 − (A1 +A2)

}

αba(α) = 0.

(4.72)

Introducing the abbreviation

Ba(α) = αB1a(α) + γ2(α)B2a +1

(p/c2

)2

E0

μ(2A1 +A3)α2ba(α), (4.73)

and in view of boundary conditions (4.70) and (4.71), together with (4.72) and (4.73), wehave the following two simultaneous dual integral equations:

∫∞

0α{ηE0f1(α)Ba(α) + f2(α)ba(α)

}sin(αx)dα = 0 (0 ≤ x < a),

∫∞

0αbs(α) cos(αx)dα = 0 (a ≤ x <∞),

(4.74)

∫∞

0α

{f3(α)Ba(α) +

E0

μf4(α)ba(α)

}sin(αx)dα = −

πqj

2μsin

(αjx cos γ

)(0 ≤ x < a),

∫∞

0Ba(α) sin(αx)dα = 0 (a ≤ x <∞).

(4.75)

Equations (4.74) and (4.75) yield the solutions

Ba(α) =πqja

2

2μ

∫1

0u1/2ga(u)J1(αau)du,

ba(α) = ηE0πqja

2

2μ

∫1

0u1/2ha(u)J1(αau)du,

(4.76)


ga(u) and ha(u) are the solutions of the following Fredholm integral equations of the secondkind:

F1ga(u) + F2ha(u) −∫1

0(su)1/2{ga(s)K1a(u, s) + ha(s)K2a(u, s)

}ds = 0, (4.77)

F3ga(u) + ηE2μF4ha(u) −

∫1

0(su)1/2{ga(s)K3a(u, s) + ha(s)K4a(u, s)

}ds = −u1/2J1

(αjau cos γ

),

(4.78)

where

K1a(u, s) = iP 2∫1

0

{M11(α)J1(αPs)H


(1)1 (ασPu)

}dα (u > s),

K2a(u, s) = iP 2ηE2μ

∫1

0

{M21(α)J1(αPs)H


(1)1 (ασPu)

}dα (u > s),

K3a(u, s) = iP 2∫1

0

{M31(α)J1(αPs)H


(1)1 (ασPu)

}dα (u > s),

K4a(u, s) = iP 2ηE2μ

∫1

0

{M41(α)J1(αPs)H


(1)1 (ασPu)

}dα (u > s),

(4.79)

and Kia(u, s) (i = 1, . . . , 4) are symmetric in u and s.

4.2.3. Mode II Dynamic Singular Stresses Near the Crack Tip

The dynamic stress intensity factor KIID is obtained as

KIID = limx→a+

{2π(x − a)}1/2{σLxys + σL

xya + σMxys + σM

xya

}

y=0

= qj(πa)1/2{−F3[gs(1) − iga(1)

]+ z2[hs(1) − iha(1)]

}

= Kg

II +KhII ,

(4.80)

where

Kg

II = −qj(πa)1/2F3

[gs(1) − iga(1)

],

KhII = qj(πa)1/2z2[hs(1) − iha(1)],

z2 =[2σ2(Ae1 +Ae2) −

(Ae2 + η + 1

)]ηE2

μ.

(4.81)


The singular parts of the dynamic local stresses and Maxwell stresses near the crack tip canbe derived as follows:

σLxx ∼ −

Kg

II

(2πr)1/2

[2 + cos

θ

2cos

3θ2

]sin

θ

2−

KhII

z2(2πr)1/2

[2E2

μηF4 + F5 cosθ

2cos

3θ2

]sin

θ

2,

σLxy ∼

Kg

II

(2πr)1/2

[1 − sin

θ

2sin

3θ2

]cos

θ

2+

KhII

z2(2πr)1/2

[E2μηF4 − F5 sin

θ

2sin

3θ2

]cos

θ

2,

σLyy ∼

Kg

II

(2πr)1/2sin

θ

2cos

θ

2cos

3θ2

+Kh

II

z2(2πr)1/2F5 sin

θ

2cos

θ

2cos

3θ2,

σMxx ∼

KhII

z2(2πr)1/2ηE2

μ sinθ

2,

σMxy ∼ −

KhII

z2(2πr)1/2ηE2

μ

(1 + η

)cos

θ

2,

σMyy ∼ −

KhII

z2(2πr)1/2ηE2

μ

(1 + 2η

)sin

θ

2.

(4.82)

The singular parts of the displacements and electric fields near the crack tip can be expressedas

ux ∼K

g

II

μ

(r

2π

)1/2( 1F3

+ cos2 θ

2

)sin

θ

2+Kh

II

μz2

(r

2π

)1/2(F5cos2 θ

2

)sin

θ

2,

uy ∼K

g

II

μz2

(r

2π

)1/2(−1 + sin2 θ

2

)cos

θ

2+Kh

II

μz2

(r

2π

)1/2(−F6 + F5sin2 θ

2

)cos

θ

2,

Ex ∼Kh

II

z2μ

1

(2πr)1/2ηE0 cos

θ

2,

Ey ∼ −Kh

II

z0μ

1

(2πr)1/2ηE0 sin

θ

2,

(4.83)

where

F5 =[2σ2(Ae1 +Ae2) −

(Ae2 − η

)]ηE2

μ,

F6 =[2σ2(Ae1 +Ae2) +

(Ae2 − η

)]ηE2

μ.

(4.84)


Table 1: Material properties of PMMA.

μ(M/m2) ν Ae1 Ae2 η εr

1.1 × 109 0.4 0 3.61 2 3

5. Dynamic Energy Release Rate

The dynamic energy release rate G is obtained as

G =∫

S

ρui,ttui,xdS +∫

Γ

{(ρΣ + Φ

)δjx −

(σLij + σM

ij

)ui,x +DiEx

}njdΓ, (5.1)

where S is the region with the contour Γ. This expression may be thought of as an extensionto the J-integral given in [3]. If all the electrical field quantities are made to vanish, then (5.1)reduces to the dynamic energy release rate for the elastic materials [8]. Writing the dynamicenergy release rate expression in terms of the mode I dynamic stress intensity factor, thereresults

G =1

(1 − 2ν)K2

I

128μz20

{C1E

4μ + C2E

2μ + 64(1 − ν)(1 − 2ν)

}, (5.2)

where

C1 = 2k22 + k2

3 + 4(1 − 2ν)k1k3 + 2(1 − 2ν)k1k2 + (1 + 4ν)k3k2 + 4(1 − ν)(1 − 2ν)η(k2 − 2ηk3

),

C2 = 4(1 − 2ν)[3k1 − 4νk2 − 3k3 + 2(1 − ν)

{12 − 16ν + (7 − 8ν)η − 8(1 − ν)η2

}η],

k1 ={

2(1 − 2ν)Ae1 + 2(1 − ν)Ae2 − η}η,

k2 ={(1 − 2ν)

(2Ae1 + η

)+ 2(1 − ν)Ae2

}η,

k3 ={(1 − 2ν)

(2Ae1 + η

)− 2(1 − ν)Ae2

}η.

(5.3)

6. Results and Discussion

To examine the effect of electroelastic interactions on the dynamic stress intensity factorand dynamic energy release rate, the solutions of the Fredholm integral equations of thesecond kind (4.17), (4.41) for Mode I and (4.65), (4.66), (4.77), (4.78) for Mode II havebeen computed numerically by the use of Gaussian quadrature formulas. We can considerpolymethylmethacrylate (PMMA), and the engineering material constants of PMMA arelisted in Table 1. The dynamic stress intensity factor KI can be found as KI = |KID| + KIS.

Figure 3 exhibits the variation of the normalized mode I dynamic stress intensity factor|KI/p0P (πa)

1/2|(p0P = μα22ϕe0) against the normalized frequency Ω = aα2 subjected to P-

waves for the normalized electric field Eμ = (η/ε0)1/2E0 = 0.0, 0.1 and the angle of incidence

γ = π/2. The dynamic stress intensity factor drops rapidly beyond the first maximum andexhibits oscillations of approximately constant period as Ω increases. The peak value of


P-wavesγ = π/2Mode I

p0P/μ = 0.01

0.02

∞

1

2

3

4

|KI/p

0P(π

a)1/

2 |

0 1 2 3 4 5

Ω

Eμ = 0Eμ = 0.1

Figure 3: Mode I dynamic stress intensity factor versus frequency (P-waves, γ = π/2).

P-wavesγ = π/2

Mode Ip0P/μ = 0.01

0.02

∞

5

10

G/G

0

0 1 2 3 4 5

Ω

Eμ = 0Eμ = 0.1

Figure 4: Mode I dynamic energy relrase rate versus frequency (P-waves, γ = π/2).


P-wavesγ = π/4Mode I

p0P/μ = 0.01

1

2

3

4

0.02

∞

|KI/p

0P(π

a)1/

2 |

0 1 2 3 4 5

Ω

Eμ = 0Eμ = 0.1

Figure 5: Mode I dynamic stress intensity factor versus frequency (P-waves, γ = π/4).

P-wavesγ = π/4

Mode II

0.1

0.2

0.3

|KII/q 0

P(π

a)1/

2 |

0 1 2 3 4 5

Ω

Eμ = 0Eμ = 0.1

Figure 6: Mode II dynamic stress intensity factor versus frequency (P-waves, γ = π/4).

|KI/p0P (πa)1/2| under Eμ = 0.0 is 1.364. Also, the peak values of |KI/p0P (πa)

1/2| underEμ = 0.1 are 1.522, 2.416, 3.310 for p0P/μ = ∞, 0.02, 0.01, respectively. As Ω → 0, thedynamic stress intensity factor tends to static stress intensity factor [5]. In the absence ofthe electric fields, the dynamic stress intensity factor becomes the solution for the elasticsolid (see e.g. [9]). Figure 4 also shows the variation of the normalized mode I dynamic


P-waves

Mode Ip0P/μ = 0.02

1

2

3

|KI/p

0P(π

a)1/

2 |

0 π/2 π

γ

Ω = 0.8

0.4

Eμ = 0Eμ = 0.1

Figure 7: Mode I dynamic stress intensity factor versus angle of incidence (P-waves).

SV-wavesγ = π/2

Mode II

0.5

1

1.5

|KII/q 0

S(π

a)1/

2 |

0 1 2 3 4 5

Ω

Eμ = 0Eμ = 0.1

Figure 8: Mode II dynamic stress intensity factor versus frequency (SV-waves, γ = π/2).

energy release rate G/G0, where G0 = πa(1 − ν)p20P/2μ is the static energy release rate. The

peak values of G/G0 under Eμ = 0.0, 0.1 for p0P/μ = ∞, 0.02, 0.01 are 1.861, 2.361, 5.838,10.96, respectively. Figure 5 shows the normalized mode I dynamic stress intensity factor|KI/p0P (πa)

1/2| versus Ω subjected to P-waves for Eμ = 0.0, 0.1 and γ = π/4. The peak valuesof |KI/p0P (πa)

1/2| under Eμ = 0.0, 0.1 are 1.078, 1.198 for p0P/μ = ∞, respectively. Figure 6


SV-wavesγ = π/4Mode I

p0S/μ = 0.01

1

2

3

4

0.02

∞|KI/p

0S(π

a)1/

2 |

0 1 2 3 4 5

Ω

Eμ = 0Eμ = 0.1

Figure 9: Mode I dynamic stress intensity factor versus frequency (SV-waves, γ = π/4).

shows the normalized mode II dynamic stress intensity factor |KII/q0P (πa)1/2|(q0P = μα2

2ϕe0)versus Ω subjected to P-waves for Eμ = 0.0, 0.1 and γ = π/4. The effect of electric fields onthe mode II dynamic stress intensity factor is small. Figure 7 displays the normalized mode Idynamic stress intensity factor |KI/p0P (πa)

1/2| against the angle of incidence γ subjected toP-waves for Eμ = 0.0, 0.1 and Ω = 0.4, 0.8 (p0P/μ = 0.02). The mode I dynamic stress intensityfactors for Ω = 0.4 and 0.8 attain its maximum values at an incident angle of approximatelyπ/2.

Figure 8 shows the variation of the normalized mode II dynamic stress intensity factor|KII/q0S(πa)

1/2|(q0S = μα22ψe0) versus Ω subjected to SV-waves for Eμ = 0.0, 0.1 and γ = π/2.

The electric fields have small effect on the mode II dynamic stress intensity factor. Figure 9shows the normalized mode I dynamic stress intensity factor |KI/p0S(πa)

1/2| (p0S = μα22ψe0)

against Ω subjected to SV-waves for Eμ = 0.0, 0.1 and γ = π/4. Similar trend to the case underP-waves is observed.

7. Conclusions

The dynamic electroelastic problem for a dielectric polymer having a finite crack has beenanalyzed theoretically. The results are expressed in terms of the dynamic stress intensityfactor and dynamic energy release rate. It is found that the dynamic stress intensity factorand dynamic energy release rate tend to increase with frequency reaching a peak and thendecrease in magnitude. These peaks depend on the angle of incidence. Also, applied electricfields increase the mode I dynamic stress intensity factor and dynamic energy release rate,whereas the mode II dynamic stress intensity factor is less dependent on the electric field.


References

[1] R. A. Toupin, “The elastic dielectric,” Journal of Rational Mechanics and Analysis, vol. 5, pp. 849–915,1956.

[2] Z. T. Kurlandzka, “Influence of electrostatic field on crack propagation in elastic dielectric,” Bulletin ofthe Polish Academy of Sciences, vol. 23, pp. 333–339, 1975.

[3] Y. E. Pak and G. Herrmann, “Conservation laws and the material momentum tensor for the elasticdielectric,” International Journal of Engineering Science, vol. 24, no. 8, pp. 1365–1374, 1986.

[4] Y. E. Pak and G. Herrmann, “Crack extension force in a dielectric medium,” International Journal ofEngineering Science, vol. 24, pp. 1375–1388, 1986.

[5] Y. Shindo and F. Narita, “The planar crack problem for a dielectric medium in a uniform electric field,”Archives of Mechanics, vol. 56, no. 6, pp. 447–463, 2004.

[6] H.-K. Hong and J.-T. Chen, “Derivations of integral equations of elasticity,” Journal of EngineeringMechanics, vol. 114, pp. 1028–1044, 1988.

[7] J. T. Chen and H.-K. Hong, “Review of dual boundary element methods with emphasis onhypersingular integrals and divergent series,” Applied Mechanics Reviews, vol. 52, pp. 17–33, 1999.

[8] G. C. Sih, “Dynamic aspects of crack propagation,” in Inelastic Behavior of Solids, pp. 607–639, McGraw-Hill, New York, NY, USA, 1968.

[9] Y. Shindo, “Dynamic singular stresses for a Griffith crack in a soft ferromagnetic elastic solid subjectedto a uniform magnetic field,” ASME Journal of Applied Mechanics, vol. 50, no. 1, pp. 50–56, 1983.


Research ArticleExistence and Exponential Stability ofPositive Almost Periodic Solutions fora Model of Hematopoiesis

J. O. Alzabut,1 J. J. Nieto,2 and G. Tr. Stamov3

1 Department of Mathematics and Physical Sciences, Prince Sultan University, P.O. Box 66833, Riyadh11586, Saudi Arabia

2 Departamento de Analisis Matematico, Universidad de Santiago de Compostela, Santiago de Compostela15782, Spain

3 Department of Mathematics, Technical University, Sofia, 8800 Sliven, Bulgaria

Correspondence should be addressed to J. O. Alzabut, [email protected]

Received 15 March 2009; Revised 28 June 2009; Accepted 30 August 2009


By employing the contraction mapping principle and applying Gronwall-Bellman’s inequality,sufficient conditions are established to prove the existence and exponential stability of positivealmost periodic solution for nonlinear impulsive delay model of hematopoiesis.

Copyright q 2009 J. O. Alzabut 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.

1. Introduction

The nonlinear delay differential equation

h′(t) = −αh(t) +β

1 + hn(t − τ) , t ≥ 0, (1.1)

where α, β, τ > 0, n ∈ N, has been proposed by Mackey and Glass [1] as an appropriatemodel of hematopoiesis that describes the process of production of all types of blood cellsgenerated by a remarkable self-regulated system that is responsive to the demands put uponit. In medical terms, h(t) denotes the density of mature cells in blood circulation at time tand τ is the time delay between the production of immature cells in the bone marrow andtheir maturation for release in circulating bloodstream. It is assumed that the cells are lostfrom the circulation at a rate α, and the flux of the cells into the circulation from the stem cellcompartment depends on the density of mature cells at the previous time t − τ .


In the real-world phenomena, the parameters can be nonlinear functions. The variationof the environment, however, plays an important role in many biological and ecologicaldynamical systems. In particular, the effects of a periodically varying environment areimportant for evolutionary theory as the selective forces on systems in a fluctuatingenvironment differ from those in a stable environment. Thus, the assumption of periodicityof the parameters are a way of incorporating the periodicity of the environment. It has beensuggested by Nicholson [2] that any periodical change of climate tends to impose its periodupon oscillations of internal origin or to cause such oscillations to have a harmonic relationto periodic climatic changes.

On the other hand, some dynamical systems which describe real phenomena arecharacterized by the fact that at certain moments in their evolution, they undergo rapidchanges. Most notably, this takes place due to certain seasonal effects such as weather,resource availability, food supplies, and mating habits. These phenomena are best describedby the so-called impulsive differential equations [3]. Thus, it is more realistic to consider thecase of combined effects: periodicity of the environment, time delays and impulse actions.Namely, an equation of the form

h′(t) = −α(t)h(t) +β(t)

1 + hn(t − τ) , t /= θk,

Δh(θk) := h(θ+k

)− h(θ−k)= γkh

(θ−k)+ δk, k ∈ N,

(1.2)

holds, where θk represent the instants at which the density suffers an increment of δk units.The density of mature cells in blood circulation decreases at prescribed instants θk by somemedication and it is proportional to the density at that time (θ−k).

Theory of impulsive delay differential equations is now being recognized not onlyto be richer than the corresponding theory of ordinary differential equations but also torepresent a more natural framework for mathematical modeling of some relevant real-worldphenomena. This justifies the intensive investigation of this type of equations in the recentyears. We refer the readers to the references [4–13]. The qualitative properties for modelof hematopoiesis have been extensively investigated in literature, see [14–22]. In the recentpaper [23], in particular, sufficient conditions have been established for the existence ofperiodic solutions, persistence, global attractivity, and oscillation of solutions of equation ofform (1.2) improving and complementing some previously obtained ones.

One can easily see, nevertheless, that most of the equations considered in theabove-mentioned papers are under periodic assumptions. In this paper, we consider thegeneralization to almost periodicity. Almost periodic functions are functions that are periodicup to a small error. Its study was initiated by Bohr in [24]. To the best of the authors’knowledge, there are a few published papers considering the notion of almost periodicity ofdelay differential equations with or without impulses, see [25–36]. Motivated by this, the aimof this paper is to establish sufficient conditions for the existence and exponential stabilityof positive almost periodic solution of nonlinear impulsive delay model of hematopoiesisof form (1.2). Our approach is based on using the contraction mapping principle as well asapplying Gronwall-Bellman’s inequality.

This paper is organized as follows. In Section 2, we present some general conceptsand results that will be used later. In Section 3, we state and prove our main results on theexistence of a unique positive almost periodic solution and then we show that it is stable.


2. Some Essential Definitions and Lemmas

Let {θk}k∈N be a fixed sequence such that σ ≤ θ1 < θ2 < . . . < θk < θk+1 < . . ., wherelimk→∞θk =∞ and σ is a positive number.

Denote by PLC([σ − τ, σ],R+) the space of all piecewise left continuous functions ϕ :[σ − τ, σ] → R

+ with points of discontinuity of the first kind at t = θk, k ∈ N. By a solution of(1.2), we mean a function h(t) defined on [σ − τ,∞) and satisfying (1.2) for t ≥ σ. For a giveninitial function ξ ∈ PLC([σ − τ, σ],R+), it is well known [37] that (1.2) has a unique solutionh(t) = h(t;σ, ξ) defined on [σ − τ,∞) and satisfying the initial condition:

h(t;σ, ξ) = ξ(t), σ − τ ≤ t ≤ σ. (2.1)

As we are interested in solutions of biomedical significance, we restrict our attention topositive ones.

To say that impulsive delay differential equations have positive almost periodicsolutions, one needs to adopt the following definitions of almost periodicity for such typeof equations.

The definitions are borrowed from the monograph [3].

Definition 2.1. The set of sequences {θp

k}, θ

p

k= θk+p − θk, k, p ∈ N, is said to be uniformly

almost periodic if for arbitrary ε > 0 there exists a relatively dense set of ε-almost periodscommon for any sequences.

Definition 2.2. A function ϕ ∈ PLC(R+,R+) is said to be almost periodic if the followingconditions hold:

(a1) the set of sequences {θp

k} is uniformly almost periodic;

(a2) for any ε > 0, there exists a real number δ = δ(ε) > 0 such that if the points t′

and t′′ belong to the same interval of continuity of ϕ(t) and satisfy the inequality|t′ − t′′| < δ, then |ϕ(t′) − ϕ(t′′)| < ε;

(a3) for any ε > 0, there exists a relatively dense set T of ε-almost periods such that ifω ∈ T , then |ϕ(t + ω) − ϕ(t)| < ε for all t ∈ R

+ satisfying the condition |t − θk| > ε,k ∈ N. The elements of T are called ε-almost periods.

Related to (1.2), we consider the linear equation:

h′(t) = −α(t)h(t), t /= θk,

Δh(θk) = γkh(θk), k ∈ N.(2.2)

It is well known [3] that (2.2) with an initial condition h(t0) = h0 has a unique solutionrepresented by the form

h(t; t0, h0) = H(t, t0)h0, t0, h0 ∈ R+, (2.3)


where H is the Cauchy matrix of (2.2) defined as follows:

H(t, s) =

⎧⎪⎪⎨

⎪⎪⎩

e−∫ tsα(r)dr , θk−1 < s ≤ t ≤ θk,

k∏

i=m

(1 + γi

)e−∫ tsα(r)dr , θm−1 < s ≤ θm ≤ θk < t ≤ θk+1.

(2.4)

Throughout this paper, we introduce the following conditions (C) for (1.2):

(C1) the function α ∈ C[R+,R+] is almost periodic in the sense of Bohr and there exists aconstant μ such that α(t) ≥ μ > 0;

(C2) the sequence {γk} is almost periodic and −1 ≤ γk ≤ 0, k ∈ N;

(C3) the set of sequences {θp

k} is uniformly almost periodic and there exists η > 0 such

that infk∈N θ1k = η > 0;

(C4) the function β(t) ∈ C[R+,R+] is almost periodic in the sense of Bohr andsupt∈R |β(t)| < ν where ν > 0 and β(0) = 0;

(C5) the sequence {δk} is almost periodic and supk∈N |δk| < κ, k ∈ N.

The following results prove helpful.

Lemma 2.3 (see [3]). Let conditions (C) hold. Then for each ε > 0, there exists ε1, 0 < ε1 < ε,relatively dense sets T of positive real numbers and Q of natural numbers such that the followingrelations are fulfilled:

(b1) |α(t +ω) − α(t)| < ε, t ∈ R+, ω ∈ T ;

(b2) |β(t +ω) − β(t)| < ε, t ∈ R+, ω ∈ T ;

(b3) |γk+p − γk| < ε, p ∈ Q, k ∈ N;

(b4) |δk+p − δk| < ε, p ∈ Q, k ∈ N;

(b5) |θp

k −ω| < ε1, ω ∈ T , p ∈ Q, k ∈ N.

Lemma 2.4 (see [3]). Let condition (C3) be fulfilled. Then for each j > 0, there exists a positiveinteger N such that on each interval of length j, there is no more than N elements of the sequence{θk}, that is,

i(s, t) ≤N(t − s) +N, (2.5)

where i(s, t) is the number of the points θk in the interval (s, t).

The following lemmas provide a bound for the Cauchy matrix H(t, s) of (2.2).

Lemma 2.5. Let conditions (C1)–(C3) be satisfied. Then for the Cauchy matrixH(t, s) of (2.2), thereexists a positive constant μ such that

H(t, s) ≤ e−μ(t−s), t ≥ s, t, s ∈ R+. (2.6)


Proof. In virtue of condition (C2), we deduce that the sequence {γk} is bounded. Further, itfollows that 1 + γk ≤ 1. Thus, from (2.4) and condition (C1), we get

H(t, s) ≤ e−μ(t−s), t ≥ s, t, s ∈ R+. (2.7)

Lemma 2.6. Let conditions (C1)–(C3) be satisfied. Then for each ε > 0, t ∈ R+, s ∈ R

+, t ≥s, |t − θk| > ε, |s − θk| > ε, k ∈ N, there exists a relatively dense set T of ε-almost periods of thefunction α(t) and a positive constant M such that for ω ∈ T, it follows that

|H(t +ω, s +ω) −H(t, s)| ≤ εMe−(μ/2)(t−s). (2.8)

Proof. Consider the sets T and Q defined as in Lemma 2.3. Let ω ∈ T . Since the matrix H(t +ω, s +ω) is a solution of (2.2), we have the following:

∂

∂tH = α(t)H(t +ω, s +ω) + [α(t) − α(t +ω)]H(t +ω, s +ω), t /= θ′k,

ΔH(θ′k, s

)= γkH(θk +ω, s +ω) +

(γk − γk+p

)H(θ′k +ω, s +ω

),

(2.9)

where θ′k = θk − p, p ∈ Q, k ∈ N. Then,

H(t +ω, s +ω) = H(t, s) +∫ t

s

H(t, r)[α(r) − α(r +ω)]H(r +ω, s +ω)dr

+∑

s<θ′k<t

H(t, θ′k + 0

)[γk+p − γk

]H(θ′k +ω, s +ω

).

(2.10)

In view of Lemma 2.3, it follows that if |t − θ′k| > ε, then θ′

k+p < t + ω < θ′k+p+1. Further, we

obtain

|H(t +ω, s +ω) −H(t, s)| ≤ ε(t − s)e−μ(t−s) + εi(s, t)e−μ(t−s), (2.11)

for |t − θ′k| > ε, |s − θ′k| > ε where i(s, t) is the number of the points θ′k in the interval (s, t).From Lemma 2.4, (2.11) becomes

|H(t +ω, s +ω) −H(t, s)|

≤ ε

[2μ

{μ2(t − s)e−(μ/2)(t−s)

}+N

2μ

{μ2(t − s)e−(μ/2)(t−s)

}+Ne−(μ/2)(t−s)

]e−(μ/2)(t−s).

(2.12)

By using the inequalities e−(μ/2)(t−s) < 1 and (μ/2)(t − s)e−(μ/2)(t−s) ≤ 1, we get

|H(t +ω, s +ω) −H(t, s)| ≤ εM, (2.13)


where

M =2μ

(1 +N +

μ

2N). (2.14)

3. The Main Result

Throughout this section, it is assumed that

ν < μ. (3.1)

Theorem 3.1. Let conditions (C) hold. Then there exists a unique positive almost periodic solutionh(t) of (1.2).

Proof. Let D ⊂ PLC(R+,R+) denote the set of all positive almost periodic functions ϕ(t) with

‖ϕ‖ ≤ K, (3.2)

where

∥∥ϕ∥∥ = sup

t∈R

∣∣ϕ(t)∣∣, K :=

1μν +

21 − e−μ κN. (3.3)

Define an operator F in D by the formula

[Fϕ](t) =

∫ t

−∞H(t, s)β(s)

11 + ϕn(s − τ)ds +

∑

θk<t

H(t, θk)δk. (3.4)

One can easily check that Fϕ is a solution of (1.2). In the following, we first show that F mapsthe set D into itself. In view of relation (2.6) and the inequality

∑

θk

e−μ(t−θk) =∞∑

m=0

∑

t−m−1≤θk<t−me−μ(t−θk) ≤ 2N

∞∑

m=0

e−μm = 2N1

1 − e−μ , (3.5)

we obtain that

∥∥Fϕ∥∥ = sup

t∈R+

{∫ t

−∞H(t, s)

∣∣β(s)∣∣ 1

1 + ϕn(s − τ)ds +∑

θk<t

H(t, θk)|δk|}

<1μν +

21 − e−μ κN = K,

(3.6)

for arbitrary ϕ ∈ D.


Now, we shall prove that Fϕ is almost periodic. Indeed, let ω ∈ T, p ∈ Q where thesets T and Q are defined as in Lemma 2.3, it follows that

∥∥Fϕ(t +ω) − Fϕ(t)

∥∥

≤ supt∈R+

{∫ t

−∞|H(t +ω, s +ω) −H(t, s)|

∣∣β(s +ω)

∣∣ 1

1 + ϕn(s +ω − τ)ds

+∫ t

−∞H(t, s)

∣∣∣∣∣∣β(s +ω)

∣∣ 1

1 + ϕn(s +ω − τ) −∣∣β(s)

∣∣ 1

1 + ϕn(s − τ)

∣∣∣∣ds

+∑

θk<t

∣∣H(t +ω, θk+p

)−H(t, θk)

∣∣∣∣δk+p

∣∣ +∑

θk<t

H(t, θk)∣∣δk+p − δk

∣∣}

(3.7)

or

∥∥Fϕ(t +ω) − Fϕ(t)∥∥

≤ supt∈R+

{∫ t

−∞|H(t +ω, s +ω) −H(t, s)|

∣∣β(s +ω)∣∣ 1

1 + ϕn(s +ω − τ)ds

+∫ t

−∞H(t, s)

{∣∣β(s +ω) − β(s)∣∣ 1

1 + ϕn(s +ω − τ)

+∣∣β(s)

∣∣∣∣∣∣

11 + ϕn(s +ω − τ) −

11 + ϕn(s − τ)

∣∣∣∣

}ds

+∑

θk<t

∣∣H(t +ω, θk+p

)−H(t, θk)

∣∣∣∣δk+p∣∣ +∑

θk<t

H(t, θk)∣∣δk+p − δk

∣∣}

≤ εC1,

(3.8)

where

C1 =2μνM +

1μ(1 + ν) + κM

2N

1 − e−(μ/2)+

2N1 − e−μ . (3.9)

In virtue of (3.6) and (3.8), we deduce that Fϕ ∈ D. Therefore, F is a self-mapping from D toD.

Finally, we prove that F is a contraction mapping on D. Let ϕ, ψ ∈ D. From (3.4), wehave

∥∥Fϕ − Fψ∥∥ ≤∫ t

−∞H(t, s)

∣∣β(s)∣∣∣∣∣∣

11 + ϕn(s − τ) −

11 + ψn(s − τ)

∣∣∣∣ds

≤ 1μν∥∥ϕ − ψ

∥∥.

(3.10)


The assumption that ν < μ implies that F is a contraction mapping on D. Then there existsa unique fixed point h ∈ D such that Fh = h. This implies that (1.2) has a unique positivealmost periodic solution h(t).

Theorem 3.2. Let conditions (C) hold. Then the unique positive almost periodic solution h(t) of (1.2)is exponentially stable.

Proof. Let g(t) be an arbitrary solution of (1.2) supplemented with the initial condition

g(t) = ζ(t), ζ ∈ PLC([σ − τ, σ],R+). (3.11)

Let h(t) be the unique positive almost periodic solution of (1.2) with the initial condition(2.1). It follows that

h(t) − g(t) = H(t, σ)(ξ − ζ) +∫ t

σ

H(t, s)β(s)(

11 + hn(s − τ) −

11 + gn(s − τ)

)ds. (3.12)

Taking the norm of both sides, we get

∥∥h(t) − g(t)∥∥ ≤ e−μ(t−σ)‖ξ − ζ‖ +

∫ t

σ

e−μ(t−s)ν∥∥h(s) − g(s)

∥∥ds. (3.13)

Setting u(t) = ‖h(t) − g(t)‖eμt and applying Gronwall-Bellman’s inequality [38] we end upwith the expression

∥∥h(t) − g(t)∥∥ ≤ ‖ξ − ζ‖e−(μ−ν)(t−σ). (3.14)

The assumption that ν < μ implies that the unique positive almost periodic solution of (1.2)is exponentially stable.

Corollary 3.3. Let conditions (C) hold. If supt∈R+ β(t) < supt∈R+ α(t) then there exists a uniquepositive almost periodic exponential stable solution h(t) of

h′(t) = −α(t)h(t) +β(t)

1 + hn(t − τ) , t /= θk,

Δh(θk) = γkh(θk), k ∈ N.

(3.15)

Acknowledgments

The authors express thier sincere thanks to the referees for their valuable comments that helpmaking the content of the paper more accurate. The research of Juan J. Nieto has been partiallysupported by Ministerio de Educacion y Ciencia and FEDER, project MTM2007-61724.


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Research ArticleExistence and Uniqueness of Positive andNondecreasing Solutions for a Class of SingularFractional Boundary Value Problems

J. Caballero Mena, J. Harjani, 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 24 April 2009; Accepted 14 June 2009

Recommended by Juan Jose Nieto

We establish the existence and uniqueness of a positive and nondecreasing solution to a singularboundary value problem of a class of nonlinear fractional differential equation. Our analysis relieson a fixed point theorem in partially ordered sets.

Copyright q 2009 J. Caballero Mena 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.

1. Introduction

Many papers and books on fractional differential equations have appeared recently. Mostof them are devoted to the solvability of the linear fractional equation in terms of a specialfunction (see, e.g., [1, 2]) and to problems of analyticity in the complex domain [3]. Moreover,Delbosco and Rodino [4] considered the existence of a solution for the nonlinear fractionaldifferential equation Dα

0+u = f(t, u), where 0 < α < 1 and f : [0, a] × R → R, 0 < a ≤ +∞is a given continuous function in (0, a) × R. They obtained results for solutions by usingthe Schauder fixed point theorem and the Banach contraction principle. Recently, Zhang [5]considered the existence of positive solution for equation Dα

0+u = f(t, u), where 0 < α < 1and f : [0, 1] × [0,∞) → [0,∞) is a given continuous function by using the sub- and super-solution methods.

In this paper, we discuss the existence and uniqueness of a positive and nondecreasingsolution to boundary-value problem of the nonlinear fractional differential equation

Dα0+u(t) + f(t, u(t)) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(1.1)


where 2 < α ≤ 3, Dα0+ is the Caputo’s differentiation and f : (0, 1] × [0,∞) → [0,∞) with

limt→ 0+f(t,−) =∞ (i.e., f is singular at t = 0).Note that this problem was considered in [6] where the authors proved the existence

of one positive solution for (1.1) by using Krasnoselskii’s fixed point theorem and nonlinearalternative of Leray-Schauder type in a cone and assuming certain hypotheses on the functionf . In [6] the uniqueness of the solution is not treated.

In this paper we will prove the existence and uniqueness of a positive andnondecreasing solution for the problem (1.1) by using a fixed point theorem in partiallyordered sets.

Existence of fixed point in partially ordered sets has been considered recently in [7–12].This work is inspired in the papers [6, 8].

For existence theorems for fractional differential equation and applications, we refer tothe survey [13]. Concerning the definitions and basic properties we refer the reader to [14].

Recently, some existence results for fractional boundary value problem have appearedin the literature (see, e.g., [15–17]).

2. Preliminaries and Previous Results

For the convenience of the reader, we present here some notations and lemmas that will beused in the proofs of our main results.

Definition 2.1. The Riemman-Liouville fractional integral of order α > 0 of a function f :(0,∞) → R is given by

Iα0+f(t) =1

Γ(α)

∫ t

0(t − s)α−1f(s)ds (2.1)

provided that the right-hand side is pointwise defined on (0,∞).

Definition 2.2. The Caputo fractional derivative of order α > 0 of a continuous function f :(0,∞) → R is given by

Dα0+f(t) =

1Γ(n − α)

∫ t

0

f (n)(s)

(t − s)α−n+1ds, (2.2)

where n − 1 < α ≤ n, provided that the right-hand side is pointwise defined on (0,∞).

The following lemmas appear in [14].

Lemma 2.3. Let n − 1 < α ≤ n, u ∈ C(n)[0, 1]. Then

Iα0+Dα0+u(t) = u(t) − c1 − c2t − · · · − cntn−1, (2.3)

where ci ∈ R, i = 1, 2, . . . , n.


Lemma 2.4. The relation

Iα0+Iβ

0+ϕ = Iα+β0+ ϕ (2.4)

is valid when Re β > 0, Re(α + β) > 0, ϕ(x) ∈ L1(0, b).

The following lemmas appear in [6].

Lemma 2.5. Givenf ∈ C[0, 1] and 2 < α ≤ 3, the unique solution of

Dα0+u(t) + f(t) = 0, 0 < t < 1,

u(0) = u′(1) = u′′(0) = 0,(2.5)

is given by

u(t) =∫1

0G(t, s)f(s)ds, (2.6)

where

G(t, s) =

⎧⎪⎪⎨

⎪⎪⎩

(α − 1)t(1 − s)α−2 − (t − s)α−1

Γ(α), 0 ≤ s ≤ t ≤ 1,

t(1 − s)α−2

Γ(α − 1), 0 ≤ t ≤ s ≤ 1.

(2.7)

Remark 2.6. Note that G(t, s) > 0 for t /= 0 and G(0, s) = 0 (see [6]).

Lemma 2.7. Let 0 < σ < 1, 2 < α ≤ 3 and F : (0, 1] → R is a continuous function withlimt→ 0+F(t) = ∞. Suppose that tσF(t) is a continuous function on [0, 1]. Then the function definedby

H(t) =∫1

0G(t, s)F(s)ds (2.8)

is continuous on [0,1], where G(t, s) is the Green function defined in Lemma 2.5.

Now, we present some results about the fixed point theorems which we will use later.These results appear in [8].

Theorem 2.8. Let (X,≤) be a partially ordered set and suppose that there exists a metric d in X suchthat (X, d) is a complete metric space. Assume that X satisfies the following condition: if {xn} is anon decreasing sequence in X such that xn → x then xn ≤ x for all n ∈ N. Let T : X → X be anondecreasing mapping such that

d(Tx, Ty

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


where ψ : [0,∞) → [0,∞) is continuous and nondecreasing function such that ψ is positive in(0,∞), ψ(0) = 0 and limt→∞ψ(t) = ∞. If there exists x0 ∈ X with x0 ≤ T(x0) then T has a fixedpoint.

If we consider that (X,≤) satisfies the following condition:

for x, y ∈ X there exists z ∈ X which is comparable to x and y, (2.10)

then we have the following theorem [8].

Theorem 2.9. Adding condition (2.10) to the hypotheses of Theorem 2.8 one obtains uniqueness ofthe fixed point of f .

In our considerations, we will work in the Banach space C[0, 1] = {x : [0, 1] →R, continuous}with the standard norm ‖x‖ = max0≤t≤1|x(t)|.

Note 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]. (2.11)

In [10] it is proved that (C[0, 1],≤) with the classic metric given by

d(x, y)= max

0≤t≤1

{∣∣x(t) − y(t)∣∣} (2.12)

satisfies condition (2) of Theorem 2.8. Moreover, for x, y ∈ C[0, 1], as the function max{x, y}is continuous in [0, 1], (C[0, 1],≤) satisfies condition (2.10).

3. Main Result

Theorem 3.1. Let 0 < σ < 1, 2 < α ≤ 3, f : (0, 1] × (0,∞) → [0,∞) is continuous andlimt→ 0+f(t,−) = ∞, tσf(t, y) is a continuous function on [0, 1] × [0,∞). Assume that there exists0 < λ ≤ Γ(α − σ)/Γ(1 − σ) such that for x, y ∈ [0,∞) with y ≥ x and t ∈ [0, 1]

0 ≤ tσ(f(t, y)− f(t, x)

)≤ λ · ln

(y − x + 1

)(3.1)

Then one’s problem (1.1) has an unique nonnegative solution.

Proof. Consider the cone

P = {u ∈ C[0, 1] : u(t) ≥ 0}. (3.2)

Note that, as P is a closed set of C[0, 1], P is a complete metric space.


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

(Tu)(t) =∫1

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

By Lemma 2.7, Tu ∈ C[0, 1]. Moreover, taking into account Remark 2.6 and as tσf(t, y) ≥ 0for (t, y) ∈ [0, 1] × [0,∞) by hypothesis, we get

(Tu)(t) =∫1

0G(t, s)s−σsσf(s, u(s))ds ≥ 0. (3.4)

Hence, T(P) ⊂ P .In what follows we check that hypotheses in Theorems 2.8 and 2.9 are satisfied.Firstly, the operator T is nondecreasing since, by hypothesis, for u ≥ v

(Tu)(t) =∫1

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

=∫1

0G(t, s)s−σsσf(s, u(s))ds

≥∫1

0G(t, s)s−σsσf(s, v(s))ds = (Tv)(t).

(3.5)

Besides, for u ≥ v

d(Tu, Tv) = maxt∈[0,1]

|(Tu)(t) − (Tv)(t)|

= maxt∈[0,1]

((Tu)(t) − (Tv)(t)) = maxt∈[0,1]

[∫1

0G(t, s)

(f(s, u(s)) − f(s, v(s))

)ds

]

= maxt∈[0,1]

[∫1

0G(t, s)s−σsσ

(f(s, u(s)) − f(s, v(s))

)ds

]

≤ maxt∈[0,1]

[∫1

0G(t, s)s−σλ · ln(u(s) − v(s) + 1)ds

]

(3.6)

As the function ϕ(x) = ln(x + 1) is nondecreasing then, for u ≥ v,

ln(u(s) − v(s) + 1) ≤ ln(‖u − v‖ + 1) (3.7)


and from last inequality we get

d(Tu, Tv) ≤ maxt∈[0,1]

[∫1

0G(t, s)s−σλ · ln(u(s) − v(s) + 1)ds

]

≤ λ · ln(‖u − v‖ + 1) · maxt∈[0,1]

∫1

0G(t, s)s−σds

= λ · ln(‖u − v‖ + 1)

· maxt∈[0,1]

[∫ t

0

(α − 1)t(1 − s)α−2 − (t − s)α−1

Γ(α)s−σds +

∫1

t

t(1 − s)α−2

Γ(α − 1)s−σds

]

≤ λ · ln(‖u − v‖ + 1)

· maxt∈[0,1]

[∫ t

0

(α − 1)t(1 − s)α−2

Γ(α)s−σds +

∫1

t

t(1 − s)α−2 · s−σΓ(α − 1)

ds

]

≤ λ · ln(‖u − v‖ + 1)

· maxt∈[0,1]

[∫ t

0

(α − 1)(1 − s)α−2

Γ(α)s−σds +

∫1

t

(1 − s)α−2 · s−σΓ(α − 1)

ds

]

= λ · ln(‖u − v‖ + 1) · maxt∈[0,1]

[∫ t

0

(1 − s)α−2s−σ

Γ(α − 1)ds +

∫1

t

(1 − s)α−2s−σ

Γ(α − 1)ds

]

=λ · ln(‖u − v‖ + 1)

Γ(α − 1)· maxt∈[0,1]

[∫1

0(1 − s)α−2s−σds

]

=λ · ln(‖u − v‖ + 1)

Γ(α − 1)·∫1

0(1 − s)α−2s−σds

=λ · ln(‖u − v‖ + 1)

Γ(α − 1)· β(1 − σ, α − 1)

=λ · ln(‖u − v‖ + 1)

Γ(α − 1)· Γ(1 − σ) · Γ(α − 1)

Γ(α − σ)

= λ · ln(‖u − v‖ + 1) · Γ(1 − σ)Γ(α − σ) ≤

Γ(α − σ)Γ(1 − σ) · λ · ln(‖u − v‖ + 1) · Γ(1 − σ)

Γ(α − σ)

= ln(‖u − v‖ + 1) = ‖u − v‖ − (‖u − v‖ − ln(‖u − v‖ + 1)).

(3.8)

Put ψ(x) = x−ln(x+1). Obviously, ψ : [0,∞) → [0,∞) is continuous, nondecreasing, positivein (0,∞), ψ(0) = 0 and limx→∞ψ(x) =∞.

Thus, for u ≥ v

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


Finally, take into account that for the zero function, 0 ≤ T0, by Theorem 2.8 our problem (1.1)has at least one nonnegative solution. Moreover, this solution is unique since (P,≤) satisfiescondition (2.10) (see comments at the beginning of this section) and Theorem 2.9.

Remark 3.2. In [6, lemma 3.2] it is proved that T : P → P is completely continuous andSchauder fixed point theorem gives us the existence of a solution to our problem (1.1).

In the sequel we present an example which illustrates Theorem 3.1.

Example 3.3. Consider the fractional differential equation (this example is inspired in [6])

D5/20+ u(t) +

(t − 1/2)2 ln(2 + u(t))√t

= 0, 0 < t < 1

u(0) = u′(1) = u′′(0) = 0

(3.10)

In this case, f(t, u) = (t − 1/2)2 ln(2 + u(t))/√t for (t, u) ∈ (0, 1] × [0,∞). Note that f is

continuous in (0, 1] × [0,∞) and limt→ 0+f(t,−) = ∞. Moreover, for u ≥ v and t ∈ [0, 1] wehave

0 ≤√t

((t − 1

2

)2

ln(2 + u) −(t − 1

2

)2

ln(2 + v)

)

(3.11)

because g(x) = ln(x + 2) is nondecreasing on [0,∞), and

√t

((t − 1

2

)2

ln(2 + u) −(t − 1

2

)2

ln(2 + v)

)

=√t ·(t − 1

2

)2

[ln(2 + u) − ln(2 + v)]

=√(t)(t − 1

2

)2[ln(

2 + u

2 + v

)]=√t

(t − 1

2

)2

ln(

2 + v + u − v2 + v

)

≤(

12

)2

ln(1 + u − v).

(3.12)

Note that Γ(α − σ)/Γ(1 − σ) = Γ(5/2 − 1/2)/Γ(1 − 1/2) = Γ(2)/Γ(1/2) = 1/√π ≥ 1/4.

Theorem 3.1 give us that our fractional differential (3.10) has an unique nonnegativesolution.

This example give us uniqueness of the solution for the fractional differential equationappearing in [6] in the particular case σ = 1/2 and α = 5/2

Remark 3.4. Note that our Theorem 3.1 works if the condition (3.1) is changed by, for x, y ∈[0,∞) with y ≥ x and t ∈ [0, 1]

0 ≤ tσ(f(t, y)− f(t, x)

)≤ λ · ψ

(y − x

)(3.13)


where ψ : [0,∞) → [0,∞) is continuous and ϕ(x) = x − ψ(x) satisfies

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

(b) ϕ(0) = 0;

(c) ϕ is positive in (0,∞);

(d) limx→∞ϕ(x) =∞.

Examples of such functions are ψ(x) = arctgx and ψ(x) = x/(1 + x).

Remark 3.5. Note that the Green function G(t, s) is strictly increasing in the first variable inthe interval (0, 1). In fact, for s fixed we have the following cases

Case 1. For t1, t2 ≤ s and t1 < t2 as, in this case,

G(t, s) =t(1 − s)α−2

Γ(α − 1). (3.14)

It is trivial that

G(t1, s) =t1(1 − s)α−2

Γ(α − 1)<

t2(1 − s)α−2

Γ(α − 1)= G(t2, s). (3.15)

Case 2. For t1 ≤ s ≤ t2 and t1 < t2, we have

G(t2, s) −G(t1, s) =

[(α − 1)t2(1 − s)α−2

Γ(α)− (t2 − s)α−1

Γ(α)

]

−[t1(1 − s)α−2

Γ(α − 1)

]

=t2(1 − s)α−2 − t1(1 − s)α−2

Γ(α − 1)− (t2 − s)α−1

Γ(α)

>(t2 − t1)(1 − s)α−2

Γ(α − 1)− (t2 − s)α−1

Γ(α − 1)

=(t2 − t1)(1 − s)α−2

Γ(α − 1)− (t2 − s)(t2 − s)α−2

Γ(α − 1).

(3.16)

Now, t2 − t1 ≥ (t2 − s) and (1 − s) ≥ (t2 − s) then

(t2 − t1)(1 − s)α−2

Γ(α − 1)>

(t2 − s)(t2 − s)α−2

Γ(α − 1). (3.17)

Hence, taking into account the last inequality and (3.16), we obtain G(t1, s) < G(t2, s).

Case 3. For s ≤ t1, t2 and t1 < t2 < 1, we have

∂G

∂t=

(α − 1)(1 − s)α−2 − (α − 1)(1 − s)α−2

Γ(α)=

α − 1Γ(α)

((1 − s)α−2 − (t − s)α−2

), (3.18)


and, as (1− s)α−2 > (t− s)α−2 for t ∈ [0, 1), it can be deduced that ∂G/∂t > 0 and consequently,G(t2, s) > G(t1, s).

This completes the proof.

Remark 3.5 gives us the following theorem which is a better result than that [6,Theorem 3.3] because the solution of our problem (1.1) is positive in (0, 1) and strictlyincreasing.

Theorem 3.6. Under assumptions of Theorem 3.1, our problem (1.1) has a unique nonnegative andstrictly increasing solution.

Proof. By Theorem 3.1 we obtain that the problem (1.1) has an unique solution u(t) ∈ C[0, 1]with u(t) ≥ 0. Now, we will prove that this solution is a strictly increasing function. Let ustake t2, t1 ∈ [0, 1] with t1 < t2, then

u(t2) − u(t1) = (Tu)(t2) − (Tu)(t1) =∫1

0(G(t2, s) −G(t1, s))f(s, u(s))ds. (3.19)

Taking into account Remark 3.4 and the fact that f ≥ 0, we get u(t2) − u(t1) ≥ 0.Now, if we suppose that u(t2)−u(t1) = 0 then

∫10(G(t2, s)−G(t1, s))f(s, u(s))ds = 0 and

as, G(t2, s) −G(t1, s) > 0 we deduce that f(s, u(s)) = 0 a.e.On the other hand, if f(s, u(s)) = 0 a.e. then

u(t) =∫1

0G(t, s)f(s, u(s))ds = 0 for t ∈ [0, 1]. (3.20)

Now, as limt→ 0+f(t, 0) = ∞, then for M > 0 there exists δ > 0 such that for s ∈ [0, 1] with0 < s < δ we get f(s, 0) > M. Observe that (0, δ) ⊂ {s ∈ [0, 1] : f(s, u(s)) > M}, consequently,

δ = μ((0, δ)) ≤ μ({

s ∈ [0, 1] : f(s, u(s)) > M})

(3.21)

and this contradicts that f(s, u(s)) = 0 a.e.Thus, u(t2) − u(t1) > 0 for t2, t1 ∈ [0, 1] with t2 > t1. Finally, as u(0) =

∫10G(0, s)f(s, u(s))ds = 0 we have that 0 < u(t) for t /= 0.

Acknowledgment

This research was partially supported by ”Ministerio de Educacion y Ciencia” Project MTM2007/65706.

References

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[10] J. J. Nieto and R. Rodrıguez-Lopez, “Contractive mapping theorems in partially ordered sets andapplications to ordinary differential equations,” Order, vol. 22, no. 3, pp. 223–239, 2005.

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Research ArticleExistence and Uniqueness ofSmooth Positive Solutions to a Class of Singularm-Point Boundary Value Problems

Xinsheng Du and Zengqin Zhao

School of Mathematics Sciences, Qufu Normal University, Qufu, Shandong 273165, China

Correspondence should be addressed to Xinsheng Du, [email protected]

Received 2 April 2009; Revised 15 September 2009; Accepted 23 November 2009


This paper investigates the existence and uniqueness of smooth positive solutions to a class ofsingular m-point boundary value problems of second-order ordinary differential equations. Anecessary and sufficient condition for the existence and uniqueness of smooth positive solutions isgiven by constructing lower and upper solutions and with the maximal theorem. Our nonlinearityf(t, u, v) may be singular at v, t = 0 and/or t = 1.

Copyright q 2009 X. Du and Z. Zhao. 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.

1. Introduction and the Main Results

In this paper, we will consider the existence and uniqueness of positive solutions to a classof second-order singular m-point boundary value problems of the following differentialequation:

−u′′(t) = f(t, u(t), u(t)), t ∈ (0, 1), (1.1)

with

u(0) =m−2∑

i=1

αiu(ηi), u(1) = 0, (1.2)

where 0 < αi < 1, i = 1, 2, . . . , m − 2, 0 < η1 < η2 < · · · < ηm−2 < 1, are constants,∑m−2

i=1 αi <1, m ≥ 3, and f satisfies the following hypothesis:


(H) f(t, u, v) : (0, 1) × (0,+∞) × (0,+∞) → [0,+∞) is continuous, nondecreasing on u,and nonincreasing on v for each fixed t ∈ (0, 1), there exists a real number b ∈ R+

such that for any r ∈ (0, 1),

f(t, u, rv) ≤ r−bf(t, u, v), ∀(t, u, v) ∈ (0, 1) × (0,+∞) × (0,+∞), (1.3)

there exists a function g : [1,∞) → (0,+∞), g(l) < l, and g(l)/l2 is integrable on(1,+∞) such that

f(t, lu, v) ≤ g(l)f(t, u, v), ∀(t, u, v) ∈ (0, 1) × (0,+∞) × (0,+∞), l ∈ [1,+∞). (1.4)

Remark 1.1. (i) Inequality (1.3) implies

f(t, u, cv) ≥ c−bf(t, u, v), if c ≥ 1. (1.5)

Conversely, (1.5) implies (1.3).(ii) Inequality (1.4) implies

f(t, cu, v) ≥(g(c−1))−1

f(t, u, v), if 0 < c < 1. (1.6)

Conversely, (1.6) implies (1.4).

Remark 1.2. It follows from (1.3), (1.4) that

f(t, u, u) ≤

⎧⎪⎪⎨

⎪⎪⎩

g(uv

)f(t, v, v), if u ≥ v > 0,

(vu

)bf(t, v, v), if v ≥ u > 0.

(1.7)

When f(t, u) is increasing with respect to u, singular nonlinear m-point boundaryvalue problems have been extensively studied in the literature, see [1–3]. However, whenf(t, u, v) is increasing on u, and is decreasing on v, the study on it has proceeded very slowly.The purpose of this paper is to fill this gap. In addition, it is valuable to point out that thenonlinearity f(t, u, v) may be singular at t = 0, t = 1 and/or v = 0.

When referring to singularity we mean that the functions f in (1.1) are allowed to beunbounded at the points v = 0, t = 0, and/or t = 1. A function u(t) ∈ C[0, 1]∩C2(0, 1) is calleda C[0, 1] (positive) solution to (1.1) and (1.2) if it satisfies (1.1) and (1.2) (u(t) > 0, for t ∈(0, 1)). A C[0, 1] (positive) solution to (1.1) and (1.2) is called a smooth (positive) solution ifu′(0+) and u′(1−) both exist (u(t) > 0 for t ∈ (0, 1)). Sometimes, we also call a smooth solutiona C1[0, 1] solution. It is worth stating here that a nontrivial C[0, 1] nonnegative solution tothe problem (1.1), (1.2) must be a positive solution. In fact, it is a nontrivial concave functionsatisfying (1.2) which, of course, cannot be equal to zero at any point t ∈ (0, 1).

To seek necessary and sufficient conditions for the existence of solutions to the aboveproblems is important and interesting, but difficult. Thus, researches in this respect are rare


up to now. In this paper, we will study the existence and uniqueness of smooth positivesolutions to the second-order singular m-point boundary value problem (1.1) and (1.2). Anecessary and sufficient condition for the existence of smooth positive solutions is given byconstructing lower and upper solutions and with the maximal principle. Also, the uniquenessof the smooth positive solutions is studied.

A function α(t) is called a lower solution to the problem (1.1), (1.2), if α(t) ∈ C[0, 1] ∩C2(0, 1) and satisfies

α′′(t) + f(t, α(t), α(t)) ≥ 0, t ∈ (0, 1),

α(0) −m−2∑

i=1

αiα(ηi)≤ 0, α(1) ≤ 0.

(1.8)

Upper solution is defined by reversing the above inequality signs. If there exist a lowersolution α(t) and an upper solution β(t) to problem (1.1), (1.2) such that α(t) ≤ β(t), then(α(t), β(t)) is called a couple of upper and lower solution to problem (1.1), (1.2).

To prove the main result, we need the following maximal principle.

Lemma 1.3 (maximal principle). Suppose that 0 < η1 < η2 < · · · < ηm−2 < bn < 1, n = 1, 2, . . .,and Fn = {u(t) ∈ C[0, bn] ∩ C2(0, bn), u(0) −

∑m−2i=1 αiu(ηi) ≥ 0, u(bn) ≥ 0}. If u ∈ Fn such that

−u′′(t) ≥ 0, t ∈ (0, bn) then u(t) ≥ 0, t ∈ [0, bn].

Proof. Let

−u′′(t) = δ(t), t ∈ (0, bn), (1.9)

u(0) −m−2∑

i=1

αiu(ηi)= r1, u(bn) = r2, (1.10)

then r1 ≥ 0, r2 ≥ 0, δ(t) ≥ 0, t ∈ (0, bn).By integrating (1.9) twice and noting (1.10), we have

u(t) =1

bn(

1 −∑m−2

i=1 αi

)+∑m−2

i=1 αiηi

[((

1 −m−2∑

i=1

αi

)

t +m−2∑

i=1

αiηi

)

r2 + (bn − t)r1

]

+∫bn

0Gn(t, s)δ(s)ds +

bn − tbn(

1 −∑m−2

i=1 αi

)+∑m−2

i=1 αiηi

m−2∑

i=1

αi

∫bn

0Gn

(ηi, s)δ(s)ds,

(1.11)

where

Gn(t, s) =1bn

⎧⎨

⎩

t(bn − s), 0 ≤ t ≤ s ≤ bn,

s(bn − t), 0 ≤ s ≤ t ≤ bn.(1.12)


In view of (1.11) and the definition of Gn(t, s), we can obtain u(t) ≥ 0, t ∈ [0, bn]. Thiscompletes the proof of Lemma 1.3.

Now we state the main results of this paper as follows.

Theorem 1.4. Suppose that (H) holds, then a necessary and sufficient condition for the problem (1.1)and (1.2) to have smooth positive solution is that

0 <

∫1

0f(s, 1 − s, 1 − s)ds <∞. (1.13)

Theorem 1.5. Suppose that (H) and (1.13) hold, then the smooth positive solution to problem (1.1)and (1.2) is also the unique C[0, 1] positive solution.

2. Proof of Theorem 1.4

2.1. The Necessary Condition

Suppose that w(t) is a smooth positive solution to the boundary value problem (1.1) and(1.2). We will show that (1.13) holds.

It follows from

w′′(t) = −f(t,w(t), w(t)) ≤ 0, t ∈ (0, 1), (2.1)

that w′(t) is nonincreasing on [0, 1]. Thus, by the Lebesgue theorem, we have

∫1

0f(t,w(t), w(t))dt = −

∫1

0w′′(t)dt = w′(0+) −w′(1−) < +∞. (2.2)

It is well known that w(t) can be stated as

w(t) =∫1

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

+1 − t

(1 −∑m−2

i=1 αi

)+∑m−2

i=1 αiηi

m−2∑

i=1

αi

∫1

0G(ηi, s)f(s,w(s), w(s))ds,

(2.3)

where

G(t, s) =

⎧⎨

⎩

t(1 − s), 0 ≤ t ≤ s ≤ 1,

s(1 − t), 0 ≤ s ≤ t ≤ 1.(2.4)


By (2.3) and (1.2) we have

1(

1 −∑m−2

i=1 αi

)+∑m−2

i=1 αiηi

m−2∑

i=1

αi

∫1

0G(ηi, s)f(s,w(s), w(s))ds =

m−2∑

i=1

αi

[w(ηi)], (2.5)

therefore because of (2.3) and (2.5),

w(t) ≥ (1 − t)m−2∑

i=1

αi

[w(ηi)], t ∈ [0, 1]. (2.6)

Since w(t) is a smooth positive solution to (1.1) and (1.2), we have

w(t) =∫1

t

(−w′(s)

)ds ≤ max

t∈[0,1]

∣∣w′(t)∣∣(1 − t), t ∈ [0, 1]. (2.7)

Let m =∑m−2

i=1 αiw[(ηi)], M = maxt∈[0,1]|w′(t)|. From (2.6), (2.7) it follows that

m(1 − t) ≤ w(t) ≤M(1 − t), t ∈ [0, 1]. (2.8)

Without loss of generality we may assume that 0 < m < 1 < M. This together with thecondition (H) implies

∫1

0f(s, 1 − s, 1 − s)ds ≤

∫1

0f

(s,

1mw(s),

1M

w(s))ds

≤ g

(1m

)Mb

∫1

0f(s,w(s), w(s))ds < +∞.

(2.9)

On the other hand, notice that w(t) is a smooth positive solution to (1.1), (1.2), wehave

f(t,w(t), w(t)) = −w′′(t)/≡ 0, t ∈ (0, 1), (2.10)

therefore, there exists a positive number t0 ∈ (0, 1) such that f(t0, w(t0), w(t0)) > 0. Obviously,w(t0) > 0 and 1 − t0 > 0. It follows from (1.7) that

0 < f(t0, w(t0), w(t0)) ≤

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

g

(w(t0)1 − t0

)f(t0, 1 − t0, 1 − t0), if w(t0) ≥ 1 − t0,

(1 − t0w(t0)

)b

f(t0, 1 − t0, 1 − t0), if w(t0) ≤ 1 − t0.(2.11)


Consequently f(t0, 1 − t0, 1 − t0) > 0, which implies that

∫1

0f(s, 1 − s, 1 − s)ds > 0. (2.12)

From (2.9) and (2.12) it follows that

0 <

∫1

0f(s, 1 − s, 1 − s) < +∞, (2.13)

which is the required inequality.

2.2. The Existence of Lower and Upper Solutions

Since g(l)/l2 is integrable on [1,+∞), thus

liml→+∞

infg(l)l

= 0. (2.14)

Otherwise, if liml→+∞ inf g(l)/l = m0 > 0, then there exists a real number X > 0, such thatg(l)/l2 ≥ m0/2l when l > X, this contradicts with the condition that g(l)/l2 is integrable on[1,+∞). By condition (H) and (2.14) we have

f(t, ru, v) ≥ h(r)f(t, u, v), r ∈ (0, 1), (2.15)

limr→ 0+

supr

h(r)= lim

p→+∞sup

p−1

h(p−1) = lim

p→+∞inf

g(p)

p= 0, (2.16)

where h(r) = (g(r−1))−1, r ∈ (0, 1).

Suppose that (1.13) holds. Let

b(t) =∫1

0G(t, s)f(s, 1 − s, 1 − s)ds

+1 − t

1 −∑m−2

i=1 αi +∑m−2

i=1 αiηi

m−2∑

i=1

αi

∫1

0G(ηi, s)f(s, 1 − s, 1 − s)ds.

(2.17)

Since by (1.13), (2.17) we obviously have

b(t) ∈ C1[0, 1] ∩ C2(0, 1), b′′(t) = −f(t, 1 − t, 1 − t), t ∈ (0, 1), (2.18)

and there exists a positive number k < 1 such that

k(1 − t) ≤ b(t) ≤ 1k(1 − t), t ∈ [0, 1]. (2.19)


By (2.14) and (2.16) we see, if 0 < l < k is sufficiently small, then

h(lk) − l ≥ 0, g

(1lk

)− 1

l≤ 0. (2.20)

Let

H(t) = lb(t), Q(t) =1lb(t), t ∈ [0, 1]. (2.21)

Then from (2.19) and (2.21) we have

lk(1 − t) ≤ H(t) ≤ 1 − t ≤ Q(t) ≤ 1lk(1 − t), t ∈ [0, 1]. (2.22)

Consequently, with the aid of (2.20), (2.22) and the condition (H) we have

H ′′(t) + f(t,H(t),H(t)) ≥ f(t, lk(1 − t), 1 − t) − lf(t, 1 − t, 1 − t)

≥ [h(lk) − l]f(t, 1 − t, 1 − t) ≥ 0,(2.23)

Q′′(t) + f(t, Q(t), Q(t)) ≤ f

(t,

1lk(1 − t), 1 − t

)− 1

lf(t, 1 − t, 1 − t)

≤[g

(1lk

)− 1

l

]f(t, 1 − t, 1 − t) ≤ 0.

(2.24)

From (2.17), (2.21) it follows that

H(0) =m−2∑

i=1

αiH(ηi), H(1) = 0, (2.25)

Q(0) =m−2∑

i=1

αiQ(ηi), Q(1) = 0, (2.26)

therefore, (2.23)–(2.26) imply that H(t), Q(t) are lower and upper solutions to the problem(1.1) and (1.2), respectively.

2.3. The Sufficient Condition

First of all, we define a partial ordering in C[0, 1] ∩ C2(0, 1) by u ≤ v, if and only if

u(t) ≤ v(t), ∀t ∈ [0, 1]. (2.27)


Then, we will define an auxiliary function. For all u(t) ∈ C[0, 1] ∩ C2(0, 1),

g(t, u(t)) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f(t,H(t),H(t)), if u(t) ≤ H(t),

f(t, u(t), u(t)), if H(t) ≤ u(t) ≤ Q(t),

f(t, Q(t), Q(t)), if u(t) ≥ Q(t).

(2.28)

By the assumption of Theorem 1.4, we have that g : (0, 1) × (−∞,+∞) → [0,+∞) iscontinuous.

Let {bn} be a sequence satisfying ηm−2 < b1 < · · · < bn < bn+1 < · · · < 1, and bn → 1 asn → +∞, and let {rn} be a sequence satisfying

H(bn) ≤ rn ≤ Q(bn), n = 1, 2, . . . . (2.29)

For each n, let us consider the following nonsingular problem:

−u′′(t) = g(t, u(t)), t ∈ [0, bn],

u(0) −m−2∑

i=1

αiu(ηi)= 0, u(bn) = rn.

(2.30)

Obviously, it follows from the proof of Lemma 1.3 that problem (2.30) is equivalent to theintegral equation

u(t) = Anu(t) =

((1 −∑m−2

i=1 αi

)t +∑m−2

i=1 αiηi)rn

bn(

1 −∑m−2

i=1 αi

)+∑m−2

i=1 αiηi+∫bn

0Gn(t, s)g(s, u(s))ds

+bn − t

bn(

1 −∑m−2

i=1 αi

)+∑m−2

i=1 αiηi

m−2∑

i=1

αi

∫bn

0Gn

(ηi, s)g(s, u(s))ds, t ∈ [0, bn],

(2.31)

where Gn(t, s) is defined in the proof of Lemma 1.3. It is easy to verify that An : Xn →Xn = C[0, bn] is a completely continuous operator and An(Xn) is a bounded set. Moreover,u ∈ C[0, bn] is a solution to (2.30) if and only if Anu = u. Using the Schauder’s fixed pointtheorem, we assert that An has at least one fixed point un ∈ C2[0, bn].

We claim that

H(t) ≤ un(t) ≤ Q(t), t ∈ [0, bn]. (2.32)

From this it follows that

−u′′(t) = f(t, u(t), u(t)), t ∈ [0, bn]. (2.33)


Suppose by contradiction that un(t) ≤ Q(t) is not satisfied on [0, bn]. Let

z(t) = Q(t) − un(t), t ∈ [0, bn], (2.34)

therefore

z(t∗) = mint∈[0,bn]

z(t) < 0. (2.35)

Since by the definition of Q(t) and (2.30) we obviously have t∗ /= 0, t∗ /= bn.Let

c = inf{t1 | z(t) < 0, t ∈ (t1, t∗]},

d = sup{t2 | z(t) < 0, t ∈ [t∗, t2)}.(2.36)

So, when t ∈ (c, d), we have Q(t) < un(t), and

g(t, un(t)) = f(t, Q(t), Q(t)),

u′′n(t) + g(t, Q(t)) = 0,

Q′′(t) + g(t, Q(t)) = Q′′(t) + f(t, Q(t), Q(t)) ≤ 0.

(2.37)

Therefore z′′(t) = Q′′(t)−u′′n(t) ≤ 0, t ∈ (c, d), that is, z(t) is an upper convex function in (c, d).By (2.30) and (2.36), for c, d we have the following two cases:

(i) z(c) = z(d) = 0,

(ii) z(c) < 0, z(d) = 0.

For case (i): it is clear that z(t) ≥ 0, t ∈ (c, d), this is a contradiction.For case (ii): in this case c = 0, z′(t∗) = 0. Since z′(t) is decreasing on [c, d], thus,

z′(t) ≤ 0, t ∈ [t∗, d], that is, z(t) is decreasing on [t∗, d]. From z(d) = 0, we see z(t∗) ≥ 0,which is in contradiction with z(t∗) < 0.

From this it follows that un(t) ≤ Q(t), t ∈ [0, bn].Similarly, we can verify that H(t) ≤ un(t), t ∈ [0, bn]. Consequently (2.32) holds.Using the method of [4] and [5, Theorem 3.2], we can obtain that there is a C[0, 1]

positive solution w(t) to (1.1), (1.2) such that H(t) ≤ w(t) ≤ Q(t), and a subsequence of{un(t)} converges to w(t) on any compact subintervals of (0, 1).



Suppose that u1(t) and u2(t) are C[0, 1] positive solutions to (1.1) and (1.2), and at least one ofthem is a smooth positive solution. If u1(t)/≡u2(t) for any t ∈ [0, 1], without loss of generality,we may assume that u2(t∗) > u1(t∗) for some t∗ ∈ (0, 1). Let

T = inf{t1 | 0 ≤ t1 < t∗, u2(t) > u1(t), t ∈ (t1, t∗]},

S = sup{t2 | t∗ ≤ t2 < 1, u2(t) > u1(t), t ∈ [t∗, t2)},

y(t) = u1(t)u′2(t) − u2(t)u′1(t), t ∈ (0, 1).

(3.1)

It follows from (3.1) that

0 ≤ T < S ≤ 1, u2(t) ≥ u1(t), t ∈ (T, S). (3.2)

By (1.2), it is easy to check that there exist the following two possible cases:

(1) u1(T) = u2(T), u1(S) = u2(S),

(2) u1(T) < u2(T), u1(S) = u2(S).

Assume that case (1) holds. By u′′i (t) ≤ 0 on (0, 1), it is easy to see that u′i(T+0) (i = 1, 2)exist (finite or∞), moreover, one of them must be finite. The same conclusion is also valid foru′i(S − 0) (i = 1, 2). It follows from (3.2) that

[u2(t) − u1(t)]|′t=T+0 ≥ 0, (3.3)

consequently

u′2(T + 0) ≥ u′1(T + 0), u′1(T + 0) is finite. (3.4)

Similarly

u′2(S − 0) ≤ u′1(S − 0), u′1(S − 0) is finite. (3.5)

From (3.1), (3.4), and (3.5) we have

lim inft−→T+0

y(t) ≥ 0 ≥ lim supt−→S−0

y(t). (3.6)

On the other hand, (3.2), (1.7), and condition (H) yield

f(t, u2(t), u2(t)) ≤ g

(u2(t)u1(t)

)f(t, u1(t), u1(t))

≤ u2(t)u1(t)

f(t, u1(t), u1(t)), t ∈ (T, S),

(3.7)


that is,

f(t, u2(t), u2(t))u2(t)

≤f(t, u1(t), u1(t))

u1(t), t ∈ (T, S), (3.8)

therefore

u′′1(t)u1(t)

≤u′′2(t)u2(t)

, t ∈ (T, S). (3.9)

From this it follows that

y′(t) = u1(t)u′′2(t) − u2(t)u′′1(t) ≥ 0, t ∈ (T, S). (3.10)

If y′(t) ≡ 0 on (T, S), then, by (3.6) we have y(t) ≡ 0, and then (u2(t)/u1(t))′ ≡ 0,

which imply that there exists a positive number c such that u2(t) = cu1(t) on (T, S). It followsfrom (3.2) that c > 1, therefore T = 0, S = 1. Substituting u2(t) = cu1(t) into (1.1) and usingcondition (H), we have

cf(t, u1(t), u1(t)) = f(t, cu1(t), cu1(t))

≤ g(c)f(t, u1(t), u1(t)), t ∈ (0, 1).(3.11)

Noticing (3.11) and f(t, u1(t), u1(t))/≡ 0, t ∈ (0, 1), we have

c ≤ g(c), (3.12)

which contradicts with the condition that g(c) < c. Therefore, y′(t) ≥ 0 and y′(t)/≡ 0 on (T, S).Thus, y(T + 0) < y(S − 0), which contradicts with (3.6). So case (1) is impossible.

By analogous methods, we can obtain a contradiction for case (2). So u1(t) ≡ u2(t) forany t ∈ [0, 1], which implies that the result of Theorem 1.5 holds.

4. Concerned Remarks and Applications

Remark 4.1. The typical function satisfying (H) is f(t, u, u) =∑n

i=1 ai(t)uλi +∑m

j=1 bj(t)u−μj ,

where ai, bj ∈ C(0, 1), 0 < λi < 1, μj > 0, (i = 1, 2, . . . , n, j = 1, 2, . . . , m).

Remark 4.2. Condition (H) includes e-concave function (see [6]) as special case. For example,Liu and Yu [7] consider the existence and uniqueness of positive solution to a class of singularboundary value problem under the following condition:

f(t, λu,

v

λ

)≥ λαf(t, u, v), ∀u, v > 0, λ ∈ (0, 1), (4.1)

where α ∈ [0, 1) and f(t, u, v) is nondecreasing on u, nonincreasing on v. Clearly, condition(H) is weaker than the above condition (4.1).


In fact, for any λ ≥ 1, from (4.1) it follows that

f(t, λu, v) ≤ f

(t, λu,

1λv

)≤ λαf(t, u, v). (4.2)

On the other hand, for any 0 < λ < 1, from (4.1) it follows that

f(t, u, v) ≥ f(t, λu, λ

v

λ

)≥ λαf(t, u, λv), (4.3)

that is, f(t, u, λv) ≤ λ−αf(t, u, v).In what follows, by using the results obtained in this paper, we study the boundary

value problem

u′′(t) + μt−γ(1 − t)−l(u−α(t) + uβ(t) +A

)= 0, 0 < t < 1,

u(0) =m−2∑

i=1

αiu(ηi), u(1) = 0,

(4.4)

where μ > 0, α > 0, β < 1, A ≥ 0. We have the following theorem.

Theorem 4.3. A necessary and sufficient condition for problem (4.4) to have smooth positive solutionis that

max{γ + α, l + α, γ − β, l − β, γ, l

}< 1. (4.5)

Moreover, when the positive solution exists, it is unique.

Remark 4.4. Consider (1.1) and the following singular m-point boundary value conditions:

u(0) = 0, u(1) =m−2∑

i=1

αiu(ηi). (4.6)

By analogous methods, we have the following results.Assume that u(t) is a C[0, 1] positive solution to (1.1) and (4.6), then u(t) can be stated

u(t) =∫1

0G(t, s)f(s, u(s), u(s)) +

t

1 −∑m−2

i=1 αiηi

m−2∑

i=1

αi

∫1

0G(ηi, s)f(s, u(s), u(s))ds, (4.7)

where G(t, s) is defined in (2.4).

Theorem 4.5. Suppose that (H) holds, then a necessary and sufficient condition for the problem (1.1)and (4.6) to have smooth positive solution is that

0 <

∫1

0f(s, s, s)ds <∞. (4.8)


Theorem 4.6. Suppose (H) and (4.8) hold, then the smooth positive solution to problem (1.1) and(4.6) is also unique C[0, 1] positive solution.

Acknowledgment

Research supported by the National Natural Science Foundation of China (10871116), theNatural Science Foundation of Shandong Province (Q2008A03) and the Doctoral ProgramFoundation of Education Ministry of China (200804460001).

References

[1] X. Du and Z. Zhao, “A necessary and sufficient condition of the existence of positive solutions tosingular sublinear three-point boundary value problems,” Applied Mathematics and Computation, vol.186, no. 1, pp. 404–413, 2007.

[2] X. Du and Z. Zhao, “On existence theorems of positive solutions to nonlinear singular differentialequations,” Applied Mathematics and Computation, vol. 190, no. 1, pp. 542–552, 2007.

[3] Z. Wei, “A necessary and sufficient condition for the existence of positive solutions of singular super-linear m-point boundary value problems,” Applied Mathematics and Computation, vol. 179, no. 1, pp.67–78, 2006.

[4] Y. Zhang, “Positive solutions of singular sublinear Emden-Fowler boundary value problems,” Journalof Mathematical Analysis and Applications, vol. 185, no. 1, pp. 215–222, 1994.

[5] P. Hartman, Ordinary Differential Equations, Brikhauser, Boston, Mass, USA, 2nd edition, 1982.[6] D. J. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones, vol. 5 of Notes and Reports in

Mathematics in Science and Engineering, Academic Press, Boston, Mass, USA, 1988.[7] Y. Liu and H. Yu, “Existence and uniqueness of positive solution for singular boundary value

problem,” Computers & Mathematics with Applications, vol. 50, no. 1-2, pp. 133–143, 2005.


Research ArticleExistence of Periodic Solution for a NonlinearFractional Differential Equation

Mohammed Belmekki,1 Juan J. Nieto,2and Rosana Rodrıguez-Lopez2

1 Departement de Mathematiques, Universite de Saıda, BP 138, 20000 Saıda, Algeria2 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Santiago de Compostela,15782 Santiago de Compostela, Spain

Correspondence should be addressed to Rosana Rodrıguez-Lopez, [email protected]

Received 2 February 2009; Revised 10 April 2009; Accepted 4 June 2009


We study the existence of solutions for a class of fractional differential equations. Due to thesingularity of the possible solutions, we introduce a new and proper concept of periodic boundaryvalue conditions. We present Green’s function and give some existence results for the linear caseand then we study the nonlinear problem.

Copyright q 2009 Mohammed Belmekki 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.

1. Introduction

Fractional calculus is a generalization of ordinary differentiation and integration to arbitrarynoninteger order. The subject is as old as the differential calculus, and goes back to time whenLeibnitz and Newton invented differential calculus. The idea of fractional calculus has been asubject of interest not only among mathematicians but also among physicists and engineers.See, for instance, [1–6].

Fractional-order models are more accurate than integer-order models, that is, there aremore degrees of freedom in the fractional-order models. Furthermore, fractional derivativesprovide an excellent instrument for the description of memory and hereditary properties ofvarious materials and processes due to the existence of a “memory” term in a model. Thismemory term insures the history and its impact to the present and future. For more details,see [7].

Fractional calculus appears in rheology, viscoelasticity, electrochemistry, electromag-netism, and so forth. For details, see the monographs of Kilbas et al. [8], Kiryakova [9],Miller and Ross [10], Podlubny [11], Oldham and Spanier [12], and Samko et al. [13], andthe papers of Diethelm et al. [14–16], Mainardi [17], Metzler et al. [18], Podlubny et al. [19],


and the references therein. For some recent advances on fractional calculus and differentialequations, see [1, 3, 20–24].

In this paper we consider the following nonlinear fractional differential equation ofthe form

Dδu(t) − λu(t) = f(t, u(t)), t ∈ J := (0, 1], 0 < δ < 1, (1.1)

where Dδ is the standard Riemann-Liouville fractional derivative, f is continuous, and λ ∈ R.This paper is organized as follows. in Section 2 we recall some definitions of fractional

integral and derivative and related basic properties which will be used in the sequel. InSection 3, we deal with the linear case where f(t, u(t)) = σ(t) is a continuous function.Section 4 is devoted to the nonlinear case.

2. Preliminary Results

In this section, we introduce notations, definitions, and preliminary facts which are usedthroughout this paper.

Let C[0, 1] the Banach space of all continuous real functions defined on [0, 1] with thenorm ‖f‖ =: sup{|f(t)| : t ∈ [0, 1]}. Define for t ∈ [0, 1], fr(t) = trf(t). Let Cr[0, 1], r ≥ 0 bethe space of all functions f such that fr ∈ C[0, 1] which turn out to be a Banach space whenendowed with the norm ‖f‖r =: sup{tr |f(t)| : t ∈ [0, 1]}.

By L1[0, 1] we denote the space of all real functions defined on [0, 1] which areLebesgue integrable.

Obviously Cr[0, 1] ⊂ L1[0, 1] if r < 1.

Definition 2.1 (see [11, 13]). The Riemann-Liouville fractional primitive of order s > 0 of afunction f : (0, 1] → R is given by

Is0f(t) =1

Γ(s)

∫ t

0(t − τ)s−1f(τ)dτ, (2.1)

provided the right side is pointwise defined on (0, 1], and where Γ is the gamma function.

For instance, Is0 exists for all s > 0, when f ∈ C((0, 1]) ∩ L1(0, 1]; note also that whenf ∈ C[0, 1], then Is0f ∈ C[0, 1] and moreover Is0f(0) = 0.

Let 0 < s < 1, if f ∈ Cr[0, 1] with r < s, then Isf ∈ C[0, 1], with Isf(0) = 0. Iff ∈ Cs[0, 1], then Isf is bounded at the origin, whereas if f ∈ Cr[0, 1] with s < r < 1, then wemay expect Isf to be unbounded at the origin.

Recall that the law of composition IrIs = Ir+s holds for all r, s > 0.

Definition 2.2 (see [11, 13]). The Riemann-Liouville fractional derivative of order s > 0 of acontinuous function f : (0, 1] → R is given by

Dsf(t) =1

Γ(1 − s)d

dt

∫ t

0(t − τ)−sf(τ)dτ =

d

dtI1−s

0 f(t). (2.2)

We have DsIsf = f for all f ∈ C(0, 1] ∩ L1(0, 1].


Lemma 2.3. Let 0 < s < 1. If one assumes u ∈ C(0, 1] ∩ L1(0, 1], then the fractional differentialequation

Dsu = 0 (2.3)

has u(t) = cts−1, c ∈ R, as unique solutions.

From this lemma we deduce the following law of composition.

Proposition 2.4. Assume that f is in C(0, 1]∩L1(0, 1] with a fractional derivative of order 0 < s < 1that belongs to C(0, 1] ∩ L1(0, 1]. Then

IsDsf(t) = f(t) + cts−1 (2.4)

for some c ∈ R.

If f ∈ Cr[0, 1] with r < 1 − s and Dsf ∈ C(0, 1] ∩ L1(0, 1], then IsDsf = f .

3. Linear Problem

In this section, we will be concerned with the following linear fractional differential equation:

Dδu(t) − λu(t) = σ(t), t ∈ J := (0, 1], 0 < δ < 1, (3.1)

where λ ∈ R, and σ is a continuous function.Before stating our main results for this section, we study the equation

Dδu(t) = σ(t), t ∈ J := (0, 1], 0 < δ < 1. (3.2)

Then

u(t) = ctδ−1 +(Iδσ)(t), t ∈ [0, 1] (3.3)

for some c ∈ R.Note that (Iδσ) ∈W1,1(0, 1) and (Iδσ)(0) = 0. However, u/∈W1,1(0, 1) since ctδ−1 has a

singularity at 0 for c /= 0.It is easy to show that u ∈ C1−δ[0, 1]. Hence we should look for solutions, not in

W1,1(0, 1) but in C1−δ[0, 1]. We cannot consider the usual initial condition u(0) = u0, butlimt�→0+ t

1−δu(t) = u0. Hence, to study the periodic boundary value problem, one has toconsider the following boundary condition of periodic type

limt�→0+

t1−δu(t) = u(1). (3.4)


From (3.3), we have

limt�→0+

t1−δu(t) = c,

u(1) = c +(Iδσ)(1)

(3.5)

that leads to the following.

Theorem 3.1. The periodic boundary value problem (3.2)–(3.4) has a unique solution u ∈ C1−δ[0, 1]if and only if

∫1

0

σ(s)ds

(1 − s)1−δ = 0. (3.6)

The previous result remains true even if δ = 1. In this case, (3.2) is reduced to theordinary differential equation

u′(t) = σ(t), (3.7)

with the periodic boundary condition

u(0) = u(1), (3.8)

and the condition (3.6) is reduced to the classical one:

∫1

0σ(s)ds = 0. (3.9)

Now, for λ different from 0, consider the homogenous linear equation

Dδu(t) − λu(t) = 0, t ∈ J := (0, 1], 0 < δ < 1. (3.10)

The solution is given by

u(t) = cΓ(δ)∞∑

i=1

λi−1

Γ(δi)tδi−1, c ∈ R. (3.11)

Indeed, we have

Dδu(t) = cΓ(δ)∞∑

i=1

λi−1

Γ(δi)Dδ(tδi−1)

(3.12)

since the series representing u is absolutely convergent.


Using the identities

Dstμ =Γ(μ + 1

)

Γ(μ + 1 − s

) tμ−s, μ > −1,

Dsts−1 = 0,

(3.13)

we get

Dδ(tδi−1)=

Γ(δi)Γ(δ(i − 1))

tδ(i−1)−1, for i > 1, Dδtδ−1 = 0. (3.14)

Then

Dδu(t) = cΓ(δ)∞∑

i=2

λi−1

Γ(δ(i − 1))tδ(i−1)−1

= λcΓ(δ)∞∑

i=2

λi−2

Γ(δ(i − 1))tδ(i−1)−1

= λcΓ(δ)∞∑

i=1

λi−1

Γ(δi)tδi−1

= λu(t).

(3.15)

Note that the solution can be expressed by means of the classical Mittag-Leffler specialfunctions [8]. Indeed

u(t) = cΓ(δ)∞∑

i=1

λi−1

Γ(δi)tδi−1

= cΓ(δ)tδ−1∞∑

i=1

(λtδ)i−1

Γ(δi)

= cΓ(δ)tδ−1∞∑

i=0

(λtδ)i

Γ(δi + δ)

= cΓ(δ)tδ−1Eδ,δ

(λtδ).

(3.16)

The previous formula remains valid for δ = 1. In this case,

Γ(1) = 1,

E1,1(λt) = E1(λt) = exp(λt).(3.17)


Then

u(t) = c exp(λt), (3.18)

which is the classical solution to the homogeneous linear differential equation

u′(t) − λu(t) = 0. (3.19)

Now, consider the nonhomogeneous problem (3.1). We seek the particular solution in thefollowing form:

up(t) =∫ t

0(t − s)δ−1Eδ,δ

(λ(t − s)δ

)σ(s)ds

=∫ t

0(t − s)δ−1

∞∑

i=0

λi(t − s)iδ

Γ(δ(i + 1))σ(s)ds.

(3.20)

It suffices to show that

up(t) = λ(Iδup

)(t) +

(Iδσ)(t). (3.21)

Indeed

(Iδup

)(t) =

1Γ(δ)

∫ t

0(t − s)δ−1up(s)ds

=1

Γ(δ)

∫ t

0

∫s

0(t − s)δ−1(s − ξ)δ−1

∞∑

i=0

λi(s − ξ)iδ

Γ(δ(i + 1))σ(ξ)dξ ds

=1

Γ(δ)

∞∑

i=0

λi

Γ(δ(i + 1))

∫ t

0

∫s

0(t − s)δ−1(s − ξ)δ−1(s − ξ)iδ σ(ξ)dξ ds

=1

Γ(δ)

∞∑

i=0

λi

Γ(δ(i + 1))

∫ t

0σ(ξ)

∫ t

ξ

(t − s)δ−1(s − ξ)δ(i+1)−1dsdξ.

(3.22)

Using the change of variable

s = (1 − θ)ξ + θt, (3.23)

we get

(Iδup

)(t) =

1Γ(δ)

∞∑

i=0

λi

Γ(δ(i + 1))

∫ t

0σ(ξ)

∫1

0(1 − θ)δ−1θδ(i+1)−1(t − ξ)δ(i+2)−1dθ dξ

=1

Γ(δ)

∞∑

i=0

λi

Γ(δ(i + 1))

∫ t

0

Γ(δ(i + 1))Γ(δ)Γ(δ(i + 2))

(t − ξ)δ(i+2)−1σ(ξ)dξ.

(3.24)


Then,

λ(Iδup

)(t) =

∞∑

i=0

λi+1

Γ(δ(i + 2))

∫ t

0(t − ξ)δ−1(t − ξ)δ(i+1)σ(ξ)dξ

=∫ t

0(t − ξ)δ−1

∞∑

i=1

λi

Γ(δ(i + 1))(t − ξ)δiσ(ξ)dξ

=∫ t

0(t − ξ)δ−1

[∞∑

i=0

λi

Γ(δ(i + 1))(t − ξ)δi − 1

Γ(δ)

]

σ(ξ)dξ

=∫ t

0(t − ξ)δ−1Eδ,δ

(λ(t − ξ)δ

)σ(ξ)dξ − 1

Γ(δ)

∫ t

0(t − ξ)δ−1σ(ξ)dξ

= up(t) −(Iδσ)(t).

(3.25)

Hence, the general solution of the nonhomogeneous equation (3.1) takes the form

u(t) = cΓ(δ)tδ−1Eδ,δ

(λtδ)+∫ t


(λ(t − s)δ

)σ(s)ds. (3.26)

Now, consider the periodic boundary value problem (3.1)–(3.4). Its unique solution isgiven by (3.26) for some c ∈ R. Also u is in C1−δ[0, 1] and

limt�→0+

t1−δu(t) = c. (3.27)

From (3.26), we have

u(1) = cΓ(δ)Eδ,δ(λ) +∫1

0(1 − s)δ−1Eδ,δ

(λ(1 − s)δ

)σ(s)ds, (3.28)

which leads to

c[1 − Γ(δ)Eδ,δ(λ)] =∫1

0(1 − s)δ−1Eδ,δ

(λ(1 − s)δ

)σ(s)ds, (3.29)

since Γ(δ)Eδ,δ(λ)/= 1 for any λ/= 0, we have

c = [1 − Γ(δ)Eδ,δ(λ)]−1∫1

0(1 − s)δ−1Eδ,δ

(λ(1 − s)δ

)σ(s)ds. (3.30)


Then the solution of the problem (3.1)–(3.4) is given by

u(t) =Γ(δ)

1 − Γ(δ)Eδ,δ(λ)tδ−1Eδ,δ

(λtδ)∫1

0(1 − s)δ−1Eδ,δ

(λ(1 − s)δ

)σ(s)ds

+∫ t


(λ(t − s)δ

)σ(s)ds.

(3.31)

Thus we have the following result.

Theorem 3.2. The periodic boundary value problem (3.1)–(3.4) has a unique solution u ∈ C1−δ[0, 1]given by

u(t) =∫1

0Gλ,δ(t, s)σ(s)ds, (3.32)

where

Gλ,δ(t, s)

=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

Γ(δ)Eδ,δ

(λtδ)Eδ,δ

(λ(1 − s)δ

)tδ−1(1 − s)δ−1

1 − Γ(δ)Eδ,δ(λ)+ (t − s)δ−1Eδ,δ

(λ(t − s)δ

), 0 ≤ s ≤ t ≤ 1,

Γ(δ)Eδ,δ

(λtδ)Eδ,δ

(λ(1 − s)δ

)tδ−1(1 − s)δ−1

1 − Γ(δ)Eδ,δ(λ), 0 ≤ t < s ≤ 1.

(3.33)

For λ, δ given, Gλ,δ is bounded on [0, 1] × [0, 1].For δ = 1, (3.1) is

u′(t) − λu(t) = σ(t), t ∈ J, (3.34)

and the boundary condition (3.4) is

u(0) = u(1). (3.35)

In this situation Green’s function is

Gλ,1(t, s) =1

1 − eλ

⎧⎨

⎩

eλ(t−s), 0 ≤ s ≤ t ≤ 1,

eλ(1+t−s), 0 ≤ t < s ≤ 1.(3.36)

which is precisely Green’s function for the periodic boundary value problem considered in[25, 26].


4. Nonlinear Problem

In this section we will be concerned with the existence and uniqueness of solution to thenonlinear problem (1.1)–(3.4). To this end, we need the following fixed point theorem ofSchaeffer.

Theorem 4.1. Assume X to be a normed linear space, and let operator F : X → X be compact. Theneither

(i) the operator F has a fixed point in X, or

(ii) the set E = {u ∈ X : u = μF(u), μ ∈ (0, 1)} is unbounded.

If u is a solution of problem (1.1)–(3.4), then it is given by

u(t) =∫1

0Gλ,δ(t, s)f(s, u(s))ds, (4.1)

where Gλ,δ is Green’s function defined in Theorem 3.2.Define the operator B : C1−δ[0, 1] → C1−δ[0, 1] by

B(u)(t) =∫1

0Gλ,δ(t, s)f(s, u(s))ds, t ∈ [0, 1]. (4.2)

Then the problem (1.1)–(3.4) has solutions if and only if the operator equation Bu = u hasfixed points.

Lemma 4.2. Suppose that the following hold:

(i) there exists a constant M > 0 such that

∣∣f(t, u)∣∣ ≤M, ∀t ∈ [0, 1], u ∈ R, (4.3)

(ii) there exists a constant k > 0 such that

∣∣f(t, u) − f(t, v)∣∣ ≤ k|u − v|, for each t ∈ [0, 1], and all u, v ∈ R. (4.4)

Then the operator B is well defined, continuous, and compact.


Proof. (a) We check, using hypothesis (4.3), that Bu ∈ C1−δ[0, 1], for every u ∈ C1−δ[0, 1].Indeed, for any t1 < t2 ∈ [0, 1], u ∈ D, we have

∣∣∣t1−δ1 B(u)(t1) − t1−δ2 B(u)(t2)

∣∣∣

=

∣∣∣∣∣t1−δ1

∫1

0Gλ,δ(t1, s)f(s, u(s))ds − t1−δ2

∫1

0Gλ,δ(t2, s)f(s, u(s))ds

∣∣∣∣∣

≤

∣∣∣∣∣∣∣

Γ(δ)Eδ,δ

(λtδ1

)

1 − Γ(δ)Eδ,δ(λ)

∫ t1

0Eδ,δ

(λ(1 − s)δ

)(1 − s)δ−1 f(s, u(s))ds

+ t1−δ1

∫ t1

0Eδ,δ

(λ(t1 − s)δ

)(t1 − s)δ−1 f(s, u(s))ds

−Γ(δ)Eδ,δ

(λtδ2

)


∫ t1

0Eδ,δ

(λ(1 − s)δ

)(1 − s)δ−1 f(s, u(s))ds

− t1−δ2

∫ t1

0Eδ,δ

(λ(t2 − s)δ

)(t2 − s)δ−1 f(s, u(s))ds

+Γ(δ)Eδ,δ

(λtδ1

)


∫ t2

t1

Eδ,δ

(λ(1 − s)δ

)(1 − s)δ−1 f(s, u(s))ds

−Γ(δ)Eδ,δ

(λtδ2

)


∫ t2

t1

Eδ,δ

(λ(1 − s)δ

)(1 − s)δ−1f(s, u(s))ds

− t1−δ2

∫ t2

t1

Eδ,δ

(λ(t2 − s)δ

)(t2 − s)δ−1 f(s, u(s))ds

+Γ(δ)Eδ,δ

(λtδ1

)


∫1

t2

Eδ,δ

(λ(1 − s)δ

)(1 − s)δ−1f(s, u(s))ds

−Γ(δ)Eδ,δ

(λtδ2

)


∫1

t2

Eδ,δ

(λ(1 − s)δ

)(1 − s)δ−1f(s, u(s))ds

∣∣∣∣∣∣∣

≤M

(Γ(δ)

|1 − Γ(δ)Eδ,δ(λ)|

∣∣∣Eδ,δ

(λtδ1

)− Eδ,δ

(λtδ2

)∣∣∣

∫ t1

0

∣∣∣Eδ,δ

(λ(1 − s)δ

)∣∣∣(1 − s)δ−1ds

+∫ t1

0

∣∣∣t1−δ1 (t1 − s)δ−1Eδ,δ

(λ(t1 − s)δ

)− t1−δ2 (t2 − s)δ−1Eδ,δ

(λ(t2 − s)δ

)∣∣∣ds

+Γ(δ)

|1 − Γ(δ)Eδ,δ(λ)|

∣∣∣Eδ,δ

(λtδ1

)− Eδ,δ

(λtδ2

)∣∣∣

∫ t2

t1

∣∣∣Eδ,δ

(λ(1 − s)δ

)∣∣∣(1 − s)δ−1ds


+ t1−δ2

∫ t2

t1

∣∣∣Eδ,δ

(λ(t2 − s)δ

)∣∣∣(t2 − s)δ−1ds

+Γ(δ)

|1 − Γ(δ)Eδ,δ(λ)|

∣∣∣Eδ,δ

(λtδ1

)− Eδ,δ

(λtδ2

)∣∣∣

∫1

t2

∣∣∣Eδ,δ

(λ(1 − s)δ

)∣∣∣(1 − s)δ−1ds

)

.

(4.5)

From the previous expression, we deduce that, if |t1 − t2| → 0, then

∣∣∣t1−δ1 B(u)(t1) − t1−δ2 B(u)(t2)

∣∣∣ −→ 0. (4.6)

Indeed, note that the integral∫ t1

0 |Eδ,δ(λ(1 − s)δ)|(1 − s)δ−1ds is bounded by

∞∑

j=0

|λ|j

Γ(δj + δ

)∫ t1

0(1 − s)δ−1ds =

∞∑

j=0

|λ|j

Γ(δj + δ

)1

δ(j + 1

) = Eδ,δ+1(|λ|). (4.7)

A similar argument is useful to study the behavior of the last three terms of the longinequality above. On the other hand, if we denote by H the second term in the right-handside of that inequality, then it is satisfied that

H =∫ t1

0

∣∣∣∣∣∣t1−δ1

∞∑

j=0

λj(t1 − s)δj+δ−1

Γ(δj + δ

) − t1−δ2

∞∑

j=0

λj(t2 − s)δj+δ−1

Γ(δj + δ

)

∣∣∣∣∣∣ds

≤∫ t1

0

∣∣∣t1−δ1 − t1−δ2

∣∣∣∞∑

j=0

|λ|j(t1 − s)δj+δ−1

Γ(δj + δ

) ds +∫ t1

0t1−δ2

∞∑

j=0

|λ|j∣∣∣(t1 − s)δj+δ−1 − (t2 − s)δj+δ−1

∣∣∣

Γ(δj + δ

) ds

≤∣∣∣t1−δ1 − t1−δ2

∣∣∣∞∑

j=0

|λ|j

Γ(δj + δ

)∫ t1

0(t1 − s)δj+δ−1ds

+ t1−δ2

∞∑

j=0

|λ|j

Γ(δj + δ

)∫ t1

0

∣∣∣(t1 − s)δj+δ−1 − (t2 − s)δj+δ−1∣∣∣ds.

(4.8)

Note that

∫ t1

0(t1 − s)δj+δ−1ds =

tδj+δ1

δj + δ≤ 1

δj + δ, (4.9)

and, concerning Rj =∫ t1

0 |(t1 − s)δj+δ−1 − (t2 − s)δj+δ−1|ds, we distinguish two cases. If j is such

that δj + δ − 1 ≥ 0, then

Rj =∫ t1

0

((t2 − s)δj+δ−1 − (t1 − s)δj+δ−1

)ds =

−(t2 − t1)δj+δ + tδj+δ2 − tδj+δ1

δj + δ, (4.10)


and, if j is such that δj + δ − 1 ≤ 0, then

Rj =∫ t1

0

((t1 − s)δj+δ−1 − (t2 − s)δj+δ−1

)ds =

(t2 − t1)δj+δ + tδj+δ1 − tδj+δ2

δj + δ. (4.11)

In consequence,

H ≤∣∣∣t1−δ1 − t1−δ2

∣∣∣∞∑

j=0

|λ|j

Γ(δj + δ

)(δj + δ

) + t1−δ2

∞∑

j=0

|λ|j

Γ(δj + δ

)(δj + δ

)Rj

=∣∣∣t1−δ1 − t1−δ2

∣∣∣Eδ,δ+1(|λ|) + t1−δ2

∞∑

j=0

|λ|j

Γ(δj + δ + 1

)Rj.

(4.12)

The first term in the right-hand side of the previous inequality clearly tends to zero as |t1 −t2| → 0. On the other hand, denoting by [·] the integer part function, we have

∞∑

j=0

|λ|j

Γ(δj + δ + 1

)Rj =[1/δ]−1∑

j=0

|λ|j

Γ(δj + δ + 1

)((t2 − t1)δj+δ + t

δj+δ1 − tδj+δ2

)

++∞∑

j=[1/δ]

|λ|j

Γ(δj + δ + 1

)(tδj+δ2 − tδj+δ1 − (t2 − t1)δj+δ

).

(4.13)

The finite sum obviously has limit zero as |t1 − t2| → 0. The infinite sum is equal to

tδ2

+∞∑

j=[1/δ]

(|λ|tδ2)j

Γ(δj + δ + 1

) − tδ1+∞∑

j=[1/δ]

(|λ|tδ1)j

Γ(δj + δ + 1

) − (t2 − t1)δ+∞∑

j=[1/δ]

(|λ|(t2 − t1)δ)j

Γ(δj + δ + 1

) , (4.14)

and its limit as |t1− t2| → 0 is zero. Note that∑+∞

j=[1/δ]((|λ|(t2 − t1)δ)

j/Γ(δj +δ+1)) is bounded

above by∑+∞

j=0(|λ|j/Γ(δj + δ + 1)) = Eδ,δ+1(|λ|).

The previous calculus shows that Bu ∈ C1−δ[0, 1], for u ∈ C1−δ[0, 1], hence we candefine B : C1−δ[0, 1] → C1−δ[0, 1].

(b) Next, we prove that B is continuous.Note that, for u, v ∈ C1−δ[0, 1] and for every t ∈ [0, 1], we have, using hypothesis (4.4),

t1−δ|B(u)(t) − B(v)(t)| ≤ t1−δ∫1

0|Gλ,δ(t, s)|

∣∣f(s, u(s)) − f(s, v(s))∣∣ds

≤ kt1−δ∫1

0|Gλ,δ(t, s)||u(s) − v(s)|ds.

(4.15)


Using the definition of ‖ · ‖1−δ, we get

‖B(u) − B(v)‖1−δ ≤ kmaxt∈[0,1]

{

t1−δ∫1

0|Gλ,δ(t, s)|sδ−1ds

}

‖u − v‖1−δ. (4.16)

Moreover,

t1−δ∫1

0|Gλ,δ(t, s)|sδ−1ds =

∫ t

0

Γ(δ)∣∣Eδ,δ

(λtδ)∣∣∣∣∣Eδ,δλ(1 − s)δ

∣∣∣

|1 − Γ(δ)Eδ,δ(λ)|(1 − s)δ−1sδ−1ds

+ t1−δ∫ t

0(t − s)δ−1

∣∣∣Eδ,δ

(λ(t − s)δ

)∣∣∣sδ−1ds

+∫1

t

Γ(δ)∣∣Eδ,δ

(λtδ)∣∣∣∣∣Eδ,δ

(λ(1 − s)δ

)∣∣∣

|1 − Γ(δ)Eδ,δ(λ)|(1 − s)δ−1sδ−1ds

≤Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|

∞∑

j=0

|λ|j

Γ(δj + δ

)∫1

0(1 − s)δj+δ−1sδ−1ds

+ t1−δ∞∑

j=0

|λ|j

Γ(δj + δ

)∫ t

0(t − s)δj+δ−1sδ−1ds.

(4.17)

Using that 0 ≤ 1 − s ≤ 1 for s ∈ [0, 1], 0 ≤ t − s ≤ 1 for s ∈ [0, t], and δj ≥ 0 for j = 0, 1, . . . , weobtain

t1−δ∫1


≤Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|

∞∑

j=0

|λ|j

Γ(δj + δ

)∫1

0(1 − s)δ−1sδ−1ds

+ t1−δ∞∑

j=0

|λ|j

Γ(δj + δ

)∫ t

0(t − s)δ−1sδ−1ds

≤Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|Eδ,δ(|λ|)

∫1

0(1 − s)δ−1sδ−1ds

+ t1−δEδ,δ(|λ|)∫ t

0(t − s)δ−1sδ−1ds.

(4.18)

Note that the Beta function, also called the Euler integral of the first kind,

β(p, q)=∫1

0xp−1(1 − x)q−1 dx, (4.19)


where p > 0 and q > 0, satisfies that β(p, q) = Γ(p)Γ(q)/Γ(p + q). In particular, β(δ, δ) =(Γ(δ))2/Γ(2δ). On the other hand, using the change of variable s = ut/(1+u), we deduce that

∫ t

0sp−1(t − s)q−1ds =

∫+∞

0

(ut

1 + u

)p−1( t

1 + u

)q−1(

t

(1 + u)2

)

du

= tp+q−1∫+∞

0

up−1

(1 + u)p+qdu = tp+q−1β

(p, q)= tp+q−1 Γ

(p)Γ(q)

Γ(p + q

) .

(4.20)

This proves that

∫ t

0sδ−1(t − s)δ−1ds = t2δ−1 (Γ(δ))

2

Γ(2δ). (4.21)

Hence,

t1−δ∫1


≤Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|Eδ,δ(|λ|)

(Γ(δ))2

Γ(2δ)+ t1−δEδ,δ(|λ|)t2δ−1 (Γ(δ))

2

Γ(2δ)

=

(Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|+ tδ)

Eδ,δ(|λ|)(Γ(δ))2

Γ(2δ)

≤(

Γ(δ)Eδ,δ(|λ|)|1 − Γ(δ)Eδ,δ(λ)|

+ 1)Eδ,δ(|λ|)

(Γ(δ))2

Γ(2δ).

(4.22)

In consequence,

‖B(u) − B(v)‖1−δ ≤ k

(Γ(δ)Eδ,δ(|λ|)|1 − Γ(δ)Eδ,δ(λ)|

+ 1)Eδ,δ(|λ|)

(Γ(δ))2

Γ(2δ)‖u − v‖1−δ. (4.23)

Finally, we check that B is compact. Let D be a bounded set in C1−δ[0, 1].(i) First, we check that {t1−δ[Bu](t) : u ∈ D} is a bounded set in C[0, 1].Indeed,

t1−δ∫1

0|Gλ,δ(t, s)|ds

≤∫ t

0

Γ(δ)∣∣∣Eδ,δ

(λtδ)Eδ,δ

(λ(1 − s)δ

)∣∣∣(1 − s)δ−1

|1 − Γ(δ)Eδ,δ(λ)|ds

+ t1−δ∫ t

0(t − s)δ−1

∣∣∣Eδ,δ

(λ(t − s)δ

)∣∣∣ds


+∫1

t

Γ(δ)∣∣∣Eδ,δ

(λtδ)Eδ,δ

(λ(1 − s)δ

)∣∣∣(1 − s)δ−1

|1 − Γ(δ)Eδ,δ(λ)|ds

=Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|

∫1

0

∣∣∣Eδ,δ

(λ(1 − s)δ

)∣∣∣(1 − s)δ−1ds

+ t1−δ∫ t

0(t − s)δ−1

∣∣∣Eδ,δ

(λ(t − s)δ

)∣∣∣ds

≤Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|

∫1

0

∞∑

j=0

(|λ|(1 − s)δ)j

Γ(δj + δ

) (1 − s)δ−1ds

+ t1−δ∫ t

0

∞∑

j=0

(|λ|(t − s)δ)j

Γ(δj + δ

) (t − s)δ−1ds

=Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|

∞∑

j=0

|λ|j

Γ(δj + δ

)∫1

0(1 − s)δ(j+1)−1ds

+ t1−δ∞∑

j=0

|λ|j

Γ(δj + δ

)∫ t

0(t − s)δ(j+1)−1ds

=Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|

∞∑

j=0

|λ|j

Γ(δj + δ

)1

δ(j + 1

) + t1−δ∞∑

j=0

|λ|j

Γ(δj + δ

)tδ(j+1)

δj + δ

=Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|

∞∑

j=0

|λ|j

Γ(δj + δ + 1

) + t∞∑

j=0

|λ|j tδj

Γ(δj + δ + 1

)

=Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|Eδ,δ+1(|λ|) + tEδ,δ+1

(|λ|tδ

).

(4.24)

Hence

t1−δ|B(u)(t)| ≤Mt1−δ∫1

0|Gλ,δ(t, s)|ds ≤M

(Γ(δ)

∣∣Eδ,δ

(λtδ)∣∣

|1 − Γ(δ)Eδ,δ(λ)|+ 1

)

Eδ,δ+1(|λ|), (4.25)

and then

‖B(u)‖1−δ ≤M

(Γ(δ)Eδ,δ(|λ|)|1 − Γ(δ)Eδ,δ(λ)|

+ 1)Eδ,δ+1(|λ|) <∞, (4.26)

which implies that {t1−δ[Bu](t) : u ∈ D} is a bounded set in C[0, 1].(ii) Now, we prove that {t1−δ[Bu](t) : u ∈ D} is an equicontinuous set in C[0, 1].

Following the calculus in (a), we show that |t1−δ1 B(u)(t1)−t1−δ2 B(u)(t2)| tends to zero as t1 �→ t2.Then {[Bu]1−δ : u ∈ D} is equicontinuous in the space C[0, 1], where v1−δ(t) = t1−δv(t),

for v ∈ C1−δ[0, 1].


As a consequence of (i) and (ii), {[Bu]1−δ : u ∈ D} is a bounded and equicontinuousset in the space C[0, 1].

Hence, for a sequence {un} in D, {[Bun]1−δ : n ∈ N} = {t1−δ[Bun](t) : n ∈ N} has asubsequence converging to ϕ ∈ C[0, T], that is,

limn→∞

∣∣∣t1−δ[Bunk](t) − ϕ(t)

∣∣∣ = 0, uniformly on t ∈ [0, 1]. (4.27)

Taking ψ(t) = tδ−1ϕ(t), we get

limn→∞

∣∣∣t1−δ[Bunk](t) − t1−δψ(t)

∣∣∣ = 0, uniformly on t ∈ [0, 1], (4.28)

which means that {Bunk}C1−δ[0,1]−−−−−−−→ ψ, which proves that B is compact.

Theorem 4.3. Assume that (4.3) and (4.4) hold. Then the problem (1.1)–(3.4) has at least onesolution in C1−δ[0, 1].

Proof. Consider the set E = {u ∈ X : u = μB(u), μ ∈ (0, 1)}.Let u be any element of E, then u = μB(u) for some μ ∈ (0, 1). Thus for each t ∈ [0, 1],

we have

|u(t)| ≤ μ

∫1

0|Gλ,δ(t, s)|

∣∣f(s, u(s))∣∣ds. (4.29)

As in Lemma 4.2, (i), we have

‖u‖1−δ ≤M

(Γ(δ)Eδ,δ(|λ|)|1 − Γ(δ)Eδ,δ(λ)|

+ 1)Eδ,δ+1(|λ|) <∞, (4.30)

which implies that the set E is bounded independently of μ ∈ (0, 1). Using Lemma 4.2 andTheorem 4.1, we obtain that the operator B has at least a fixed point.

Remark 4.4. In Lemma 4.2, condition (4.3) is used to prove that the operator B is continuous.Hence, in Lemma 4.2 and, in consequence, in Theorem 4.3, we can assume the weakercondition.

(i) For each u0 ∈ C1−δ[0, 1] fixed, there exists ku0 > 0 such that

∣∣f(t, u) − f(t, u0(t))∣∣ ≤ ku0 |u − u0(t)|, for each t ∈ [0, 1], and all u ∈ R, (4.31)

instead of (4.3).

However, to prove the existence and uniqueness of solution given in the followingtheorem, we need to assume the Lipschitzian character of f (condition (4.3)).


Theorem 4.5. Assume that (4.4) holds. Then the problem (1.1)–(3.4) has a unique solution inC1−δ[0, 1], provided that

k(Γ(δ))2

Γ(2δ)Eδ,δ(|λ|)

(Γ(δ)Eδ,δ(|λ|)|1 − Γ(δ)Eδ,δ(λ)|

+ 1)

< 1. (4.32)

Proof. We use the Banach contraction principle to prove that the operator B has a unique fixedpoint.

Using the calculus in (b) Lemma 4.2, B is a contraction by condition (4.32). As aconsequence of Banach fixed point theorem, we deduce that B has a unique fixed point whichgives rise to a unique solution of problem (1.1)–(3.4).

Remark 4.6. If δ = 1, condition (4.32) is reduced to

ke|λ|(

e|λ|∣∣1 − eλ

∣∣ + 1

)

< 1. (4.33)

Acknowledgment

The research of J. J. Nieto and R. Rodrıguez-Lopez has been partially supported by Ministeriode Educacion y Ciencia and FEDER, project MTM2007-61724, and by Xunta de Galicia andFEDER, project PGIDIT06PXIB207023PR.

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Research ArticleExistence of Positive Solutions forMultipoint Boundary Value Problem onthe Half-Line with Impulses

Jianli Li1 and Juan J. Nieto2

1 Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China2 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Santiago de Compostela,15782 Santiago de Compostela, Spain

Correspondence should be addressed to Jianli Li, [email protected]

Received 7 March 2009; Revised 18 April 2009; Accepted 25 April 2009


We consider a multi-point boundary value problem on the half-line with impulses. By using afixed-point theorem due to Avery and Peterson, the existence of at least three positive solutions isobtained.

Copyright q 2009 J. Li and J. J. Nieto. 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.

1. Introduction

Impulsive differential equations are a basic tool to study evolution processes that aresubjected to abrupt changes in their state. For instance, many biological, physical, andengineering applications exhibit impulsive effects (see [1–3]). It should be noted that recentprogress in the development of the qualitative theory of impulsive differential equations hasbeen stimulated primarily by a number of interesting applied problems [4–24].

In this paper, we consider the existence of multiple positive solutions of the followingimpulsive boundary value problem (for short BVP) on a half-line:

u ′′(t) + q(t)f(t, u) = 0, 0 < t <∞, t /= tk,

Δu(tk) = Ik(u(tk)), k = 1, . . . , p,

u(0) =m−2∑

i=1

αiu(ξi), u ′(∞) = 0,

(1.1)


where u ′(∞) = limt→+∞u′(t), 0 < ξ1 < ξ2 < · · · < ξm−2 < +∞, 0 < t1 < t2 < · · · < tp < +∞,

Δu(tk) = u(t+k) − u(t−

k), and αi, f, q, and Ik satisfy

(H1) 0 <∑m−2

i=1 αi < 1;

(H2) f(t, u) ∈ C([0,∞)×[0,+∞), [0,+∞)), Ik(u) ∈ C([0,+∞), [0,+∞)), and when u/(1+t)is bounded, f(t, u) and Ik(u) are bounded on [0,+∞);

(H3) q(t) ∈ C([0,∞), [0,+∞)) and q(t) is not identically zero on any compact subintervalof (0,+∞). Furthermore q(t) satisfies

supt∈[0,+∞)

∫+∞

0G(t, s)q(s)ds < +∞, (1.2)

where

G(t, s) =

⎧⎨

⎩

s, 0 ≤ s ≤ t < +∞,

t, 0 ≤ t ≤ s < +∞.(1.3)

Boundary value problems on the half-line arise quite naturally in the study of radiallysymmetric solutions of nonlinear elliptic equations and there are many results in this area,see [8, 13, 14, 20, 25–27], for example.

Lian et al. [25] studied the following boundary value problem of second-orderdifferential equation with a p-Laplacian operator on a half-line:

(ϕp(u ′(t))

)′ + φ(t)f(t, u, u ′

)= 0, 0 < t < +∞,

αu(0) − βu ′(0) = 0, u ′(∞) = 0.(1.4)

They showed the existence at least three positive solutions for (1.4) by using a fixed pointtheorem in a cone due to Avery-Peterson [28].

Yan [20], by using Leray-Schauder theorem and fixed point index theory presentssome results on the existence for the boundary value problems on the half-line with impulsesand infinite delay.

However to the best knowledge of the authors, there is no paper concerned with theexistence of three positive solutions to multipoint boundary value problems of impulsivedifferential equation on infinite interval so far. Motivated by [20, 25], in this paper, we aim toinvestigate the existence of triple positive solutions for BVP (1.1). The method chosen in thispaper is a fixed point technique due to Avery and Peterson [28].

2. Preliminaries

In this section, we give some definitions and results that we will use in the rest of the paper.

Definition 2.1. Suppose P is a cone in a Banach. The map α is a nonnegative continuousconcave functional on P provided α : P → [0,∞) is continuous and

α(tx + (1 − t)y

)≥ tα(x) + (1 − t)α

(y)

(2.1)


for all x, y ∈ P , and t ∈ [0, 1]. Similarly, the map β is a nonnegative continuous convexfunctional on P provided β : P → [0,∞) is continuous and

β(tx + (1 − t)y

)≤ tβ(x) + (1 − t)β

(y)

(2.2)

for all x, y ∈ P , and t ∈ [0, 1].

Let γ, θ be nonnegative, continuous, convex functionals on P and α be a nonnegative,continuous, concave functionals on P , and ψ be a nonnegative continuous functionals on P .Then, for positive real numbers a, b, c, and d, we define the convex sets

P(γ, d

)={x ∈ P : γ(x) < d

},

P(γ, α, b, d

)={x ∈ P : b ≤ α(x), γ(x) ≤ d

},

P(γ, θ, α, b, c, d

)={x ∈ P : b ≤ α(x), θ(x) ≤ c, γ(x) ≤ d

},

(2.3)

and the closed set

R(γ, ψ, a, d

)={x ∈ P : a ≤ ψ(x), γ(x) ≤ d

}. (2.4)

To prove our main results, we need the following fixed point theorem due to Averyand Peterson in [28].

Theorem 2.2. Let P be a cone in a real Banach space E. Let γ and θ be nonnegative continuousconvex functionals on a cone P , α be a nonnegative continuous concave functional on P , and ψ be anonnegative continuous functional on P satisfying ψ(λx) ≤ λψ(x) for 0 ≤ λ ≤ 1, such that for somepositive numbersM and d

α(x) ≤ ψ(x), ‖x‖ ≤Mγ(x) (2.5)

for all x ∈ P(γ, d). Suppose

Φ : P(γ, d) −→ P(γ, d) (2.6)

is completely continuous and there exist positive numbers a, d, and c with a < b such that

(i) {x ∈ P(γ, θ, α, b, c, d) : α(x) > b}/= ∅ and α(Φx) > b for x ∈ P(γ, θ, α, b, c, d);

(ii) α(Φx) > b for x ∈ P(γ, α, b, d) with θ(Φx) > c;

(iii) 0/∈R(γ, ψ, a, d) and ψ(Tx) < a for x ∈ R(γ, ψ, a, d), with ψ(Φx) = a.

Then Φ has at least three fixed points x1, x2, x3 ∈ P(γ, d) such that

γ(xi) ≤ d, for i = 1, 2, 3, ψ(x1) < a,

a < ψ(x2) with α(x2) < b, α(x3) > b.(2.7)


3. Some Lemmas

Define PC[0,+∞) = {u : [0,+∞) → R | u(t) is continuous at each t /= tk, left continuous att = tk, u(t+k) exists, k = 1, . . . , p}.

By a solution of (1.1) we mean a function u in PC[0,∞) satisfying the relations in (1.1).

Lemma 3.1. u(t) is a solution of (1.1) if and only if u(t) is a solution of the following equation:

u(t) =∫+∞

0G(t, s)q(s)f(s, u(s))ds +

∑

0<tk<t

Ik(u)

+∑m−2

i=1 αi

1 −∑m−2

i=1 αi

⎡

⎣∫+∞

0G(ξi, s)q(s)f(s, u(s))ds +

∑

0<tk<ξi

Ik(u)

⎤

⎦

:= Tu(t),

(3.1)

where G(t, s) is defined as (1.3).

The proof is similar to Lemma 3 in [9], and here we omit it.For tp < a∗ < b∗ < +∞, let c∗ = min{a∗/(1 + a∗), 1/(1 + b∗)}. Then

G(t, s)1 + t

≥ c∗G(r, s)1 + r

,1

1 + t≥ c∗

1 + r, for t ∈ [a∗, b∗], r ∈ [0,+∞), s ∈ [0,+∞). (3.2)

It is clear that 0 < c∗ < 1. Consider the space E defined by

E =

{

u ∈ PC[0,+∞) : supt∈[0,∞)

|u(t)|1 + t

< +∞}

. (3.3)

E is a Banach space, equipped with the norm ‖u‖ = sup0≤t<+∞(|u(t)|/(1 + t)) < +∞.Define the cone P ⊂ E by

P ={u ∈ E : u(t) ≥ 0, t ∈ [0,+∞), min

t∈[a∗,b∗]

u(t)1 + t

≥ c∗‖u‖}. (3.4)

Lemma 3.2 (see [20, Theorem 2.2]). Let M ⊂ PC[0,+∞). Then M is compact in PC[0,+∞), ifthe following conditions hold:

(a) M is bounded in PC[0,+∞);

(b) the functions belonging to M are piecewise equicontinuous on any interval of [0,+∞);

(c) the functions from M are equiconvergent, that is, given ε > 0, there corresponds τ(ε) > 0such that |f(t) − f(+∞)| < ε for any t ≥ τ(ε) and f ∈M.

Lemma 3.3. T : P → P is completely continuous.


Proof. Firstly, for u ∈ P , from (H1)–(H3), it is easy to check that Tu is well defined, andTu(t) ≥ 0 for all t ∈ [0,+∞). For t ∈ [a∗, b∗]

11 + t

(Tu)(t) =1

1 + t

∫+∞

0G(t, s)q(s)f(s, u(s))ds +

11 + t

p∑

k=1

Ik(u)

+1

1 + t

∑m−2i=1 αi

1 −∑m−2

i=1 αi

⎡

⎣∫+∞

0G(ξi, s)q(s)f(s, u(s))ds +

∑

0<tk<ξi

Ik(u)

⎤

⎦

≥ c∗[∫+∞

0

G(r, s)1 + r

q(s)f(s, u(s))ds +1

1 + r

p∑

k=1

Ik(u)

]

+ c∗∑m−2

i=1 αi

1 −∑m−2

i=1 αi

⎡

⎣∫+∞

0

G(ξi, s)1 + r

q(s)f(s, u(s))ds +1

1 + r

∑

0<tk<ξi

Ik(u)

⎤

⎦

≥ c∗Tu(r)1 + r

, for r ∈ [0,+∞)

(3.5)

so

mint∈[a∗,b∗]

Tu(t)1 + t

≥ c∗‖Tu‖, (3.6)

which shows TP ⊆ P .Now we prove that T is continuous and compact, respectively. Let un → u as n → +∞

in P . Then there exists r0 such that supn∈N\{0}‖un‖ < r0. By (H2) we have f(t, u) is boundedon [0,+∞) × [0, r0]. Set B0 = sup{f(t, u) : (t, u/(1 + t)) ∈ [0,+∞) × [0, r0]}, and we have

∫+∞

0

G(t, s)1 + t

q(s)∣∣f(s, un) − f(s, u)

∣∣ds ≤ 2B0

∫+∞

0

G(t, s)1 + t

q(s)ds. (3.7)

Therefore by the Lebesgue dominated convergence theorem and continuity of f and Ik, onearrives at

‖Tun − Tu‖

≤ supt∈[0,+∞)

11 + t

{∫+∞

0G(t, s)q(s)

∣∣f(s, un) − f(s, u)∣∣ds +

∑

0<tk<t|Ik(un) − Ik(u)| +

∑m−2i=1 αi

1 −∑m−2

i=1 αi

×

⎡

⎣∫+∞

0G(ξi, s)q(s)

∣∣f(s, un) − f(s, u)∣∣ds +

∑

0<tk<ξi

|Ik(un) − Ik(u)|

⎤

⎦

⎫⎬

⎭

−→ 0 as n −→ +∞.

(3.8)

Therefore T is continuous.


Let Ω be any bounded subset of P . Then there exists r > 0 such that ‖u‖ ≤ r for allu ∈ Ω. Set B1 = sup{f(t, u) : (t, u/(1+t)) ∈ [0,+∞)×[0, r], B2k = sup{Ik(u) : u/(1+t) ∈ [0, r]},then

‖Tu‖ = supt∈[0,+∞)

11 + t

{∫+∞

0G(t, s)q(s)

∣∣f(s, u)

∣∣ds +

∑

0<tk<t|Ik(u)|

+∑m−2

i=1 αi

1 −∑m−2

i=1 αi

⎡

⎣∫+∞

0G(ξi, s)q(s)

∣∣f(s, u)

∣∣ds +

∑

0<tk<ξi

|Ik(u)|

⎤

⎦

⎫⎬

⎭

≤ B1

[∫+∞

0G(t, s)q(s)ds +

∑m−2i=1 αi

1 −∑m−2

i=1 αi

∫+∞

0G(ξi, s)q(s)ds

]

+

(

1 +∑m−2

i=1 αi

1 −∑m−2

i=1 αi

)p∑

k=1

B2k.

(3.9)

So TΩ is bounded.Moreover, for any ν ∈ (0,+∞) and t′, t′′ ∈ (tk, tk+1] ⊂ [0, ν](t′ < t′′), and u ∈ Ω, then

∣∣∣∣Tu(t′′)1 + t′′

− Tu(t′)1 + t′

∣∣∣∣ ≤∑m−2

i=1 αi

1 −∑m−2

i=1 αi

⎡

⎣B1

∫+∞

0G(ξi, s)q(s)ds +

∑

0<tk<ξi

B2k

⎤

⎦∣∣∣∣

11 + t′′

− 11 + t′

∣∣∣∣

+ B1

∫+∞

0

∣∣∣∣G(t′′, s)1 + t′′

− G(t′, s)1 + t′

∣∣∣∣q(s)ds +∑

0<tk<t′B2k

∣∣∣∣1

1 + t′′− 1

1 + t′

∣∣∣∣

−→ 0, uniformly as t′ −→ t′′.

(3.10)

So TΩ is quasi-equicontinuous on any compact interval of [0,+∞).Finally, we prove for any ε, there exists sufficiently large N1 > 0 such that

∣∣∣∣Tu(t′′)1 + t′′

− Tu(t′)1 + t′

∣∣∣∣ < ε, ∀t′, t′′ ≥N1, u ∈ Ω. (3.11)

Since∫+∞

0 G(t, s)q(s)ds < +∞, we can choose N1 > 0 such that

∑m−2i=1 αi

N1

(1 −

∑m−2i=1 αi

)

⎡

⎣B1

∫+∞

0G(ξi, s)q(s)ds +

∑

0<tk<ξi

B2k

⎤

⎦ <ε

6,

B1∫+∞

0 G(t, s)q(s)dsN1

<ε

6,

p∑

k=1

B2k

N1≤ ε

6.

(3.12)


For t′, t′′ ≥N1, it follows that

∣∣∣∣(Tu)(t′)

1 + t′− (Tu)(t′′)

1 + t′′

∣∣∣∣ ≤

∑m−2i=1 αi

1 −∑m−2

i=1 αi

⎡

⎣B1

∫+∞

0G(ξi, s)q(s)ds +

∑

0<tk<ξi

B2k

⎤

⎦(

11 + t′′

+1

1 + t′

)

+ B1

∫+∞

0

G(t′, s)1 + t′

q(s)ds + B1

∫+∞

0

G(t′′, s)1 + t′′

q(s)ds

+p∑

k=1

B2k

(1

1 + t′′+

11 + t′

)

<ε

3+ε

6+ε

6+ε

3= ε.

(3.13)

That is (3.11) holds. By Lemma 3.2, TΩ is relatively compact. In sum, T : P → P is completelycontinuous.

4. Existence of Three Positive Solutions

Let the nonnegative continuous concave functional α, the nonnegative continuous convexfunctionals γ and θ, and the nonnegative continuous functionals ψ be defined on the cone Pby

γ(u) = ψ(u) = θ(u) = supt∈[0,+∞)

u(t)1 + t

, α(u) = mint∈[a∗,b∗]

u(t)1 + t

. (4.1)

For notational convenience, we denote by

M = mint∈[a∗,b∗]

∫+∞

0

G(t, s)1 + t

q(s)ds,

M1 =∑m−2

i=1 αi

1 −∑m−2

i=1 αi

∫+∞

0G(ξi, s)q(s)ds.

(4.2)

The main result of this paper is the following.

Theorem 4.1. Assume (H1)–(H3) hold. Let ak ≥ 0, 0 < a < b/c∗ < c = d, b/M < c∗d/2(M+M1)and suppose that f, Ik satisfy the following conditions:

(A1) f(t, u) ≤ c∗d/2(M +M1), Ik(u) ≤ dc∗/2M2 for (t, u/(1 + t) ∈ [0,+∞) × [0, d],

(A2) f(t, u) > b/M for (t, u/(1 + t)) ∈ [a∗, b∗] × [b, c],

(A3) f(t, u) < c∗a/2(M+M1), Ik(u) ≤ ac∗ak/2M2 for t ∈ (t, u/(1+ t)) ∈ [0,+∞)× [0, a],


where M2 =∑p

k=1 ak/(1 −∑m−2

i=1 αi). Then (1.1) has at least three positive solutions u1, u2 and u3

such that

γ(ui) ≤ d, for i = 1, 2, 3, ψ(u1) < a, a < ψ(u2) with α(u2) < b, α(u3) > b. (4.3)

Proof.Step 1. From the definition α, ψ, and γ , we easily show that

α(u) ≤ ψ(u), ‖u‖ ≤ γ(u) for u ∈ P(γ, d). (4.4)

Next we will show that

T : P(γ, d) −→ P(γ, d). (4.5)

In fact, for u ∈ P(γ, d), then

supt∈[0,+∞)

u(t)1 + t

≤ d. (4.6)

From condition (A1), we obtain

f(t, u) ≤ dc∗

2(M +M1), Ik(u) ≤

dc∗

2M2. (4.7)

It follows that

γ(Tu) = supt∈[0,+∞)

Tu(t)1 + t

≤ 1c∗

mint∈[a∗,b∗]

Tu(t)1 + t

≤ 1c∗

mint∈[a∗,b∗]

[1

1 + t

∫+∞

0G(t, s)q(s)f(s, u)ds +

11 + t

p∑

k=1

Ik(u)

+1

1+t·∑m−2

i=1 αi

1−∑m−2

i=1 αi

⎛

⎝∫+∞

0G(ξi, s)q(s)f(s, u)ds +

∑

0<tk<ξi

Ik(u)

⎞

⎠

⎤

⎦

≤ 1c∗· c∗d

2(M +M1)

[

mint∈[a∗,b∗]

∫+∞

0

G(t, s)1 + t

q(s)ds +∑m−2

i=1 αi

1 −∑m−2

i=1 αi

∫+∞

0G(ξi, s)q(s)ds

]

+1c∗·c∗d

∑p

k=1 ak

2M2

[

1 +∑m−2

i=1 αi

1 −∑m−2

i=1 αi

]

≤ d

2+d

2= d.

(4.8)

Thus (4.5) holds.


Step 2. We show that condition (i) in Theorem 2.2 holds. Taking u(t) = (1+ t)((b+d)/2), thenu ∈ P(γ, θ, α, b, c, d) and α(u) > b, which shows {u ∈ P(γ, θ, α, b, c, d) | α(u) > b}/= ∅. Thus foru ∈ P(γ, θ, α, b, c, d) , there is

b ≤ u(t)1 + t

≤ for t ∈ [a∗, b∗]. (4.9)

Hence by (A2), we have

α(Tu) = mint∈[a∗,b∗]

Tu(t)1 + t

≥ mint∈[a∗,b∗]

∫+∞

0

G(t, s)1 + t

q(s)f(s, u)ds

>b

M· mint∈[a∗,b∗]

∫+∞

0

G(t, s)1 + t

q(s)ds = b.

(4.10)

Therefore we have

α(Tu) > b, ∀u ∈ P(γ, θ, α, b, c, d

). (4.11)

This shows the condition (i) in Theorem 2.2 is satisfied.Step 3. We now prove (ii) in Theorem 2.2 holds. For u ∈ P(γ, α, b, d) with θ(Tu) > c, we have

α(Tu) = mint∈[a∗,b∗]

Tu(t)1 + t

≥ c∗‖Tu‖ = c∗θ(Tu) > c∗c > b. (4.12)

Hence, condition (ii) in Theorem 2.2 is satisfied.Step 4. Finally, we prove (iii) in Theorem 2.2 is satisfied. Since ψ(0) = 0 < a, so 0/∈R(γ, ψ, a, d).Suppose that u ∈ R(γ, θ, a, d) with ψ(u) = a, then

0 ≤ u(t)1 + t

≤ a, (4.13)

by the condition (A3) of this theorem,

ψ(Tu) = supt∈[0,+∞)

Tu(t)1 + t

≤ 1c∗

mint∈[a∗,b∗]

Tu(t)1 + t

≤ 1c∗· c∗a

2(M+M1)

[

mint∈[a∗,b∗]

∫+∞

0

G(t, s)1 + t

q(s)ds +∑m−2

i=1 αi

1−∑m−2

i=1 αi

∫+∞

0G(ξi, s)q(s)ds

]

+1c∗·c∗a

∑p

k=1 ak

2M2

[

1 +∑m−2

i=1 αi

1 −∑m−2

i=1 αi

]

≤ a

2+a

2= a.

(4.14)


Thus condition (iii) in Theorem 2.2 holds. Therefore an application of Theorem 2.2 impliesthe boundary value problem (1.1) has at least three positive solutions such that

supt∈[0,+∞)

ui(t)1 + t

≤ d, i = 1, 2, 3,

supt∈[0,+∞)

u1(t)1 + t

< a, a < supt∈[0,+∞)

u2(t)1 + t

with mint∈[a∗,b∗]

u2(t)1 + t

< b,

mint∈[a∗,b∗]

u3(t)1 + t

> b.

(4.15)

5. An Example

Now we consider the following boundary value problem

u ′′(t) + q(t)f(t, u) = 0, 0 < t < +∞, t /= t1,

Δu(t1) = I1(u(t1)), t1 = 1,

u(0) =14u

(14

)+

14u(4), u ′(∞) = 0

f(t, u) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

1100

e−t + 4( u

1 + t

)7,

u

1 + t≤ 1,

1100

e−t + 4,u

1 + t≥ 1,

I1(u) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

12

( u

1 + t

)4,

u

1 + t≤ 1,

12,

u

1 + t≥ 1,

(5.1)

q(t) = e−t. Choose a1 = 1/2, a = 1/2, b = 3/4, c = d = 48. If taking a∗ = 2, b∗ = 3, then c∗ = 1/4,and M = (1 − e−3)/4, M1 = 2 − e−1/4 − e−4, M2 = 1. Consequently, f(t, u), Ik(u) satisfies thefollowing:

(1) f(t, u) ≤ 1/100+4 < c∗d/2(M+M1), I1(u) ≤ 1/2 < 3 = c∗a1d/2M2, for (t, u/(1+t)) ∈[0,+∞) × [0, 48];

(2) f(t, u) > 4 > b/M, for (t, u/(1 + t)) ∈ [2, 3] × [3/4, 48];

(3) f(t, u) < 1/100+ (1/2)5 ≤ c∗a/2(M+M1), I1(u) ≤ 1/32 = c∗a1a/2M2, for (t, u/(1+t)) ∈ [0,+∞) × [0, 1/2].

Then all conditions of Theorem 4.1 hold, so by Theorem 4.1, boundary value problem (5.1)has at least three positive solutions.


Acknowledgments

This work is supported by the NNSF of China (no. 60671066), A project supported byScientific Research Fund of Hunan Provincial Education Department (07B041) and Programfor Young Excellent Talents in Hunan Normal University, The research of J. J. Nieto has beenpartially supported by Ministerio de Educacion y Ciencia and FEDER, Project MTM2007-61724, and by Xunta de Galicia and FEDER, Project PGIDIT06PXIB207023PR.

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Research ArticleExistence of Solutions for Fractional DifferentialInclusions with Antiperiodic Boundary Conditions

Bashir Ahmad1 and Victoria Otero-Espinar2

1 Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203,Jeddah 21589, Saudi Arabia


Correspondence should be addressed to Victoria Otero-Espinar, [email protected]

Received 21 January 2009; Revised 6 March 2009; Accepted 18 March 2009


We study the existence of solutions for a class of fractional differential inclusions with anti-periodicboundary conditions. The main result of the paper is based on Bohnenblust-Karlins fixed pointtheorem. Some applications of the main result are also discussed.

Copyright q 2009 B. Ahmad and V. Otero-Espinar. 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.

1. Introduction

In some cases and real world problems, fractional-order models are found to be moreadequate than integer-order models as fractional derivatives provide an excellent tool forthe description of memory and hereditary properties of various materials and processes.The mathematical modelling of systems and processes in the fields of physics, chemistry,aerodynamics, electro dynamics of complex medium, polymer rheology, and so forth,involves derivatives of fractional order. In consequence, the subject of fractional differentialequations is gaining much importance and attention. For details and examples, see [1–14]and the references therein.

Antiperiodic boundary value problems have recently received considerable attentionas antiperiodic boundary conditions appear in numerous situations, for instance, see[15–22].

Differential inclusions arise in the mathematical modelling of certain problems ineconomics, optimal control, and so forth. and are widely studied by many authors, see [23–27] and the references therein. For some recent development on differential inclusions, werefer the reader to the references [28–32].


Chang and Nieto [33] discussed the existence of solutions for the fractional boundaryvalue problem:

c0D

δt y(t) ∈ F

(t, y(t)

), t ∈ [0, 1], δ ∈ (1, 2),

y(0) = α, y(1) = β, α, β /= 0.(1.1)

In this paper, we consider the following fractional differential inclusions withantiperiodic boundary conditions

cDqx(t) ∈ F(t, x(t)), t ∈ [0, T], T > 0, 1 < q ≤ 2,

x(0) = −x(T), x′(0) = −x′(T),(1.2)

where cDq denotes the Caputo fractional derivative of order q, F : [0, T] × R → 2R \ {∅}.Bohnenblust-Karlin fixed point theorem is applied to prove the existence of solutions of (1.2).

2. Preliminaries

Let C([0, T]) denote a Banach space of continuous functions from [0, T] into R with the norm‖x‖ = supt∈[0,T]{|x(t)|}. Let L1([0, T],R) be the Banach space of functions x : [0, T] → R

which are Lebesgue integrable and normed by ‖x‖L1=∫T

0 |x(t)|dt.Now we recall some basic definitions on multivalued maps [34, 35].Let (X, ‖ · ‖) be a Banach space. Then a multivalued map G : X → 2X is convex

(closed) valued if G(x) is convex (closed) for all x ∈ X. The map G is bounded on boundedsets if G(B) =

⋃x∈BG(x) is bounded in X for any bounded set B of X (i.e., supx∈B{sup{|y| : y ∈

G(x)}} < ∞). G is called upper semicontinuous (u.s.c.) on X if for each x0 ∈ X, the set G(x0)is a nonempty closed subset of X, and if for each open set B of X containing G(x0), there existsan open neighborhood N of x0 such that G(N) ⊆ B. G is said to be completely continuousif G(B) is relatively compact for every bounded subset B of X. If the multivalued map G iscompletely continuous with nonempty compact values, then G is u.s.c. if and only if G has aclosed graph, that is, xn → x∗, yn → y∗, yn ∈ G(xn) imply y∗ ∈ G(x∗). In the following study,BCC(X) denotes the set of all nonempty bounded, closed, and convex subset of X. G has afixed point if there is x ∈ X such that x ∈ G(x).

Let us record some definitions on fractional calculus [8, 11, 13].

Definition 2.1. For a function g : [0,∞) → R, the Caputo derivative of fractional order q > 0is defined as

cDqg(t) =1

Γ(n − q

)∫ t

0(t − s)n−q−1g(n)(s)ds, n − 1 < q < n, n =

[q]+ 1, (2.1)

where [q] denotes the integer part of the real number q and Γ denotes the gamma function.


Definition 2.2. The Riemann-Liouville fractional integral of order q > 0 for a function g isdefined as

Iqg(t) =1

Γ(q)∫ t

0

g(s)

(t − s)1−q ds, q > 0, (2.2)

provided the right-hand side is pointwise defined on (0,∞).

Definition 2.3. The Riemann-Liouville fractional derivative of order q > 0 for a function g isdefined by

Dqg(t) =1

Γ(n − q

)(

d

dt

)n∫ t

0

g(s)

(t − s)q−n+1ds, n =

[q]+ 1, (2.3)

provided the right-hand side is pointwise defined on (0,∞).In passing, we remark that the Caputo derivative becomes the conventional nth

derivative of the function as q → n and the initial conditions for fractional differentialequations retain the same form as that of ordinary differential equations with integerderivatives. On the other hand, the Riemann-Liouville fractional derivative could hardlyproduce the physical interpretation of the initial conditions required for the initial valueproblems involving fractional differential equations (the same applies to the boundary valueproblems of fractional differential equations). Moreover, the Caputo derivative for a constantis zero while the Riemann-Liouville fractional derivative of a constant is nonzero. For moredetails, see [13].

For the forthcoming analysis, we need the following assumptions:

(A1) let F : [0, T] × R → BCC(R); (t, x) → f(t, x) be measurable with respect to t foreach x ∈ R, u.s.c. with respect to x for a.e. t ∈ [0, T], and for each fixed x ∈ R, theset SF,y := {f ∈ L1([0, T],R) : f(t) ∈ F(t, x) for a.e. t ∈ [0, T]} is nonempty,

(A2) for each r > 0, there exists a function mr ∈ L1([0, T],R+) such that ‖F(t, x)‖ =sup{|v| : v(t) ∈ F(t, x)} ≤ mr(t) for each (t, x) ∈ [0, T] × R with |x| ≤ r, and

lim infr→+∞

⎛

⎝∫T

0mr(t)dtr

⎞

⎠ = γ <∞, (2.4)

where mr depends on r. For example, for F(t, x) = x, we have mr(t) = r and hence γ = T. IfF(t, x) = x2, then γ is not finite.

Definition 2.4 ([16, 33]). A function x ∈ C([0, T]) is a solution of the problem (1.2) if thereexists a function f ∈ L1([0, T],R) such that f(t) ∈ F(t, x(t)) a.e. on [0, T] and

x(t) =∫ t

0

(t − s)q−1

Γ(q) f(s)ds − 1

2

∫T

0

(T − s)q−1

Γ(q) f(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) f(s)ds,

(2.5)


which, in terms of Green’s function G(t, s), can be expressed as

x(t) =∫T

0G(t, s)f(s)ds, (2.6)

where

G(t, s) =

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

− (T − s)q−1

2Γ(q) +

(T − 2t)(T − s)q−2

4Γ(q − 1

) , 0 < t < s < T,

(t − s)q−1 − 12(T − s)q−1

Γ(q) +

(T − 2t)(T − s)q−2

4Γ(q − 1

) , 0 < s < t < T.

(2.7)

Here, we remark that the Green’s function G(t, s) for q = 2 takes the form (see [22])

G(t, s) =

⎧⎪⎪⎨

⎪⎪⎩

14(−T − 2t + 2s), 0 < t < s < T,

14(−T + 2t − 2s), 0 < s < t < T.

(2.8)

Now we state the following lemmas which are necessary to establish the main resultof the paper.

Lemma 2.5 (Bohnenblust-Karlin [36]). Let D be a nonempty subset of a Banach space X, which isbounded, closed, and convex. Suppose that G : D → 2X \{0} is u.s.c. with closed, convex values suchthat G(D) ⊂ D and G(D) is compact. Then G has a fixed point.

Lemma 2.6 ([37]). Let I be a compact real interval. Let F be a multivalued map satisfying (A1)and let Θ be linear continuous from L1(I,R) → C(I), then the operator Θ ◦ SF : C(I) →BCC(C(I)), x �→ (Θ ◦ SF)(x) = Θ(SF,x) is a closed graph operator in C(I) × C(I).

3. Main Result

Theorem 3.1. Suppose that the assumptions (A1) and (A2) are satisfied, and

γ <4Γ(q)

(5 + q

)Tq−1

. (3.1)

Then the antiperiodic problem (1.2) has at least one solution on [0, T].


Proof. To transform the problem (1.2) into a fixed point problem, we define a multivaluedmap Ω : C([0, T]) → 2C([0,T]) as

Ω(x) =

{

h ∈ C([0, T]) : h(t) =∫ t

0

(t − s)q−1

Γ(q) f(s)ds − 1

2

∫T

0

(T − s)q−1

Γ(q) f(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) f(s)ds, f ∈ SF,x

}

.

(3.2)

Now we prove that Ω satisfies all the assumptions of Lemma 2.6, and thus Ω has a fixed pointwhich is a solution of the problem (1.2). As a first step, we show that Ω(x) is convex for eachx ∈ C([0, T]). For that, let h1, h2 ∈ Ω(x). Then there exist f1, f2 ∈ SF,x such that for eacht ∈ [0, T], we have

hi(t) =∫ t

0

(t − s)q−1

Γ(q) fi(s)ds −

12

∫T

0

(T − s)q−1

Γ(q) fi(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) fi(s)ds, i = 1, 2.

(3.3)

Let 0 ≤ λ ≤ 1. Then, for each t ∈ J, we have

[λh1 + (1 − λ)h2](t) =∫ t

0

(t − s)q−1

Γ(q)[λf1(s) + (1 − λ)f2(s)

]ds

− 12

∫T

0

(T − s)q−1

Γ(q)[λf1(s) + (1 − λ)f2(s)

]ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

)[λf1(s) + (1 − λ)f2(s)

]ds.

(3.4)

Since SF,x is convex (F has convex values), therefore it follows that λh1 + (1 − λ)h2 ∈ Ω(x).In order to show that Ω(x) is closed for each x ∈ C([0, T]), let {un}n≥0 ∈ Ω(x) be such

that un → u(n → ∞) in C([0, T]). Then u ∈ C([0, T]) and there exists a vn ∈ SF,x such that

un(t) =∫ t

0

(t − s)q−1

Γ(q) vn(s)ds −

12

∫T

0

(T − s)q−1

Γ(q) vn(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) vn(s)ds.

(3.5)


As F has compact values, we pass onto a subsequence to obtain that vn converges to v inL1([0, T],R+). Thus, v ∈ SF,x and

un(t) −→ u(t) =∫ t

0

(t − s)q−1

Γ(q) v(s)ds − 1

2

∫T

0

(T − s)q−1

Γ(q) v(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) v(s)ds.

(3.6)

Hence u ∈ Ω(x).Next we show that there exists a positive number r such that Ω(Br) ⊆ Br, where Br =

{x ∈ C([0, T]) : ‖x‖ ≤ r}. Clearly Br is a bounded closed convex set in C([0, T]) for eachpositive constant r. If it is not true, then for each positive number r, there exists a functionxr ∈ Br, hr ∈ Ω(xr) with ‖Ω(xr)‖ > r, and

hr(t) =∫ t

0

(t − s)q−1

Γ(q) fr(s)ds −

12

∫T

0

(T − s)q−1

Γ(q) fr(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) fr(s)ds, for some fr ∈ SF,xr .

(3.7)

On the other hand, in view of (A2), we have

r < ‖Ω(xr)‖

≤∫ t

0

|t − s|q−1

Γ(q)∣∣fr(s)

∣∣ds +12

∫T

0

|T − s|q−1

Γ(q)∣∣fr(s)

∣∣ds

+14|T − 2t|

∫T

0

|T − s|q−2

Γ(q − 1

)∣∣fr(s)

∣∣ds

≤ Tq−1

Γ(q)∫T

0mr(s)ds +

Tq−1

2Γ(q)∫T

0mr(s)ds +

Tq−1

4Γ(q − 1

)∫T

0mr(s)ds

=Tq−1(5 + q

)

4Γ(q)∫T

0mr(s)ds,

(3.8)

where we have used the fact that

∣∣∣(T − 2t)(T − s)q−2∣∣∣ ≤∣∣∣(T − t)(T − s)q−2

∣∣∣ ≤ |(T − t)|q−1, for t < s,

∣∣∣(T − 2t)(T − s)q−2∣∣∣ ≤∣∣∣(T − t)(T − s)q−2

∣∣∣ ≤ |(T − s)|q−1, for t ≥ s.

(3.9)

Dividing both sides of (3.8) by r and taking the lower limit as r → ∞, we find that γ ≥4Γ(q)/(5 + q)Tq−1, which contradicts (3.1). Hence there exists a positive number r ′ such thatΩ(Br ′) ⊆ Br ′.


Now we show that Ω(Br ′) is equicontinuous. Let t′, t′′ ∈ [0, T] with t′ < t′′. Let x ∈ Br ′

and h ∈ Ω(x), then there exists f ∈ SF,x such that for each t ∈ [0, T], we have

h(t) =∫ t

0

(t − s)q−1

Γ(q) f(s)ds − 1

2

∫T

0

(T − s)q−1

Γ(q) f(s)ds +

14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) f(s)ds. (3.10)

Using (3.8), we obtain

∣∣h(t′′)− h(t′)∣∣ =

∣∣∣∣∣

∫ t′′

0

(t′′ − s)q−1

Γ(q) f(s)ds +

14(T − 2t′′

)∫T

0

(T − s)q−2

Γ(q − 1

) f(s)ds

−∫ t′

0

(t′ − s)q−1

Γ(q) f(s)ds − 1

4(T − 2t′

)∫T

0

(T − s)q−2

Γ(q − 1

) f(s)ds

∣∣∣∣∣

≤

∣∣∣∣∣∣∣

∫ t′

0

[(t′′ − s)q−1 − (t′ − s)q−1

]

Γ(q) f(s)ds

∣∣∣∣∣∣∣+

∣∣∣∣∣

∫ t′′

t′

(t′′ − s)q−1

Γ(q) f(s)ds

∣∣∣∣∣

+

∣∣∣∣∣− (t

′′ − t′)2

∫T

0

(T − s)q−2

Γ(q − 1

) f(s)ds

∣∣∣∣∣

≤ 1Γ(q)∫ t′

0

∣∣∣(t′′ − s

)q−1 − (t′ − s)q−1∣∣∣mr ′(s)ds +

Tq−1

Γ(q)∫ t′′

t′mr ′(s)ds

+(t′′ − t′)Tq−2

2Γ(q − 1

)∫T

0mr ′(s)ds.

(3.11)

Obviously the right-hand side of the above inequality tends to zero independently of x ∈ Br ′

as t′′ → t′. Thus, Ω is equicontinuous.As Ω satisfies the above assumptions, therefore it follows by Ascoli-Arzela theorem

that Ω is a compact multivalued map.Finally, we show that Ω has a closed graph. Let xn → x∗, hn ∈ Ω(xn) and hn → h∗.

We will show that h∗ ∈ Ω(x∗). By the relation hn ∈ Ω(xn), we mean that there exists fn ∈ SF,xn

such that for each t ∈ [0, T],

hn(t) =∫ t

0

(t − s)q−1

Γ(q) fn(s)ds −

12

∫T

0

(T − s)q−1

Γ(q) fn(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) fn(s)ds.

(3.12)


Thus we need to show that there exists f∗ ∈ SF,x∗ such that for each t ∈ [0, T],

h∗(t) =∫ t

0

(t − s)q−1

Γ(q) f∗(s)ds −

12

∫T

0

(T − s)q−1

Γ(q) f∗(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) f∗(s)ds.

(3.13)

Let us consider the continuous linear operator Θ : L1([0, T],R) → C([0, T]) so that

f �−→ Θ(f)(t) =

∫ t

0

(t − s)q−1

Γ(q) f(s)ds − 1

2

∫T

0

(T − s)q−1

Γ(q) f(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) f(s)ds.

(3.14)

Observe that

‖hn(t) − h∗(t)‖ =∥∥∥∥∥

∫ t

0

(t − s)q−1

Γ(q)(fn(s) − f∗(s)

)ds − 1

2

∫T

0

(T − s)q−1

Γ(q)(fn(s) − f∗(s)

)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

)(fn(s) − f∗(s)

)ds

∥∥∥∥∥−→ 0 as n −→ ∞.

(3.15)

Thus, it follows by Lemma 2.6 that Θ◦SF is a closed graph operator. Further, we have hn(t) ∈Θ(SF,xn). Since xn → x∗, therefore, Lemma 2.6 yields

h∗(t) =∫ t

0

(t − s)q−1

Γ(q) f∗(s)ds −

12

∫T

0

(T − s)q−1

Γ(q) f∗(s)ds

+14(T − 2t)

∫T

0

(T − s)q−2

Γ(q − 1

) f∗(s)ds, for some f∗ ∈ SF,x∗ .

(3.16)

Hence, we conclude that Ω is a compact multivalued map, u.s.c. with convex closed values.Thus, all the assumptions of Lemma 2.6 are satisfied and so by the conclusion of Lemma 2.6,Ω has a fixed point x which is a solution of the problem (1.2).

Remark 3.2. If we take F(t, y) = {f(t, y)}, where f : [0, T] × R → R is a continuous function,then our results correspond to a single-valued problem (a new result).

Applications

As an application of Theorem 3.1, we discuss two cases in relation to the nonlinearity Fin (1.2), namely, F has (a) sublinear growth in its second variable (b) linear growth in itssecond variable (state variable). In case of sublinear growth, there exist functions η(t), ρ(t) ∈L1([0, T],R+), μ ∈ [0, 1) such that ‖F(t, x)‖ ≤ η(t)|x|μ + ρ(t) for each (t, x) ∈ [0, T] × R.


In this case, mr(t) = η(t)rμ+ρ(t). For the linear growth, the nonlinearity F satisfies the relation‖F(t, x)‖ ≤ η(t)|x| + ρ(t) for each (t, x) ∈ [0, T] × R. In this case mr(t) = η(t)r + ρ(t) andthe condition (3.1) modifies to ‖η‖L1 < 4Γ(q)/(5 + q)Tq−1. In both the cases, the antiperiodicproblem (1.2) has at least one solution on [0, T].

Examples

(a) We consider ‖F(t, x)‖ ≤ η(t)|x|1/3 + ρ(t) and T = 1 in (1.2). Here, η(t), ρ(t) ∈ L1([0, 1],R+).Clearly F(t, x) satisfies the assumptions of Theorem 3.1 with 0 < 4Γ(q)/(5 + q) (condition(3.1)). Thus, by the conclusion of Theorem 3.1, the antiperiodic problem (1.2) has at least onesolution on [0, 1].

(b) As a second example, let F(t, x) be such that ‖F(t, x)‖ ≤ (1/(1 + t)2)|x| + e−t andT = 1 in (1.2). In this case, (3.1) takes the form 1/2 < 4Γ(q)/(5 + q). As all the assumptions ofTheorem 3.1 are satisfied, the antiperiodic problem (1.2) has at least one solution on [0, 1].

Acknowledgments

The authors are grateful to the referees for their valuable suggestions that led to theimprovement of the original manuscript. The research of V. Otero-Espinar has been partiallysupported by Ministerio de Educacion y Ciencia and FEDER, Project MTM2007-61724, andby Xunta de Galicia and FEDER, Project PGIDIT06PXIB207023PR.

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Research ArticleExistence Results for Nonlinear Boundary ValueProblems of Fractional IntegrodifferentialEquations with Integral Boundary Conditions

Bashir Ahmad1 and Juan J. Nieto2

1 Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203,Jeddah 21589, Saudi Arabia


Correspondence should be addressed to Bashir Ahmad, bashir [email protected]

Received 9 December 2008; Revised 13 January 2009; Accepted 23 January 2009


This paper deals with some existence results for a boundary value problem involving a nonlinearintegrodifferential equation of fractional order q ∈ (1, 2] with integral boundary conditions. Ourresults are based on contraction mapping principle and Krasnosel’skiı’s fixed point theorem.

Copyright q 2009 B. Ahmad and J. J. Nieto. 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.

1. Introduction

In the last few decades, fractional-order models are found to be more adequate than integer-order models for some real world problems. Fractional derivatives provide an excellenttool for the description of memory and hereditary properties of various materials andprocesses. This is the main advantage of fractional differential equations in comparison withclassical integer-order models. Fractional differential equations arise in many engineeringand scientific disciplines as the mathematical modelling of systems and processes in thefields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymerrheology, and so forth, involves derivatives of fractional order. In consequence, the subjectof fractional differential equations is gaining much importance and attention. For examplesand details, see [1–22] and the references therein. However, the theory of boundary valueproblems for nonlinear fractional differential equations is still in the initial stages and manyaspects of this theory need to be explored.

Integrodifferential equations arise in many engineering and scientific disciplines, oftenas approximation to partial differential equations, which represent much of the continuum


phenomena. Many forms of these equations are possible. Some of the applications areunsteady aerodynamics and aero elastic phenomena, visco elasticity, visco elastic panel insuper sonic gas flow, fluid dynamics, electrodynamics of complex medium, many modelsof population growth, polymer rheology, neural network modeling, sandwich systemidentification, materials with fading memory, mathematical modeling of the diffusion ofdiscrete particles in a turbulent fluid, heat conduction in materials with memory, theory oflossless transmission lines, theory of population dynamics, compartmental systems, nuclearreactors, and mathematical modeling of a hereditary phenomena. For details, see [23–29] andthe references therein.

Integral boundary conditions have various applications in applied fields such as bloodflow problems, chemical engineering, thermoelasticity, underground water flow, populationdynamics, and so forth. For a detailed description of the integral boundary conditions, werefer the reader to a recent paper [30]. For more details of nonlocal and integral boundaryconditions, see [31–37] and references therein.

In this paper, we consider the following boundary value problem for a nonlinearfractional integrodifferential equation with integral boundary conditions

cDqx(t) = f(t, x(t), (χx)(t)), 0 < t < 1, 1 < q ≤ 2,

αx(0) + βx′(0) =∫1

0q1(x(s))ds, αx(1) + βx′(1) =

∫1

0q2(x(s))ds,

(1.1)

where cD is the Caputo fractional derivative, f : [0, 1] × X × X → X, for γ : [0, 1] × [0, 1] →[0,∞),

(χx)(t) =∫ t

0γ(t, s)x(s)ds, (1.2)

q1, q2 : X → X and α > 0, β ≥ 0 are real numbers. Here, (X, ‖ · ‖) is a Banach space and C =C([0, 1], X) denotes the Banach space of all continuous functions from [0, 1] → X endowedwith a topology of uniform convergence with the norm denoted by ‖ · ‖C.

2. Preliminaries

First of all, we recall some basic definitions [15, 18, 20].

Definition 2.1. For a function f : [0,∞) → R, the Caputo derivative of fractional order q isdefined as

cDqf(t) =1

Γ(n − q)

∫ t

0(t − s)n−q−1f (n)(s)ds, n − 1 < q < n, n = [q] + 1, (2.1)

where [q] denotes the integer part of the real number q.


Definition 2.2. The Riemann-Liouville fractional integral of order q is defined as

Iqf(t) =1

Γ(q)

∫ t

0

f(s)

(t − s)1−q ds, q > 0, (2.2)

provided the integral exists.

Definition 2.3. The Riemann-Liouville fractional derivative of order q for a function f(t) isdefined by

Dqf(t) =1

Γ(n − q)

(d

dt

)n∫ t

0

f(s)

(t − s)q−n+1ds, n = [q] + 1, (2.3)

provided the right hand side is pointwise defined on (0,∞).In passing, we remark that the definition of Riemann-Liouville fractional derivative,

which did certainly play an important role in the development of theory of fractionalderivatives and integrals, could hardly produce the physical interpretation of the initialconditions required for the initial value problems involving fractional differential equations.The same applies to the boundary value problems of fractional differential equations. It wasCaputo definition of fractional derivative which solved this problem. In fact, the Caputoderivative becomes the conventional nth derivative of the function f(t) as q → n andthe initial conditions for fractional differential equations retain the same form as that ofordinary differential equations with integer derivatives. Another difference is that the Caputoderivative for a constant is zero while the Riemann-Liouville fractional derivative of aconstant is nonzero. For more details, see [20].

Lemma 2.4 (see [22]). For q > 0, the general solution of the fractional differential equationcDqx(t) = 0 is given by

x(t) = c0 + c1t + c2t2 + · · · + cn−1t

n−1, (2.4)

where ci ∈ R, i = 0, 1, 2, . . . , n − 1 (n = [q] + 1).

In view of Lemma 2.4, it follows that

IqcDqx(t) = x(t) + c0 + c1t + c2t2 + · · · + cn−1t

n−1, (2.5)

for some ci ∈ R, i = 0, 1, 2, . . . , n − 1 (n = [q] + 1).Now, we state a known result due to Krasnosel’skiı [38] which is needed to prove the

existence of at least one solution of (1.1).

Theorem 2.5. Let M be a closed convex and nonempty subset of a Banach space X. Let A,B be theoperators such that (i)Ax +By ∈M whenever x, y ∈M, (ii)A is compact and continuous, (iii) B isa contraction mapping. Then there exists z ∈M such that z = Az + Bz.


Lemma 2.6. For any ζ, η1, η2 ∈ C[0, 1], the unique solution of the boundary value problem

cDqx(t) = ζ(t), 0 < t < 1, 1 < q ≤ 2,

αx(0) + βx′(0) =∫1

0η1(s)ds, αx(1) + βx′(1) =

∫1

0η2(s)ds,

(2.6)

is given by

x(t) =∫1

0G(t, s)ζ(s)ds +

1α2

[

(α(1 − t) + β)∫1

0η1(s)ds + (β + αt)

∫1

0η2(s)ds

]

, (2.7)

where G(t, s) is the Green’s function given by

G(t, s) =

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

α(t − s)q−1 + (β − αt)(1 − s)q−1

αΓ(q)+β(β − αt)(1 − s)q−2

α2Γ(q − 1), s ≤ t,

(β − αt)(1 − s)q−1

αΓ(q)+β(β − αt)(1 − s)q−2

α2Γ(q − 1), t ≤ s.

(2.8)

Proof. Using (2.5), for some constants b1, b2 ∈ R, we have

x(t) = Iqζ(t) − b1 − b2t =∫ t

0

(t − s)q−1

Γ(q)ζ(s)ds − b1 − b2t. (2.9)

In view of the relations cDqIqx(t) = x(t) and IqIpx(t) = Iq+px(t) for q, p > 0, x ∈ L(0, 1), weobtain

x′(t) =∫ t

0

(t − s)q−2

Γ(q − 1)ζ(s)ds − b2. (2.10)

Applying the boundary conditions for (2.6), we find that

b1 =1α2

[

β

∫1

0η2(s)ds − (β + α)

∫1

0η1(s)ds

]

−β

αΓ(q)

∫1

0(1 − s)q−1ζ(s)ds

−β2

α2Γ(q − 1)

∫1

0(1 − s)q−2ζ(s)ds,

b2 =1α

[∫1

0η1(s)ds −

∫1

0η2(s)ds

]

+1

Γ(q)

∫1

0(1 − s)q−1ζ(s)ds

+β

αΓ(q − 1)

∫1

0(1 − s)q−2ζ(s)ds.

(2.11)


Thus, the unique solution of (2.6) is

x(t) =∫ t

0

[α(t − s)q−1 + (β − αt)(1 − s)q−1

αΓ(q)+β(β − αt)(1 − s)q−2

α2Γ(q − 1)

]ζ(s)ds

+∫1

t

[(β − αt)(1 − s)q−1

αΓ(q)+β(β − αt)(1 − s)q−2

α2Γ(q − 1)

]ζ(s)ds

+1α2

[

(α(1 − t) + β)∫1

0η1(s)ds + (β + αt)

∫1

0η2(s)ds

]

=∫1

0G(t, s)ζ(s)ds +

1α2

[

(α(1 − t) + β)∫1

0η1(s)ds + (β + αt)

∫1

0η2(s)ds

]

,

(2.12)

where G(t, s) is given by (2.8). This completes the proof.

3. Main Results

Theorem 3.1. Assume that f : [0, 1]×X ×X → X is jointly continuous and maps bounded subsetsof [0, 1] ×X ×X into relatively compact subsets of X, γ : [0, 1] × [0, 1] → [0,∞) is continuous withγ0 = max{γ(t, s) : (t, s) ∈ [0, 1] × [0, 1]}, and q1, q2 : X → X are continuous functions. Further,there exist positive constants L1, L1, L2, L3,M2,M3 such that

(A1) ‖f(t, x(t), (χx)(t)) − f(t, y(t), (χy)(t))‖ ≤ L1‖x − y‖ + L1‖χx − χy‖, for all t ∈ [0, 1],x, y ∈ X,

(A2) ‖q1(x)−q1(y)‖ ≤ L2‖x−y‖, ‖q2(x)−q2(y)‖ ≤ L3‖x−y‖ with ‖q1(x)‖ ≤M2, ‖q2(x)‖ ≤M3, for all x, y ∈ X.

Then the boundary value problem (1.1) has a unique solution provided

(L1 + γ0L1

)[

β + 2ααΓ(q + 1)

+β2 + αβ

α2Γ(q)

]+β + α

α2(L2 + L3) < 1 (3.1)

with

L1 + γ0L1 ≤12

[β + 2α

αΓ(q + 1)+β2 + αβ

α2Γ(q)

]−1

. (3.2)

Proof. Define � : C → C by

(�x)(t) =1

Γ(q)

∫ t

0(t − s)q−1f(s, x(s), (χx)(s))ds

+∫1

0

[(β − αt)(1 − s)q−1

αΓ(q)+β(β − αt)(1 − s)q−2

α2Γ(q − 1)

]f(s, x(s), (χx)(s))ds

+1α2

[(α(1 − t) + β)

∫1

0q1(x(s))ds + (β + αt)

∫1

0q2(x(s))ds

], t ∈ [0, 1].

(3.3)


Setting supt∈[0,1]‖f(t, 0, 0)‖ = M1 (by the assumption on f) and Choosing

r ≥ 2[M1

(β + 2α

αΓ(q + 1)+β2 + αβ

α2Γ(q)

)+α + β

α2(M2 +M3)

], (3.4)

we show that �Br ⊂ Br, where Br = {x ∈ C : ‖x‖ ≤ r}. For x ∈ Br, we have

‖(�x)(t)‖ ≤ 1Γ(q)

∫ t

0(t − s)q−1‖f(s, x(s), (χx)(s))‖ds

+∫1

0|β − αt|

[(1 − s)q−1

αΓ(q)+β(1 − s)q−2

α2Γ(q − 1)

]‖f(s, x(s), (χx)(s))‖ds

+α + β

α2(M2 +M3)

≤ 1Γ(q)

∫ t

0(t − s)q−1[‖f(s, x(s), (χx)(s)) − f(s, 0, 0)‖ + ‖f(s, 0, 0)‖

]ds

+∫1

0|β − αt|

[(1 − s)q−1

αΓ(q)+β(1 − s)q−2

α2Γ(q − 1)

]

×[‖f(s, x(s), (χx)(s)) − f(s, 0, 0)‖ + ‖f(s, 0, 0)‖

]ds +

α + β

α2(M2 +M3)

≤((L1 + γ0L1

)r +M1

)[

1Γ(q)

∫ t

0(t − s)q−1ds

+∫1

0|β − αt|

((1 − s)q−1

αΓ(q)+β(1 − s)q−2

α2Γ(q − 1)

)ds

]

+α + β

α2(M2 +M3)

=((L1 + γ0L1

)r +M1

)[

tq

Γ(q + 1)+ |β − αt|

(1

αΓ(q + 1)+

β

α2Γ(q)

)]

+α + β

α2(M2 +M3)

≤(L1 + γ0L1

)[

2α + β

αΓ(q + 1)+β2 + αβ

α2Γ(q)

]r

+M1

[2α + β

αΓ(q + 1)+β2 + αβ

α2Γ(q)

]+α + β

α2(M2 +M3) ≤ r.

(3.5)


Now, for x, y ∈ C and for each t ∈ [0, 1], we obtain

‖(�x)(t) − (�y)(t)‖

≤ 1Γ(q)

∫ t

0(t − s)q−1‖f(s, x(s), (χx)(s)) − f(s, y(s), (χy)(s))‖ds

+∫1

0|β − αt|

[(1 − s)q−1

αΓ(q)+β(1 − s)q−2

α2Γ(q − 1)

]‖f(s, x(s), (χx)(s)) − f(s, y(s), (χy)(s))‖ds

+α + β

α2

[∫1

0‖q1(x(s)) − q1(y(s))‖ds +

∫1

0‖q2(x(s)) − q2(y(s))‖ds

]

≤(L1 + γ0L1

)‖x − y‖C

[tq

Γ(q + 1)+ |β − αt|

(1

αΓ(q + 1)+

β

α2Γ(q)

)]

+α + β

α2(L2 + L3)‖x − y‖C

≤{(L1 + γ0L1

)[

2α + β

αΓ(q + 1)+β2 + αβ)α2Γ(q)

]+α + β

α2(L2 + L3)

}‖x − y‖C

≤ Λα,β,q,γ0,L1,L1,L2,L3‖x − y‖C,

(3.6)

where

Λα,β,q,γ0,L1,L1,L2,L3=(L1 + γ0L1

)[

2α + β

αΓ(q + 1)+β2 + αβ

α2Γ(q)

]+α + β

α2(L2 + L3), (3.7)

which depends only on the parameters involved in the problem. As Λα,β,q,L1,L1,L2,L3< 1,

therefore � is a contraction. Thus, the conclusion of the theorem follows by the contractionmapping principle.

Theorem 3.2. Assume that (A1)-(A2) hold with ‖f(t, x(t), (χx)(t))‖ ≤ μ(t), for all (t, x, χx) ∈[0, 1] ×X ×X, where μ ∈ L1([0, 1], R+) and

(L1 + γ0L1

)(

α + β

αΓ(q + 1)+β2 + αβ

α2Γ(q)

)+α + β

α2(L2 + L3) < 1. (3.8)

Then the boundary value problem (1.1) has at least one solution on [0, 1].

Proof. Let us fix

r ≥ ‖μ‖L1

[2α + β

αΓ(q + 1)+β2 + αβ

α2Γ(q)

]+α + β

α2(M2 +M3), (3.9)


and consider Br = {x ∈ C : ‖x‖ ≤ r}. We define the operators Φ and Ψ on Br as

(Φx)(t) =1

Γ(q)

∫ t

0(t − s)q−1f(s, x(s), (χx)(s))ds,

(Ψx)(t) =∫1

0

[(β − αt)(1 − s)q−1

αΓ(q)+β(β − αt)(1 − s)q−2

α2Γ(q − 1)

]f(s, x(s), (χx)(s))ds

+1α2

[

(α(1 − t) + β)∫1

0q1(x(s))ds + (β + αt)

∫1

0q2(x(s))ds

]

.

(3.10)

For x, y ∈ Br, we find that

‖Φx + Ψy‖ ≤ ‖μ‖L1

[2α + β

αΓ(q + 1)+β2 + αβ

α2Γ(q)

]+α + β

α2(M2 +M3) ≤ r. (3.11)

Thus, Φx+Ψy ∈ Br. It follows from the assumption (A1), (A2) that Ψ is a contraction mappingfor

(L1 + γ0L1

)(

α + β

αΓ(q + 1)+β2 + αβ

α2Γ(q)

)+α + β

α2(L2 + L3) < 1. (3.12)

Continuity of f implies that the operator Φ is continuous. Also, Φ is uniformly bounded onBr as

‖Φx‖ ≤‖μ‖L1

Γ(q + 1). (3.13)

Now we prove the compactness of the operator Φ. In view of (A1), we definesup(t,x,χx)∈Ω‖f(s, x(s), (χx)(s))‖ = fmax,Ω = [0, 1] × Br × Br, and consequently we have

∥∥(Φx)(t1) − (Φx)(t2)∥∥

=

∥∥∥∥∥1

Γ(q)

∫ t1

0

[(t2 − s)q−1 − (t1 − s)q−1]f(s, x(s), (χx)(s))ds

+∫ t2

t1

(t2 − s)q−1f(s, x(s))ds

∥∥∥∥∥

≤fmax

Γ(q + 1)∣∣2(t2 − t1)q + t

q

1 − tq

2

∣∣,

(3.14)

which is independent of x. So Φ is relatively compact on Br. Hence, By Arzela AscoliTheorem, Φ is compact on Br. Thus all the assumptions of Theorem 2.5 are satisfied and


the conclusion of Theorem 2.5 implies that the boundary value problem (1.1) has at least onesolution on [0, 1].

Example 3.3. Consider the following boundary value problem:

cDqx(t) =1

(t + 7)2

|x|1 + |x| +

∫ t

0

e−(s−t)

49x(s)ds, t ∈ [0, 1], 1 < q ≤ 2,

x(0) + x′(0) =∫1

0

|x(s)|5 + |x(s)|ds, x(1) + x′(1) =

∫1

0

|x(s)|7 + |x(s)|ds.

(3.15)

Here, f(t, x) = (1/(t + 7)2)(|x|/(1 + |x|)), γ(t, s) = e−(s−t)/49, q1(x) = |x|/(5 + |x|), q2(x) =|x|/(7 + |x|), α = 1, β = 1. As ‖f(t, x, χx) − f(t, y, χy)‖ ≤ (1/49)‖x − y‖ + ‖χx − χy‖, ‖q1(x) −q1(y)‖ ≤ (1/5)‖x − y‖, ‖q2(x) − q2(y)‖ ≤ (1/7)‖x − y‖, therefore, (A1) and (A2) are satisfiedwith L1 = 1/49, L1 = 1, γ0 = ((e − 1)/49)L2 = 1/5, L3 = 1/7. Further,

(L1 + γ0L1

)[

β + 2ααΓ(q + 1)

+β2 + αβ

α2Γ(q)

]+β + α

α2(L2 + L3) < 1⇐⇒ e

49

(3

Γ(q + 1)+

2Γ(q)

)<

1135

.

(3.16)

Thus, by Theorem 3.1, the boundary value problem (3.15) has a unique solution on [0, 1].

Acknowledgments

The authors are grateful to the anonymous referee for his/her valuable suggestions that ledto the improvement of the original manuscript. The research of J. J. Nieto has been partiallysupported by Ministerio de Educacion y Ciencia and FEDER, Project MTM2007-61724, andby Xunta de Galicia and FEDER, project PGIDIT05PXIC20702PN.

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Research ArticleFirst-Order Singular and DiscontinuousDifferential Equations

Daniel C. Biles1 and Rodrigo Lopez Pouso2

1 Department of Mathematics, Belmont University, 1900 Belmont Blvd., Nashville, TN 37212, USA2 Department of Mathematical Analysis, University of Santiago de Compostela,15782 Santiago de Compostela, Spain

Correspondence should be addressed to Rodrigo Lopez Pouso, [email protected]

Received 10 March 2009; Accepted 4 May 2009


We use subfunctions and superfunctions to derive sufficient conditions for the existence ofextremal solutions to initial value problems for ordinary differential equations with discontinuousand singular nonlinearities.

Copyright q 2009 D. C. Biles and R. Lopez Pouso. 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.

1. Introduction

Let t0, x0 ∈ R and L > 0 be fixed and let f : [t0, t0 + L] × R → R be a given mapping. We areconcerned with the existence of solutions of the initial value problem

x′ = f(t, x), t ∈ I := [t0, t0 + L], x(t0) = x0. (1.1)

It is well-known that Peano’s theorem ensures the existence of local continuouslydifferentiable solutions of (1.1) in case f is continuous. Despite its fundamental importance,it is probably true that Peano’s proof of his theorem is even more important than the resultitself, which nowadays we know can be deduced quickly from standard fixed point theorems(see [1, Theorem 6.2.2] for a proof based on the Schauder’s theorem). The reason for believingthis is that Peano’s original approach to the problem in [2] consisted in obtaining the greatestsolution as the pointwise infimum of strict upper solutions. Subsequently this idea wasimproved by Perron in [3], who also adapted it to study the Laplace equation by meansof what we call today Perron’s method. For a more recent and important revisitation of themethod we mention the work by Goodman [4] on (1.1) in case f is a Caratheodory function.For our purposes in this paper, the importance of Peano’s original ideas is that they can


be adapted to prove existence results for (1.1) under such weak conditions that standardfunctional analysis arguments are no longer valid. We refer to differential equations whichdepend discontinuously on the unknown and several results obtained in papers as [5–9], seealso the monographs [10, 11].

On the other hand, singular differential equations have been receiving a lot of attentionin the last years, and we can quote [7, 12–19]. The main objective in this paper is to establishan existence result for (1.1) with discontinuous and singular nonlinearities which generalizesin some aspects some of the previously mentioned works.

This paper is organized as follows. In Section 2 we introduce the relevant definitionstogether with some previously published material which will serve as a basis for provingour main results. In Section 3 we prove the existence of the greatest and the smallestCaratheodory solutions for (1.1) between given lower and upper solutions and assuming theexistence of a L1-bound for f on the sector delimited by the graphs of the lower and uppersolutions (regular problems), and we give some examples. In Section 4 we show that lookingfor piecewise continuous lower and upper solutions is good in practice, but once we havefound them we can immediately construct a pair of continuous lower and upper solutionswhich provide better information on the location of the solutions. In Section 5 we provetwo existence results in case f does not have such a strong bound as in Section 3 (singularproblems), which requires the addition of some assumptions over the lower and uppersolutions. Finally, we prove a result for singular quasimonotone systems in Section 6 andwe give some examples in Section 7. Comparison with the literature is provided throughoutthe paper.

2. Preliminaries

In the following definition AC(I) stands for the set of absolutely continuous functions on I.

Definition 2.1. A lower solution of (1.1) is a function l ∈ AC(I) such that l(t0) ≤ x0 andl′(t) ≤ f(t, l(t)) for almost all (a.a.) t ∈ I; an upper solution is defined analogously reversingthe inequalities. One says that x is a (Caratheodory) solution of (1.1) if it is both a lower andan upper solution. On the other hand, one says that a solution x∗ is the least one if x∗ ≤ x onI for any other solution x, and one defines the greatest solution in a similar way. When boththe least and the greatest solutions exist, one calls them the extremal solutions.

It is proven in [8] that (1.1) has extremal solutions if f is L1-bounded for all x ∈R, f(·, x) is measurable, and for a.a. t ∈ I f(t, ·) is quasi-semicontinuous, namely, for all x ∈ R

we have

lim supy→x−

f(t, y

)≤ f(t, x) ≤ lim inf

y→x+f(t, y

). (2.1)

A similar result was established in [20] assuming moreover that f is superpositionallymeasurable, and the systems case was considered in [5, 8]. The term “quasi-semicontinuous”in connection with (2.1) was introduced in [5] for the first time and it appears to beconveniently short and descriptive. We note however that, rigorously speaking, we shouldsay that f(t, ·) is left upper and right lower semicontinuous.


On the other hand, the above assumptions imply that the extremal solutions of (1.1)are given as the infimum of all upper solutions and the supremum of all lower solutions, thatis, the least solution of (1.1) is given by

uinf(t) = inf{u(t) : u upper solution of (1.1)

}, t ∈ I, (2.2)

and the greatest solution is

lsup(t) = sup{l(t) : l lower solution of (1.1)}, t ∈ I. (2.3)

The mappings uinf and lsup turn out to be the extremal solutions even under moregeneral conditions. It is proven in [9] that solutions exist even if (2.1) fails on the points of acountable family of curves in the conditions of the following definition.

Definition 2.2. An admissible non-quasi-semicontinuity (nqsc) curve for the differentialequation x′ = f(t, x) is the graph of an absolutely continuous function γ : [a, b] ⊂ [t0, t0+L] →R such that for a.a. t ∈ [a, b] one has either γ ′(t) = f(t, γ(t)), or

γ ′(t) ≥ f(t, γ(t)

)whenever γ ′(t) ≥ lim inf

y→ (γ(t))+f(t, y

), (2.4)

γ ′(t) ≤ f(t, γ(t)

)whenever γ ′(t) ≤ lim sup

y→ (γ(t))−f(t, y

). (2.5)

Remark 2.3. The condition (2.1) cannot fail over arbitrary curves. As an example note that(1.1) has no solution for t0 = 0 = x0 and

f(t, x) =

⎧⎨

⎩

1, if x < 0,

−1, if x ≥ 0.(2.6)

In this case (2.1) only fails over the line x = 0, but solutions coming from above that linecollide with solutions coming from below and there is no way of continuing them to the rightonce they reach the level x = 0. Following Binding [21] we can say that the equation “jams”at x = 0.

An easily applicable sufficient condition for an absolutely continuous function γ :[a, b] ⊂ I → R to be an admissible nqsc curve is that either it is a solution or there existε > 0 and δ > 0 such that one of the following conditions hold:

(1) γ ′(t) ≥ f(t, y) + ε for a.a. t ∈ [a, b] and all y ∈ [γ(t) − δ, γ(t) + δ],

(2) γ ′(t) ≤ f(t, y) − ε for a.a. t ∈ [a, b] and all y ∈ [γ(t) − δ, γ(t) + δ].

These conditions prevent the differential equation from exhibiting the behavior of theprevious example over the line x = 0 in several ways. First, if γ is a solution of x′ = f(t, x) thenany other solution can be continued over γ once they contact each other and independentlyof the definition of f around the graph of γ . On the other hand, if (1) holds then solutionsof x′ = f(t, x) can cross γ from above to below (hence at most once), and if (2) holds then


solutions can cross γ from below to above, so in both cases the equation does not jam overthe graph of γ .

For the convenience of the reader we state the main results in [9]. The next resultestablishes the fact that we can have “weak” solutions in a sense just by assuming verygeneral conditions over f .

Theorem 2.4. Suppose that there exists a null-measure set N ⊂ I such that the following conditionshold:

(1) condition (2.1) holds for all (t, x) ∈ (I \N) ×R except, at most, over a countable family ofadmissible non-quasi-semicontinuity curves;

(2) there exists an integrable function g = g(t), t ∈ I, such that

∣∣f(t, x)∣∣ ≤ g(t) ∀(t, x) ∈ (I \N) × R. (2.7)

Then the mapping

u∗inf(t) = inf{u(t) : u upper solution of (1.1),

∣∣u′∣∣ ≤ g + 1 a.e.

}, t ∈ I (2.8)

is absolutely continuous on I and satisfies u∗inf(t0) = x0 and u∗inf′(t) = f(t, u∗inf(t)) for a.a. t ∈ I \ J ,

where J = ∪n,m∈NJn,m and for all n,m ∈ N the set

Jn,m :={t ∈ I : u∗inf

′(t) − 1n> sup

{f(t, y

): u∗inf(t) −

1m

< y < u∗inf(t)}}

(2.9)

contains no positive measure set.Analogously, the mapping

l∗sup(t) = sup{l(t) : l lower solution of (1.1),

∣∣l′∣∣ ≤ g + 1 a.e.

}, t ∈ I, (2.10)

is absolutely continuous on I and satisfies l∗sup(t0) = x0 and l∗sup′(t) = f(t, l∗sup(t)) for a.a. t ∈ I \K,

where K = ∪n,m∈NKn,m and for all n,m ∈ N the set

Kn,m :={t ∈ I : l∗sup

′(t) +1n< inf

{f(t, y

): l∗sup(t) < y < l∗sup(t) +

1m

}}(2.11)

contains no positive measure set.

Note that if the sets Jn,m and Kn,m are measurable then u∗inf and l∗sup immediatelybecome the extremal Caratheodory solutions of (1.1). In turn, measurability of those setscan be deduced from some measurability assumptions on f . The next lemma is a slightgeneralization of some results in [8] and the reader can find its proof in [9].


Lemma 2.5. Assume that for a null-measure set N ⊂ I the mapping f satisfies the followingcondition.

For each q ∈ Q, f(·, q) is measurable, and for (t, x) ∈ (I \N) × R one has

min

{

lim supy→x−

f(t, y

), lim sup

y→x+f(t, y

)}

≤ f(t, x) ≤ max{

lim infy→x−

f(t, y

), lim inf

y→x+f(t, y

)}.

(2.12)

Then the mappings t ∈ I �→ sup{f(t, y) : x1(t) < y < x2(t)} and t ∈ I �→ inf{f(t, y) :x1(t) < y < x2(t)} are measurable for each pair x1, x2 ∈ C(I) such that x1(t) < x2(t) forall t ∈ I.

Remark 2.6. A revision of the proof of [9, Lemma 2] shows that it suffices to impose (2.12) forall (t, x) ∈ (I \N) × R such that x1(t) < x < x2(t). This fact will be taken into account in thispaper.

As a consequence of Theorem 2.4 and Lemma 2.5 we have a result about existenceof extremal Caratheodory solutions for (1.1) and L1-bounded nonlinearities. Note that theassumptions in Lemma 2.5 include a restriction over the type of discontinuities that can occurover the admissible nonqsc curves, but remember that such a restriction only plays the role ofimplying that the sets Jn,m and Kn,m in Theorem 2.4 are measurable. Therefore, only using theaxiom of choice one can find a mapping f in the conditions of Theorem 2.4 which does notsatisfy the assumptions in Lemma 2.5 and for which the corresponding problem (1.1) lacksthe greatest (or the least) Caratheodory solution.

Theorem 2.7 ([9, Theorem 4]). Suppose that there exists a null-measure set N ⊂ I such that thefollowing conditions hold:

(i) for every q ∈ Q, f(·, q) is measurable;

(ii) for every t ∈ I \N and all x ∈ R one has either (2.1) or

lim infy→x−

f(t, y

)≥ f(t, x) ≥ lim sup

y→x+f(t, y

), (2.13)

and (2.1) can fail, at most, over a countable family of admissible nonquasisemicontinuitycurves;

(iii) there exists an integrable function g = g(t), t ∈ I, such that

∣∣f(t, x)∣∣ ≤ g(t) ∀(t, x) ∈ (I \N) × R. (2.14)

Then the mapping uinf defined in (2.2) is the least Caratheodory solution of (1.1) and the mappinglsup defined in (2.3) is the greatest one.


Remark 2.8. Theorem 4 in [9] actually asserts that u∗inf, as defined in (2.8), is the leastCaratheodory solution, but it is easy to prove that in that case u∗inf = uinf, as defined in (2.2).Indeed, let U be an arbitrary upper solution of (1.1), let g = max{|U′|, g} and let

v∗inf(t) = inf{u(t) : u upper solution of (1.1),

∣∣u′

∣∣ ≤ g + 1 a.e.

}, t ∈ I. (2.15)

Theorem 4 in [9] implies that also v∗inf is the least Caratheodory solution of (1.1), thus u∗inf =v∗inf ≤ U on I. Hence u∗inf = uinf.

Analogously we can prove that l∗sup can be replaced by lsup in the statement of [21,Theorem 4].

3. Existence between Lower and Upper Solutions

Condition (iii) in Theorem 2.7 is rather restrictive and can be relaxed by assumingboundedness of f between a lower and an upper solution.

In this section we will prove the following result.

Theorem 3.1. Suppose that (1.1) has a lower solution α and an upper solution β such that α(t) ≤ β(t)for all t ∈ I and let E = {(t, x) ∈ I × R : α(t) ≤ x ≤ β(t)}.

Suppose that there exists a null-measure setN ⊂ I such that the following conditions hold:

(iα,β) for every q ∈ Q ∩ (mint∈Iα(t),maxt∈Iβ(t)), the mapping f(·, q) with domain {t ∈ I :α(t) ≤ q ≤ β(t)} is measurable;

(iiα,β) for every (t, x) ∈ E, t /∈N, one has either (2.1) or (2.13), and (2.1) can fail, at most, over acountable family of admissible non-quasisemicontinuity curves contained in E;

(iiiα,β) there exists an integrable function g = g(t), t ∈ I, such that

∣∣f(t, x)∣∣ ≤ g(t) ∀(t, x) ∈ E, t /∈N. (3.1)

Then (1.1) has extremal solutions in the set

[α, β

]:=

{z ∈ AC(I) : α(t) ≤ z(t) ≤ β(t) ∀t ∈ I

}. (3.2)

Moreover the least solution of (1.1) in [α, β] is given by

x∗(t) = inf{u(t) : u upper solution of (1.1), u ∈

[α, β

]}, t ∈ I, (3.3)

and the greatest solution of (1.1) in [α, β] is given by

x∗(t) = sup{l(t) : l lower solution of (1.1), l ∈

[α, β

]}, t ∈ I. (3.4)


Proof. Without loss of generality we suppose that α′ and β′ exist and satisfy |α′| ≤ g, |β′| ≤ g,α′ ≤ f(t, α), and β′ ≥ f(t, β) on I \N. We also may (and we do) assume that every admissiblenqsc curve in condition (iiα,β), say γ : [a, b] → R, satisfies for all t ∈ [a, b] \N either γ ′(t) =f(t, γ(t)) or (2.4)-(2.5).

For each (t, x) ∈ I × R we define

F(t, x) :=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

f(t, α(t)), if x < α(t),

f(t, x), if α(t) ≤ x ≤ β(t),

f(t, β(t)

), if x > β(t).

(3.5)

Claim 1. The modified problem

x′ = F(t, x), t ∈ I, x(t0) = x0, (3.6)

satisfies conditions (1) and (2) in Theorem 2.4 with f replaced by F. First we note that (2) isan immediate consequence of (iiiα,β) and the definition of F.

To show that condition (1) in Theorem 2.4 is satisfied with f replaced by F, let (t, x) ∈(I \N) × R be fixed. The verification of (2.1) for F at (t, x) is trivial in the following cases:α(t) < x < β(t) and f satisfies (2.1) at (t, x), x < α(t), x > β(t) and α(t) = x = β(t). Let usconsider the remaining situations: we start with the case x = α(t) < β(t) and f satisfies (2.1)at (t, x), for which we have F(t, x) = f(t, x) and

lim supy→x−

F(t, y

)= f(t, α(t)) = f(t, x) ≤ lim inf

y→x+f(t, y

)= lim inf

y→x+F(t, y

), (3.7)

and an analogous argument is valid when α(t) < β(t) = x and f satisfies (2.1).The previous argument shows that F satisfies (2.1) at every (t, x) ∈ (I \N) ×R except,

at most, over the graphs of the countable family of admissible nonquasisemicontinuity curvesin condition (iiα,β) for x′ = f(t, x). Therefore it remains to show that if γ : [a, b] ⊂ I → R isone of those admissible nqsc curves for x′ = f(t, x) then it is also an admissible nqsc curve forx′ = F(t, x). As long as the graph of γ remains in the interior of E we have nothing to provebecause f and F are the same, so let us assume that γ = α on a positive measure set P ⊂ [a, b],P ∩N = ∅. Since α and γ are absolutely continuous there is a null measure set N such thatα′(t) = γ ′(t) for all t ∈ P \ N, thus for t ∈ P \ N we have

γ ′(t) ≤ f(t, γ(t)

)= lim sup

y→ (γ(t))−F(t, y

), γ ′(t) ≤ F

(t, γ(t)

), (3.8)

so condition (2.5) with f replaced by F is satisfied on P \ N. On the other hand, we have tocheck whether γ ′(t) ≥ F(t, γ(t)) for those t ∈ P \ N at which we have

γ ′(t) ≥ lim infy→ (γ(t))+

F(t, y

). (3.9)


We distinguish two cases: α(t) < β(t) and α(t) = β(t). In the first case (3.9) is equivalent to

γ ′(t) ≥ lim infy→ (γ(t))+

f(t, y

), (3.10)

and therefore either γ ′(t) = f(t, γ(t)) or condition (2.4) holds, yielding γ ′(t) ≥ f(t, γ(t)) =F(t, γ(t)). If α(t) = β(t) then we have γ ′(t) = α′(t) = β′(t) ≥ f(t, β(t)) = F(t, γ(t)).

Analogous arguments show that either γ ′ = F(t, γ) or (2.4)-(2.5) hold for F at almostevery point where γ coincides with β, so we conclude that γ is an admissible nqsc curve forx′ = F(t, x).

By virtue of Claim 1 and Theorem 2.4 we can ensure that the functions x∗ and x∗

defined as

x∗(t) = inf{u(t) : u upper solution of (3.6),

∣∣u′

∣∣ ≤ g + 1 a.e.

}, t ∈ I,

x∗(t) = sup{l(t) : l lower solution of (3.6),

∣∣∣l′∣∣∣ ≤ g + 1 a.e.

}, t ∈ I,

(3.11)

are absolutely continuous on I and satisfy x∗(t0) = x∗(t0) = x0 and x′∗(t) = F(t, x∗(t)) for a.a.t ∈ I \ J , where J = ∪n,m∈NJn,m and for all n,m ∈ N the set

Jn,m :={t ∈ I : x′∗(t) −

1n> sup

{F(t, y

): x∗(t) −

1m

< y < x∗(t)}}

(3.12)

contains no positive measure set, and x∗′(t) = F(t, x∗(t)) for a.a. t ∈ I \ K, where K =∪n,m∈NKn,m and for all n,m ∈ N the set

Kn,m :={t ∈ I : x∗′(t) +

1n< inf

{F(t, y

): x∗(t) < y < x∗(t) +

1m

}}(3.13)

contains no positive measure set.

Claim 2. For all t ∈ I we have

x∗(t) = inf{u(t) : u upper solution of (1.1), u ∈

[α, β

],∣∣u′

∣∣ ≤ g + 1 a.e.}, (3.14)

x∗(t) = sup{l(t) : l lower solution of (1.1), l ∈

[α, β

],∣∣l′

∣∣ ≤ g + 1 a.e.}. (3.15)

Let u be an upper solution of (3.6) and let us show that u(t) ≥ α(t) for all t ∈ I. Reasoning bycontradiction, assume that there exist t1, t2 ∈ I such that t1 < t2, u(t1) = α(t1) and

u(t) < α(t) ∀t ∈ (t1, t2]. (3.16)

For a.a. t ∈ (t1, t2] we have

u′(t) ≥ F(t, u(t)) = f(t, α(t)) ≥ α′(t), (3.17)


which together with u(t1) = α(t1) imply u ≥ α on [t1, t2], a contradiction with (3.16). Thereforeevery upper solution of (3.6) is greater than or equal to α, and, on the other hand, β is an uppersolution of (3.6) with |β′| ≤ g a.e., thus x∗ satisfies (3.14).

One can prove by means of analogous arguments that x∗ satisfies (3.15).

Claim 3. x∗ is the least solution of (1.1) in [α, β] and x∗ is the greatest one. From (3.14) and(3.15) it suffices to show that x∗ and x∗ are actually solutions of (3.6). Therefore we only haveto prove that J and K are null measure sets.

Let us show that the set J is a null measure set. First, note that

J =

⎧⎨

⎩t ∈ I : x′∗(t) > lim sup

y→ (x∗(t))−F(t, y

)⎫⎬

⎭, (3.18)

and we can split J = A ∪ B, where A = {t ∈ J : x∗(t) > α(t)} and B = J \A = {t ∈ J : x∗(t) =α(t)}.

Let us show that B is a null measure set. Since α and x∗ are absolutely continuous theset

C ={t ∈ I : α′(t) does not exist

}

∪{t ∈ I : x′∗(t) does not exist

}

∪{t ∈ I : α(t) = x∗(t), α′(t)/=x′∗(t)

}(3.19)

is null. If B /⊂C then there is some t0 ∈ B such that α(t0) = x∗(t0) and α′(t0) = x′∗(t0), but thenthe definitions of B and F yield

α′(t0) > lim supy→ (α(t0))

−F(t0, y

)= f(t0, α(t0)). (3.20)

Therefore B \ C ⊂N and thus B is a null measure set.The set A can be expressed as A = ∪∞k=1Ak, where for each k ∈ N

Ak =

⎧⎨

⎩t ∈ I : x∗(t) > α(t) +

1k, x′∗(t) > lim sup

y→ (x∗(t))−F(t, y

)⎫⎬

⎭

=∞⋃

n,m=1

Ak ∩ Jn,m.

(3.21)

For k,m ∈ N, k < m, we have x∗(t) − 1/m > x∗(t) − 1/k, so the definition of F impliesthat

Ak ∩ Jn,m ={t ∈ I : x∗(t) > α(t) +

1k, x′∗(t) −

1n> sup

{f(t, y

): x∗(t) −

1m

< y < x∗(t)}}

(3.22)

which is a measurable set by virtue of Lemma 2.5 and Remark 2.6.


Since Jn,m contains no positive measure subset we can ensure that Ak ∩ Jn,m is a nullmeasure set for all m ∈ N, m > k, and since Jn,m increases with n and m, we conclude thatAk = ∪∞n,m=1(Ak ∩ Jn,m) is a null measure set. Finally A is null because it is the union ofcountably many null measure sets.

Analogous arguments show that K is a null measure set, thus the proof of Claim 3 iscomplete.

Claim 4. x∗ satisfies (3.3) and x∗ satisfies (3.4). Let U ∈ [α, β] be an upper solution of (1.1), letg = max{|U′|, g}, and for all t ∈ I let

y∗(t) = inf{u(t) : u upper solution of (3.6),

∣∣u′

∣∣ ≤ g + 1 a.e.

}. (3.23)

Repeating the previous arguments we can prove that also y∗ is the least Caratheodorysolution of (1.1) in [α, β], thus x∗ = y∗ ≤ U on I. Hence x∗ satisfies (3.3).

Analogous arguments show that x∗ satisfies (3.4).

Remark 3.2. Problem (3.6) may not satisfy condition (i) in Theorem 2.7 as the compositionsf(·, α(·)) and f(·, β(·)) need not be measurable. That is why we used Theorem 2.4, instead ofTheorem 2.7, to establish Theorem 3.1.

Next we show that even singular problems may fall inside the scope of Theorem 3.1 ifwe have adequate pairs of lower and upper solutions.

Example 3.3. Let us denote by [z] the integer part of a real number z. We are going to showthat the problem

x′ =[

1t + |x|

]x +

sgn(x)2

, for a.a. t ∈ [0, 1], x(0) = 0 (3.24)

has positive solutions. Note that the limit of the right hand side as (t, x) tends to the origindoes not exist, so the equation is singular at the initial condition.

In order to apply Theorem 3.1 we consider (1.1) with t0 = 0 = x0, L = 1, and

f(t, x) =

⎧⎪⎨

⎪⎩

[1

t + x

]x +

12, if x > 0,

12, if x ≤ 0.

(3.25)

It is elementary matter to check that α(t) = 0 and β(t) = t, t ∈ I, are lower and uppersolutions for the problem. Condition (2.1) only fails over the graphs of the functions

γn(t) =1n− t, t ∈

[0,

1n

], n ∈ N, (3.26)

which are a countable family of admissible nqsc curves at which condition (2.13) holds.


Finally note that

∣∣f(t, x)

∣∣ ≤ 3

2∀(t, x) ∈ I × R, 0 ≤ x ≤ t, (3.27)

so condition (iiiα,β) is satisfied.Theorem 3.1 ensures that our problem has extremal solutions between α and β which,

obviously, are different from zero almost everywhere. Therefore (3.24) has positive solutions.

The result of Theorem 3.1 may fail if we assume that condition (iiα,β) is satisfied onlyin the interior of the set E. This is shown in the following example.

Example 3.4. Let us consider problem (1.1) with t0 = x0 = 0, L = 1 and f : [0, 1] × R → R

defined as

f(t, x) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

1, if x < 0,

12, if x = 0,

−1, if x > 0.

(3.28)

It is easy to check that α(t) = 0 and β(t) = t for all t ∈ [0, 1] are lower and upper solutionsfor this problem and that all the assumptions of Theorem 3.1 are satisfied in the interior of E.However this problem has no solution at all.

In order to complete the previous information we can say that condition (iiα,β) in theinterior of E is enough if we modify the definitions of lower and upper solutions in thefollowing sense.

Theorem 3.5. Suppose that α and β are absolutely continuous functions on I such that α(t) < β(t)for all t ∈ (t0, t0 + L], α(t0) ≤ x0 ≤ β(t0),

α′(t) ≤ lim infy→ (α(t))+

f(t, y

)for a.a. t ∈ I,

β′(t) ≥ lim supy→ (β(t))−

f(t, y

)for a.a. t ∈ I,

(3.29)

and let E = {(t, x) ∈ I × R : α(t) ≤ x ≤ β(t)}.Suppose that there exists a null-measure setN ⊂ I such that conditions (iα,β) and (iiiα,β) hold

and, moreover,

(ii◦α,β) for every (t, x) ∈◦E, t /∈N, one has either (2.1) or (2.13), and (2.1) can fail, at most, over a

countable family of admissible non-quasisemicontinuity curves contained in◦E.

Then the conclusions of Theorem 3.1 hold true.


Proof (Outline)

It follows the same steps as the proof of Theorem 3.1 but replacing F by

F(t, x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

0, if t = t0,

lim infy→ (α(t))+

f(t, y

), if t > t0, x ≤ α(t),

f(t, x), if t > t0, α(t) < x < β(t),

lim supy→ (β(t))−

f(t, y

), if t > t0, x ≥ β(t).

(3.30)

Note that condition (2.1) with f replaced by F is immediately satisfied over the graphs of αand β thanks to the definition of F.

Remarks

(i) The function α in Example 3.4 does not satisfy the conditions in Theorem 3.5.

(ii) When f(t, ·) satisfies (2.1) everywhere or almost all t ∈ I then every couple of lowerand upper solutions satisfies the conditions in Theorem 3.5, so this result is notreally new in that case (which includes the Caratheodory and continuous cases).

4. Discontinuous Lower and Upper Solutions

Another modification of the concepts of lower and upper solutions concerns the possibility ofallowing jumps in their graphs. Since the task of finding a pair of lower and upper solutionsis by no means easy in general, and bearing in mind that constant lower and upper solutionsare the first reasonable attempt, looking for lower and upper solutions “piece by piece” mightmake it easier to find them in practical situations. Let us consider the following definition.

Definition 4.1. One says that α : I → R is a piecewise continuous lower solution of (1.1) ifthere exist t0 < t1 < · · · < tn = t0 + L such that

(a) for all i ∈ {1, 2, . . . , n}, one has α ∈ AC(ti−1, ti) and for a.a. t ∈ I

α′(t) ≤ f(t, α(t)), (4.1)

(b) limt→ t+0α(t) = α(t0) ≤ x0, for all i ∈ {1, 2, . . . , n}

limt→ t−i

α(t) = α(ti) > limt→ t+i

α(t), (4.2)

and limt→ t−n α(t) = α(tn).

A piecewise continuous upper solution of (1.1) is defined reversing the relevant inequalities.


The existence of a pair of well-ordered piecewise continuous lower and uppersolutions implies the existence of a better pair of continuous lower and upper solutions. Weestablish this more precisely in our next proposition. Note that the proof is constructive.

Proposition 4.2. Assume that all the conditions in Theorem 3.1 hold with piecewise continuous lowerand upper solutions α and β.

Then the following statements hold:

(i) there exist a lower solution α and an upper solution β such that

α ≤ α ≤ β ≤ β on I; (4.3)

(ii) if u is an upper solution of (1.1) with α ≤ u ≤ β then α ≤ u, and if l is a lower solutionwith α ≤ l ≤ β then l ≤ β.

In particular, the conclusions of Theorem 3.1 remain valid and, moreover, every solution of (1.1)between α and β lies between α and β.

Proof. We will only prove the assertions concerning α because the proofs for β are analogous.To construct α we simply have to join the points (tk, α(tk)), k ∈ {1, . . . , n − 1} with the

graph of α|(tk ,tk+1] by means of an absolutely continuous curve with derivative less than orequal to −g a.e., g being the function given in (iiiα,β). It can be easily proven that this α is alower solution of (1.1) that lies between α and β.

Moreover, if u is an upper solution of (1.1) between α and β then we have

u′(t) ≥ f(t, u(t)) ≥ −g(t) a.e. on [t0, t0 + L], (4.4)

so it cannot go below α.

Piecewise continuous lower and upper solutions in the sense of Definition 4.1 werealready used in [15, 22]. It is possible to generalize further the concept of lower andupper solutions, as a piecewise continuous lower solution is a particular case of a boundedvariation function that has a nonincreasing singular part. Bounded variation lower and uppersolutions with monotone singular parts were used in [23, 24], but it is not clear whetherTheorem 3.1 is valid with this general type of lower and upper solutions. Anyway, piecewisecontinuous lower and upper solutions are enough in practical situations, and since these canbe transformed into continuous ones which provide better information we will only considerfrom now on continuous lower and upper solutions as defined in Definition 2.1.

5. Singular Differential Equations

It is the goal of the present section to establish a theorem on existence of solutions for (1.1)between a pair of well-ordered lower and upper solutions and in lack of a local L1 bound.Solutions will be weak, in the sense of the following definition. By ACloc((t0, t0+L]) we denotethe set of functions ξ such that ξ|[t0+ε,t0+L] ∈ AC([t0 + ε, t0 +L]) for all ε ∈ (0, L), and in a similarway we define L1

loc((t0, t0 + L]).


Definition 5.1. We say that α ∈ C(I) ∩ ACloc((t0, t0 + L]) is a weak lower solution of (1.1) ifα(t0) ≤ x0 and α′(t) ≤ f(t, α(t)) for a.a. t ∈ I. A weak upper solution is defined analogouslyreversing inequalities. A weak solution of (1.1) is a function which is both a weak lowersolution and a weak upper solution.

We will also refer to extremal weak solutions with obvious meaning.Note that (lower/upper) solutions, as defined in Definition 2.1, are weak

(lower/upper) solutions but the converse is false in general. For instance the singular linearproblem

x′ =x

t− cos(1/t)

t, t ∈ (0, 1], x(0) = 0, (5.1)

has exactly the following weak solutions:

xa(t) = t sin(

1t

)+ at, t ∈ (0, 1], xa(0) = 0 (a ∈ R), (5.2)

and none of them is absolutely continuous on [0, 1]. Another example, which uses lower andupper solutions, can be found in [15, Remark 2.4].

However weak (lower/upper) solutions are of Caratheodory type provided they havebounded variation. We establish this fact in the next proposition.

Proposition 5.2. Let a, b ∈ R be such that a < b and let h : [a, b] → R be continuous on [a, b] andlocally absolutely continuous on (a, b].

A necessary and sufficient condition for h to be absolutely continuous on [a, b] is that h be ofbounded variation on [a, b].

Proof. The necessary part is trivial. To estalish the sufficiency of our condition we use Banach-Zarecki’s theorem, see [18, Theorem 18.25]. Let N ⊂ [a, b] be a null measure set, we have toprove that h(N) is also a null measure set. To do this let n0 ∈ N be such that a + 1/n0 < b.Since h is absolutely continuous on [a + 1/n0, b] the set h(N ∩ [a + 1/n, b]) is a null measureset for each n ≥ n0. Therefore h(N) is also a null measure set because

h(N) ⊂ {h(a)} ∪(

∞⋃

n=n0

h

(N ∩

[a +

1n, b

]))

. (5.3)

Next we present our main result on existence of weak solutions for (1.1) in absence ofintegrable bounds.

Theorem 5.3. Suppose that (1.1) has a weak lower solution α and a weak upper solution β such thatα(t) ≤ β(t) for all t ∈ I and α(t0) = x0 = β(t0).

Suppose that there is a null-measure set N ⊂ I such that conditions (iα,β) and (iiα,β) inTheorem 3.1 hold for E = {(t, x) ∈ I × R : α(t) ≤ x ≤ β(t)}, and assume moreover that the followingcondition holds:


(iii∗α,β) there exists g ∈ L1loc((t0, t0 +L]) such that for all (t, x) ∈ E, t /∈N, one has |f(t, x)| ≤ g(t).

Then (1.1) has extremal weak solutions in the set

[α, β]w :={z ∈ C(I) ∩ ACloc((t0, t0 + L]) : α(t) ≤ z(t) ≤ β(t) ∀t ∈ I

}. (5.4)

Moreover the least weak solution of (1.1) in [α, β] is given by

x∗(t) = inf{u(t) : u weak upper solution of (1.1), u ∈

[α, β

]w

}, t ∈ I, (5.5)

and the greatest weak solution of (1.1) in [α, β]w is given by

x∗(t) = sup{l(t) : l weak lower solution of (1.1), l ∈

[α, β

]w

}, t ∈ I. (5.6)

Proof. We will only prove that (5.6) defines the greatest weak solution of (1.1) in [α, β]w, asthe arguments to show that (5.5) is the least one are analogous.

First note that α is a weak lower solution between α and β, so x∗ is well defined.Let {tn}n be a decreasing sequence in (t0, t0 + L) such that lim tn = t0. Theorem 2.7

ensures that for every n ∈ N the problem

y′ = f(t, y

), t ∈ [tn, t0 + L] =: In, y(tn) = x∗(tn), (5.7)

has extremal Caratheodory solutions between α|In and β|In . Let yn denote the greatest solutionof (5.7) between α|In and β|In . By virtue of Theorem 2.7 we also know that yn is the greatestlower solution of (5.7) between α|In and β|In .

Next we prove in several steps that x∗ = yn on In for each n ∈ N.

Step 1 (yn ≥ x∗ on In for each n ∈ N). The restriction to In of each weak lower solutionbetween α and β is a lower solution of (5.7) between α|In and β|In , thus yn is, on the intervalIn, greater than or equal to any weak lower solution of (1.1) between α and β. The definitionof x∗ implies then that yn ≥ x∗ on In.

Step 2 (yn+1 ≥ yn on In for all n ∈ N). First, since yn+1 ≥ x∗ on In+1 we have yn+1(tn) ≥x∗(tn) = yn(tn). Reasoning by contradiction, assume that there exists s ∈ (tn, t0 + L) such thatyn+1(s) < yn(s). Then there is some r ∈ [tn, s) such that yn+1(r) = yn(r) and yn+1 < yn on (r, s),but then the mapping

y(t) =

⎧⎨

⎩

yn+1(t), if t ∈ [tn+1, r],

yn(t), if t ∈ [r, t0 + L],(5.8)

would be a solution of (5.7) (with n replaced by n+1) between α|In+1 and β|In+1 which is greaterthan yn+1 on (r, s), a contradiction.


The above properties of {yn}n imply that the following function is well defined:

y∞(t) =

⎧⎨

⎩

x0, if t = t0,

limyn(t), if t ∈ (t0, t0 + L].(5.9)

Step 3 (y∞ ∈ C(I) ∩ ACloc((t0, t0 + L])). Let ε ∈ (0, L) be fixed. Condition (iii∗α,β) implies thatfor all n ∈ N such that tn < t0 + ε we have

∣∣y′n(t)

∣∣ =

∣∣f

(t, yn(t)

)∣∣ ≤ g(t) for a. a. t ∈ [t0 + ε, t0 + L], (5.10)

with g ∈ L1(t0 + ε, t0 + L). Hence for s, t ∈ [t0 + ε, t0 + L], s ≤ t, we have

∣∣y∞(t) − y∞(s)∣∣ = lim

n→∞

∣∣∣∣∣

∫ t

s

y′n

∣∣∣∣∣≤

∫ t

s

g, (5.11)

and therefore y∞ ∈ AC([t0 + ε, t0 + L]). Since ε ∈ (0, L) was fixed arbitrarily in the previousarguments, we conclude that y∞ ∈ ACloc((t0, t0 + L]).

The continuity of y∞ at t0 follows from the continuity of α and β at t0, the assumptionα(t0) = x0 = β(t0), and the relation

α(t) ≤ y∞(t) ≤ β(t) ∀t ∈ (t0, t0 + L]. (5.12)

Step 4 (y∞ is a weak lower solution of (1.1)). For ε ∈ (0, L) and n ∈ N such that tn < t0 + εwe have (5.10) with g ∈ L1(t0 + ε, t0 + L), hence lim supy′n ∈ L1(t0 + ε, t0 + L), and for s, t ∈(t0 + ε, t0 + L), s < t, Fatou’s lemma yields

y∞(t) − y∞(s) = limn→∞

∫ t

s

y′n ≤∫ t

s

lim sup y′n. (5.13)

Hence for a.a. t ∈ [t0 + ε, t0 + L] we have

y′∞(t) ≤ lim sup y′n(t) = lim sup f(t, yn(t)

). (5.14)

Let J1 = ∪n∈NAn where An = {t ∈ [t0+ε, t0+L] : y∞(t) = yn(t)} and J2 = [t0+ε, t0+L]\J1.For n ∈ N and a.a. t ∈ An we have y′∞(t) = y′n(t) = f(t, yn(t)) = f(t, y∞(t)), thus

y′∞(t) = f(t, y∞(t)) for a.a. t ∈ J1.


On the other hand, for a.a. t ∈ J2 the relation (5.14) and the increasingness of {yn(t)}yield

y′∞(t) ≤ lim sup f(t, yn(t)

)≤ lim sup

y→ (y∞(t))−f(t, y

). (5.15)

Let t0 ∈ J2 \N be such that (5.15) holds. We have two possibilities: either (2.1) holdsfor f at (t0, y∞(t0)) and then from (5.15) we deduce y′∞(t0) ≤ f(t0, y∞(t0)), or y∞(t0) = γ(t0),where γ is an admissible curve of non quasisemicontinuity. In the last case we have that eithert0 belongs to a null-measure set or y′∞(t0) = γ ′(t0), which, in turn, yields two possibilities:either γ ′(t0) = f(t, γ(t0)) and then y′∞(t0) = f(t, y∞(t0)), or γ ′(t0)/= f(t, γ(t0)) and then (5.15),with y∞(t0) = γ(t0) and y′∞(t0) = γ ′(t0), and the definition of admissible curve of nonquasisemicontinuity imply that y′∞(t0) ≤ f(t0, y∞(t0)).

The above arguments prove that y′∞(t) ≤ f(t, y∞(t)) a.e. on [t0 + ε, t0 + L], and sinceε ∈ (0, L) was fixed arbitrarily, the proof of Step 4 is complete.

Conclusion

The construction of y∞ and Step 1 imply that y∞ ≥ x∗ and the definition of x∗ and Step 4imply that x∗ ≥ y∞. Therefore for all n ∈ N we have x∗ = yn on In and then x∗ is a weaksolution of (1.1). Since every weak solution is a weak lower solution, x∗ is the greatest weaksolution of (1.1) in [α, β]w.

The assumption α(t0) = β(t0) in Theorem 5.3 can be replaced by other types ofconditions. The next theorem generalizes the main results in [7, 12–14] concerning existenceof solutions of singular problems of the type of (1.1).

Theorem 5.4. Suppose that (1.1) with x0 = 0 has a weak lower solution α and a weak upper solutionβ such that α(t) ≤ β(t) for all t ∈ I and α > 0 on (t0, t0 + L].

Suppose that there is a null-measure set N ⊂ I such that conditions (iα,β) and (iiα,β) inTheorem 3.1 hold for E = {(t, x) ∈ I × R : α(t) ≤ x ≤ β(t)}, and assume moreover that the followingcondition holds:

(iii+α,β) for every r ∈ (0, 1) there exists gr ∈ L1(I) such that for all (t, x) ∈ E, t /∈N, and r ≤ x ≤1/r one has |f(t, x)| ≤ gr(t).

Then the conclusions of Theorem 5.3 hold true.

Proof. We start observing that there exists a weak upper solution β such that α ≤ β ≤ β onI and α(t0) = 0 = β(t0). If β(t0) = 0 then it suffices to take β as β. If β(t0) > 0 we proceedas follows in order to construct β: let {xn}n be a decreasing sequence in (0, β(t0)) such thatlim xn = 0 and for every n ∈ N let yn be the greatest solution between α and β of

y′ = f(t, y

), t ∈ I, y(t0) = xn. (5.16)


claim [yn exists]

Let ε ∈ (0, L) be so small that α(t) < xn − ε < xn + ε < β(t) for all t ∈ [t0, t0 + ε]. Condition(iii+α,β) implies that there exists gε ∈ L1(t0, t0 + ε) such that for a.a. t ∈ [t0, t0 + ε] and all

x ∈ [xn − ε, xn + ε] we have |f(t, x)| ≤ gε(t). Let v±(t) = xn ±∫ t

0gε(s)ds, t ∈ [t0, t0 + ε] and letδ ∈ (0, ε] be such that xn − ε ≤ v− ≤ v+ ≤ xn + ε on [t0, t0 + δ]. We can apply Theorem 2.7 tothe problem

ξ′ = f(t, ξ), t ∈ [t0, t0 + δ], ξ(t0) = xn, (5.17)

and with respect to the lower solution v− and the upper solution v+, so there exists ξn thegreatest solution between v− and v+ of (5.17). Notice that if x is a solution of (5.17) thenv− ≤ x ≤ v+, so ξn is also the greatest solution between α and β of (5.17).

Now condition (iii+α,β) ensures that Theorem 2.7 can be applied to the problem

z′ = f(t, z), t ∈ [t0 + δ, t0 + L], z(t0 + δ) = ξn(t0 + δ), (5.18)

with respect to the lower solution α and the upper solution β (both functions restricted to[t0 + δ, t0 + L]). Hence there exists zn the greatest solution of (5.18) between α and β.

Obviously we have

yn(t) =

⎧⎨

⎩

ξn(t), if t ∈ [t0, t0 + δ],

zn(t), if t ∈ [t0 + δ, t0 + L].(5.19)

Analogous arguments to those in the proof of Theorem 5.3 show that β = lim yn is aweak upper solution and it is clear that β(t0) = 0 = α(t0).

Finally we show that (iii∗α,β) holds with β replaced by β. We consider a decreasingsequence (an)n such that a0 = t0 + L and liman = t0. As α and β are positive on (t0, t0 + L],we can find ri > 0 such that ri ≤ α ≤ β ≤ 1/ri on [ai+1, ai]. We deduce then from (iii+α,β) theexistence of ψi ∈ L1(ai+1, ai) so that |f(t, x)| ≤ ψi(t) for a.e. t ∈ [ai+1, ai] and all x ∈ [α(t), β(t)].The function g defined by g(t) = ψi(t) for t ∈ (ai+1, ai] works.

Theorem 5.3 implies that (1.1) has extremal weak solutions in [α, β]w which, moreover,satisfy (5.6) and (5.5) with β replaced by β. Furthermore if x is a weak solution of (1.1) in[α, β]w then x ≤ β on I. Assume, on the contrary, that x(s) > β(s) for some s ∈ (t0, t0 + L),then there would exist n ∈ N such that yn(s) < x(s) and then y = max{x, yn} would be asolution of (5.16) between α and β which is strictly greater than yn on some subinterval, acontradiction. Hence (1.1) has extremal weak solutions in [α, β]w which, moreover, satisfy(5.6) and (5.5).


6. Systems

Let us consider the following system of n ∈ N ordinary differential equations:

−→x ′(t) =−→f(t,−→x

)for a. a. t ∈ I = [t0, t0 + L], −→x(t0) = −→x0, (6.1)

where t0 ∈ R, L > 0, −→x = (x1, . . . , xn),−→x0 = (x0,1, . . . , x0,n), and

−→f = (f1, f2, . . . , fn) : I × R

n →R

n.Our goal is to extend Theorem 5.3 to this multidimensional setting, which, as usual,

requires the right hand side−→f to be quasimonotone, as we will define later.

We start extending to the vector case the definitions given before for scalar problems.To do so, let ACloc((t0, t0 + L],Rn) denote the set of functions −→y = (y1, y2, . . . , yn) : (t0, t0 +L] → R

n such that for each i ∈ {1, 2, . . . , n} the component yi is absolutely continuous on[t0 + ε, t0 + L] for each ε ∈ (0, L). Also, C(I,Rn) stands for the class of R

n-valued functionswhich are defined and continuous on I.

A weak lower solution of (6.1) is a function −→α = (α1, α2, . . . , αn) ∈ C(I,Rn) ∩ACloc((t0, t0 + L],Rn) such that for each i ∈ {1, 2, . . . , n} we have αi(0) ≤ x0,i and for a.a.t ∈ I we have α′i(t) ≤ fi(t,

−→α(t)). Weak upper solutions are defined similarly by reversing therelevant inequalities, and weak solutions of (6.1) are functions which are both weak lowerand weak upper solutions.

In the set C(I,Rn) ∩ ACloc((t0, t0 + L],Rn) we define a partial ordering as follows: let−→α,−→β ∈ C(I,Rn) ∩ ACloc((t0, t0 + L],Rn), we write −→α ≤

−→β if every component of −→α is less

than or equal to the corresponding component of−→β on the whole of I. If −→α,

−→β ∈ C(I,Rn) ∩

ACloc((t0, t0 + L],Rn) are such that −→α ≤−→β then we define

[−→α,−→β]

:={−→η ∈ C(I,Rn) ∩ACloc((t0, t0 + L],Rn) : −→α ≤ −→η ≤

−→β}. (6.2)

Extremal (least and greatest) weak solutions of (6.1) in a certain subset of C(I,Rn) ∩ACloc((t0, t0 + L],Rn) are defined in the obvious way considering the previous ordering.

Now we are ready to extend Theorem 5.3 to the vector case. We will denote by −→ei theith canonical vector. The proof follows the line of that of [8, Theorem 5.1].

Theorem 6.1. Suppose that (6.1) has weak lower and upper solutions −→α and−→β such that −→α ≤

−→β ,

−→α(t0) = −→x0 =−→β (t0), and let E∗ = {(t,−→x) : t ∈ I, −→α(t) ≤ −→x ≤

−→β (t)}.

Suppose that−→f is quasimonotone nondecreasing in E∗, that is, for i ∈ {1, 2, . . . , n} and

(t,−→x), (t,−→y) ∈ E∗ the relations −→x ≤ −→y and xi = yi imply fi(t,−→x) ≤ fi(t,

−→y).Suppose, moreover, that for each −→η = (η1, . . . , ηn) ∈ [−→α,

−→β ] the following conditions hold:

(H1) the function−→f (·,−→η(·)) is measurable;

(H2) for all i ∈ {1, . . . , n} and a.a. t ∈ I one has either

lim sups→x−

fi(t,−→η(t) +

(s − ηi(t)

)−→ei)≤ fi

(t,−→η(t) +

(x − ηi(t)

)−→ei)

≤ lim infs→x+

fi(t,−→η(t) +

(s − ηi(t)

)−→ei),

(6.3)


or

lim infs→x−

fi(t,−→η(t) +

(s − ηi(t)

)−→ei)≥ fi

(t,−→η(t) +

(x − ηi(t)

)−→ei)

≥ lim sups→x+

fi(t,−→η(t) +

(s − ηi(t)

)−→ei),

(6.4)

and (6.3) fails, at most, over a countable family of admissible nqsc curves of the scalardifferential equation x′ = fi(t,

−→η(t) + (x − ηi(t))−→ei) contained in the sector Ei := {(t, x) :t ∈ I, αi(t) ≤ x ≤ βi(t)};

(H3) there exists g ∈ L1loc((t0, t0 + L]) such that for i ∈ {1, 2, . . . , n} and a.a. t ∈ I one has

|fi(t,−→η(t))| ≤ g(t).

Then (6.1) has extremal weak solutions in [α, β]. Moreover the least weak solution −→x∗ = (x∗,1, . . . , x∗,n)is given by

x∗,i(t) = inf{ui(t) : (u1, . . . , un) weak upper solution of (6.1) in

[−→α,−→β]}

(6.5)

and the greatest weak solution−→x∗ = (x∗1, . . . , x

∗n) is given by

x∗i (t) = sup{li(t) : (l1, . . . , ln) weak lower solution of (6.1) in

[−→α,−→β]}

. (6.6)

Proof. Let−→L = (L1, . . . , Ln) ∈ [−→α,

−→β ] be a weak lower solution of (6.1), and let g ∈ L1

loc((t0, t0 +

L]) be as in (H3) and such that |L′i| ≤ g a.e. on I for all i ∈ {1, . . . , n}. Now let−→ξ∗ = (ξ∗1, . . . , ξ

∗n)

be defined for i ∈ {1, . . . , n} as

ξ∗i (t) = sup{li(t) : (l1, . . . , ln) weak lower solution in

[−→α,−→β],∣∣l′i

∣∣ ≤ g a.e. on I}. (6.7)

In particular,−→ξ∗ ≥ −→L. Further, every possible solution of (6.1) in [−→α,

−→β ] is less than or equal to

−→ξ∗ by (6.7) and (H3), independently of

−→L.

Claim 1 (−→ξ∗ ∈ C(I,Rn)∩ACloc((t0, t0 +L],Rn)). If (l1, . . . , ln) is a weak lower solution in [−→α,

−→β ]

with |l′i| ≤ g a.e. on I then for s, t ∈ (t0, t0 + L], s < t, we have

|li(t) − li(s)| ≤∫ t

s

g(r)dr, (6.8)

which implies

∣∣ξ∗i (t) − ξ∗i (s)

∣∣ ≤∫ t

s

g(r)dr, (6.9)

and, therefore, ξ∗i ∈ ACloc((t0, t0+L]). Further ξ∗i is continuous at t0 because αi(t) ≤ ξ∗i (t) ≤ βi(t)for all t ∈ I, αi and βi are continuous at t0 and αi(t0) = βi(t0).


Claim 2.−→ξ∗ is the greatest weak solution of (6.1) in [−→α,

−→β ]. For each weak lower solution

−→l ∈ [−→α,

−→β ] such that |l′i| ≤ g a.e., the quasimonotonicity of

−→f yields

l′i(t) ≤ fi(t,−→ξ∗(t) +

(li(t) − ξ∗i (t)

)−→ei)

for a. a. t ∈ I. (6.10)

Hence li is a weak lower solution between αi and βi of the scalar problem

x′ = fi(t,−→ξ∗(t) +

(x − ξ∗i (t)

)−→ei)

for a. a. t ∈ I, x(t0) = x0,i, (6.11)

and then Theorem 5.3 implies that li ≤ y∗i , where y∗i is the greatest weak solution of (6.11) in

[αi, βi]. Then−→ξ∗ ≤ −→y∗ = (y∗1, . . . , y

∗n).

On the other hand, we have

y∗i′(t) = fi

(t,−→ξ∗(t) +

(y∗i (t) − ξ

∗i (t)

)−→e i

)≤ fi

(t,−→y∗

)for a. a. t ∈ I, (6.12)

hence−→y∗ is a weak lower solution of (6.1) in [−→α,

−→β ] with |y′i| ≤ g a.e. on I, thus

−→ξ∗ ≥ −→y∗.

Therefore−→ξ∗ is a weak solution of (6.1), and, by (6.7) and (H3), it is the greatest one in [−→α,

−→β ].

In particular, the greatest weak solution of (6.1) in [−→α,−→β ] exists and it is greater than or equal

to−→L.

Claim 3. The greatest weak solution of (6.1) in [−→α,−→β ],−→x∗, satisfies (6.6). The weak lower

solution−→L ∈ [−→α,

−→β ] was fixed arbitrarily, so

−→x∗ is greater than or equal to any weak lower

solution in [−→α,−→β ]. On the other hand,

−→x∗ is a weak lower solution.

Analogously, the least weak solution of (6.1) in [−→α,−→β ] is given by (6.5).

7. Examples

Example 7.1. Let us show that the following singular and non-quasisemicontinous problemhas a unique positive Caratheodory solution:

x′ = f(x) =[

1x

], t ∈ I =

[0,

12

], x(0) = 0. (7.1)

Here square brackets mean integer part, and by positive solution we mean a solution whichis positive on (0, 1/2].

First note that (7.1) has at most one positive weak solution because the right hand sidein the differential equation is nonincreasing with respect to the unknown on (0,+∞), thus atno point can solutions bifurcate.

For all x > 0 we have [1/x] ≤ 1/x and therefore β(t) =√

2t, t ∈ I, is an upper solutionof (7.1) as it solves the majorant problem

x′ =1x, x(0) = 0. (7.2)


On the other hand it is easy to check that for 0 < x ≤ 1 we have [1/x] ≥ 1/(2x) andthen α(t) =

√t, t ∈ I, is a lower solution.

The function f is continuous between the graphs of α and β except over the linesγn(t) = 1/n, t ∈ [n−2/2, n−2], which are admissible nqsc curves for all n ∈ N, n ≥ 2 (note thatγ1 is not an admissible nqsc curve but it does not lie between α and β).

Finally, for r ∈ (0, 1) we have

∣∣f(x)

∣∣ ≤ n − 1 for max{r, α(t)} ≤ x ≤ β(t), (7.3)

where n ∈ N is such that 1/n < r.Therefore Theorem 5.4 implies the existence of a weak solution of (7.1) between α and

β. Moreover, this weak solution between α and β is increasing, so Proposition 5.2 ensures thatit is, in fact, a Caratheodory solution on I.

It is possible to extend the solution on the right of t = 1/2 to some t∗ where the solutionwill assume the value 1. The solution cannot be extended further on the right of t∗, as (7.1)with x(0) = 0 replaced by x(0) = 1 has no solution on the right of 0.

We owe to the anonymous referee the following remarks. Problem (7.1) isautonomous, so it falls inside the scope of the results in [21], which ensure that if we findα > 0 such that

∫α

0

ds

[1/s]< +∞, (7.4)

then (7.1) has a positive absolutely continuous solution defined implicitly by

∫x(t)

0

ds

[1/s]= t ∀t ∈

(0,

∫α

0

ds

[1/s]

). (7.5)

Since

∫1

0

ds

[1/s]=

π2 − 66

≈ 0.644934, (7.6)

we deduce that the solution x(t) is defined at least on [0, t∗], where t∗ = (π2 − 6)/6 andx(t∗) = 1.

Example 7.2. Let ε : [0, 1] → R be measurable and 0 < ε(t) ≤ 1 for a.a. t ∈ [0, 1]. We will provethat for each k ∈ R, k ≥ 1, the problem

x′ = f(t, x) =[

1xk

]−

[1tk

]+ ε(t), t ∈ I = [0, 1], x(0) = 0 (7.7)

has a unique positive Caratheodory solution.Note that the equation is not separable and f assumes positive and negative values on

every neighborhood of the initial condition. Moreover the equation is singular at the initialcondition with respect to both of its variables.


Once again the right hand side in the differential equation is nonincreasing withrespect to the unknown x on (0,+∞), thus we have at most one positive weak solution.

Lower and upper solutions are given by, respectively, α(t) = t/3 and β(t) = t for t ∈ I.For each t ∈ (0, 1] the function f(t, ·) is continuous between the graphs of α and β

except over the lines γn(t) = n−1/k, t ∈ [n−1/k, T], where T = min{3n−1/k, 1}, n ∈ N. Let usshow that f is positive between α and β, thus γn will be an admissible nqsc curve for eachn ∈ N. For t ∈ (0, 1] and (n + 1)−1/k < x ≤ n−1/k, n ∈ N, we have

f(t, x) = n −[

1tk

]+ ε(t), (7.8)

and if, moreover, we restrict our attention to those t > 0 such that α(t) ≤ x ≤ β(t) then wehave (n + 1)−1/k < t ≤ 3n−1/k which implies

[n

3

]≤

[1tk

]≤ n, (7.9)

and thus for t ∈ (0, 1], (n + 1)−1/k < x ≤ n−1/k, and α(t) ≤ x ≤ β(t), we have

ε(t) ≤ f(t, x) ≤ n −[n

3

]+ 1 ≤ 2

3n + 2. (7.10)

This shows that f is positive between α and β and, moreover, we can say that for r ∈ (0, 1)it suffices to take n ∈ N such that (n + 1)−1/k < r to have |f(t, x)| ≤ (2/3)n + 2 for all (t, x)between the graphs of α and β and r ≤ x ≤ 1/r.

Therefore Theorem 5.4 implies the existence of a weak solution of (7.7) between α andβ. Moreover, since f is positive between α and β the solution is increasing and, therefore, it isa Caratheodory solution.

The previous two examples fit the conditions of Theorems 5.3 and 5.4. Next we showan example where Theorem 5.3 can be used but it is not clear whether or not we can alsoapply Theorem 5.4.

Example 7.3. Let a > 0 be fixed and consider the problem

x′ = f(t, x) =[

1x + at

], t ∈ I = [0, 1], x(0) = 0. (7.11)

Lower and upper solutions are given by α ≡ 0 and β(t) =√

2t, t ∈ I. Since f isnonnegative between α and β the lines γn(t) = −at+1/n, n ∈ N, are admissible nqsc curves forthe differential equation. Finally it is easy to check that 0 ≤ f(t, x) ≤ n+1 if a−1(n+1)−1 ≤ t ≤ 1and x ≥ 0, thus one can construct g ∈ L1

loc((0, 1]) such that |f(t, x)| ≤ g(t) for a.a. t ∈ I and0 ≤ x ≤

√2t.

Theorem 5.3 ensures that (7.11) has extremal weak solutions between α and β.Moreover (7.11) has a unique solution between α and β as f is nonincreasing with respectto the unknown. Further, the unique solution is monotone and therefore it is a Caratheodorysolution.


Acknowledgments

The research of Rodrigo Lopez Pouso is partially supported by Ministerio de Educaciony Ciencia, Spain, Project MTM2007-61724, and by Xunta de Galicia, Spain, ProjectPGIDIT06PXIB207023PR.

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[8] E. R. Hassan and W. Rzymowski, “Extremal solutions of a discontinuous scalar differential equation,”Nonlinear Analysis: Theory, Methods & Applications, vol. 37, no. 8, pp. 997–1017, 1999.

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Research ArticleHomoclinic Solutions of Singular NonautonomousSecond-Order Differential Equations

Irena Rachunkova and Jan Tomecek

Department of Mathematical Analysis and Applications of Mathematics, Faculty of Science,Palacky University, 17 listopadu 12, 771 46 Olomouc, Czech Republic

Correspondence should be addressed to Irena Rachunkova, [email protected]

Received 27 April 2009; Revised 1 September 2009; Accepted 15 September 2009


This paper investigates the singular differential equation (p(t)u′)′ = p(t)f(u), having a singularityat t = 0. The existence of a strictly increasing solution (a homoclinic solution) satisfying u′(0) = 0,u(∞) = L > 0 is proved provided that f has two zeros and a linear behaviour near −∞.

Copyright q 2009 I. Rachunkova and J. Tomecek. 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.

1. Introduction

Having a positive parameter L, we consider the problem

(p(t)u′

)′ = p(t)f(u), (1.1)

u′(0) = 0, u(∞) = L, (1.2)

under the following basic assumptions for f and p

f ∈ Liploc((−∞, L]), f(0) = f(L) = 0, (1.3)

f(x) < 0 for x ∈ (0, L), (1.4)

there exists B < 0 such that f(x) > 0 for x ∈[B, 0), (1.5)

F(B)= F(L), where F(x) = −

∫x

0f(z) dz, (1.6)

p ∈ C([0,∞)) ∩ C1((0,∞)), p(0) = 0, (1.7)

p′(t) > 0, t ∈ (0,∞), limt→∞

p′(t)p(t)

= 0. (1.8)


Then problem (1.1), (1.2) generalizes some models arising in hydrodynamics or in thenonlinear field theory (see [1–5]). However (1.1) is singular at t = 0 because p(0) = 0.

Definition 1.1. If c > 0, then a solution of (1.1) on [0, c] is a function u ∈ C1([0, c]) ∩ C2((0, c])satisfying (1.1) on (0, c]. If u is a solution of (1.1) on [0, c] for each c > 0, then u is a solutionof (1.1) on [0,∞).

Definition 1.2. Let u be a solution of (1.1) on [0,∞). If u moreover fulfils conditions (1.2), it iscalled a solution of problem (1.1), (1.2).

Clearly, the constant function u(t) ≡ L is a solution of problem (1.1), (1.2). Animportant question is the existence of a strictly increasing solution of (1.1), (1.2) becauseif such a solution exists, many important physical properties of corresponding models can beobtained. Note that if we extend the function p(t) in (1.1) from the half–line onto R (as aneven function), then any solution of (1.1), (1.2) has the same limit L as t → −∞ and t → ∞.Therefore we will use the following definition.

Definition 1.3. A strictly increasing solution of problem (1.1), (1.2) is called a homoclinicsolution.

Numerical investigation of problem (1.1), (1.2), where p(t) = t2 and f(u) = 4λ2(u +1)u(u − L), λ > 0, can be found in [1, 4–6]. Problem (1.1), (1.2) can be also transformed ontoa problem about the existence of a positive solution on the half-line. For p(t) = tk, k ∈ N andfor p(t) = tk, k ∈ (1,∞), such transformed problem was solved by variational methods in[7, 8], respectively. Some additional assumptions imposed on f were needed there. Relatedproblems were solved, for example, in [9, 10].

Here, we deal directly with problem (1.1), (1.2) and continue our earlier considerationsof papers [11, 12], where we looked for additional conditions which together with (1.3)–(1.8)would guarantee the existence of a homoclinic solution.

Let us characterize some results reached in [11, 12] in more details. Both these papersassume (1.3)–(1.8). In [11] we study the case that f has at least three zeros L0 < 0 < L. Moreprecisely, the conditions,

f(L0) = 0, there exists δ > 0 such that f ∈ C1((−δ, 0)), limx→ 0−

f ′(x) < 0,

p ∈ C2((0,∞)), limt→∞

p′′(t)p(t)

= 0,(1.9)

are moreover assumed. Then there exist c > 0, B ∈ (L0, 0), and a solution u of (1.1) on [0, c]such that

u(0) = B, u′(0) = 0, (1.10)

u′(t) > 0 for t ∈ (0, c], u(c) = L. (1.11)

We call such solution an escape solution. The main result of [11] is that (under (1.3)–(1.8),(1.9)) the set of solutions of (1.1), (1.10) for B ∈ (L0, 0) consists of escape solutions andof oscillatory solutions (having values in (L0, L)) and of at least one homoclinic solution.


In [12] we omit assumptions (1.9) and prove that assumptions (1.3)–(1.8) are sufficient for theexistence of an escape solution and also for the existence of a homoclinic solution providedthe p fulfils

∫1

0

dsp(s)

<∞. (1.12)

If (1.12) is not valid, then the existence of both an escape solution and a homoclinic solutionis proved in [12], provided that f satisfies moreover

f(x) > 0 for x < 0, (1.13)

limx→−∞

|x|f(x)

=∞. (1.14)

Assumption (1.13) characterizes the case that f has just two zeros 0 and L in the interval(−∞, L]. Further, we see that if (1.14) holds, then f is either bounded on (−∞, L] or f isunbounded earlier and has a sublinear behaviour near −∞.

This paper also deals with the case that f satisfies (1.13) and is unbounded above on(−∞, L]. In contrast to [12], here we prove the existence of a homoclinic solution for f havinga linear behaviour near −∞. The proof is based on a full description of the set of all solutionsof problem (1.1), (1.10) for B < 0 and on the existence of an escape solutions in this set.

Finally, we want to mention the paper [13], where the problem

1p(t)(p(t)u′(t)

)′ = f(t, u(t), p(t)u′(t)

),

u(0) = ρ0 ∈ (−1, 0), limt→∞

u(t) = ξ ∈ (0, 1),

limt→∞

p(t)u′(t) = 0

(1.15)

is investigated under the assumptions that f is continuous, it has three distinct zeros andsatisfies the sign conditions similar to those in [11, (3.4)]. In [13], an approach quite differentfrom [11, 12] is used. In particular, by means of properties of the associated vector field(u(t), p(t)u′(t)) together with the Kneser’s property of the cross sections of the solutions’funnel, the authors provide conditions which guarantee the existence of a strictly increasingsolution of (1.15). The authors apply this general result to problem

1tn−1

(tn−1u′

)′= 4λ2(u + 1)u(u − ξ),

limt→ 0+

tn−1u′(t) = 0, limt→∞

u(t) = ξ,

(1.16)

and get a strictly increasing solution of (1.16) for a sufficiently small ξ. This corresponds tothe results of [11], where ξ ∈ (0, 1) may be arbitrary.


2. Initial Value Problem

In this section, under the assumptions (1.3)–(1.8) and (1.13) we prove some basic propertiesof solutions of the initial value problem (1.1), (1.10), where B < 0.

Lemma 2.1. For each B < 0 there exists a maximal c∗ ∈ (0,∞] such that problem (1.1), (1.10) has aunique solution u on [0, c∗) and

u(t) ≥ B for t ∈ [0, c∗). (2.1)

Further, for each b ∈ (0, c∗), there existsMb > 0 such that

|u(t)| +∣∣u′(t)

∣∣ ≤Mb, t ∈ [0, b],

∫b

0

p′(s)p(s)

∣∣u′(s)

∣∣ds ≤Mb. (2.2)

Proof. Let u be a solution of problem (1.1), (1.10) on [0, c) ⊂ [0,∞). By (1.1), we have

u′′(t) +p′(t)p(t)

u′(t) − f(u(t)) = 0 for t ∈ (0, c), (2.3)

and multiplying by u′ and integrating between 0 and t, we get

u′2(t)2

+∫ t

0

p′(s)p(s)

u′2(s)ds + F(u(t)) = F(B), t ∈ (0, c). (2.4)

Let u(t1) < B for some t1 ∈ (0, c). Then (2.4) yields F(u(t1)) ≤ F(B), which is not possible,because F is decreasing on (−∞, 0). Therefore u(t) ≥ B for t ∈ [0, c).

Let η > 0. Consider the Banach space C([0, η]) (with the maximum norm) and anoperator F : C([0, η]) → C([0, η]) defined by

(Fu)(t) = B +∫ t

0

1p(s)

∫ s

0p(τ)f(u(τ))dτ ds. (2.5)

A function u is a solution of problem (1.1), (1.2) on [0, η] if and only if it is a fixed point of theoperator F. Using the Lipschitz property of f we can prove that the operator is contractivefor each sufficiently small η and from the Banach Fixed Point Theorem we conclude that thereexists exactly one solution of problem (1.1), (1.2) on [0, η]. This solution u has the form

u(t) = B +∫ t

0

1p(s)

∫ s

0p(τ)f(u(τ))dτ ds (2.6)

for t ∈ [0, η]. Hence, u can be extended onto each interval [0, b] where u is bounded. So, wecan put c∗ = sup{b > 0 : u is bounded on [0, b]}.


Let b ∈ (0, c∗). Then there exists M ∈ (0,∞) such that |f(u(t))| ≤ M for t ∈ [0, b]. So,(2.6) yields

∣∣u′(t)

∣∣ ≤ M

1p(t)

∫ t

0p(s)ds, t ∈ (0, b]. (2.7)

Put

ϕ(t) =1

p(t)

∫ t

0p(s)ds, ψ(t) =

∫b

t

p′(s)p2(s)

∫s

0p(τ)dτ ds, t ∈ (0, b]. (2.8)

Then

0 < ϕ(t) ≤ t for t ∈ (0, b], (2.9)

and, by “per partes” integration we derive limt→ 0+ψ(t) = b − ϕ(b). Multiplying (2.7) byp′(t)/p(t) and integrating it over (0, b), we get

∫b

0

p′(t)p(t)

∣∣u′(t)∣∣dt ≤ M

∫b

0

p′(t)p2(t)

∫ t

0p(s)dsdt = M

(b − ϕ(b)

). (2.10)

Estimates (2.2) follow from (2.7)–(2.10) for

Mb = Mb + |B| + Mb2. (2.11)

Remark 2.2. The proof of Lemma 2.1 yields that if c∗ <∞, then limt→ c∗u(t) =∞.Let us put

f(x) =

⎧⎨

⎩

0 for x > L,

f(x) for x ≤ L,(2.12)

and consider an auxiliary equation

(p(t)u′

)′ = p(t)f(u). (2.13)

Similarly as in the proof of Lemma 2.1 we deduce that problem (2.13), (1.10) has a uniquesolution on [0,∞). Moreover the following lemma is true.

Lemma 2.3 ([12]). For each B0 < 0, b > 0 and each ε > 0, there exists δ > 0 such that for any B1,B2 ∈ [B0, 0)

|B1 − B2| < δ =⇒ |u1(t) − u2(t)| +∣∣u′1(t) − u

′2(t)∣∣ < ε, t ∈ [0, b]. (2.14)

Here ui is a solution of problem (2.13), (1.10) with B = Bi, i = 1, 2.


Proof. Choose B0 < 0, b > 0, ε > 0. Let K > 0 be the Lipschitz constant for f on [B0, L]. By (2.6)for f = f , B = Bi, u = ui, i = 1, 2,

|u1(t) − u2(t)| ≤ |B1 − B2| +∫ t

0

1p(s)

∫ s

0p(τ)

∣∣∣f(u1(τ)) − f(u2(τ))

∣∣∣dτ ds

≤ |B1 − B2| +Kt

∫ t

0|u1(τ) − u2(τ)|dτ

≤ |B1 − B2| +Kb

∫ t

0|u1(τ) − u2(τ)|dτ, t ∈ [0, b].

(2.15)

From the Gronwall inequality, we get

|u1(t) − u2(t)| ≤ |B1 − B2|eKb2, t ∈ [0, b]. (2.16)

Similarly, by (2.6), (2.9), and (2.16),

∣∣u′1(t) − u′2(t)∣∣ ≤ 1

p(t)

∫ t

0p(s)

∣∣∣f(u1(s)) − f(u2(s))∣∣∣ds

≤ K1

p(t)

∫ t

0p(s)|u1(s) − u2(s)|ds

≤ Kb|B1 − B2|eKb2, t ∈ [0, b].

(2.17)

If we choose δ > 0 such that

δ <ε

(1 +Kb)eKb2 , (2.18)

we get (2.14).

Remark 2.4. Choose a ≥ 0 and C ≤ L, and consider the initial conditions

u(a) = C, u′(a) = 0. (2.19)

Arguing as in the proof of Lemma 2.1, we get that problem (2.13), (2.19) has a unique solutionon [a,∞). In particular, for C = 0 and C = L, the unique solution of problem (2.13), (2.19)(and also of problem (1.1), (2.19)) is u ≡ 0 and u ≡ L, respectively.

Lemma 2.5. Let u be a solution of problem (1.1), (1.10). Assume that there exists a ≥ 0 such that

u(t) < 0 for t ≥ a, u′(a) = 0. (2.20)


Then u′(t) > 0 for t > a and

limt→∞

u(t) = 0, limt→∞

u′(t) = 0. (2.21)

Proof. By (1.13) and (2.20), f(u(t)) > 0 on [a,∞) and thus p(t)u′(t) and u′(t) are positive on(a,∞). Consequently, there exists limt→∞u(t) = B1 ∈ (u(a), 0]. Further, by (1.1),

u′′(t) +p′(t)p(t)

u′(t) = f(u(t)), t > 0, (2.22)

and, by multiplication and integration over [a, t],

u′2(t)2

+∫ t

a

p′(s)p(s)

u′2(s)ds = F(u(a)) − F(u(t)), t > a. (2.23)

Therefore,

0 ≤ limt→∞

∫ t

a

p′(s)p(s)

u′2(s)ds ≤ F(u(a)) − F(B1) <∞, (2.24)

and hence limt→∞u′2(t) exists. Since u is bounded on [0,∞), we get

limt→∞

u′2(t) = limt→∞

u′(t) = 0. (2.25)

By (1.3), (1.8), and (2.22), limt→∞u′′(t) exists and, since u′ is bounded on [0,∞), we get

limt→∞u′′(t) = 0. Hence, letting t → ∞ in (2.22), we obtain f(B1) = 0. Therefore, B1 = 0

and (2.21) is proved.

Lemma 2.6. Let u be a solution of problem (1.1), (1.10). Assume that there exist a1 > 0 and A1 ∈(0, L) such that

u(t) > 0 ∀t > a1, u(a1) = A1, u′(a1) = 0. (2.26)

Then u′(t) < 0 for all t > a1 and (2.21) holds.

Proof. Since u fulfils (2.26), we can find a maximal b > a1 such that 0 < u(t) < L for t ∈ [a1, b)and consequently f(u(t)) = f(u(t)) for t ∈ [a1, b). By (4.23) and (2.26), f(u(t)) < 0 on [a1, b)and thus p(t)u′(t) and u′(t) are negative on (a1, b). So, u is positive and decreasing on [a1, b)which yields b =∞ (otherwise, we get u(b) = 0, contrary to (2.26)). Consequently there existslimt→∞u(t) = L1 ∈ [0, A1). By multiplication and integration (2.22) over [a1, t], we obtain

u′2(t)2

+∫ t

a1

p′(s)p(s)

u′2(s)ds = F(A1) − F(u(t)), t > a1. (2.27)


By similar argument as in the proof of Lemma 2.5 we get that limt→∞u′(t) = 0 and L1 = 0.

Therefore (2.21) is proved.

3. Damped Solutions

In this section, under assumptions (1.3)–(1.8) and (1.13) we describe a set of all dampedsolutions which are defined in the following way.

Definition 3.1. A solution of problem (1.1), (1.10) (or of problem (2.13), (1.10)) on [0,∞) iscalled damped if

sup{u(t) : t ∈ [0,∞)} < L. (3.1)

Remark 3.2. We see, by (2.12), that u is a damped solution of problem (1.1), (1.10) if and onlyif u is a damped solution of problem (2.13), (1.10). Therefore, we can borrow the argumentsof [12] in the proofs of this section.

Theorem 3.3. If u is a damped solution of problem (1.1), (1.10), then u has a finite number of isolatedzeros and satisfies (2.21); or u is oscillatory (it has an unbounded set of isolated zeros).

Proof. Let u be a damped solution of problem (1.1), (1.10). By Remark 2.2, we have c∗ =∞ inLemma 2.1 and hence

u(t) ≥ B for t ∈ [0,∞). (3.2)

Step 1. If u has no zero in (0,∞), then u(t) < 0 for t ≥ 0 and, by Lemma 2.5, u fulfils (2.21).

Step 2. Assume that θ > 0 is the first zero of u on (0,∞). Then, due to Remark 2.4, u′(θ) > 0.Let u(t) > 0 for t ∈ (θ,∞). By virtue of (1.4), f(u(t)) < 0 for t ∈ (θ,∞) and thus p(t)u′(t)is decreasing. Let u′ be positive on (θ,∞). Then u′ is also decreasing, u is increasing andlimt→∞u(t) = L ∈ (0, L), due to (3.1). Consequently, limt→∞u

′(t) = 0. Letting t → ∞ in (2.22),we get limt→∞u

′′(t) = f(L) < 0, which is impossible because u′ is bounded below. Thereforethere are a1 > θ and A1 ∈ (0, L) satisfying (2.26) and, by Lemma 2.6, either u fulfils (2.21) oru has the second zero θ1 > a1 with u′(θ1) < 0. So u is positive on (θ, θ1) and has just one localmaximum A1 = u(a1) in (θ, θ1). Moreover, putting a = 0 and t = a1 in (2.23), we have

0 <

∫a1

0

p′(s)p(s)

u′2(s)ds = F(B) − F(A1), (3.3)

and hence

F(A1) < F(B). (3.4)


Step 3. Let u have no other zeros. Then u(t) < 0 for t ∈ (θ1,∞). Assume that u′ is negativeon [θ1,∞). Then, due to (2.1), limt→∞u(t) = L ∈ [B, 0). Putting a = a1 in (2.23) and lettingt → ∞, we obtain

0 < limt→∞

[u′2(t)

2+∫ t

a1

p′(s)p(s)

u′2(s)ds

]

= F(A1) − F(L). (3.5)

Therefore, limt→∞u′2(t) exists and, since u is bounded, we deduce that

limt→∞

u′(t) = 0. (3.6)

Letting t → ∞ in (2.22), we get limt→∞u′′(t) = f(L) > 0, which contradicts the fact that u′

is bounded above. Therefore, u′ cannot be negative on the whole interval [θ1,∞) and thereexists b1 > θ1 such that u′(b1) = 0. Moreover, according to (3.2), u(b1) ∈ [B, 0).

Then, Lemma 2.5 yields that u fulfils (2.21). Since u′ is positive on (b1,∞), u has justone minimum B1 = u(b1) on (θ1,∞). Moreover, putting a = a1 and t = b1 in (2.23), we have

0 <

∫b1

a1

p′(s)p(s)

u′2(s)ds = F(A1) − F(B1), (3.7)

which together with (3.4) yields

F(B1) < F(A1) < F(B). (3.8)

Step 4. Assume that u has its third zero θ2 > θ1. Then we prove as in Step 2 that u has just onenegative minimum B1 = u(b1) in (θ1, θ2) and (3.8) is valid. Further, as in Step 2, we deducethat either u fulfils (2.21) or u has the fourth zero θ3 > θ2, u is positive on (θ2, θ3) with justone local maximum A2 = u(a2) < L on (θ2, θ3), and F(A2) < F(B1). This together with (3.8)yields

F(A2) < F(B1) < F(A1) < F(B). (3.9)

If u has no other zeros, we deduce as in Step 3 that u has just one negative minimum B2 =u(b2) in (θ3,∞), F(B2) < F(A2) and u fulfils (2.21).

Step 5. If u has other zeros, we use the previous arguments and get that either u has a finitenumber of zeros and then fulfils (2.21) or u is oscillatory.

Remark 3.4. According to the proof of Theorem 3.3, we see that if u is oscillatory, it has justone positive local maximum between the first and the second zero, then just one negativelocal minimum between the second and the third zero, and so on. By (3.8), (3.9), (1.4)–(1.6)and (1.13), these maxima are decreasing (minima are increasing) for t increasing.


Lemma 3.5. A solution u of problem (1.1), (1.10) fulfils the condition

sup{u(t) : t ∈ [0,∞)} = L (3.10)

if and only if u fulfils the condition

limt→∞

u(t) = L, u′(t) > 0 for t ∈ (0,∞). (3.11)

Proof. Assume that u fulfils (3.10). Then there exists θ ∈ (0,∞) such that u(θ) = 0, u′(t) > 0for t ∈ (0, θ]. Otherwise sup{u(t) : t ∈ [0,∞)} = 0, due to Lemma 2.5. Let a1 ∈ (θ,∞) be suchthat u′(t) > 0 on (θ, a1), u′(a1) = 0. By Remark 2.4 and (3.10), u(a1) ∈ (0, L). Integrating (1.1)over (a1, t), we get

u′(t) =1

p(t)

∫ t

a1

p(s)f(u(s))ds, ∀t > a1. (3.12)

Due to (1.4), we see that u is strictly decreasing for t > a1 as long as u(t) ∈ (0, L). Thus,there are two possibilities. If u(t) > 0 for all t > a1, then from Lemma 2.6 we get (2.21),which contradicts (3.10). If there exists θ1 > a1 such that u(θ1) = 0, then in view Remark 2.4we have u′(θ1) < 0. Using the arguments of Steps 3–5 of the proof of Theorem 3.3, we getthat u is damped, contrary to (3.10). Therefore, such a1 cannot exist and u′ > 0 on (0,∞).Consequently, limt→∞u(t) = L. So, u fulfils (3.11). The inverse implication is evident.

Remark 3.6. According to Definition 1.3 and Lemma 3.5, u is a homoclinic solution of problem(1.1), (1.10) if and only if u is a homoclinic solution of problem (2.13), (1.10).

Theorem 3.7 (on damped solutions). Let B satisfy (1.5) and (1.6). Assume that u is a solution ofproblem (1.1), (1.10) with B ∈ [B, 0). Then u is damped.

Proof. Let u be a solution of (1.1), (1.10) with B ∈ [B, 0). Then, by (1.4)–(1.6),

F(B) ≤ F(L). (3.13)

Assume on the contrary that u is not damped. Then u is defined on the interval [0,∞) andsup{u(t) : t ∈ [0,∞)} = L or there exists b ∈ (0,∞) such that u(b) = L, u′(b) > 0, and u(t) < Lfor t ∈ [0, b). If the latter possibility occurs, (2.22) and (3.13) give by integration

0 <u′2(b)

2+∫b

0

p′(s)p(s)

u′2(s)ds = F(B) − F(L) ≤ 0, (3.14)

a contradiction. If sup{u(t) : t ∈ [0,∞)} = L, then, by Lemma 3.5, u fulfils (3.11). So u has aunique zero θ > 0. Integrating (2.22) over [0, θ], we get

u′2(θ)2

+∫θ

0

p′(s)p(s)

u′2(s)ds = F(B), (3.15)


and so

u′2(θ) < 2F(B). (3.16)

Integrating (2.22) over [θ, t], we obtain for t > θ

u′2(t)2− u′2(θ)

2+∫ t

θ

p′(s)p(s)

u′2(s)ds = F(u(θ)) − F(u(t)) = −F(u(t)). (3.17)

Therefore, u′2(θ) > 2F(u(t)) on (θ,∞), and letting t → ∞, we get u′2(θ) ≥ 2F(L). This togetherwith (3.16) contradicts (3.13). We have proved that u is damped.

Theorem 3.8. Let Md be the set of all B < 0 such that corresponding solutions of problem (1.1),(1.10) are damped. ThenMd is open in (−∞, 0).

Proof. Let B0 ∈ Md and u0 be a solution of (1.1), (1.10) with B = B0. So, u0 is damped and u0

is also a solution of (2.13).(a) Let u0 be oscillatory. Then its first local maximum belongs to (0, L). Lemma 2.3

guarantees that if B is sufficiently close to B0, the corresponding solution u of (2.13), (1.10)has also its first local maximum in (0, L). This means that there exist a1 > 0 and A1 ∈ (0, L)such that u satisfies (2.26). Now, we can continue as in the proof of Theorem 3.3 using thearguments of Steps 2–5 and Remark 3.2 to get that u is damped.

(b) Let u0 have at most a finite number of zeros. Then, by Theorem 3.3, u0 fulfils (2.21).Choose c0 ∈ (0, F(L)/3). Since u0 fulfils (2.22), we get by integration over [0, t]

u′20 (t)2

+∫ t

0

p′(s)p(s)

u′20 (s)ds = F(B0) − F(u0(t)), t > 0. (3.18)

For t → ∞, we get, by (2.21),

∫∞

0

p′(s)p(s)

u′20 (s)ds = F(B0). (3.19)

Therefore, we can find b > 0 such that

∫∞

b

p′(s)p(s)

u′20 (s)ds < c0. (3.20)

Let Mb be the constant of Lemma 2.1. Choose ε ∈ (0, c0/2Mb). Assume that B < 0 and uis a corresponding solution of problem (2.13), (1.10). Using Lemma 2.1, Lemma 2.3 and thecontinuity of F, we can find δ > 0 such that if |B − B0| < δ, then

|F(B) − F(B0)| < c0, (3.21)


moreover |u′0(t) − u′(t)| < ε for t ∈ [0, b] and

∫b

0

p′(s)p(s)

∣∣∣u′20 (s) − u

′2(s)∣∣∣ds ≤ max

t∈[0,b]

∣∣u′0(t) − u′(t)

∣∣∫b

0

p′(s)p(s)

(∣∣u′0(s)∣∣ +∣∣u′(s)

∣∣)ds

≤ ε · 2Mb <c0

2Mb2Mb = c0.

(3.22)

Therefore, we have

∫b

0

p′(s)p(s)

∣∣∣u′20 (s) − u

′2(s)∣∣∣ds < c0. (3.23)

Consequently, integrating (2.13) over [0, t] and using (3.19)–(3.23), we get for t ≥ b

F(B) − F(u(t)) =∫ t

0

p′(s)p(s)

u′2(s)ds +u′2(t)

2≥∫ t

0

p′(s)p(s)

u′2(s)ds

≥∫b

0

p′(s)p(s)

u′2(s)ds =∫b

0

p′(s)p(s)

(u′2(s) − u′20 (s)

)ds

+∫b

0

p′(s)p(s)

u′20 (s)ds > −c0 +∫b

0

p′(s)p(s)

u′20 (s)ds

= −c0 +∫∞

0

p′(s)p(s)

u′20 (s)ds −∫∞

b

p′(s)p(s)

u′20 (s)ds

> −c0 + F(B0) − c0 = −2c0 + F(B0) − F(B) + F(B)

> −3c0 + F(B).

(3.24)

We get F(u(t)) < 3c0 < F(L) for t ≥ b. Therefore, F(u(t)) = F(u(t)) for t ≥ b and, due to(1.4)–(1.6),

sup{u(t) : t ∈ [b,∞)} < L. (3.25)

Assume that there is b0 ∈ (0, b) such that u(b0) = L, u′(b0) > 0. Then, since (p(t)u′(t))′ = 0 ift > b0 and u(t) > L, we get u′(t) > 0 and u(t) > L for t > b0, contrary to (3.25). Hence we getthat u fulfils (3.1).

4. Escape Solutions

During the whole section, we assume (1.3)–(1.8) and (1.13). We prove that problem (1.1),(1.10) has at least one escape solution. According to Section 1 and Remark 2.2, we work withthe following definitions.


Definition 4.1. Let c > 0. A solution of problem (1.1), (1.10) on [0, c] is called an escape solutionif

u(c) = L, u′(t) > 0 for t ∈ (0, c]. (4.1)

Definition 4.2. A solution u of problem (2.13), (1.10) is called an escape solution, if there existsc > 0 such that

u(c) = L, u′(t) > 0 for t ∈ (0,∞). (4.2)

Remark 4.3. If u is an escape solution of problem (2.13), (1.10), then u is an escape solution ofproblem (1.1), (1.10) on some interval [0, c].

Theorem 4.4 (on three types of solutions.). Let u be a solution of problem (1.1), (1.10). Then u isjust one of the following three types

(I) u is damped;

(II) u is homoclinic;

(III) u is escape.

Proof. By Definition 3.1, u is damped if and only if (3.1) holds. By Lemma 3.5 andDefinition 1.3, u is homoclinic if and only if (3.10) holds. Let u be neither damped norhomoclinic. Then there exists c > 0 such that u is bounded on [0, c], u(c) = L, u′(c) > 0. So, uhas its first zero θ ∈ (0, c) and u′(t) > 0 on (0, θ]. Assume that there exist a1 ∈ (θ, c) such thatu(a1) ∈ (0, L) and u′(a1) = 0. Then, by Lemma 2.6, either u fulfils (2.21) or u has its secondzero and, arguing as in Steps 2–5 of the proof of Theorem 3.3, we deduce that u is a dampedsolution. This contradiction implies that u′(t) > 0 on (0, c]. Therefore, by Definition 4.1, u isan escape solution.

Theorem 4.5. LetMe ⊂ (−∞, 0) be the set of all B such that the corresponding solutions of (1.1),(1.10) are escape solutions. The setMe is open in (−∞, 0).

Proof. Let B0 ∈ Me and u0 be a solution of problem (1.1), (1.10) with B = B0. So, u0 fulfils (4.1)for some c > 0. Let u0 be a solution of problem (2.13), (1.10) with B = B0. Then u0 = u0 on[0, c] and u0 is increasing on [c,∞). There exists ε > 0 and c0 > c such that u0(c0) = L + ε. Letu1 be a solution of problem (2.13), (1.10) for some B1 < 0. Lemma 2.3 yields δ > 0 such thatif |B1 − B0| < δ, then u1(c0) > u0(c0) − ε = L. Therefore, u1 is an escape solution of problem(2.13), (1.10). By Remark 4.3, u1 is also an escape solution of problem (1.1), (1.10) on someinterval [0, c1] ⊂ [0, c0].

To prove that the setMe of Theorem 4.5 is nonempty we will need the following twolemmas.

Lemma 4.6. Let B < 0. Assume that u is a solution of problem (1.1), (1.10) on [0, b) and [0, b) is amaximal interval where u is increasing and u(t) ∈ [B, L] for t ∈ [0, b). Then

∫ t

02F(u(s))p(s)p′(s)ds = F(u(t))p2(t) +

12p2(t)u′2(t), t ∈ (0, b). (4.3)


Proof.

Step 1. We show that the interval (0, b) is nonempty. Since u(0) = B < 0 and f satisfies (1.3),(1.13), we can find θ > 0 such that

u(t) < 0, f(u(t)) > 0 for t ∈ (0, θ). (4.4)

Integrating (1.1) over (0, t), we obtain

u′(t) =1

p(t)

∫ t

0p(s)f(u(s))ds > 0 for t ∈ (0, θ]. (4.5)

So, u is an increasing solution of problem (1.1), (1.10) on [0, θ] and u(t) ∈ [B, 0] for t ∈ [0, θ].Therefore the nonempty interval [0, b) exists.

Step 2. By multiplication of (1.1) by pu′ and integration over (0, t), we obtain

12p2(t)u′2(t) =

∫ t

0f(u(s))u′(s)p2(s)ds, t ∈ (0, b). (4.6)

Using the “per partes” integration, we get for t ∈ (0, b)

∫ t

0f(u(s))u′(s)p2(s)ds = −F(u(t))p2(t) +

∫ t

02F(u(s))p(s)p′(s)ds. (4.7)

This relation together with (4.6) implies (4.3).

Remark 4.7. Consider a solution u of Lemma 4.6. If u is an escape solution, then b < ∞.Assume that u is not an escape solution. Then both possibilities b < ∞ and b = ∞ can occur.Let b < ∞. By Theorem 4.4 and Lemma 2.5, u(b) ∈ (0, L), u′(b) = 0. Let b = ∞. We writeu(b) = limt→∞u(t), u′(b) = limt→∞u

′(t). Using Lemmas 3.5 and 2.5 and Theorem 4.4, weobtain u′(b) = 0 and either u(b) = 0 or u(b) = L.

Lemma 4.8. Let C < B and let {Bn}∞n=1 ⊂ (−∞, C). Then for each n ∈ N :

(i) there exists a solution un of problem (1.1), (1.10) with B = Bn,

(ii) there exists bn > 0 such that [0, bn) is the maximal interval on which the solution un isincreasing and its values in this interval are contained in [Bn, L],

(iii) there exists γn ∈ (0, bn) satisfying un(γn) = C.

If the sequence {γn}∞n=1 is unbounded, then there exists � ∈ N such that u� is an escape solution.


Proof. Similar arugmets can be found in [12]. By Lemma 2.1, the assertion (i) holds. Thearguments in Step 1 of the proof of Lemma 4.6 imply (ii). The strict monotonicity of un andRemark 4.7 yields a unique γn. Assume that {γn}∞n=1 is unbounded. Then

limn→∞

γn =∞, γn < bn, n ∈ N (4.8)

(otherwise, we take a subsequence). Assume on the contrary that for any n ∈ N, un is not anescape solution. Choose n ∈ N. Then, by Remark 4.7,

un(bn) ∈ [0, L], u′n(bn) = 0. (4.9)

Due to (4.9), (1.2) and (ii) there exists γn ∈ [γn, bn) satisfying

u′n(γn)= max

{u′n(t) : t ∈

[γn, bn

)}. (4.10)

By (i) and (ii), un satisfies

u′′n(t) +p′(t)p(t)

u′n(t) = f(un(t)), t ∈ (0, bn). (4.11)

Integrating it over [0, t], we get

u′2n (t)2

+ F(un(t)) = F(Bn) −∫ t

0

p′(s)p(s)

u′2n (s)ds, t ∈ (0, bn). (4.12)

Put

En(t) =u′2n (t)

2+ F(un(t)), t ∈ (0, bn). (4.13)

Then, by (4.12),

dEn(t)dt

= −p′(t)p(t)

u′2n (t) < 0, t ∈ (0, bn). (4.14)

We see that En is decreasing. From (1.4) and (1.6) we get that F is increasing on [0, L] andconsequently by (4.9) and (4.13), we have

En

(γn)> F(un

(γn))

= F(C), En(bn) = F(un(bn)) ≤ F(L). (4.15)

Integrating (4.14) over (γn, bn) and using (4.10), we obtain

En

(γn)− En(bn) =

∫bn

γn

p′(t)p(t)

u′2n (t)dt ≤ u′n(γn)(L − C)Kn, (4.16)


where

Kn = sup{p′(t)p(t)

: t ∈[γn, bn

)}∈ (0,∞). (4.17)

Further, by (4.15),

F(C) < En

(γn)≤ F(L) + u′n

(γn)(L − C)Kn, (4.18)

F(C) − F(L)L − C · 1

Kn< u′n

(γn). (4.19)

Conditions (1.8) and (4.8) yield limn→∞Kn = 0, which implies

limn→∞

u′n(γn)=∞. (4.20)

By (4.13) and (4.18),

u′2n(γn)

2≤ En

(γn)≤ En

(γn)≤ F(L) + u′n

(γn)(L − C)Kn, (4.21)

and consequently

u′n(γn)(

12u′n(γn)− (L − C)Kn

)≤ F(L) <∞, n ∈ N, (4.22)

which contradicts (4.20). Therefore, at least one escape solution of (1.1), (1.10) with B < Bmust exist.

Theorem 4.9 (on escape solution.). Assume that (1.3)–(1.8) and (1.13) hold and let

0 < lim infx→−∞

|x|f(x)

<∞. (4.23)

Then there exists B < B such that the corresponding solution of problem (1.1), (1.10) is an escapesolution.

Proof. Let C < B and let {Bn}∞n=1, {un}∞n=1, {bn}∞n=1, and {γn}∞n=1 be sequences from Lemma 4.8.Moreover, let

limn→∞

Bn = −∞. (4.24)

By (4.24) we can find n0 ∈ N such that Bn < 2C for n ≥ n0. We assume that for any n ∈ N, un

is not an escape solution and we construct a contradiction.


Step 1. We derive some inequality for u′n. By Remark 4.7, we have

un(bn) ∈ [0, L], u′n(bn) = 0, n ∈ N, (4.25)

and, by Lemma 4.8, the sequence {γn}∞n=1 is bounded. Therefore there exists Γ ∈ (0,∞) suchthat

γn ≤ Γ, n ∈ N. (4.26)

Choose an arbitrary n ≥ n0. According to Lemma 4.6, un satisfies equality (4.3), that is

∫ t

02F(un(s))p(s)p′(s)ds = F(un(t))p2(t) +

12p2(t)u′2n (t), t ∈ (0, bn). (4.27)

Since un(0) = Bn < 2C < 0 and un is increasing on [0, bn), there exists a unique γn ∈ (0, γn)such that

un

(γn)=

12Bn < C = un

(γn). (4.28)

Having in mind, due to (1.4)–(1.8), that the inequality

F(un(t))p(t)p′(t) ≥ 0 for t ∈ [0, bn) (4.29)

holds, we get

∫ t

02F(un(s))p(s)p′(s)ds >

∫ γn

02F(un(s))p(s)p′(s)ds, t ∈

[γn, bn

). (4.30)

By virtue of (1.6) and (1.13), we see that F is decreasing on (−∞, 0), which yields

min{F(un(t)) : t ∈

[0, γn

]}= F(un

(γn))

= F

(Bn

2

). (4.31)

Hence,

∫ t

02F(un(s))p(s)p′(s)ds > F

(Bn

2

)p2(γn

), t ∈

[γn, bn

). (4.32)

Since un(γn) = C and un(bn) ∈ [0, L], the monotonicity of un yields un(t) ∈ [C, L] for t ∈[γn, bn], and consequently

max{F(un(t)) : t ∈

[γn, bn

)}= F(C). (4.33)


Therefore (4.27) and (4.32) give

F

(Bn

2

)p2(γn

)

p2(t)< F(C) +

12u′2n (t), t ∈

[γn, bn

). (4.34)

Step 2. We prove that the sequence {γn}∞n=1 is bounded below by some positive number. Since

un is a solution of (1.1) on [0, bn), we have

(p(t)u′n(t)

)′ = p(t)f(un(t)), t ∈(0, γn

). (4.35)

Integrating it, we get

u′n(t) =1

p(t)

∫ t

0p(s)f(un(s))ds ≤ f(σnBn)

P(t)p(t)

, t ∈(0, γn

), (4.36)

where σn ∈ [1/2, 1] satisfies f(σnBn) = max{f(x) : x ∈ [Bn, (1/2)Bn]} and P(t) =∫ t

0p(s) ds.Having in mind (1.8), we see that p is increasing and 0 < P(t)/p(t) ≤ t for t ∈ (0,∞).Consequently

limt→ 0+

∫ t

0

P(s)p(s)

ds = 0. (4.37)

Integrating (4.36) over (0, γn), we obtain

12Bn − Bn ≤ f(σnBn)

∫ γn

0

P(s)p(s)

ds, (4.38)

and hence

∫ γn

0

P(s)p(s)

ds ≥ 12|Bn|

f(σnBn). (4.39)

By (4.23) we get

lim infn→∞

12|Bn|

f(σnBn)= lim inf

n→∞

12σn

|σnBn|f(σnBn)

> 0, (4.40)

which, due to (4.39), yields

lim infn→∞

∫ γn

0

P(s)p(s)

ds > 0. (4.41)

So, by virtue of (4.37), there exists γ0 > 0 such that γn ≥ γ0 for n ≥ n0.


Step 3. We construct a contradiction. Putting γ0 in (4.34), we have

F

(Bn

2

)p2(γ0

)

p2(t)− F(C) < 1

2u′2n (t), t ∈

[γn, bn

). (4.42)

Due to (4.23), limx→−∞f(x) =∞. Therefore, limx→−∞F(x) =∞, and consequently, by (4.24),

limn→∞

F

(Bn

2

)=∞. (4.43)

In order to get a contradiction, we distinguish two cases.

Case 1. Let lim supn→∞bn <∞, that is, we can find b0 > 0, n1 ∈ N, n1 ≥ n0, such that

bn ≤ b0 for n ∈ N, n ≥ n1. (4.44)

Then, by (4.43), for each sufficiently large n ∈ N, we get

F

(Bn

2

)>

p2(b0)p2(γ0)(F(C) +

12

). (4.45)

Putting it to (4.42), we have

12< F

(Bn

2

)p2(γ0

)

p2(b0)− F(C) < 1

2u′2n (t), t ∈

[γn, bn

). (4.46)

Therefore 1 ≤ u′n(bn), contrary to (4.25).

Case 2. Let lim supn→∞bn = ∞. We may assume limn→∞bn = ∞ (otherwise we take asubsequence). Then there exists n2 ∈ N, n2 ≥ n0, such that

Γ + 1 ≤ bn for n ∈ N, n ≥ n2. (4.47)

Due to (4.43), for each sufficiently large n ∈ N, we get

F

(Bn

2

)>

p2(Γ + 1)p2(γ0)(F(C) +

12(L − C)2

). (4.48)

Putting it to (4.42), we have

12(L − C)2 < F

(Bn

2

)p2(γ0

)

p2(Γ + 1)− F(C) < 1

2u′2n (t), t ∈

[γn,Γ + 1

]. (4.49)


Therefore, L − C < u′n(t) for t ∈ [γn,Γ + 1]. Integrating it over [γn,Γ + 1], we obtain

(L − C)(Γ + 1 − γn

)< un(Γ + 1) − un

(γn)= un(Γ + 1) − C, (4.50)

which yields, by (4.26), L < un(Γ + 1) and also L < un(bn), contrary to (4.25). Thesecontradictions obtained in both cases imply that there exists � ∈ N such that u� is an escapesolution.

5. Homoclinic Solution

The following theorem provides the existence of a homoclinic solution under the assumptionthat the function f in (1.1) has a linear behaviour near −∞. According to Definition 1.2, ahomoclinic solution is a strictly increasing solution of problem (1.1), (1.2).

Theorem 5.1 (on homoclinic solution). Let the assumptions of Theorem 4.9 be satisfied. Then thereexists B < B such that the corresponding solution of problem (1.1), (1.10) is a homoclinic solution.

Proof. For B < 0 denote by uB the corresponding solution of problem (1.1), (1.10). Let Md

and Me be the set of all B < 0 such that uB is a damped solution and an escape solution,respectively. By Theorems 3.7, 3.8, 4.5, and 4.9, the setsMd andMe are nonempty and openin (−∞, 0). Therefore, the setMh = (−∞, 0) \ (Md ∪Me) is nonempty. Choose B∗ ∈ Mh. Then,by Theorem 4.4, uB∗ is a homoclinic solution. Moreover, due to Theorem 3.7, B∗ < B.

Example 5.2. The function

f(x) =

⎧⎨

⎩

c0x for x < 0,

x(x − L) for x ∈ [0, L],(5.1)

where c0 is a negative constant, satisfies the conditions (1.3)–(1.6), (1.13), and (4.23).

Acknowledgments

The authors thank the referee for valuable comments. This work was supported by theCouncil of Czech Government MSM 6198959214.

References

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[5] P. M. Lima, N. B. Konyukhova, A. I. Sukov, and N. V. Chemetov, “Analytical-numerical investigationof bubble-type solutions of nonlinear singular problems,” Journal of Computational and AppliedMathematics, vol. 189, no. 1-2, pp. 260–273, 2006.

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[7] D. Bonheure, J. M. Gomes, and L. Sanchez, “Positive solutions of a second-order singular ordinarydifferential equation,” Nonlinear Analysis: Theory, Methods & Applications, vol. 61, no. 8, pp. 1383–1399,2005.

[8] M. Conti, L. Merizzi, and S. Terracini, “Radial solutions of superlinear equations on RN . I. A global

variational approach,” Archive for Rational Mechanics and Analysis, vol. 153, no. 4, pp. 291–316, 2000.[9] H. Berestycki, P.-L. Lions, and L. A. Peletier, “An ODE approach to the existence of positive solutions

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equations: theoretical analysis and numerical simulations of ground states,” Boundary Value Problems,vol. 2006, Article ID 28719, 28 pages, 2006.


Research ArticleLimit Properties of Solutions of SingularSecond-Order Differential Equations

Irena Rachunkova,1 Svatoslav Stanek,1 Ewa Weinmuller,2and Michael Zenz2

1 Department of Mathematical Analysis, Faculty of Science, Palacky University,Tomkova 40, 779 00 Olomouc, Czech Republic

2 Institute for Analysis and Scientific Computing, Vienna University of Technology,Wiedner Hauptstrasse 8-10, 1040 Wien, Austria

Correspondence should be addressed to Irena Rachunkova, [email protected]

Received 23 April 2009; Accepted 28 May 2009


We discuss the properties of the differential equation u′′(t) = (a/t)u′(t) + f(t, u(t), u′(t)), a.e. on(0, T], where a ∈ R\{0}, and f satisfies the Lp-Caratheodory conditions on [0, T] × R

2 for somep > 1. A full description of the asymptotic behavior for t → 0+ of functions u satisfying theequation a.e. on (0, T] is given. We also describe the structure of boundary conditions whichare necessary and sufficient for u to be at least in C1[0, T]. As an application of the theory, newexistence and/or uniqueness results for solutions of periodic boundary value problems are shown.

Copyright q 2009 Irena Rachunkova 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.

1. Motivation

In this paper, we study the analytical properties of the differential equation

u′′(t) =a

tu′(t) + f

(t, u(t), u′(t)

), a.e. on (0, T], (1.1)

where a ∈ R \ {0}, u : [0, T] → R, and the function f is defined for a.e. t ∈ [0, T] and forall (x, y) ∈ D ⊂ R × R. The above equation is singular at t = 0 because of the first term inthe right-hand side, which is in general unbounded for t → 0. In this paper, we will alsoalow the function f to be unbounded or bounded but discontinuous for certain values ofthe time variable t ∈ [0, T]. This form of f is motivated by a variety of initial and boundaryvalue problems known from applications and having nonlinear, discontinuous forcing terms,such as electronic devices which are often driven by square waves or more complicated


discontinuous inputs. Typically, such problems are modelled by differential equations wheref has jump discontinuities at a discrete set of points in (0, T), compare [1].

This study serves as a first step toward analysis of more involved nonlinearities, wheretypically, f has singular points also in u and u′. Many applications, compare [2–12], showingthese structural difficulties are our main motivation to develop a framework on existenceand uniqueness of solutions, their smoothness properties, and the structure of boundaryconditions necessary for u to have at least continuous first derivative on [0, T]. Moreover,using new techniques presented in this paper, we would like to extend results from [13, 14](based on ideas presented in [15]) where problems of the above form but with appropriatelysmooth data functionf have been discussed.

Here, we aim at the generalization of the existence and uniqueness assertions derivedin those papers for the case of smooth f . We are especially interested in studying the limitproperties of u for t → 0 and the structure of boundary conditions which are necessary andsufficient for u to be at least in C1[0, T].

To clarify the aims of this paper and to show that it is necessary to develop anew technique to treat the nonstandard equation given above, let us consider a modelproblem which we designed using the structure of the boundary value problem describing amembrane arising in the theory of shallow membrane caps and studied in [10]; see also [6, 9],

(t3u′(t)

)′+ t3(

18u2(t)

− a0

u(t)+ b0t

2γ−4)

= 0, 0 < t < 1, (1.2)

subject to boundary conditions

limt→ 0+

t3u′(t) = 0, u(1) = 0, (1.3)

where a0 ≥ 0, b0 < 0, γ > 1. Note that (1.2) can be written in the form

u′′(t) = −3tu′(t) −

(1

8u2(t)− a0

u(t)+ b0t

2γ−4)

= 0, 0 < t < 1, (1.4)

which is of form (1.1) with

T = 1, a = −3, f(t, u, u′

)= −(

18u2− a0

u+ b0t

2γ−4). (1.5)

Function f is not defined for u = 0 and for t = 0 if γ ∈ (1, 2). We now briefly discuss asimplified linear model of (1.4),

u′′(t) = −3tu′(t) − b0t

β, 0 < t < 1, (1.6)

where β = 2γ − 4 and γ > 1. Clearly, this means that β > −2.


The question which we now pose is the role of the boundary conditions (1.3), moreprecisely, are these boundary conditions necessary and sufficient for the solution u of (1.6) to beunique and at least continuously differentiable, u ∈ C1[0, 1]? To answer this question, we canuse techniques developed in the classical framework dealing with boundary value problems,exhibiting a singularity of the first and second kind; see [15, 16], respectively. However, inthese papers, the analytical properties of the solution u are derived for nonhomogeneousterms being at least continuous. Clearly, we need to rewrite problem (1.6) first and obtain itsnew form stated as,

(t3u′(t)

)′+ t3(b0t

β)= 0, 0 < t < 1, (1.7)

which suggest to introduce a new variable, v(t) := t3u′(t). In a general situation, especially forthe nonlinear case, it is not straightforward to provide such a transformation, however. Wenow introduce z(t) := (u(t), v(t))T and immediately obtain the following system of ordinarydifferential equations:

z′(t) =1t3

(0 1

0 0

)

z(t) −(

0

b0tβ+3

)

, 0 < t < 1, (1.8)

where β + 3 > 1, or equivalently,

z′(t) =1t3Mz(t) + g(t), M :=

(0 1

0 0

)

, g(t) := −(

0

b0tβ+3

)

, (1.9)

where g ∈ C[0, 1]. According to [16], the latter system of equations has a continuous solutionif and only if the regularity condition Mz(0) = 0 holds. This results in

v(0) = 0⇐⇒ limt→ 0+

t3u′(t) = 0, (1.10)

compare conditions (1.3). Note that the Euler transformation, ζ(t) := (u(t), tu′(t))T which isusually used to transform (1.6) to the first-order form would have resulted in the followingsystem:

ζ′(t) =1tNζ(t) +w(t), N :=

(0 1

0 −2

)

, w(t) := −(

0

b0tβ+1

)

. (1.11)

Here, w may become unbounded for t → 0, the condition Nζ(0) = 0, or equivalentlylimt→ 0+tu

′(t) = 0 is not the correct condition for the solution u to be continuous on [0, 1].From the above remarks, we draw the conclusion that a new approach is necessary to

study the analytical properties of (1.1).


2. Introduction

The following notation will be used throughout the paper. Let J ⊂ R be an interval.Then, we denote by L1(J) the set of functions which are (Lebesgue) integrable on J . Thecorresponding norm is ‖u‖1 :=

∫J |u(t)|dt. Let p > 1. By Lp(J), we denote the set of functions

whose pth powers of modulus are integrable on J with the corresponding norm given by‖u‖p := (

∫J |u(t)|

pdt)1/p.Moreover, let us by C(J) and C1(J) denote the sets of functions being continuous on J

and having continuous first derivatives on J , respectively. The norm on C[0, T] is defined as‖u‖∞ := maxt∈[0,T]{|u(t)|}.

Finally, we denote by AC(J) and AC1(J) the sets of functions which are absolutelycontinuous on J and which have absolutely continuous first derivatives on J , respectively.Analogously, ACloc(J) and AC1

loc(J) are the sets of functions being absolutely continuous oneach compact subinterval I ⊂ J and having absolutely continuous first derivatives on eachcompact subinterval I ⊂ J , respectively.

As already said in the previous section, we investigate differential equations of theform

u′′(t) =a

tu′(t) + f

(t, u(t), u′(t)

), a.e. on (0, T], (2.1)

where a ∈ R \ {0}. For the subsequent analysis we assume that

f satisfies the Lp-Caratheodory conditions on [0, T] × R × R, for some p > 1 (2.2)

specified in the following definition.

Definition 2.1. Let p > 1. A function f satisfies the Lp-Caratheodory conditions on the set [0, T]×R × R if

(i) f(·, x, y) : [0, T] → R is measurable for all (x, y) ∈ R × R,

(ii) f(t, ·, ·) : R × R → R is continuous for a.e. t ∈ [0, T],

(iii) for each compact set K ⊂ R × R there exists a function mK(t) ∈ Lp[0, T] such that|f(t, x, y)| ≤ mK(t) for a.e. t ∈ [0, T] and all (x, y) ∈ K.

We will provide a full description of the asymptotical behavior for t → 0+ of functionsu satisfying (2.1) a.e. on (0, T]. Such functions u will be called solutions of (2.1) if theyadditionally satisfy the smoothness requirement u ∈ AC1[0, T]; see next definition.

Definition 2.2. A function u : [0, T] → R is called a solution of (2.1) if u ∈ AC1[0, T] andsatisfies

u′′(t) =a

tu′(t) + f

(t, u(t), u′(t)

)a.e. on (0, T]. (2.3)

In Section 3, we consider linear problems and characterize the structure of boundaryconditions necessary for the solution to be at least continuous on [0, 1]. These results aremodified for nonlinear problems in Section 4. In Section 5, by applying the theory developed


in Section 4, we provide new existence and/or uniqueness results for solutions of singularboundary value problems (2.1) with periodic boundary conditions.

3. Linear Singular Equation

First, we consider the linear equation, a ∈ R \ {0},

u′′(t) =a

tu′(t) + h(t), a.e. on (0, T], (3.1)

where h ∈ Lp[0, T] and p > 1.As a first step in the analysis of (3.1), we derive the necessary auxiliary estimates used

in the discussion of the solution behavior. For c ∈ [0, T], let us denote by

ϕa(c, t) := ta∫ c

t

h(s)sa

ds, t ∈ (0, T]. (3.2)

Assume that a < 0. Then

0 <

(∫ t

0

dssaq

)1/q

=

(t1−aq

1 − aq

)1/q

, t ∈ (0, T]. (3.3)

Now, let a > 0, c > 0. Without loss of generality, we may assume that 1/p /= 1 − a. For 1/p =1 − a, we choose p∗ ∈ (1, p), and we have h ∈ Lp∗[0, T] and 1/p∗ > 1 − a.

First, let a ∈ (0, 1 − 1/p). Then 1/q = 1 − 1/p > a, 1 − aq > 0, and

0 <

∣∣∣∣

∫ c

t

dssaq

∣∣∣∣

1/q

=

∣∣∣∣∣c1−aq − t1−aq

1 − aq

∣∣∣∣∣

1/q

<

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(c1−aq

1 − aq

)1/q

, if c ≥ t > 0,

(t1−aq

1 − aq

)1/q

, if c < t ≤ T.

(3.4)

Now, let a > 1 − 1/p. Then 1/q = 1 − 1/p < a, 1 − aq < 0, and

0 ≤∣∣∣∣

∫ c

t

dssaq

∣∣∣∣

1/q

=

∣∣∣∣∣c1−aq − t1−aq

1 − aq

∣∣∣∣∣

1/q

<

⎧⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎩

(c1−aq

aq − 1

)1/q

, if c < t ≤ T,

(t1−aq

aq − 1

)1/q

, if c ≥ t > 0.

(3.5)

Hence, for a > 0, c > 0,

0 ≤∣∣∣∣

∫ c

t

dssaq

∣∣∣∣

1/q

<∣∣1 − aq

∣∣−1/q(c1/q−a + t1/q−a

), t ∈ (0, T]. (3.6)


Consequently, (3.3), (3.6), and the Holder inequality yield, t ∈ (0, T],

∣∣ϕa(c, t)

∣∣ ≤ ta

(c1/q−a + t1/q−a

)∣∣1 − aq

∣∣−1/q‖h‖p, if a > 0, c > 0,

∣∣ϕa(0, t)

∣∣ ≤ tat1/q−a(1 − aq)−1/q‖h‖p, if a < 0.

(3.7)

Therefore

ϕa(c, t) ∈ C(0, T], limt→ 0+

ϕa(c, t) = 0, if a > 0, c > 0, (3.8)

ϕa(0, t) ∈ C(0, T], limt→ 0+

ϕa(0, t) = 0, if a < 0, (3.9)

which means that ϕa ∈ C[0, 1]. We now use the properties of ϕa to represent all functionsu ∈ AC1

loc(0, T] satisfying (3.1) a.e. on [0, T]. Remember that such function u does not need tobe a solution of (3.1) in the sense of Definition 2.2.

Lemma 3.1. Let a ∈ R \ {0}, c ∈ (0, T], and let ϕa(c, t) be given by (3.2).

(i) If a/= − 1, then

{c1 + c2t

a+1 +∫ c

t

ϕa(c, s)ds, c1, c2 ∈ R, t ∈ (0, T]}

(3.10)

is the set of all functions u ∈ AC1loc(0, T] satisfying (3.1) a.e. on (0, T].

(ii) If a = −1, then

{c1 + c2 ln t +

∫ c

t

ϕ−1(c, s)ds, c1, c2 ∈ R, t ∈ (0, T]}

(3.11)


Proof. Let a/= − 1. Note that (3.1) is linear and regular on (0, T]. Since the functions u1h(t) = 1

and u2h(t) = ta+1 are linearly independent solutions of the homogeneous equation u′′(t) −

(a/t)u′(t) = 0 on (0, T], the general solution of the homogeneous problem is

uh(t) = c1 + c2ta+1, c1, c2 ∈ R. (3.12)

Moreover, the function up(t) =∫ct ϕa(c, s)ds is a particular solution of (3.1) on (0, T]. Therefore,

the first statement follows. Analogous argument yields the second assertion.

We stress that by (3.8), the particular solution up =∫ct ϕa(c, s)ds of (3.1) belongs to

C1[0, T]. For a < 0, we can see from (3.9) that it is useful to find other solution representationswhich are equivalent to (3.10) and (3.11), but use ϕa(0, t) instead of ϕa(c, t), if c > 0.

Lemma 3.2. Let a < 0 and let ϕa(0, t) be given by (3.2).


(i) If a/= − 1, then

{

c1 + c2ta+1 −

∫ t

0ϕa(0, s)ds, c1, c2 ∈ R, t ∈ (0, T]

}

(3.13)


(ii) If a = −1, then

{

c1 + c2 ln t −∫ t

0ϕ−1(0, s)ds, c1, c2 ∈ R, t ∈ (0, T]

}

(3.14)


Proof. Let us fix c ∈ (0, T] and define

p(t) :=∫ c

t

ϕa(c, s)ds +∫ t

0ϕa(0, s)ds, t ∈ (0, T]. (3.15)

In order to prove (i) we have to show that p(t) = d1 + d2ta+1 for t ∈ (0, T], where d1, d2 ∈ R.

This follows immediately from (3.9), since

p(c) =∫ c

0ϕa(0, s)ds,

p′(t) = −ϕa(c, t) + ϕa(0, t)

= −ta∫ c

0

h(s)sa

ds, t ∈ (0, T],

(3.16)

and hence we can define di as follows:

d2 := − 1a + 1

∫ c

0

h(s)sa

ds, d1 := p(c) − d2ca+1. (3.17)

For a = −1 we have

d2 := −∫ c

0sh(s)ds, d1 :=

∫ c

0ϕ−1(0, s)ds − d2 ln c, (3.18)

which completes the proof.

Again, by (3.9), the particular solution,

up(t) = −∫ t

0ϕa(0, s)ds, (3.19)


of (3.1) for a < 0 satisfies up ∈ C1[0, 1]. Main results for the linear singular equation (3.1) arenow formulated in the following theorems.

Theorem 3.3. Let a > 0 and let u ∈ AC1loc(0, T] satisfy equation (3.1) a.e. on [0, T]. Then

limt→ 0+

u(t) ∈ R, limt→ 0+

u′(t) = 0. (3.20)

Moreover, u can be extended to the whole interval [0, T] in such a way that u ∈ AC1[0, T].

Proof. Let a function u be given. Then, by (3.10), there exist two constants c1, c2 ∈ R such thatfor t ∈ (0, T],

u(t) = c1 + c2ta+1 +

∫ c

t

ϕa(c, s)ds,

u′(t) = c2(a + 1)ta − ϕa(c, t).

(3.21)

Using (3.8), we conclude

limt→ 0+

u(t) = c1 +∫ c

0ϕa(c, s)ds =: c3 ∈ R, lim

t→ 0+u′(t) = 0. (3.22)

For u(0) := c3 and u′(0) = 0, we have u ∈ C1[0, T]. Furthermore, for a.e. t ∈ (0, T],

u′′(t) = c2(a + 1)ata−1 − h(t) + ata−1∫ c

t

h(s)sa

ds. (3.23)

By the Holder inequality and (3.6) it follows that

∣∣u′′(t)∣∣ ≤ c2(a + 1)ata−1 + |h(t)| +Mta−1

(c1/q−a + t1/q−a

)‖h‖p ∈ L1[0, T], (3.24)

where

M = a∣∣1 − aq

∣∣−1/q. (3.25)

Therefore u′′ ∈ L1[0, T], and consequently u ∈ AC1[0, T].

It is clear from the above theorem, that u ∈ AC1[0, T] given by (3.21) is a solution of(3.1) for a > 0. Let us now consider the associated boundary value problem,

u′′(t) =a

tu′(t) + h(t), a.e. on (0, T], (3.26a)

B0U(0) + B1U(T) = β, U(t) := (u(t), u′(t))T , (3.26b)


where B0, B1 ∈ R2×2 are real matrices, and β ∈ R

2 is an arbitrary vector. Then the followingresult follows immediately from Theorem 3.3.

Theorem 3.4. Let a > 0, p > 1. Then for any h(t) ∈ Lp[0, T] and any β ∈ R2 there exists a unique

solution u ∈ AC1[0, 1] of the boundary value problem (3.26a) and (3.26b) if and only if the followingmatrix,

B0

(1 0

0 0

)

+ B1

(1 Ta+1

0 (a + 1)Ta

)

∈ R2×2, (3.27)

is nonsingular.

Proof. Let u be a solution of (3.1). Then u satisfies (3.21), and the result follows immediatelyby substituting the values,

u(0) = c1 +∫ c

0ϕa(c, s)ds, u(T) = c1 + c2T

a+1 +∫ c

T

ϕa(c, s)ds,

u′(0) = 0, u′(T) = c2(a + 1)Ta − ϕa(c, T),

(3.28)

into the boundary conditions (3.26b).

Theorem 3.5. Let a < 0 and let a function u ∈ AC1loc(0, T] satisfy equation (3.1) a.e. on (0, T]. For

a ∈ (−1, 0), only one of the following properties holds:

(i) limt→ 0+u(t) ∈ R, limt→ 0+u′(t) = 0,

(ii) limt→ 0+u(t) ∈ R, limt→ 0+u′(t) = ±∞.

For a ∈ (−∞,−1], u satisfies only one of the following properties:

(i) limt→ 0+u(t) ∈ R, limt→ 0+u′(t) = 0,

(ii) limt→ 0+u(t) = ∓∞, limt→ 0+u′(t) = ±∞.

In particular, u can be extended to the whole interval [0, T] with u ∈ AC1[0, T] if and only iflimt→ 0+u

′(t) = 0.

Proof. Let a ∈ (−1, 0), and let u be given. Then, by (3.13), there exist two constants c1, c2 ∈ R

such that

u(t) = c1 + c2ta+1 −

∫ t

0ϕa(0, s)ds for t ∈ (0, T]. (3.29)

Hence

u′(t) = c2(a + 1)ta − ϕa(0, t) for t ∈ (0, T]. (3.30)


Let c2 = 0, then it follows from (3.9) limt→ 0+u′(t) = 0. Also, by (3.29), limt→ 0+u(t) = c1 ∈ R.

Let c2 /= 0. Then (3.9), (3.29), and (3.30) imply that

limt→ 0+

u(t) = c1 ∈ R, limt→ 0+

u′(t) = +∞, if c2 > 0,

limt→ 0+

u(t) = c1 ∈ R, limt→ 0+

u′(t) = −∞, if c2 < 0.(3.31)

Let a = −1. Then, by (3.14), for any c1, c2 ∈ R,

u(t) = c1 + c2 ln t −∫ t

0ϕ−1(0, s)ds for t ∈ (0, T], (3.32)

u′(t) = c21t− ϕ−1(0, t) for t ∈ (0, T]. (3.33)

If c2 = 0, then limt→ 0+u′(t) = 0 by (3.9), and it follows from (3.32) that limt→ 0+u(t) = c1 ∈ R.

Let c2 /= 0. Then we deduce from (3.9), (3.32), and (3.33) that

limt→ 0+

u(t) = −∞, limt→ 0+

u′(t) = +∞, if c2 > 0,

limt→ 0+

u(t) = +∞, limt→ 0+

u′(t) = −∞, if c2 < 0.(3.34)

Let a < −1. Then on (0, T], u satisfies (3.29) and (3.30), with c1, c2 ∈ R. If c2 = 0, then, by (3.9),limt→ 0+u

′(t) = 0 and limt→ 0+u(t) = c1 ∈ R. Let c2 /= 0. Then

limt→ 0+

u(t) = +∞, limt→ 0+

u′(t) = −∞, if c2 > 0,

limt→ 0+

u(t) = −∞, limt→ 0+

u′(t) = +∞, if c2 < 0.(3.35)

In particular, for a < 0, u can be extended to [0, T] in such a way that u ∈ C1[0, T] if and onlyif c2 = 0. Then, the associated boundary conditions read u(0) = c1 and u′(0) = 0. Finally, fora.e. t ∈ (0, T],

u′′(t) = −h(t) − ata−1∫ t

0

h(s)sa

ds, (3.36)

and by the Holder inequality, (3.3), and (3.25),

∣∣u′′(t)∣∣ ≤ |h(t)| +Mta−1t1/q−a‖h‖p ∈ L1[0, T]. (3.37)

Therefore u′′ ∈ L1[0, T], and consequently u ∈ AC1[0, T].


Again, it is clear that u given by (3.29) for a ∈ (−1, 0) and a < −1, and u given by (3.32)for a = −1 is a solution of (3.1), and u ∈ AC1[0, 1] if and only if u′(0) = 0. Let us now considerthe boundary value problem

u′′(t) =a

tu′(t) + h(t), a.e. on (0, T], (3.38a)

u′(0) = 0, b0u(0) + b1u(T) + b2u′(T) = β, (3.38b)

where b0, b1, b2, β ∈ R are real constants. Then the following result follows immediately fromTheorem 3.5.

Theorem 3.6. Let a < 0, p > 1. Then for any h(t) ∈ Lp[0, T] and any b2, β ∈ R there exists a uniquesolution u ∈ AC1[0, 1] of the boundary value problem (3.38a) and (3.38b) if and only if b0 + b1 /= 0.

Proof. Let u be a solution of (3.1). Then u satisfies (3.29) for a ∈ (−1, 0) and a < −1, and (3.32)for a = −1. We first note that, by (3.9), for all a < 0,

u′(0) = limt→ 0+

u′(t) = 0⇐⇒ c2 = 0. (3.39)

Therefore, c2 = 0 in both, (3.29) and (3.32), and the result now follows by substituting thevalues,

u(0) = c1, u(T) = c1 −∫T

0ϕa(0, s)ds, u′(T) = −ϕa(0, T), (3.40)

into the boundary conditions (3.38b).

To illustrate the solution behaviour, described by Theorems 3.3 and 3.5, we havecarried out a series of numerical calculations on a MATLAB software package bvpsuitedesigned to solve boundary value problems in ordinary differential equations. The solveris based on a collocation method with Gaussian collocation points. A short description ofthe code can be found in [17]. This software has already been used for a variety of singularboundary value problems relevant for applications; see, for example, [18].

The equations being dealt with are of the form

u′′(t) =a

tu′(t) +

13√

1 − t, t ∈ (0, 1), (3.41)

subject to initial or boundary conditions specified in the following graphs. All solutions werecomputed on the unit interval [0, 1].

Finally, we expect limt→ 0+u(t) = ±∞, and therefore we solve (3.41) subject to theterminal conditions u(1) = α, u′(1) = β. See Figures 1, 2, and 3.


−4

−3

−2

−1

0

1

2

3

4

0 0.2 0.4 0.6 0.8 1

u (0) = 1, u (1) = 3u (0) = −1, u (1) = −3

Figure 1: Illustrating Theorem 3.3: solutions of differential equation (3.41) with a = 1, subject to boundaryconditions u(0) = α, u(1) = β. See graph legend for the values of α and β. According to Theorem 3.3 itholds that u′(0) = 0 for each choice of α and β.

4. Limit Properties of Functions Satisfying NonlinearSingular Equations

In this section we assume that the function u ∈ AC1loc(0, T] satisfying differential equation

(2.1) a.e. on [0, T] is given. The first derivative of such a function does not need to becontinuous at t = 0 and hence, due to the lack of smoothness, u does not need to be a solutionof (2.1) in the sense of Definition 2.2. In the following two theorems, we discuss the limitproperties of u for t → 0.

Theorem 4.1. Let us assume that (2.2) holds. Let a > 0 and let u ∈ AC1loc(0, T] satisfy equation

(2.1) a.e. on [0, T]. Finally, let us assume that that

sup{|u(t)| +

∣∣u′(t)∣∣ : t ∈ (0, T]

}<∞. (4.1)

Then

limt→ 0+

u(t) ∈ R, limt→ 0+

u′(t) = 0, (4.2)

and u can be extended on [0, T] in such a way that u ∈ AC1[0, T].

Proof. Let h(t) := f(t, u(t), u′(t)) for a.e. t ∈ [0, T]. By (2.2), there exists a function mK ∈Lp[0, T] such that |f(t, u(t), u′(t))| ≤ mK(t) for a.e. t ∈ [0, T]. Therefore, h ∈ Lp[0, T]. Since theequality u′′(t) = (a/t)u′(t) + h(t) holds a.e. on [0, T], the result follows immediately due toTheorem 3.3.


−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

0 0.2 0.4 0.6 0.8 1

u (0) = 2, u (1) = 0.75u (0) = −2, u (1) = 0u (0) = 0, u’ (0) = 0

Figure 2: Illustrating Theorem 3.5 for a ∈ (−1, 0): solutions of differential equation (3.41) with a = −1/2,subject to boundary conditions u(0) = α, u(1) = β. See graph legend for the values of α and β. Accordingto Theorem 3.5 a solution u satisfies u′(0) = +∞ or u′(0) = −∞ or u′(0) = 0 in dependence of values α andβ. In order to precisely recover a solution satisfying u′(0) = 0, the respective simulation was carried out asan initial value problem with u(0) = 0 and u′(0) = 0.

−100

−50

0

50

100

0 0.2 0.4 0.6 0.8 1

u (1) = 1, u′(1) = −1u (1) = 0, u′(1) = 0

u (1) = −1, u′(1) = 1u (1) = −2, u′(1) = 2

Figure 3: Illustrating Theorem 3.5 for a ∈ (−∞,−1): solutions of differential equation (3.41) with a = −2,subject to boundary conditions u(1) = α, u′(1) = β. See graph legend for the values of α and β. Here,limt→ 0+u(t) = ±∞, and limt→ 0+u

′(t) = ∓∞, or u(0) ∈ R, u′(0) = 0.


Theorem 4.2. Let us assume that condition (2.2) holds. Let a < 0 and let u ∈ AC1loc(0, T] satisfy

equation (2.1) a.e. on (0, T]. Let us also assume that (4.1) holds. Then

limt→ 0+

u(t) ∈ R, limt→ 0+

u′(t) = 0, (4.3)

and u can be extended on [0, T] in such a way that u ∈ AC1[0, T].

Proof. Let h ∈ Lp[0, T] be as in the proof of Theorem 4.1. According to Theorem 3.5 and (4.1),u satisfies (4.3) both for a ∈ (−1, 0) and a ∈ (−∞,−1].

5. Applications

Results derived in Theorems 4.1 and 4.2 constitute a useful tool when investigating thesolvability of nonlinear singular equations subject to different types of boundary conditions.In this section, we utilize Theorem 4.1 to show the existence of solutions for periodicproblems. The rest of this section is devoted to the numerical simulation of such problems.

Periodic Problem

We deal with a problem of the following form:

u′′(t) =a

tu′(t) + f

(t, u(t), u′(t)

), a.e. on (0, T], (5.1a)

u(0) = u(T), u′(0) = u′(T). (5.1b)

Definition 5.1. A function u ∈ AC1[0, T] is called a solution of the boundary value problem (5.1a)and (5.1b), if u satisfies equation (5.1a) for a.e. t ∈ [0, T] and the periodic boundary conditions(5.1b).

Conditions (5.1b) can be written in the form (3.26b) with B0 = I, B1 = −I, and β = 0.Then, matrix (3.27) has the form

I

(1 0

0 0

)

− I(

1 Ta+1

0 (a + 1)Ta

)

=

(0 −Ta+1

0 −(a + 1)Ta

)

, (5.2)

and we see that it is singular. Consequently, the assumption of Theorem 3.4 is notsatisfied, and the linear periodic problem (3.26b) subject to (5.1b) is not uniquely solvable.However this is not true for nonliner periodic problems. In particular, Theorem 5.6 givesa characterization of a class of nonlinear periodic problems (5.1a) and (5.1b) which haveonly one solution. We begin the investigation of problem (5.1a) and (5.1b) with a uniquenessresult.

Theorem 5.2 (uniqueness). Let a > 0 and let us assume that condition (2.2) holds. Further, assumethat for each compact setK ⊂ R × R there exists a nonnegative function hK ∈ L1[0, T] such that

x1 > x2, y1 ≥ y2 =⇒ f(t, x1, y1

)− f(t, x2, y2

)> −hK(t)

(y1 − y2

)(5.3)


for a.e. t ∈ [0, T] and all (x1, y1), (x2, y2) ∈ K. Then problem (5.1a) and (5.1b) has at most onesolution.

Proof. Let u1 and u2 be different solutions of problem (5.1a) and (5.1b). Since u1, u2 ∈AC1[0, T], there exists a compact set K ⊂ R × R such that (ui(t), u′i(t)) ∈ K for t ∈ [0, T].Let us define the difference function v(t) := u1(t) − u2(t) for t ∈ [0, T]. Then

v(0) = v(T), v′(0) = v′(T). (5.4)

First, we prove that there exists an interval [α, β] ⊂ (0, T] such that

v(t) > 0 for t ∈[α, β], v′(t) > 0 for t ∈

[α, β), v′

(β)= 0. (5.5)

We consider two cases.

Case 1. Assume that u1 and u2 have an intersection point, that is, there exists t0 ∈ [0, T) suchthat v(t0) = 0. Since u1 and u2 are different, there exists t1 ∈ [0, T], t1 /= t0, such that v(t1)/= 0.

(i) Let t1 > t0. We can assume that v(t1) > 0. (Otherwise we choose v := u2 − u1.) Thenwe can find a0 ∈ (t0, t1) satisfying v(t) > 0 for t ∈ [a0, t1] and v′(a0) > 0. Let b0 ∈ (a0, T] bethe first zero of v′. Then, if we set [α, β] := [a0, b0], we see that [α, β] satisfies (5.5). Let v′ haveno zeros on [a0, T]. Then v > 0, v′ > 0 on [a0, T], and, due to (5.4), v(0) > 0, v′(0) > 0. Sincev(t0) = 0, we can find α ∈ (0, t0) and β ∈ (a, t0) such that [α, β] satisfies (5.5).

(ii) Let v = 0 on [t0, T]. By (5.4), v(0) = 0, v′(0) = 0 and t1 ∈ (0, t0). We may againassume that v(t1) > 0. It is possible to find α ∈ (0, t1) such that v(α) > 0, v′(α) > 0, v(t) > 0 on[α, t1]. Since v(t0) = 0, v′ has at least one zero in (α, t0). If β ∈ (α, t0) is the first zero of v′, then[α, β] satisfies (5.5).

Case 2. Assume that u1 and u2 have no common point, that is, v(t)/= 0 on [0, T]. We mayassume that v > 0 on [0, T]. By (5.4), there exists a point t0 ∈ (0, T) satisfying v′(t0) = 0.

(i) Let v′ = 0 on [0, t0]. Then, by (5.1a) and (5.3),

v′′(t) =a

tv′(t) + f

(t, u1(t), u′1(t)

)− f(t, u2(t), u′2(t)

)>(at− hK(t)

)v′(t) = 0 (5.6)

for a.e. t ∈ [0, t0], which is a contradiction to v′′ = 0 on [0, t0].(ii) Let v′(t1)/= 0 for some t1 ∈ [0, t0). If v′(t1) > 0, then we can find an interval [α, β] ⊂

(t1, t0] satisfying (5.5). If v′(t1) < 0 and v′(t) ≤ 0 on [0, t0], then v(0) > v(t0) and, by (5.4),v(T) > v(t0), v′(T) ≤ 0. Hence, there exists an interval [α, β] ⊂ (t0, T] satisfying (5.5).

To summarize, we have shown that in both, the case of intersecting solutions u1 andu2 and the case of separated u1 and u2, there exists an interval [α, β] ⊂ (0, T] satisfying (5.5).

Now, by (5.1a), (5.3), and (5.5), we obtain

v′′(t) >(at− hK(t)

)v′(t) for a.e. t ∈

[α, β]. (5.7)


Denote by h∗(t) := a/t − hK(t). Then h∗ ∈ L1[α, β], and v′′(t) − h∗(t)v′(t) > 0 for a.e. t ∈ [α, β].Consequently,

(

v′(t) exp

(

−∫ t

α

h∗(s)ds

))′> 0 for a.e. t ∈

[α, β]. (5.8)

Integrating the last inequality in [α, β], we obtain

v′(β)

exp

(

−∫β

α

h∗(s)ds

)

> v′(α) > 0, (5.9)

which contradicts v′(β) = 0. Consequently, we have shown that u1 ≡ u2, and the resultfollows.

In the following theorem we formulate sufficient conditions for the existence of at leastone solution of problem (5.1a) and (5.1b) with a > 0. In the proof of this theorem, we workalso with auxiliary two-point boundary conditions:

u(0) = u(T), u′(T) = 0. (5.10)

Under the assumptions of Theorem 4.1 any solution of (5.1a) satisfies u′(0) = 0. Therefore,we can investigate (5.1a) subject to the auxiliary conditions (5.10) instead of the equivalentoriginal problem (5.1a) and (5.1b). This change of the problem setting is useful for obtainingof a priori estimates necessary for the application of the Fredholm-type existence theorem(Lemma 5.5) during the proof.

Theorem 5.3 (existence). Let a > 0 and let (2.2) hold. Further, assume that there exist A,B ∈ R,c > 0, ω ∈ C[0,∞), and ψ ∈ L1[0, T] such that A ≤ B,

f(t, A, 0) ≤ 0, f(t, B, 0) ≥ 0 (5.11)

for a.e. t ∈ [0, T],

f(t, x, y

)signy ≥ −ω

(∣∣y∣∣)(∣∣y

∣∣ + ψ(t))

(5.12)

for a.e. t ∈ [0, T] and all x ∈ [A,B], y ∈ R, where

ω(x) ≥ c, x ∈ [0,∞),∫∞

0

dsω(s)

=∞. (5.13)

Then problem (5.1a) and (5.1b) has a solution u such that

A ≤ u(t) ≤ B, t ∈ [0, T]. (5.14)


Proof.Step 1 (existence of auxiliary solutions un). By (5.13), there exists ρ∗ > 0 such that

∫ρ∗

0

dsω(s)

> ‖ψ‖1 +(

1 +T

c

)(B −A) =: r. (5.15)

For y ∈ R, let

χ(y)

:=

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

1, if∣∣y∣∣ ≤ ρ∗,

2 −∣∣y∣∣

ρ∗, if ρ∗ <

∣∣y∣∣ < 2ρ∗,

0, if∣∣y∣∣ ≥ 2ρ∗.

(5.16)

Motivated by [19], we choose n ∈ N, n > 1/T , and, for a.e. t ∈ [0, T], all (x, y) ∈ R × R, andε ∈ [0, 1], we define the following functions:

hn

(t, x, y

)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

χ(y)(a

ty + f

(t, x, y

))− A

n, if t ≥ 1

n,

−An, if t <

1n,

(5.17)

wA(t, ε) = sup{∣∣hn(t, A, 0) − hn

(t, A, y

)∣∣ :∣∣y∣∣ ≤ ε

},

wB(t, ε) = sup{∣∣hn(t, B, 0) − hn

(t, B, y

)∣∣ :∣∣y∣∣ ≤ ε

},

(5.18)

fn(t, x, y

)=

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

hn

(t, B, y

)+wB

(t,

x − Bx − B + 1

), if x > B,

hn

(t, x, y

), if A ≤ x ≤ B,

hn

(t, A, y

)−wA

(t,

A − xA − x + 1

), if x < A.

(5.19)

Due to (5.11),

A

n+ hn(t, A, 0) ≤ 0,

B

n+ hn(t, B, 0) ≥ 0 (5.20)

for a.e. t ∈ [0, T]. It can be shown that wA and wB which satisfy the Lp-Caratheodoryconditions on [0, T] × [0, 1] are nondecreasing in their second argument and wA(t, 0) =wB(t, 0) = 0 a.e. on [0, T]; see [19]. Therefore, fn also satisfies the Lp-Caratheodory conditionson [0, T] × R × R, and there exists a function mn ∈ Lp[0, T] such that |fn(t, x, y)| ≤ mn(t) fora.e. t ∈ [0, T] and all (x, y) ∈ R × R.


We now investigate the auxiliary problem

u′′(t) =u(t)n

+ fn(t, u(t), u′(t)

), u(0) = u(T), u′(T) = 0. (5.21)

Since the homogeneous problem u′′(t) = (1/n)u(t), u(0) = u(T), u′(T) = 0, has only the trivialsolution, we conclude by the Fredholm-type Existence Theorem (see Lemma 5.5) that thereexists a solution un ∈ AC1[0, T] of problem (5.21).Step 2 (estimates of un). We now show that

A ≤ un(t) ≤ B, t ∈ [0, T], n ∈ N, n >1T. (5.22)

Let us define v(t) := A − un(t) for t ∈ [0, T] and assume

max{v(t) : t ∈ [0, T]} = v(t0) > 0. (5.23)

By (5.21), we can assume that t0 ∈ (0, T]. Since v′(t0) = 0, we can find δ > 0 such that

v(t) > 0,∣∣v′(t)

∣∣ =∣∣u′n(t)

∣∣ <v(t)

v(t) + 1< 1 on (t0 − δ, t0] ⊂ (0, T]. (5.24)

Then, by (5.19), (5.20), and (5.21), we have

u′′n = fn(t, un(t), u′n(t)

)+un(t)n

= hn

(t, A, u′n(t)

)−wA

(t,

v(t)v(t) + 1

)+un(t)n

≤ hn(t, A, 0) + hn

(t, A, u′n(t)

)− hn(t, A, 0) −wA

(t,∣∣u′n(t)

∣∣) +un(t)n

≤ hn(t, A, 0) +A

n− v(t)

n< 0

(5.25)

for a.e. t ∈ (t0 − δ, t0]. Hence,

0 >

∫ t0

t

u′′n(s)ds = −u′n(t) = v′(t), t ∈ (t0 − δ, t0), (5.26)

which contradicts (5.23), and thus A ≤ un(t) on [0, T]. The inequality un(t) ≤ B on [0, T] canbe proved in a very similar way.Step 3 (estimates of u′n). We now show that

∣∣u′n(t)∣∣ ≤ ρ∗, t ∈ [0, T], n ∈ N, n >

1T. (5.27)


By (5.19) and (5.22) we have fn(t, un(t), u′n(t)) = hn(t, un(t), u′n(t)) for a.e. t ∈ [0, T], and so,due to (5.17) and (5.21), we have for a.e. t ∈ [0, T],

(u′′n(t) −

1n(un(t) −A)

)sign u′n(t)

=

⎧⎪⎪⎨

⎪⎪⎩

χ(u′n(t))(atu′n(t) + f(t, un(t), u′n(t))

)sign u′n(t), if t ≥ 1

n,

0, if t <1n.

(5.28)

Denote ρ := ‖u′n‖∞ = |u′n(t0)|. If ρ > 0, then t0 ∈ [0, T).

Case 1. Let u′n(t0) = ρ. Then there exists t1 ∈ (t0, T] such that u′n(t) > 0 on [t0, t1), u′n(t1) = 0.By (5.12), (5.22), (5.28), and a > 0, it follows for a.e. t ∈ [t0, t1],

u′′n(t) ≥ χ(u′n(t)

)f(t, un(t), u′n(t)

)sign u′n(t)

≥ −χ(u′n(t)

)ω(u′n(t)

)(u′n(t) + ψ(t)

)

≥ −ω(u′n(t)

)(u′n(t) + ψ(t)

).

(5.29)

Consequently,

∫ t1

t0

u′′n(t)ω(u′n(t))

dt ≥ −∫ t1

t0

(u′n(t) + ψ(t)

)dt,

∫ρ

0

dsω(s)

≤ un(t1) − un(t0) + ‖ψ‖1 < r,

(5.30)

where r is given by (5.15). Therefore ρ < ρ∗.

Case 2. Let u′n(t0) = −ρ. Then there exists t1 ∈ (t0, T] such that u′n(t) < 0 on [t0, t1), u′n(t1) = 0.By (5.12), (5.13), (5.22), (5.28), and a > 0, we obtain for a.e. t ∈ [t0, t1]

−u′′n(t) ≥ −χ(u′n(t)

)f(t, un(t), u′n(t)

)sign u′n(t) −

1n(un(t) −A)

≥ −χ(u′n(t)

)ω(∣∣u′n(t)

∣∣)(∣∣u′n(t)∣∣ + ψ(t)

)− 1n(B −A)

≥ −ω(∣∣u′n(t)

∣∣)(∣∣u′n(t)

∣∣ + ψ(t) +1c(B −A)

).

(5.31)


−0.94

−0.935

−0.93

−0.925

−0.92

−0.915

−0.91

−0.905

−0.9

−0.895

−0.89

0 0.2 0.4 0.6 0.8 1

a = 0.5a = 1a = 2

Figure 4: Illustrating Theorem 5.6: solutions of differential equation (5.43), subject to periodic boundaryconditions (5.1a). See graph legend for the values of a.

Consequently,

−∫ t1

t0

u′′n(t)ω(−u′n(t))

dt ≥ −∫ t1

t0

(−u′n(t) + ψ(t) +

1c(B −A)

)dt,

∫ρ

0

dsω(s)

≤ un(t0) − un(t1) + ‖ψ‖1 +T

c(B −A) < r.

(5.32)

Hence, according to (5.15), we again have ρ < ρ∗.

Step 4 (convergence of {un}). Consider the sequence {un} of solutions of problems (5.21),n ∈ N, n > 1/T . It has been shown in Steps 2 and 3 that (5.22) and (5.27) hold, which meansthat the sequences {un} and {u′n} are bounded in C[0, T]. Therefore {un} is equicontinuouson [0, T]. According to (5.17), (5.19), and (5.21), we obtain for t ∈ [1/n, T],

u′n(t) = −∫T

t

(fn(s, un(s), u′n(s)

)+un(s)n

)ds

= −∫T

t

(a

su′n(s) + f

(s, un(s), u′n(s)

)+un(s) −A

n

)ds.

(5.33)

Let us now choose an arbitrary compact subinterval [a0, T] ⊂ (0, T]. Then there exists n0 ∈ N

such that [1/n, T] ⊂ [a0, T] for each n ≥ n0. By (5.33), the sequence {u′n} is equicontinuous on


6

6.5

7

7.5

8

8.5

9

9.5×10−4

0 0.2 0.4 0.6 0.8 1

(a)

−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4×10−5

0 0.2 0.4 0.6 0.8 1

(b)

Figure 5: Error estimate (a) and residual (b) for (5.43)-(5.1a), a = 1.

[a0, T]. Therefore, we can find a subsequence {um} such that {um} converges uniformly on[0, T], and {u′m} converges uniformly on [a0, T]. By the diagonalization theorem; see [11], wecan find a subsequence {u } such that there exists u ∈ C[0, T] ∩ C1(0, T] with

lim →∞

u (t) = u(t) uniformly on [0, T],

lim →∞

u′ (t) = u′(t) locally uniformly on (0, T].(5.34)

Therefore u(0) = u(T) and u′(T) = 0. For → ∞ in (5.33), Lebesgue’s dominated


−0.06

−0.04

−0.02

0

0.02

0.04

0.06

0.08

0 0.2 0.4 0.6 0.8 1

Figure 6: First derivative of the numerical solution to (5.43)-(5.1a) with a = 1.

convergence theorem yields

u′(t) = −∫T

t

(a

su′(s) + f

(s, u(s), u′(s)

))

ds, t ∈ (0, T]. (5.35)

Consequently, u ∈ AC1loc(0, T] satisfies equation (5.1a) a.e. on [0, T]. Moreover, due to (5.22)

and (5.27), we have

A ≤ u(t) ≤ B for t ∈ [0, T],∣∣u′(t)

∣∣ ≤ ρ∗ for t ∈ (0, T]. (5.36)

Hence (4.1) is satisfied. Applying Theorem 4.1, we conclude that u ∈ AC1[0, T] and u′(0) = 0.Therefore u satisfies the periodic conditions on [0, T]. Thus u is a solution of problem (5.1a)and (5.1b) and A ≤ u ≤ B on [0, T].

Example 5.4. Let T = 1, k ∈ N, ε = ±1, h ∈ Lp[0, 1] for some p > 1, and c0 ∈ C(0, 1). Moreover,let h be nonnegative, and let c0 be bounded on [0, 1]. Then in Theorem 5.3 the following classof functions f is covered:

f(t, x, y

)= h(t)

(x2k+1 + εexyn + c0(t) cos

(√|x|))

(5.37)

for a.e. t ∈ [0, 1] and all x, y ∈ R, provided n = 2m+ 1 if ε = 1 and n = 1 if ε = −1. In particular,for t ∈ (0, 1], x, y ∈ R

f1(t, x, y

)=

1√1 − t

(x3 + exy5 + cos

1t

cos√|x|), (5.38)


−0.939

−0.9385

−0.938

−0.9375

−0.937

0 0.01 0.02 0.03 0.04 0.05 0.06

n = 7n = 8

n = 9n = 10

(a)

−0.939

−0.9385

−0.938

−0.9375

−0.937

0.94 0.95 0.96 0.97 0.98 0.99 1

n = 7n = 8

n = 9n = 10

(b)

Figure 7: Numerical solutions of (5.43)-(5.1a) and a = 1 in the vicinity of t = 0 (a) and t = 1 (b). The stepsize is decreasing according to h = 1/2n.

or

f2(t, x, y

)=

1√1 − t

(x3 − exy + cos

1t

cos√|x|). (5.39)

In order to show the existence of solutions to the periodic boundary value problem (5.1a)and (5.1b), the Fredholm-type Existence Theorem is used, see for example, in [20, Theorem4], [11, Theorem 2.1] or [21, page 25]. For convenience, we provide its simple formulationsuitable for our purpose below.


1.55

1.6

1.65

1.7

1.75

1.8

1.85

1.9

0 0.2 0.4 0.6 0.8 1

a = 0.1a = 0.5a = 1

a = 2a = 5

Figure 8: Illustrating Theorem 5.6: solutions of the boundary value problem (5.44)-(5.1a). See graph legendfor the values of a.

Lemma 5.5 (Fredholm-type existence theorem). Let f satisfy (2.2), let matrices B0, B1 ∈ R2×2,

vector β ∈ R2 be given, and let c1, c2 ∈ L1[0, T]. Let us denote by U(t) := (u(t), u′(t))T , and assume

that the linear homogeneous boundary value problem

u′′ + c1(t)u′ + c2(t)u = 0, B0U(0) + B1U(T) = 0 (5.40)

has only the trivial solution. Moreover, let us assume that there exists a function m ∈ Lp[0, T] suchthat

∣∣f(t, x, y

)∣∣ ≤ m(t) for a.e. t ∈ [0, T] and all x, y ∈ R. (5.41)

Then the problem

u′′ + c1(t)u′ + c2(t)u = f(t, u, u′

), B0U(0) + B1U(T) = β (5.42)

has a solution u ∈ AC1[0, T].

If we combine Theorems 5.2 and 5.3, we obtain conditions sufficient for the solution of(5.1a) and (5.1b) to be unique.

Theorem 5.6 (existence and uniqueness). Let all assumptions of Theorems 5.2 and 5.3 hold. Thenproblem (5.1a) and (5.1b) has a unique solution u. Moreover u satisfies (5.14).


0.8

1

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8×10−3

0 0.2 0.4 0.6 0.8 1

(a)

−2.5

−2

−1.5

−1

−0.5

0

0.5×10−3

0 0.2 0.4 0.6 0.8 1

(b)

Figure 9: Error estimate (a) and residual (b) for (5.44)-(5.1a), a = 1.

Example 5.7. Functions satisfying assumptions of Theorem 5.6 can have the form

f(t, x, y

)=

a√1 − t

(x3 + exy5 + t

), (5.43)

f(t, x, y

)=

a√1 − t

(x3 − e−xy

)− 16√t, (5.44)

for t ∈ (0, 1], x, y ∈ R.

We now illustrate the above theoretical findings by means of numerical simulations.Figure 4 shows graphs of solutions of problem (5.43), (5.1a). In Figure 5 we display the errorestimate for the global error of the numerical solution and the so-called residual (defect)


−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0 0.2 0.4 0.6 0.8 1

Figure 10: First derivative of the numerical solution to (5.44)-(5.1a) with a = 1.

Table 1: Estimated convergence order for the periodic boundary value problem (5.43)-(5.1a) and a = 1.

i Error estimate Conv. order

1 5.042446e–003 —

2 2.850171e–003 0.823075

3 1.681410e–003 0.761377

4 1.029876e–003 0.707200

5 6.514046e–004 0.660845

6 4.231359e–004 0.622433

7 2.807926e–004 0.591616

8 1.894611e–004 0.567604

9 1.294654e–004 0.549335

10 8.930836e–005 0.535699

obtained from the substitution of the numerical solution into the differential equation. Bothquantities are rather small and indicate that we have found a solution to the analyticalproblem (5.43)-(5.1a).

We now pose that question about the values of the first derivative at the end points ofthe interval of integration, t = 0 and t = 1. According to the theory, it holds that u′(0) = u′(1) =0. Therefore, we approximate the values of the first derivative of the numerical solution andshow these values in Figure 6. One can see that indeed u′(0) ≈ 0, u′(1) ≈ 0. Also, to supportthis observation, we plotted in Figure 7 the numerical solutions obtained for the step size htending to zero, or equivalently, grids becoming finer.

We finally observe experimentally the order of convergence of the numerical method(collocation). Clearly, we do not expect very hight order to hold, since the analytical solutionhas nonsmooth higher derivatives. However, the method is convergent and, according toTable 1, we observe that its order tends to 1/2.

The results of the numerical simulation for the boundary value problem (5.44)-(5.1a),can be found in Figures 8, 9, 10, and 11.


1.564

1.566

1.568

1.57

1.572

1.574

1.576

1.578

1.58

1.582

1.584

0 0.01 0.02 0.03 0.04 0.05 0.06

n = 7n = 8

n = 9n = 10

(a)

1.566

1.568

1.57

1.572

1.574

1.576

1.578

1.58

1.582

1.584

1.586

0.94 0.95 0.96 0.97 0.98 0.99 1

n = 7n = 8

n = 9n = 10

(b)

Figure 11: Numerical solutions of (5.44)-(5.1a) and a = 1 in the vicinity of t = 0 (a) and t = 1 (b). The stepsize is decreasing according to h = 1/2n.

Acknowledgments

This research was supported by the Council of Czech Goverment MSM6198959214 and by theGrant no. A100190703 of the Grant Agency of the Academy of Sciences of the Czech Republic.

References

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[6] R. W. Dickey, “Rotationally symmetric solutions for shallow membrane caps,” Quarterly of AppliedMathematics, vol. 47, no. 3, pp. 571–581, 1989.

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[9] K. N. Johnson, “Circularly symmetric deformation of shallow elastic membrane caps,” Quarterly ofApplied Mathematics, vol. 55, no. 3, pp. 537–550, 1997.

[10] I. Rachunkova, O. Koch, G. Pulverer, and E. Weinmuller, “On a singular boundary value problemarising in the theory of shallow membrane caps,” Journal of Mathematical Analysis and Applications,vol. 332, no. 1, pp. 523–541, 2007.

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[12] J. Y. Shin, “A singular nonlinear differential equation arising in the Homann flow,” Journal ofMathematical Analysis and Applications, vol. 212, no. 2, pp. 443–451, 1997.

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[14] O. Koch, “Asymptotically correct error estimation for collocation methods applied to singularboundary value problems,” Numerische Mathematik, vol. 101, no. 1, pp. 143–164, 2005.

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[16] F. R. de Hoog and R. Weiss, “The numerical solution of boundary value problems with an essentialsingularity,” SIAM Journal on Numerical Analysis, vol. 16, no. 4, pp. 637–669, 1979.

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[20] A. Lasota, “Sur les problemes lineaires aux limites pour un systeme d’equations differentiellesordinaires,” Bulletin de l’Academie Polonaise des Sciences. Serie des Sciences Mathematiques, Astronomiqueset Physiques, vol. 10, pp. 565–570, 1962.

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Research ArticleMultiplicity Results Using BifurcationTechniques for a Class of Fourth-Order m-PointBoundary Value Problems

Yansheng Liu1 and Donal O’Regan2

1 Department of Mathematics, Shandong Normal University, Jinan 250014, China2 Department of Mathematics, National University of Ireland, Galway, Ireland

Correspondence should be addressed to Yansheng Liu, [email protected]

Received 13 March 2009; Accepted 12 April 2009


By using bifurcation techniques, this paper investigates the existence of nodal solutions for a classof fourth-order m-point boundary value problems. Our results improve those in the literature.

Copyright q 2009 Y. Liu and D. O’Regan. 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.

1. Introduction

Consider the following fourth order m-point boundary value problem (BVP, for short)

u(4)(t) = f(u(t), u′′(t)

), t ∈ (0, 1)

u′(0) = 0, u(1) =m−2∑

i=1

αiu(ηi)

u′′′(0) = 0, u′′(1) =m−2∑

i=1

αiu′′(ηi),

(1.1)

where f : R × R → R is a given sign-changing continuous function, m ≥ 3, ηi ∈ (0, 1), andαi > 0 for i = 1, . . . , m − 2 with

m−2∑

i=1

αi < 1. (1.2)


Multi-point boundary value problems for ordinary differential equations arise indifferent areas of applied mathematics and physics. The existence of solutions of the secondorder multi-point boundary value problems has been studied by many authors and themethods used are the nonlinear alternative of Leray-Schauder, coincidence degree theory,fixed point theorems in cones and global bifurcation techniques (see [1–9], and the referencestherein). In [5], Ma investigated the existence and multiplicity of nodal solutions for

u′′(t) + f(u(t)) = 0, t ∈ (0, 1);

u′(0) = 0, u(1) =m−2∑

i=1

αiu(ηi) (1.3)

when

ηi ∈ Q(i = 1, 2, . . . , m − 2) with 0 < η1 < η2 < · · · < ηm−2 < 1, (1.4)

and αi > 0 for i = 1, . . . , m − 2 satisfying (1.2). He obtained some results on the spectrum ofthe linear operator corresponding to (1.1). It should be pointed out that the main tool used in[5] is results on bifurcation coming from the trivial solutions and we note no use was madeof global results on bifurcation from infinity.

Recently [10] Wei and Pang studied the existence and multiplicity of nontrivialsolutions for the fourth order m-point boundary value problems:

u(4)(t) = f(u(t), u

′′(t)), t ∈ (0, 1)

u(0) = 0, u(1) =m−2∑

i=1

αiu(ηi)

u′′(0) = 0, u

′′(1) =

m−2∑

i=1

αiu′′(ηi),

(1.5)

where f : R × R → R is a given sign-changing continuous function, m ≥ 3, ηi ∈ (0, 1), andαi > 0 for i = 1, . . . , m − 2 satisfies (1.2).

Motivated by [5, 10], in this paper we consider the existence and multiplicity of nodalsolutions for BVP (1.1). The method used here is Rabinowitz’s global bifurcation theorem. Tothe best of our best knowledge, only [10] seems to have considered the existence of nontrivialor positive solutions of the nonlinear multi-point boundary value problems for fourth orderdifferential equations. As in [5, 10] we suppose (1.2) is satisfied throughout.

The paper is organized as follows. Section 2 gives some preliminaries. Section 3 isdevoted to the existence of multiple solutions for BVP (1.1). To conclude this section we givesome notation and state three lemmas, which will be used in Section 3. Following the notationof Rabinowitz, let E be a real Banach space and L : E → E be a compact linear map. If thereexists μ ∈ R = [0,+∞) and 0/=v ∈ E such that v = μLv, μ is said to be a real characteristic


value of L. The set of real characteristic values of L will be denoted by σ(L). The multiplicityof μ ∈ σ(L) is

dim∞⋃

j=1

N((I − μL)j

), (1.6)

where N(A) denotes the null space of A. Suppose that H : R × E → E is compact andH(λ, u) = o(‖u‖) at u = 0 uniformly on bounded λ intervals. Then

u = λLu +H(λ, u) (1.7)

possesses the line of trivial solutions Θ = {(λ, 0) | λ ∈ R}. It is well known that if μ ∈ R, anecessary condition for (μ, 0) to be a bifurcation point of (1.7) with respect to Θ is that μ ∈σ(L). If μ is a simple characteristic value of L, let v denote the eigenvector of L correspondingto μ normalized so ‖v‖ = 1. By Σ we denote the closure of the set of nontrivial solutions of(1.7). A component of Σ is a maximal closed connected subset. It was shown in (Rabinowitz[11, Theorems 1.3, 1.25, 1.27]), the following.

Lemma 1.1. If μ ∈ σ(L) is simple, then Σ contains a component Cμ which can be decomposed intotwo subcontinua C+

μ, C−μ such that for some neighborhood B of (μ, 0),

(λ, u) ∈ C+μ

(C−μ

)∩ B, (λ, u)/=

(μ, 0)

(1.8)

implies (λ, u) = (λ, αv +w) where α > 0(α < 0) and |λ − μ| = o(1), ‖w‖ = o(|α|) at α = 0.Moreover, each of C+

μ, C−μ either

(i) meets infinity in Σ, or(ii) meets (μ, 0) where μ/= μ ∈ σ(L), or(iii) contains a pair of points (λ, u), (λ,−u), u/= 0.

The following are global results for (1.7) on bifurcation from infinity (see, Rabinowitz[9, Theorem 1.6 and Corollary 1.8]).

Lemma 1.2. Suppose L is compact and linear,H(λ, u) is continuous on R × E,H(λ, u) = o(‖u‖) atu = ∞ uniformly on bounded λ intervals, and ‖u‖2H(λ, u/‖u‖2) is compact. If μ ∈ σ(L) is of oddmultiplicity, then Σ possesses an unbounded component Dμ which meets (μ,∞). Moreover if Λ ⊂ R

is an interval such that Λ ∩ σ(L) = {μ} and ℘ is a neighborhood of (μ,∞) whose projection on R liesin Λ and whose projection on E is bounded away from 0, then either

(i)Dμ \ ℘ is bounded in R × E in which case Dμ \ ℘ meets Θ = {(λ, 0) | λ ∈ R} or(ii)Dμ \ ℘ is unbounded.If (ii) occurs and Dμ \ ℘ has a bounded projection on R, then Dμ \ ℘ meets (μ,∞) where

μ/= μ ∈ σ(L).

Lemma 1.3. Suppose the assumptions of Lemma 1.2 hold. If μ ∈ σ(L) is simple, then Dμ can bedecomposed into two subcontinua D+

μ , D−μ and there exists a neighborhood I ⊂ ℘ of (μ,∞) such that

(λ, u) ∈ D+μ(D

−μ) ∩ I and (λ, u)/= (μ,∞) implies (λ, u) = (λ, αv + w) where α > 0(α < 0) and

|λ − μ| = o(1), ‖w‖ = o(|α|) at |α| =∞.


2. Preliminaries

Let X = C[0, 1] with the norm ‖u‖ = maxt∈[0,1]|u(t)|, Y = {u ∈ C1[0, 1] : u′(0) = 0, u(1) =∑m−2i=1 αiu(ηi)} with the norm ‖u‖1 = max{‖u‖, ‖u′‖}, Z = {u ∈ C2[0, 1] : u′(0) = 0, u(1) =∑m−2i=1 αiu(ηi)} with the norm ‖u‖2 = max{‖u‖, ‖u′‖, ‖u′′ ‖}. Then X, Y , and Z are Banach

spaces.For any C1 function u, if u(t0) = 0, then t0 is a simple zero of u if u′(t0)/= 0. For any

integer k ∈ N and any ν ∈ {±}, as in [6], define sets Tνk ⊂ Z consisting of the set of functions

u ∈ Z satisfying the following conditions:

(i) u′(0) = 0, νu(0) > 0 and u′(1)/= 0;

(ii) u′ has only simple zeros in (0, 1), and has exactly k − 1 such zeros;

(iii) u has a zero strictly between each two consecutive zeros of u′.

Note T−k = −T+k and let Tk = T−k ∪ T

+k . It is easy to see that the sets T−k and T+

k are disjointand open in Z. Moreover, if u ∈ Tν

k, then u has at least k − 2 zeros in (0, 1), and at most k − 1

zeros in (0, 1].Let E = R×Y under the product topology. As in [12], we add the points {(λ,∞) : λ ∈ R}

to the space E. Let Φ+k= R × T+

k, Φ−

k= R × T−

k, and Φk = R × Tk.

We first convert BVP (1.1) into another form. Suppose u(t) is a solution of BVP (1.1).Let v(t) = −u′′(t). Notice that

u′′(t) + v(t) = 0, t ∈ I;

u′(0) = 0, u(1) =m−2∑

i=1

αiu(ηi).

(2.1)

Thus u(t) can be written as

u(t) = Lv(t), (2.2)

where the operator L is defined by

Lv(t) :=∫1

0H(t, s)v(s)ds, ∀v ∈ Y, (2.3)

where

H(t, s) = G(t, s) +∑m−2

i=1 αiG(ηi, s)

1 −∑m−2

i=1 αiηi,

G(t, s) =

⎧⎨

⎩

1 − t, 0 ≤ s ≤ t ≤ 1;

1 − s, 0 ≤ t ≤ s ≤ 1.

(2.4)


Therefore we obtain the following equivalent form of (1.1)

v′′(t) + f((Lv)(t),−v(t)) = 0, t ∈ (0, 1);

v′(0) = 0, v(1) =m−2∑

i=1

αiv(ηi).

(2.5)

For the rest of this paper we always suppose that the initial value problem

v′′(t) + f((Lv)(t),−v(t)) = 0, t ∈ (0, 1);

v(t0) = v′(t0) = 0(2.6)

has the unique trivial solution v ≡ 0 on [0, 1] for any t0 ∈ [0, 1]; in fact some suitableconditions such as a Lipschitz assumption or f ∈ C1 guarantee this.

Define two operators on Y by

(Av)(t) := (LFv)(t), (Fv)(t) := f((Lv)(t),−v(t)), t ∈ I, v ∈ Y. (2.7)

Then it is easy to see the following lemma holds.

Lemma 2.1. The linear operator L and operator A are both completely continuous from Y to Y and

‖Lv‖1 ≤M‖v‖ ≤M‖v‖1, ∀v ∈ Y, (2.8)

whereM = max{1, (1/8)(1 +∑m−2

i=1 αi/(1 −∑m−2

i=1 αiηi))}.Moreover, u ∈ C4[0, 1] is a solution of BVP (1.1) if and only if v = −u′′ is a solution of the

operator equation v = Av.

Let the function Γ(s) be defined by

Γ(s) = cos s −m−2∑

i=1

αi cosηis, s ∈ R. (2.9)

Then we have the following lemma.

Lemma 2.2. (i) For each k ≥ 1, Γ(s) has exactly one zero sk ∈ Ik := ((k − 1)π, kπ), so

s1 < s2 < · · · < sk −→ +∞ (k −→ +∞); (2.10)

(ii) the characteristic value of L is exactly given by μk = s2k, k = 1, 2, . . ., and the eigenfunction

φk corresponding to μk is φk(t) = cos skt;(iii) the algebraic multiplicity of each characteristic value μk of L is 1;(iv)φk ∈ T+

kfor k = 1, 2, 3, . . ., and φ1 is strictly positive on (0, 1).


Proof. From [5] and by a similar analysis as in the proof of [6, Lemma 3.3] we obtain (i) and(ii).

Now we assert (iii) holds. Suppose, on the contrary, there exists y ∈ Y such that (I −μkL)y = μ−1

k φk. Then y ∈ Z and

−y′′ − s2ky = cos skt. (2.11)

From y′(0) = 0 we know the general solution of this differential equation is

y = C cos skt −1

2skt sin skt. (2.12)

From (i) and (ii) of this lemma, C cos skt satisfies the boundary condition. Thus

cos sk =m−2∑

i=1

αi cosηisk, sin sk =m−2∑

i=1

αiηi sinηisk. (2.13)

Then, by (1.2),

1 =

(m−2∑

i=1

αi cosηisk

)2

+

(m−2∑

i=1

αiηi sinηisk

)2

≤m−2∑

i,j=1

αiαj

(∣∣cosηisk cosηjsk∣∣ +∣∣sinηisk sinηjsk

∣∣)

≤(

m−2∑

i=1

αi

)2

< 1,

(2.14)

a contradiction. Thus the algebraic multiplicity of each characteristic value μk of L is 1.Finally, from sk ∈ ((k − 1)π, kπ) and s1 ∈ (0, π/2), it is easy to see that (iv) holds.

Lemma 2.3. For d = (d1, d2) ∈ R+ × R

+ \ {(0, 0)}, define a linear operator

Ldv(t) =(d1L

2 + d2L)v(t), ∀t ∈ I, v ∈ Y, (2.15)

where L is defined as in (2.3). Then the generalized eigenvalues of Ld are simple and are given by

0 < λ1(Ld) < λ2(Ld) < · · · < λk(Ld) −→ +∞ (k −→ +∞), (2.16)

where

λk(Ld) =μ2k

d1 + d2μk. (2.17)


The generalized eigenfunction corresponding to λk(Ld) is

φk(t) = cos skt, (2.18)

where μk, sk, φk are as in Lemma 2.2.

Proof. Suppose there exist λ and v /= 0 such that v = λLdv. Set u(t) = Lv(t). Then from (2.2)–(2.7) and (2.15) it is easy to see that u/= 0 and

u(4)(t) = λ(d1u(t) − d2u

′′(t)), t ∈ (0, 1);

u′(0) = 0, u(1) =m−2∑

i=1

αiu(ηi);

u”’(0) = 0, u′′(1) =

m−2∑

i=1

αiu′′(ηi).

(2.19)

Denote L−1u = −u′′ for u ∈ Z. Then there exist two complex numbers r1 and r2 suchthat

u(4)(t) − λ(d1u(t) − d2u

′′(t))=(L−1 − r2I

)(L−1 − r1I

)u(t) = 0. (2.20)

Now if there exists some ri(i = 1, 2) such that

(L−1 − riI

)u(t) = 0, (2.21)

then by Lemma 2.2 we know ri = s2k = μk for some k ∈ N, and consequently

u(t) = cos skt (2.22)

is a nontrivial solution. Substituting (2.22) into (2.19), we have

λ =μ2k

d1 + d2μk. (2.23)

On the other hand, suppose, for example,

(L−1 − r1I

)u(t)/= 0,

(L−1 − r2I

)(L−1 − r1I

)u(t) = 0. (2.24)


Let w(t) := (L−1 − r1I)u(t). Then (L−1 − r2I)w(t) = 0. Reasoning as previouslymentioned, we have r2 = s2

kfor some k ∈ N, and consequently w(t) = a cos skt(a/= 0) is a

nontrivial solution. Therefore,

(L−1 − r1I

)u(t) = a cos skt. (2.25)

If r1 = s2k, then the general solution of the differential equation (2.25), satisfying u′(0) =

0, is

u(t) = C cos skt −a

2skt sin skt, (2.26)

which is similar to (2.12). Reasoning as in the proof of Lemma 2.2 we can get a contradiction.Thus r1 /= s2

k and the general solution of (2.25), satisfying u′(0) = 0, is

u(t) = u(t) +a cos skts2k− r1

, (2.27)

where u(t) is the general solution of homogeneous differential equation corresponding to(2.25)

(L−1 − r1I

)u(t) = 0. (2.28)

Notice the term a cos skt/(s2k − r1) in (2.27) satisfies the boundary condition of (1.1) at

t = 1, so u(t) also satisfies

u′(0) = 0, u(1) =m−2∑

i=1

αiu(ηi). (2.29)

Therefore, by Lemma 2.2 we know u(t) = C cos sjt for some j ∈ N, and consequently

r1 = s2j /= s2

k, u(t) = C cos sjt +a cos skts2k − s

2j

. (2.30)

By substituting this into (2.19), we have

aλ(d1 + d2μk

)= aμ2

k, Cλ(d1 + d2μj

)= Cμ2

j . (2.31)

Since μj /=μk, if there exists some λ such that (2.31) holds, then

d1 + d2μk

d1 + d2μj=

μ2k

μ2j

, (2.32)


which implies

d1d2 /= 0, d1

(1μk

+1μj

)

= −d2, (2.33)

a contradiction with d1 > 0 and d2 > 0.Consequently, (2.24) does not hold. This together with (2.20)–(2.23) and Lemma 2.2

guarantee that the generalized eigenvalues of Ld are given by

0 < λ1(Ld) < λ2(Ld) < · · · < λk(Ld) −→ +∞ (k −→ +∞), (2.34)

where λk(Ld) = μ2k/(d1 + d2μk). The generalized eigenfunction corresponding to λk(Ld) is

φk(t) = cos skt.Now we are in a position to show the generalized eigenvalues of Ld are simple.Clearly, from above we know for λk := λk(Ld), (I−λkLd)φk = 0 and dimN(I−λkLd) = 1.

Suppose there exists an v ∈ C2 such that

(I − λkLd)v =1μk

φk(t). (2.35)

This together with (2.3) and (2.15) guarantee that v ∈ Y . If we let u(t) = (Lv)(t) as above,then we have

u(4)(t) − λk(d1u(t) − d2u

′′(t))= cos skt, t ∈ (0, 1), (2.36)

u′(0) = 0, u(1) =m−2∑

i=1

αiu(ηi); u”’(0) = 0, u

′′(1) =

m−2∑

i=1

αiu′′(ηi). (2.37)

Consider the following homogeneous equation corresponding to (2.36):

u(4)(t) −μ2k

d1 + d2μk

(d1u(t) − d2u

′′(t))= 0. (2.38)

The characteristic equation associated with (2.38) is

λ4 −μ2k

d1 + d2μk

(d1 − d2λ

2)= 0. (2.39)

Then there exists a real number η such that

(λ2 + μk

)(λ2 − η

)= λ4 −

μ2k

d1 + d2μk

(d1 − d2λ

2)= 0. (2.40)

Notice that −ημk = −d1μ2k/(d1 + d2μk) < 0 if d1 > 0. So η > 0 if d1 > 0, and η = 0 if d1 = 0.


First we consider the case d1 > 0. In this case the general solution of (2.38) is

c1e√ηt + c2e

−√ηt + c3 cos skt + c4 sin skt. (2.41)

After computation we obtain that the general solution of (2.36) is

u(t) = c1e√ηt + c2e

−√ηt + c3 cos skt + c4 sin skt + at sin skt, (2.42)

where a = −(d1 + d2μk)/2sk(2d1μk + d2μ2k). From boundary condition u′(0) = u”’(0) = 0 in

(2.37) it follows that

√η(c1 − c2)skc4 = 0;

η√η(c1 − c2) − s3

kc4 = 0.(2.43)

By η > 0 and μk > 0, we know c1 − c2 = 0 and c4 = 0. Then (2.42) can be rewritten as

u(t) = c1[e√ηt + e−

√ηt] + c3 cos skt + at sin skt. (2.44)

Notice that the term c3 cos skt satisfies (2.37). From the boundary condition

u(1) =m−2∑

i=1

αiu(ηi), u

′′(1) =

m−2∑

i=1

αiu′′(ηi),

(t sin skt)′′= 2sk cos skt − s2

kt sin skt

(2.45)

we have

c1[e√η + e−

√η] + a sin sk =

m−2∑

i=1

αi

[c1(e√ηηi + e−

√ηηi)+ aηi sin skηi

], (2.46)

c1η[e√η + e−

√η] − as2

k sin sk =m−2∑

i=1

αi

[c1η(e√ηηi + e−

√ηηi)− aηis2

k sin skηi]. (2.47)

Multiply (2.46) by s2k

and then add to (2.47) to obtain

c1

(η + s2

k

)[e√η + e−

√η] = c1

(η + s2

k

)m−2∑

i=1

αi

(e√ηηi + e−

√ηηi). (2.48)

On the other hand, from (1.2) it can be seen that

e√η + e−

√η >

m−2∑

i=1

αi

(e√ηηi + e−

√ηηi). (2.49)


This together with (2.48) guarantee that c1 = 0. Therefore, (2.42) reduces to

u(t) = c3 cos skt + at sin skt. (2.50)

Similar to (2.12), a contradiction can be derived.Next consider the case d1 = 0. Then η = 0 from above. In this case the general solution

of (2.38) is

c1 + c2t + c3 cos skt + c4 sin skt. (2.51)

By a similar process, one can easily get a contradiction.To sum up, the generalized eigenvalues of Ld are simple, and the proof of this lemma

is complete.

3. Main Results

We now list the following hypotheses for convenience.

(H1) There exists a = (a1, a2) ∈ R+ × R

+ \ {(0, 0)} such that

f(x, y)= a1x − a2y + o

(∣∣(x, y)∣∣), as

∣∣(x, y)∣∣ −→ 0, (3.1)

where (x, y) ∈ R × R, and |(x, y)| := max{|x|, |y|}.(H2) There exists b = (b1, b2) ∈ R

+ × R+ \ {(0, 0)} such that

f(x, y)= b1x − b2y + o

(∣∣(x, y)∣∣), as

∣∣(x, y)∣∣ −→ ∞. (3.2)

(H3) There exists R > 0 such that

∣∣f(x, y)∣∣ <

R

M, for

(x, y)∈{(

x, y)

: |x| ≤MR,∣∣y∣∣ ≤ R

}, (3.3)

where M is defined as in Lemma 2.1.

(H4) There exist two constants r1 < 0 < r2 such that f(x,−r1) ≥ 0 and f(x,−r2) ≤ 0for x ∈ [−Mr,Mr], and f(x,−y) satisfies a Lipschitz condition in y for (x, y) ∈[−Mr,Mr] × [r1, r2], where r = max{|r1|, r2}.

Now we are ready to give our main results.

Theorem 3.1. Suppose (H1)-(H2) hold. Suppose there exists two integers i0 ≥ 0 and k > 0 such thateither

μ2i0+k

a1 + a2μi0+k< 1 <

μ2i0+1

b1 + b2μi0+1(3.4)


or

μ2i0+k

b1 + b2μi0+k< 1 <

μ2i0+1

a1 + a2μi0+1(3.5)

holds. Then BVP (1.1) has at least 2k nontrivial solutions.

Theorem 3.2. Suppose (H1) and (H2) hold and one of (H3) and (H4) hold. Suppose there exists twointegers i0 and j0 such that

μ2i0

a1 + a2μi0

< 1,μ2j0

b1 + b2μj0

< 1. (3.6)

Then BVP(1.1) has at least 2(i0 + j0) solutions.

To set it up we first consider global results for the equation

v = λAv, (3.1λ)

on Y , where λ ∈ R, and the operator A is defined as in (2.7). Under the condition (H1), (3.1λ)can be rewritten as

v = λLav +Ha(λ, v), (3.7)

here Ha(λ, v) = λAv−λLav, La is defined as in (2.12) (replacing d with a). Obviously, by (H1)and Lemma 2.1–2.3, it can be seen that Ha(λ, v) is o(‖v‖1) for v near 0 uniformly on boundedλ intervals and La is a compact linear map on Y . A solution of (3.1λ) is a pair (λ, v) ∈ E. By(H1), the known curve of solutions {(λ, 0) | λ ∈ R}will henceforth be referred to as the trivialsolutions. The closure of the set on nontrivial solutions of (3.1λ) will be denoted by Σ as inLemma 1.1.

If Ha(λ, v) ≡ 0, then (3.7) becomes a linear system

v = λLav. (3.8)

By Lemma 2.3, (3.8) possesses an increasing sequence of simple eigenvalues

0 < λ1 < λ2 < · · · < λk < · · · , with λk =μ2k

a1 + a2μkas k −→ +∞. (3.9)

Any eigenfunction φk = cos skt corresponding to λk is in T+k .

A similar analysis as in [6, Proposition 4.1] yields the following results.

Lemma 3.3. Suppose that (λ, v) is a solution of (3.1λ) and v /≡ 0. Then v ∈ ∪∞i=1Ti.


Lemma 3.4. Assume that (H1) holds and λk is defined by (3.9). Then for each integer k > 0 and eachν = +, or −, there exists a continua Cν

kof solutions of (3.1λ) in Φν

k∪ {(λk, 0)}, which meets {(λk, 0)}

and∞ in Σ.

Proof. Consider (3.7) as a bifurcation problem from the trivial solution. From Lemma 1.1and condition (H1) it follows that for each positive integer k ∈ N, Σ contains a componentCk ⊆ E = R × Y which can be decomposed into two subcontinua C+

k, C−

ksuch that for some

neighborhood B of (λk, 0),

(λ, v) ∈ C+k

(C−k)∩ B, (λ, v)/= (λk, 0) (3.10)

imply (λ, v) = (λ, αφk +w), where α > 0(α < 0) and |λ − λk| = o(1), ‖w‖1 = o(|α|) at α = 0.By (3.7) and the continuity of the operator A : Y → Z, the set Cν

klies in R ×Z and the

injection Cνk → R × Z is continuous. Thus, Cν

k is also a continuum in R × Z, and the aboveproperties hold in R × Z.

Since Tk is open in Z and φk ∈ T+k

, we know

v

α= φk +

w

α∈ T+

k (3.11)

for 0/=α sufficiently small. Then there exists ε0 > 0 such that for ε ∈ (0, ε0), we have

(λ, v) ∈ Φk,

(Ck

{(λk, 0)}

)∩ Bε ⊂ Φk, (3.12)

where Bε is an open ball in R × Z of radius ε centered at (λk, 0). It follows from the proof of[6, Proposition 4.1] that

(λ, v) ∈ Ck ∩ (R × ∂Tk) =⇒ u = 0, (3.13)

which means Ck \ {(λk, 0)} ∩ ∂Φk = ∅. Consequently, Ck lies in Φk ∪ {(λk, 0)}.Similarly we have that Cν

k lies in Φνk ∪ {(λk, 0)} (ν = + or −).

Next we show alternative (ii) of Lemma 1.1 is impossible. If not, without loss ofgenerality, assume that C+

kmeets (λi, 0) with λk /=λi ∈ σ(La). Then there exists a sequence

(ξm, zm) ∈ C+k with ξm → λi and zm → 0 as m → +∞. Notice that

zm = ξmLazm +H(ξm, zm). (3.14)

Dividing this equation by ‖zm‖1 and using Lemma 2.1 and H(ξm, zm) = o(‖zm‖1) as m →+∞, we may assume without loss of generality that zm/‖zm‖1 → z as m → +∞. Thus from(3.14) it follows that

z = λiLaz. (3.15)

Since z/= 0, by Lemmas 2.2 and 2.3, z belongs to T+i or T−i . By (3.1λ) and the continuity of the

operator A : Y → Z, from ‖zm − z‖1 → 0 it follows that ‖zm − z‖2 → 0. Notice that T+i and


T−i are open in Z. Therefore, zm ∈ T+i or T−i for m sufficiently large, which is a contradiction

with zm ∈ T+k(m ≥ 1), i /= k. Hence alternative (ii) of Lemma 1.1 is not possible.

Finally it remains to show alternative (iii) of Lemma 1.1 is impossible. In fact, noticethat T−k = −T+

k , and T−k ∩ T+k = ∅. If u ∈ T+

k , then −u ∈ T−k . This guarantees that Cνk does not

contain a pair of points (λ, v), (λ,−v), v /= 0.Therefore alternative (i) of Lemma 1.1 holds. This implies that for each integer k ∈ N

and each ν = +, or −, there exists a continua Cνk of solutions of (3.1λ) in Φν

k ∪ {(λk, 0)}, whichmeets {(λk, 0)} and∞ in Σ.

Under the condition (H2), (3.1λ) can be rewritten as

v = λLbv +Hb(λ, v), (3.16)

here Hb(λ, v) = λAv − λLbv, Lb is defined as in (2.12) (replacing d with b).Let h(x, y) := f(x, y) − b1x + b2y. Then from (H2) it follows that lim|(x,y)|→∞h(x, y)/

|(x, y)| = 0. Define a function

h(r) := max{∣∣h(x, y)∣∣ :∣∣(x, y

)∣∣ ≤ r}. (3.17)

Then h(r) is nondecreasing and

limr→∞

h(r)r

= 0. (3.18)

Obviously, by (3.18) and Lemma 2.1, it can be seen that Hb(λ, v) is o(‖v‖1) for v near ∞uniformly on bounded λ intervals and Lb is a compact linear map on Y .

Similar to (3.8), by Lemma 2.3, Lb possesses an increasing sequence of simpleeigenvalues

0 < λ1 < λ2 < · · · < λk < · · · , with λk =μ2k

b1 + b2μkas k −→ +∞. (3.19)

Note φk = cos skt is an eigenfunction corresponding to λk. Obviously, it is in T+k

.

Lemma 3.5. Assume that (H1)-(H2) holds. Then for each integer k > 0 and each ν = +, or −, thereexists a continua Dν

kof Σ in Φν

k∪ {(λk,∞)} coming from {(λk,∞)}, which meets (λk, 0) or has an

unbounded projection on R.

Proof. From (2.7), (3.16), and (3.18) it follows that Hb(λ, v) is continuous on E, Hb(λ, v) =o(‖v‖1) at v =∞ uniformly on bounded λ intervals. Moreover, as in the proof of [12, Theorem2.4], one can see that ‖v‖2

1Hb(λ, v/‖v‖21) is compact. From Lemma 2.3 we know λk is a simple

characteristic value of Lb for each integer k ∈ N. Thus by Lemmas 1.2 and 1.3, Σ contains acomponent Dk which can be decomposed into two subcontinua D+

k, D−

kwhich meet {(λk,∞)}.

Now we show that for a smaller neighborhood I ⊂ ℘ of (λk,∞), (λ, v) ∈ D+k(D−

k) ∩ I

and (λ, v)/= (λk,∞) imply that v ∈ T+k(T−

k). In fact, by Lemma 1.3 we already know that there


exists a neighborhood I ⊂ ℘ of (λk,∞) satisfying (λ, v) ∈ D+k(D

−k) ∩ I and (λ, v)/= (λk,∞)

imply (λ, v) = (λ, αvk +w) where α > 0(α < 0) and |λ − λk| = o(1), ‖w‖1 = o(|α|) at |α| =∞.As in the proof of Lemma 3.4, Dν

k is also a continuum in R×Z, and the above propertieshold in R × Z. Since Tν

kis open in Z and w/α is smaller compared to φk ∈ T+

knear α = +∞,

φk +w/α and therefore v = αφk +w ∈ T+k

for α near +∞. Here and in the following the sameargument works if + is replaced by −.

Therefore, D+k∩I ⊂ (R×T+

k )∪(λk,∞). Now we have two cases to consider, that is, D+k \I

is bounded or unbounded. First suppose D+k\ I is bounded. Then there exists (λ, v) ∈ ∂D+

k

with v ∈ ∂T+k

. If v /= 0, by Lemma 3.3 we know v ∈ Tνj for some positive integer j /= k and

ν ∈ {+,−}. As in the proof of Lemma 3.4, we get a contradiction, which means v = 0. Thusthere exists a sequence (ξm, zm) ∈ D+

k with zm → v ≡ 0 as m → +∞. This together with (H1)guarantee that (ξm, zm) satisfies (3.14).

As in the proof of Lemma 3.4, we may assume without loss of generality thatzm/‖zm‖1 → z and ξm → ξ as m → +∞. Then we have

z = ξLaz. (3.20)

Since z/= 0, ξ /= 0 is an eigenvalue of operator La. From this, (3.9), and (3.19) it follows thatξ = λj for some positive integer j. Then by Lemma 2.3, z belongs to T+

j or T−j . Notice that‖zm − z‖1 → 0 and so ‖zm − z‖2 → 0 as in the proof of Lemma 3.4. Thus zm ∈ T+

j or T−j form sufficiently large since T+

j and T−j are open. This together with zm ∈ T+k(m ≥ 1) guarantee

that k = j. This means D+k

meets (λk, 0) if D+k\ I is bounded.

Next suppose D+k\ I is unbounded. In this case we show D+

k\ I has an unbounded

projection on R. If not, then there exists a sequence (ζm, ym) ∈ D+k\ I with ζm → ζ and

‖ym‖1 → +∞ as m → +∞. Let xm := ym/‖ym‖1, m ≥ 1. From the fact that

ym = ζmLbym +Hb

(ζm, ym

)(3.21)

it follows that

xm = ζmLbxm +Hb

(ζm, ym

)

∥∥ym

∥∥1

. (3.22)

Notice that Lb : Y → Y is completely continuous. We may assume that there exists w ∈ Ywith ‖w‖1 = 1 such that ‖xm −w‖1 → 0 as m → +∞.

Letting m → +∞ in (3.22) and noticing Hb(ζm, ym)/‖ym‖1 → 0 as m → +∞ oneobtains

w = ζLbw. (3.23)

Since w/= 0, ζ /= 0 is an eigenvalue of operator Lb, that is, ζ = λk0 for some positive integerk0 /= k. Then by Lemma 2.3 w belongs to T+

k0or T−

k0. Notice the fact that ‖xm −w‖1 → 0 and

so ‖xm −w‖2 → 0 as in the proof of Lemma 3.4. Thus xm ∈ T+k0

or T−k0for m sufficiently large

since T+k0

and T−k0

are open. This is a contradiction with xm ∈ T+k(m ≥ 1). Thus D+

k\ I has an

unbounded projection on R.


Proof of Theorem 3.1. Suppose first that

μ2i0+k

a1 + a2μi0+k< 1 <

μ2i0+1

b1 + b2μi0+1. (3.24)

Using the notation of (3.9) and (3.19), this means λi0+k < 1 < λi0+1 and so from Lemma 2.3 weknow

λi0+1 < λi0+2 < · · · < λi0+k < 1 < λi0+1 < λi0+2 < · · · < λi0+k. (3.25)

Consider (3.7) as a bifurcation problem from the trivial solution. We need only showthat

Cνi0+j

⋂({1} × Y )/= ∅, j = 1, 2, . . . , k; ν = +,−. (3.26)

Suppose, on the contrary and without loss of generality, that

C+i0+i

⋂({1} × Y ) = ∅, for some i, 1 ≤ i ≤ k. (3.27)

By Lemma 3.4 we know that C+i0+i

joins (λi0+i, 0) to infinity in Σ and (λ, v) = (0, 0) is the uniquesolution of (3.1λ) (in which λ = 0) in E. This together with λi0+i < 1 guarantee that there existsa sequence {(ζm, ym)} ⊂ C+

i0+isuch that ζm ∈ (0, 1) and ‖ym‖1 → ∞ as m → +∞. We may

assume that ζm → ζ ∈ [0, 1] as m → +∞. Let xm := ym/‖ym‖1, m ≥ 1, then (3.22) holds.Similarly, we may assume that there exists w ∈ Y with ‖w‖1 = 1 such that ‖xm −w‖1 → 0as m → +∞ and (3.23) holds. From the proof of Lemma 3.5 one can see ζ = λi0+i, whichcontradicts λi0+i > 1. Thus (3.27) is not true, which means (3.26) holds.

Next suppose that

μ2i0+k

b1 + b2μi0+k< 1 <

μ2i0+1

a1 + a2μi0+1. (3.28)

This means

λi0+1 < λi0+2 < · · · < λi0+k < 1 < λi0+1 < λi0+2 < · · · < λi0+k. (3.29)

Consider (3.16) as a bifurcation problem from infinity. As above we need only to provethat

Dνi0+j

⋂({1} × Y )/= ∅, j = 1, 2, . . . , k; ν = +,−. (3.30)

From Lemma 3.5, we know that Dνi0+j

comes from {(λi0+j ,∞)}, meets (λi0+j , 0) or has anunbounded projection on R. If it meets (λi0+j , 0), then the connectedness of Dν

i0+jand λi0+j > 1


guarantees that (3.30) is satisfied. On the other hand, if Dνi0+j

has an unbounded projection onR, notice that (λ, v) = (0, 0) is the unique solution of (3.1λ) (in which λ = 0) in E, so (3.30)also holds.

Proof of Theorem 3.2. First suppose that (H3) holds. Then there exists ε > 0 such that

(1 + ε)∣∣f(x, y)∣∣ <

R

M, for

(x, y)∈{(

x, y)

: |x| ≤MR,∣∣y∣∣ ≤ R

}. (3.31)

Let (λ, v) be a solution of (3.1λ) such that 0 ≤ λ < 1 + ε and ‖v‖1 ≤ R. Then by (2.7),(3.1λ), (3.31) and Lemma 2.1 it is easy to see

‖v‖1 = λ‖Av‖1 = λ‖LFv‖1 ≤ λM‖Fv‖ = Mmaxt∈[0,1]

∣∣λf((Lv)(t),−v(t))

∣∣ < M

R

M= R. (3.32)

Therefore,

Σ ∩([0, 1 + ε] × ∂BR

)= ∅. (3.33)

This together with (3.32) and Lemmas 3.4 and 3.5 implies that

Cνk ∩([0, 1 + ε] × BR

)⊂ [0, 1 + ε] × BR, k = 1, 2, . . . , i0; (3.34)

Dνj ∩([0, 1 + ε] × ∂BR

)= ∅, j = 1, 2, . . . , j0; (3.35)

where BR = {v ∈ Y | ‖v‖1 < R} and BR = {v ∈ Y | ‖v‖1 ≤ R}, Cνk and Dν

j are obtained fromLemmas 3.4 and 3.5, respectively.

Since Cνk

is a unbounded component of solutions of (3.1λ) joining (λk, 0) in E, it followsfrom (3.33) and (3.34) that Cν

kcrosses the hyperplane {1} × Y with (1, vν) such that ‖vν‖1 < R

(ν = + or −, k = 1, 2, . . . , i0). This means BVP (2.5) has 2i0 nontrivial solutions {vνi }

i01 in which

v+1 and v−1 are positive and negative solutions, respectively.

On the other hand, by (3.33), (3.35), and Lemma 3.5 one can obtain

Dνj ∩({1} ×

(Y \ BR

))/= ∅, j = 1, 2, . . . , j0. (3.36)

This means BVP (2.5) has 2j0 nontrivial solutions {wνi }

j01 in which w+

1 and w−1 are positive andnegative solutions, respectively.

Now it remains to show this theorem holds when the condition (H4) is satisfied.From the above we need only to prove that

(i) for (λ, v) ∈ Cνi (ν = + or −, i = 1, 2, . . . , i0),

r1 < v(t) < r2, t ∈ [0, 1]. (3.37)


(ii) for (λ, v) ∈ Dνj (ν = + or −, j = 1, 2, . . . , j0), we have that either

maxt∈[0,1]

v(t) > r2, t ∈ [0, 1] (3.38)

or

mint∈[0,1]

v(t) < r1, t ∈ [0, 1]. (3.39)

In fact, like in [13], suppose on the contrary that there exists (λ, v) ∈ Cνi

⋃Dν

j such thateither

max{v(t) : t ∈ [0, 1]} = r2 (3.40)

or

min{v(t) : t ∈ [0, 1]} = r1 (3.41)

for some i, j.First consider the case max{v(t) : t ∈ [0, 1]} = r2. Then there exists t ∈ [0, 1] such that

v(t) = r2. Let

τ0 =: inf{t ∈[0, t]

: v(s) ≥ for s ∈[t, t]}

,

τ1 =: sup{t ∈[t, 1]

: v(s) ≥ 0 for s ∈[t, t]}

.

(3.42)

Then

max{v(t) : t ∈ [τ0, τ1]} = r2, (3.43)

0 ≤ v(t) ≤ r2, t ∈ [τ0, τ1]. (3.44)

Therefore, v(t) is a solution of the following equation

−v′′(t) = λf((Lv)(t),−v(t)), t ∈ (τ0, τ1) (3.45)

with v(τ0) = v(τ1) = 0 if 0 < τ0 < τ1 < 1 and v′(τ0) = 0 if τ0 = 0.By (H4), there exists M ≥ 0 such that f(x,−y) + My is strictly increasing in y for

(x, y) ∈ [r1, r2] × [−Mr,Mr], where r = max{|r1|, r2}. Then

−v′′ + λMv = λ(f((Lv)(t),−v(t)) +Mv

), t ∈ (τ0, τ1). (3.46)


Using (H4) and Lemma 2.1 again, we can obtain

− (r2 − v(t))′′+ λM(r2 − v(t))

= −λ[f((Lv)(t),−v(t)) +Mv(t) −Mr2

]

= −λ[f((Lv)(t),−v(t)) +Mv(t) −

(f((Lv)(t),−r2) +Mr2

)]− λf((Lv)(t),−r2)

≥ 0, t ∈ (τ0, τ1)

(3.47)

and if τ1 = 1, by (1.2) we know v(1) < r2. Therefore,

r2 − v(τ0) > 0, r2 − v(τ1) > 0 if 0 < τ0 < τ1 < 1;

(r2 − v(τ0))′ = 0 if τ0 = 0;

r2 − v(τ1) > 0 if τ1 = 1.

(3.48)

This together with (3.47) and the maximum principle [14] imply that r2 − v(t) > 0 in[τ0, τ1], which contradicts (3.43).

The proof in the case min{v(t) : t ∈ [0, 1]} = r1 is similar, so we omit it.

Acknowledgments

The Project Supported by NNSF of P. R. China (10871120), the Key Project of Chinese Ministryof Education (No: 209072), and the Science & Technology Development Funds of ShandongEducation Committee (J08LI10).

References

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[2] C. P. Gupta, “A generalized multi-point boundary value problem for second order ordinarydifferential equations,” Applied Mathematics and Computation, vol. 89, no. 1-3, pp. 133–146, 1998.

[3] R. Ma and N. Castaneda, “Existence of solutions of nonlinear m-point boundary-value problems,”Journal of Mathematical Analysis and Applications, vol. 256, no. 2, pp. 556–567, 2001.

[4] R. Ma and D. O’Regan, “Nodal solutions for second-order m-point boundary value problems withnonlinearities across several eigenvalues,” Nonlinear Analysis: Theory, Methods & Applications, vol. 64,no. 7, pp. 1562–1577, 2006.

[5] R. Ma, “Nodal solutions for a second-order m-point boundary value problem,” CzechoslovakMathematical Journal, vol. 56(131), no. 4, pp. 1243–1263, 2006.

[6] B. P. Rynne, “Spectral properties and nodal solutions for second-order, m-point, boundary valueproblems,” Nonlinear Analysis: Theory, Methods & Applications, vol. 67, no. 12, pp. 3318–3327, 2007.

[7] J. Sun, X. Xu, and D. O’Regan, “Nodal solutions for m-point boundary value problems usingbifurcation methods,” Nonlinear Analysis: Theory, Methods & Applications, vol. 68, no. 10, pp. 3034–3046, 2008.

[8] J. R. L. Webb, “Positive solutions of some three point boundary value problems via fixed point indextheory,” Nonlinear Analysis: Theory, Methods & Applications, vol. 47, no. 7, pp. 4319–4332, 2001.

[9] X. Xu, “Multiple sign-changing solutions for some m-point boundary-value problems,” ElectronicJournal of Differential Equations, vol. 2004, no. 89, pp. 1–14, 2004.


[10] Z. Wei and C. Pang, “Multiple sign-changing solutions for fourth order m-point boundary valueproblems,” Nonlinear Analysis: Theory, Methods & Applications, vol. 66, no. 4, pp. 839–855, 2007.

[11] P. H. Rabinowitz, “Some global results for nonlinear eigenvalue problems,” Journal of FunctionalAnalysis, vol. 7, no. 3, pp. 487–513, 1971.

[12] P. H. Rabinowitz, “On bifurcation from infinity,” Journal of Differential Equations, vol. 14, no. 3, pp.462–475, 1973.

[13] R. Ma, “Global behavior of the components of nodal solutions of asymptotically linear eigenvalueproblems,” Applied Mathematics Letters, vol. 21, no. 7, pp. 754–760, 2008.

[14] M. H. Protter and H. F. Weinberger, Maximum Principles in Differential Equations, Springer, New York,NY, USA, 1984.


Research ArticleMultipoint Singular Boundary-Value Problem forSystems of Nonlinear Differential Equations

Jaromır Bastinec,1 Josef Diblık,1, 2 and Zdenek Smarda1

1 Department of Mathematics, Faculty of Electrical Engineering and Communication,Brno University of Technology, Technicka 8, 616 00 Brno, Czech Republic

2 Department of Mathematics and Descriptive Geometry, Faculty of Civil Engineering,Brno University of Technology, Zizkova 17, 662 37 Brno, Czech Republic

Correspondence should be addressed to Josef Diblık, [email protected]

Received 14 April 2009; Revised 9 July 2009; Accepted 16 August 2009


A singular Cauchy-Nicoletti problem for a system of nonlinear ordinary differential equationsis considered. With the aid of combination of Wazewski’s topological method and Schauder’sprinciple, the theorem concerning the existence of a solution of this problem (having the graphin a prescribed domain) is proved.

Copyright q 2009 Jaromır Bastinec 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.

1. Introduction

In the present paper the following Cauchy-Nicoletti problem

y′i = fi(x, y), i = 1, . . . , n, (1.1)

yp

(x+p

)= Ap, yq

(x±q

)= Aq, yr

(x−r)= Ar,

p = 1, . . . , k; q = k + 1, . . . , s; r = s + 1, . . . , n(1.2)

is considered, where y = (y1, . . . , yn), x ∈ I = [a, b] and a = x1 = · · · = xk < xk+1 ≤ · · · ≤xs < xs+1 = · · · = xn = b; Ai, i = 1, . . . , n are real constants. Denote Ii = I \ {xi}, i = 1, . . . , nand J =

⋂ni=1Ii. We will suppose fi ∈ C(Θi,R), i = 1, . . . , n where the domain Θi ⊂ Ii × R

n

(satisfying a relation Θi∩{x = x∗}/= ∅ for every x∗ ∈ Ii) is more precisely specified in Section 2.The continuity of the function fi is not required at the point xi, i = 1, . . . , n. Solution of theproblem (1.1), (1.2) is defined in the following sense.


Definition 1.1. A vector-function y(x) = (y1(x), . . . , yn(x)) ∈ C(I,Rn) with yi ∈ C1(Ii,R),i = 1, . . . , n, is said to be a solution of the problem (1.1), (1.2) if it satisfies the system (1.1) onJ and if, moreover, conditions (1.2) hold.

Although singular boundary value problems were widely considered by using variousmethods (see, e.g., [1–11]), the method used here is based on a different approach. Namely,it uses simultaneously the topological method of Wazewski and Schauder’s principle.Note that the method of Wazewski (see, e.g., [12–14]) was used for the investigation ofvarious asymptotic and singular problems, for example, in [3–6, 11, 12, 15]. For successfulgeneralization of this to multipoint boundary-value problems, the basic obstacle must beovercome: the applying of topological method assumes that every intersection of so-calledregular polyfacial set and the plane x = x∗ = const, x∗ ∈ (a, b) is an open set in the spaceof dependent variables. Nature of the problems considered, as followed from problem (1.1),(1.2) does not permit straightforward generalization of this approach since the cross-sectionby the plane x = xi, i = 1, . . . , n is not an open set in the space y. The above mentioned obstacleis overcome in the present paper by connecting the topological method and the fixed pointtheorem.

Let us explain the main idea of this approach. Each equation of the system (1.1) isconsidered separately (as a scalar equation) under the supposition that nondiagonal variablesare changed by functions taken from a prescribed set M of vector functions. For everyscalar equation (together with the corresponding Cauchy initial condition which is subtractedfrom (1.2)) it is, with the aid of Wazewski’s method and qualitative properties of solutionsof differential equations, showed that there exists its solution having the same propertieswhich were supposed for the corresponding coordinate of vector functions from M. Inthis way an operator T is defined. For verification of conditions of Schauder’s principle(namely, the continuity of operator T), Wazewski’s method is used again. Stationary pointof operator T defines a solution of the problem (1.1), (1.2). The paper is organized as follows.In Section 2 the main result is formulated. Illustrative examples are contained in Section 3.Auxiliary results are stated in Section 4. In Section 5 we prove results concerning scalarsingular problems and the last section contains the proof of the main result.

2. Existence of Solutions of the Problem (1.1), (1.2)Let αi, βi ∈ C1(I,R), i = 1, 2, . . . , n be functions satisfying αi(xi) = βi(xi) = Ai and αi(x) < βi(x)on Ii. Define

Ω ={(

x, y1, . . . , yn

): x ∈ I, αi(x) ≤ yi ≤ βi(x), i = 1, . . . , n

},

Ωi ={(

x, y1, . . . , yn

): x ∈ Ii,

(x, y1, . . . , yn

)∈ Ω}.

(2.1)

Let us suppose that there exists a domain Θi, i = 1, 2 . . . , n such that Ωi ⊂ Θi ⊂ Ii × Rn; cross

section Si(x) = {(x, y) ∈ Θi} is an open set for every fixed x ∈ Ii and fi ∈ C(Θi,R). Theseassumptions are supposed in the sequel. Define, moreover,

Γi ={(

x, yi1 , . . . , yin−1

): x ∈ Ii, {i1, . . . , in−1} = {1, . . . , n} \ {i},

αs(x) ≤ ys ≤ βs(x), s = i1, . . . , in−1},

Fi

(x, y)≡ fi(x, y)− y′i, i = 1, . . . , n.

(2.2)


Result of the paper is given in the following theorem.

Theorem 2.1. Assume that

n∑

j=1

Mij(x)∣∣yj − zj

∣∣ ≤ fi

(x, y)− fi(x, z) ≤

n∑

j=1

Nij(x)∣∣yj − zj

∣∣ (2.3)

for every (x, y1, . . . , yn), (x, z1, . . . , zn) ∈ Ωi with yi > zi where Mij(x), Nij(x), i, j = 1, . . . , n, arecontinuous on Ii functions, such that for a constant ξ > 0

|Mii(x)| > ξn∑

j=1,j /= i

∣∣Mij(x)

∣∣, |Nii(x)| > ξ

n∑

j=1,j /= i

∣∣Nij(x)

∣∣. (2.4)

Let, moreover,

Fi(x, y)∣∣yi=αi(x)

· Fi(x, y)∣∣yi=βi(x)

< 0,

signMii(x) = signNii(x) = signFi(x, y)∣∣yi=βi(x)

,(2.5)

if (x, y1, . . . , yi−1, yi+1, . . . , yn) ∈ Γi, i = 1, . . . , n. Then there exists at least one solution y(x) =(y1(x), . . . , yn(x)) of the problem (1.1), (1.2) such that on Ii, i = 1, . . . , n:

αi(x) < yi(x) < βi(x). (2.6)

Remark 2.2. In the formulation of the problem (1.1), (1.2) the inequalities xk < xk+1 and xs <xs+1 were supposed. Analyzing the method of proof of Theorem 2.1 we conclude that theresult remains valid in the cases when xk ≤ xk+1 and xs ≤ xs+1 too. This means, for example,that a singular Cauchy problem

y′i = fi(x, y), yi

(x+

1

)= Ai, i = 1, . . . , n (2.7)

is a partial case of given result as well as a two-point boundary-value problem:

y′i = fi(x, y), i = 1, . . . , n

yp

(x+

1

)= Ap, p = 1, . . . , k, yr

(x−n)= Ar, r = k + 1, . . . , n.

(2.8)

Remark 2.3. In [9] a technique based on Kneser’s theorem is introduced to extend thetopological method of Wazewski for Caratheodory systems. It has, for example, been usedto study the asymptotic behavior of the solutions of a perturbed linear system:

x = [A(t) + B(t)]x + g(t, x), (2.9)


where the n × n matrices A (diagonal) and B are locally integrable, g ∈ Carloc((t0,∞) × Cn),

and the solutions are unique with respect to their initial values. The existence of solutionsxp = (xp1, xp2, . . . , xpn), p = 1, 2, . . . , n such that for any i /= p

limt→∞

xpi(t)xpp(t)

= 0 (2.10)

is studied. This is accomplished with the above-mentioned extension of the topological retractmethod for Carathedory systems which can be applied due to the construction of a suitableregular polyfacial sets. This technique makes it possible to extend the results initially proved(by Wazewski’s method) for ordinary differential equations with continuous right-hand sidesto Carathedory systems. Similar method has, for example, been used in a recent paper [6]where the technique developed in [9] is utilized. Along these lines, we can analyse our resultin terms of its possible extension to systems (1.1) with Caratheodory right-hand sides. Sincethe Lipschitz-type condition (2.3) is necessary in the proof of Theorem 2.1 for verifying thecontinuity of the operatorT, it cannot be omitted. Therefore, our result seems to be extendablefor Caratheodory systems (1.1) if the uniqueness of the solutions is ensured with respect totheir initial values (save at singular points).

3. Examples

Let us consider two illustrative nonlinear systems. The first one has a linear part whichdetermines the existence of the solution of the problem considered. The second one is aperturbation of a system for which we know analytic solution of singular problem.

Example 3.1. Let us consider a singular problem:

y′1 = 2y1

x−x(y2

3 + 1)

10,

y′2 =y2

(x − 1/2)2−

y1 + y3 + 120(x − 1/2)

,

y′3 = −3y3

1 − x +(1 − x)

(y2

2 + 1)

10,

y1(0+) = 0, y2

(12±

)= 0, y3

(1−)= 0.

(3.1)

For this problem all conditions of Theorem 2.1 are valid for

α1(x) = 0, β1(x) = 3x,

α2(x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

3(x − 1

2

), if x ∈

[0,

12

],

0, if x ∈(

12, 1],


β2(x) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

0, if x ∈[

0,12

],

3(x − 1

2

), if x ∈

(12, 1],

α3(x) = 0, β3(x) = 3(1 − x)2

(3.2)

and, for example, for ξ = 1, M11(x) = N11(x) = 2/x, M12(x) = N12(x) = M31(x) = N31(x) = 0,M13(x) = −N13(x) = −3x(1 − x)2/5, M21(x) = M23(x) = −N21(x) = −N23(x) = −1/(20|x −1/2|), M22(x) = N22(x) = 1/(x − 1/2)2, M32(x) = −N32(x) = −|x − 1/2|, and M33(x) =−N33(x) = −3/(1 − x). Consequently, there is at least one solution to this problem y(x) =(y1(x), y2(x), y3(x)) such that

0 < y1(x) < 3x for x ∈ (0, 1],

min{

3(x − 1

2

); 0}

< y2(x) < max{

0; 3(x − 1

2

)}for x ∈

[0,

12

)∪(

12, 1],

0 < y3(x) < 3(1 − x)2 for x ∈ [0, 1).

(3.3)

Example 3.2. Let us consider a singular problem:

y′1 =2x3

y21 + εx2y2

2 ,

y′2 =3

(x − 1/2)4y2

2 + ε

(x − 1

2

)3

y23 ,

y′3 =−2

(1 − x)3y2

3 + ε(1 − x)2y21 ,

y1(0+) = 0, y2

(12±

)= 0, y3

(1−)= 0,

(3.4)

where ε is a real constant, |ε| < 0.01. For this problem all conditions of Theorem 2.1 are valid,for example, for

α1(x) = x2 − 0.1x3, β1(x) = x2 + 0.1x3,

α2(x) =(x − 1

2

)3

− 0.1(x − 1

2

)4

, β2(x) =(x − 1

2

)3

+ 0.1(x − 1

2

)4

,

α3(x) = (1 − x)2 − 0.1(1 − x)3, β3(x) = (1 − x)2 + 0.1(1 − x)3

(3.5)

and for ξ = 1, M11(x) = 2/x, M12(x) = M31(x) = −N12(x) = −N31(x) = −M23(x) = N23(x) =−0.1, N11(x) = 6/x, M13(x) = N13(x) = M21(x) = N21(x) = M32(x) = N32(x) = 0, M22(x) =−12/(1/2 − x) if x < 1/2,N22(x) = −5/(1/2 − x) if x < 1/2,M22(x) = 3/(x − 1/2) if x >1/2, N22(x) = 12/(x − 1/2) if x > 1/2, M33(x) = −5/(1 − x), and N33(x) = −3/(1 − x).


Consequently, there is at least one solution to this problem y(x) = (y1(x), y2(x), y3(x)) suchthat

x2 − 0.1x3 < y1(x) < x2 + 0.1x3 for x ∈ (0, 1],

(x − 1

2

)3

− 0.1(x − 1

2

)4

< y2(x) <(x − 1

2

)3

+ 0.1(x − 1

2

)4

for x ∈ [0, 1] \{

12

},

(1 − x)2 − 0.1(1 − x)3 < y3(x) < (1 − x)2 + 0.1(1 − x)3 for x ∈ [0, 1).

(3.6)

If ε = 0, then the considered system turns into system

y′1 =2x3

y21 , y′2 =

3

(x − 1/2)4y2

2 , y′3 =−2

(1 − x)3y2

3 , (3.7)

having solution

y1 = x2, y2 =(

x − 12

)3

, y3 = (1 − x)2, (3.8)

which satisfies (3.4).

4. Preliminaries

In the sequel we will apply topological method of Wazewski (see, e.g., [12–14]). Thereforewe give a short summary of it. Let us consider the system of ordinary differential equations

y′ = g(x, y)

(4.1)

with y ∈ Rn. Below, it will be assumed that the right-hand sides of the system (4.1) are

continuous functions defined on an open (x, y)-set Ω∗ ⊂ R × Rn.

Definition 4.1 (see [12]). An open subset Ω0 of the set Ω∗ is called an (n, p)-subset of Ω∗ withrespect to the system (4.1) if the following conditions are satisfied.

(1) There exist continuously differentiable functions ni : Ω∗ → R, i = 1, . . . , � andpj : Ω∗ → R, j = 1, . . . , m; � +m > 0 such that

Ω0 ={(

x, y)∈ Ω∗ : ni

(x, y)< 0, pj

(x, y)< 0 ∀i, j

}. (4.2)

(2) nα(x, y) < 0 holds for the derivatives of the functions nα(x, y), α = 1, . . . , � alongtrajectories of the system (4.1) on the set

Nα ={(

x, y)∈ Ω∗, nα

(x, y)= 0, ni

(x, y)≤ 0, pj

(x, y)≤ 0 ∀i /=α and j

}. (4.3)


(3) pβ(x, y) > 0 holds for the derivatives of the functions pβ(x, y), β = 1, . . . , m alongtrajectories of the system (4.1) on the set

Pβ ={(

x, y)∈ Ω∗, pβ

(x, y)= 0, ni

(x, y)≤ 0, pj

(x, y)≤ 0 ∀i and j /= β

}. (4.4)

As usual, if ω ⊂ R×Rn, then intω, ∂ω and ω denote the interior, the boundary, and the closure

of ω, respectively.

Definition 4.2. The point (x0, y0) ∈ Ω∗∩∂Ω0 is called an egress point (or ingress point) of Ω0 withrespect to the system (4.1) if, for every fixed solution of the problem y(x0) = y0, there existsan ε > 0 such that (x, y(x)) ∈ Ω0 for x0 − ε ≤ x < x0 (x0 < x ≤ x0 + ε). An egress point (ingresspoint) (x0, y0) of Ω0 is called a strict egress point (strict ingress point) of Ω0 if (x, y(x))/∈Ω 0 oninterval x0 < x ≤ x0 + ε1 (x0 − ε1 ≤ x < x0) for an ε1 > 0.

The set of all points of egress (strict egress) is denoted by Ω0e (Ω0

se). It is proved in [12,page 281], that when a set Ω0 is an (n, p)-subset of Ω∗ then Ω0

e ≡ Ω0se.

Theorem 4.3 (see [12, page 282]). Let Ω0 be some (n, p)-subset of Ω∗ with respect to the system(4.1). Let S be a nonempty compact subset ofΩ0 ∪Ω0

e such that the set S∩Ω0e is not a retract of S but

is a retract of Ω0e. Then there is at least one point (x0, y0) ∈ S ∩Ω0 such that the graph of a solution

y(x) of the Cauchy problem y(x0) = y0 for (4.1) lies in Ω0 on its right-hand maximal interval ofexistence.

5. Partial Singular Problems

At this part we are interested in existence of solutions of some auxiliary singular problemsfor one scalar equation. We consider two cases below with respect to the location of singularpoint (at the left end or at the right end of the interval considered).

5.1. Singular Point Coincides with the Left End of Interval

Consider the initial problem

y′ = B(x, y), (5.1)

y(u+) = K, (K ∈ R) (5.2)

on an interval (u, v] with u < v. By a solution of problem (5.1), (5.2) on interval (u, v] wemean the function y ∈ C([u, v],R) ∩ C1((u, v],R) which satisfies (5.1) on (u, v] and thecondition (5.2). Let functions λ(x), μ(x) be continuously differentiable on (u, v], λ(u+) =μ(u+) = K and λ(x) < μ(x) on (u, v]. Denote

Θ+ ={(

x, y)

: x ∈ (u, v], λ(x) < y < μ(x)}. (5.3)


Let us suppose that there exists a domain Θ ⊂ (u, v] × R, such that Θ+ ⊂ Θ and the crosssection S+(x) = {(x, y) ∈ Θ} is an open set for every x ∈ (u, v]. Define an auxiliary function

H(x, y)≡ B(x, y)− y′. (5.4)

Lemma 5.1. Suppose that B ∈ C(Θ,R) satisfies the local Lipschitz condition with respect to thevariable y in Θ and, moreover,

H(x, λ(x)) < 0 < H(x, μ(x)

)if x ∈ (u, v]. (5.5)

Then each point (v, y∗) where y∗ ∈ [λ(v), μ(v)] defines a solution y = y∗(x) of (5.1) on (u, v] suchthat (5.2) holds, y∗(v) = y∗, and

λ(x) < y∗(x) < μ(x), x ∈ (u, v]. (5.6)

Proof. Let us evaluate the derivative of the function w(x, y) ≡ (y − λ(x))(y − μ(x)) along thetrajectories of (5.1) if (x, y) ∈ N, where

N ={(

x, y)

: x ∈ (u, v], w(x, y)= 0}. (5.7)

We get

dw(x, y)

dx=(y′ − λ′(x)

)·(y − μ(x)

)+(y − λ(x)

)·(y′ − μ′(x)

)

=[B(x, y)− λ′(x)

](y − μ(x)

)+(y − λ(x)

)[B(x, y)− μ′(x)

].

(5.8)

Since (x, y) ∈ N, then either y = μ(x) or y = λ(x). In the first case we have

dw(x, y)dx

∣∣∣∣y=μ(x)

=(μ(x) − λ(x)

)·H(x, μ(x)

)(5.9)

and in the second one

dw(x, y)dx

∣∣∣∣y=λ(x)

= −H(x, λ(x)) ·(μ(x) − λ(x)

). (5.10)

Thus, in view of condition (5.5),

dw(x, y)dx

∣∣∣∣(x,y)∈N

> 0, (5.11)

and, consequently, all points of the setN = Θ∩∂Θ+ are for x ∈ (u, v) the points of strict egressof Θ+ with respect to (5.1).


Let us consider behaviour of a solution y = y∗(x) of the problem y∗(v) = y∗ ∈[λ(v), μ(v)] for decreasing values of x ∈ (u, v]. Let us suppose that this solution leavesthe domain Θ+ passing through a boundary point (x0, y∗(x0)) ∈ N where x0 ∈ (u, v) and(x, y(x)) ∈ Θ+ for x ∈ (x0, v]. In this is case this point a point of ingress (for increasing x)with respect to (5.1) and this contradicts the fact that each point of the setN is for x ∈ (u, v)a point of strict egress. Only one possibility remains valid: solution y∗(x) is simultaneously asolution of the problem (5.1), (5.2). The lemma is proved.

Lemma 5.2. Let all assumptions of Lemma 5.1 hold except condition (5.5) which is replaced by thecondition:

H(x, μ(x)

)< 0 < H(x, λ(x)) if x ∈ (u, v]. (5.12)

Then there is at least one solution y = y∗(x) of the problem (5.1), (5.2) on (u, v] such that inequalities(5.6) hold.

Proof. Let us define the set N and the function w(x, y) in the same way as in the proof ofLemma 5.1. Then the derivative of w(x, y) along the trajectories of (5.1) satisfies, in view ofcondition (5.12), the inequality

dw(x, y)dx

∣∣∣∣(x,y)∈N

< 0. (5.13)

This means that all points of the setN are for x ∈ (u, v) the points of strict ingress of Θ+ withrespect to (5.1).

Let us change the orientation of the x-axis into reverse. Then all points of the setN arefor x ∈ (u, v) the points of strict egress of Θ+ with respect to (5.1).

Is it easy to see that the two-point set {λ(v − ), μ(v − )}, where is a small positivenumber, is a retract of the setN in view of existence of the retraction

r(x, y)=(v − , μ

(v −

)+[λ(v −

)− μ(v −

)] y − μ(x)λ(x) − μ(x)

), (5.14)

where (x, y) ∈ N. Clearly, the nonempty compact set S = [λ(v − ), μ(v − )] is not a retractof its boundary ∂S = {λ(v − ), μ(v − )} (see, e.g., [16]). All assumptions of topologicalprinciple of Wazewski are valid, and, by Theorem 4.3 (in its formulation we put Ω0 ≡ intΘ+,p1(x, y) ≡ w(x, y), j = 1, n1 ≡ x−v+ and � = 1), there exists at least one solution y = y∗(x)of the problem (5.1), (5.2) with graph belonging to the domain Θ+ on (u, v − ]. By the samearguments, as in the proof of Lemma 5.1, this solution can be continued on the interval (u, v].The lemma is proved.

5.2. Singular Point Coincides with the Right End of Interval

Let us consider the initial problem (5.1), (5.15) where

y(v−)= K, (K ∈ R) (5.15)


on an interval [u, v), with u < v. By a solution of (5.1), (5.15) on interval [u, v) we mean thefunction y ∈ C([u, v],R) ∩ C1([u, v),R) which satisfies (5.1) on interval [u, v) and condition(5.15). Let λ(x), μ(x) be continuously differentiable functions on [u, v), λ(v−) = μ(v−) = Kand λ(x) < μ(x) on [u, v). Denote

Θ− ={(

x, y)

: x ∈ [u, v), λ(x) < y < μ(x)}. (5.16)

Let us suppose that there exists a domain Θ ⊂ [u, v) × R, such that Θ− ⊂ Θ and the crosssection S−(x) = {(x, y) ∈ Θ} is an open set for every x ∈ [u, v). The proofs of followingLemmas 5.3 and 5.4 can be made by the similar manner as the proofs of Lemmas 5.1 and 5.2.Hence, they are omitted.

Lemma 5.3. Suppose that B ∈ C(Θ,R) satisfies the local Lipschitz condition with respect to thevariable y in Θ and, moreover,

H(x, λ(x)) < 0 < H(x, μ(x)

)if x ∈ [u, v). (5.17)

Then there is at least one solution y = y∗∗(x) of the problem (5.1), (5.15) on (u, v] such that

λ(x) < y∗∗(x) < μ(x). (5.18)

Lemma 5.4. Let all assumptions of Lemma 5.3 hold except condition (5.17) which is replaced by thecondition:

H(x, μ(x)

)< 0 < H(x, λ(x)) if x ∈ [u, v). (5.19)

Then each point (u, y∗∗) where y∗∗ ∈ [λ(u), μ(u)] defines a solution y = y∗∗(x) of (5.1) on [u, v),y∗∗(u) = y∗∗ and the inequalities (5.18) hold.


6.1. Construction of Operator

Let us consider the system

y′i = fi(x, ϕ1(x), . . . , ϕi−1(x), yi, ϕi+1(x), . . . , ϕn(x)

), i = 1, 2, . . . , n (6.1)

with (ϕ1(x), . . . , ϕn(x)) ∈M, where

M ={(

ϕ1(x), ϕ2(x), . . . , ϕn(x)), x ∈ I, ϕi ∈ C(I,R),

αi(x)≤ ϕi(x) ≤ βi(x), i = 1, 2, . . . , n} (6.2)

This system, strictly speaking, consists of separated scalar equations. Therefore in thefollowing we will consider equations of system (6.1) separately.


y1

x1 xn x

y∗1(v)β1(x)

y1(x)

α1(x)

Figure 1: Using Lemma 5.1.

y1

x1 xn x

y∗1(v)

β1(x)

y1(x)

α1(x)

Figure 2: Using Lemma 5.2.

(a) Let us consider the first equation of system (6.1) (which corresponds to the valuei = 1) together with corresponding initial value which is subtracted from (1.2):

y′1(x) = f1(x, y1, ϕ2(x), . . . , ϕn(x)

),

y1(x+

1

)= A1.

(6.3)

Let us put B(x, y) ≡ f1(x, y, ϕ2(x), . . . , ϕn(x)), λ(x) ≡ α1(x), μ(x) ≡ β1(x), u = x1, v = xn andK = A1. In view of condition (2.5) we see that either condition (5.5) or condition (5.12) holdsfor H(x, y) ≡ F1(x, y). From Lemmas 5.1 and 5.2 (it is easy to see that their assumptions arevalid) the existence of a solution of the problem (6.3) satisfying inequalities (5.6) follows. Anillustration to the cases where Lemma 5.1 or Lemma 5.2 is used is given in Figures 1 and 2.


In the sequel we will consider a solution y1(x) = y∗1(x) of problem (6.3) chosen in aunique way. We define this solution (in the case when Lemma 5.1 was used) by means of theadditional condition

y1(xn) = y∗1(v) = y∗1 =12(α1(xn) + β1(xn)

). (6.4)

If Lemma 5.2 was used, then denote the set of all solutions of problem (6.3) with the indicatedproperties as a set Y and put y1(xn) = y∗1(v) = min{y(v) : y ∈ Y}. Obviously this minimumexists and y∗1(v) > λ(v).

Define the first coordinate T1 of operator T by relation

T1(ϕ2, . . . , ϕn

)= y∗1. (6.5)

From inequalities (5.6) it follows that (y∗1, ϕ2, . . . , ϕn) ∈ M. The same reasoning can berepeated for i = 2, . . . , k.

(b) Now consider the last equation of system (6.1) (which corresponds to the valuei = n) together with the corresponding initial value which is subtracted from (1.2):

y′n = fn(x, ϕ1(x), . . . , ϕn−1(x), yn

),

yn

(x−n)= An.

(6.6)

Let us put B(x, y) ≡ fn(x, ϕ1(x), . . . , ϕn−1(x), y), λ(x) ≡ αn(x), μ(x) ≡ βn(x), u = x1, v = xn andK = An. In view of condition (2.5) we see that either condition (5.17) or condition (5.19) holdsfor H(x, y) ≡ Fn(x, y). From Lemmas 5.3 and 5.4 the existence of a solution of the problem(6.6) satisfying inequalities (5.18) follows. Similarly as in the part (a) above, we chose thesolution yn(x) = y∗∗n (x) of problem (6.6), which is uniquely defined.

Define the last coordinate Tn of operator T by relation

Tn(ϕ1, . . . , ϕn−1

)= y∗n. (6.7)

From inequalities (5.18) it follows that (ϕ1, . . . , ϕn−1, y∗n) ∈ M. The same reasoning can be

repeated for i = s + 1, . . . , n − 1.(c) Let us consider the equation of system (6.1) which corresponds to the value i = s

together with corresponding initial value which follows from (1.2):

y′s = fs(x, ϕ1(x), . . . , ϕs−1(x), ys, ϕs+1(x), . . . , ϕn(x)

),

ys

(x±)= As.

(6.8)

Let us put B(x, y) ≡ fs(x, ϕ1(x), . . . , ϕs−1(x), y, ϕs+1(x), . . . , ϕn(x), λ(x) ≡ αs(x), μ(x) ≡ βs(x)and K = As. Consider, at first, the problem (6.8) on interval [x1, xs). For this, let us put u = x1,v = xs. In view of condition (2.5) we see that either condition (5.17) or condition (5.19) holdsfor H(x, y) = Fs(x, y) and with the aid of Lemmas 5.3 and 5.4 (as in the part (b)) we candefine the unique solution ys(x) = yΔΔ(x) of (6.8) on interval [x1, xs). Now consider the


problem (6.8) on interval (xs, xn]. Put u = xs, v = xn. In view of condition (2.5) we see thateither condition (5.5) or condition (5.12) holds for H(x, y) ≡ Fs(x, y) and with the aid ofLemmas 5.1 and 5.2 (as in part (a)) we define the unique solution ys(x) = yΔ(x) of (6.8) oninterval (xs, xn].

At the end we define, by a unique manner, the solution y∗s(x) of the problem (6.8) as

y∗s(x) =

⎧⎨

⎩

yΔΔ(x), x ∈ [x1, xs),

yΔ(x), x ∈ (xs, xn].(6.9)

Define the sth coordinate Ts of operator T by relation

Ts(ϕ1, . . . , ϕs−1, ϕs+1, . . . , ϕn

)= y∗s. (6.10)

It is easy to see that (ϕ1, . . . , ϕs−1, y∗s, ϕs+1, . . . , ϕn) ∈ M. The same reasoning can be repeated

for i = k + 1, . . . , s − 1.(d) Now we are able to define operator T. For every ϕ = (ϕ1, . . . , ϕn) ∈ M define

Tϕ = y∗ with

T = (T1, T2, . . . , Tn), (6.11)

where y∗ = (y∗1, . . . , y∗n) ∈ M. Note that y∗ is defined in the unique way. Obviously, T(M) ⊂

M.

6.2. Verification of Schauder’s Assumptions

Let us consider the Banach space Ψ of functions ψ(x) = (ψ1(x), ψ2(x), . . . , ψn(x)), continuouson I, with the norm

∥∥ψ∥∥ = max

i=1,2,...,n

{max

I

∣∣ψi(x)

∣∣}. (6.12)

Clearly M ⊂ Ψ and, as follows from the properties of the functions αi(x), βi(x), i = 1, 2, . . . , n,M is a closed, bounded and convex set. It remains to prove that T is a continuous mappingsuch that T(M) is a relatively compact subset of Ψ. Then all the assumptions of Schauder’sfixed-point theorem will be satisfied (e.g., [15, page 29]). With respect to the relativecompactness of T(M) it is sufficient to prove in accordance with Arzela-Ascoli theorem thatT(M) is uniformly bounded and equicontinuous on I.

(α) The uniform boundedness follows from the inequality

∥∥ϕ∥∥ < L, (6.13)

where L = maxI{|αi(x)|, |βi(x)|, i = 1, 2, . . . , n}, which holds for every ϕ ∈M.


(β) Let us prove the equicontinuity of each function ϕ ∈ T(M). On I1 the first coordinateϕ1 of ϕ satisfies (as it follows from the construction of T) an equation of the type

ϕ′1 = f1(x, ϕ1, ν2(x), . . . , νn(x)

)(6.14)

with (ϕ1, ν2, . . . , νn) ∈M. Since f1 ∈ C(Θ1,R), (6.14) yields

∣∣ϕ′1(x)

∣∣ < Kδ, x ∈ [x1 + δ, xn], x1 + δ < xn, 0 < δ = const, (6.15)

where the constant Kδ exists and depends on δ. Let us put δ1 = min(δ/2, ε∗/Kδ/2) where ε∗

is an arbitrary positive number and δ is so small that

max[x1,x1+δ]

∣∣β1(x) −A1∣∣ <

ε∗

2, max

[x1,x1+δ]|α1(x) −A1| <

ε∗

2. (6.16)

Let us suppose that |z1 − z2| < δ1, z1, z2 ∈ [x1, xn]. Then either z1, z2 ∈ [x1, x1 + δ] or z1, z2 ∈[x1 + δ/2, xn]. In the first case

∣∣ϕ1(z1) − ϕ1(z2)∣∣ ≤∣∣ϕ1(z1) −A1

∣∣ +∣∣ϕ1(z2) −A1

∣∣ <ε∗

2+ε∗

2= ε∗ (6.17)

and in the second one (by Lagrange’s mean value theorem)

∣∣ϕ1(z1) − ϕ1(z2)∣∣ ≤ Kδ/2|z1 − z2| < ε∗. (6.18)

So, for each positive ε∗ there is a δ1 > 0 such that |ϕ1(z1) − ϕ1(z2)| < ε∗ for |z1 − z2| < δ1

and each function of the type of ϕ1(x) is equicontinuous. By analogy we can show that thefunctions of the type ϕj(x), j = 2, . . . , n are equicontinuous too. Finally, for |z1 − z2| < δ1, weget ‖ϕ(z1) − ϕ(z2)‖ < ε∗ and the equicontinuity of the set T(M) is proved.

(γ) Continuity of operator T. Let us suppose that y0, y ∈M and

Y 0 = Ty0, Y = Ty. (6.19)

In the sequel we prove that the operator T is continuous. We prove that

∥∥∥Y 0 − Y∥∥∥ < ε if

∥∥∥y0 − y∥∥∥ < δ, (6.20)

where δ ≤ εξ and ξ was defined in formulation of Theorem 2.1. The following constructionwill show that operator T is continuous. All expressions in the following will be well defined(supposing, if necessary, ε sufficiently small).


Consider the identity (see the definition of T)

Y 0′i (x) ≡ fi

(x, η0

1(x), η02(x), . . . , η

0n(x)), (6.21)

where i = 1, 2, . . . , n, η0i (x) ≡ Y 0

i (x), η0j (x) ≡ y0

j (x), j /= i, (x, η01, η

02, . . . , η

0n) ∈ Ωi and the

equation

Y ′i = fi(x, η1(x), η2(x), . . . , ηn(x)

), (6.22)

where i = 1, 2, . . . , n, ηi = Yi, ηj = ηj(x) ≡ yj(x), j /= i, (x, η1(x), η2(x), . . . , ηn(x)) ∈ Ωi.Note that in view of definition of T a solution of (6.22) is given by Yi ≡ Yi(x). Define, fori = 1, 2, . . . , n, the functions

Wi

(x, Yi

)=(Yi − Y 0

i (x) − ε)(

Yi − Y 0i (x) + ε

)(6.23)

and the sets

Pi ={(

x, Yi

):(x, Yi

)∈ Ωi,Wi

(x, Yi

)= 0}. (6.24)

(γ1) Let us evaluate the derivative of W1(x, Y1) along the trajectories of (6.22) for i = 1if (x, Y1) ∈ P1. We get,

dW1

(x, Y1

)

dx=[Y ′1 − Y

0′1 (x)](

Y1 − Y 01 (x) + ε

)+(Y1 − Y 0

1 (x) − ε)[

Y ′1 − Y0′1 (x)]. (6.25)

Since (x, Y1) ∈ P1, then either Y1 = Y 01 (x) + ε or Y1 = Y 0

1 (x) − ε. So,

dW1(x, Y1)dx

∣∣∣∣∣Y1=Y 0

1 (x)±ε

= ±2ε[f1

(x, Y 0

1 (x) ± ε, y2(x), . . . , yn(x))− f1

(x, Y 0

1 (x), y02(x), . . . , y

0n(x))]

.

(6.26)


According to (2.3) and (6.20):

ε

⎛

⎝M11(x) − ξn∑

j=2

∣∣M1j(x)

∣∣

⎞

⎠ ≤ εM11(x) +n∑

j=2

M1j(x)∣∣∣yj(x) − y0

j (x)∣∣∣

≤ f1

(x, Y 0

1 (x) + ε, y2(x), . . . , yn(x))− f1

(x, Y 0

1 (x), y02(x), . . . , y

0n(x))

≤ εN11(x) +n∑

j=2

N1j(x)∣∣∣yj(x) − y0

j (x)∣∣∣ ≤ ε

⎛

⎝N11(x) + ξn∑

j=2

∣∣N1j(x)

∣∣

⎞

⎠,

ε

⎛

⎝−N11(x) − ξn∑

j=2

∣∣N1j(x)∣∣

⎞

⎠ ≤ −εN11(x) −n∑

j=2

N1j(x)∣∣∣yj(x) − y0

j (x)∣∣∣

≤ f1

(x, Y 0

1 (x) − ε, y2(x), . . . , yn(x))− f1

(x, Y 0

1 (x), y02(x), . . . , y

0n(x))

≤ −εM11(x) −n∑

j=2

M1j(x)∣∣∣yj(x) − y0

j (x)∣∣∣ ≤ ε

⎛

⎝−M11(x) + ξn∑

j=2

∣∣M1j(x)∣∣

⎞

⎠.

(6.27)

Therefore (in view of (2.4), (6), and (6.20))

dW1(x, Y1)dx

∣∣∣∣∣(x,Y1)∈P1

> 0 if N11(x) > 0 on I1, (6.28)

dW1(x, Y1)dx

∣∣∣∣∣(x,Y1)∈P1

< 0 if N11(x) < 0 on I1. (6.29)

If inequality (6.28) and suppositions of Lemma 5.1 (in the situation, described in Section 6.1,(a)) hold simultaneously, then points of the set ∂Q1, where

Q1 = {(x, Y1) : x ∈ (x1, xn], w(x, Y1) < 0,W1(x, Y1) < 0} (6.30)

with w defined in the proof of Lemma 5.1, are (for all x ∈ (x1, xn)) the points of strict egressfor Q1 with respect to (6.22) with i = 1 (this equation is at the same time an equation ofthe type (6.1) for i = 1). Since Y 0

1 (x+1 ) = Y1(x+

1 ) and (in view of construction of operator T)Y 0

1 (x−n) = Y1(x−n), then |Y 0

1 (x) − Y1(x)| < ε (see Figure 3).Indeed, if this inequality does not hold then there is a x∗ ∈ I1 such that |Y 0

1 (x∗) −

Y1(x∗)| = ε and by (6.28): |Y 01 (x) − Y1(x)| > ε on (x∗, xn]. This is impossible.

If inequality (6.29) and suppositions of Lemma 5.2 (in the situation, described inSection 6.1, (a)) hold simultaneously, then all points of the set ∂Q1 are, for all x ∈ (x1, xn),the points of strict ingress for Q1 with respect to (6.22) with i = 1 (see Figure 4).

In view of construction (x, Y 01 (x)) ∈ Ω1 and (x, Y1(x)) ∈ Ω1. If inequality |Y 0

1 (x) −Y1(x)| < ε does not hold, then there is a x∗ ∈ I1 such that |Y 0

1 (x∗) − Y1(x∗)| = ε and |Y 0

1 (x) −Y1(x)| < ε on (x1, x

∗). This is impossible, since point (x∗, Y1(x∗) is the point of strict ingress.


Y1

a b x

β1(x)

Y 01 (x) + ε

Y 01 (x) − ε

Y 01 (x)

Q1

Y1(x)

α1(x)

Figure 3: Continuity of T (the first case).

Y1

a b x

β1(x)

Y 01 (x) + ε

Y 01 (x) − ε

Y 01 (x)

Q1

Y1(x)

α1(x)

Figure 4: Continuity of T (the second case).

In both considered cases, |Y 01 (x) − Y1(x)| < ε on I1 and, consequently, on I too. We

conclude that

∣∣∣Y1(x) − Y 01 (x)∣∣∣ < ε on I if

∥∥∥y0 − y∥∥∥ < δ. (6.31)

If inequality (6.29) and suppositions of Lemma 5.1 (in the situation, described in part 6.1, (a))hold simultaneously, then for small ε : N1 ∩ P1 /= ∅, where N1 ≡ N (N was defined in theproof of Lemma 5.1), and there exist a point (xΔ, yΔ) ∈ N1 ∩ P1 which is at the same timea point of strict egress and a point of strict ingress for Q1. This is excluded by condition (6).


For the same reason is the case when (6.28) and Lemma 5.2 hold simultaneously impossible.Analogously we can investigate (6.22) if i = 2, . . . , k.

(γ2) Let us evaluate the derivative of Wn(x, Yn) along the trajectories of (6.22) for i = n

if (x, Yn) ∈ Pn. The similar computations as above lead to inequalities

dWn(x, Yn)dx

∣∣∣∣∣(x,Yn)∈Pn

> 0 if Nnn(x) > 0 on In, (6.32)

dWn(x, Yn)dx

∣∣∣∣∣(x,Yn)∈Pn

< 0 if Nnn(x) < 0 on In. (6.33)

If inequality (6.32) and suppositions of Lemma 5.3 (in the situation described in Section 6.1,(b)) hold simultaneously, then all points of the set ∂Qn, where

Qn = {(x, Yn) : x ∈ (x1, xn), w(x, Yn) < 0,Wn(x, Yn) < 0} (6.34)

with w defined as in the proof of Lemma 5.1, are (for x ∈ (x1, xn)) the points of strict egressfor Qn with respect to (6.22) with i = n (since this equation is at the same time an equation ofthe type (6.1) for i = n).

If inequality (6.33) and suppositions of Lemma 5.4 (in the situation described inSection 6.1, (b)) hold simultaneously, then all points of the set ∂Qn for x ∈ (x1, xn) are pointsof strict ingress.

In both of these cases we conclude similarly, as in part (γ1), that |Y 0n(x) − Y 0

n(x)| < εon I if ‖y0 − y‖ < δ. The cases when inequality (6.32) and suppositions of Lemma 5.4hold simultaneously or when inequality (6.33) and suppositions of Lemma 5.3 holdsimultaneously are impossible according to (6).

Analogously we can proceed if i = s + 1, . . . , n − 1.(γ3) Let us evaluate the derivative of Wq along the trajectories of (6.22) for q = k +

1, . . . , s if (x, Yq) ∈ Pq. It is easy to see (by analogy with γ1)) that the following four cases(6.35)–(6.38) are possible:

dWq

(x, Yq

)

dx> 0 if Nqq(x) > 0, on Iq, (6.35)

dWq

(x, Yq

)

dx> 0 if Nqq(x) > 0 on

[x1, xq

),

dWq

(x, Yq

)

dx< 0 if Nqq(x) < 0 on

(xq, xn

],

(6.36)


dWq

(x, Yq

)

dx< 0 if Nqq(x) < 0 on Iq, (6.37)

dWq

(x, Yq

)

dx< 0 if Nqq(x) < 0 on

[x1, xq

),

dWq

(x, Yq

)

dx> 0 if Nqq(x) > 0 on

(xq, xn

].

(6.38)

Each of the admissible cases (i.e., if suppositions of Lemmas 5.1 , 5.3 and inequality (6.35)hold; or if suppositions of Lemmas 5.2 , 5.3 and inequalities (6.36) hold; or if suppositionsof Lemmas 5.2 , 5.4 and inequality (6.37) hold; or if suppositions of Lemmas 5.1 , 5.4 andinequalities (6.38) hold) can be considered as above (see parts (γ1) and (γ2)) and, therefore,for q = k + 1, . . . , s : |Y 0

q (x) − Yq(x)| < ε on I if ‖y0 − y‖ < δ. The remaining cases areimpossible in view of (6). Connecting all parts (γ1)–(γ3) we conclude that (6.20) holds and,consequently, operator T is continuous. All conditions of Schauder’s principle are valid.Therefore, the operator T has a fixed point, that is, the problem (1.1), (1.2) has a solutionwith indicated properties which follow from the form of the set M. Strong inequalities in(2.6) are a consequence of the fact that boundaries of considered sets are transversal withrespect to integral curves. The proof is complete.

Acknowledgments

This research was supported by the Councils of Czech Government MSM 0021630503, MSM0021630519, and MSM 0021630529, and by the Grant 201/08/0469 of Czech Grant Agency.

References

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[2] V. A. Cecik, “Investigation of systems of ordinary differential equations with a singularity,” TrudyMoskovskogo Matematiceskogo Obscestva, vol. 8, pp. 155–198, 1959 (Russian).

[3] J. Diblık, “The singular Cauchy-Nicoletti problem for the system of two ordinary differentialequations,” Mathematica Bohemica, vol. 117, no. 1, pp. 55–67, 1992.

[4] J. Diblık and C. Nowak, “A nonuniqueness criterion for a singular system of two ordinary differentialequations,” Nonlinear Analysis: Theory, Methods & Applications, vol. 64, no. 4, pp. 637–656, 2006.

[5] J. Diblık and M. Ruzickova, “Existence of positive solutions of a singular initial problem for anonlinear system of differential equations,” The Rocky Mountain Journal of Mathematics, vol. 34, no.3, pp. 923–944, 2004.

[6] J. Diblık and M. Ruzickova, “Inequalities for solutions of singular initial problems for Caratheodorysystems via Wazewski’s principle,” Nonlinear Analysis: Theory, Methods & Applications, vol. 69, no. 12,pp. 4482–4495, 2008.

[7] I. T. Kiguradze, Some Singular Boundary Value Problems for Ordinary Differential Equations, TbilisiUniversity Press, Tbilisi, Russia, 1975.

[8] N. B. Konyukhova, “Singular cauchy problems for systems of ordinary differential equations,” USSRComputational Mathematics and Mathematical Physics, vol. 23, no. 3, pp. 72–82, 1983.

[9] P. K. Palamides, Y. G. Sficas, and V. A. Staıkos, “Wazewski’s topological method for Caratheodorysystems,” Mathematical Systems Theory, vol. 17, no. 3, pp. 243–261, 1984.


[10] I. Rachunkova, O. Koch, G. Pulverer, and E. Weinmuller, “On a singular boundary value problemarising in the theory of shallow membrane caps,” Journal of Mathematical Analysis and Applications,vol. 332, no. 1, pp. 523–541, 2007.

[11] B. Vrdoljak, “On solutions of the Lagerstrom equation,” Archivum Mathematicum, vol. 24, no. 3, pp.111–122, 1988.

[12] P. Hartman, Ordinary Differential Equations, vol. 38 of Classics in Applied Mathematics, SIAM,Philadelphia, Pa, USA, 2nd edition, 2002.

[13] R. Srzednicki, “Wazewski method and Conley index,” in Handbook of Differential Equations: OrdinaryDifferential Equations, A. Canada, P. Drabek, and A. Fonda, Eds., vol. 1, pp. 591–684, Elsevier/North-Holland, Amsterdam, The Netherlands, 2004.

[14] T. Wazewski, “Sur un principe topologique de l’examen de l’allure asymptotique des integrales desequations differentielles ordinaires,” Annales Polonici Mathematici, vol. 20, pp. 279–313, 1947.

[15] I. Gyori and G. Ladas, Oscillation Theory of Delay Differential Equations, Oxford MathematicalMonographs, The Clarendon Press, Oxford University Press, New York, NY, USA, 1991.

[16] K. Borsuk, Theory of Retracts, Monografie Matematyczne, Tom 44, PWN, Warsaw, Poland, 1967.


Research ArticleNew Results on Multiple Solutions forNth-Order Fuzzy Differential Equations underGeneralized Differentiability

A. Khastan,1, 2 F. Bahrami,1, 2 and K. Ivaz1, 2

1 Department of Applied Mathematics, University of Tabriz, Tabriz 51666 16471, Iran2 Research Center for Industrial Mathematics, University of Tabriz, Tabriz 51666 16471, Iran

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

Received 30 April 2009; Accepted 1 July 2009


We firstly present a generalized concept of higher-order differentiability for fuzzy functions.Then we interpret Nth-order fuzzy differential equations using this concept. We introduce newdefinitions of solution to fuzzy differential equations. Some examples are provided for whichboth the new solutions and the former ones to the fuzzy initial value problems are presented andcompared. We present an example of a linear second-order fuzzy differential equation with initialconditions having four different solutions.

Copyright q 2009 A. Khastan 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.

1. Introduction

The term “fuzzy differential equation” was coined in 1987 by Kandel and Byatt [1] andan extended version of this short note was published two years later [2]. There are manysuggestions to define a fuzzy derivative and in consequence, to study fuzzy differentialequation [3]. One of the earliest was to generalize the Hukuhara derivative of a set-valuedfunction. This generalization was made by Puri and Ralescu [4] and studied by Kaleva [5].It soon appeared that the solution of fuzzy differential equation interpreted by Hukuharaderivative has a drawback: it became fuzzier as time goes by [6]. Hence, the fuzzy solutionbehaves quite differently from the crisp solution. To alleviate the situation, Hullermeier[7] interpreted fuzzy differential equation as a family of differential inclusions. The mainshortcoming of using differential inclusions is that we do not have a derivative of a fuzzy-number-valued function.

The strongly generalized differentiability was introduced in [8] and studied in [9–11]. This concept allows us to solve the above-mentioned shortcoming. Indeed, the strongly


generalized derivative is defined for a larger class of fuzzy-number-valued functions thanthe Hukuhara derivative. Hence, we use this differentiability concept in the present paper.Under this setting, we obtain some new results on existence of several solutions for Nth-order fuzzy differential equations. Higher-order fuzzy differential equation with Hukuharadifferentiability is considered in [12] and the existence and uniqueness of solution fornonlinearities satisfying a Lipschitz condition is proved. Buckley and Feuring [13] presentedtwo different approaches to the solvability of Nth-order linear fuzzy differential equations.

Here, using the concept of generalized derivative and its extension to higher-orderderivatives, we show that we have several possibilities or types to define higher-orderderivatives of fuzzy-number-valued functions. Then, we propose a new method to solvehigher-order fuzzy differential equations based on the selection of derivative type covering allformer solutions. With these ideas, the selection of derivative type in each step of derivationplays a crucial role.

2. Preliminaries

In this section, we give some definitions and introduce the necessary notation which will beused throughout this paper. See, for example, [6].

Definition 2.1. Let X be a nonempty set. A fuzzy set u in X is characterized by its membershipfunction u : X → [0, 1]. Thus, u(x) is interpreted as the degree of membership of an elementx in the fuzzy set u for each x ∈ X.

Let us denote by RF the class of fuzzy subsets of the real axis (i.e., u : R → [0, 1])satisfying the following properties:

(i) u is normal, that is, there exists s0 ∈ R such that u(s0) = 1,

(ii) u is convex fuzzy set (i.e., u(ts+(1− t)r) ≥ min{u(s), u(r)}, for all t ∈ [0, 1], s, r ∈ R),

(iii) u is upper semicontinuous on R,

(iv) cl{s ∈ R | u(s) > 0} is compact where cl denotes the closure of a subset.

Then RF is called the space of fuzzy numbers. Obviously, R ⊂ RF . For 0 < α ≤ 1 denote[u]α = {s ∈ R | u(s) ≥ α} and [u]0 = cl{s ∈ R | u(s) > 0}. If u belongs to RF, then α-level set[u]α is a nonempty compact interval for all 0 ≤ α ≤ 1. The notation

[u]α =[uα, uα], (2.1)

denotes explicitly the α-level set of u. One refers to u and u as the lower and upper branchesof u, respectively. The following remark shows when [uα, uα] is a valid α-level set.

Remark 2.2 (see [6]). The sufficient conditions for [uα, uα] to define the parametric form of afuzzy number are as follows:

(i) uα is a bounded monotonic increasing (nondecreasing) left-continuous function on(0, 1] and right-continuous for α = 0,

(ii) uα is a bounded monotonic decreasing (nonincreasing) left-continuous function on(0, 1] and right-continuous for α = 0,

(iii) uα ≤ uα, 0 ≤ α ≤ 1.


For u, v ∈ RF and λ ∈ R, the sum u + v and the product λ · u are defined by [u + v]α =[u]α + [v]α, [λ · u]α = λ[u]α, for allα ∈ [0, 1], where [u]α + [v]α means the usual addition oftwo intervals (subsets) of R and λ[u]α means the usual product between a scalar and a subsetof R.

The metric structure is given by the Hausdorff distance:

D : RF × RF −→ R+ ∪ {0}, (2.2)

by

D(u, v) = supα∈[0,1]

max{∣∣uα − vα

∣∣,∣∣uα − vα∣∣}. (2.3)

The following properties are wellknown:

(i) D(u +w,v +w) = D(u, v), for allu, v,w ∈ RF,

(ii) D(k · u, k · v) = |k|D(u, v), for all k ∈ R, u, v ∈ RF,

(iii) D(u + v,w + e) ≤ D(u,w) +D(v, e), for allu, v,w, e ∈ RF,

and (RF,D) is a complete metric space.

Definition 2.3. Let x, y ∈ RF . If there exists z ∈ RF such that x = y + z, then z is called theH-difference of x, y and it is denoted x � y.

In this paper the sign “�” stands always for H-difference and let us remark that x �y /=x + (−1)y in general. Usually we denote x + (−1)y by x − y, while x � y stands for theH-difference.

3. Generalized Fuzzy Derivatives

The concept of the fuzzy derivative was first introduced by Chang and Zadeh [14]; it wasfollowed up by Dubois and Prade [15] who used the extension principle in their approach.Other methods have been discussed by Puri and Ralescu [4], Goetschel and Voxman [16],Kandel and Byatt [1, 2]. Lakshmikantham and Nieto introduced the concept of fuzzydifferential equation in a metric space [17]. Puri and Ralescu in [4] introduced H-derivative(differentiability in the sense of Hukuhara) for fuzzy mappings and it is based on the H-difference of sets, as follows. Henceforth, we suppose I = (T1, T2) for T1 < T2, T1, T2 ∈ R.

Definition 3.1. Let F : I → RF be a fuzzy function. One says, F is differentiable at t0 ∈ I ifthere exists an element F ′(t0) ∈ RF such that the limits

limh→ 0+

F(t0 + h) � F(t0)h

, limh→ 0+

F(t0) � F(t0 − h)h

(3.1)

exist and are equal to F ′(t0). Here the limits are taken in the metric space (RF,D).


The above definition is a straightforward generalization of the Hukuhara differen-tiability of a set-valued function. From [6, Proposition 4.2.8], it follows that Hukuharadifferentiable function has increasing length of support. Note that this definition of derivativeis very restrictive; for instance, in [9], the authors showed that if F(t) = c · g(t), where c is afuzzy number and g : [a, b] → R

+ is a function with g ′(t) < 0, then F is not differentiable.To avoid this difficulty, the authors [9] introduced a more general definition of derivative forfuzzy-number-valued function. In this paper, we consider the following definition [11].

Definition 3.2. Let F : I → RF and fix t0 ∈ I. One says F is (1)-differentiable at t0, if thereexists an element F ′(t0) ∈ RF such that for all h > 0 sufficiently near to 0, there exist F(t0 +h) � F(t0), F(t0) � F(t0 − h), and the limits (in the metric D)

limh→ 0+

F(t0 + h) � F(t0)h

= limh→ 0+

F(t0) � F(t0 − h)h

= F ′(t0). (3.2)

F is (2)-differentiable if for all h < 0 sufficiently near to 0, there exist F(t0 + h) � F(t0), F(t0) �F(t0 − h) and the limits (in the metric D)

limh→ 0−

F(t0 + h) � F(t0)h

= limh→ 0−

F(t0) � F(t0 − h)h

= F ′(t0). (3.3)

If F is (n)-differentiable at t0, we denote its first derivatives by D(1)n F(t0), for n = 1, 2.

Example 3.3. Let g : I → R+ and define f : I → RF by f(t) = c · g(t), for all t ∈ I. If g is

differentiable at t0 ∈ I, then f is generalized differentiable on t0 ∈ I and we have f ′(t0) = c ·g ′(t0). For instance, if g ′(t0) > 0, f is (1)-differentiable. If g ′(t0) < 0, then f is (2)-differentiable.

Remark 3.4. In the previous definition, (1)-differentiability corresponds to the H-derivativeintroduced in [4], so this differentiability concept is a generalization of the H-derivative andobviously more general. For instance, in the previous example, for f(t) = c ·g(t) with g ′(t0) <0, we have f ′(t0) = c · g ′(t0).

Remark 3.5. In [9], the authors consider four cases for derivatives. Here we only consider thetwo first cases of [9, Definition 5]. In the other cases, the derivative is trivial because it isreduced to crisp element (more precisely, F ′(t0) ∈ R. For details, see [9, Theorem 7]).

Theorem 3.6. Let F : I → RF be fuzzy function, where [F(t)]α = [fα(t), gα(t)] for each α ∈ [0, 1].

(i) If F is (1)-differentiable, then fα and gα are differentiable functions and [D11F(t)]

α =[f ′α(t), g

′α(t)].

(ii) If F is (2)-differentiable, then fα and gα are differentiable functions and [D12F(t)]

α =[g ′α(t), f

′α(t))].

Proof. See [11].

Now we introduce definitions for higher-order derivatives based on the selection ofderivative type in each step of differentiation. For the sake of convenience, we concentrate onthe second-order case.


For a given fuzzy function F, we have two possibilities (Definition 3.2) to obtain thederivative of F ot t: D(1)

1 F(t) and D(1)2 F(t). Then for each of these two derivatives, we have

again two possibilities: D(1)1 (D(1)

1 F(t)), D(1)2 (D(1)

1 F(t)), and D(1)1 (D(1)

2 F(t)), D(1)2 (D(1)

2 F(t)),respectively.

Definition 3.7. Let F : I → RF and n,m = 1, 2. One says say F is (n,m)-differentiable at t0 ∈ I,if D(1)

n F exists on a neighborhood of t0 as a fuzzy function and it is (m)-differentiable at t0.The second derivatives of F are denoted by D

(2)n,mF(t0) for n,m = 1, 2.

Remark 3.8. This definition is consistent. For example, if F is (1, 2) and (2, 1)-differentiablesimultaneously at t0, then F is (1)- and (2)-differentiable around t0. By remark in [9], F is acrisp function in a neighborhood of t0.

Theorem 3.9. Let D(1)1 F : I → RF or D

(1)2 F : I → RF be fuzzy functions, where [F(t)]α =

[fα(t), gα(t)].

(i) If D(1)1 F is (1)-differentiable, then f ′α and g ′α are differentiable functions and [D(2)

1,1F(t)]α=

[f ′′α(t), g′′α(t)].

(ii) If D(1)1 F is (2)-differentiable, then f ′α and g ′α are differentiable functions and [D(2)

1,2F(t)]α=

[g ′′α(t), f′′α(t)].

(iii) If D(1)2 F is (1)-differentiable, then f ′α and g ′α are differentiable functions and [D(2)

2,1F(t)]α=

[g ′′α(t), f′′α(t)].

(iv) If D(1)2 F is (2)-differentiable, then f ′α and g ′α are differentiable functions and [D(2)

2,2F(t)]α=

[f ′′α(t), g′′α(t)].

Proof. We present the details only for the case (i), since the other cases are analogous.If h > 0 and α ∈ [0, 1], we have

[D

(1)1 F(t + h) �D(1)

1 F(t)]α

=[f ′α(t + h) − f ′α(t), g ′α(t + h) − g ′α(t)

], (3.4)

and multiplying by 1/h, we have

[D

(1)1 F(t + h) �D(1)

1 F(t)]α

h=[f ′α(t + h) − f ′α(t)

h,g ′α(t + h) − g ′α(t)

h

]. (3.5)

Similarly, we obtain

[D

(1)1 F(t) �D(1)

1 F(t − h)]α

h=[f ′α(t) − f ′α(t − h)

h,g ′α(t) − g ′α(t − h)

h

]. (3.6)


Passing to the limit, we have

[D

(2)1,1F(t)

]α=[f ′′α(t), g

′′α(t)]. (3.7)

This completes the proof of the theorem.

Let N be a positive integer number, pursuing the above-cited idea, we writeD

(N)k1,...,kN

F(t0) to denote the Nth-derivatives of F at t0 with ki = 1, 2 for i = 1, . . . ,N. Nowwe intend to compute the higher derivatives (in generalized differentiability sense) of theH-difference of two fuzzy functions and the product of a crisp and a fuzzy function.

Lemma 3.10. If f, g : I → RF are Nth-order generalized differentiable at t ∈ I in the same case ofdifferentiability, then f + g is generalized differentiable of orderN at t and (f + g)(N)(t) = f (N)(t) +g(N)(t). (The sum of two functions is defined pointwise.)

Proof. By Definition 3.2 the statement of the lemma follows easily.

Theorem 3.11. Let f, g : I → RF be second-order generalized differentiable such that f is (1,1)-differentiable and g is (2,1)-differentiable or f is (1,2)-differentiable and g is (2,2)-differentiable or f is(2,1)-differentiable and g is (1,1)-differentiable or f is (2,2)-differentiable and g is (1,2)-differentiableon I. If theH-difference f(t)�g(t) exists for t ∈ I, then f�g is second-order generalized differentiableand

(f � g

)′′(t) = f ′′(t) + (−1) · g ′′(t), (3.8)

for all t ∈ I.

Proof. We prove the first case and other cases are similar. Since f is (1)-differentiable andg is (2)-differentiable on I, by [10, Theorem 4], (f � g)(t) is (1)-differentiable and we have(f � g)′(t) = f ′(t) + (−1) · g ′(t). By differentiation as (1)-differentiability in Definition 3.2 andusing Lemma 3.10, we get (f � g)(t) is (1,1)-differentiable and we deduce

(f � g

)′′(t) =(f ′(t) + (−1) · g ′(t)

)′ = f ′′(t) + (−1) · g ′′(t). (3.9)

The H-difference of two functions is understood pointwise.

Theorem 3.12. Let f : I → R and g : I → RF be two differentiable functions (g is generalizeddifferentiable as in Definition 3.2).

(i) If f(t) · f ′(t) > 0 and g is (1)-differentiable, then f · g is (1)-differentiable and

(f · g

)′(t) = f ′(t) · g(t) + f(t) · g ′(t). (3.10)


(ii) If f(t) · f ′(t) < 0 and g is (2)-differentiable, then f · g is (2)-differentiable and

(f · g

)′(t) = f ′(t) · g(t) + f(t) · g ′(t). (3.11)

Proof. See [10].

Theorem 3.13. Let f : I → R and g : I → RF be second-order differentiable functions (g isgeneralized differentiable as in Definition 3.7).

(i) If f(t)·f ′(t) > 0, f ′(t)·f ′′(t) > 0, and g is (1,1)-differentiable then f ·g is (1,1)-differentiableand

(f · g

)′′(t) = f ′′(t) · g(t) + 2f ′(t) · g ′(t) + f(t) · g ′′(t). (3.12)

(ii) If f(t)·f ′(t) < 0, f ′(t)·f ′′(t) < 0 and g is (2,2)-differentiable then f ·g is (2,2)-differentiableand

(f · g

)′′(t) = f ′′(t) · g(t) + 2f ′(t) · g ′(t) + f(t) · g ′′(t). (3.13)

Proof. We prove (i), and the proof of another case is similar. If f(t) · f ′(t) > 0 and g is (1)-differentiable, then by Theorem 3.12 we have

(f · g

)′(t) = f ′(t) · g(t) + f(t) · g ′(t). (3.14)

Now by differentiation as first case in Definition 3.2, since g ′(t) is (1)-differentiable and f ′(t) ·f ′′(t) > 0, then we conclude the result.

Remark 3.14. By [9, Remark 16], let f : I → R, γ ∈ RF and define F : I → RF by F(t) = γ ·f(t),for all t ∈ I. If f is differentiable on I, then F is differentiable on I, with F ′(t) = γ · f ′(t). ByTheorem 3.12, if f(t) · f ′(t) > 0, then F is (1)-differentiable on I. Also if f(t) · f ′(t) < 0, thenF is (2)-differentiable on I. If f(t) · f ′(t) = 0, by [9, Theorem 10], we have F ′(t) = γ · f ′(t). Wecan extend this result to second-order differentiability as follows.

Theorem 3.15. Let f : I → R be twice differentiable on I, γ ∈ RF and define F : I → RF byF(t) = γ · f(t), for all t ∈ I.

(i) If f(t) · f ′(t) > 0 and f ′(t) · f ′′(t) > 0, then F(t) is (1,1)-differentiable and its secondderivative, D(2)

1,1F, is F′′(t) = γ · f ′′(t),

(ii) If f(t) · f ′(t) > 0 and f ′(t) · f ′′(t) < 0, then F(t) is (1,2)-differentiable with D(2)1,2F) =

γ · f ′′(t),

(iii) If f(t)·f ′(t) < 0 and f ′(t)·f ′′(t) > 0, then F(t) is (2,1)-differentiable withD(2)2,1F = γ ·f ′′(t),

(iv) If f(t)·f ′(t) < 0 and f ′(t)·f ′′(t) < 0, then F(t) is (2,2)-differentiable withD(2)2,2F = γ ·f ′′(t).


Proof. Cases (i) and (iv) follow from Theorem 3.13. To prove (ii), since f(t) · f ′(t) > 0, byRemark 3.14, F is (1)-differentiable and we have D

(1)1 F = γ ·f ′(t) on I. Also, since f ′(t) ·f ′′(t) <

0, then D(1)1 F is (2)-differentiable and we conclude the result. Case (iii) is similar to previous

one.

Example 3.16. If γ is a fuzzy number and φ : (0, 3) → R, where

φ(t) = t2 − 3t + 2 (3.15)

is crisp second-order polynomial, then for

F(t) = γ · φ(t), (3.16)

we have the following

(i) for 0 < t < 1: φ(t) · φ′(t) < 0 and φ′(t) · φ′′(t) < 0 then by (iv), F(t) is (2-2)-differentiable and its second derivative, D(2)

2,2F is F ′′(t) = 2 · γ ,

(ii) for 1 < t < 3/2: φ(t) · φ′(t) > 0 and φ′(t) · φ′′(t) < 0 then by (ii), F(t) is (1-2)-differentiable with D

(2)1,2F = 2 · γ ,

(iii) for 3/2 < t < 2: φ(t) · φ′(t) < 0 and φ′(t) · φ′′(t) > 0 then by (iii), F(t) is (2-1)-differentiable and D

(2)2,1F = 2 · γ ,

(iv) for 2 < t < 3: φ(t)·φ′(t) > 0 and φ′(t)·φ′′(t) > 0 then by (i), F(t) is (1-1)-differentiableand D

(2)1,1F = 2 · γ ,

(v) for t = 1, 3/2, 2: we have φ′(t) · φ′′(t) = 0, then by [9, Theorem 10] we have F ′(t) =γ · φ′(t), again by applying this theorem, we get F ′′(t) = 2 · γ.

4. Second-Order Fuzzy Differential Equations

In this section, we study the fuzzy initial value problem for a second-order linear fuzzydifferential equation:

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

y′′(t) + a · y′(t) + b · y(t) = σ(t),

y(0) = γ0,

y′(0) = γ1,

(4.1)

where a, b > 0, γ0, γ1 ∈ RF, and σ(t) is a continuous fuzzy function on some interval I. Theinterval I can be (0, A) for some A > 0 or I = (0,∞). In this paper, we suppose a, b > 0. Ourstrategy of solving (4.1) is based on the selection of derivative type in the fuzzy differentialequation. We first give the following definition for the solutions of (4.1).

Definition 4.1. Let y : I → RF be a fuzzy function and n,m ∈ {1, 2}. One says y is an (n,m)-solution for problem (4.1) on I, if D1

nyD2n,my exist on I and D2

n,my(t) + a ·D1ny(t) + b · y(t) =

σ(t), y(0) = γ0, D1ny(0) = γ1.


Let y be an (n,m)-solution for (4.1). To find it, utilizing Theorems 3.6 and 3.9and considering the initial values, we can translate problem (4.1) to a system of second-order linear ordinary differential equations hereafter, called corresponding (n,m)-system forproblem (4.1).

Therefore, four ODEs systems are possible for problem (4.1), as follows:

(1, 1)-system

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y(0;α) = γ0α, y(0;α) = γ0

α,

y′(0;α) = γ1α, y′(0;α) = γ1

α,

(4.2)

(1, 2)-system

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y(0;α) = γ0α, y(0;α) = γ0

α,

y′(0;α) = γ1α, y′(0;α) = γ1

α,

(4.3)

(2, 1)-system

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎩

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y(0;α) = γ0α, y(0;α) = γ0

α,

y′(0;α) = γ1, y′(0;α) = γ1,

(4.4)

(2, 2)-system

⎧⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎩

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y′′(t;α) + ay′(t;α) + by(t;α) = σ(t;α),

y(0;α) = γ0α, y(0;α) = γ0

α,

y′(0;α) = γ1α, y′(0;α) = γ1

α.

(4.5)


Theorem 4.2. Let n,m ∈ {1, 2} and y = [y, y] be an (n,m)-solution for problem (4.1) on I. Then y

and y solve the associated (n,m)-systems.

Proof. Suppose y is the (n,m)-solution of problem (4.1). According to the Definition 4.1, thenD1

ny and D2n,my exist and satisfy problem (4.1). By Theorems 3.6 and 3.9 and substituting

y, y and their derivatives in problem (4.1), we get the (n,m)-system corresponding to (n,m)-solution. This completes the proof.

Theorem 4.3. Let n,m ∈ {1, 2} and fα(t) and gα(t) solve the (n,m)-system on I, for every α ∈[0, 1]. Let [F(t)]α = [fα(t), gα(t)]. If F has valid level sets on I and D2

n,mF exists, then F is an(n,m)-solution for the fuzzy initial value problem (4.1).

Proof. Since [F(t)]α = [fα(t), gα(t)] is (n,m)-differentiable fuzzy function, by Theorems 3.6and 3.9 we can compute D1

nF and D2n,mF according to f ′α, g

′α, f

′′α, g

′′α. Due to the fact that fα, gα

solve (n,m)-system, from Definition 4.1, it comes that F is an (n,m)-solution for (4.1).

The previous theorems illustrate the method to solve problem (4.1). We first choose thetype of solution and translate problem (4.1) to a system of ordinary differential equations.Then, we solve the obtained ordinary differential equations system. Finally we find such adomain in which the solution and its derivatives have valid level sets and using StackingTheorem [5] we can construct the solution of the fuzzy initial value problem (4.1).

Remark 4.4. We see that the solution of fuzzy differential equation (4.1) depends upon theselection of derivatives. It is clear that in this new procedure, the unicity of the solution islost, an expected situation in the fuzzy context. Nonetheless, we can consider the existence offour solutions as shown in the following examples.

Example 4.5. Let us consider the following second-order fuzzy initial value problem

y′′(t) = σ0, y(0) = γ0, y′(0) = γ1, t ≥ 0, (4.6)

where σ0 = γ0 = γ1 are the triangular fuzzy number having α-level sets [α − 1, 1 − α].If y is (1,1)-solution for the problem, then

[y′(t)

]α =[y′(t;α), y′(t;α)

],

[y′′(t)

]α =[y′′(t;α), y′′(t;α)

], (4.7)

and they satisfy (1,1)-system associated with (4.1). On the other hand, by ordinary differentialtheory, the corresponding (1,1)-system has only the following solution:

y(t;α) = (α − 1)

(t2

2+ t + 1

)

, y(t;α) = (1 − α)(

t2

2+ t + 1

)

. (4.8)

We see that [y(t)]α = [y(t;α), y(t;α)] are valid level sets for t ≥ 0 and

y = [α − 1, 1 − α] ·(

t2

2+ t + 1

)

. (4.9)


By Theorem 3.15, y is (1,1)-differentiable for t ≥ 0. Therefore, y defines a (1,1)-solution fort ≥ 0.

For (1,2)-solution, we get the following solutions for (1,2)-system:

y(t;α) = (α − 1)

(

− t2

2+ t + 1

)

, y(t;α) = (1 − α)(

− t2

2+ t + 1

)

, (4.10)

where y(t) has valid level sets for t ∈ [0, 1]. How ever-also [y(t)]α = [α−1, 1−α]·(−(t2/2)+t+1)where y is (1,2)-differentiable. Then y gives us a (1,2)-solution on (0, 1).

(2,1)-system yields

y(t;α) = (α − 1)

(

− t2

2− t + 1

)

, y(t;α) = (1 − α)(

− t2

2− t + 1

)

, (4.11)

where y(t) has valid level sets for t ∈ [0,√

3−1]. We can see y is a (2,1)-solution on (0,√

3−1)Finally, (2-2)-system gives

y(t;α) = (α − 1)

(t2

2− t + 1

)

, y(t;α) = (1 − α)(

t2

2− t + 1

)

, (4.12)

where y(t) has valid level sets for all t ∈ [0, 1], and defines a (2,2)-solution on (0, 1).

Then we have an example of a second-order fuzzy initial value problem with fourdifferent solutions.

Example 4.6. Consider the fuzzy initial value problem:

y′′(t) + y(t) = σ0, y(0) = γ0, y′(0) = γ1 ∀t ≥ 0, (4.13)

where σ0 is the fuzzy number having α-level sets = [α, 2 − α] and [γ0]α = [γ1]

α = [α − 1, 1 − α].To find (1,1)-solution, we have

y(t;α) = α(1 + sin t) − sin t − cos t, y(t;α) = (2 − α)(1 + sin t) − sin t − cos t, (4.14)

where y(t) has valid level sets for t ≥ 0 and y(t) = σ0·(1+sin t)−sin t−cos t. From Theorem 3.15,y is (1,2)-differentiable on (0, π/2), then by Remark 3.8, y is not (1, 1)-differentiable on(0, π/2). Hence, no (1,1)-solution exists for t > 0.

For (1,2)-solutions we deduce

y(t;α) = α(1 + sinh t) − sinh t − cos t,

y(t;α) = (2 − α)(1 + sinh t) − sinh t − cos t,(4.15)

we see that y(t) has valid level sets and is (1,1)-differentiable for t > 0. Since the (1,2)-systemhas only the above solution, then (1,2)-solution does not exist.


For (2,1)-solutions we get

y(t;α) = α(1 − sinh t) + sinh t − cos t,

y(t;α) = (2 − α)(1 − sinh t) + sinh t − cos t,(4.16)

we see that the fuzzy function y(t) has valid level sets for t ∈ [0, ln(1 +√

2)] and define a(2,1)-solution for the problem on (0, ln(1 +

√2)).

Finally, to find (2,2)-solution, we find

y(t;α) = α(1 − sin t) + sin t − cos t, y(t;α) = (2 − α)(1 − sin t) + sin t − cos t, (4.17)

that y(t) has valid level sets for t ≥ 0 and y is (2,2)-differentiable on (0, π/2).

We then have a linear fuzzy differential equation with initial condition and twosolutions.

5. Higher-Order Fuzzy Differential Equations

Selecting different types of derivatives, we get several solutions to fuzzy initial value problemfor second-order fuzzy differential equations. Theorem 4.2 has a crucial role in our strategy.To extend the results to Nth-order fuzzy differential equation, we can follow the proofof Theorem 4.2 to get the same results for derivatives of higher order. Therefore, we canextend the presented argument for second-order fuzzy differential equation to Nth-order.Under generalized derivatives, we would expect at most 2N solutions for an Nth-order fuzzydifferential equation by choosing the different types of derivatives.

Acknowledgments

We thank Professor J. J. Nieto for his valuable remarks which improved the paper. Thisresearch is supported by a grant from University of Tabriz.

References

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[2] A. Kandel and W. J. Byatt, “Fuzzy processes,” Fuzzy Sets and Systems, vol. 4, no. 2, pp. 117–152, 1980.[3] J. J. Buckley and T. Feuring, “Fuzzy differential equations,” Fuzzy Sets and Systems, vol. 110, no. 1, pp.

43–54, 2000.[4] M. L. Puri and D. A. Ralescu, “Differentials of fuzzy functions,” Journal of Mathematical Analysis and

Applications, vol. 91, no. 2, pp. 552–558, 1983.[5] O. Kaleva, “Fuzzy differential equations,” Fuzzy Sets and Systems, vol. 24, no. 3, pp. 301–317, 1987.[6] P. Diamond and P. Kloeden, Metric Spaces of Fuzzy Sets, World Scientific, Singapore, 1994.[7] E. Hullermeier, “An approach to modelling and simulation of uncertain dynamical systems,”

International Journal of Uncertainty, Fuzziness and Knowledge-Based Systems, vol. 5, no. 2, pp. 117–137,1997.

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[10] B. Bede, I. J. Rudas, and A. L. Bencsik, “First order linear fuzzy differential equations undergeneralized differentiability,” Information Sciences, vol. 177, no. 7, pp. 1648–1662, 2007.

[11] Y. Chalco-Cano and H. Roman-Flores, “On new solutions of fuzzy differential equations,” Chaos,Solitons & Fractals, vol. 38, no. 1, pp. 112–119, 2008.

[12] D. N. Georgiou, J. J. Nieto, and R. Rodrıguez-Lopez, “Initial value problems for higher-order fuzzydifferential equations,” Nonlinear Analysis: Theory, Methods & Applications, vol. 63, no. 4, pp. 587–600,2005.

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[16] R. Goetschel Jr. and W. Voxman, “Elementary fuzzy calculus,” Fuzzy Sets and Systems, vol. 18, no. 1,pp. 31–43, 1986.

[17] V. Lakshmikantham and J. J. Nieto, “Differential equations in metric spaces: an introduction andan application to fuzzy differential equations,” Dynamics of Continuous, Discrete & Impulsive Systems.Series A, vol. 10, no. 6, pp. 991–1000, 2003.


Research ArticleOn Some Generalizations Bellman-Bihari Resultfor Integro-Functional Inequalities forDiscontinuous Functions and Their Applications

Angela Gallo1 and Anna Maria Piccirillo2

1 Department of Mathematics and Applications, “R.Caccioppoli” University of Naples “Federico II”,Claudio street 21, 80125 Naples, Italy

2 Department of Civil Engineering, Second University of Naples, Roma, street 21, 81100 Caserta, Italy

Correspondence should be addressed to Angela Gallo, [email protected]

Received 22 December 2008; Revised 21 April 2009; Accepted 28 May 2009


We present some new nonlinear integral inequalities Bellman-Bihari type with delay for discontin-uous functions (integro-sum inequalities; impulse integral inequalities). Some applications of theresults are included: conditions of boundedness (uniformly), stability by Lyapunov (uniformly),practical stability by Chetaev (uniformly) for the solutions of impulsive differential and integro-differential systems of ordinary differential equations.

Copyright q 2009 A. Gallo and A. M. Piccirillo. 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.

1. Introduction

The first generalizations of the Bihari result for discontinuous functions which satisfynonlinear impulse inequality (integro-sum inequality) are connected with such types ofinequalities:

(a)

v(t) ≤ c +∫ t

t0

p(τ) vm(τ) dτ +∑

t0<ti<t

βi v(ti − 0), m > 0, m/= 1, (1.1)

(b)

v(t) ≤ c +∫ t

t0

p(τ)ϕ(v(τ))dτ +∑

t0<ti<t

βi v(ti − 0), (1.2)


Which are studied in the publications by Bainov, Borysenko, Iovane, Laksmikantham, Leela,Martynyuk, Mitropolskiy, Samoilenko ([1–13]), and in many others. In these investigationsthe method of integral inequalities for continuous functions is generalized to the case ofpiecewise continuous (one-dimensional inequalities) and discontinuous (multidimensionalinequalities) functions.

For the generalization of the integral inequalities method for discontinuous functionsand for their applications to qualitative analysis of impulsive systems: existence, uniqueness,boundedness, comparison, stability, and so forth. We refer to the results [2–5, 12, 14] andfor periodic boundary value problems we cite [15–17]. More recently, a novel variationalapproach appeared in [18]. This approach to impulsive differential equations also used thecritical point theory for the existence of solutions of a nonlinear Dirichlet impulsive problemand in [19] some new comparison principles and the monotone iterative technique toestablish a more general existence theorem for a periodic boundary value problem. Reference[20] is very interesting in that it gives a complete overview of the state-of-the-art of theimpulsive differential, inclusions.

In this paper, in Section 2, we investigate new analogies Bihari results for piece-wisecontinuous functions and, in Section 3, the conditions of boundedness, stability, pract-icalstability of the solutions of nonlinear impulsive differential and integro-differential systems.

2. General Bihari Theorems for Integro-Functional Inequalitiesfor Discontinuous Functions

Let us consider the class ℘ of continuous functions p : R → R, p(t) ≤ t, lim|t|→∞

p(t) = ∞ (p =

p(t) is the delaying argument). The following holds.

Theorem 2.1. (a) Let one suppose that for x ≥ x0 the following integro-sum functional inequalityholds:

u(x) ≤ ϕ(x) + q(x)∫x

xi

f(τ) W(u(p(τ)))dτ +

∑

x0<xi<x

βi um(xi − 0), (2.1)

where q(x) ≥ 1, ϕ(x) is a positive nondecreasing function, βi = const ≥ 0, f : R+ → R+, m =const > 0; function u(x) is a nonnegative piecewise-continuous,with I-st kind of discontinuities inthe points xi : x0 < x1 < · · · limn→∞xn =∞, p(t) belongs to the class ℘.

(b) Function W(x) satisfies such conditions:

(i) W(γβ) ≤W(γ)W(β);

(ii) W : R+ → R+, W(0) = 0;

(iii) W is nondecreasing.

Then for arbitrary x ∈ ]x0 ,∞[ the next estimate holds:

u(x) ≤ ϕ(x)q(x)G−1i

[∫x

xi

f(τ)ϕ(τ)

W[ϕ(p(τ))q(p(τ))]dτ

]

for x ∈ ]xi, xi+1[

∫x

xi

f(τ)ϕ(τ)

W[ϕ(p(τ))q(p(τ))]dτ ∈ Dom

(G−1

i

),

(2.2)


G0(u) =∫u

1

dσ

W(σ), (2.3)

Gi(u) =∫u

ci

dσ

W(σ), i = 1, 2, . . . , (2.4)

ci =(

1 + βi ϕm−1(xi)qm(xi − 0)

)G−1

i−1

(∫xi

xi−1

f(τ)ϕ(τ)


)

,

i = 1, 2, . . . if m ∈ ]0, 1], ∀x ≥ x0,

ci =(

1 + βi ϕm−1(xi)qm(xi − 0)

)[

G−1i−1

(∫xi

xi−1

f(τ)ϕ(τ)

W[ϕ(p(τ))q(p(τ))]

dτ

)]

,

m

i = 1, 2, . . . if m ≥ 1, ∀x ≥ x0.

(2.5)

Proof. It follows from inequality (2.1)

u(x)ϕ(x)

≤ 1 + q(x)∫x

x0

f(τ) W(u(p(τ)))

ϕ(τ)dτ +

∑

x0<xi<x

βium(xi − 0)

ϕ(x)

≤ q(x)

{

1 +∫x

x0

f(τ)ϕ(τ)

W(u(p(τ)))dτ +

∑

x0<xi<x

βi ϕm−1(xi − 0)

[u(xi − 0)ϕ(xi − 0)

]m}

.

(2.6)

Denoting by

u∗(x) = 1 +∫x

x0

f(τ)ϕ(τ)

W(u(p(τ)))dτ +

∑

x0<xi<x

βi ϕm−1(xi − 0)

[u(xi − 0)ϕ(xi − 0)

]m,

u∗(x) = 1 for x = x0,

(2.7)

then

u(xi − 0) ≤ ϕ(xi − 0) q(xi − 0)u∗(xi − 0),

u(x) ≤ ϕ(x) q(x)u∗(x),

u(p(τ))≤ ϕ(p(τ))q(p(τ))u∗(p(τ))≤ ϕ(p(τ))q(p(τ))u∗(τ),

=⇒ u∗(x) ≤ 1 +∫x

x0

f(τ)ϕ(τ)

W[ϕ(p(τ))q(p(τ))]W(u∗(τ))dτ

+∑

x0<xi<x

βiϕm−1(xi − 0)qm(xi − 0)u∗m(xi − 0).

(2.8)


Let us consider the interval I1 = [x0, x1[. Then

u∗(x) ≤ G−10

(∫x

x0

f(τ)ϕ(τ)


)

,

if only∫x

x0

f(τ)ϕ(τ)


(G−1

0

),

(2.9)

where G0(ξ) =∫ ξ

1(dτ/W(τ)). So it results in

u(x) ≤ ϕ(x)q(x)G−10

[∫x

x0

f(τ)ϕ(τ)


]

, (2.10)

and estimate (2.2) is valid in I1.Let us suppose that for x ∈ Ik = xk−1, xk , k = 2, 3, . . . estimate (2.2) is fulfilled. Then

for every x ∈ Ik+1 we have

u∗(x) ≤ G−1k

(∫x

xk

f(τ)ϕ(τ)


)

with∫x

xk

f(τ)ϕ(τ)


(G−1

k

),

(2.11)

where Gk(ξ) is determined from (2.3)–(2.5).Taking into account such inequality

u(x) ≤ ϕ(x)q(x)u∗(x), (2.12)

we obtain estimate (2.2) for every x ∈ [x0 ,∞[.Let us consider the class I of functions f such that

(i) f(x)-positive, continuous, nondecreasing for x > 0;

(ii) ∀u ≥ 1, v > 0 ⇒ u−1 f(v) < f(u−1 v);

(iii) f(0) = 0.

The following result is proved.

Theorem 2.2. Suppose that the part (a) of Theorem 2.1 is valid and functionW : [0 ,∞[→ [0 ,∞[belongs to the class I. Then for arbitrary x0 ≤ x ≤ x∗ such estimate holds:

u(x) ≤ ϕ(x)q(x)G∗i−1

[∫x

xi

f(τ) q(p(τ))dτ

]

for Ii = [xi , xi+1 [, i = 0, 1, . . . , (2.13)


where

G∗0(η)=∫η

1

dσ

W(σ), G∗i

(η)=∫η

c∗i

dσ

W(σ)i = 1, 2, . . . ,

c∗i =(

1 + βi ϕm−1(xi)qm(xi)

)G∗i−1

−1

(∫xi

xi−1

f(τ)q(p(τ))dτ

)

if m ∈ ]0, 1],

c∗i =(

1 + βi ϕm−1(xi)qm(xi)

)[

G∗i−1−1

(∫xi

xi−1

f(τ)q(p(τ))dτ

)]mif m ≥ 1,

(2.14)

and x∗ = supx{∫xxi−1

f(τ)q(p(τ))dτ ∈ Dom(G∗i−1−1)}, i = 1, 2, . . . .

Proof. By using the previous theorem we have u(x) ≤ ϕ(x)g(x)u∗(x), u∗(x) = 1 x = x0. Onthe interval I1

du∗(x)dx

=f(x)ϕ(x)

W(u(p(x))). (2.15)

Then

u(p(x))≤ ϕ(p(x))q(p(x))u∗(p(x))≤ ϕ(x)q

(p(x))u∗(x),

du∗(x)dx

≤f(x)ϕ(x)

W(q(p(x))ϕ(x)u∗(x)

)

≤f(x)q

(p(x))

ϕ(x)q(p(x))W(q(p(x))ϕ(x)u∗(x)

)

≤ f(x)q(p(x))W(u∗(x)).

(2.16)

Taking into account estimate (2.16), we obtain

∫x

x0

u∗′(σ)W(u∗(σ))

dσ ≤∫x

x0

f(τ)q(p(τ))dτ,

∫x

x0

u∗′(σ)W(u∗(σ))

dσ =∫u∗(x)

u∗(x0)

du

W(u)= G∗0(u

∗(x)) −G∗0(u∗(x0)),

u∗(x0) = 1, u∗(x) ≥ 1, G∗0(u∗(x0)) = G∗0(1) = 0,

G∗0(u∗(x)) ≤

∫x

x0

f(τ) q(p(τ))dτ.

(2.17)


Then in I1 we have

u(x) ≤ ϕ(x)q(x) G∗0−1

[∫x

x0

f(τ)q(p(τ))dτ

]

if only∫x

x0

f(τ)q(p(τ))dτ ∈ Dom

(G∗0−1).

(2.18)

As in the previously theorem, the proof is completed by using the inductive method.

The following result is easily to obtain

Theorem 2.3. Suppose that for x ≥ x0 the next inequality holds:

u(x) ≤ u0 + q(x)

[∫x

x0

f(s)u(p(s))ds +∫x

x0

f(s)

(∫x

x0

g(τ)u(p(τ))dτ

)

ds

]

+∫x

x0

h(s)W(u(σ(s)))ds +∑

x0<xi<x

βi um(xi − 0),

(2.19)

where functions u(x), f(x), q(x), g(x), h(x), p(x), σ(x) are real nonnegative for x ≥ x0 >0, p(x), σ(x) ∈ I, q(x) ≥ 1, βi ≥ 0, function W satisfies conditions (i),. . .,(iii) of Theorem 2.1.

Then for x ≥ x0 it results in

u(x) ≤∏

x0<xi<x

(1 + βiq

m(xi)um−10

)exp

(∫x

x0

q(p(τ))[f(τ) + g(τ)

]dτ

)

· ψ−10

⎛

⎝∫x

x0

h(τ) W

⎡

⎣∏

x0<xi<σ(τ)

(1 + βiq

m(xi)um−10

)⎤

⎦W

×[

q(σ(τ)) exp

(∫σ(τ)

x0

q(p(s))[f(s) + g(s)

]ds

)]

dτ

)

, if m ∈ ]0, 1]

∫x

x0

h(τ)W

⎡

⎣∏

x0<xi<σ(τ)

(1 + βiq

m(xi)um−10

)⎤

⎦W

×[

q(σ(τ)) exp

(∫σ(τ)

x0

q(p(s))[f(s) + g(s)

]ds

)]

dτ ∈ Dom(ψ−1

0

),

(2.20)


where ψ0(u) =∫uu0(dv/W(v));

u(x) ≤∏

x0<xi<x

(1 + βiq

m(xi)um−10

)exp

(

m

∫x

x0

q(p(τ))[f(τ) + g(τ)

])

· ψ−10

⎛

⎝∫x

x0

h(τ)

⎡

⎣∏

x0<xi<σ(τ)

(1 + βiq

m(xi)um−10

)⎤

⎦

·W[

q(σ(τ)) exp

(

m

∫σ(τ)

x0

q(p(s))[f(s) + g(s)

]ds

)]

dτ

)

, if m ≥ 1,

∫x

x0

h(τ)W

⎡

⎣∏

x0<xi<σ(τ)

(1 + βiq

m(xi)um−10

)⎤

⎦W

×[

q(σ(τ)) exp

(

m

∫σ(τ)

x0

q(p(s))[f(s) + g(s)

]ds

)]

dτ ∈ Dom(ψ−1

0

).

(2.21)

The proof the same procedure as that of (Iovane [21, Theorems 2.1 and 3.1]).

Corollary 2.4. Suppose that

(a) m = 1, then the result of Theorem 2.1 coincides with the result [22, Theorem 3.7.1, page232];

(b) m = 1, ϕ(x) = c, q(x) = 1, p(t) = t, then the result of Theorem 2.1 coincides with result[12, Proposition 2.3, page 2143];

(c) q(x) = 1, W(u) = u, p(t) = t, then one obtains the analogy of Gronwall- Bellman resultfor discontinuous functions [23, Lemma 1] and estimate (2.2) reduces in the following form:

u(x) ≤ ϕ(x)∏

x0<xi<x

(1 + βi ϕ

m−1(xi))

exp

(∫x

x0

f(τ) dτ

)

if m ∈ ]0, 1], ∀x ≥ x0,

u(x) ≤ ϕ(x)∏

x0<xi<x

(1 + βi ϕ

m−1(xi))

exp

(

m

∫x

x0

f(τ) dτ

)

if m ≥ 1, ∀x ≥ x0.

(2.22)

(d) q(x) = 1, W(u) = u, then one obtains the result [21, Theorem 2.1] and estimate (2.2) areas follows:

u(x) ≤ ϕ(x)∏

x0<xi<x

(1 + βi ϕ

m−1(xi))

exp

(∫x

x0

f(τ)ϕ(p(τ))

ϕ(τ)dτ

)

, if m ∈ ]0, 1], ∀x ≥ x0;

u(x) ≤ ϕ(x)∏

x0<xi<x

(1 + βi ϕ

m−1(xi))

exp

(

m

∫x

x0

f(τ)ϕ(p(τ))

ϕ(τ)dτ

)

if m ≥ 1, ∀x ≥ x0.

(2.23)


(e) q(x) = 1,W(u) = um,m > 0, p(t) = t, then one obtains the analogy of Bihari result fordiscontinuous functions [23, Lemma 2] and estimate (2.2) reduces as follows are reduced:

u(x) ≤ ϕ(x)∏

x0<xi<x

(1 + βi ϕ

m−1(xi))[

1 + (1 −m)∫x

x0

ϕm−1(τ)f(τ)dτ

]1/(1−m)

,

if 0 < m < 1, ∀x ≥ x0,

u(x) ≤ ϕ(x)∏

x0<xi<x

(1 + βimϕm−1(xi)

)⎡

⎣1 − (m − 1)

[∏

x0<xi<x

(1 + βi mϕm−1(xi)

)]m−1

×∫x

x0

ϕm−1(τ) f(τ) dτ

]− 1/(m−1)

∀x ≥ x0,

(2.24)

such that

∫x

x0

ϕm−1(τ)f(τ) dτ ≤ 1m, m > 1,

∏

x0<xi<x

(1 + βi ϕ

m−1(xi))<

(1 +

1m − 1

)1/(m−1)

.

(2.25)

(f) W(u) = um,, m> 0, then estimate (2.2) reduces as follows (see [21, Theorem 2.2]):

u(x) ≤ ϕ(x) q(x)∏

x0<xi<x

(1 + βi ϕ

m−1(xi) qm(xi))

×[

1 + (1 −m)∫x

x0

ϕm−1(τ)f(τ)qm(p(τ))[ϕ(p(τ))

ϕ(τ)

]mdτ

]1/(1−m)

if 0 < m < 1, ∀x ≥ x0,

u(x) ≤ ϕ(x) q(x)∏

x0<xi<x

(1 + βimϕm−1(xi) qm(xi)

)

×

⎧⎨

⎩1 − (m − 1)

[∏

x0<xi<x

(1 + βi mϕm−1(xi)qm(xi)

)]m−1

×∫x

x0

ϕm−1(τ) f(τ)qm(p(τ))[ϕ(p(τ))

ϕ(τ)

]mdτ

}− 1/(m−1)

∀x ≥ x0


such that

∫x

x0

ϕm−1(τ) f(τ)qm(p(τ))[ϕ(p(τ))

ϕ(τ)

]mdτ ≤ 1

m, m > 1,

∏

x0<xi<x

(1 + βimϕm−1(xi)qm(xi)

)<

(1 +

1m − 1

)−1/(m−1)

.

(2.26)

(g) Suppose that in Theorem 2.3 q(x) = 1,W(u) = u, σ(s) = p(s) = s, then estimates (2.20),(2.21) reduce as shown:

u(x) ≤ u0

∏

x0<xi<x

(1 + βi u

m−10 (xi)

)exp

[∫x

x0

[f(ξ) + g(ξ) + h(ξ)

]dξ

]

if m ∈ ]0, 1], ∀x ≥ x0;

u(x) ≤ u0

∏

x0<xi<x

(1 + βi u

m−10 (xi)

)exp

[

m

∫x

x0

[f(ξ) + g(ξ) + h(ξ)

]dξ

]

if m ≥ 1, ∀x ≥ x0,

(2.27)

which coincide with result of [21, Theorem 3.1] for h(t) = u0.

3. Applications

Let us consider the following system of differential equations

dx

dt= F(t, x), t /= ti,

Δx|t=ti = Ii(x)(3.1)

where x ∈ Rn, F ∈ Rn, Ii(x) ∈ Rn (i = 1, 2, . . .), t ≥ t0 ≥ 0, limi→∞ ti = ∞, ti−1 < ti for all i =1, 2, . . ..

Let us assume that F(t, x) and Ii(x) are defined in the domain D = {(t, x) : t ∈ I =[t0, T], T ≤ ∞, ‖x‖ ≤ h } and satisfy such conditions:

(a) ‖F(t, x)‖ ≤ f(t)W(‖x‖), f : R+ → R+,

W satisfies conditions (i)–(iii) of Theorem 2.1;

(b) ‖Ii(x) ‖ ≤ βi‖x‖m, βi = const > 0, m > 0.

Consider x(t) = x( t, t0, x0) the solution of Cauchy problem for system (3.1). Then

x(t, t0, x0 ) = x0 +∫ t

t0

F(τ, x(τ, t0, x0))dτ +∑

t0<ti<t

Ii(x(ti − 0, t0, x0)), (3.2)


from which it follows

‖x(t, t0, x0 )‖ ≤ ‖x0‖ +∫ t

t0

f(τ) W(‖x(τ, t0, x0)‖)dτ +∑

t0<ti<t

βi‖x(ti − 0, t0, x0)‖m. (3.3)

By using the result of Theorem 2.1 and estimate (2.2) we obtain

‖x(t, t0, x0)‖ ≤ ‖x0‖G−1i

[∫x

xi

f(τ)W(‖x0‖)‖x0‖

dτ

]

for x ∈ ]xi, xi+1[ ,

∫x

xi

f(τ)W(‖x0‖)‖x0‖

dτ ∈ Dom(G−1

i

),

(3.4)

where

G0(u) =∫u

1

dσ

W(σ), Gi(u) =

∫u

ci

dσ

W(σ), i = 1, 2, ...,

ci =(

1 + βi‖x0‖m−1)G−1

i−1

(∫xi

xi−1

f(τ)W(‖x0‖)‖x0‖

dτ

)

,

i = 1, 2, . . . if m ∈ ]0, 1], ∀x ≥ x0,

ci =(

1 + βi‖x0‖m−1)[

G−1i−1

(∫xi

xi−1

f(τ)W(‖x0‖)‖x0‖

dτ

)]m,

i = 1, 2, . . . if m ≥ 1, ∀x ≥ x0.

(3.5)

Let us consider some particular cases of W .If W(u) = u, m = 1, estimate (3.4) is reduced in such form

‖x(t, t0, x0)‖ ≤ ‖x0‖∏

t0<ti<t

(1 + βi

)exp

[∫ t

t0

f(τ)dτ

]

. (3.6)

Then such result holds.

Proposition 3.1. Let the following conditions be fulfilled for system (3.1) :

(i) ‖F(t, x)‖ ≤ f(t)‖x‖;

(ii) ‖Ii(x)‖ ≤ βi‖x‖;

(iii) ∃m1(t0) = const. > 0 :∏

t0<ti<t(1 + βi) ≤ m1(t0) <∞;

(iv) ∃m2(t0) = const. > 0 :∫ tt0f(τ) dτ ≤ m2(t0) <∞, ∀t ≥ t0 .


Then one has:

(a) All solutions of system (3.1) are bounded (uniformly, if mi(t0) are independent of t0) andsuch estimate is valid:

‖x(t, t0, x0)‖ ≤ m1(t0) exp[m2(t0)]‖x0‖. (3.7)

(b) The trivial solution of system (3.1) is stable by Lyapunov (uniformly stable relative t0, ifmi(t0) = mi, i = 1, 2).

Remark 3.2. If conditions I–IV of Proposition 3.1 are valid and λ/Λ < (m1(t0) exp[m2(t0)])−1,

then the trivial solution is (λ,Λ, I)-stable by Chetaev (uniformly (λ,Λ, I)-stable, if mi(t0),i = 1, 2 is independent of t0).

If W(u) = ul, l /= 1, m = 1 the estimate (3.4) is reduced in such form

‖x(t, t0, x0)‖ ≤∏

t0<ti<t

(1 + βi

)[

‖x0‖1−l + (1 − l)∫ t

t0

f(τ)dτ

]1/(1−l)

∀t ≥ t0, if 0 < l < 1, (3.8)

‖x(t, t0, x0)‖ ≤ ‖x0‖∏

t0<ti<t

(1 + βi

)

×

⎡

⎣1 − (l − 1) ‖x0‖l−1 ·[∏

t0<ti<t

(1 + βi

)]l−1∫ t

t0

f(τ)dτ

⎤

⎦

− 1/(l−1)

∀t ≥ t0,

(3.9)

∫ t

t0

f(τ)dτ <

⎛

⎝(l − 1)

[

‖x0‖∏

t0<ti<t

(1 + βi

)]l−1⎞

⎠

−1

, if l > 1. (3.10)

From estimate (3.8) the next propositions follow.

Proposition 3.3. Suppose that such conditions occur:

(a) ‖F(t, x) − F(t, y)‖ ≤ f(t) ‖x − y‖l, 0 < l < 1 for allx, y ∈ D

(b) estimates ii–iv of Proposition 3.1 be fulfilled.

Then all the solutions of system (3.1) are bounded (uniformly if mi(t0) = mi, i = 1, 2).

Remark 3.4. Suppose that conditions (a), (b) of Proposition 3.3 are valid and

λ1−l + (1 − l) m2(t0) <[

Λm1(t0)

]1−l. (3.11)


Then trivial solution of system (3.1) is (λ,Λ, I)-stable by Chetaev (uniformly if mi(t0) isindependent of t0).

Proposition 3.5. Let conditions ii–iv of Proposition 3.1 be fulfilled for system (3.1), inequality (3.10)holds and

‖F(t, x)‖ ≤ f(t)‖x ‖l, l > 1. (3.12)

Then trivial solution of system (3.1) is stable by Lyapunov (uniformly ifmi(t0) = mi, i = 1, 2).

Remark 3.6. If W(u) = u1, 1 > 0, and m/= 1 the conditions of boundedness, stability, (λ,Λ, I)-stability is investigated in [14, see Theorems 3.4–3.6]; the estimates of the solutions of system(3.1) with non-Lipschitz type of discontinuities are investigated in [23, see Proposition 1,Proposition 2].

Let us consider the following impulsive system of integro-differential equations:

dx

dt= F(t, x,K[x(t)]), t /= ti,

Δx|t=ti = Ii(x),(3.13)

where x ∈ Rn, F ∈ Rn, Ii(x) ∈ Rn (i = 1, 2, . . .) and defined in the domain D, K[x(t)] =∫ tt0k(t, τ, x(τ)) dτ .

We suppose that such conditions are valid:

(i) ‖F(t, x, y)‖ ≤ f(t)[‖x‖ + ‖y‖] for allx, y ∈ D, f : R+ → R+;

(ii) ‖k(t, s, x)‖ ≤ g(t)‖x‖ for all s ∈ [t0, t], g : R+ → R+;

(iii) ‖Ii(x)‖ ≤ βi‖x‖m for allx, y ∈ D, βi = const > 0, m > 0 m/= 1.

It is easy to see that

‖x(t, t0, x0 )‖ ≤ ‖x0‖ +∫ t

t0

f(τ)‖x(τ, t0, x0)‖dτ

+∫ t

t0

f(τ)

(∫ τ

t0

g(ξ)‖x(ξ, t0, x0)‖dξ)

dτ +∑

t0<ti<t

βi‖x(ti − 0, t0, x0)‖m

=⇒ ‖x(t, t0, x0)‖ ≤ ‖x0‖∏

t0<ti<t

(1 + βi‖x0‖m−1

)exp∫ t

t0

[f(ξ) + g(ξ)

]dξ,

if 0 < m ≤ 1, t ≥ t0

(3.14)

‖x(t)‖ ≤ ‖ x0‖∏

t0<ti<t

(1 + βi‖x0‖m−1

)exp

(

m

∫ t

t0

[f(ξ) + g(ξ)

]dξ

)

,

if 0m ≥ 1, t ≥ t0.

(3.15)

From estimate (3.15) such result follows.


Proposition 3.7. Let one suppose that for system (3.13) conditions (i)–(iii) take place for m > 1 andthe following estimates are fulfilled:

(a) ∃ m3(t0) = const. > 0 :∏

t0<ti<t(1 + βi‖x0‖m−1) ≤ m3(t0) <∞;

(b) ∃ m4(t0) = const. > 0 :∫ tt0[f(ξ) + g(ξ)] dξ ≤ m4(t0) <∞ for all t ≥ t0.

Then we have:

(i) All solutions of system (3.13) are bounded and satisfy the estimate:

‖x(t)‖ ≤ m3(t0) exp[m4(t0)]‖x0‖. (3.16)

(ii) The trivial solution of system (3.13) is stable by Lyapunov (uniformly, if mi(t0) = mi, i =3, 4).

(iii) The trivial solution of system (3.13) is (λ,Λ, I)-stable by Chetaev (uniformly if mi(t0) isindependent of t0) and m3(t0) exp[m4(t0)] < Λ/λ.

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Research ArticleOn Step-Like Contrast Structure of SingularlyPerturbed Systems

Mingkang Ni and Zhiming Wang

Department of Mathematics, East China Normal University, Shanghai 200241, China

Correspondence should be addressed to Zhiming Wang, [email protected]

Received 15 April 2009; Revised 5 July 2009; Accepted 14 July 2009


The existence of a step-like contrast structure for a class of high-dimensional singularly perturbedsystem is shown by a smooth connection method based on the existence of a first integral foran associated system. In the framework of this paper, we not only give the conditions underwhich there exists an internal transition layer but also determine where an internal transition timeis. Meanwhile, the uniformly valid asymptotic expansion of a solution with a step-like contraststructure is presented.

Copyright q 2009 M. Ni and Z. 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.

1. Introduction

The problem of contrast structures is a singularly perturbed problem whose solutions withboth internal transition layers and boundary layers. In recent years, the study of contraststructures is one of the hot research topics in the study of singular perturbation theory. Inwestern society, most works on internal layer solutions concentrate on singularly perturbedparabolic systems by geometric method (see [1] and the references therein). In Russia, theworks on singularly perturbed ordinary equations are concerned by boundary functionmethod [2–5]. One of the basic difficulties for such a problem is unknown of where an internaltransition layer is in advance.

Butuzov and Vasil’eva initiated the concept of contrast structures in 1987 [6] andstudied the following boundary value problem of a second-order semilinear equation with astep-like contrast structure, which is called a monolayer solution in [1]

μ2y′′ = F(y, t), 0 ≤ t ≤ 1,

y(0, μ)= y0, y

(1, μ)= y1,

(1.1)

where μ > 0 is a small parameter and F has a desired smooth scalar function on its arguments.


Suppose that the reduced equation F(y, t) = 0 has two isolated solutions y = ϕi(t) (i =1, 2) on 0 ≤ t ≤ 1, which satisfy the following condition:

ϕ1(t) < ϕ2(t), Fy

(ϕi(t), t

)> 0, i = 1, 2. (1.2)

The condition (1.2) indicates that there exist two saddle equilibria Mi(ϕi(t), 0) (i =1, 2) in the phase plane (y, z) of the associated equations given by

dy

dτ= z,

dz

dτ= F(y, t), 0 < t < 1, (1.3)

where t is fixed and −∞ < τ < +∞.It is shown in [6] that the existence of an internal transition layer for the problem (1.1)

is closely related to the existence of a heteroclinic orbit connecting M1 and M2. The principalvalue t0 of an internal transition time t∗ is determined by an equation as follows:

∫ϕ2(t0)

ϕ1(t0)F(y, t0)dy = 0. (1.4)

In [7], Vasil’eva further studied the existence of step-like contrast structures for a classof singularly perturbed equations given by

μdu

dt= f(u, v, t),

μdv

dt= g(u, v, t), 0 ≤ t ≤ 1,

(1.5)

where f and g are scalar functions. For (1.5), we may impose either a first class of boundarycondition or a second class of boundary condition.

Suppose that there exist two solutions {ϕi(t), ψi(t)} (i = 1, 2) of the reduced equationsf(u, v, t) = 0; g(u, v, t) = 0, and Mi(ϕi(t), ψi(t)) (i = 1, 2) are two saddle equilibria in thephase plane (u, v) of the associated equations given by

du

dτ= f(u, v, t

);

dv

dτ= f(u, v, t

),

(1.6)

where t is fixed with 0 < t < 1. This indicates that the eigenvalues λik(t) (i, k = 1, 2) of theJacobian matrix

A(t)=

(fu fv

gu gv

)

u=ϕi(t), v=ψi(t)

(1.7)


satisfy the condition as follows:

either λi1(t)> 0, λi2

(t)< 0, or λi1

(t)< 0, λi2

(t)> 0. (1.8)

If (1.6) is a Hamilton equation, that is, gu = −fv, it implies that gdu−fdv = dH(u, v, t).Then, the equation to determine t0 is given by

H(ϕ1(t0), ψ1(t0), t0

)= H

(ϕ2(t0), ψ2(t0), t0

). (1.9)

Geometrically, (1.9) is also a condition for the existence of a heteroclinic orbit connectingM1(ϕ1(t), ψ1(t)) and M2(ϕ2(t), ψ2(t)).

Unfortunately, for a high dimensional singularly perturbed system, we cannot alwaysfind such an equation like (1.9) to determine t0 at which there exists a heteroclinic orbit. This isone difficulty to further study the problem on step-like contrast structures. On the other hand,we know that the existence of a spike-like or a step-like contrast structure of high dimensionis closely related to the existence of a homoclinic or heteroclinic orbit in its correspondingphase space, respectively. However, the existence of a homoclinic or heteroclinic orbit in highdimension space and how to construct such an orbit are themselves open in general in thequalitative analysis (geometric method) theory [8–10]. To explore these high dimensionalcontrast structure problems, we just start from some particular class of singularly perturbedsystem and are trying to develop some approach to construct a desired heteroclinic orbitby using a first integral method for such a class of the system and determine its internaltransition time t0.

2. Problem Formulation

We consider a class of semilinear singularly perturbed system as follows:

μ2y′′1 = f1(y1, y2, . . . , yn, t

);

μ2y′′2 = f2(y1, y2, . . . , yn, t

);

...

μ2y′′n = fn(y1, y2, . . . , yn, t

),

(2.1)

with a first class of boundary condition given by

yk

(0, μ)= y0

k, k = 1, 2, . . . , n;

y′j(0, μ)= z0

j j = 1, 2, . . . , n − 1;

y′n(1, μ)= z1

n,

(2.2)

where μ > 0 is a small parameter.


The class of system (2.1) in question has a strong application background inengineering. For example, in the study of smart materials of variated current of liquid [11, 12],its math model is a kind of such a system like (2.1), where the small parameter μ indicates aparticle. The given boundary condition (2.2) corresponds the stability condition [H3] listedlater to ensure that there exists a solution for the problem in question.

For our convenience, the system (2.1) can also be written in the following equivalentform,

μy′1 = z1;

μy′2 = z2;

...

μy′n = zn;

μz′1 = f1(y1, y2, . . . , yn, t

);

μz′2 = f2(y1, y2, . . . , yn, t

);

...

μz′n = fn(y1, y2, . . . , yn, t

).

(2.3)

Then, the corresponding boundary condition (2.2) is now written as

yk

(0, μ)= y0

k, zj(0, μ)= μz0

j , zn(1, μ)= μz1

n; k = 1, 2, . . . , n, j = 1, 2, . . . , n − 1. (2.4)

The following assumptions are fundamental in theory for the problem in question.[H1] Suppose that the functions fi (i = 1, 2, . . . , n) are sufficiently smooth on the

domain D = {(y1, y2, . . . , yn, t) | |yi| ≤ li, 0 ≤ t ≤ 1, i = 1, 2, . . . , n}, where li > 0 are realnumbers.

[H2] Suppose that the reduced system of (2.1) given by

f1(y1, y2, . . . , yn, t

)= 0;

f2(y1, y2, . . . , yn, t

)= 0;

...

fn(y1, y2, . . . , yn, t

)= 0

(2.5)

has two isolated solutions on D:

{y1 = a1

1(t), y2 = a12(t), . . . , yn = a1

n(t)},

{y1 = a2

1(t), y2 = a22(t), . . . , yn = a2

n(t)}. (2.6)


[H3] Suppose that the characteristic equation of the system (2.3) given by

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

−λ · · · 0 1 · · · 0

.... . .

......

. . ....

0 · · · −λ 0 · · · 1

f1y1 · · · f1yn −λ. . .

...

.... . .

......

. . . 0

fny1 · · · fnyn 0 · · · −λ

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣y1=ai1(t),y2=ai2(t),...,yn=ain(t)

= 0; (i = 1, 2) (2.7)

has 2n real valued solutions λk(t), k = 1, 2, . . . , 2n, where

Re λk(t) < 0, k = 1, 2, . . . , 2n − 1;

Re λ2n(t) > 0.(2.8)

Remark 2.1. [H3] is called as a stability condition. For a more general stability condition givenby

Re λ1(t) < 0, . . . ,Re λk(t) < 0;

Re λk+1(t) > 0, . . . ,Re λ2n(t) > 0, 1 < k < 2n,(2.9)

it will be studied in the other paper because of more complicated dynamic performancepresented.

Under the assumption of [H3], there may exist a solution y(t, μ) with only twoboundary layers that occurred at t = 0 and t = 1, for which the detailed discussion hasbeen given by [13, Theorem 4.2], or it may consults [5, Theorem 2.4, page 49]. We are onlyinterested in a solution y(t, μ) with a step-like contrast structure in this paper. That is, thereexists t∗ ∈ (0, 1) such that the following limit holds:

limμ→ 0

y(t, μ)=

⎧⎨

⎩

a1(t), 0 < t < t∗,

a2(t), t∗ < t < 1.(2.10)

We regard the solution y(t, μ) defined above with such a step-like contrast structure asbeing smoothly connected by two pure boundary solutions: y(−)(t, μ), 0 ≤ t < t∗ and y(+)(t, μ),t∗ < t ≤ 1. That is,

y(−)(t∗, μ)= y(+)(t∗, μ

); z(−)

(t∗, μ)= z(+)

(t∗, μ). (2.11)


The assumption [H3] ensures that the corresponding associated system given by

dyk

dτ= zk, 0 < t < 1,

dzkdτ

= fk(y1, y2, . . . , yn, t

), k = 1, 2, . . . , n,

(2.12)

has two equilibria Mi(ai1(t), a

i2(t), . . . , a

in(t), 0, . . . , 0) (i = 1, 2), where t is fixed. They are both

hyperbolic saddle points. From [13, Theorem 4.2] (or [5, Theorem 2.4]), it yields that thereexists a stable manifold Ws(Mi) of 2n − 1 dimensions and an unstable manifold Wu(Mi) ofone-dimension in a neighborhood of Mi. To get a heteroclinic orbit connecting M1 and M2 inthe corresponding phase space, we need some more assumptions as follows.

[H4] Suppose that the associated system (2.12) has a first integral

Φ(y1, . . . , yn, z1, . . . , zn, t

)= C, (2.13)

where C is an arbitrary constant and Φ is a smooth function on its arguments.Then, the first integral passing through Mi (i = 1, 2) can be represented by

Φ(y1, . . . , yn, z1, . . . , zn, t

)= Φ(Mi, t

). (2.14)

[H5] Suppose that (2.14) is solvable with respect to zn, which is denoted by

zn = h(y1, . . . , yn, z1, . . . , zn−1, t,Mi

), (i = 1, 2). (2.15)

Let z(−)n = h(−)(y(−)1 , . . . , y

(−)n , z

(−)1 , . . . , z

(−)n−1, t,M1) and z

(+)n = h(+)(y(+)

1 , . . . , y(+)n , z

(+)1 , . . .,

z(+)n−1, t,M2) be the parametric expressions of orbit passing through the hyperbolic saddle

points M1 and M2, respectively.Corresponding to the given boundary condition (2.2), we consider the following initial

value relation at τ = 0

y(−)k (0) = y

(+)k (0), k = 1, . . . , n; z

(−)j (0) = z

(+)j (0), j = 1, . . . , n − 1. (2.16)

Let

H(t)= h(−) − h(+) = 0, (2.17)

where

h(−) = h(−)(y(−)1 (0), . . . , y(−)

n (0), z(−)1 (0), . . . , z(−)n−1(0), t,M1

);

h(+) = h(+)(y(+)1 (0), . . . , y(+)

n (0), z(+)1 (0), . . . , z(+)n−1(0), t,M2

).

(2.18)


[H6] Suppose that (2.17) is solvable with respect to t and it yields a solution t = t0.That is, H(t0) = 0 and H ′(t0)/= 0.

Remark 2.2. It is easy to see from (2.14) and (2.17) that the necessary condition of the existenceof a heteroclinic orbit connecting M1 and M2 can also be expressed as “the equation

Φ(M1, t

)= Φ(M2, t

)(2.19)

is solvable with respect to t = t0.”

3. Construction of Asymptotic Solution

We seek an asymptotic solution of the problem (2.1)-(2.2) of the form

yk

(t, μ)=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

∞∑

l=0

μl[y(−)kl (t) + Πlyk(τ0) +Q

(−)l

yk(τ)], 0 ≤ t ≤ t∗;

∞∑

l=0

μl[y(+)kl (t) + Rlyk(τ1) +Q

(+)l yk(τ)

], t∗ ≤ t ≤ 1,

zk(t, μ)=

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

∞∑

l=0

μl[z(−)kl (t) + Πlzk(τ0) +Q

(−)l zk(τ)

], 0 ≤ t ≤ t∗;

∞∑

l=0

μl[z(+)kl (t) + Rlzk(τ1) +Q

(+)l

zk(τ)], t∗ ≤ t ≤ 1,

(3.1)

where τ0 = t/μ, τ = (t − t∗)/μ, τ1 = (t − 1)/μ; and for k = 1, 2, . . . , n, y(∓)kl (t) (0 < t < 1)

are coefficients of regular terms; Πlyk(τ0) (τ0 ≥ 0) are coefficients of boundary layer termsat t = 0; Rlyk(τ1) (τ1 ≤ 0) are coefficients of boundary layer terms at t = 1; and Q

(∓)l

yk(τ)(−∞ < τ < +∞) are left and right coefficients of internal transition terms at t = t∗. Meanwhile,similar definitions are for z(∓)

kl(t), Πlzk(τ0), Rlzk(τ1), and Q

(∓)l

zk(τ).The position of a transition time t∗ ∈ (0, 1) is unknown in advance. It needs being

determined during the construction of an asymptotic solution. Suppose that t∗ has also anasymptotic expression of the form

t∗ = t0 + μt1 + μ2t2 + · · · , (3.2)

where tl (l = 0, 1, 2, . . .) are temporarily unknown at the moment and will be determined later.Meanwhile, let

y1(t∗) = y∗10 + μy∗11 + μ2y∗12 + · · · , (3.3)

where y∗1l (l = 0, 1, 2, . . .) are all constants, independent of μ, and y1(t∗) takes value betweena1

1(t∗) and a2

1(t∗). For example, y1(t∗) = (1/2)(a1

1(t∗) + a2

1(t∗)).


Then, we will determine the asymptotic solution (3.1) step by step using “a smoothconnection method” based on the boundary function method [13] or [5]. The smoothconnection condition (2.11) can be written as

Q(−)0 yk(0) + a1

k(t0) = Q(+)0 yk(0) + a2

k(t0), Q(−)0 zk(0) = Q

(+)0 zk(0);

Q(−)l

yk(0) +[y′(−)1l (t0)tl + ξ

(−)1l

]= Q

(−)0 yk(0) +

[y′(+)1l (t0)tl + ξ

(+)1l

];

Q(−)l

zk(0) +[z′(−)1l (t0)tl + η

(−)1l

]= Q

(−)0 zk(0) +

[z′(+)1l (t0)tl + η

(+)1l

],

(3.4)

where k = 1, 2, . . . , n, l = 1, 2, . . .; ξ(∓)1l = ξ(∓)1l (t0, . . . , tl−1), and ξ

(∓)1l = ξ

(∓)1l (t0, . . . , tl−1) are all the

known functions depending only on t0, . . . , tl−1.Substituting (3.1) into (2.1)-(2.2) and equating separately the terms depending on

t, τ0, τ1, and τ by the boundary function method, we can obtain the equations to deter-mine {y(∓)

kl(t), z(∓)

kl(t)}; {Πlyk(τ0),Πlzk(τ0)}, {Rlyk(τ1), Rlzk(τ1)}, and {Q(∓)

lyk(τ), Q

(∓)l

zk(τ)},respectively. The equations to determine the zero-order coefficients of regular terms{y(∓)

k0 (t), z(∓)k0 (t)} (k = 1, 2, . . . , n) are given by

z(∓)10 (t) = z

(∓)20 (t) = · · · = z

(∓)n0 (t) = 0;

f1

(y(∓)10 , y

(∓)20 , . . . , y

(∓)n0 , t)= 0;

f2

(y(∓)10 , y

(∓)20 , . . . , y

(∓)n0 , t)= 0;

...

fn(y(∓)10 , y

(∓)20 , . . . , y

(∓)n0 , t)= 0.

(3.5)

It is clear to see that (3.6) coincides with the reduced system (2.11). Therefore, by [H2],(3.6) has the solution

{y(∓)10 , y

(∓)20 , . . . , y

(∓)n0

}={ai

1(t), ai2(t), . . . , a

in(t)}, i = 1, 2. (3.6)

The equations to determine {y(∓)kl (t),z

(∓)kl (t)} (k = 1, 2, . . . , n; l = 1, 2, . . .) are given by

y′1l−1 = z1l, y′2l−1 = z2l, . . . , y

′nl−1 = znl;

z′1l−1 = f1y1(t)y1l + f1y2

(t)y2l + · · · + f1yn(t)ynl + h1l(t);

z′2l−1 = f2y1(t)y1l + f2y2

(t)y2l + · · · + f2yn(t)ynl + h2l(t);

...

z′nl−1 = fny1(t)y1l + fny2

(t)y2l + · · · + fnyn(t)ynl + hnl(t).

(3.7)


Here the superscript (∓) is omitted for the variables y(∓)kl and z

(∓)kl in (3.8) for simplicity in

notation. To understand ykl and zkl, we agree that they take (−) when 0 ≤ t ≤ t0; while theytake (+) when t0 ≤ t ≤ 1. The terms hkl(t) (k = 1, 2, . . . , n; l = 1, 2, . . .) are expressed in termsof ykm and zkm (k = 1, 2, . . . , n; m = 0, 1, . . . , l − 1). Also f (·)(t) are known functions that takevalue at (ai

1(t), ai2(t), . . . , a

in(t)), where i = 1 when 0 ≤ t ≤ t0 and i = 2 when t0 ≤ t ≤ 1.

Since (3.8) is an algebraic linear system, the solution {y(∓)kl

(t), z(∓)kl

(t)} (k = 1, 2, . . . , n;l = 1, 2, . . .) is uniquely solvable by [H3].

Next, we give the equations and their conditions for determining the zero-ordercoefficient of an internal transition layer {Q(−)

0 yk(τ), Q(−)0 zk(τ)} as follows:

d

dτQ

(−)0 yk = Q

(−)0 zk, −∞ < τ ≤ 0;

d

dτQ

(−)0 zk = fk

(a1

1(t0) +Q(−)0 y1, . . . , a

1n(t0) +Q

(−)0 yn, t0

),

Q(−)0 y1(0) = y∗10 − a

11(t0);

Q(−)0 yk(−∞) = 0, Q(−)

0 zk(−∞) = 0, k = 1, 2, . . . , n.

(3.8)

We rewrite (3.9) in a different form by making the change of variables

y(−)k = a1

k(t0) +Q(−)0 yk, z

(−)k = Q

(−)0 zk, k = 1, 2, . . . , n. (3.9)

Then, (3.9) is further written in these new variables as

dy(−)k

dτ= z

(−)k , −∞ < τ ≤ 0;

dz(−)k

dτ= fk

(y(−)1 , y

(−)2 , . . . , y

(−)n , t0

),

(3.10)

y(−)0 (0) = y∗10;

y(−)k (−∞) = a1

k(t0), z(−)k (−∞) = 0, k = 1, 2, . . . , n.

(3.11)

From [H3], it yields that the equilibrium (a11(t0), . . . , a

1n(t0), 0, . . . , 0) of the autonomous

system (3.11) is a hyperbolic saddle point. Therefore, there exists an unstable one-dimensional manifold Wu(M1). For the existence of a solution of (3.11) satisfying (3.12), weneed the following assumption.

[H7] Suppose that the hyperplane y(−)1 (0) = y∗10 intersects the manifold Wu(M1) in the

phase space (y(−)1 (t0), y

(−)2 (t0), . . . , y

(−)n (t0)) × (z(−)1 (t0), z

(−)2 (t0), . . . , z

(−)n (t0)), where t0 ∈ (0, 1) is

a parameter.


Then, {y(−)k (0), z(−)k (0)} (k = 1, 2, . . . , n) are known values after {y(−)

k (τ), z(−)k (τ)} beingsolved by [H7]. We can get the equations and the corresponding boundary conditions todetermine {Q(+)

0 yk(τ), Q(+)0 zk(τ)} as follows:

d

dτQ

(+)0 yk = Q

(+)0 zk, 0 ≤ τ < +∞;

d

dτQ

(+)0 zk = fk

(a2

1(t0) +Q(+)0 y1, . . . , a

2n(t0) +Q

(+)0 yn, t0

),

(3.12)

Q(+)0 yk(0) = y

(−)k (0) − a2

k(t0), k = 1, 2, . . . , n;

Q(+)0 zj(0) = z

(−)j (0), j = 1, 2, . . . , n − 1;

Q(+)0 yk(+∞) = 0, Q

(+)0 zk(+∞) = 0, k = 1, 2, . . . , n.

(3.13)

Introducing a similar transformation as doing for (3.9), we can get

dy(+)k

dτ= z

(+)k , 0 ≤ τ < +∞;

dz(+)k

dτ= fk

(y(+)1 , y

(+)2 , . . . , y

(+)n , t0

),

(3.14)

y(+)k (0) = y

(−)k (0), k = 1, 2, . . . , n;

z(+)j (0) = z

(−)j (0), j = 1, 2, . . . , n − 1;

y(+)k (+∞) = a2

k(t0), z(+)k (+∞) = 0, k = 1, 2, . . . , n.

(3.15)

To ensure that the existence of a solution of (3.15)-(3.16), we need the followingassumption.

[H8] Suppose that the hypercurve {y(+)1 (0) = y

(−)1 (0), . . . , y(+)

n (0) = y(−)n (0), z(+)1 (0) =

z(−)1 (0), . . . , z(+)n (0) = z

(−)n (0)} intersects the manifold Ws(M2) in the phase space

(y(+)1 (t0), y

(+)2 (t0), . . . , y

(+)n (t0)) × (z(+)1 (t0), z

(+)2 (t0), . . . , z

(+)n (t0)), where t0 ∈ (0, 1) is a parameter.

Here it should be emphasized that under the conditions of [H7] and [H8], the solutions{Q(∓)

0 yk(τ), Q(∓)0 zk(τ)} (k = 1, 2, . . . , n) not only exist but also decay exponentially [13], or [5].

If the parameter t0 is determined, {Q(∓)0 yk(τ), Q

(∓)0 zk(τ)} (k = 1, 2, . . . , n) are

completely known. To determine t0, it is closely related to the existence of a heteroclinic orbitconnecting M1 and M2 in the phase space.

By the given initial values (3.14) or (3.16), we have already obtained

y(+)k (0) = y

(−)k (0), k = 1, 2, . . . , n,

z(+)j (0) = z

(−)j (0), j = 1, 2, . . . , n − 1.

(3.16)


If we show z(+)n (0) = z

(−)n (0), the smooth connection condition (2.11) for the zero-order is

satisfied. By [H4] and [H5], we have

z(∓)n = h(∓)

(y(∓)1 , . . . , y

(∓)n , z

(∓)1 , . . . , z

(∓)n−1, t0,M1,2

). (3.17)

Since y(∓)k

(k = 1, 2, . . . , n) and z(∓)j (j = 1, 2, . . . , n − 1) only depend on y∗10, while y∗10 only

depends on t0, the necessary condition for existence of a heteroclinic orbit connecting M1

and M2 at τ = 0 is given by

h(−)(y∗10(t0), t0,M1)= h(+)(y∗10(t0), t0,M2

), (3.18)

or

Φ(M1, t0) = Φ(M2, t0). (3.19)

However (3.19) or (3.20) is the one to determine t0. Then, by [H6], there exists ant0 = t∗0 from (3.19) or (3.20). We can see that the process of determining t0 is the one ofa smooth connection. Therefore, all the zero-order terms {Q(∓)

0 yk(τ), Q(∓)0 zk(τ)} have now

been completely determined by the smooth connection for the zero-order coefficients of theasymptotic solution.

For the high-order terms {Q(∓)l

yk(τ), Q(∓)l

zk(τ)} (l = 1, 2, . . .), we have the equationsand their boundary conditions as follows:

d

dτQ

(−)l yk = Q

(−)l zk, −∞ < τ ≤ 0;

d

dτQ

(−)l

zk = f(−)ky1

(τ)Q(−)l

y1 + f(−)ky2

(τ)Q(−)l

y2 + · · · f (−)kyn

(τ)Q(−)l

yn + G(−)lk (τ),

Q(−)l y1(0) = y∗1l −

[y1l(t0)tl + ξ1l(t0, . . . , tl−1)

];

Q(−)l yk(−∞) = 0, Q

(−)l zk(−∞) = 0, k = 1, 2, . . . , n,

d

dτQ

(+)l yk = Q

(+)l zk, 0 ≤ τ < +∞;

d

dτQ

(+)l zk = f

(+)ky1

(τ)Q(+)l y1 + f

(+)ky2

(τ)Q(+)l y2 + · · · f (+)

kyn(τ)Q(−)

l yn + G(+)lk (τ),

Q(+)l yk(0) = y∗kl(0) −

[y′kl(t0)tl + ξkl(t0, . . . , tl−1)

], k = 1, 2, . . . , n;

Q(+)l

zj(0) = z∗jl −[z′kl(t0)tl + ηkl(t0, . . . , tl−1)

], j = 1, 2, . . . , n − 1;

Q(+)l yk(+∞) = 0, Q

(+)l zk(+∞) = 0, k = 1, 2, . . . , n,

(3.20)

where f(∓)(·) (τ) represent known functions that take value at (ai

1(t0) + Q(∓)0 y1(τ), . . . , ai

n(t0) +

Q(∓)0 yn(τ)); G

(∓)lk

(τ) are the known functions that only depend on those asymptotic terms


whose subscript is 0, 1, . . . , l − 1; and ξkl and ηkl are all the known functions. Since (3.21) areall linear boundary value problems, it is not difficult to prove the existence of solution andthe exponential decaying of solution without imposing any extra condition.

As for boundary functions Πy(τ0) and Ry(τ1), it is easy to obtain their constructionsby using the normal boundary function method. So we would not discuss the details on themhere [13] or [5]. However, it is worth mentioning that the coefficients tl (l = 1, 2, . . .) in (2.3)will be determined by an equation as follows:

H ′(t0)tl = ξ(t0, . . . , tl−1). (3.21)

Then, tl can be solved from (3.22) by [H6]. Then, we have so far constructed the asymptoticexpansion of a solution with an internal transition layer for the problem (2.1)-(2.2) and theasymptotic expansion of an internal transition time t∗.

4. Existence of Step-Like Solution and Its Limit Theorem

We mentioned in Section 2that the solution with a step-like contrast structure can be regardedas a smooth connection by two solutions of pure boundary value problem from left and right,respectively. To this end, we establish the following two associated problems.

For the left associated problem,

μ(y(−)k

)′= z

(−)k

;

μ(z(−)k

)′= fk

(y(−)1 , . . . , y

(−)n , t

),

(4.1)

y(−)k

(0, μ)= y0

k, k = 1, 2, . . . , n;

z(−)j

(0, μ)= μz0

j , j = 1, 2, . . . , n − 1;

y(−)1

(t∗, μ)= y1(t∗),

(4.2)

where 0 ≤ t ≤ t∗ < 1, t∗ is a parameter, such a solution {y(−)k

(t, μ), z(−)k

(t, μ)} of (4.1) and (4.2)exists by [H1]–[H3] [14, 15]. Then, we have {y(−)

k (t∗, μ), z(−)k (t∗, μ)}, k = 1, 2, . . . , n.For the right associated problem,

μ(y(+)k

)′= z

(+)k

;

μ(z(+)k

)′= fk

(y(+)1 , . . . , y

(+)n , t

),

(4.3)

y(+)k

(t∗, μ)= y

(−)k (t∗), k = 1, 2, . . . , n;

z(+)j

(t∗, μ)= z

(−)j (t∗), j = 1, 2, . . . , n − 1;

z(+)n

(1, μ)= μz1

n,

(4.4)


where 0 < t∗ ≤ t ≤ 1, t∗ is still a parameter, the similar reason is for the existence of{y(+)

k(t, μ), z(+)

k(t, μ)} of (4.3) and (4.4) [14, 15].

Then, we write the asymptotic expansion of {y(∓)k

(t, μ), z(∓)k

(t, μ)} as follows:

yk

(t, μ)=

⎧⎪⎨

⎪⎩

y(−)k

(t, μ)= a1

k(t) + Π0yk(τ0) +Q(−)0 yk(τ) +O

(μ), 0 ≤ t ≤ t∗;

y(+)k

(t, μ)= a2

k(t) + R0yk(τ1) +Q(+)0 yk(τ) +O

(μ), t∗ ≤ t ≤ 1,

(4.5)

zk(t, μ)=

⎧⎪⎨

⎪⎩

z(−)k

(t, μ)= Π0zk(τ0) +Q

(−)0 zk(τ) +O

(μ), 0 ≤ t ≤ t∗;

z(+)k

(t, μ)= R0zk(τ1) +Q

(+)0 zk(τ) +O

(μ), t∗ ≤ t ≤ 1,

(4.6)

where τ0 = t/μ, τ1 = (t − 1)/μ and τ = (t − t∗)/μ.We proceed to show that there exists an t∗ indeed in the neighborhood of t0 such

that the solution {y(−)k

(t, μ), z(−)k

(t, μ)} of the left associated problem (4.1) and (4.2) and thesolution {y(+)

k(t, μ), z(+)

k(t, μ)} of the right associated problem (4.3) and (4.4) smoothly connect

at t∗, from which we obtain the desired step-like solution.From the asymptotic expansion of (4.5) and (4.6), we know that {a1

1(t), a12(t), . . . , a

1n(t)}

and {a21(t), a

22(t), . . . , a

2n(t)} are the solutions of the reduced system (2.5). In the neighborhood

of t0, the boundary functions {Π0yk(τ0),Π0zk(τ0)} and {R0yk(τ1), R0zk(τ1)} are bothexponentially small. Thus, they can be omitted in the neighborhood of t0.

We are now concerned with the equations and the boundary conditions forwhich {Q(−)

0 yk(τ), Q(−)0 zk(τ)} satisfy. They can be obtained easily from (3.9) just with the

replacement of t0 by t∗,

d

dτQ

(−)0 yk = Q

(−)0 zk, −∞ < τ ≤ 0;

d

dτQ

(−)0 zk = fk

(a1

1(t∗) +Q

(−)0 y1, . . . , a

1n(t∗) +Q

(−)0 yn, t

∗),

Q(−)0 y1(0) = y∗10 − a

11(t∗);

Q(−)0 yk(−∞) = 0, Q

(−)0 zk(−∞) = 0, k = 1, 2, . . . , n.

(4.7)

After the change of variables given by

y(−)k

= a1k(t∗) +Q

(−)0 yk(τ); z

(−)k

= Q(−)0 zk(τ); k = 1, 2, . . . , n, (4.8)


then (4.7) can be written as

dy(−)k

dτ= z

(−)k

, −∞ < τ ≤ 0;

dz(−)k

dτ= fk

(y(−)1 , . . . , y

(−)n , t∗

),

(4.9)

y(−)1 = y1(t∗);

y(−)k (−∞) = a1

k(t∗), z

(−)k (−∞) = 0, k = 1, 2, . . . , n.

(4.10)

It is similar to get the equations and the boundary conditions for which{Q(+)

0 yk(τ), Q(+)0 zk(τ)} satisfy

d

dτQ

(+)0 yk = Q

(+)0 zk, 0 ≤ τ < +∞;

d

dτQ

(+)0 zk = fk

(a2

1(t∗) +Q

(+)0 y1, . . . , a

2n(t∗) +Q

(+)0 yn, t

∗),

Q(+)0 yk(0) = y

(−)k (0) − a2

k(t∗), k = 1, 2, . . . , n;

Q(+)0 zj(0) = z

(−)j (0), j = 1, 2, . . . , n − 1;

Q(+)0 yk(+∞) = 0, Q

(+)0 zk(+∞) = 0.

(4.11)

After the transformation given by

y(+)k = a2

k(t∗) +Q

(+)0 yk(τ); z

(+)k = Q

(+)0 zk(τ); k = 1, 2, . . . , n, (4.12)

then (4.11) can be written as

dy(+)k

dτ= z

(+)k , 0 ≤ τ < +∞;

dz(+)k

dτ= fk

(y1, . . . , yn, t

∗),

(4.13)

y(+)k (0) = y

(−)k (0), k = 1, 2, . . . , n;

z(+)k (0) = z

(−)j (0), j = 1, 2, . . . , n − 1;

y(+)k (+∞) = 0, z

(+)k (+∞) = 0.

(4.14)

[H3] and [H4] imply that there exists a first integral

Φ(y(−)1 , . . . , y

(−)n , z

(−)1 , . . . , z

(−)n , t∗

)= Φ(M1, t

∗) (4.15)


of the system (4.9) that approaches M1(a11(t∗), a1

2(t∗), . . . , a1

n(t∗), 0, . . . , 0) as τ → −∞; and

there exists a first integral

Φ(y(+)1 , . . . , y

(+)n , z

(+)1 , . . . , z

(+)n , t∗

)= Φ(M2, t

∗) (4.16)

of the system (4.13) that approaches M2(a21(t∗), a2

2(t∗), . . . , a2

n(t∗), 0, . . . , 0) as τ → +∞.

In views of [H5], from (4.15) and (4.16) we have

z(∓)n = h(∓)

(y(∓)1 , . . . , y

(∓)n , z

(∓)1 , . . . , z

(∓)n−1, t

∗,M1,2

). (4.17)

Then, we know from (4.2) and (4.4) that y(−)k

(t∗, μ) = y(+)k

(t∗, μ), k = 1, 2, . . . , n; and z(−)j (t, μ) =

z(+)j (t, μ), j = 1, 2, . . . , n−1 for the solutions {y(∓)

k (t, μ), z(∓)k (t, μ)} of the left and right associatedproblems. For a smooth connection of the solutions, the remaining is to prove

z(−)n

(t∗, μ)= z

(+)n

(t∗, μ). (4.18)

Let

Δ(t∗) = z(−)n

(t∗, μ)− z(+)n

(t∗, μ). (4.19)

Substituting (4.6) into (4.19), we have

Δ(t∗) = Q(−)0 zn(0) −Q(+)

0 zn(0) +O(μ)= h(−) − h(+) +O

(μ)

= H(t∗) +O(μ)= H(t0) +H ′(t0)(t∗ − t0) +O(t∗ − t0)2 +O

(μ),

(4.20)

where O(μ) can be regarded as Cμ for simplicity.If we take t∗ = t0 ±Kμ in (4.20), we have

Δ(t∗) = ±H ′(t0)Kμ +O(μ). (4.21)

Since the sign of H ′(t0) is fixed, (4.21) has an opposite sign when K is sufficiently large, forexample, K > C, and μ is sufficiently small. That is,

(H ′(t0)Kμ +O

(μ))(−H ′(t0)Kμ +O

(μ))

< 0, as K > C, μ 1. (4.22)

Then, there exists t ∈ [t0 −Kμ, t0 +Kμ] such that Δ(t) = 0 by applying the intermediate valuetheorem to (4.21). This implies in turn that (4.18) holds.

Therefore, we have shown that there exists a step-like contrast structure for theproblem (2.1)-(2.2). We summarize it as the following main theorem of this paper.


Theorem 4.1. Suppose that [H1]–[H8] hold. Then, there exists an μ0 > 0 such that there exists a step-like contrast structure solution yk(t, μ) (k = 1, 2, . . . , n) of the problem (2.1)-(2.2) when 0 < μ < μ0.Moreover, the following asymptotic expansion holds

yk

(t, μ)=

⎧⎨

⎩

a1k(t) + Π0yk(τ0) +Q

(−)0 yk(τ) +O

(μ), 0 ≤ t ≤ t;

a2k(t) + R0yk(τ1) +Q

(+)0 yk(τ) +O

(μ), t ≤ t ≤ 1.

(4.23)

Remark 4.2. Only existence of solution with a step-like contrast structures is guarantiedunder the conditions of [H1]–[H8]. There may exists a spike-like contrast structure, or thecombination of them [16] for the problem (2.1)-(2.2). They need further study.

5. Conclusive Remarks

The existence of solution with step-like contrast structures for a class of high-dimensionalsingular perturbation problem investigated in this paper shows that how to get a heteroclinicorbit connecting saddle equilibria M1 and M2 in the corresponding phase space is a keyto find a step-like internal layer solution. Using only one first integral of the associatedsystem, this demands only bit information on solution, is our first try to construct adesired heteroclinic orbit in high-dimensional phase space. It needs surely further study forthis interesting connection between the existence of a heteroclinic orbit of high-dimensionin qualitative theory and the existence of a step-like contrast structure (internal layersolution) in a high-dimensional singular perturbation boundary value problem of ordinarydifferential equations. The particular boundary condition we adopt in this paper is just forthe corresponding stability condition, which ensures the existence of solution of the problemin this paper. For the other type of boundary condition, we need some different stabilitycondition to ensure the existence of solution of the problem in question, which we also needto study separately. Finally, if we want to construct a higher-order asymptotic expansion, it issimilar with obvious modifications in which only more complicated techniques involved.

Acknowledgments

The authors are grateful for the referee’s suggestion that helped to improve the presentationof this paper. This work was supported in part by E-Institutes of Shanghai MunicipalEducation Commission (N.E03004); and in part by the NSFC with Grant no. 10671069. Thiswork was also supported by Shanghai Leading Academic Discipline Project with Project no.B407.

References

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[2] A. B. Vasil’eva and V. F. Butuzov, The Asymptotic Method of Singularly Perturbed Theory, Nauka,Moscow, Russia, 1990.

[3] V. F. Butuzov, A. B. Vasil’eva, and N. N. Nefedov, “Asymptotic theory of contrast structures,”Automation and Remote Control, no. 7, pp. 4–32, 1997.


[4] A. B. Vasil’eva, V. F. Butuzov, and N. N. Nefedov, “Contrast structures in singularly perturbedproblems,” Pure Mathematics and Applied Mathematics, vol. 4, no. 3, pp. 799–851, 1998.

[5] A. B. Vasil’eva, V. F. Butuzov, and L. V. Kalachev, The Boundary FunctionMethod for Singular PerturbationProblems, vol. 14 of SIAM Studies in Applied Mathematics, SIAM, Philadelphia, Pa, USA, 1995.

[6] V. F. Butuzov and A. B. Vasil’eva, “Asymptotic behavior of a solution of contrasting structure type,”Mathematical Notes, vol. 42, no. 6, pp. 956–961, 1987.

[7] A. B. Vasil’eva, “On contrast structures of step type for a system of singularly perturbed equations,”Mathematics and Mathematical Physics, vol. 34, no. 10, pp. 1401–1411, 1994.

[8] F. Battelli and K. J. Palmer, “Singular perturbations, transversality, and Sil’nikov saddle-focushomoclinic orbits,” Journal of Dynamics and Differential Equations, vol. 15, no. 2-3, pp. 357–425, 2003.

[9] K. Palmer and F. Battelle, “Heteroclinic orbits in singularly perturbed systems,” to appear in Discreteand Continuous Dynamical Systems.

[10] F. Battelli and M. Feckan, “Global center manifolds in singular systems,” Nonlinear DifferentialEquations and Applications, vol. 3, no. 1, pp. 19–34, 1996.

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and devices,” Journal of Intelligent Material Systems and Structures, vol. 7, no. 2, pp. 123–130, 1996.[13] A. B. Vasil’eva and V. F. Butuzov, Asymptotic Expansions of the Singularly Perturbed Equations, Nauka,

Moscow, Russia, 1973.[14] V. A. Esipova, “Asymptotic properties of solutions of general boundary value problems for singularly

perturbed conditionally stable systems of ordinary differential equations,” Differential Equations, vol.11, no. 11, pp. 1457–1465, 1975.

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Research ArticlePositive Solutions for a Class of Coupled System ofSingular Three-Point Boundary Value Problems

Naseer Ahmad Asif and Rahmat Ali Khan

Centre for Advanced Mathematics and Physics, Campus of College of Electrical and MechanicalEngineering, National University of Sciences and Technology, Peshawar Road, Rawalpindi 46000, Pakistan

Correspondence should be addressed to Rahmat Ali Khan, rahmat [email protected]

Received 27 February 2009; Accepted 15 May 2009


Existence of positive solutions for a coupled system of nonlinear three-point boundary valueproblems of the type −x′′(t) = f(t, x(t), y(t)), t ∈ (0, 1), −y′′(t) = g(t, x(t), y(t)), t ∈ (0, 1),x(0) = y(0) = 0, x(1) = αx(η), y(1) = αy(η), is established. The nonlinearities f , g : (0, 1) ×(0,∞) × (0,∞) → [0,∞) are continuous and may be singular at t = 0, t = 1, x = 0, and/or y = 0,while the parameters η, α satisfy η ∈ (0, 1), 0 < α < 1/η. An example is also included to show theapplicability of our result.

Copyright q 2009 N. A. Asif and R. A. Khan. 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.

1. Introduction

Multipoint boundary value problems (BVPs) arise in different areas of applied mathematicsand physics. For example, the vibration of a guy wire composed of N parts with a uniformcross-section and different densities in different parts can be modeled as a Multipointboundary value problem [1]. Many problems in the theory of elastic stability can also bemodeled as Multipoint boundary value problem [2].

The study of Multipoint boundary value problems for linear second order ordinarydifferential equations was initiated by Il’in and Moiseev, [3, 4], and extended to nonlocallinear elliptic boundary value problems by Bitsadze et al. [5, 6]. Existence theory for nonlinearthree-point boundary value problems was initiated by Gupta [7]. Since then the study ofnonlinear three-point BVPs has attracted much attention of many researchers, see [8–11] andreferences therein for boundary value problems with ordinary differential equations and also[12] for boundary value problems on time scales. Recently, the study of singular BVPs hasattracted the attention of many authors, see for example, [13–18] and the recent monographby Agarwal et al. [19].


The study of system of BVPs has also fascinated many authors. System of BVPs withcontinuous nonlinearity can be seen in [20–22] and the case of singular nonlinearity can beseen in [8, 21, 23–26]. Wei [25], developed the upper and lower solutions method for theexistence of positive solutions of the following coupled system of BVPs:

−x′′ (t) = f(t, x (t) , y (t)

), t ∈ (0, 1) ,

−y′′ (t) = g(t, x (t) , y (t)

), t ∈ (0, 1) ,

x (0) = 0, x (1) = 0,

y (0) = 0, y (1) = 0,

(1.1)

where f, g ∈ C((0, 1)×(0,∞)×(0,∞), [0,∞)), and may be singular at t = 0, t = 1, x = 0 and/ory = 0.

By using fixed point theorem in cone, Yuan et al. [26] studied the following coupledsystem of nonlinear singular boundary value problem:

x(4) (t) = f(t, x (t) , y (t)

), t ∈ (0, 1) ,

−y′′ (t) = g(t, x (t) , y (t)

), t ∈ (0, 1) ,

x (0) = x (1) = x′′ (0) = x′′ (1) = 0,

y (0) = y (1) = 0,

(1.2)

f, g are allowed to be superlinear and are singular at t = 0 and/or t = 1. Similarly, results arestudied in [8, 21, 23].

In this paper, we generalize the results studied in [25, 26] to the following more generalsingular system for three-point nonlocal BVPs:

−x′′ (t) = f(t, x (t) , y (t)

), t ∈ (0, 1) ,

−y′′ (t) = g(t, x (t) , y (t)

), t ∈ (0, 1) ,

x (0) = 0, x (1) = αx(η),

y (0) = 0, y (1) = αy(η),

(1.3)

where η ∈ (0, 1), 0 < α < 1/η, f, g ∈ C((0, 1) × (0,∞) × (0,∞), [0,∞)). We allow f and g to besingular at t = 0, t = 1, and also x = 0 and/or y = 0. We study the sufficient conditions forexistence of positive solution for the singular system (1.3) under weaker hypothesis on f andg as compared to the previously studied results. We do not require the system (1.3) to havelower and upper solutions. Moreover, the cone we consider is more general than the conesconsidered in [20, 21, 26].

By singularity, we mean the functions f(t, x, y) and g(t, x, y) are allowed to beunbounded at t = 0, t = 1, x = 0, and/or y = 0. To the best of our knowledge, existenceof positive solutions for a system (1.3) with singularity with respect to dependent variable(s)has not been studied previously. Moreover, our conditions and results are different from those


studied in [21, 24–26]. Throughout this paper, we assume that f, g : (0, 1) × (0,∞) × (0,∞) →[0,∞) are continuous and may be singular at t = 0, t = 1, x = 0, and/or y = 0. We also assumethat the following conditions hold:

(A1) f(·, 1, 1), g(·, 1, 1) ∈ C((0, 1), (0,∞)) and satisfy

a :=∫1

0t (1 − t) f (t, 1, 1)dt < +∞, b :=

∫1

0t (1 − t) g (t, 1, 1)dt < +∞. (1.4)

(A2) There exist real constants αi, βi such that αi ≤ 0 ≤ βi < 1, i = 1, 2, β1 + β2 < 1 and forall t ∈ (0, 1), x, y ∈ (0,∞),

c β1f(t, x, y

)≤ f(t, cx, y

)≤ cα1f

(t, x, y

), if 0 < c ≤ 1,

cα1f(t, x, y

)≤ f(t, cx, y

)≤ c β1f

(t, x, y

), if c ≥ 1,

c β2f(t, x, y

)≤ f(t, x, cy

)≤ cα2f

(t, x, y

), if 0 < c ≤ 1,

cα2f(t, x, y

)≤ f(t, x, cy

)≤ c β2f

(t, x, y

), if c ≥ 1.

(1.5)

(A3) There exist real constants γi, ρi such that γi ≤ 0 ≤ ρi < 1, i = 1, 2, ρ1 + ρ2 < 1 and forall t ∈ (0, 1), x, y ∈ (0,∞),

c ρ1g(t, x, y

)≤ g(t, cx, y

)≤ cγ1g

(t, x, y

), if 0 < c ≤ 1,

cγ1g(t, x, y

)≤ g(t, cx, y

)≤ c ρ1g

(t, x, y

), if c ≥ 1,

c ρ2g(t, x, y

)≤ g(t, x, cy

)≤ cγ2g

(t, x, y

), if 0 < c ≤ 1,

cγ2g(t, x, y

)≤ g(t, x, cy

)≤ c ρ2g

(t, x, y

), if c ≥ 1,

(1.6)

for example, a function that satisfies the assumptions (A2) and (A3) is

h(t, x, y

)=

m∑

i=1

n∑

j=1

pij (t)xriysj , (1.7)

where pij ∈ C((0, 1), (0,∞)), ri, sj < 1, i = 1, 2, . . . , m; j = 1, 2, . . . , n such that

max1≤i≤m

ri + max1≤j≤n

sj < 1. (1.8)

The main result of this paper is as follows.

Theorem 1.1. Assume that (A1)–(A3) hold. Then the system (1.3) has at least one positive solution.


2. Preliminaries

For each u ∈ E := C[0, 1], we write ‖u‖ = max{u(t) : t ∈ [0, 1]}. Let P = {u ∈ E : u(t) ≥ 0, t ∈[0, 1]}. Clearly, (E, ‖ · ‖) is a Banach space and P is a cone. Similarly, for each (x, y) ∈ E × E,we write ‖(x, y)‖1 = ‖x‖ + ‖y‖. Clearly, (E × E, ‖ · ‖1) is a Banach space and P × P is a cone inE × E. For any real constant r > 0, define Ωr = {(x, y) ∈ E × E : ‖(x, y)‖1 < r}. By a positivesolution of (1.3), we mean a vector (x, y) ∈ C[0, 1]∩C2(0, 1)×C[0, 1]∩C2(0, 1) such that (x, y)satisfies (1.3) and x > 0, y > 0 on (0, 1). The proofs of our main result (Theorem 1.1) is basedon the Guo’s fixed-point theorem.

Lemma 2.1 (Guo’s Fixed-Point Theorem [27]). Let K be a cone of a real Banach space E, Ω1, Ω2

be bounded open subsets of E and θ ∈ Ω1 ⊂ Ω2. Suppose that T : K ∩ (Ω2 \Ω1) → K is completelycontinuous such that one of the following condition hold:

(i) ‖Tx‖ ≤ ‖x‖ for x ∈ ∂Ω1 ∩K and ‖Tx‖ ≥ ‖x‖ for x ∈ ∂Ω2 ∩K;

(ii) ‖Tx‖ ≤ ‖x‖ for x ∈ ∂Ω2 ∩K and ‖Tx‖ ≥ ‖x‖ for x ∈ ∂Ω1 ∩K.

Then, T has a fixed point in K ∩ (Ω2 \Ω1).

The following result can be easily verified.

Result 1. Let t1, t2 ∈ R such that t1 < t2. Let x ∈ C[t1, t2], x ≥ 0 and concave on [t1, t2]. Then,x(t) ≥ min{t − t1, t2 − t}maxs∈[t1,t2]x(s) for all t ∈ [t1, t2].

Choose n0 ∈ {3, 4, 5, . . .} such that n0 > max{1/η, 1/(1 − η), (2 − α)/(1 − αη)}. For fixedn ∈ {n0, n0 + 1, n0 + 2, . . .} and z ∈ C[0, 1], the linear three-point BVP

−u′′ (t) = z (t) , t ∈[

1n, 1 − 1

n

],

u

(1n

)= 0, u

(1 − 1

n

)= αu

(η),

(2.1)

has a unique solution

u (t) =∫1−1/n

1/nHn (t, s) z (s)ds, (2.2)

where Hn : [1/n, 1 − 1/n] × [1/n, 1 − 1/n] → [0,∞) is the Green’s function and is given by

Hn (t, s)=

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

(t−1/n) (1− 1/n−s)1 − 2/n + α/n − αη −

α (t − 1/n)(η − s

)

1 − 2/n + α/n − αη − (t − s) ,1n≤ s ≤ t ≤ 1 − 1

n, s ≤ η,

(t − 1/n) (1 − 1/n − s)1 − 2/n + α/n − αη −

α (t − 1/n)(η − s

)

1 − 2/n + α/n − αη ,1n≤ t ≤ s ≤ 1 − 1

n, s ≤ η,

(t − 1/n) (1 − 1/n − s)1 − 2/n + α/n − αη ,

1n≤ t ≤ s ≤ 1 − 1

n, s ≥ η,

(t − 1/n) (1 − 1/n − s)1 − 2/n + α/n − αη − (t − s) , 1

n≤ s ≤ t ≤ 1 − 1

n, s ≥ η.

(2.3)


We note that Hn(t, s) → H(t, s) as n → ∞, where

H (t, s) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

t (1 − s)1 − αη −

αt(η − s

)

1 − αη − (t − s) , 0 ≤ s ≤ t ≤ 1, s ≤ η,

t (1 − s)1 − αη −

αt(η − s

)

1 − αη , 0 ≤ t ≤ s ≤ 1, s ≤ η,

t (1 − s)1 − αη , 0 ≤ t ≤ s ≤ 1, s ≥ η,

t (1 − s)1 − αη − (t − s) , 0 ≤ s ≤ t ≤ 1, s ≥ η,

(2.4)

is the Green’s function corresponding the boundary value problem

−u′′ (t) = z (t) , t ∈ [0, 1] ,

u (0) = 0, u (1) = αu(η) (2.5)

whose integral representation is given by

u (t) =∫1

0H (t, s) z (s)ds. (2.6)

Lemma 2.2 (see [9]). Let 0 < α < 1/η. If z ∈ C[0, 1] and z ≥ 0, then then unique solution u of theproblem (2.5) satisfies

mint∈[η,1]

u (t) ≥ γ‖u‖, (2.7)

where γ = min{αη, α(1 − η)/(1 − αη), η}.

We need the following properties of the Green’s function Hn in the sequel.

Lemma 2.3 (see [11]). The function Hn can be written as

Hn (t, s) = Gn (t, s) +α (t − 1/n)

1 − 2/n + α/n − αηGn

(η, s), (2.8)

where

Gn (t, s) =n

n − 2

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

(s − 1

n

)(1 − 1

n− t),

1n≤ s ≤ t ≤ 1 − 1

n,

(t − 1

n

)(1 − 1

n− s),

1n≤ t ≤ s ≤ 1 − 1

n.

(2.9)


Following the idea in [10], we calculate upper bound for the Green’s function Hn inthe following lemma.

Lemma 2.4. The function Hn satisfies

Hn (t, s) ≤ μn

(s − 1

n

)(1 − 1

n− s), (t, s) ∈

[1n, 1 − 1

n

]×[

1n, 1 − 1

n

], (2.10)

where μn = max{1, α}/(1 − 2/n + α/n − αη).

Proof. For (t, s) ∈ [1/n, 1 − 1/n] × [1/n, 1 − 1/n], we discuss various cases.

Case 1. s ≤ η, s ≤ t; using (2.3), we obtain

Hn (t, s) = s − 1n+ (α − 1)

(t − 1/n) (s − 1/n)1 − 2/n + α/n − αη . (2.11)

If α > 1, the maximum occurs at t = 1 − 1/n, hence

Hn (t, s) ≤ Hn

(1 − 1

n, s

)= α

(s − 1/n)(1 − 1/n − η

)

1 − 2/n + α/n − αη

≤ α(s − 1/n) (1 − 1/n − s)

1 − 2/n + α/n − αη ≤ μn

(s − 1

n

)(1 − 1

n− s),

(2.12)

and if α ≤ 1, the maximum occurs at t = s, hence

Hn (t, s) ≤ Hn (s, s) =(s − 1/n)

(1 − 1/n − s + α

(s − η

))

1 − 2/n + α/n − αη

≤ (s − 1/n) (1 − 1/n − s)1 − 2/n + α/n − αη ≤ μn

(s − 1

n

)(1 − 1

n− s).

(2.13)

Case 2. s ≤ η, s ≥ t; using (2.3), we have

Hn (t, s) =(t − 1/n) (1 − 1/n − s)

1 − 2/n + α/n − αη − α(t − 1/n)

(η − s

)

1 − 2/n + α/n − αη ≤(t − 1/n) (1 − 1/n − s)

1 − 2/n + α/n − αη

≤ (s − 1/n) (1 − 1/n − s)1 − 2/n + α/n − αη ≤ μn

(s − 1

n

)(1 − 1

n− s).

(2.14)

Case 3. s ≥ η, t ≤ s; using (2.3), we have

Hn (t, s) =(t − 1/n) (1 − 1/n − s)

1 − 2/n + α/n − αη ≤ (s − 1/n) (1 − 1/n − s)1 − 2/n + α/n − αη ≤ μn

(s − 1

n

)(1 − 1

n− s).

(2.15)


Case 4. s ≥ η, t ≥ s; using (2.3), we have

Hn (t, s) = s − 1n+(t − 1

n

)α(η − 1/n

)− (s − 1/n)

1 − 2/n + α/n − αη . (2.16)

For α(η − 1/n) > s − 1/n, the maximum occurs at t = 1 − 1/n, hence

Hn (t, s) ≤ Hn

(1 − 1

n, s

)= α

(η − 1/n

)(1 − 1/n − s)

1 − 2/n + α/n − αη ≤ α(s − 1/n) (1 − 1/n − s)

1 − 2/n + α/n − αη

≤ μn

(s − 1

n

)(1 − 1

n− s).

(2.17)

For α(η − 1/n) ≤ s − 1/n, the maximum occurs at t = s, so

Hn (t, s) ≤ Hn (s, s) =(s − 1/n) (1 − 1/n − s)

1 − 2/n + α/n − αη ≤ μn

(s − 1

n

)(1 − 1

n− s). (2.18)

Now, we consider the nonlinear nonsingular system of BVPs

−x′′ (t) = f

(t,max

{x (t) +

1n,

1n

},max

{y (t) +

1n,

1n

}), t ∈

[1n, 1 − 1

n

],

−y′′ (t) = g

(t,max

{x (t) +

1n,

1n

},max

{y (t) +

1n,

1n

}), t ∈

[1n, 1 − 1

n

],

x

(1n

)= 0, x

(1 − 1

n

)= αx

(η),

y

(1n

)= 0, y

(1 − 1

n

)= αy

(η).

(2.19)

We write (2.19) as an equivalent system of integral equations

x (t) =∫1−1/n

1/nHn (t, s) f

(s,max

{x (s) +

1n,

1n

},max

{y (s) +

1n,

1n

})ds,

y (t) =∫1−1/n

1/nHn (t, s) g

(s,max

{x (s) +

1n,

1n

},max

{y (s) +

1n,

1n

})ds.

(2.20)

By a solution of the system (2.19), we mean a solution of the corresponding system of integralequations (2.20). Define a retraction σn : [0, 1] → [1/n, 1−1/n] by σn(t) = max{1/n,min{t, 1−1/n}} and an operator Tn : E × E → P × P by

Tn(x, y)=(An

(x, y), Bn

(x, y))

, (2.21)

where operators An, Bn : E × E → P are defined by


An

(x, y)(t) =

∫1−1/n

1/nHn (σn (t) , s) f

(s,max

{x (s) +

1n,

1n

},max

{y (s) +

1n,

1n

})ds,

Bn

(x, y)(t) =

∫1−1/n

1/nHn (σn (t) , s) g

(s,max

{x (s) +

1n,

1n

},max

{y (s) +

1n,

1n

})ds.

(2.22)

Clearly, if (xn, yn) ∈ E×E is a fixed point of Tn, then (xn, yn) is a solution of the system (2.19).

Lemma 2.5. Assume that (A1)–(A3) holds. Then Tn : P × P → P × P is completely continuous.

Proof. Clearly, for any (x, y) ∈ P × P , An(x, y), Bn(x, y) ∈ P . We show that the operator An :P × P → P is uniformly bounded. Let d > 0 be fixed and consider

D ={(

x, y)∈ P × P :

∥∥(x, y)

∥∥

1 ≤ d}. (2.23)

Choose a constant c ∈ (0, 1] such that c(x + 1/3) ≤ 1, c(y + 1/3) ≤ 1, (x, y) ∈ D. Then, forevery (x, y) ∈ D, using (2.22), Lemma 2.4, (A1) and (A2), we have

An

(x, y)(t) =

∫1−1/n


(s, x (s) +

1n, y (s) +

1n

)ds

=∫1−1/n


(s, c

x (s) + 1/nc

, cy (s) + 1/n

c

)ds

≤(

1c

)β1∫1−1/n


(s, c

(x (s) +

1n

), c

y (s) + 1/nc

)ds

≤(

1c

)β1(

1c

)β2∫1−1/n


(s, c

(x (s) +

1n

), c

(y (s) +

1n

))ds

≤ cα1−β1−β2

∫1−1/n

1/nHn (σn (t) , s)

(x(s) +

1n

)α1

f

(s, 1, c

(y (s) +

1n

))ds

≤ cα1−β1+α2−β2

∫1−1/n

1/nHn (σn (t) , s)

(x(s) +

1n

)α1(y(s) +

1n

)α2

f (s, 1, 1)ds

≤ cα1−β1+α2−β2

∫1−1/n

1/nHn (σn (t) , s)

(1n

)α1(

1n

)α2

f (s, 1, 1)ds

≤ μncα1−β1+α2−β2n−α1−α2

∫1−1/n

1/n

(s − 1

n

)(1 − 1

n− s)f (s, 1, 1)ds


∫1−1/n

1/ns (1 − s) f (s, 1, 1)ds


∫1

0s (1 − s) f (s, 1, 1)ds = aμnc

α1−β1+α2−β2n−α1−α2,

(2.24)


which implies that

‖An

(x, y)‖ ≤ aμnc

α1−β1+α2−β2n−α1−α2, (2.25)

that is, An(D) is uniformly bounded. Similarly, using (2.22), Lemma 2.4, (A1) and (A3), wecan show that Bn(D) is also uniformly bounded. Thus, Tn(D) is uniformly bounded. Now weshow that An(D) is equicontinuous. Define

ω = max{

max(t,x,y)∈[1/n,1−1/n]×[0,d]×[0,d]

f

(t, x +

1n, y +

1n

),

max(t,x,y)∈[1/n,1−1/n]×[0,d]×[0,d]

g

(t, x +

1n, y +

1n

)}.

(2.26)

Let t1, t2 ∈ [0, 1] such that t1 ≤ t2. Since Hn(t, s) is uniformly continuous on [1/n, 1 − 1/n] ×[1/n, 1 − 1/n], for any ε > 0, there exist δ = δ(ε) > 0 such that |t1 − t2| < δ implies

|Hn (σn (t1) , s) −Hn (σn (t2) , s)| <ε

ω (1 − 2/n)for s ∈

[1n, 1 − 1

n

]. (2.27)

For (x, y) ∈ D, using (2.22)–(2.27), we have

∣∣An

(x, y)(t1) −An

(x, y)(t2)∣∣

=

∣∣∣∣∣

∫1−1/n

1/n

(Hn (σn (t1) , s) −Hn (σn (t2) , s) f

(s, x (s) +

1n, y (s) +

1n

))ds

∣∣∣∣∣

≤∫1−1/n

1/n|Hn (σn (t1) , s) −Hn (σn (t2) , s)| f

(s, x (s) +

1n, y (s) +

1n

)ds

≤ ω

∫1−1/n

1/n|Hn (σn (t1) , s) −Hn (σn (t2) , s)|ds

< ωε

ω (1 − 2/n)

∫1−1/n

1/nds =

ε

(1 − 2/n)

(1 − 2

n

)= ε.

(2.28)

Hence,

∣∣An

(x, y)(t1) −An

(x, y)(t2)∣∣ < ε, ∀

(x, y)∈ D, |t1 − t2| < δ, (2.29)

which implies that An(D) is equicontinuous. Similarly, using (2.22)–(2.27), we can showthat Bn(D) is also equicontinuous. Thus, Tn(D) is equicontinuous. By Arzela-Ascoli theorem,Tn(D) is relatively compact. Hence, Tn is a compact operator.


Now we show that Tn is continuous. Let (xm, ym), (x, y) ∈ P × P such that ‖(xm, ym) −(x, y)‖1 → 0 as m → +∞. Then by using (2.22) and Lemma 2.4, we have

∣∣An

(xm, ym

)(t) −An

(x, y)(t)∣∣

=

∣∣∣∣∣

∫1−1/n

1/nHn (σn (t) , s)

(f

(s, xm (s) +

1n, ym (s) +

1n

)− f(s, x (s) +

1n, y (s) +

1n

))ds

∣∣∣∣∣

≤∫1−1/n

1/nHn (σn (t) , s)

∣∣∣∣f(s, xm (s) +

1n, ym (s) +

1n

)− f(s, x (s) +

1n, y (s) +

1n

)∣∣∣∣ds

≤ μn

∫1−1/n

1/n

(s− 1

n

)(1− 1

n−s)∣∣∣∣f(s, xm (s)+

1n, ym (s)+

1n

)−f(s, x (s)+

1n, y (s)+

1n

)∣∣∣∣ds.

(2.30)

Consequently,

∥∥An

(xm, ym

)−An

(x, y)∥∥

≤ μn

∫1−1/n

1/n

(s − 1

n

)(1 − 1

n− s)

×∣∣∣∣f(s, xm (s) +

1n, ym (s) +

1n

)− f(s, x (s) +

1n, y (s) +

1n

)∣∣∣∣ds.

(2.31)

By Lebesgue dominated convergence theorem, it follows that

∥∥An

(xm, ym

)−An

(x, y)∥∥ −→ 0 as m −→ +∞. (2.32)

Similarly, by using (2.22) and Lemma 2.4, we have

∥∥Bn

(xm, ym

)− Bn

(x, y)∥∥ −→ 0 as m −→ +∞. (2.33)

From (2.32) and (2.33), it follows that

∥∥Tn(xm, ym) − Tn(x, y)∥∥

1 −→ 0 as m −→ +∞, (2.34)

that is, Tn : P ×P → P ×P is continuous. Hence, Tn : P ×P → P ×P is completely continuous.

3. Main Results

Proof of Theorem 1.1. Let M = max{μn0 ,max{1, α}/(1 − αη)}. Choose a constant R > 0 suchthat

R ≥ max{(2aM)1/(1−α1−α2), (2bM)1/(1−γ1−γ2)

}. (3.1)


Choose a constant c1 ∈ (0, 1] such that c1 (x(t) + 1/n0) ≤ 1, c1(y(t) + 1/n0) ≤ 1, (x, y) ∈∂ΩR ∩ (P ×P), t ∈ (0, 1). For any (x, y) ∈ ∂ΩR ∩ (P ×P), using (2.22), (3.1), (A1), and (A2), wehave

An

(x, y)(t) =

∫1−1/n


(s, x (s) +

1n, y (s) +

1n

)ds

=∫1−1/n


(s, c1

x (s) + 1/nc1

, c1y (s) + 1/n

c1

)ds

≤(

1c1

)β1∫1−1/n


(s, c1

(x (s) +

1n

), c1

y (s) + 1/nc1

)ds

≤(

1c1

)β1(

1c1

)β2∫1−1/n


(s, c1

(x (s) +

1n

), c1

(y (s) +

1n

))ds

≤ cα1−β1−β2

1

∫1−1/n

1/nHn (σn (t) , s)

(x(s) +

1n

)α1

f

(s, 1, c1

(y (s) +

1n

))ds

≤ cα1−β1+α2−β2

1

∫1−1/n

1/nHn (σn (t) , s)

(x(s) +

1n

)α1(y(s) +

1n

)α2

f (s, 1, 1)ds

≤ cα1−β1+α2−β2

1

∫1−1/n

1/nHn (σn (t) , s) (x(s))

α1(y(s)

)α2f (s, 1, 1)ds

≤ μncα1−β1+α2−β2

1

∫1−1/n

1/n

(s − 1

n

)(1 − 1

n− s)(x(s))α1

(y(s)

)α2f (s, 1, 1)ds


1

∫1−1/n

1/ns (1 − s) (x(s))α1

(y(s)

)α2f (s, 1, 1)ds


1

∫1

0s (1 − s) (x(s))α1

(y(s)

)α2f (s, 1, 1)ds

≤Mcα1−β1+α2−β2

1

∫1

0s (1 − s) (x(s))α1

(y(s)

)α2f (s, 1, 1)ds.

(3.2)

Since,

Mcα1−β1+α2−β2

1

∫1

0s (1 − s) (x(s))α1

(y(s)

)α2f (s, 1, 1)ds

≤M

∫1

0s (1 − s) (x(s))α1

(y(s)

)α2f (s, 1, 1)ds

≤ aMR α1+α2 ≤ R

2,

(3.3)


it follows that

∥∥An

(x, y)∥∥ ≤ R

2=

∥∥(x, y)

∥∥

1

2, ∀

(x, y)∈ ∂ΩR ∩ (P × P) . (3.4)

Similarly, using (2.22), (3.1), (A1), and (A3), we have

∥∥Bn

(x, y)∥∥ ≤

∥∥(x, y)

∥∥

1

2, ∀

(x, y)∈ ∂ΩR ∩ (P × P) . (3.5)

From (3.4), and (3.5), it follows that

∥∥Tn(x, y)∥∥

1 ≤∥∥(x, y)

∥∥1, ∀

(x, y)∈ ∂ΩR ∩ (P × P) . (3.6)

Choose a real constant r ∈ (0, R) such that

r ≤ min{(2aM)1/(1−β1−β2), (2bM)1/(1−ρ1−ρ2)

}. (3.7)

Choose a constant c2 ∈ (0, 1] such that c2(x(t) + 1/n0) ≤ 1, c2(y(t) + 1/n0) ≤ 1, (x, y) ∈∂Ωr ∩ (P × P), t ∈ (0, 1). For any (x, y) ∈ ∂Ωr ∩ (P × P), using (2.22), (3.7), (A1), and (A2), wehave

An

(x, y)(t) =

∫1−1/n


(s, x (s) +

1n, y (s) +

1n

)ds

=∫1−1/n


(s, c2

x (s) + 1/nc2

, c2y (s) + 1/n

c2

)ds

≥(

1c2

)α1∫1−1/n


(s, c2

(x (s) +

1n

), c2

y (s) + 1/nc2

)ds

≥(

1c2

)α1(

1c2

)α2∫1−1/n


(s, c2

(x (s) +

1n

), c2

(y (s) +

1n

))ds

≥ cβ1−α1−α2

2

∫1−1/n

1/nHn (σn (t) , s)

(x(s) +

1n

)β1

f

(s, 1, c2

(y (s) +

1n

))ds

≥ cβ1−α1+β2−α2

2

∫1−1/n

1/nHn (σn (t) , s)

(x(s) +

1n

)β1(y(s) +

1n

)β2

f (s, 1, 1)ds

≥ cβ1−α1+β2−α2

2

∫1−1/n

1/nHn (σn (t) , s) (x(s))

β1(y(s)

)β2f (s, 1, 1)ds ≥ r

2.

(3.8)


We used the fact that

cβ1−α1+β2−α2

2

∫1−1/n

1/nHn (σn (t) , s) (x(s))

β1(y(s)

)β2f (s, 1, 1)ds

≥∫1−1/n

1/nHn (σn (t) , s) (x(s))

β1(y(s)

)β2f (s, 1, 1)ds

≥ aMrβ1+β2 .

(3.9)

Thus,

∥∥An

(x, y)∥∥ ≥

∥∥(x, y)

∥∥

1

2, ∀

(x, y)∈ ∂Ωr ∩ (P × P) . (3.10)

Similarly, using (2.22), (3.7), (A1) and (A3), we have,

∥∥Bn

(x, y)∥∥ ≥

∥∥(x, y)∥∥

1

2, ∀

(x, y)∈ ∂Ωr ∩ (P × P) . (3.11)

From (3.10) and (3.11), it follows that

∥∥Tn(x, y)∥∥

1 ≥∥∥(x, y)

∥∥1, ∀

(x, y)∈ ∂Ωr ∩ (P × P) . (3.12)

Hence by Lemma 2.1, Tn has a fixed point (xn, yn) ∈ (P × P) ∩ (ΩR \Ωr), that is,

xn = An

(xn, yn

), yn = Bn

(xn, yn

). (3.13)

Moreover, by (3.4), (3.5), (3.10) and (3.11), we have

r

2≤ ‖xn‖ ≤

R

2,

r

2≤∥∥yn

∥∥ ≤ R

2.

(3.14)

Since (xn, yn) is a solution of the system (2.19), hence xn and yn are concave on [1/n, 1−1/n].Moreover, maxt∈[1/n,1−1/n]xn(t) = ‖xn‖ and maxt∈[1/n,1−1/n]yn(t) = ‖yn‖. For h ∈ (1/n, 1/2),using result (2.2) and (3.14), we have

rh

2≤xn (t) ≤

R

2, ∀t ∈ [h, 1 − h] ,

rh

2≤yn (t) ≤

R

2, ∀t ∈ [h, 1 − h] ,

(3.15)


which implies that {(xn, yn)} is uniformly bounded on [h, 1−h]. Now we show that {(xn, yn)}is equicontinuous on [h, 1−h]. Choose η ∈ (h, 1−h) and 0 < α < (1−2h)/(η−h) and considerthe integral equation

xn (t) =xn (1 − h) − α xn

(η)− (1 − α)xn (h)

1 − 2h + αh − αη (t − h) + xn (h)

+∫1−h

h

Hh−1 (t, s) f(s, xn (s) +

1n, yn (s) +

1n

)ds, t ∈ [h, 1 − h] .

(3.16)

Using Lemma 2.3, we have

xn (t) =xn (1 − h) − α xn

(η)− (1 − α)xn (h)

1 − 2h + αh − αη (t − h) + xn (h)

+1 − h − t1 − 2h

∫ t

h

(s − h) f(s, xn (s) +

1n, yn (s) +

1n

)ds

+t − h

1 − 2h

∫1−h

t

(1 − h − s) f(s, xn (s) +

1n, yn (s) +

1n

)ds

+α (t − h)

1 − 2h + αh − αη

∫1−h

h

Gh−1(η, s)f

(s, xn (s) +

1n, yn (s) +

1n

)ds, t ∈ [h, 1 − h] .

(3.17)

Differentiating with respect to t, we obtain

x′n (t) =xn (1 − h) − α xn

(η)− (1 − α)xn (h)

1 − 2h + αh − αη

− 11 − 2h

∫ t

h

(s − h) f(s, xn (s) +

1n, yn (s) +

1n

)ds

+1

1 − 2h

∫1−h

t

(1 − h − s) f(s, xn (s) +

1n, yn (s) +

1n

)ds

+α

1 − 2h + αh − αη

∫1−h

h

Gh−1(η, s)f

(s, xn (s) +

1n, yn (s) +

1n

)ds, t ∈ [h, 1 − h] ,

(3.18)

which implies that

∣∣x′n (t)∣∣ ≤ (1 + α)R

1 − 2h + αh − αη +∫1−h

h

f

(s, xn (s) +

1n, yn (s) +

1n

)ds

+α

1 − 2h + αh − αη

∫1−h

h

Gh−1(η, s)f

(s, xn (s) +

1n, yn (s) +

1n

)ds, t ∈ [h, 1 − h] ,

(3.19)


In view of (A2) and (3.15), we have

∣∣x′n (t)

∣∣ ≤ (1 + α)R

1 − 2h + αh − αη + cα1−β1+α2−β2

1

(hr

2

)α1+α2∫1−h

h

f (s, 1, 1)ds

+α

1 − 2h + αh − αηcα1−β1+α2−β2

1

(hr

2

)α1+α2∫1−h

h

Gh−1(η, s)f (s, 1, 1)ds, t ∈ [h, 1 − h] ,

(3.20)

which implies that

‖x′n‖ ≤(1 + α)R

1 − 2h + αh − αη + cα1−β1+α2−β2

1

(hr

2

)α1+α2∫1−h

h

f (s, 1, 1)ds

+α

1 − 2h + αh − αηcα1−β1+α2−β2

1

(hr

2

)α1+α2∫1−h

h

Gh−1(η, s)f (s, 1, 1)ds, t ∈ [h, 1 − h] .

(3.21)

Similarly, consider the integral equation

yn (t) =yn (1 − h) − α yn

(η)− (1 − α)yn (h)

1 − 2h + αh − αη (t − h) + yn (h)

+∫1−h

h

Hh−1 (t, s) g(s, xn (s) +

1n, yn (s) +

1n

)ds, t ∈ [h, 1 − h] ,

(3.22)

using (A3) and (3.15), we can show that

‖y′n‖ ≤(1 + α)R

1 − 2h + αh − αη + cγ1−ρ1+γ2−ρ2

1

(hr

2

)γ1+γ2∫1−h

h

g (s, 1, 1)ds

+α

1 − 2h + αh − αηcγ1−ρ1+γ2−ρ2

1

(hr

2

)γ1+γ2∫1−h

h

Gh−1(η, s)g (s, 1, 1)ds, t ∈ [h, 1 − h] .

(3.23)

In view of (3.21) and (3.23), {(xn, yn)} is equicontinuous on [h, 1 − h]. Hence by Arzela-Ascoli theorem, the sequence {(xn, yn)} has a subsequence {(xnk , ynk)} converging uniformlyon [h, 1 − h] to (x, y) ∈ (P × P) ∩ (ΩR \Ωr). Let us consider the integral equation

xnk (t) =xnk (1 − h) − α xnk

(η)− (1 − α)xnk (h)

1 − 2h + αh − αη (t − h) + xnk (h)

+∫1−h

h

Hh−1 (t, s) f(s, xnk (s) +

1nk

, ynk (s) +1nk

)ds, t ∈ [h, 1 − h] .

(3.24)


Letting nk → ∞, we have

x (t) =x (1 − h) − α x

(η)− (1 − α)x (h)

1 − 2h + αh − αη (t − h) + x (h)

+∫1−h

h

Hh−1 (t, s) f(s, x (s) , y (s)

)ds, t ∈ [h, 1 − h] .

(3.25)

Differentiating twice with respect to t, we have

−x′′ (t) = f(t, x (t) , y (t)

), t ∈ [h, 1 − h] . (3.26)

Letting h → 0, we have

−x′′ (t) = f(t, x (t) , y (t)

), t ∈ (0, 1) . (3.27)

Similarly, consider the integral equation

ynk (t) =ynk (1 − h) − α ynk

(η)− (1 − α)ynk (h)

1 − 2h + αh − αη (t − h) + ynk (h)

+∫1−h

h

Hh−1 (t, s) g(s, xnk (s) +

1nk

, ynk (s) +1nk

)ds, t ∈ [h, 1 − h] ,

(3.28)

we can show that

−y′′ (t) = g(t, x (t) , y (t)

), t ∈ (0, 1) . (3.29)

Now, we show that (x, y) also satisfies the boundary conditions. Since,

x (0) = limnk→∞

x

(1nk

)= lim

nk→∞xnk

(1nk

)= 0,

x (1) = limnk→∞

x

(1 − 1

nk

)= lim

nk→∞xnk

(1 − 1

nk

)= lim

nk→∞αxnk

(η)= αx

(η).

(3.30)

Similarly, we can show that

y (t) = 0, y (1) = αy(η). (3.31)

Equations (3.27)–(3.31) imply that (x, y) is a solution of the system (1.3). Moreover, (x, y) ispositive. In fact, by (3.27) x is concave and by Lemma 2.2

x (1) ≥ mint∈[η,1]

x (t) ≥ γ ‖x‖ > 0, (3.32)


implies that x(t) > 0 for all t ∈ (0, 1). Similarly, y(t) > 0 for all t ∈ (0, 1). The proof ofTheorem 1.1 is complete.

Example 3.1. Let

f(t, x, y

)=

m∑

i=1

n∑

j=1

tpi(1 − t)qj xriysj ,

g(t, x, y

)=

m′∑

k=1

n′∑

l=1

tp′k(1 − t)q

′lxr ′

k ys′l ,

(3.33)

where the real constants pi, qj , ri, sj satisfy pi, qj > −2, ri, sj < 1, i = 1, 2, . . . , m; j = 1, 2, . . . , n,with max1≤i≤mri + max1≤j≤nsj < 1 and the real constants p′k, q

′l, r′k, s

′l satisfy p′k, q

′l > −2, r ′k, s

′l <

1, k = 1, 2, . . . , m′; l = 1, 2, . . . , n′, with max1≤k≤m′r′k +max1≤l≤n′s

′l < 1. Clearly, f and g satisfy the

assumptions (A1)–(A3). Hence, by Theorem 1.1, the system (1.3) has a positive solution.

Acknowledgment

Research of R. A. Khan is supported by HEC, Pakistan, Project 2- 3(50)/PDFP/HEC/2008/1.

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Research ArticlePositive Solutions of Singular Multipoint BoundaryValue Problems for Systems of NonlinearSecond-Order Differential Equations on InfiniteIntervals in Banach Spaces

Xingqiu Zhang

School of Mathematics, Liaocheng University, Liaocheng, Shandong 252059, China

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

Received 27 April 2009; Accepted 12 June 2009


The cone theory together with Monch fixed point theorem and a monotone iterative techniqueis used to investigate the positive solutions for some boundary problems for systems ofnonlinear second-order differential equations with multipoint boundary value conditions oninfinite intervals in Banach spaces. The conditions for the existence of positive solutions areestablished. In addition, an explicit iterative approximation of the solution for the boundary valueproblem is also derived.

Copyright q 2009 Xingqiu 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.

1. Introduction

In recent years, the theory of ordinary differential equations in Banach space has become anew important branch of investigation (see, e.g., [1–4] and references therein). By employinga fixed point theorem due to Sadovskii, Liu [5] investigated the existence of solutions forthe following second-order two-point boundary value problems (BVP for short) on infiniteintervals in a Banach space E:

x′′(t) = f(t, x(t), x′(t)

), t ∈ J,

x(0) = x0, x′(∞) = y∞,(1.1)

where f ∈ C[J×E×E, E], J = [0,+∞), x′(∞) = limt→∞x′(t). On the other hand, the multipoint

boundary value problems arising naturally from applied mathematics and physics have been


studied so extensively in scalar case that there are many excellent results about the existenceof positive solutions (see, i.e., [6–12] and references therein). However, to the best of ourknowledge, only a few authors [5, 13, 14] have studied multipoint boundary value problemsin Banach spaces and results for systems of second-order differential equation are rarely seen.Motivated by above papers, we consider the following singular m-point boundary valueproblem on an infinite interval in a Banach space E

x′′(t) + f(t, x(t), x′(t), y(t), y′(t)

)= 0,

y′′(t) + g(t, x(t), x′(t), y(t), y′(t)

)= 0, t ∈ J+,

x(0) =m−2∑

i=1

αix(ξi), x′(∞) = x∞,

y(0) =m−2∑

i=1

βiy(ξi), y′(∞) = y∞,

(1.2)

where J = [0,∞), J+ = (0,∞), αi, βi ∈ [0,+∞) and ξi ∈ (0,+∞) with 0 < ξ1 < ξ2 < · · · <ξm−2 < +∞, 0 <

∑m−2i=1 αi < 1, 0 <

∑m−2i=1 βi < 1,

∑m−2i=1 αiξi/(1 −

∑m−2i=1 αi) > 1,

∑m−2i=1 βiξi/(1 −∑m−2

i=1 βi) > 1. In this paper, nonlinear terms f and g may be singular at t = 0, x, y = θ, and/orx′, y′ = θ, where θ denotes the zero element of Banach space E. By singularity, we mean that‖f(t, x0, x1, y0, y1)‖ → ∞ as t → 0+ or xi, yi → θ (i = 0, 1).

Very recently, by using Shauder fixed point theorem, Guo [15] obtained the existenceof positive solutions for a class of nth-order nonlinear impulsive singular integro-differentialequations in a Banach space. Motivated by Guo’s work, in this paper, we will use the conetheory and the Monch fixed point theorem combined with a monotone iterative technique toinvestigate the positive solutions of BVP (1.2). The main features of the present paper are asfollows. Firstly, compared with [5], the problem we discussed here is systems of multipointboundary value problem and nonlinear term permits singularity not only at t = 0 but alsoat x, y, x′, y′ = θ. Secondly, compared with [15], the relative compact conditions we usedare weaker. Furthermore, an iterative sequence for the solution under some normal typeconditions is established which makes it very important and convenient in applications.

The rest of the paper is organized as follows. In Section 2, we give some preliminariesand establish several lemmas. The main theorems are formulated and proved in Section 3.Then, in Section 4, an example is worked out to illustrate the main results.

2. Preliminaries and Several Lemmas

Let

FC[J, E] =

{

x ∈ C[J, E] : supt∈J

‖x(t)‖t + 1

<∞}

,

DC1[J, E] =

{

x ∈ C1[J, E] : supt∈J

‖x(t)‖t + 1

<∞, supt∈J‖x′(t)‖ <∞

}

.

(2.1)


Evidently, C1[J, E] ⊂ C[J, E], DC1[J, E] ⊂ FC[J, E]. It is easy to see that FC[J, E] is a Banachspace with norm

‖x‖F = supt∈J

‖x(t)‖t + 1

, (2.2)

and DC1[J, E] is also a Banach space with norm

‖x‖D = max{‖x‖F, ‖x′‖C

}, (2.3)

where

‖x′‖C = supt∈J‖x′(t)‖. (2.4)

Let X = DC1[J, E] ×DC1[J, E] with norm

‖(x, y)‖X = max{‖x‖D, ‖y‖D

}, ∀

(x, y

)∈ X. (2.5)

Then (X, ‖·, ·‖X) is also a Banach space. The basic space using in this paper is (X, ‖·, ·‖X).Let P be a normal cone in E with normal constant N which defines a partial ordering

in E by x ≤ y. If x ≤ y and x /=y, we write x < y. Let P+ = P \ {θ}. So, x ∈ P+ if and only ifx > θ. For details on cone theory, see [4].

In what follows, we always assume that x∞ ≥ x∗0, y∞ ≥ y∗0, x∗0, y

∗0 ∈ P+. Let P0λ = {x ∈

P : x ≥ λx∗0}, P1λ = {y ∈ P : y ≥ λy∗0} (λ > 0). Obviously, P0λ, P1λ ⊂ P+ for any λ > 0. Whenλ = 1, we write P0 = P01, P1 = P11, that is, P0 = {x ∈ P : x ≥ x∗0}, P 1 = {y ∈ P : y ≥ y∗0}.Let P(F) = {x ∈ FC[J, E] : x(t) ≥ θ, ∀t ∈ J}, and P(D) = {x ∈ DC1[J, E] : x(t) ≥ θ, x′(t) ≥θ, ∀t ∈ J}. It is clear, P(F), P(D) are cones in FC[J, E] and DC1[J, E], respectively. A map(x, y) ∈ DC1[J, E]∩C2[J ′+, E] is called a positive solution of BVP (1.2) if (x, y) ∈ P(D)×P(D)and (x(t), y(t)) satisfies (1.2).

Let α, αF, αD, αX denote the Kuratowski measure of noncompactness inE, FC[J, E], DC1[J, E] and X, respectively. For details on the definition and propertiesof the measure of noncompactness, the reader is referred to [1–4]. Let L[J+, J] be all Lebesguemeasurable functions from J+ to J . Denote

D0 =1

1 −∑m−2

i=1 αi

m−2∑

i=1

αiξi, D1 =1

1 −∑m−2

i=1 βi

m−2∑

i=1

βiξi. (2.6)


Let us list some conditions for convenience.

(H1) f, g ∈ C[J+ × P0λ × P0λ × P1λ × P1λ, P] for any λ > 0 and there exist ai, bi, ci ∈ L[J+, J]and hi ∈ C[J+ × J+, J] (i = 0, 1) such that

‖f(t, x0, x1, y0, y1

)‖ ≤ a0(t) + b0(t)h0

(‖x0‖, ‖x1‖, ‖y0‖, ‖y1‖

),

∀t ∈ J+, xi ∈ P0, yi ∈ P1 (i = 0, 1),

‖g(t, x0, x1, y0, y1

)‖ ≤ a1(t) + b1(t)h1

(‖x0‖, ‖x1‖, ‖y0‖, ‖y1‖

),

∀t ∈ J+, xi ∈ P0, yi ∈ P1 (i = 0, 1),

‖f(t, x0, x1, y0, y1

)‖

c0(t)(‖x0‖ + ‖x1‖ + ‖y0‖ + ‖y1‖

) −→ 0,‖g

(t, x0, x1, y0, y1

)‖

c1(t)(‖x0‖ + ‖x1‖ + ‖y0‖ + ‖y1‖

) −→ 0

as xi ∈ P0, yi ∈ P1 (i = 0, 1), ‖x0‖ + ‖x1‖ + ‖y0‖ + ‖y1‖ −→ ∞,

(2.7)

uniformly for t ∈ J+, and

∫∞

0ai(t)dt = a∗i <∞,

∫∞

0bi(t)dt = b∗i <∞,

∫∞

0ci(t)(1 + t)dt = c∗i <∞ (i = 0, 1). (2.8)

(H2) For any t ∈ J+, R > 0 and countable bounded set Vi ⊂ DC1[J, P ∗0R], Wi ⊂DC1[J, P ∗1R] (i = 0, 1), there exist Lij(t), Kij(t) ∈ L[J, J] (i, j = 0, 1) such that

α(f(t, V0(t), V1(t),W0(t),W1(t))

)≤

1∑

i=0

L0i(t)α(Vi(t)) +K0i(t)α(Wi(t)),

α(g(t, V0(t), V1(t),W0(t),W1(t))

)≤

1∑

i=0

L1i(t)α(Vi(t)) +K1i(t)α(Wi(t)),

(2.9)

with

(Di + 1)∫+∞

0[(Li0(s) +Ki0(s))(1 + s) + Li1(s) +Ki1(s)]ds <

12

(i = 0, 1), (2.10)

where P ∗0R = {x ∈ P, x ≥ x∗0, ‖x‖ ≤ R}, P ∗1R = {y ∈ P, y ≥ y∗0, ‖y‖ ≤ R}.(H3) t ∈ J+, x

∗0 ≤ xi ≤ xi, y

∗0 ≤ yi ≤ yi (i = 0, 1) imply

f(t, x0, x1, y0, y1

)≤ f

(t, x0, x1, y0, y1

),

g(t, x0, x1, y0, y1

)≤ g

(t, x0, x1, y0, y1

).

(2.11)

In what follows, we write Q1 = {x ∈ DC1[J, P] : x(i)(t) ≥ x∗0, ∀t ∈ J, i = 0, 1}, Q2 = {y ∈DC1[J, P] : y(i)(t) ≥ y∗0, ∀t ∈ J, i = 0, 1}, and Q = Q1 ×Q2. Evidently, Q1, Q2, and Q are closedconvex sets in DC1[J, E] and X, respectively.


We will reduce BVP (1.2) to a system of integral equations in E. To this end, we firstconsider operator A defined by

A(x, y

)(t) =

(A1

(x, y

)(t), A2

(x, y

)(t)

), (2.12)

where

A1(x, y

)(t) =

1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds + tx∞,

(2.13)

A2(x, y

)(t) =

1

1 −∑m−2

i=1 βi

[(m−2∑

i=1

βiξi

)

y∞ +m−2∑

i=1

βi

∫ ξi

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds + ty∞.

(2.14)

Lemma 2.1. If condition (H1) is satisfied, then operator A defined by (2.12) is a continuous operatorfrom Q into Q.

Proof. Let

ε0 = min

⎧⎨

⎩1

8c∗0(

1 +∑m−2

i=1 αiξm−2/(

1 −∑m−2

i=1 αi

)) ,1

8c∗1(

1 +∑m−2

i=1 βiξm−2/(

1 −∑m−2

i=1 βi))

⎫⎬

⎭,

(2.15)

r = min

{‖x∗0‖N

,‖y∗0‖N

}

> 0. (2.16)

By virtue of condition (H1), there exists an R > r such that

∥∥f(t, x0, x1, y0, y1

)∥∥ ≤ ε0c0(t)(‖x0‖ + ‖x1‖ +

∥∥y0∥∥ +

∥∥y1∥∥), ∀t ∈ J+, xi ∈ P0, yi ∈ P1 (i = 0, 1),

‖x0‖ + ‖x1‖ +∥∥y0

∥∥ +∥∥y1

∥∥ > R,∥∥f

(t, x0, x1, y0, y1

)∥∥ ≤ a0(t) +M0b0(t), ∀t ∈ J+, xi ∈ P0, yi ∈ P1 (i = 0, 1),

‖x0‖ + ‖x1‖ +∥∥y0

∥∥ +∥∥y1

∥∥ ≤ R,

(2.17)


where

M0 = max{h0(u0, u1, v0, v1) : r ≤ ui, vi ≤ R (i = 0, 1)}. (2.18)

Hence

‖f(t, x0, x1, y0, y1

)‖ ≤ ε0c0(t)

(‖x0‖ + ‖x1‖ + ‖y0‖ + ‖y1‖

)+ a0(t) +M0b0(t),

∀t ∈ J+, xi ∈ P0, yi ∈ P1 (i = 0, 1).(2.19)

Let (x, y) ∈ Q, we have, by (2.19)

‖f(t, x(t), x′(t), y(t), y′(t)

)‖

≤ ε0c0(t)(1 + t)(‖x(t)‖

t + 1+‖x′(t)‖t + 1

+‖y(t)‖t + 1

+‖y′(t)‖t + 1

)+ a0(t) +M0b0(t)

≤ ε0c0(t)(1 + t)(‖x‖F + ‖x′‖C + ‖y‖F + ‖y′‖C

)+ a0(t) +M0b0(t)

≤ 2ε0c0(t)(1 + t)(‖x‖D + ‖y‖D

)+ a0(t) +M0b0(t)

≤ 4ε0c0(t)(1 + t)‖(x, y)‖X + a0(t) +M0b0(t), ∀t ∈ J+,

(2.20)

which together with condition (H2) implies the convergence of the infinite integral

∫∞

0‖f

(s, x(s), x′(s), y(s), y′(s)

)‖ds. (2.21)

Thus, we have

∥∥∥∥∥

∫ t

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

∥∥∥∥∥≤∫ t

0

∫+∞

s

‖f(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dτ ds

≤∫+∞

0

∫ t

0‖f

(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dsdτ

≤ t

∫∞

0‖f

(s, x(s), x′(s), y(s), y′(s)

)‖ds, ∀t ∈ J+,

(2.22)


which together with (2.13) and (H1) implies that

‖A1(x, y

)(t)‖ ≤

∫ t

0

∫+∞

s

‖f(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dτ ds + t‖x∞‖ +

∑m−2i=1 αiξi

1 −∑m−2

i=1 αi

‖x∞‖

+1

1 −∑m−2

i=1 αi

m−2∑

i=1

αi

∫ ξm−2

0

∫+∞

s

‖f(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dτ ds

≤ t

(∫+∞

0‖f

(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dτ + ‖x∞‖

)+

∑m−2i=1 αiξi

1 −∑m−2

i=1 αi

‖x∞‖

+1

1 −∑m−2

i=1 αi

m−2∑

i=1

αiξm−2

(∫+∞

0‖f

(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dτ

).

(2.23)

Therefore, by (2.15) and (2.20), we get

‖A1(x, y

)(t)‖

1 + t≤∫+∞

0‖f

(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dτ + ‖x∞‖ +

∑m−2i=1 αiξi

1 −∑m−2

i=1 αi

‖x∞‖

+1

1 −∑m−2

i=1 αi

m−2∑

i=1

αiξm−2

(∫+∞

0‖f

(τ, x(τ), x′(τ), y(τ), y′(τ)

)‖dτ

)

≤(

1 +1

1 −∑m−2

i=1 αi

m−2∑

i=1

αiξm−2

)[4ε0c

∗0‖(x, y

)‖X + a∗0 +Mb∗0

]

+

(

1 +∑m−2

i=1 αiξi

1 −∑m−2

i=1 αi

)

‖x∞‖

≤ 12‖(x, y)‖X +

(

1 +1

1 −∑m−2

i=1 αi

m−2∑

i=1

αiξm−2

)(a∗0 +Mb∗0

)

+

(

1 +∑m−2

i=1 αiξi

1 −∑m−2

i=1 αi

)

‖x∞‖.

(2.24)

Differentiating (2.13), we obtain

A′1(x, y

)(t) =

∫+∞

t

f(s, x(s), x′(s), y(s), y′(s)

)ds + x∞. (2.25)


Hence,

∥∥A′1

(x, y

)(t)

∥∥ ≤

∫+∞

0‖f

(s, x(s), x′(s), y(s), y′(s)

)‖ds + ‖x∞‖

≤ 4ε0c∗0‖(x, y)‖X + a∗0 +M0b

∗0 + ‖x∞‖

≤ 12‖(x, y)‖X + a∗0 +M0b

∗0 + ‖x∞‖, ∀t ∈ J.

(2.26)

It follows from (2.24) and (2.25) that

‖A1(x, y)‖D ≤12‖(x, y)‖X +

(

1 +∑m−2

i=1 αiξm−2

1 −∑m−2

i=1 αi

)(a∗0 +M0b

∗0)+

(

1 +∑m−2

i=1 αiξi

1 −∑m−2

i=1 αi

)

‖x∞‖.

(2.27)

So, A1(x, y) ∈ DC1[J, E]. On the other hand, it can be easily seen that

A1(x, y

)(t) ≥

∑m−2i=1 αiξi

1 −∑m−2

i=1 αi

x∞ ≥ x∞ ≥ x∗0, ∀t ∈ J,

A′1(x, y

)(t) ≥ x∞ ≥ x∗0, ∀t ∈ J.

(2.28)

So, A1(x, y) ∈ Q1. In the same way, we can easily get that

‖A2(x, y)‖D ≤12‖(x, y)‖X +

(

1 +∑m−2

i=1 βiξm−2

1 −∑m−2

i=1 βi

)(a∗1 +M1b

∗1

)+

(

1 +∑m−2

i=1 βiξi

1 −∑m−2

i=1 βi

)

‖y∞‖,

A2(x, y

)(t) ≥

∑m−2i=1 βiξi

1 −∑m−2

i=1 βiy∞ ≥ y∞ ≥ y∗0, ∀t ∈ J,

A′2(x, y

)(t) ≥ y∞ ≥ y∗0, ∀t ∈ J,

(2.29)

where M1 = max{h1(u0, u1, v0, v1) : r ≤ ui, vi ≤ R (i = 0, 1)}. Thus, A maps Q into Q and weget

‖A(x, y)‖X ≤12‖(x, y)‖X + γ, (2.30)

where

γ = max

{(

1 +∑m−2

i=1 αiξm−2

1 −∑m−2

i=1 αi

)(a∗0 +Mb∗0

)+

(

1 +∑m−2

i=1 αiξi

1 −∑m−2

i=1 αi

)

‖x∞‖,

(

1 +∑m−2

i=1 βiξm−2

1 −∑m−2

i=1 βi

)(a∗1 +M1b

∗1

)+

(

1 +∑m−2

i=1 βiξi

1 −∑m−2

i=1 βi

)∥∥y∞

∥∥}

.

(2.31)


Finally, we show that A is continuous. Let (xm, ym), (x, y) ∈ Q, ‖(xm, ym)− (x, y)‖X →0 (m → ∞). Then {(xm, ym)} is a bounded subset of Q. Thus, there exists r > 0 such thatsupm‖(xm, ym)‖X < r for m ≥ 1 and ‖(x, y)‖X ≤ r + 1. Similar to (2.24) and (2.26), it is easy tohave

‖A1(xm, ym) −A1(x, y)‖X

≤∫+∞

0

∥∥∥f

(s, xm(s), x′m(s), ym(s), y′m(s)

)− f

(s, x(s), x′(s), y(s), y′(s)

)∥∥∥ds

+∑m−2

i=1 αiξm−2

1 −∑m−2

i=1 αi

∫+∞

0

∥∥f

(s, xm(s), x′m(s), ym(s), y′m(s)

)− f

(s, x(s), x′(s), y(s), y′(s)

)∥∥ds.

(2.32)

It is clear,

f(t, xm(t), x′m(t), ym(t), y′m(t)

)−→ f

(t, x(t), x′(t), y(t), y′(t)

)as m −→ ∞, ∀t ∈ J+, (2.33)

and by (2.20),

∥∥f(t, xm(t), x′m(t), ym(t), y′m(t)

)− f

(t, x(t), x′(t), y(t), y′(t)

)∥∥

≤ 8ε0c0(t)(1 + t)r + 2a0(t) + 2M0b0(t)

= σ0(t) ∈ L[J, J], m = 1, 2, 3, . . . , ∀t ∈ J+.

(2.34)

It follows from (2.33) and (2.34) and the dominated convergence theorem that

limm→∞

∫∞

0

∥∥f(s, xm(s), x′m(s), ym(s), y′m(s)

)− f

(s, x(s), x′(s), y(s), y′(s)

)∥∥ds = 0. (2.35)

It follows from (2.32) and (2.35) that ‖A1(xm, ym) −A1(x, y)‖D → 0 as m → ∞. By the samemethod, we have ‖A2(xm, ym)−A2(x, y)‖D → 0 as m → ∞. Therefore, the continuity of A isproved.

Lemma 2.2. If condition (H1) is satisfied, then (x, y) ∈ Q ∩ (C2[J+, E] × C2[J+, E]) is a solution ofBVP (1.2) if and only if (x, y) ∈ Q is a fixed point of operator A.

Proof. Suppose that x ∈ Q ∩ (C2[J+, E] × C2[J+, E]) is a solution of BVP (1.2). For t ∈ J,integrating (1.2) from t to +∞, we have

x′(t) = x∞ +∫+∞

t

f(s, x(s), x′(s), y(s), y′(s)

)ds,

y′(t) = y∞ +∫+∞

t

g(s, x(s), x′(s), y(s), y′(s)

)ds.

(2.36)


Integrating (2.36) from 0 to t, we get

x(t) = x(0) + tx∞ +∫ t

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds, (2.37)

y(t) = y(0) + ty∞ +∫ t

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds. (2.38)

Thus, we obtain

x(ξi) = x(0) + ξix∞ +∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds,

y(ξi) = y(0) + ξiy∞ +∫ ξi

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds,

(2.39)

which together with the boundary value conditions imply that

x(0) =1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξ

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

, (2.40)

y(0) =1

1 −∑m−2

i=1 βi

[(m−2∑

i=1

βiξi

)

y∞ +m−2∑

i=1

βi

∫ ξ

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

. (2.41)

Substituting (2.40) and (2.41) into (2.37) and (2.38), respectively, we have

x(t) =1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds + tx∞,

y(t) =1

1 −∑m−2

i=1 βi

[(m−2∑

i=1

βiξi

)

y∞ +m−2∑

i=1

βi

∫ ξi

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds + ty∞.

(2.42)

It follows from Lemma 2.1 that the integral∫ t

0

∫+∞s f(τ, x(τ), x′(τ), y(τ), y′(τ))dτ ds and the

integral∫ t

0

∫+∞s g(τ, x(τ), x′(τ), y(τ), y′(τ))dτ ds are convergent. Thus, (x, y) is a fixed point of

operator A.


Conversely, if (x, y) is fixed point of operator A, then direct differentiation gives theproof.

Lemma 2.3. Let (H1) be satisfied, V ⊂ Q is a bounded set. Then (AiV )(t)/(1 + t) and (A′iV )(t) areequicontinuous on any finite subinterval of J and for any ε > 0, there existsNi > 0 such that

∥∥∥∥∥Ai

(x, y

)(t1)

1 + t1−Ai

(x, y

)(t2)

1 + t2

∥∥∥∥∥< ε,

∥∥A′i

(x, y

)(t1) −A′i

(x, y

)(t2)

∥∥ < ε (2.43)

uniformly with respect to (x, y) ∈ V as t1, t2 ≥Ni (i = 1, 2).

Proof. We only give the proof for operator A1, the proof for operator A2 can be given in asimilar way. By (2.13), we have

A1(x, y

)(t) =

1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds + tx∞

=1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+ tx∞ + t

∫+∞

t

f(s, x(s), x′(s), y(s), y′(s)

)ds +

∫ t

0sf

(s, x(s), x′(s), y(s), y′(s)

)ds.

(2.44)

For (x, y) ∈ V, t2 > t1, we obtain by (2.44)

∥∥∥∥∥A1

(x, y

)(t1)

1 + t1−A1

(x, y

)(t2)

1 + t2

∥∥∥∥∥

≤∣∣∣∣

11 + t1

− 11 + t2

∣∣∣∣ ·1

1 −∑m−2

i=1 αi

×[(

m−2∑

i=1

αiξi

)

‖x∞‖ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∣∣∣∣

t11 + t1

− t21 + t2

∣∣∣∣ · ‖x∞‖

+

∥∥∥∥∥t1

1 + t1

∫+∞

t1

f(s, x(s), x′(s), y(s), y′(s)

)ds − t2

1 + t2

∫+∞

t2

f(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥


+

∥∥∥∥∥

∫ t1

0

s

1 + t1f(s, x(s), x′(s), y(s), y′(s)

)ds −

∫ t2

0

s

1 + t2f(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥

≤∣∣∣∣

11 + t1

− 11 + t2

∣∣∣∣ ·

1

1 −∑m−2

i=1 αi

×[(

m−2∑

i=1

αiξi

)

‖x∞‖ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∣∣∣∣

t11 + t1

− t21 + t2

∣∣∣∣ · ‖x∞‖ +

∣∣∣∣

t11 + t1

− t21 + t2

∣∣∣∣ ·

∥∥∥∥

∫+∞

0f(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥

+∣∣∣∣

t11 + t1

− t21 + t2

∣∣∣∣ ·

∥∥∥∥∥

∫ t1

0f(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥

+t2

1 + t2

∥∥∥∥∥

∫ t2

t1

f(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥

+∣∣∣∣

11 + t1

− 11 + t2

∣∣∣∣ ·∥∥∥∥∥

∫ t1

0sf

(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥

+

∥∥∥∥∥

∫ t2

t1

sf(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥.

(2.45)

Then, it is easy to see by (2.45) and (H1) that {A1V (t)/(1+ t)} is equicontinuous on any finitesubinterval of J .

Since V ⊂ Q is bounded, there exists r > 0 such that for any (x, y) ∈ V, ‖(x, y)‖X ≤ r.By (2.25), we get

∥∥A′1(x, y

)(t1) −A′1

(x, y

)(t2)

∥∥ =

∥∥∥∥∥

∫ t2

t1

f(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥

≤∫ t2

t1

[4ε0rc(s)(1 + s) + a0(s) +M0b0(s)]ds.

(2.46)

It follows from (2.46) and (H1) and the absolute continuity of Lebesgue integral that {A′1V (t)}is equicontinuous on any finite subinterval of J .

In the following, we are in position to show that for any ε > 0, there exists N1 > 0 suchthat

∥∥∥∥∥A1

(x, y

)(t1)

1 + t1−A1

(x, y

)(t2)

1 + t2

∥∥∥∥∥< ε,

∥∥A′1(x, y

)(t1) −A′1

(x, y

)(t2)

∥∥ < ε (2.47)

uniformly with respect to x ∈ V as t1, t2 ≥N.


Combining with (2.45), we need only to show that for any ε > 0, there existssufficiently large N > 0 such that

∥∥∥∥∥

∫ t1

0

s

1 + t1f(s, x(s), x′(s), y(s), y′(s)

)ds −

∫ t2

0

s

1 + t2f(s, x(s), x′(s), y(s), y′(s)

)ds

∥∥∥∥∥< ε (2.48)

for all x ∈ V as t1, t2 ≥ N. The rest part of the proof is very similar to Lemma 2.3 in [5], weomit the details.

Lemma 2.4. Let V be a bounded set in DC1[J, E] ×DC1[J, E]. Assume that (H1) holds. Then

αD(AiV ) = max

{

supt∈J

α

((AiV )(t)

1 + t

), sup

t∈Jα((AiV )′(t)

)}

. (2.49)

Proof. The proof is similar to that of Lemma 2.4 in [5], we omit it.

Lemma 2.5 (see [1, 2]). Monch Fixed-Point Theorem. Let Q be a closed convex set of E and u ∈ Q.Assume that the continuous operator F : Q → Q has the following property: V ⊂ Q countable,V ⊂ co({u} ∪ F(V ))⇒ V is relatively compact. Then F has a fixed point in Q.

Lemma 2.6. If (H3) is satisfied, then for x, y ∈ Q, x(i) ≤ y(i), t ∈ J (i = 0, 1) imply that (Ax)(i) ≤(Ay)(i), t ∈ J (i = 0, 1).

Proof. It is easy to see that this lemma follows from (2.13), (2.25), and condition (H3). Theproof is obvious.

Lemma 2.7 (see [16]). Let E and F are bounded sets in E, then

α(D × F) = max{α(D), α(F)}, (2.50)

where α and α denote the Kuratowski measure of noncompactness in E × E and E, respectively.

Lemma 2.8 (see [16]). Let P be normal (fully regular) in E, P = P × P, then P is normal (fullyregular) in E × E.

3. Main Results

Theorem 3.1. If conditions (H1) and (H2) are satisfied, then BVP (1.2) has a positive solution(x, y) ∈ (DC1[J, E] ∩C2[J ′+, E]) × (DC1[J, E] ∩C2[J ′+, E]) satisfying (x)

(i)(t) ≥ x∗0, (y)(i)(t) ≥ y∗0

for t ∈ J (i = 0, 1).

Proof. By Lemma 2.1, operator A defined by (2.13) is a continuous operator from Q into Q,and, by Lemma 2.2, we need only to show that A has a fixed point (x, y) in Q. Choose R > 2γand let Q∗ = {(x, y) ∈ Q : ‖(x, y)‖X ≤ R}. Obviously, Q∗ is a bounded closed convex set inspace DC1[J, E]×DC1[J, E]. It is easy to see that Q∗ is not empty since ((1+ t)x∞, (1+ t)y∞) ∈Q∗. It follows from (2.27) and (3.6) that (x, y) ∈ Q∗ implies A(x, y) ∈ Q∗, that is, A maps Q∗


into Q∗. Let V = {(xm, ym) : m = 1, 2, . . .} ⊂ Q∗ satisfying V ⊂ co{{(u0, v0)} ∪ AV } for some(u0, v0) ∈ Q∗. Then ‖(xm, ym)‖X ≤ R. We have, by (2.13) and (2.25),

A1(xm, ym

)(t)

=1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, xm(τ), x′m(τ), ym(τ), y′m(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

f(τ, xm(τ), x′m(τ), ym(τ), y′m(τ)

)dτ ds + tx∞,

A′1(xm, ym

)(t) =

∫+∞

t

f(s, xm(s), x′m(s), ym(s), y′m(s)

)ds + x∞.

(3.1)

By Lemma 2.4, we have

αD(A1V ) = max

{

supt∈J

α((A1V )′(t)

), sup

t∈Jα

((A1V )(t)

1 + t

)}

, (3.2)

where A1V (t) = {A1(xm, ym)(t) : m = 1, 2, 3, . . .}, and (A1V )′(t) = {A′1(xm, ym)(t) : m =1, 2, 3, . . .}.

By (2.21), we know that the infinite integral∫+∞

0 ‖f(t, x(t), x′(t), y(t), y′(t))‖dt is

convergent uniformly for m = 1, 2, 3, . . . . So, for any ε > 0, we can choose a sufficiently largeT > 0 such that

∫+∞

T

‖f(t, x(t), x′(t), y(t), y′(t)

)‖dt < ε. (3.3)

Then, by [1, Theorem 1.2.3], (2.44), (3.1), (3.3), (H2), and Lemma 2.7, we obtain

α

((A1V )(t)

1 + t

)≤ 2

D0

1 + t

∫T

0α({


):(xm, ym

)∈ V

})ds + 2ε

+ 2∫T

0

t

1 + tα({


):(xm, ym

)∈ V

})ds + 2ε

≤ 2D0

∫T

0α({


):(xm, ym

)∈ V

})ds

+ 2∫T

0α({


):(xm, ym

)∈ V

})ds + 4ε

≤ (2D0 + 2)∫+∞

0α({


):(xm, ym

)∈ V

})ds + 4ε

≤ (2D0 + 2)αX(V )∫+∞

0(L00(s) +K00(s))(1 + s) + (L01(s) +K01(s))ds + 4ε,


α((A′1V

)(t)

)≤ 2

∫+∞

0α({


):(xm, ym

)∈ V

})ds + 2ε

≤ 2αX(V )∫+∞

0(L00(s) +K00(s))(1 + s) + (L01(s) +K01(s))ds + 2ε.

(3.4)

It follows from (3.2) and (3.4) that

αD(A1V ) ≤ (2D0 + 2)αX(V )∫+∞

0(L00(s) +K00(s))(1 + s) + (L01(s) +K01(s))ds. (3.5)

In the same way, we get

αD(A2V ) ≤ (2D1 + 2)αX(V )∫+∞

0(L10(s) +K10(s))(1 + s) + (L11(s) +K11(s))ds. (3.6)

On the other hand, αX(V ) ≤ αX{co({u} ∪ (AV ))} = αX(AV ). Then, (3.5), (3.6), (H2), andLemma 2.7 imply αX(V ) = 0, that is, V is relatively compact in DC1[J, E] × DC1[J, E].Hence, the Monch fixed point theorem guarantees that A has a fixed point (x, y) in Q∗. Thus,Theorem 3.1 is proved.

Theorem 3.2. Let cone P be normal and conditions (H1)–(H3) be satisfied. Then BVP (1.2) has apositive solution (x, y) ∈ Q ∩ (C2[J ′+, E] × C2[J ′+, E]) which is minimal in the sense that u(i)(t) ≥x(i)(t), v(i)(t) ≥ y(i)(t), t ∈ J (i = 0, 1) for any positive solution (u, v) ∈ Q∩ (C2[J ′+, E]×C2[J ′+, E])of BVP (1.2). Moreover, ‖((x, y))‖X ≤ 2γ+‖(u0, v0)‖X, and there exists a monotone iterative sequence{(un(t), vn(t))} such that u(i)

n (t) → x(i)(t), v(i)n (t) → y(i)(t) as n → ∞ (i = 0, 1) uniformly on J

and u′′n(t) → x′′(t), v′′n(t) → y′′(t) as n → ∞ for any t ∈ J+, where

u0(t) =1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x∗0, x

∗0, y

∗0, y

∗0)dτ ds

]

+∫ t

0

∫+∞

s

f(τ, x∗0, x

∗0, y

∗0, y

∗0)dτ ds + tx∞,

(3.7)

v0(t) =1

1 −∑m−2

i=1 βi

[(m−2∑

i=1

βiξi

)

y∞ +m−2∑

i=1

βi

∫ ξi

0

∫+∞

s

g(τ, x∗0, x

∗0, y

∗0, y

∗0)dτ ds

]

+∫ t

0

∫+∞

s

g(τ, x∗0, x

∗0, y

∗0, y

∗0)dτ ds + ty∞,

(3.8)


un(t) =1

1 −∑m−2

i=1 αi

×[(

m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, un−1(τ), u′n−1(τ), vn−1(τ), v′n−1(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

f(τ, un−1(τ), u′n−1(τ), vn−1(τ), v′n−1(τ)

)dτ ds + tx∞, ∀t ∈ J (n = 1, 2, 3, . . .),

(3.9)

vn(t) =1

1 −∑m−2

i=1 βi

×[(

m−2∑

i=1

βiξi

)

y∞ +m−2∑

i=1

βi

∫ ξi

0

∫+∞

s

g(τ, un−1(τ), u′n−1(τ), vn−1(τ), v′n−1(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

g(τ, un−1(τ), u′n−1(τ), vn−1(τ), v′n−1(τ)

)dτ ds + ty∞, ∀t ∈ J (n = 1, 2, 3, . . .).

(3.10)

Proof. From (3.7), one can see that (u0, v0) ∈ C[J, E] × C[J, E] and

u′0(t) =∫+∞

t

f(s, x∗0, x

∗0, y

∗0, y

∗0)ds + x∞. (3.11)

By (3.7) and (3.11), we have that u(i)0 ≥ x∞ ≥ x∗0 (i = 0, 1) and

‖u0(t)‖ ≤∫ t

0

∫+∞

s

‖f(τ, x∗0, x

∗0, y

∗0, y

∗0)‖dτ ds + t‖x∞‖ +

∑m−2i=1 αiξi

1 −∑m−2

i=1 αi

‖x∞‖

+1

1 −∑m−2

i=1 αi

m−2∑

i=1

αi

∫ ξm−2

0

∫+∞

s

‖f(τ, x∗0, x

∗0, y

∗0, y

∗0)‖dτ ds

≤ t

(∫+∞

0‖f

(τ, x∗0, x

∗0, y

∗0, y

∗0)‖dτ + ‖x∞‖

)+

∑m−2i=1 αiξi

1 −∑m−2

i=1 αi

‖x∞‖

+1

1 −∑m−2

i=1 αi

m−2∑

i=1

αiξm−2

(∫+∞

0‖f

(τ, x∗0, x

∗0, y

∗0, y

∗0)‖dτ

)

≤ t

[∫+∞

0a0(s) + b0(s)h0

(‖x∗0‖, ‖x∗0‖, ‖y∗0‖, ‖y∗0‖

)ds + ‖x∞‖

]+

∑m−2i=1 αiξi

1 −∑m−2

i=1 αi

‖x∞‖

+1

1 −∑m−2

i=1 αi

m−2∑

i=1

αiξm−2

(∫+∞

0a0(s) + b0(s)h0

(‖x∗0‖, ‖x∗0‖, ‖y∗0‖, ‖y∗0‖

)ds

),


‖u′0(t)‖ ≤∫+∞

t

‖f(τ, x∗0, x

∗0, y

∗0, y

∗0)‖dτ + ‖x∞‖

≤∫+∞

0a0(s) + b0(s)h0

(‖x∗0‖, ‖x∗0‖, ‖y∗0‖, ‖y∗0‖

)ds + ‖x∞‖,

(3.12)

which imply that ‖u0‖D < ∞. Similarly, we have ‖v0‖D < ∞. Thus, (u0, v0) ∈ DC1[J, E] ×DC1[J, E]. It follows from (2.13) and (3.9) that

(un, vn)(t) = A(un−1, vn−1)(t), ∀t ∈ J, n = 1, 2, 3, . . . . (3.13)

By Lemma 2.1, we get (un, vn) ∈ Q and

‖(un, vn)‖X = ‖A(un−1, vn−1)‖X ≤12‖(un−1, vn−1)‖X + γ. (3.14)

By Lemma 2.6 and (3.13), we have

(x∗0, y

∗0)≤(u(i)0 (t), v(i)

0 (t))≤(u(i)1 (t), v(i)

1 (t))≤ · · · ≤

(u(i)n (t), v(i)

n (t))≤ · · · , ∀t ∈ J (i = 0, 1).

(3.15)

It follows from (3.14), by induction, that

‖(un, vn)‖X ≤ γ +(

12

)γ + · · · +

(12

)n−1

γ +(

12

)n

‖(u0, v0)‖X

≤γ(1 − (1/2)n

)

1 − 1/2+ ‖(u0, v0)‖X

≤ 2γ + ‖(u0, v0)‖X (n = 1, 2, 3, . . .).

(3.16)

Let K = {(x, y) ∈ Q : ‖(x, y)‖X ≤ 2γ + ‖(u0, v0)‖X}. Then, K is a bounded closed convexset in space DC1[J, E] × DC1[J, E] and operator A maps K into K. Clearly, K is not emptysince (u0, v0) ∈ K. Let W = {(un, vn) : n = 0, 1, 2, . . .}, AW = {A(un, vn) : n = 0, 1, 2, . . .}.Obviously, W ⊂ K and W = {(u0, v0)}∪A(W). Similar to above proof of Theorem 3.1, we canobtain αX(AW) = 0, that is, W is relatively compact in DC1[J, E] ×DC1[J, E]. So, there existsan (x, y) ∈ DC1[J, E] × DC1[J, E] and a subsequence {(unj , vnj ) : j = 1, 2, 3, . . .} ⊂ W suchthat {(unj , vnj )(t) : j = 1, 2, 3, . . .} converges to (x(i)(t), y(i)(t)) uniformly on J (i = 0, 1). Since

that P is normal and {(u(i)n (t), v(i)

n (t)) : n = 1, 2, 3, . . .} is nondecreasing, it is easy to see thatthe entire sequence {(u(i)

n (t), v(i)n (t)) : n = 1, 2, 3, . . .} converges to (x(i)(t), y(i)(t)) uniformly on

J (i = 0, 1). Since (un, vn) ∈ K and K are closed convex sets in space DC1[J, E] × DC1[J, E],we have (x, y) ∈ K. It is clear,

f(s, un(s), u′n(s), vn(s), v′n(s)

)−→ f

(s, x(s), x′(s), y(s), y′(s)

), as n −→ ∞, ∀s ∈ J+. (3.17)


By (H1) and (3.16), we have

‖f(s, un(s), u′n(s), vn(s), v′n(s)

)− f

(s, x(s), x′(s), y(s), y′(s)

)‖

≤ 8ε0c(s)(1 + s)‖(un, vn)‖X + 2a0(s) + 2M0b0(s)

≤ 8ε0c(s)(1 + s)(2γ + ‖(u0, v0)‖X

)+ 2a0(s) + 2M0b0(s).

(3.18)

Noticing (3.17) and (3.18) and taking limit as n → ∞ in (3.9), we obtain

x(t) =1

1 −∑m−2

i=1 αi

[(m−2∑

i=1

αiξi

)

x∞ +m−2∑

i=1

αi

∫ ξi

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

f(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds + tx∞.

(3.19)

In the same way, taking limit as n → ∞ in (3.10), we get

y(t) =1

1 −∑m−2

i=1 βi

[(m−2∑

i=1

βiξi

)

y∞ +m−2∑

i=1

βi

∫ ξi

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds

]

+∫ t

0

∫+∞

s

g(τ, x(τ), x′(τ), y(τ), y′(τ)

)dτ ds + ty∞,

(3.20)

which together with (3.19) and Lemma 2.2 implies that (x, y) ∈ K ∩C2[J+, E] ×C2[J+, E] and(x(t), y(t)) is a positive solution of BVP (1.2). Differentiating (3.9) twice, we get

u′′n(t) = −f(t, un−1(t), u′n−1(t), vn−1(t), v′n−1(t)

), ∀t ∈ J ′+, n = 1, 2, 3, . . . . (3.21)

Hence, by (3.17), we obtain

limn→∞

u′′n(t) = −f(t, x(t), x′(t), y(t), y′(t)

)= x′′(t), ∀t ∈ J ′+. (3.22)

Similarly, we have

limn→∞

v′′n(t) = −g(t, x(t), x′(t), y(t), y′(t)

)= y′′(t), ∀t ∈ J ′+. (3.23)

Let (p(t), q(t)) be any positive solution of BVP (1.2). By Lemma 2.2, we have (p, q) ∈ Qand (p(t), q(t)) = (A(p, q))(t), for t ∈ J. It is clear that p(i)(t) ≥ x∗0 > θ, q(i)(t) ≥ y∗0 > θ for any


t ∈ J (i = 0, 1). So, by Lemma 2.6, we have p(i)(t) ≥ u(i)0 (t), q(i)(t) ≥ v

(i)0 (t) for any t ∈ J (i =

0, 1). Assume that p(i)(t) ≥ u(i)n−1(t), q

(i)(t) ≥ v(i)n−1(t) for t ∈ J, n ≥ 1 (i = 0, 1). Then, it follows

from Lemma 2.6 that (A(i)1 (p, q)(t), A(i)

2 (p, q)(t)) ≥ (A(i)1 (un−1, vn−1))(t), A

(i)2 (un−1, vn−1))(t)) for

t ∈ J (i = 0, 1), that is, (p(i)(t), q(i)(t)) ≥ (u(i)n (t), v(i)

n (t)) for t ∈ J (i = 0, 1). Hence, by induction,we get

p(i)(t) ≥ x(i)n (t), q(i)(t) ≥ y(i)

n (t) ∀t ∈ J (i = 0, 1; m = 0, 1, 2, . . .). (3.24)

Now, taking limits in (3.24), we get p(i)(t) ≥ x(i)(t), q(i)(t) ≥ y(i)(t) for t ∈ J (i = 0, 1), and thetheorem is proved.

Theorem 3.3. Let cone P be fully regular and conditions (H1) and (H3) be satisfied. Then theconclusion of Theorem 3.2 holds.

Proof. The proof is almost the same as that of Theorem 3.2. The only difference is that, insteadof using condition (H2), the conclusion αX(W) = 0 is implied directly by (3.15) and (3.16),the full regularity of P and Lemma 2.4.

4. An Example

Consider the infinite system of scalar singular second order three-point boundary valueproblems:

−x′′n(t) =1

3n2√t(1 + t)

(

2 + xn(t) + yn(t) + x′2n(t) + y′3n(t) +1

2n2xn(t)+

18n3x′2n(t)

)1/3

+1

3e2t(1 + t)ln(1 + xn(t)),

−y′′n(t) =1

6n3√t2(1 + t)

(

1 + x3n(t) + x′4n(t) +1

3n2y3n(t)+

14n3y′2n(t)

)1/5

+1

6e3t2(1 + t)ln(1 + y′2n(t)

),

xn(0) =23xn(1), x′n(∞) =

1n, yn(0) =

34yn(1), y′n(∞) =

12n

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

(4.1)

Proposition 4.1. Infinite system (4.1) has a minimal positive solution (xn(t), yn(t)) satisfyingxn(t), x′n(t) ≥ 1/n, yn(t), y′n(t) ≥ 1/2n for 0 ≤ t < +∞ (n = 1, 2, 3, . . .).

Proof. Let E = c0 = {x = (x1, . . . , xn, . . .) : xn → 0} with the norm ‖x‖ = supn|xn|. Obviously,(E, ‖ · ‖) is a real Banach space. Choose P = {x = (xn) ∈ c0 : xn ≥ 0, n = 1, 2, 3, . . .}. It iseasy to verify that P is a normal cone in E with normal constants 1. Now we consider infinite


system (4.1), which can be regarded as a BVP of form (1.2) in E with α1 = 2/3, β1 = 3/4, ξ1 =1, x∞ = (1, 1/2, 1/3, . . .), y∞ = (1/2, 1/4, 1/6, . . .). In this situation, x = (x1, . . . , xn, . . .), u =(u1, . . . , un, . . .), y = (y1, . . . , yn, . . .), v = (v1, . . . , vn, . . .), f = (f1, . . . , fn, . . .), in which

fn(t, x, u, y, v

)=

1

3n2√t(1 + t)

(2 + xn + yn + u2n + v3n +

12n2xn

+1

8n3u2n

)1/3

+1

3e2t(1 + t)ln(1 + xn),

gn(t, x, u, y, v

)=

1

6n3 3√t2(1 + t)

(1 + x3n + u4n +

13n2y3n

+1

4n3v2n

)1/5

+1

6e3t(1 + t)ln(1 + v2n).

(4.2)

Let x∗0 = x∞ = (1, 1/2, 1/3, . . .), y∗0 = y∞ = (1/2, 1/4, 1/6, . . .). Then P0λ = {x =(x1, x2, . . . , xn, . . .) : xn ≥ λ/n, n = 1, 2, 3, . . .}, P1λ = {y = (y1, y2, . . . , yn, . . .) : yn ≥ λ/2n, n =1, 2, 3, . . .}, for λ > 0. It is clear, f, g ∈ C[J+ × P0λ × P0λ × P1λ × P1λ, P] for any λ > 0. Notice thate3t >

3√t2, e2t >

√t for t > 0, by (4.2), we get

‖f(t, x, u, y, v

)‖ ≤ 1

3√t

[(114

+ ‖x‖ + ‖u‖ + ‖v‖ + ‖y‖)1/3

+ ln(1 + ‖x‖)]

,

‖g(t, x, u, y, v

)‖ ≤ 1

6 3√t2

[(4 + ‖x‖ + ‖u‖)1/5 + ln(1 + ‖v‖)

],

(4.3)

which imply (H1) is satisfied for a0(t) = 0, b0(t) = c0(t) = 1/3√t, a1(t) = 0, b1(t) = c1(t) =

1/6 3√t2 and

h0(u0, u1, u2, u3) =(

114

+ u0 + u1 + u2 + u3

)1/3

+ ln(1 + u0),

h1(u0, u1, u2, u3) = (4 + u0 + u1)1/5 + ln(1 + u3).

(4.4)

Let f1 = {f11 , f

12 , . . . , f

1n, . . .}, f2 = {f2

1 , f22 , . . . , f

2n, . . .}, and g1 = {g1

1 , g12 , . . . , g

1n, . . .}, g2 =

{g21 , g

22 , . . . , g

2n, . . .}, where

f1n

(t, x, u, y, v

)=

1

3n2√t(1 + t)

(2 + xn + yn + u2n + v3n +

12n2xn

+1

8n3u2n

)1/3

, (4.5)

f2n

(t, x, u, y, v

)=

13e2t(1 + t)

ln(1 + xn), (4.6)

g1n

(t, x, u, y, v

)=

1

6n3 3√t2(1 + t)

(1 + x3n + u4n +

13n2y3n

+1

4n3v2n

)1/5

, (4.7)

g2n

(t, x, u, y, v

)=

16e3t(1 + t)

ln(1 + v2n). (4.8)


Let t ∈ J+, R > 0 be given, and {z(m)} be any sequence in f1(t, P ∗0R, P∗0R, P

∗1R, P

∗1R), where z(m) =

(z(m)1 , . . . , z

(m)n , . . .). By (4.5), we have

0 ≤ z(m)n ≤ 1

3n2√t

(114

+ 4R)1/3

(n,m = 1, 2, 3, . . .). (4.9)

So, {z(m)n } is bounded and by the diagonal method together with the method of constructing

subsequence, we can choose a subsequence {mi} ⊂ {m} such that

{z(m)n

}−→ zn as i −→ ∞ (n = 1, 2, 3, . . .), (4.10)

which implies by virtue of (4.9)

0 ≤ zn ≤1

3n2√t

(114

+ 4R)1/3

(n = 1, 2, 3, . . .). (4.11)

Hence z = (z1, . . . , zn, . . .) ∈ c0. It is easy to see from (4.9)–(4.11) that

∥∥∥z(mi) − z∥∥∥ = sup

n

∣∣∣z(mi)n − zn

∣∣∣ −→ 0 as i −→ ∞. (4.12)

Thus, we have proved that f1(t, P ∗0R, P∗0R, P

∗1R, P

∗1R) is relatively compact in c0.

For any t ∈ J+, R > 0, x, y, x, y ∈ D ⊂ P ∗0R, we have by (4.6)

∣∣∣f2n

(t, x, u, y, v

)− f2

n

(t, x, u, y, v

)∣∣∣ =1

3e2t(1 + t)|ln(1 + xn) − ln(1 + xn)|

≤ 13e2t(1 + t)

|xn − xn|1 + ξn

,

(4.13)

where ξn is between xn and xn. By (4.13), we get

∥∥∥f2(t, x, u, y, v)− f2(t, x, u, y, v

)∥∥∥ ≤1

3e2t(1 + t)‖x − x‖, x, y, x, y ∈ D. (4.14)

In the same way, we can prove that g1(t, P ∗0R, P∗0R, P

∗1R, P

∗1R) is relatively compact in c0, and we

can also get

∥∥∥g2(t, x, u, y, v)− g2(t, x, u, y, v

)∥∥∥ ≤1

6e3t(1 + t)‖v − v‖, x, y, x, y ∈ D. (4.15)

Thus, by (4.14) and (4.15), it is easy to see that (H2) holds for L00(t) = 1/3e2t(1 + t), K11(t) =1/6e3t(1 + t). Thus, our conclusion follows from Theorem 3.1. This completes the proof.


Acknowledgment

The project is supported financially by the National Natural Science Foundation of China(10671167) and the Natural Science Foundation of Liaocheng University (31805).

References

[1] D. J. Guo, V. Lakshmikantham, and X. Z. Liu, Nonlinear Integral Equations in Abstract Spaces, vol. 373of Mathematics and Its Applications, Kluwer Academic Publishers, Dordrecht, The Netherlands, 1996.

[2] K. Deimling, Ordinary Differential Equations in Banach Spaces, vol. 596 of Lecture Notes in Mathematics,Springer, Berlin, Germany, 1977.

[3] V. Lakshmikantham and S. Leela, Nonlinear Differential Equations in Abstract Spaces, vol. 2 ofInternational Series in Nonlinear Mathematics: Theory, Methods and Applications, Pergamon Press, Oxford,UK, 1981.


[5] Y. Liu, “Boundary value problems for second order differential equations on unbounded domains ina Banach space,” Applied Mathematics and Computation, vol. 135, no. 2-3, pp. 569–583, 2003.

[6] C. P. Gupta, “Solvability of a three-point nonlinear boundary value problem for a second orderordinary differential equation,” Journal of Mathematical Analysis and Applications, vol. 168, no. 2, pp.540–551, 1992.

[7] W. Feng and J. R. L. Webb, “Solvability of m-point boundary value problems with nonlinear growth,”Journal of Mathematical Analysis and Applications, vol. 212, no. 2, pp. 467–480, 1997.

[8] J. X. Sun, X. Xu, and D. O’Regan, “Nodal solutions for m-point boundary value problems usingbifurcation methods,” Nonlinear Analysis: Theory, Methods & Applications, vol. 68, no. 10, pp. 3034–3046, 2008.

[9] X. Xu, “Positive solutions for singular m-point boundary value problems with positive parameter,”Journal of Mathematical Analysis and Applications, vol. 291, no. 1, pp. 352–367, 2004.

[10] R. Ma and N. Castaneda, “Existence of solutions of nonlinear m-point boundary-value problems,”Journal of Mathematical Analysis and Applications, vol. 256, no. 2, pp. 556–567, 2001.

[11] J. Zhao, Z. Liu, and L. Liu, “The existence of solutions of infinite boundary value problems forfirst-order impulsive differential systems in Banach spaces,” Journal of Computational and AppliedMathematics, vol. 222, no. 2, pp. 524–530, 2008.

[12] G. Zhang and J. X. Sun, “Positive solutions of m-point boundary value problems,” Journal ofMathematical Analysis and Applications, vol. 291, no. 2, pp. 406–418, 2004.

[13] Y.-L. Zhao and H.-B. Chen, “Existence of multiple positive solutions for m-point boundary valueproblems in Banach spaces,” Journal of Computational and Applied Mathematics, vol. 215, no. 1, pp. 79–90, 2008.

[14] B. Liu, “Positive solutions of a nonlinear four-point boundary value problems in Banach spaces,”Journal of Mathematical Analysis and Applications, vol. 305, no. 1, pp. 253–276, 2005.

[15] D. J. Guo, “Existence of positive solutions for nth-order nonlinear impulsive singular integro-differential equations in Banach spaces,” Nonlinear Analysis: Theory, Methods & Applications, vol. 68,no. 9, pp. 2727–2740, 2008.

[16] D. J. Guo and V. Lakshmikantham, “Coupled fixed points of nonlinear operators with applications,”Nonlinear Analysis: Theory, Methods & Applications, vol. 11, no. 5, pp. 623–632, 1987.


Research ArticlePositive Solutions to Singular andDelay Higher-Order Differential Equations onTime Scales

Liang-Gen Hu,1 Ti-Jun Xiao,2 and Jin Liang3

1 Department of Mathematics, University of Science and Technology of China, Hefei 230026, China2 School of Mathematical Sciences, Fudan University, Shanghai 200433, China3 Department of Mathematics, Shanghai Jiao Tong University, Shanghai 200240, China

Correspondence should be addressed to Jin Liang, [email protected]

Received 21 March 2009; Accepted 1 July 2009


We are concerned with singular three-point boundary value problems for delay higher-orderdynamic equations on time scales. Theorems on the existence of positive solutions are obtainedby utilizing the fixed point theorem of cone expansion and compression type. An example is givento illustrate our main result.

Copyright q 2009 Liang-Gen Hu 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.

1. Introduction

In this paper, we are concerned with the following singular three-point boundary valueproblem (BVP for short) for delay higher-order dynamic equations on time scales:

(−1)nuΔ2n(t) = w(t)f(t, u(t − c)), t ∈ [a, b],

u(t) = ψ(t), t ∈ [a − c, a),

uΔ2i(a) − βi+1u

Δ2i+1(a) = αi+1u

Δ2i(�),

γi+1uΔ2i

(�) = uΔ2i(b), 0 ≤ i ≤ n − 1,

(1.1)

where c ∈ [0, (b − a)/2], � ∈ (a, b), βi ≥ 0, 1 < γi < (b − a + βi)/(� − a + βi), 0 ≤ αi <(b − γi� + (γi − 1)(a − βi))/(b − �), i = 1, 2, . . . , n and ψ ∈ C([a − c, a]). The functionalw : (a, b) → [0,+∞) is continuous and f : [a, b] × (0,+∞) → [0,+∞) is continuous. Our


nonlinearity w may have singularity at t = a and/or t = b, and f may have singularity atu = 0.

To understand the notations used in (1.1), we recall the following definitions whichcan be found in [1, 2].

(a) A time scale T is a nonempty closed subset of the real numbers R. T has the topologythat it inherits from the real numbers with the standard topology. It follows that thejump operators σ, ρ : T → T,

σ(t) = inf{τ ∈ T : τ > t}, ρ(t) = sup{τ ∈ T : τ < t} (1.2)

(supplemented by inf ∅ := sup T and sup ∅ := inf T) are well defined. The pointt ∈ T is left-dense, left-scattered, right-dense, right-scattered if ρ(t) = t, ρ(t) < t,σ(t) = t, σ(t) < t, respectively. If T has a left-scattered maximum t1 (right-scatteredminimum t2), define T

k = T − {t1} (Tk = T − {t2}); otherwise, set Tk = T (Tk = T).

By an interval [a, b] we always mean the intersection of the real interval [a, b] withthe given time scale, that is, [a, b]∩T. Other types of intervals are defined similarly.

(b) For a function f : T → R and t ∈ Tk, the Δ-derivative of f at t, denoted by fΔ(t),

is the number (provided it exists) with the property that, given any ε > 0, there is aneighborhood U ⊂ T of t such that

∣∣∣f(σ(t)) − f(s) − fΔ(t)[σ(t) − s]∣∣∣ ≤ ε|σ(t) − s|, ∀s ∈ U. (1.3)

(c) For a function f : T → R and t ∈ Tk, the ∇-derivative of f at t, denoted by f∇(t),is the number (provided it exists) with the property that, given any ε > 0, there is aneighborhood U ⊂ T of t such that

∣∣∣f(ρ(t)

)− f(s) − f∇(t)

[ρ(t) − s

]∣∣∣ ≤ ε∣∣ρ(t) − s

∣∣, ∀s ∈ U. (1.4)

(d) If FΔ(t) = f(t)(Φ∇(t) = g(t)), then we define the integral

∫ t

a

f(�)Δ� = F(t) − F(a)(∫ t

a

g(�)∇� = Φ(t) −Φ(a)

)

. (1.5)

Theoretically, dynamic equations on time scales can build bridges between continuousand discrete mathematics. Practically, dynamic equations have been proposed as models inthe study of insect population models, neural networks, and many physical phenomenawhich include gas diffusion through porous media, nonlinear diffusion generated bynonlinear sources, chemically reacting systems as well as concentration in chemical ofbiological problems [2]. Hence, two-point and multipoint boundary value problems fordynamic equations on time scales have attracted many researchers’ attention (see, e.g., [1–19]and references therein). Moreover, singular boundary value problems have also been treatedin many papers (see, e.g., [4, 5, 12–14, 18] and references therein).


In 2004, J. J. DaCunha et al. [13] considered singular second-order three-pointboundary value problems on time scales

uΔΔ(t) + f(t, u(t)) = 0, (0, 1] ∩ T,

u(0) = 0, u(p)= u(σ2(1)

) (1.6)

and obtained the existence of positive solutions by using a fixed point theorem due to Gaticaet al. [14], where f : (0, 1] × (0,∞) → (0,∞) is decreasing in u for every t ∈ (0, 1] and mayhave singularity at u = 0.

In 2006, Boey and Wong [11] were concerned with higher-order differential equationon time scales of the form

(−1)n−1yΔn

(t) = (−1)p+1F(t, y(σn−1(t)

)), t ∈ [a, b],

yΔi

(a) = 0, 0 ≤ i ≤ p − 1,

yΔi

(σ(b)) = 0, p ≤ i ≤ n − 1,

(1.7)

where p, n are fixed integers satisfying n ≥ 2, 1 ≤ p ≤ n − 1. They obtained some existencetheorems of positive solutions by using Krasnosel’skii fixed point theorem.

Recently, Anderson and Karaca [8] studied higher-order three-point boundary valueproblems on time scales and obtained criteria for the existence of positive solutions.

The purpose of this paper is to investigate further the singular BVP for delay higher-order dynamic equation (1.1). By the use of the fixed point theorem of cone expansionand compression type, results on the existence of positive solutions to the BVP (1.1) areestablished.

The paper is organized as follows. In Section 2, we give some lemmas, which will berequired in the proof of our main theorem. In Section 3, we prove some theorems on theexistence of positive solutions for BVP (1.1). Moreover, we give an example to illustrate ourmain result.

2. Lemmas

For 1 ≤ i ≤ n, let Gi(t, s) be Green’s function of the following three-point boundary valueproblem:

−uΔΔ(t) = 0, t ∈ [a, b],

u(a) − βiuΔ(a) = αiu(�), γiu(�) = u(b),(2.1)

where � ∈ (a, b) and αi, βi, γi satisfy the following condition:

(C)

βi ≥ 0, 1 < γi <b − a + βi� − a + βi

, 0 ≤ αi <b − γi� +

(γi − 1

)(a − βi

)

b −� . (2.2)


Throughout the paper, we assume that σ(b) = b.

From [8], we know that for any (t, s) ∈ [a, b] × [a, b] and 1 ≤ i ≤ n,

Gi(t, s) =

⎧⎨

⎩

Gi1(t, s), s ∈ [a,�],

Gi2(t, s), s ∈ [�, b],(2.3)

where

Gi1(t, s) =1di

⎧⎨

⎩

[γi(t −�) + b − t

](σ(s) + βi − a

), σ(s) ≤ t,

[γi(σ(s) −ω) + b − σ(s)

](t + βi − a

)+ αi(� − b)(t − σ(s)), t ≤ s,

Gi2(t, s) =1di

⎧⎨

⎩

[σ(s)(1 − αi) + αi� + βi − a

](b − t) + γi

(� − a + βi

)(t − σ(s)), σ(s) ≤ t,

[t(1 − αi) + αi� + βi − a

](b − σ(s)), t ≤ s,

di =(γi − 1

)(a − βi

)+ (1 − αi)b +�

(αi − γi

).

(2.4)

The following four lemmas can be found in [8].

Lemma 2.1. Suppose that the condition (C) holds. Then the Green function ofGi(t, s) in (2.3) satisfies

Gi(t, s) > 0, (t, s) ∈ (a, b) × (a, b). (2.5)

Lemma 2.2. Assume that the condition (C) holds. Then Green’s function Gi(t, s) in (2.3) satisfies

Gi(t, s) ≤ max{Gi(b, s), Gi(σ(s), s)}, (t, s) ∈ [a, b] × [a, b]. (2.6)

Remark 2.3. (1) If s ∈ ((γi(� − a + βi) − αi� − βi + a)/(1 − αi), b], s ≤ t, we know that Gi(t, s) isnonincreasing in t and

Gi(b, s)Gi(σ(s), s)

=γi(� − a + βi

)(b − σ(s))

(σ(s)(1 − αi) + αi� + βi − a

)(b − σ(s))

≥γi(� − a + βi

)

b(1 − αi) + αi� + βi − a> 0.

(2.7)

Therefore, we have

Gi(b, s) ≤ Gi(t, s) ≤ Gi(σ(s), s) ≤ δiGi(b, s), (2.8)

where

δi =b(1 − αi) + αi� + βi − a

γi(� − a + βi

) > 1. (2.9)


(2) If t and s satisfy the other cases, then we get that Gi(t, s) is nondecreasing in t and

Gi(t, s) ≤ Gi(b, s). (2.10)

Lemma 2.4. Assume that (C) holds. Then Green’s function Gi(t, s) in (2.3) verifies the followinginequality:

Gi(t, s) ≥ min{t − ab − a ,

b − tγi(b − a)

}Gi(b, s)

≥ min{

t − aδi(b − a)

,b − t

γi(b − a)

}max{Gi(b, s), Gi(σ(s), s)}.

(2.11)

Remark 2.5. If s ∈ [�, (γi(� − a + βi) − αi� − βi + a)/(1 − αi)), s ≤ t, then we find

(γi − 1

)(a − βi

)+ (1 − αi)σ(s) +�

(αi − γi

)< 0. (2.12)

So there exists a misprint on [8, Page 2431, line 23]. From (2.3), it follows that

Gi(t, s)Gi(b, s)

=

[σ(s)(1 − αi) + αi� + βi − a

](b − t) + γi

(� − a + βi

)(t − σ(s))

γi(� − a + βi

)(b − σ(s))

≥(� + βi − a

)(b − t) + γi

(� − a + βi

)(t − σ(s))

γi(� − a + βi

)(b − a)

≥ b − tγi(b − a)

.

(2.13)

Consequently, we get

Gi(t, s) ≥b − t

γi(b − a)Gi(b, s). (2.14)

If s ∈ ((γi(� − a + βi) − αi� − βi + a)/(1 − αi), b], s ≤ t, then, from (2.8), we obtain

Gi(t, s) ≥t − ab − aGi(b, s) ≥

t − aδi(b − a)

Gi(σ(s), s). (2.15)


Remark 2.6. If we set hi(t) := min{(t − a)/δi(b − a), (b − t)/γi(b − a)}, then we have

Gi(t, s) ≥ hi(t) max{Gi(b, s), Gi(σ(s), s)}, (t, s) ∈ [a, b] × [a, b]. (2.16)

Denote

‖Gi(·, s)‖ = maxt∈[a,b]

|Gi(t, s)|, s ∈ [a, b]. (2.17)

Thus we have

Gi(t, s) ≥ hi(t)‖Gi(·, s)‖, (t, s) ∈ [a, b] × [a, b]. (2.18)

Lemma 2.7. Assume that condition (C) is satisfied. For Gi(t, s) as in (2.3), put H1(t, s) := G1(t, s)and recursively define

Hj(t, s) =∫b

a

Hj−1(t, r)Gj(r, s)Δr (2.19)

for 2 ≤ j ≤ n. Then Hn(t, s) is Green’s function for the homogeneous problem

(−1)nuΔ2n(t) = 0, t ∈ [a, b],

uΔ2i(a) − βi+1u

Δ2i+1(a) = αi+1u

Δ2i(�),

γi+1uΔ2i

(�) = uΔ2i(b), 0 ≤ i ≤ n − 1.

(2.20)

Lemma 2.8. Assume that (C) holds. Denote

K :=n−1∏

j=1

kj , L :=n−1∏

j=1

lj , (2.21)

then Green’s function Hn(t, s) in Lemma 2.7 satisfies

h1(t)L‖Gn(·, s)‖ ≤ Hn(t, s) ≤ K‖Gn(·, s)‖, (t, s) ∈ [a, b] × [a, b], (2.22)

where

kj =∫b

a

∥∥Gj(·, s)∥∥Δs > 0, lj =

∫b

a

∥∥Gj(·, s)∥∥hj+1(s)Δs, 1 ≤ j ≤ n − 1. (2.23)


Proof. We proceed by induction on n ≥ 2. We denote the statement by P(n). From Lemma 2.7,it follows that

‖H2(t, s)‖ =∥∥∥∥∥

∫b

a

H1(t, r)G2(r, s)Δr

∥∥∥∥∥

≤∫b

a

‖G1(·, r)‖‖G2(·, s)‖Δr = k1‖G2(·, s)‖,

(2.24)

and from (2.18), we have

H2(t, s) =∫b

a

H1(t, r)G2(r, s)Δr

≥∫b

a

h1(t)‖G1(·, r)‖ × h2(r)‖G2(·, s)‖Δr

= h1(t)l1‖G2(·, s)‖.

(2.25)

So P(2) is true.

We now assume that P(m) is true for some positive integer m ≥ 2. From Lemma 2.7, itfollows that

‖Hm+1(t, s)‖ =∥∥∥∥∥

∫b

a

Hm(t, r)Gm+1(r, s)Δr

∥∥∥∥∥

≤∫b

a

Hm(t, r)‖Gm+1(r, s)‖Δr

≤

⎛

⎝∫b

a

m−1∏

j=1

kj × ‖Gm(·, r)‖Δr

⎞

⎠‖Gm+1(·, s)‖

=m∏

j=1

kj‖Gm+1(·, s)‖,

Hm+1(t, s) =∫b

a

Hm(t, r)Gm+1(r, s)Δr

≥∫b

a

h1(t) ×m−1∏

j=1

ljGm(·, r)hm+1(r)‖Gm+1(·, s)‖Δr

= h1(t)m∏

j=1

lj‖Gm+1(·, s)‖.

(2.26)

So P(m + 1) holds. Thus P(n) is true by induction.


Lemma 2.9 (see [20]). Let (E, ‖ · ‖) be a real Banach space and P ⊂ E a cone. Assume that T :Pζ,η → P is completely continuous operator such that

(i) Tu � u for u ∈ ∂Pζ and Tu � u for u ∈ ∂Pη,

(ii) Tu � u for u ∈ ∂Pζ and Tu � u for u ∈ ∂Pη.

Then T has a fixed point u∗ ∈ P with ζ ≤ ‖u∗‖ ≤ η.

3. Main Results

We assume that {am}m≥1 and {bm}m≥1 are strictly decreasing and strictly increasing sequences,respectively, with limm→∞am = a, limm→∞bm = b and a1 < b1. A Banach space E = C([a, b]) isthe set of real-valued continuous (in the topology of T) functions u(t) defined on [a, b] withthe norm

‖u‖ = maxt∈[a,b]

|u(t)|. (3.1)

Define a cone by

P ={u ∈ E : u(t) ≥ h1(t)L

K‖u‖, t ∈ [a, b]

}. (3.2)

Set

Pξ = {u ∈ P : ‖u‖ < ξ}, ∂Pξ = {u ∈ P : ‖u‖ = ξ}, ξ > 0,

Pζ,η ={u ∈ P : ζ < ‖u‖ < η

}, 0 < ζ < η,

Y1 = {t ∈ [a, b] : t − c < a}, Y2 = {t ∈ [a, b] : t − c ≥ a},

Ym = {t ∈ Y2 : t − c ∈ [a, am] ∪ [bm, b]}.

(3.3)

Assume that

(C1) ψ : [a − c, a] → (0,∞) is continuous;

(C2) we have

0 < K

∫q

p

‖Gn(·, s)‖w(s)Δs, K

∫b

a

‖Gn(·, s)‖w(s)Δs < +∞, (3.4)

for constants p and q with a + c < p < q < b;

(C3) the function f : [a, b] × (0,+∞) → R+ is continuous and w : (a, b) → R

+ iscontinuous satisfying

limm→∞

supu∈Pζ,η

K

∫

Ym

‖Gn(·, s)‖w(s)f(s, u(s − c))Δs = 0, ∀0 < ζ < η. (3.5)


We seek positive solutions u : [a, b] → R+, satisfying (1.1). For this end, we transform

(1.1) into an integral equation involving the appropriate Green function and seek fixed pointsof the following integral operator.

Define an operator T : C+[a, b] → C[a, b] by

(Tu)(t) =∫b

a

Hn(t, s)w(s)f(s, u(s − c))Δs, ∀u ∈ C+([a, b]), (3.6)

where C+[a, b] = {u ∈ C[a, b] | u(t) ≥ 0, t ∈ [a, b]}.

Proposition 3.1. Let (C1), (C2), and (C3) hold, and let ζ, η be fixed constants with 0 < ζ < η. ThenT : Pζ,η → P is completely continuous.

Proof. We separate the proof into four steps.

Step 1. For each u ∈ Pζ,η, Tu is bounded.By condition (C3), there exists some positive integer m0 satisfying

supu∈Pζ,η

K

∫

Ym0

‖Gn(·, s)‖w(s)f(s, u(s − c))Δs ≤ 1, (3.7)

where

Ym0 = {t ∈ Y2 : t − c ∈ [a, am0] ∪ [bm0 , b]}; (3.8)

here, we used the fact that for each u ∈ Pζ,η and t ∈ [am0 + c, bm0 + c] ∩ [a, b],

η ≥ u(t − c) ≥ h1(t − c)LK

‖u‖ ≥ ζmin{h1(am0)L

K,h1(bm0)L

K,h1(b − c)L

K

}= ζh > 0, (3.9)

where

h = min{h1(am0)L

K,h1(bm0)L

K,h1(b − c)L

K

}. (3.10)

Set

D := max{f(t, ψ(t − c)

): t ∈ Y1

},

Q := max{f(t, u(t − c)) : t ∈ Y2, ζh ≤ u(t − c) ≤ η

}.

(3.11)


Then we obtain

Tu(t) ≤ supt∈[a,b]

supu∈Pζ,η

∫b

a

Hn(t, s)w(s)f(s, u(s − c))Δs

≤ K supu∈Pζ,η

∫

Y1

‖Gn(·, s)‖w(s)f(s, u(s − c))Δs

+ supu∈Pζ,η

K

∫

Ym0


+ supu∈Pζ,η

K

∫

Y2\Ym0


≤ 1 + max{D,Q}K∫b

a

‖Gn(·, s)‖w(s)Δs < +∞.

(3.12)

Consequently, Tu is bounded and well defined.

Step 2. T : Pζ,η → P . For every u ∈ Pζ,η, we get from (2.22)

‖Tu‖ = supt∈[a,b]

∫b

a


≤ K

∫b

a

‖Gn(·, s)‖w(s)f(s, u(s − c))Δs.

(3.13)

Then by the above inequality

(Tu)(t) =∫b

a


≥∫b

a

h1(t)L‖Gn(·, s)‖w(s)f(s, u(s − c))Δs

≥ h1(t)LK

‖Tu‖.

(3.14)

This leads to Tu ∈ P .


Step 3. We will show that T : Pζ,η → P is continuous. Let {um}m≥1 be any sequence in Pζ,η

such that limm→∞um = u ∈ Pζ,η. Notice also that as m → ∞,

φm(s) =∣∣f(s, um(s − c)) − f(s, u(s − c))

∣∣w(s) −→ 0, for s ∈ (a + c, b),

∣∣f(s, um(s − c)) − f(s, u(s − c))

∣∣w(s)

=∣∣f(s, ψ(s − c)

)− f(s, ψ(s − c)

)∣∣w(s) = 0, for s ∈ (a, a + c),

∫

Y2

Hn(t, s)φm(s)Δs ≤ supx∈Pζ,η

2K∫

Y2

‖Gn(·, s)‖w(s)f(s, x(s))Δs < +∞.

(3.15)

Now these together with (C2) and the Lebesgue dominated convergence theorem [10] yieldthat as m → ∞,

‖Tum − Tu‖ = supt∈[a,b]

∫b

a

Hn(t, s)w(s)∣∣f(s, um(s − c)) − f(s, u(s − c))

∣∣Δs −→ 0. (3.16)

Step 4. T : Pζ,η → P is compact.Define

wm(t) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

min{w(t), w(am)}, a ≤ t ≤ am,

w(t), am ≤ t ≤ bm,

min{w(t), w(bm)}, bm ≤ t ≤ b,

fm(t, u(t − c)) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

f(t, ψ(t − c)

), a ≤ t < a + c,

min{f(t, u(t − c)), f(t, u(am))

}, a + c ≤ t ≤ am + c,

f(t, u(t − c)), t ∈ [am + c, bm + c] ∩ [a, b],

min{f(t, u(t − c)), f(t, u(bm))

}, t ∈ [bm + c, b] ∩ [a, b],

(3.17)

and an operator sequence {Tm} for a fixed m by

(Tmu)(t) =∫b

a

Hn(t, s)wm(s)fm(s, u(s − c))Δs, ∀t ∈ [a, b]. (3.18)

Clearly, the operator sequence {Tm} is compact by using the Arzela-Ascoli theorem[3], for each m ∈ N. We will prove that Tm converges uniformly to T on Pζ,η. For any u ∈ Pζ,η,


we obtain

‖Tmu − Tu‖ = supt∈[a,b]

∣∣∣∣∣

∫b

a

Hn(t, s)(wm(s)fm(s, u(s − c)) −w(s)f(s, u(s − c))

)Δs

∣∣∣∣∣

≤ K

∫b

a

‖Gn(·, s)‖∣∣wm(s)fm(s, u(s − c)) −w(s)f(s, u(s − c))

∣∣Δs

≤ K

∫

Y1

‖Gn(·, s)‖|wm(s) −w(s)|f(s, ψ(s − c)

)Δs

+K

∫

Y2

‖Gn(·, s)‖∣∣wm(s)fm(s, u(s − c)) −w(s)f(s, u(s − c))

∣∣Δs.

(3.19)

From (C1), (C2), and the Lebesgue dominated convergence theorem [10], we see that theright-hand side (3.19) can be sufficiently small for mbeing big enough. Hence the sequence{Tm} of compact operators converges uniformly to T on Pζ,η so that operator T is compact.Consequently, T : Pζ,η → P is completely continuous by using the Arzela-Ascoli theorem[3].

Proposition 3.2. It holds that v ∈ Pζ,η is a solution of (1.1) if and only if Tv = v.

Proof. If v ∈ Pζ,η and Tv = v, then we have

(−1)nvΔ2n(t) = (−1)nTvΔ2n

(t) = w(t)f(t, v(t − c)), (3.20)

and for any 0 ≤ i ≤ n − 1,

vΔ2i(a) − βi+1v

Δ2i+1(a) = αi+1v

Δ2i(�), γi+1v

Δ2i(�) = vΔ2i

(b). (3.21)

From [8, Lemma 3.1], we know that v(t) ≥ 0 on [a, b]. So we conclude that v is the solutionof BVP (1.1).

For convenience, we list the following notations and assumptions:

R =

(

μK

∫q

p

‖Gn(·, s)‖w(s)Δs

)−1

, μ = min

{h1(p)L

K,h1(q)L

K

}

;

κ =

[

K

∫b

a


]−1

;

(3.22)


fξμξ :=

f(t, u(t − c))u(t)

, t ∈[p, q], u ∈

[μξ, ξ

]; (3.23)

fζρ :=

f(t, u(t − c))u(t)

, t ∈ Y2, u ∈[ρ, ζ]; (3.24)

S(ρ)= sup

u∈∂Pρ

K

∫

Y2

‖Gn(·, s)‖w(s)f(s, u(s − c))Δs, ρ > 0. (3.25)

From condition (C2) and (3.12), we have S(ρ) < +∞.

Theorem 3.3. Assume that there exist positive constants ρ, ζ, ξ, r with ζ < μξ, r < κ and ζ ≥κS(ρ)/(κ − r) such that

(i) fξμξ

> R and fζρ < r;

(ii) f(t, ψ(t − c))/u(t) < r, for all t ∈ Y1 and u ∈ [ρ, ζ].

If (C1), (C2), and (C3) hold, then the boundary value problem (1.1) has at least one positive solutionu such that

u(t) =

⎧⎨

⎩

ψ(t), if t ∈ [a − c, a),

u∗(t), if t ∈ [a, b],

ζ ≤ ‖u∗‖ ≤ ξ.

(3.26)

Proof. Define the operator T : Pζ,ξ → P by (3.6). From (i) and (3.23), it follows that thereexists ε1 > 0 such that

f(t, u(t − c)) ≥ (R + ε1)u(t), for t ∈[p, q], u ∈

[μξ, ξ

]. (3.27)

We claim that

Tu � u, ∀u ∈ ∂Pξ. (3.28)

If it is false, then there exists some u1 ∈ ∂Pξ with Tu1 ≤ u1, that is, u1 − Tu1 ∈ P which impliesthat u1(t) ≥ Tu1(t) for t ∈ [a, b].

Set

λ = min{u1(t) : t ∈

[p, q]}≥ min

{h1(p)L

K,h1(q)L

K

}

‖u1‖ = μξ. (3.29)


We know from (2.22) and (3.27) that for t ∈ [p, q],

u1(t) ≥ Tu1(t)

=∫b

a

Hn(t, s)w(s)f(s, u1(s − c))Δs

=∫

Y1

Hn(t, s)w(s)f(s, u1(s − c))Δs +∫

Y2


≥∫q

p


≥ min{h1(p), h1(q)}

L

∫q

p

‖Gn(·, s)‖w(s)f(s, u1(s − c))Δs

≥ (R + ε1) mint∈[p,q]

u1(t)μK∫q

p


≥ λR

[

μK

∫q

p


]

+ λε1μK

∫q

p


= λ + λε1μK

∫q

p

‖Gn(·, s)‖w(s)Δs,

(3.30)

the first inequality of (C2) implies that

u1(t) > λ, ∀t ∈[p, q]. (3.31)

Clearly, (3.31) contradicts (3.29). This means that (3.28) holds.Next we will show that

Tu � u, ∀u ∈ Pζ. (3.32)

Suppose on the contrary that there exists some u2 ∈ ∂Pζ with u2 ≤ Tu2 for all t ∈ [a, b].For (t, u) ∈ Y2 × [ρ, ζ], from (i) and (3.24), there exists ε2 > 0 such that

f(t, u(t − c)) ≤ (r − ε2)u(t). (3.33)

and for (t, u) ∈ Y1 × [ρ, ζ], there exists ε2 > 0, from (ii), such that

f(t, ψ(t − c)

)≤ (r − ε2)u(t). (3.34)

Put

Y3 :={t ∈ Y2 : u2(t) > ρ

}, u2(t) =

⎧⎨

⎩

min{u2(t), ρ

}, t ∈ Y2,

ρ, t ∈ Y1.(3.35)


If Y3 = ∅, then we take u2(t) = ρ. It is easy to see that (h1(t − c)Lζ)/K ≤ u2(t − c) ≤ ‖u2‖ = ζfor t ∈ Y2 and u2(t) ∈ C+[a, b], ‖u2‖ = ρ, that is, u2 ∈ ∂Pρ. From (3.33) and (3.34), we find that

‖Tu2‖ = supt∈[a,b]

∫b

a


≤ K

∫b

a

‖Gn(·, s)‖w(s)f(s, u2(s − c))Δs

= K

∫

Y1

‖Gn(·, s)‖w(s)f(s, ψ(s − c)

)Δs +K

∫

Y3

‖Gn(·, s)‖w(s)f(s, u2(s − c))Δs

+K

∫

Y2\Y3

‖Gn(·, s)‖w(s)f(s, u2(s − c))Δs

≤ (r − ε2)maxt∈Y1

u2(t)∫

Y1


+ sup(t,u2)∈Y3×[ρ,ζ]

f(t, u2(t − c))K∫

Y3


+ supu2∈∂Pρ

K

∫

Y2

‖Gn(·, s)‖w(s)f(s, u2(s − c))Δs

≤ ζrK

∫b

a

‖Gn(·, s)‖w(s)Δs + S(ρ)− ζε2K

∫b

a


= ζrκ−1 − ζε2κ−1 + S

(ρ)

< ζ = ‖u2‖

(3.36)

yielding a contradiction with u2 ≤ Tu2 for all t ∈ [a, b]. This means that (3.32) holds.Therefore, from (3.28), (3.32) and Lemma 2.9, we conclude that the operator T has at leastone fixed point u∗ ∈ Pζ,ξ. From the definition of the cone P and (2.18), we see that u∗(t) > 0for all t ∈ (a, b). Thus, Proposition 3.2 implies that u∗ is a solution of BVP (1.1). So we obtainthe desired result.

Adopting the same argument as in Theorem 3.3 , we obtain the following results.

Corollary 3.4. Let ρ, ζ, r, fζρ be as in Theorem 3.3. Suppose that (ii) of Theorem 3.3 holds and

limξ→∞fξμξ

= +∞. If (C1), (C2), and (C3) holds , then boundary value problem (1.1) has at leastone positive solution u ∈ Pζ,η such that

u(t) =

⎧⎨

⎩

ψ(t), if t ∈ [a − c, a),

u∗∗(t), if t ∈ [a, b],

ζ ≤ ‖u∗∗‖ ≤ η, ζ < μη.

(3.37)


Theorem 3.5. Assume that there exist positive constants ρi, ζi, ξi, r with ζi < μξi, r < κ and ζi ≥κS(ρi)/(κ − r), i = 1, 2, . . . , m such that

(iii) fξiμξi

> R and fζiρi < r;

(iv) f(t, ψ(t − c))/u(t) < r, for all t ∈ Y1 and u ∈ [ρi, ζi].

If (C1), (C2), and (C3) hold, then boundary value problem (1.1) has at least m positive solutionsui ∈ Pζi,ξi such that for i = 1, 2, . . . , m

ui(t) =

⎧⎨

⎩

ψ(t), if t ∈ [a − c, a),

u∗i (t), if t ∈ [a, b]

ζ ≤∥∥u∗i

∥∥ ≤ ξ.

(3.38)

Example 3.6. Let T = R. Consider the following singular three-point boundary value problemsfor delay four-order dynamic equations:

u(4)(t) + f(t, u(t − 1)) = 0, t ∈ [0, 4],

u(0) =12u(1), 2u(1) = u(4),

u′′(0) =12u′′(1), 2u′′(1) = u′′(4),

u(t) = et, t ∈ [−1, 0),

(3.39)

where, for any t ∈ [0, 4], ρ = 1, ζ = 1480, μ = 0.112, ξ = 13500, M1 = 1 and M2 = 1/502,

f(t, u(t − 1)) =

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

2M1u(t), (t, u) ∈ (1, 4]×[ξ,+∞),

M1u(t)

(

1+sinπ(u(t) − ϑφ

)

2(ϑ − ϑφ

) +cosπ(u(t) − ϑφ

)

2(ϑ − ϑφ

)

)

, (t, u) ∈ (1, 4]×[μξ, ξ

],

12M2u(t) cos

π(u(t) − μ

)

2(ϑφ − μ

) +2M1ϑφ sinπ(u(t) − μ

)

2(ϑφ − μ

) , (t, u) ∈ (1, 4]×[ζ, μξ

],

12M2u(t)

[

2 − sinπ(u(t) − �

)

2(μ − �

) − cosπ(u(t) − �

)

2(μ − �

)

]

, (t, u) ∈ (1, 4] ×[ρ, ζ],

ρu(t)−1/2 − u(t)1/2 +12M2ρ, (t, u) ∈ (1, 4] ×

(0, ρ],

12M2u(t), (t, u) ∈ [0, 1) × R.

(3.40)


Clearly, we know that

α =12, β = 0, γ = 2, η = 1, δ =

54, d =

12,

p =32, q =

72, hi(t) = min

{t

5,

4 − t8

}, i = 1, 2,

G(4, s) = 12s (s ∈ [0, 1]), G(4, s) = 4(4 − s) (s ∈ [1, 3]),

G(s, s) = (4 − s)(1 + s) (s ∈ [3, 4]).

(3.41)

Simple computations yield

K =∫4

0‖G1(·, s)‖ds =

∫1

012sds +

∫3

14(4 − s)ds +

∫4

3(1 + s)(4 − s)ds = 24.17,

L =∫4

0‖G1(·, s)h2(s)ds

=∫1

012s

s

5ds +

∫20/13

14(4 − s)s

5ds +

∫3

20/134(4 − s)4 − s

8ds +

∫4

3

(4 − s)2(1 + s)8

ds

= 4.695,

μ = min

{h1(p)L

K,h1(q)L

K

}

= 0.112,

R =

(

μK

∫7/2

3/2‖G2(·, s)‖ds

)−1

= 0.282,

κ =

[

K

∫4

0‖G2(·, s)‖ds

]−1

=1

24.172.

(3.42)

Obviously,

limm→∞

supu∈Pζ,η

K

∫

Ym

‖G2(·, s)‖ f(s, u(s − c))ds = 0, ∀ 0 < ζ < η. (3.43)

If (t, u) ∈ (1, 4] × (0, 1], then we have

h(t) = h(t)ρ ≤ u(t) ≤ ρ = 1. (3.44)

Therefore, we get

f(t, u(t − 1)) ≤ h(t)−1/2 − h(t)1/2 +12M2, for (t, u) ∈ (1, 4] × (0, 1]. (3.45)


From (3.25), it follows that

S(1) = supu∈∂P1

K

∫4

1‖G2(·, s)‖f(s, u(s − 1))ds

≤ K

∫20/13

112s

((5s

)1/2

−(s

5

)1/2

+12M2

)

ds

+K

∫3

20/134(4 − s)

((8

4 − s

)1/2

−(

4 − s8

)1/2

+12M2

)

ds

+K

∫4

3(1 + s)(4 − s)

((8

4 − s

)1/2

−(

4 − s8

)1/2

+12M2

)

ds

≤ 1120.

(3.46)

Thus,

ζ = 1480 ≥ κS(1)κ − r ≈ 1461.37, ξ = 13500, ζ < μξ. (3.47)

Therefore, by Theorem 3.3 , the BVP (3.39) has at least one positive solution u such that

u(t) =

⎧⎨

⎩

et, if t ∈ [−1, 0),

u∗(t), if t ∈ [0, 4],

1480 ≤ ‖u∗‖ ≤ 13500.

(3.48)

Acknowledgments

The authors would like to thank the referees for helpful comments and suggestions. The workwas supported partly by the NSF of China (10771202), the Research Fund for Shanghai KeyLaboratory of Modern Applied Mathematics (08DZ2271900), and the Specialized ResearchFund for the Doctoral Program of Higher Education of China (2007035805).

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Research ArticleRecent Existence Results for Second-OrderSingular Periodic Differential Equations

Jifeng Chu1, 2 and Juan J. Nieto3

1 Department of Mathematics, College of Science, Hohai University, Nanjing 210098, China2 Department of Mathematics, Pusan National University, Busan 609-735, South Korea3 Departamento de Analisis Matematico, Facultad de Matematicas, Universidad de Santiago de Compostela,15782 Santiago de Compostela, Spain

Correspondence should be addressed to Jifeng Chu, [email protected]

Received 12 February 2009; Accepted 29 April 2009


We present some recent existence results for second-order singular periodic differential equations.A nonlinear alternative principle of Leray-Schauder type, a well-known fixed point theorem incones, and Schauder’s fixed point theorem are used in the proof. The results shed some light onthe differences between a strong singularity and a weak singularity.

Copyright q 2009 J. Chu and J. J. Nieto. 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.

1. Introduction

The main aim of this paper is to present some recent existence results for the positive T -periodic solutions of second order differential equation

x′′ + a(t)x = f(t, x) + e(t), (1.1)

where a(t), e(t) are continuous and T -periodic functions. The nonlinearity f(t, x) iscontinuous in (t, x) and T -periodic in t. We are mainly interested in the case that f(t, x) has arepulsive singularity at x = 0:

limx→ 0+

f(t, x) = +∞, uniformly in t. (1.2)

It is well known that second order singular differential equations describe manyproblems in the applied sciences, such as the Brillouin focusing system [1] and nonlinearelasticity [2]. Therefore, during the last two decades, singular equations have attractedmany researchers, and many important results have been proved in the literature; see, for


example, [3–10]. Recently, it has been found that a particular case of (1.1), the Ermakov-Pinney equation

x′′ + a(t)x =1x3

(1.3)

plays an important role in studying the Lyapunov stability of periodic solutions ofLagrangian equations [11–13].

In the literature, two different approaches have been used to establish the existenceresults for singular equations. The first one is the variational approach [14–16], andthe second one is topological methods. Because we mainly focus on the applications oftopological methods to singular equations in this paper, here we try to give a brief sketchof this problem. As far as the authors know, this method was started with the pioneeringpaper of Lazer and Solimini [17]. They proved that a necessary and sufficient condition forthe existence of a positive periodic solution for equation

x′′ =1xλ

+ e(t) (1.4)

is that the mean value of e is negative, e < 0, here λ ≥ 1, which is a strong forcecondition in a terminology first introduced by Gordon [18]. Moreover, if 0 < λ < 1, whichcorresponds to a weak force condition, they found examples of functions e with negativemean values and such that periodic solutions do not exist. Since then, the strong forcecondition became standard in the related works; see, for instance, [2, 8–10, 13, 19–21], andthe recent review [22]. With a strong singularity, the energy near the origin becomes infinityand this fact is helpful for obtaining the a priori bounds needed for a classical application ofthe degree theory. Compared with the case of a strong singularity, the study of the existenceof periodic solutions under the presence of a weak singularity by topological methods ismore recent but has also attracted many researchers [4, 6, 23–28]. In [27], for the first timein this topic, Torres proved an existence result which is valid for a weak singularity whereasthe validity of such results under a strong force assumption remains as an open problem.Among topological methods, the method of upper and lower solutions [6, 29, 30], degreetheory [8, 20, 31], some fixed point theorems in cones for completely continuous operators[25, 32–34], and Schauder’s fixed point theorem [27, 35, 36] are the most relevant tools.

In this paper, we select several recent existence results for singular equation (1.1)via different topological tools. The remaining part of the paper is organized as follows. InSection 2, some preliminary results are given. In Section 3, we present the first existence resultfor (1.1) via a nonlinear alternative principle of Leray-Schauder. In Section 4, the secondexistence result is established by using a well-known fixed point theorem in cones. Thecondition imposed on a(t) in Sections 3 and 4 is that the Green function G(t, s) associatedwith the linear periodic equations is positive, and therefore the results cannot cover thecritical case, for example, when a is a constant, a(t) = k2, 0 < k <

√λ1 = π/T , and λ1 is

the first eigenvalue of the linear problem with Dirichlet conditions x(0) = x(T) = 0. Differentfrom Sections 3 and 4, the results obtained in Section 5, which are established by Schauder’sfixed point theorem, can cover the critical case because we only need that the Green functionG(t, s) is nonnegative. All results in Sections 3–5 shed some lights on the differences betweena strong singularity and a weak singularity.


To illustrate our results, in Sections 3–5, we have selected the following singularequation:

x” + a(t)x = x−α + μxβ + e(t), (1.5)

here a, e ∈ C[0, T], α, β > 0, and μ ∈ R is a given parameter. The corresponding results arealso valid for the general case

x” + a(t)x =b(t)xα

+ μc(t)xβ + e(t), (1.6)

with b, c ∈ C[0, T]. Some open problems for (1.5) or (1.6) are posed.In this paper, we will use the following notation. Given ψ ∈ L1[0, T], we write ψ � 0

if ψ ≥ 0 for a.e. t ∈ [0, T], and it is positive in a set of positive measure. For a given functionp ∈ L1[0, T] essentially bounded, we denote the essential supremum and infimum of p by p∗

and p∗, respectively.

2. Preliminaries

Consider the linear equation

x′′ + a(t)x = p(t) (2.1)

with periodic boundary conditions

x(0) = x(T), x′(0) = x′(T). (2.2)

In Sections 3 and 4, we assume that

(A) the Green function G(t, s), associated with (2.1)–(2.2), is positive for all (t, s) ∈[0, T] × [0, T].

In Section 5, we assume that

(B) the Green function G(t, s), associated with (2.1)–(2.2), is nonnegative for all (t, s) ∈[0, T] × [0, T].

When a(t) = k2, condition (A) is equivalent to 0 < k2 < λ1 = (π/T)2 and condition (B)is equivalent to 0 < k2 ≤ λ1. In this case, we have

G(t, s) =

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

sin k(t − s) + sin k(T − t + s)2k(1 − cos kT)

, 0 ≤ s ≤ t ≤ T,

sin k(s − t) + sin k(T − s + t)2k(1 − cos kT)

, 0 ≤ t ≤ s ≤ T.

(2.3)


For a nonconstant function a(t), there is an Lp-criterion proved in [37], which is givenin the following lemma for the sake of completeness. Let K(q) denote the best Sobolevconstant in the following inequality:

C‖u‖2q ≤∥∥u′∥∥2

2, ∀u ∈ H10(0, T). (2.4)

The explicit formula for K(q) is

K(q)=

⎧⎪⎪⎪⎨

⎪⎪⎪⎩

2πqT1+2/q

(2

2 + q

)1−2/q( Γ(1/q)Γ(1/2 + 1/q)

)2

if 1 ≤ q <∞,

4T, if q =∞,

(2.5)

where Γ is the Gamma function; see [21, 38]

Lemma 2.1. Assume that a(t) � 0 and a ∈ Lp[0, T] for some 1 ≤ p ≤ ∞. If

‖a‖p < K(2p), (2.6)

then the condition (A) holds. Moreover, condition (B) holds if

‖a‖p ≤ K(2p). (2.7)

When the hypothesis (A) is satisfied, we denote

m = min0≤s,t≤T

G(t, s), M = max0≤s,t≤T

G(t, s), σ =m

M. (2.8)

Obviously, M > m > 0 and 0 < σ < 1.Throughout this paper, we define the function γ : R → R by

γ(t) =∫T

0G(t, s)e(s)ds, (2.9)

which corresponds to the unique T -periodic solution of

x′′ + a(t)x = e(t). (2.10)

3. Existence Result (I)

In this section, we state and prove the first existence result for (1.1). The proof is based on thefollowing nonlinear alternative of Leray-Schauder, which can be found in [39]. This part canbe regarded as the scalar version of the results in [4].


Lemma 3.1. Assume Ω is a relatively compact subset of a convex set K in a normed space X. LetT : Ω → K be a compact map with 0 ∈ Ω. Then one of the following two conclusions holds:

(a) T has at least one fixed point in Ω;

(b) thereexist x ∈ ∂Ω and 0 < λ < 1 such that x = λTx.

Theorem 3.2. Suppose that a(t) satisfies (A) and f(t, x) satisfies the following.

(H1) There exist constants σ > 0 and ν ≥ 1 such that

f(t, x) ≥ σx−ν, ∀t ∈ [0, T], ∀0 < x � 1. (3.1)

(H2) There exist continuous, nonnegative functions g(x) and h(x) such that

0 ≤ f(t, x) ≤ g(x) + h(x) ∀(t, x) ∈ [0, T] × (0,∞), (3.2)

g(x) > 0 is nonincreasing and h(x)/g(x) is nondecreasing in x ∈ (0,∞).

(H3) There exists a positive number r such that σr + γ∗ > 0, and

r

g(σr + γ∗

){1 + h

(r + γ∗

)/g(r + γ∗

)} > ω∗, here ω(t) =∫T

0G(t, s)ds. (3.3)

Then for each e ∈ C(R/TZ,R), (1.1) has at least one positive periodic solution x withx(t) > γ(t) for all t and 0 < ‖x − γ‖ < r.

Proof. The existence is proved using the Leray-Schauder alternative principle, together witha truncation technique. The idea is that we show that

x′′ + a(t)x = f(t, x(t) + γ(t)

)(3.4)

has a positive periodic solution x satisfying x(t) + γ(t) > 0 for t and 0 < ‖x‖ < r. If this istrue, it is easy to see that u(t) = x(t) + γ(t) will be a positive periodic solution of (1.1) with0 < ‖u − γ‖ < r since

u′′ + a(t)u = x′′ + γ” + a(t)x + a(t)γ = f(t, x + γ

)+ e(t) = f(t, u) + e(t). (3.5)

Since (H3) holds, we can choose n0 ∈ {1, 2, · · · } such that 1/n0 < σr + γ∗ and

ω∗g(σr + γ∗

){

1 +h(r + γ∗

)

g(r + γ∗

)

}

+1n0

< r. (3.6)

Let N0 = {n0, n0 + 1, · · · }. Consider the family of equations

x′′ + a(t)x = λfn(t, x(t) + γ(t)

)+a(t)n

, (3.7)


where λ ∈ [0, 1], n ∈N0, and

fn(t, x) =

⎧⎪⎪⎨

⎪⎪⎩

f(t, x), if x ≥ 1n,

f

(t,

1n

), if x ≤ 1

n.

(3.8)

Problem (3.7) is equivalent to the following fixed point problem:

x = λTnx +1n, (3.9)

where Tn is defined by

(Tnx)(t) = λ

∫T

0G(t, s)fn

(s, x(s) + γ(s)

)ds +

1n. (3.10)

We claim that any fixed point x of (3.9) for any λ ∈ [0, 1] must satisfy ‖x‖/= r.Otherwise, assume that x is a fixed point of (3.9) for some λ ∈ [0, 1] such that ‖x‖ = r.Note that

x(t) − 1n= λ

∫T

0G(t, s)fn

(s, x(s) + γ(s)

)ds

≥ λm

∫T

0fn(s, x(s) + γ(s)

)ds

= σMλ

∫T

0fn(s, x(s) + γ(s)

)ds

≥ σ maxt∈[0,T]

{

λ

∫T

0G(t, s)fn

(s, x(s) + γ(s)

)ds

}

= σ

∥∥∥∥x −1n

∥∥∥∥.

(3.11)

By the choice of n0, 1/n ≤ 1/n0 < σr + γ∗. Hence, for all t ∈ [0, T], we have

x(t) ≥ σ

∥∥∥∥x −1n

∥∥∥∥ +1n≥ σ

(‖x‖ − 1

n

)+

1n≥ σr. (3.12)

Therefore,

x(t) + γ(t) ≥ σr + γ∗ >1n. (3.13)


Thus we have from condition (H2), for all t ∈ [0, T],

x(t) = λ

∫T

0G(t, s)fn

(s, x(s) + γ(s)

)ds +

1n

= λ

∫T

0G(t, s)f

(s, x(s) + γ(s)

)ds +

1n

≤∫T

0G(t, s)f

(s, x(s) + γ(s)

)ds +

1n

≤∫T

0G(t, s)g

(x(s) + γ(s)

){

1 +h(x(s) + γ(s)

)

g(x(s) + γ(s)

)

}

ds +1n

≤ g(σr + γ∗

){

1 +h(r + γ∗

)

g(r + γ∗

)

}∫T

0G(t, s)ds +

1n

≤ g(σr + γ∗

){

1 +h(r + γ∗

)

g(r + γ∗

)

}

ω∗ +1n0

.

(3.14)

Therefore,

r = ‖x‖ ≤ g(σr + γ∗

){

1 +h(r + γ∗

)

g(r + γ∗

)

}

ω∗ +1n0

. (3.15)

This is a contradiction to the choice of n0, and the claim is proved.From this claim, the Leray-Schauder alternative principle guarantees that

x = Tnx +1n

(3.16)

has a fixed point, denoted by xn, in Br = {x ∈ X : ‖x‖ < r}, that is, equation

x′′ + a(t)x = fn(t, x(t) + γ(t)

)+a(t)n

(3.17)

has a periodic solution xn with ‖xn‖ < r. Since xn(t) ≥ 1/n > 0 for all t ∈ [0, T] and xn isactually a positive periodic solution of (3.17).

In the next lemma, we will show that there exists a constant δ > 0 such that

xn(t) + γ(t) ≥ δ, ∀t ∈ [0, T], (3.18)

for n large enough.In order to pass the solutions xn of the truncation equations (3.17) to that of the original

equation (3.4), we need the following fact:

∥∥x′n∥∥ ≤ H (3.19)


for some constant H > 0 and for all n ≥ n0. To this end, by the periodic boundary conditions,x′n(t0) = 0 for some t0 ∈ [0, T]. Integrating (3.17) from 0 to T , we obtain

∫T

0a(t)xn(t)dt =

∫T

0

[fn(t, xn(t) + γ(t)

)+a(t)n

]dt. (3.20)

Therefore

∥∥x′n∥∥ = max

0≤t≤T

∣∣x′n(t)

∣∣ = max

0≤t≤T

∣∣∣∣∣

∫ t

t0

x′′n(s)ds

∣∣∣∣∣

= max0≤t≤T

∣∣∣∣∣

∫ t

t0

[fn(s, xn(s) + γ(s)

)+a(s)n− a(s)xn(s)

]ds

∣∣∣∣∣

≤∫T

0

[fn(s, xn(s) + γ(s)

)+a(s)n

]ds +

∫T

0a(s)xn(s)ds

= 2∫T

0a(s)xn(s)ds < 2r‖a‖1 = H.

(3.21)

The fact ‖xn‖ < r and (3.19) show that {xn}n∈N0is a bounded and equicontinuous

family on [0, T]. Now the Arzela-Ascoli Theorem guarantees that {xn}n∈N0has a subsequence,

{xnk}k∈N, converging uniformly on [0, T] to a function x ∈ X. Moreover, xnk satisfies theintegral equation

xnk(t) =∫T

0G(t, s)f

(s, xnk(s) + γ(s)

)ds +

1nk

. (3.22)

Letting k → ∞, we arrive at

x(t) =∫T

0G(t, s)f

(s, x(s) + γ(s)

)ds, (3.23)

where the uniform continuity of f(t, x) on [0, T]× [δ, r + γ∗] is used. Therefore, x is a positiveperiodic solution of (3.4).

Lemma 3.3. There exist a constant δ > 0 and an integer n2 > n0 such that any solution xn of (3.17)satisfies (3.18) for all n ≥ n2.

Proof. The lower bound in (3.18) is established using the strong force condition (H1) of f(t, x).By condition (H1), there exists c0 ∈ (0, 1) small enough such that

f(t, x) ≥ σc−ν0 > max{r‖a‖1, a

∗(r + γ∗)+ e∗}, ∀0 ≤ t ≤ T, 0 < x ≤ c0. (3.24)


Take n1 ∈N0 such that 1/n1 ≤ c0 and let N1 = {n1, n1 + 1, · · · }. For n ∈N1, let

αn = min0≤t≤T

[xn(t) + γ(t)

], βn = max

0≤t≤T

[xn(t) + γ(t)

]. (3.25)

We claim first that βn > c0 for alln ∈ N1. Otherwise, suppose that βn ≤ c0 for somen ∈N1. Then from (3.24), it is easy to verify

fn(t, xn(t) + γ(t)

)> r‖a‖1. (3.26)

Integrating (3.17) from 0 to T , we deduce that

0 =∫T

0

[x”n(t) + a(t)xn(t) − fn

(t, xn(t) + γ(t)

)− a(t)

n

]dt

=∫T

0a(t)xn(t)dt −

(1n

)∫T

0a(t)dt −

∫T

0fn(t, xn(t) + γ(t)

)dt

<

∫T

0a(t)xn(t)dt − r‖a‖1 ≤ 0.

(3.27)

This is a contradiction. Thus βn > c0 for n ∈N1.Now we consider the minimum values αn. Let n ≥ n1. Without loss of generality, we

assume that αn < c0, otherwise we have (3.18). In this case,

αn = min0≤t≤T

[xn(t) + γ(t)

]= xn(tn) + γ(tn) < c0 (3.28)

for some tn ∈ [0, T]. As βn > c0, there exists cn ∈ [0, 1] (without loss of generality, we assumetn < cn) such that xn(cn) + γ(cn) = c0 and xn(t) + γ(t) ≤ c0 for tn ≤ t ≤ cn. By (3.24), it can bechecked that

fn(t, xn(t) + γ(t)

)> a(t)

(xn(t) + γ(t)

)+ e(t), ∀t ∈ [tn, cn]. (3.29)

Thus for t ∈ (tn, cn], we have x”n(t) + γ”(t) > 0. As x′n(tn) + γ ′(tn) = 0, x′n(t) + γ ′(t) > 0

for all t ∈ (tn, cn] and the function yn := xn + γ is strictly increasing on [tn, cn]. We use ξn todenote the inverse function of yn restricted to [tn, cn].

In order to prove (3.18) in this case, we first show that, for n ∈N1,

xn(t) + γ(t) ≥ 1n. (3.30)


Otherwise, suppose that αn < 1/n for some n ∈N1. Then there would exist bn ∈ (tn, cn)such that xn(bn) + γ(bn) = 1/n and

xn(t) + γ(t) ≤ 1n

for tn ≤ t ≤ bn,1n≤ xn(t) + γ(t) ≤ c0 for bn ≤ t ≤ cn. (3.31)

Multiplying (3.17) by x′n(t) + γ ′(t) and integrating from bn to cn, we obtain

∫R1

1/nf(ξn(y), y)dy =

∫ cn

bn

f(t, xn(t) + γ(t)

)(x′n(t) + γ ′(t)

)dt

=∫ cn

bn

fn(t, xn(t) + γ(t)

)(x′n(t) + γ ′(t)

)dt

=∫ cn

bn

(x′′n(t) + a(t)xn(t) −

a(t)n

)(x′n(t) + γ ′(t)

)dt

=∫ cn

bn

x′′n(t)(x′n(t) + γ ′(t)

)dt

+∫ cn

bn

(a(t)xn(t) −

a(t)n

)(x′n(t) + γ ′(t)

)dt.

(3.32)

By the facts ‖xn‖ < r and ‖x′n‖ ≤ H, one can easily obtain that the right side of the aboveequality is bounded. As a consequence, there exists L > 0 such that

∫R1

1/nf(ξn(y), y)dy ≤ L. (3.33)

On the other hand, by the strong force condition (H1), we can choose n2 ∈ N1 largeenough such that

∫ c0

1/nf(ξn(y), y)dy ≥ σ

∫ c0

1/ny−νdy > L (3.34)

for all n ∈N2 = {n2, n2 + 1, · · · }. So (3.30) holds for n ∈N2.Finally, multiplying (3.17) by x′n(t) + γ ′(t) and integrating from tn to cn, we obtain

∫ c0

αn

f(ξn(y), y)dy =

∫ cn

tn

f(t, xn(t) + γ(t)

)(x′n(t) + γ ′(t)

)dt

=∫ cn

tn

fn(t, xn(t) + γ(t)

)(x′n(t) + γ ′(t)

)dt

=∫ cn

tn

(x′′n(t) + a(t)xn(t) − a(t)

n

)(x′n(t) + γ ′(t)

)dt.

(3.35)


(We notice that the estimate (3.30) is used in the second equality above). In the same way,one may readily prove that the right-hand side of the above equality is bounded. On theother hand, if n ∈N2, by (H1),

∫ c0

αn

f(ξn(y), y)dy ≥ σ

∫ c0

αn

y−νdy −→ +∞ (3.36)

if αn → 0+. Thus we know that αn ≥ δ for some constant δ > 0.

From the proof of Theorem 3.2 and Lemma 3.3, we see that the strong force condition(H1) is only used when we prove (3.18). From the next theorem, we will show that, for thecase γ∗ ≥ 0, we can remove the strong force condition (H1), and replace it by one weak forcecondition.

Theorem 3.4. Assume that (A) and (H2)–( H3) are satisfied. Suppose further that

(H4) for each constant L > 0, there exists a continuous function φL � 0 such that f(t, x) ≥ φL(t)for all (t, x) ∈ [0, T] × (0, L].

Then for each e(t) with γ∗ ≥ 0, (1.1) has at least one positive periodic solution x with x(t) > γ(t) forall t and 0 < ‖x − γ‖ < r.

Proof. We only need to show that (3.18) is also satisfied under condition (H4) and γ∗ ≥ 0. Therest parts of the proof are in the same line of Theorem 3.2. Since (H4) holds, there exists acontinuous function φr+γ∗ � 0 such that f(t, x) ≥ φr+γ∗(t) for all (t, x) ∈ [0, T] × (0, r + γ∗]. Letxr+γ∗ be the unique periodic solution to the problems (2.1)–(2.2) with h = φr+γ∗ . That is

xr+γ∗(t) =∫T

0G(t, s)φr+γ∗(s)ds. (3.37)

Then we have

xr+γ∗(t) + γ(t) =∫T

0G(t, s)φr+γ∗(s)ds + γ(t) ≥ Φ∗ + γ∗ > 0, (3.38)

here

Φ(t) =∫T

0G(t, s)φr+γ∗(s)ds. (3.39)

Corollary 3.5. Assume that a(t) satisfies (A) and α > 0, β ≥ 0, μ > 0. Then

(i) if α ≥ 1, β < 1, then for each e ∈ C(R/TZ,R), (1.5) has at least one positive periodicsolution for all μ > 0;

(ii) if α ≥ 1, β ≥ 1, then for each e ∈ C(R/TZ,R), (1.5) has at least one positive periodicsolution for each 0 < μ < μ1, here μ1 is some positive constant.


(iii) if α > 0, β < 1, then for each e ∈ C(R/TZ,R) with γ∗ ≥ 0, (1.5) has at least one positiveperiodic solution for all μ > 0;

(iv) if α > 0, β ≥ 1, then for each e ∈ C(R/TZ,R) with γ∗ ≥ 0, (1.5) has at least one positiveperiodic solution for each 0 < μ < μ1.

Proof. We apply Theorems 3.2 and 3.4. Take

g(x) = x−α, h(x) = μxβ, (3.40)

then (H2) is satisfied, and the existence condition (H3) becomes

μ <r(σr + γ∗

)α −ω∗

ω∗(r + γ∗

)α+β (3.41)

for some r > 0. Note that condition (H1) is satisfied when α ≥ 1, while (H4) is satisfied whenα > 0. So (1.5) has at least one positive periodic solution for

0 < μ < μ1 := supr>0

r(σr + γ∗

)α −ω∗

ω∗(r + γ∗

)α+β . (3.42)

Note that μ1 =∞ if β < 1 and μ1 <∞ if β ≥ 1. Thus we have (i)–(iv).

4. Existence Result (II)

In this section, we establish the second existence result for (1.1) using a well-known fixedpoint theorem in cones. We are mainly interested in the superlinear case. This part isessentially extracted from [24].

First we recall this fixed point theorem in cones, which can be found in [40]. Let K bea cone in X and D is a subset of X, we write DK = D ∩K and ∂KD = (∂D) ∩K.

Theorem 4.1 (see [40]). Let X be a Banach space and K a cone in X. Assume Ω1,Ω2 are open

bounded subsets of X with Ω1K /= ∅,Ω

1K ⊂ Ω2

K. Let

T : Ω2K −→ K (4.1)

be a completely continuous operator such that

(a) ‖Tx‖ ≤ ‖x‖ for x ∈ ∂KΩ1,

(b) There exists υ ∈ K \ {0} such that x /= Tx + λυ for all x ∈ ∂KΩ2 and all λ > 0.

Then T has a fixed point in Ω2K \Ω1

K.


In applications below, we take X = C[0, T] with the supremum norm ‖ · ‖ and define

K ={x ∈ X : x(t) ≥ 0 ∀t ∈ [0, T], min

0≤t≤Tx(t) ≥ σ‖x‖

}. (4.2)

Theorem 4.2. Suppose that a(t) satisfies (A) and f(t, x) satisfies (H2)–(H3). Furthermore, assumethat

(H5) there exist continuous nonnegative functions g1(x), h1(x) such that

f(t, x) ≥ g1(x) + h1(x), ∀(t, x) ∈ [0, T] × (0,∞), (4.3)

g1(x) > 0 is nonincreasing and h1(x)/g1(x) is nondecreasing in x;

(H6) there exists R > 0 with σR > r such that

σR

g1(R + γ∗

){1 + h1

(σR + γ∗

)/g1(σR + γ∗

)} ≤ ω∗. (4.4)

Then (1.1) has one positive periodic solution x with r < ‖x − γ‖ ≤ R.

Proof. As in the proof of Theorem 3.2, we only need to show that (3.4) has a positive periodicsolution u ∈ X with u(t) + γ(t) > 0 and r < ‖u‖ ≤ R.

Let K be a cone in X defined by (4.2). Define the open sets

Ω1 = {x ∈ X : ‖x‖ < r}, Ω2 = {x ∈ X : ‖x‖ < R}, (4.5)

and the operator T : Ω2K → K by

(Tx)(t) =∫T

0G(t, s)f

(s, x(s) + γ(s)

)ds, 0 ≤ t ≤ T. (4.6)

For each x ∈ Ω2K \ Ω1

K, we have r ≤ ||x|| ≤ R. Thus 0 < σr + γ∗ ≤ x(t) + γ(t) ≤ R + γ∗

for all t ∈ [0, T]. Since f : [0, T] × [σr + γ∗, R + γ∗] → [0,∞) is continuous, then the operator

T : Ω2K \ Ω1

K → K is well defined and is continuous and completely continuous. Next weclaim that:

(i) ‖Tx‖ ≤ ‖x‖ for x ∈ ∂KΩ1, and

(ii) there exists υ ∈ K \ {0} such that x /= Tx + λυ for all x ∈ ∂KΩ2 and all λ > 0.


We start with (i). In fact, if x ∈ ∂KΩ1, then ‖x‖ = r and σr + γ∗ ≤ x(t) + γ(t) ≤ r + γ∗ forall t ∈ [0, T]. Thus we have

(Tx)(t) =∫T

0G(t, s)f

(s, x(s) + γ(s)

)ds

≤∫T

0G(t, s)g

(x(s) + γ(s)

){

1 +h(x(s) + γ(s)

)

g(x(s) + γ(s)

)

}

ds

≤ g(σr + γ∗

){

1 +h(r + γ∗

)

g(r + γ∗

)

}∫T

0G(t, s)ds

≤ g(σr + γ∗

){

1 +h(r + γ∗

)

g(r + γ∗

)

}

ω∗ < r = ‖x‖.

(4.7)

Next we consider (ii). Let υ(t) ≡ 1, then υ ∈ K \ {0}. Next, suppose that there existsx ∈ ∂KΩ2 and λ > 0 such that x = Tx + λυ. Since x ∈ ∂KΩ2, then σR + γ∗ ≤ x(t) + γ(t) ≤ R + γ∗

for all t ∈ [0, T]. As a result, it follows from (H5) and (H6) that, for all t ∈ [0, T],

x(t) = (Tx)(t) + λ =∫T

0G(t, s)f

(s, x(s) + γ(s)

)ds + λ

≥∫T

0G(t, s)g1

(x(s) + γ(s)

){

1 +h1(x(s) + γ(s)

)

g1(x(s) + γ(s)

)

}

ds + λ

≥ g1(R + γ∗

){

1 +h1(σR + γ∗

)

g1(σR + γ∗

)

}∫T

0G(t, s)ds + λ

≥ g1(R + γ∗

){

1 +h1(σR + γ∗

)

g1(σR + γ∗

)

}

ω∗ + λ

> g1(R + γ∗

){

1 +h1(σR + γ∗

)

g1(σR + γ∗

)

}

ω∗ ≥ σR.

(4.8)

Hence min0≤t≤T

x(t) > σR, this is a contradiction and we prove the claim.

Now Theorem 4.1 guarantees that T has at least one fixed point x ∈ Ω2K \ Ω1

K withr ≤ ‖x‖ ≤ R. Note ‖x‖/= r by (4.7).

Combined Theorem 4.2 with Theorems 3.2 or 3.4, we have the following twomultiplicity results.

Theorem 4.3. Suppose that a(t) satisfies (A) and f(t, x) satisfies (H1)–( H3) and (H5)–( H6). Then(1.1) has two different positive periodic solutions x and x with 0 < ‖x − γ‖ < r < ‖x − γ‖ ≤ R.

Theorem 4.4. Suppose that a(t) satisfies (A) and f(t, x) satisfies (H2)–( H6). Then (1.1) has twodifferent positive periodic solutions x and x with 0 < ‖x − γ‖ < r < ‖x − γ‖ ≤ R.


Corollary 4.5. Assume that a(t) satisfies (A) and α > 0, β > 1, μ > 0. Then

(i) if α ≥ 1, then for each e ∈ C(R/TZ,R), (1.5) has at least two positive periodic solutionsfor each 0 < μ < μ1;

(ii) if α > 0, then for each e ∈ C(R/TZ,R) with γ∗ ≥ 0, (1.5) has at least two positive periodicsolutions for each 0 < μ < μ1.

Proof. Take g1(x) = x−α, h1(x) = μxβ. Then (H5) is satisfied and the existence condition (H6)becomes

μ ≥σR(R + γ∗

)α −ω∗ω∗(σR + γ∗

)α+β . (4.9)

Since β > 1, it is easy to see that the right-hand side goes to 0 as R → +∞. Thus, for any given0 < μ < μ1, it is always possible to find such R � r that (4.9) is satisfied. Thus, (1.5) has anadditional positive periodic solution x.

5. Existence Result (III)

In this section, we prove the third existence result for (1.1) by Schauder’s fixed point theorem.We can cover the critical case because we assume that the condition (B) is satisfied. This partcomes essentially from [35], and the results for the vector version can be found in [4].

Theorem 5.1. Assume that conditions (B) and (H2), (H4) are satisfied. Furthermore, suppose that

(H7) there exists a positive constant R > 0 such that R > Φ∗,Φ∗+γ∗ > 0 and R ≥ g(Φ∗+γ∗){1+h(R + γ∗)/g(R + γ∗)}ω∗, here Φ∗ = mintΦ(t), Φ(t) =

∫T0G(t, s)φR+γ∗(s)ds.

Then (1.1) has at least one positive T -periodic solution.

Proof. A T -periodic solution of (1.1) is just a fixed point of the map T : X → X defined by(4.6). Note that T is a completely continuous map.

Let R be the positive constant satisfying (H7) and r = Φ∗ > 0. Then we have R > r > 0.Now we define the set

Ω = {x ∈ X : r ≤ x(t) ≤ R ∀t}. (5.1)

Obviously, Ω is a closed convex set. Next we prove T(Ω) ⊂ Ω.In fact, for each x ∈ Ω, using that G(t, s) ≥ 0 and condition (H4),

(Tx)(t) ≥∫T

0G(t, s)φR+γ∗(s)ds ≥ Φ∗ = r > 0. (5.2)


On the other hand, by conditions (H2) and (H7), we have

(Tx)(t) ≤∫T

0G(t, s)g

(x(s) + γ(s)

){

1 +h(x(s) + γ(s)

)

g(x(s) + γ(s)

)

}

ds

≤ g(Φ∗ + γ∗

){

1 +h(R + γ∗

)

g(R + γ∗

)

}

ω∗ ≤ R.

(5.3)

In conclusion, T(Ω) ⊂ Ω. By a direct application of Schauder’s fixed point theorem, the proofis finished.

As an application of Theorem 5.1, we consider the case γ∗ = 0. The following corollaryis a direct result of Theorem 5.1.

Corollary 5.2. Assume that conditions (B) and (H2), (H4) are satisfied. Furthermore, assume that

(H8) there exists a positive constant R > 0 such that R > Φ∗ and

R ≥ g(Φ∗)

{

1 +h(R + γ∗

)

g(R + γ∗

)

}

ω∗. (5.4)

If γ∗ = 0, then (1.1) has at least one positive T -periodic solution.

Corollary 5.3. Suppose that a satisfies (B) and 0 < α < 1, β ≥ 0, then for each e(t) ∈ C(R/TZ,R)with γ∗ = 0,one has the following:

(i) if α + β < 1 − α2, then (1.5) has at least one positive periodic solution for each μ ≥ 0.

(ii) if α+β ≥ 1−α2, then (1.5) has at least one positive T -periodic solution for each 0 ≤ μ < μ2,where μ2 is some positive constant.

Proof. We apply Corollary 3.5 and follow the same notation as in the proof of Corollary 3.5.Then (H2) and (H4) are satisfied, and the existence condition (H8) becomes

μ <RΦα

∗ −ω∗

ω∗(R + γ∗

)α+β , (5.5)

for some R > 0 with R > Φ∗. Note that

Φ∗ =(R + γ∗

)−αω∗. (5.6)

Therefore, (5.5) becomes

μ <R(R + γ∗

)−α2

ωα∗ −ω∗

ω∗(R + γ∗

)α+β , (5.7)

for some R > 0.


So (1.5) has at least one positive T -periodic solution for

0 < μ < μ2 = supR>0

R(R + γ∗

)−α2

ωα∗ −ω∗

ω∗(R + γ∗

)α+β . (5.8)

Note that μ2 =∞ if α + β < 1 − α2 and μ2 <∞ if α + β ≥ 1 − α2. We have the desired results (i)and (ii).

Remark 5.4. The validity of (ii) in Corollary 5.3 under strong force conditions remains stillopen to us. Such an open problem has been partially solved by Corollary 3.5. However, we donot solve it completely because we need the positivity of G(t, s) in Corollary 3.5, and thereforeit is not applicable to the critical case. The validity for the critical case remains open to theauthors.

The next results explore the case when γ∗ > 0.

Theorem 5.5. Suppose that a(t) satisfies (B) and f(t, x) satisfies condition (H2). Furthermore,assume that

(H9) there exists R > γ∗ such that

g(γ∗){

1 +h(R + γ∗

)

g(R + γ∗

)

}

ω∗ ≤ R. (5.9)

If γ∗ > 0, then (1.1) has at least one positive T -periodic solution.

Proof. We follow the same strategy and notation as in the proof of Theorem 5.1. Let R be thepositive constant satisfying (H9) and r = γ∗, then R > r > 0 since R > γ∗. Next we proveT(Ω) ⊂ Ω.

For each x ∈ Ω, by the nonnegative sign of G(t, s) and f(t, x), we have

(Tx)(t) =∫T

0G(t, s)f(s, x(s))ds + γ(t) ≥ γ∗ = r > 0. (5.10)

On the other hand, by (H2) and (H9), we have

(Tx)(t) ≤∫T

0G(t, s)g

(x(s) + γ(s)

){

1 +h(x(s) + γ(s)

)

g(x(s) + γ(s)

)

}

ds

≤ g(γ∗){

1 +h(R + γ∗

)

g(R + γ∗

)

}

ω∗ ≤ R.

(5.11)

In conclusion, T(Ω) ⊂ Ω, and the proof is finished by Schauder’s fixed point theorem.


Corollary 5.6. Suppose that a(t) satisfies (B) and α, β ≥ 0, then for each e ∈ C(R/TZ,R) withγ∗ > 0, one has the following:

(i) if α + β < 1, then (1.5) has at least one positive T -periodic solution for each μ ≥ 0;

(ii) if α + β ≥ 1, then (1.5) has at least one positive T -periodic solution for each 0 ≤ μ < μ3,where μ3 is some positive constant.

Proof. We apply Theorem 5.5 and follow the same notation as in the proof of Corollary 3.5.Then (H2) is satisfied, and the existence condition (H9) becomes

μ <Rγ∗α −ω∗

ω∗(R + γ∗

)α+β (5.12)

for some R > 0. So (1.5) has at least one positive T -periodic solution for

0 < μ < μ3 = supR>0

Rγ∗α −ω∗

ω∗(R + γ∗

)α+β . (5.13)

Note that μ3 = ∞ if α + β < 1 and μ3 < ∞ if α + β ≥ 1. We have the desired results (i) and(ii).

Acknowledgments

The authors express their thanks to the referees for their valuable comments and suggestions.The research of J. Chu is supported by the National Natural Science Foundation of China(Grant no. 10801044) and Jiangsu Natural Science Foundation (Grant no. BK2008356).The research of J. J. Nieto is partially supported by Ministerio de Education y Cienciaand FEDER, Project MTM2007-61724, and by Xunta de Galicia and FEDER, projectPGIDIT06PXIB207023PR.

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Research ArticleRobust Monotone Iterates for Nonlinear SingularlyPerturbed Boundary Value Problems

Igor Boglaev

Institute of Fundamental Sciences, Massey University, Private Bag 11-222,4442 Palmerston North, New Zealand

Correspondence should be addressed to Igor Boglaev, [email protected]

Received 8 April 2009; Accepted 11 May 2009


This paper is concerned with solving nonlinear singularly perturbed boundary value problems.Robust monotone iterates for solving nonlinear difference scheme are constructed. Uniformconvergence of the monotone methods is investigated, and convergence rates are estimated.Numerical experiments complement the theoretical results.

Copyright q 2009 Igor Boglaev. 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.

1. Introduction

We are interested in numerical solving of two nonlinear singularly perturbed problems ofelliptic and parabolic types.

The first one is the elliptic problem

−μ2u′′ + f(x, u) = 0, x ∈ ω = (0, 1), u(0) = 0, u(1) = 0,

fu ≥ c∗ = const > 0, (x, u) ∈ ω × (−∞,∞), fu = ∂f/∂u,(1.1)

where μ is a positive parameter, and f is sufficiently smooth function. For μ� 1 this problemis singularly perturbed, and the solution has boundary layers near x = 0 and x = 1 (see [1]for details).

The second problem is the one-dimensional parabolic problem

−μ2uxx + ut + f(x, t, u) = 0, (x, t) ∈ Q = ω × (0, T], ω = (0, 1),

u(0, t) = 0, u(1, t) = 0, u(x, 0) = u0(x), x ∈ ω,

fu ≥ 0, (x, t, u) ∈ Q × (−∞,∞),

(1.2)


where μ is a positive parameter. Under suitable continuity and compatibility conditions onthe data, a unique solution of this problem exists. For μ � 1 problem (1.2) is singularlyperturbed and has boundary layers near the lateral boundary of Q (see [2] for details).

In the study of numerical methods for nonlinear singularly perturbed problems, thetwo major points to be developed are: (i) constructing robust difference schemes (this meansthat unlike classical schemes, the error does not increase to infinity, but rather remainsbounded, as the small parameter approaches zero); (ii) obtaining reliable and efficientcomputing algorithms for solving nonlinear discrete problems.

Our goal is to construct a μ-uniform numerical method for solving problem (1.1), thatis, a numerical method which generates μ-uniformly convergent numerical approximationsto the solution. We use a numerical method based on the classical difference scheme and thepiecewise uniform mesh of Shishkin-type [3]. For solving problem (1.2), we use the implicitdifference scheme based on the piecewise uniform mesh in the x-direction, which convergesμ-uniformly [4].

A major point about the nonlinear difference schemes is to obtain reliable and efficientcomputational methods for computing the solution. The reliability of iterative techniquesfor solving nonlinear difference schemes can be essentially improved by using component-wise monotone globally convergent iterations. Such methods can be controlled every time.A fruitful method for the treatment of these nonlinear schemes is the method of upper andlower solutions and its associated monotone iterations [5]. Since an initial iteration in themonotone iterative method is either an upper or lower solution, which can be constructeddirectly from the difference equation without any knowledge of the exact solution, thismethod simplifies the search for the initial iteration as is often required in the Newtonmethod. In the context of solving systems of nonlinear equations, the monotone iterativemethod belongs to the class of methods based on convergence under partial ordering (see [5,Chapter 13] for details).

The purpose of this paper is to construct μ-uniformly convergent monotone iterativemethods for solving μ-uniformly convergent nonlinear difference schemes.

The structure of the paper is as follows. In Section 2, we prove that the classicaldifference scheme on the piecewise uniform mesh converges μ-uniformly to the solutionof problem (1.1). A robust monotone iterative method for solving the nonlinear differencescheme is constructed. In Section 3, we construct a robust monotone iterative method forsolving problem (1.2). In the final Section 4, we present numerical experiments whichcomplement the theoretical results.

2. The Elliptic Problem

The following lemma from [1] contains necessary estimates of the solution to (1.1).

Lemma 2.1. If u(x) ∈ C0(ω) ∩ C2(ω) is the solution to (1.1), the following estimates hold:

maxx∈ω|u(x)| ≤ c−1

∗ maxx∈ω

∣∣f(x, 0)∣∣,

∣∣u′(x)∣∣ ≤ C

[1 + μ−1Π(x)

],

Π(x) = exp(−√c∗μ

)+ exp

(−√c∗(1 − x)

μ

),

(2.1)

where constant C is independent of μ.


For μ� 1, the boundary layers appear near x = 0 and x = 1.

2.1. The Nonlinear Difference Scheme

Introduce a nonuniform mesh ωh

ωh = {xi, 0 ≤ i ≤N; x0 = 0, xN = 1; hi = xi+1 − xi}. (2.2)

For solving (1.1), we use the classical difference scheme

Lhv(x) + f(x, v) = 0, x ∈ ωh, v(0) = 0, v(1) = 0,

Lhvi = −μ2(�i)−1[(vi+1 − vi)(hi)

−1 − (vi − vi−1)(hi−1)−1],

(2.3)

where vi = v(xi) and �i = (hi−1 + hi)/2. We introduce the linear version of this problem

(Lh + c

)w(x) = f0(x), x ∈ ωh, w(0) = 0, w(1) = 0, (2.4)

where c(x) ≥ 0. We now formulate a discrete maximum principle for the difference operatorLh + c and give an estimate of the solution to (2.4).

Lemma 2.2. (i) If a mesh function w(x) satisfies the conditions

(Lh + c

)w(x) ≥ 0 (≤ 0), x ∈ ωh, w(0), w(1) ≥ 0 (≤ 0), (2.5)

then w(x) ≥ 0 (≤ 0), x ∈ ωh.(ii) If c(x) ≥ c∗ = const > 0, then the following estimate of the solution to (2.4) holds true:

‖w‖ωh ≤ max ‖f0‖ωh/c∗, (2.6)

where ‖w‖ωh = maxx∈ωh |w(x)|, ‖f0‖ωh = maxx∈ωh |f0(x)|.

The proof of the lemma can be found in [6].

2.2. Uniform Convergence on the Piecewise Uniform Mesh

We employ a layer-adapted mesh of a piecewise uniform type [3]. The piecewise uniformmesh is formed in the following manner. We divide the interval ω = [0, 1] into three parts[0, ς], [ς, 1−ς], and [1−ς, 1]. Assuming that N is divisible by 4, in the parts [0, ς], [1−ς, 1] weuse the uniform mesh with N/4 + 1 mesh points, and in the part [ς, 1 − ς] the uniform meshwith N/2 + 1 mesh points is in use. The transition points ς and 1 − ς are determined by

ς = min{

4−1,μ lnN√c∗

}. (2.7)


This defines the piecewise uniform mesh. If the parameter μ is small enough, then theuniform mesh inside of the boundary layers with the step size hμ = 4ςN−1 is fine, and theuniform mesh outside of the boundary layers with the step size h = 2(1 − 2ς)N−1 is coarse,such that

N−1 < h < 2N−1, hμ = 4μ(√

c∗N)−1 lnN. (2.8)

In the following theorem, we give the convergence property of the difference scheme(2.3).

Theorem 2.3. The difference scheme (2.3) on the piecewise uniform mesh (2.8) converges μ-uniformly to the solution of (1.1):

maxx∈ωh|v(x) − u(x)| ≤ CN−1 lnN, (2.9)

where constant C is independent of μ and N.

Proof. Using Green’s function Gi of the differential operator μ2d2/dx2 on [xi, xi+1], werepresent the exact solution u(x) in the form

u(x) = u(xi)φIi (x) + u(xi+1)φII

i (x) +∫xi+1

xi

Gi(x, s)f(s, u)ds,(2.10)

where the local Green function Gi is given by

Gi(x, s) =1

μ2wi(s)

⎧⎨

⎩

φIi (s)φ

IIi (x), x ≤ s,

φIi (x)φ

IIi (s), x ≥ s,

wi(s) = φIIi (s)

[φIi (x)]′

x=s− φI

i (s)[φIIi (x)

]′

x=s,

(2.11)

and φIi (x), φII

i (x) are defined by

φIi (x) =

xi+1 − xhi

, φIIi (x) =

x − xi

hi, xi ≤ x ≤ xi+1. (2.12)

Equating the derivatives du(xi − 0)/dx and du(xi + 0)/dx, we get the following integral-difference formula:

Lhu(xi) =1�i

∫xi

xi−1

φIIi−1(s)f(s)ds +

1�i

∫xi+1

xi

φIi (s)f(s)ds, (2.13)


where here and below we suppress variable u in f . Representing f(x) on [xi−1, xi] and[xi, xi+1] in the forms

f(x) = f(xi − 0) +∫x

xi

df

dsds, f(x) = f(xi + 0) +

∫x

xi

df

dsds, (2.14)

the above integral-difference formula can be written as

Lhu(x) + f(x, u) = Tr(x), x ∈ ωh, (2.15)

where the truncation error Tr(x) of the exact solution u(x) to (1.1) is defined by

Tr(xi) ≡ −1�i

∫xi

xi−1

φIIi−1(s)

(∫s

xi

df

dξdξ

)

ds − 1�i

∫xi+1

xi

φIi (s)

(∫ s

xi

df

dξdξ

)

ds. (2.16)

From here, it follows that

|Tr(xi)| ≤1�i

∫xi

xi−1

φIIi−1(s)

(∫xi

xi−1

∣∣∣∣df

dξ

∣∣∣∣dξ

)

ds +1�i

∫xi+1

xi

φIi (s)

(∫xi+1

xi

∣∣∣∣df

dξ

∣∣∣∣dξ

)

ds. (2.17)

From Lemma 2.1, the following estimate on df/dx holds:

∣∣∣∣df

dx

∣∣∣∣ ≤ C[1 + μ−1Π(x)

]. (2.18)

We estimate the truncation error Tr in (2.17) on the interval (0, 1/2]. Consider the followingthree cases: xi ∈ (0, ς), xi = ς, and xi ∈ (ς, 1/2]. If xi ∈ (0, ς), then hi−1 = hi = hμ, and takinginto account that Π(x) < 2 in (2.18), we have

|Tr(xi)| ≤ Chμ

(1 + 2μ−1

), xi ∈ (0, ς), (2.19)

where here and throughout C denotes a generic constant that is independent of μ and N.If xi = ς, then hi−1 = hμ, hi = h. Taking into account that ς = μ lnN/

√c∗, Π(x) < 2, and

Π(x) ≤ 2 exp(−√c∗x/μ), we have

|Tr(ς)| ≤ C

hμ + h

[h2μ

(1 + 2μ−1

)+ h2 + 2

(√c∗N

)−1]

≤ C[hμ

(1 + 2μ−1

)+ h + 2

(√c∗N

)−1].

(2.20)

If xi ∈ (ς, 1/2], then hi−1 = hi = h, and we have

|Tr(xi)| ≤ C[h + 2

(√c∗N

)−1], xi ∈ (ς, 1/2]. (2.21)


Thus,

|Tr(xi)| ≤ C[hμ

(1 + 2μ−1

)+ h + 2

(√c∗N

)−1], xi ∈ (0, 1/2]. (2.22)

In a similar way we can estimate Tr on [1/2, 1) and conclude that

|Tr(xi)| ≤ C[hμ

(1 + 2μ−1

)+ h + 2

(√c∗N

)−1], xi ∈ ωh. (2.23)

From here and (2.8), we conclude that

maxxi∈ωh

|Tr(xi)| ≤ CN−1 lnN. (2.24)

From (2.3), (2.15), by the mean-value theorem, we conclude that w = v−u satisfies thedifference problem

Lhw(x) + fuw(x) = −Tr(x), x ∈ ωh, w(0) = w(1) = 0. (2.25)

Using the assumption on fu from (1.1) and (2.24), by (2.6), we prove the theorem.

2.3. The Monotone Iterative Method

In this section, we construct an iterative method for solving the nonlinear difference scheme(2.3) which possesses monotone convergence.

Additionally, we assume that f(x, u) from (1.1) satisfies the two-sided constraint

0 < c∗ ≤ fu ≤ c∗, c∗, c∗ = const. (2.26)

The iterative method is constructed in the following way. Choose an initial meshfunction v(0), then the iterative sequence {v(n)}, n ≥ 1, is defined by the recurrence formulae

(Lh + c∗

)z(n)(x) = −Rh

(x, v(n−1)

), x ∈ ωh,

z(1)(0) = −v(0)(0), z(1)(1) = −v(0)(1), z(n)(0) = z(n)(1) = 0, n ≥ 2,

v(n)(x) = v(n−1)(x) + z(n)(x), x ∈ ωh,

Rh(x, v(n−1)

)= Lhv(n−1)(x) + f

(x, v(n−1)

),

(2.27)

where Rh(x, v(n−1)) is the residual of the difference scheme (2.3) on v(n−1).We say that v(x) is an upper solution of (2.3) if it satisfies the inequalities

Lhv(x) + f(x, v) ≥ 0, x ∈ ωh, v(0), v(1) ≥ 0. (2.28)


Similarly, v(x) is called a lower solution if it satisfies the reversed inequalities. Upper andlower solutions satisfy the inequality

v(x) ≤ v(x), x ∈ ωh. (2.29)

Indeed, by the definition of lower and upper solutions and the mean-value theorem, for δv =v − v we have

Lhδv + fu(x)δv(x) ≥ 0, x ∈ ωh, δv(x) ≥ 0, x = 0, 1, (2.30)

where fu(x) = cu[x, v(x) + ϑ(x)δv(x)], 0 < ϑ(x) < 1. In view of the maximum principle inLemma 2.2, we conclude the required inequality.

The following theorem gives the monotone property of the iterative method (2.27).

Theorem 2.4. Let v(0), v(0) be upper and lower solutions of (2.3) and f satisfy (2.26). Then theupper sequence {v(n)} generated by (2.27) converges monotonically from above to the unique solutionv of (2.3), the lower sequence {v(n)} generated by (2.27) converges monotonically from below to v:

v(n)(x) ≤ v(n+1)(x) ≤ v(x) ≤ v(n+1)(x) ≤ v(n)(x), x ∈ ωh, (2.31)

and the sequences converge at the linear rate q = 1 − c∗/c∗.

Proof. We consider only the case of the upper sequence. If v(0) is an upper solution, then from(2.27) we conclude that

(Lh + c∗

)z(1)(x) ≤ 0, x ∈ ωh, z(1)(0), z(1)(1) ≤ 0. (2.32)

From Lemma 2.2, by the maximum principle for the difference operatorLh+c∗, it follows thatz(1)(x) ≤ 0, x ∈ ωh. Using the mean-value theorem and the equation for z(1), we representRh(x, v(1)) in the form

Rh(x, v(1)

)= −(c∗ − f (1)

u (x))z(1)(x), x ∈ ωh, (2.33)

where f(1)u (x) = fu[x, v

(0)(x) + ϑ(1)(x)z(1)(x)], 0 < ϑ(1)(x) < 1. Since the mesh function z(1) isnonpositive on ωh and taking into account (2.26), we conclude that v(1) is an upper solution.By induction on n, we obtain that z(n)(x) ≤ 0, x ∈ ωh, n ≥ 1, and prove that {v(n)} is amonotonically decreasing sequence of upper solutions.

We now prove that the monotone sequence {v(n)} converges to the solution of (2.3).Similar to (2.33), we obtain

R(x, v(n−1)

)= −(c∗ − f (n−1)

u (x))z(n−1)(x), x ∈ ωh, n ≥ 2, (2.34)


and from (2.27), it follows that z(n+1) satisfies the difference equation

(L + c∗)z(n)(x) =(c∗ − f (n−1)

u (x))z(n−1)(x), x ∈ ωh. (2.35)

Using (2.26) and (2.6), we have

‖z(n)‖ωh ≤ qn−1‖z(1)‖ωh . (2.36)

This proves the convergence of the upper sequence at the linear rate q. Now by linearity ofthe operator Lh and the continuity of f , we have also from (2.27) that the mesh function vdefined by

v(x) = limn→∞

v(n)(x), x ∈ ωh, (2.37)

is the exact solution to (2.3). The uniqueness of the solution to (2.3) follows from estimate(2.6). Indeed, if by contradiction, we assume that there exist two solutions v1 and v2 to (2.3),then by the mean-value theorem, the difference δv = v1 − v2 satisfies the difference problem

Lhδv + fuδv = 0, x ∈ ωh, δv(0) = δv(1) = 0. (2.38)

By (2.6), δv = 0 which leads to the uniqueness of the solution to (2.3). This proves thetheorem.

Consider the following approach for constructing initial upper and lower solutionsv(0) and v(0). Introduce the difference problems

(Lh + c∗

)v(0)ν = ν

∣∣f(x, 0)∣∣, x ∈ ωh,

v(0)ν (0) = v

(0)ν (1) = 0, ν = 1,−1,

(2.39)

where c∗ from (2.26). Then the functions v(0)1 , v(0)

−1 are upper and lower solutions, respectively.We check only that v(0)

1 is an upper solution. From the maximum principle in Lemma 2.2, itfollows that v(0)

1 ≥ 0 on ωh. Now using the difference equation for v(0)1 and the mean-value

theorem, we have

Rh(x, v

(0)1

)= f(x, 0) +

∣∣f(x, 0)∣∣ +(f(0)u − c∗

)v(0)1 . (2.40)

Since f(0)u ≥ c∗ and v

(0)1 is nonnegative, we conclude that v(0)

1 is an upper solution.


Theorem 2.5. If the initial upper or lower solution v(0) is chosen in the form of (2.39), then themonotone iterative method (2.27) converges μ-uniformly to the solution v of the nonlinear differencescheme (2.3)

∥∥∥v(n) − v

∥∥∥ωh≤

c0qn

1 − q∥∥f(x, 0)

∥∥ωh ,

q = 1 − c∗/c∗ < 1, c0 = (3c∗ + c∗)/(c∗c∗).

(2.41)

Proof. From (2.27), (2.39), and the mean-value theorem, by (2.6),

∥∥∥z(1)

∥∥∥ωh≤ 1

c∗

∥∥∥Lhv(0)

∥∥∥ωh

+1c∗

∥∥∥f(x, v(0))

∥∥∥ωh

≤ 1c∗

(c∗∥∥∥v(0)

∥∥∥ωh

+∥∥f(x, 0)

∥∥ωh

)

+1c∗∥∥f(x, 0)

∥∥ωh +

∥∥∥v(0)∥∥∥ωh.

(2.42)

From here and estimating v(0) from (2.39) by (2.6),

∥∥∥v(0)∥∥∥ωh≤ 1

c∗

∥∥f(x, 0)∥∥ωh, (2.43)

we conclude the estimate on z(1) in the form

∥∥∥z(1)∥∥∥ωh≤ c0∥∥f(x, 0)

∥∥ωh , (2.44)

where c0 is defined in the theorem. From here and (2.36), we conclude that

∥∥∥z(n)∥∥∥ωh≤ c0q

n−1∥∥f(x, 0)∥∥ωh . (2.45)

Using this estimate, we have

∥∥∥v(n+k) − v(n)∥∥∥ωh≤

n+k−1∑

i=n

∥∥∥v(i+1) − v(i)∥∥∥ωh

=n+k−1∑

i=n

∥∥∥z(i+1)∥∥∥ωh

≤q

1 − q

∥∥∥z(n)∥∥∥ωh≤

c0qn

1 − q∥∥f(x, 0)

∥∥ωh .

(2.46)

Taking into account that limv(n+k) = v as k → ∞, where v is the solution to (2.3), we concludethe theorem.

From Theorems 2.3 and 2.5 we conclude the following theorem.


Theorem 2.6. Suppose that the initial upper or lower solution v(0) is chosen in the form of (2.39).Then the monotone iterative method (2.27) on the piecewise uniformmesh (2.8) converges μ-uniformlyto the solution of problem (1.1):

∥∥∥v(n) − u

∥∥∥ωh≤ C(N−1 lnN + qn

), (2.47)

where q = 1 − c∗/c∗, and constant C is independent of μ and N.

3. The Parabolic Problem

3.1. The Nonlinear Difference Scheme

Introduce uniform mesh ωτ on [0, T]

ωτ = {tk = kτ, 0 ≤ k ≤Nτ, Nττ = T}. (3.1)

For approximation of problem (1.2), we use the implicit difference scheme

Lv(x, t) − τ−1v(x, t − τ) = −f(x, t, v), (x, t) ∈ ωh ×ωτ \ {∅},

v(0, t) = 0, v(1, t) = 0, v(x, 0) = u0(x), x ∈ ωh,

L = Lh + τ−1,

(3.2)

where ωh and Lh are defined in (2.2) and (2.3), respectively. We introduce the linear versionof problem (3.2)

(L + c)w(x, t) = f0(x, t), x ∈ ωh,

w(0, t) = 0, w(1, t) = 0, c(x, t) ≥ 0, x ∈ ωh.(3.3)

We now formulate a discrete maximum principle for the difference operator L + c and givean estimate of the solution to (3.3).

Lemma 3.1. (i) If a mesh function w(x, t) on a time level t ∈ ωτ \ {∅} satisfies the conditions

(L + c)w(x, t) ≥ 0 (≤ 0), x ∈ ωh, w(0, t), w(1, t) ≥ 0 (≤ 0), (3.4)

then w(x, t) ≥ 0 (≤ 0), x ∈ ωh.


(ii) If c(x, t) ≥ c∗ = const > 0, then the following estimate of the solution to (3.3) holds true:

‖w(t)‖ωh ≤∥∥f0(t)

∥∥ωh/(c∗ + τ−1

), (3.5)

where ‖w(t)‖ωh = maxx∈ωh |w(x, t)|, ‖f0(t)‖ωh = maxx∈ωh |f0(x, t)|.

The proof of the lemma can be found in [6].

3.2. The Monotone Iterative Method

Assume that f(x, t, u) from (3.2) satisfies the two-sided constraint

0 ≤ fu(x, t, u) ≤ c∗, c∗ = const. (3.6)

We consider the following iterative method for solving (3.2). Choose an initial meshfunction v(0)(x, t). On each time level, the iterative sequence {v(n)(x, t)}, n = 1, . . . , n∗, isdefined by the recurrence formulae

(L + c∗)z(n)(x, t) = −R(x, t, v(n−1)

), x ∈ ωh,

z(1)(0, t) = −v(0)(0, t), z(1)(1, t) = −v(0)(1, t),

z(n)(0, t) = z(n)(1, t) = 0, n ≥ 2, v(n)(x, t) = v(n−1)(x, t) + z(n)(x, t),

R(x, t, v(n−1)

)= Lv(n−1)(x, t) + f

(x, t, v(n−1)

)− τ−1v(x, t − τ),

v(x, t) = v(n∗)(x, t), x ∈ ωh, v(x, 0) = u0(x), x ∈ ωh,

(3.7)

where R(x, t, v(n−1)) is the residual of the difference scheme (3.2) on v(n−1).On a time level t ∈ ωτ \{∅}, we say that v(x, t) is an upper solution of (3.2) with respect

to v(x, t − τ) if it satisfies the inequalities

Lv(x, t) + f(x, t, v) − τ−1v(x, t − τ) ≥ 0, x ∈ ωh,

v(0, t) ≥ 0, v(1, t) ≥ 0.(3.8)

Similarly, v(x, t) is called a lower solution if it satisfies all the reversed inequalities. Upperand lower solutions satisfy the inequality

v(x, t) ≤ v(x, t), p ∈ ωh. (3.9)

This result can be proved in a similar way as for the elliptic problem.The following theorem gives the monotone property of the iterative method (3.7).


Theorem 3.2. Assume that f(x, t, u) satisfies (3.6). Let v(x, t − τ) be given and v(0)(x, t), v(0)(x, t)be upper and lower solutions of (3.2) corresponding v(x, t − τ). Then the upper sequence {v(n)(x, t)}generated by (3.7) converges monotonically from above to the unique solution v(x, t) of the problem

Lv(x, t) + f(x, t, v) − τ−1v(x, t − τ) = 0, x ∈ ∂ωh,

v(0, t) = 0, v(1, t) = 0,(3.10)

the lower sequence {v(n)(x, t)} generated by (3.7) converges monotonically from below to v(x, t) andthe following inequalities hold

v(n−1)(x, t) ≤ v(n)(x, t) ≤ v(x, t) ≤ v(n)(x, t) ≤ v(n−1)(x, t), x ∈ ωh. (3.11)

Proof. We consider only the case of the upper sequence, and the case of the lower sequencecan be proved in a similar way.

If v(0) is an upper solution, then from (3.7) we conclude that

Lz(1)(x, t) ≤ 0, x ∈ ωh, z(1)(0, t) ≤ 0, z(1)(1, t) ≤ 0. (3.12)

From Lemma 3.1, it follows that

z(1)(x, t) ≤ 0, x ∈ ωh, (3.13)

and from (3.7), it follows that v(1) satisfies the boundary conditions.Using the mean-value theorem and the equation for z(1) from (3.7), we represent

R(x, t, v(1)) in the form

R(x, t, v(1)

)= −(c∗ − f (1)

u (x, t))z(1)(x, t), (3.14)

where f(1)u (x, t) = fu[x, t, v

(0)(x, t) + ϑ(1)(x, t)z(1)(x, t)], 0 < ϑ(1)(x, t) < 1. Since the meshfunction z(1) is nonpositive on ωh and taking into account (3.6), we conclude that v(1) is anupper solution to (3.2). By induction on n, we obtain that z(n)(x, t) ≤ 0, x ∈ ωh, n ≥ 1, andprove that {v(n)(x, t)} is a monotonically decreasing sequence of upper solutions.

We now prove that the monotone sequence {v(n)} converges to the solution of (3.2).The sequence {v(n)} is monotonically decreasing and bounded below by v, where v is anylower solution (3.9). Now by linearity of the operatorL and the continuity of f , we have alsofrom (3.7) that the mesh function v defined by

v(x, t) = limn→∞

v(n)(x, t), x ∈ ωh, (3.15)


is an exact solution to (3.2). If by contradiction, we assume that there exist two solutions v1

and v2 to (3.2), then by the mean-value theorem, the difference δv = v1 − v2 satisfies thesystem

Lδv(x, t) + fuδv(x, t) = 0, x ∈ ωh, δv(0, t) = v(1, t) = 0. (3.16)

By Lemma 3.1, δv = 0 which leads to the uniqueness of the solution to (3.2). This proves thetheorem.

Consider the following approach for constructing initial upper and lower solutionsv(0)(x, t) and v(0)(x, t). Introduce the difference problems

Lv(0)ν (x, t) = ν

∣∣∣f(x, t, 0) − τ−1v(x, t − τ)∣∣∣, x ∈ ωh,

v(0)ν (0, t) = v(0)(1, t) = 0, ν = 1,−1.

(3.17)

The functions v(0)1 (x, t), v(0)

−1 (x, t) are upper and lower solutions, respectively. This result canbe proved in a similar way as for the elliptic problem.

Theorem 3.3. Let initial upper or lower solution be chosen in the form of (3.17), and let f satisfy(3.6). Suppose that on each time level the number of iterates n∗ ≥ 2. Then for the monotone iterativemethods (3.7), the following estimate on convergence rate holds:

max1≤k≤Nτ

‖v(tk) − v∗(tk)‖ωh ≤ Cηn∗−1, η =c∗

(c∗ + τ−1

) , (3.18)

where v∗(x, t) is the solution to (3.2), v(x, t) = v(n∗)(x, t), and constant C is independent of μ, N,and τ .

Proof. Similar to (3.14), using the mean-value theorem and the equation for z(n) from (3.7),we have

Lv(n)(x, t) + f(x, t, v(n)

)− τ−1v(x, t − τ) = −

[c∗ − f (n)

u (x, t)]z(n)(x, t),

f(n)u (x, t) ≡ fu

[x, t, v(n−1)(x, t) + ϑ(n)(x, t)z(n)(x, t)

], 0 < ϑ(n)(x, t) < 1.

(3.19)

From here and (3.7), we have

(L + c∗)z(n)(x, t) =(c∗ − f (n)

u

)z(n−1)(x, t), x ∈ ωh. (3.20)


Using (3.5) and (3.6), we have

∥∥∥z(n)

∥∥∥ωh≤ ηn−1

∥∥∥z(1)

∥∥∥ωh, (3.21)

where η is defined in (3.18).Introduce the notation

w(x, t) = v∗(x, t) − v(x, t), (3.22)

where v(x, t) = v(n∗)(x, t). Using the mean-value theorem, from (3.2) and (3.19), we concludethat w(x, τ) satisfies the problem

Lw(x, τ) + fu(x, τ)w(x, τ) =(c∗ − f (n∗)

u (x, τ))z(n∗)(x, τ), x ∈ ωh,

w(0, τ) = w(1, τ) = 0, x ∈ ∂ωh,

(3.23)

where f(n∗)u (x, τ) = fu[x, τ, v(x, τ) + ϑ(x, τ)w(x, τ)], 0 < ϑ(x, τ) < 1, and we have taken into

account that v(x, 0) = v∗(x, 0) = u0(x). By (3.5), (3.6), and (3.21),

‖w(τ)‖ωh ≤ c∗τηn∗−1∥∥∥z(1)(τ)

∥∥∥ωh. (3.24)

Using (3.6), (3.17), and the mean-value theorem, estimate z(1)(x, τ) from (3.7) by (3.5),

∥∥∥z(1)(τ)∥∥∥ωh≤ τ∥∥∥Lv(0)(τ)

∥∥∥ωh

+ c∗τ∥∥∥v(0)(τ)

∥∥∥ωh

+ τ∥∥∥f(x, τ, 0) − τ−1u0

∥∥∥ωh

≤(

2τ + c∗τ2)∥∥∥f(x, τ, 0) − τ−1u0

∥∥∥ωh

≤ (2 + c∗τ)(τ∥∥f(x, τ, 0)

∥∥ωh +

∥∥∥u0∥∥∥ωh

)≤ C1,

(3.25)

where C1 is independent of τ (τ ≤ T), μ, and N. Thus,

‖w(τ)‖ωh ≤ c∗C1τηn∗−1. (3.26)

Similarly, from (3.2) and (3.19), it follows that

Lw(x, 2τ) + fu(x, 2τ)w(x, 2τ) =(c∗ − f (n∗)

u (x, 2τ))z(n∗)(x, 2τ) + τ−1w(x, τ), x ∈ ωh. (3.27)

Using (3.21), by (3.5),

‖w(2τ)‖ωh ≤ ‖w(τ)‖ωh + c∗τηn∗−1∥∥∥z(1)(2τ)

∥∥∥ωh. (3.28)


Using (3.17), estimate z(1)(x, 2τ) from (3.7) by (3.5),

∥∥∥z(1)(2τ)

∥∥∥ωh≤ (2 + c∗τ)

(τ∥∥f(x, 2τ, 0)

∥∥ωh + ‖v(τ)‖ωh

)≤ C2, (3.29)

where v(x, τ) = v(n∗)(x, τ). As follows from Theorem 3.2, the monotone sequences {v(n)(x, τ)}and {v(n)(x, τ)} are bounded from above and below by, respectively, v(0)(x, τ) and v(0)(x, τ).Applying (3.5) to problem (3.17) at t = τ , we have

∥∥∥v(0)(τ)

∥∥∥ωh≤ τ∥∥∥f(x, τ, 0) − τ−1u0(x)

∥∥∥ωh≤ K1, (3.30)

where constant K1 is independent of μ, N, and τ . Thus, we prove that C2 is independent ofμ, N, and τ . From (3.26) and (3.28), we conclude

‖w(2τ)‖ωh ≤ c∗(C1 + C2)τηn∗−1. (3.31)

By induction on k, we prove

‖w(tk)‖ωh ≤ c∗(

k∑

l=1

Cl

)

τηn∗−1, k = 1, . . . ,Nτ , (3.32)

where all constants Cl are independent of μ, N, and τ . Taking into account that Nττ = T , weprove the estimate (3.18) with C = c∗Tmax1≤l≤NτCl.

In [4], we prove that the difference scheme (3.2) on the piecewise uniform mesh (2.8)converges μ-uniformly to the solution of problem (1.2):

max1≤k≤Nτ

‖v∗(tk) − u(tk)‖ωh ≤ C(N−1 lnN + τ

), (3.33)

where v∗(x, t) is the exact solution to (3.2), and constant C is independent of μ, N, and τ .From here and Theorem 3.3, we conclude the following theorem.

Theorem 3.4. Suppose that on each time level the initial upper or lower solution v(0) is chosen in theform of (3.17) and n∗ ≥ 2. Then the monotone iterative method (3.7) on the piecewise uniform mesh(2.8) converges μ-uniformly to the solution of problem (1.2):

‖v(tk) − u(tk)‖ωh ≤ C(N−1 lnN + τ + ηn∗−1

), (3.34)

where η = c∗/(c∗ + τ−1), and constant C is independent of μ,N, and τ .

4. Numerical Experiments

It is found that in all numerical experiments the basic feature of monotone convergence ofthe upper and lower sequences is observed. In fact, the monotone property of the sequences


Table 1: Numbers of iterations for the Newton iterative method.

v(0) \N 128 256 512 1024−1 7 7 9 ∗1 8 11 18 ∗3 73 ∗ ∗ ∗

holds at every mesh point in the domain. This is, of course, to be expected from the analyticalconsideration.

4.1. The Elliptic Problem

Consider problem (1.1) with f(u) = (u − 3)/(4 − u). We mention that ur = 3 is the solutionof the reduced problem, where μ = 0. This problem gives c∗ = 1/25, c∗ = 1, and initial lowerand upper solutions are chosen in the form of (2.39). The stopping criterion for the monotoneiterative method (2.27) is

∥∥∥v(n) − v(n−1)∥∥∥ωh≤ 10−5. (4.1)

Our numerical experiments show that for 10−1 ≤ μ ≤ 10−6 and 32 ≤N ≤ 1024, iterationcounts for monotone method (2.27) on the piecewise uniform mesh are independent of μ andN, and equals 12 and 8 for the lower and upper sequences, respectively. These numericalresults confirm our theoretical results stated in Theorem 2.5.

In Table 1, we present numbers of iterations for solving the test problem by the Newtoniterative method with the initial iterations v(0)(x) = −1, 1, 3, x ∈ ωh. Here μ = 10−3 is inuse, and we denote by an “∗” if more than 100 iterations is needed to satisfy the stoppingcriterion, or if the method diverges. The numerical results indicate that the Newton methodcannot be used successfully for this test problem.

4.2. The Parabolic Problem

For the parabolic problem (1.2), we consider the test problem with f(u) = exp(−1) − exp(−u)and u0 = 0. This problem gives c∗ = exp(−1), c∗ = 1, and the initial lower and upper solutionsare chosen in the form of (3.17).

The stopping test for the monotone method (3.7) is defined by

∥∥∥v(n)(t) − v(n−1)(t)∥∥∥ωh≤ 10−5. (4.2)

Our numerical experiments show that for 10−1 ≤ μ ≤ 10−6 and 32 ≤ N ≤ 1024, oneach time level the number of iterations for monotone method (3.7) on the piecewise uniformmesh is independent of μ and N and equal 4, 4, and 3 for τ = 0.1, 0.05, 0.01, respectively.These numerical results confirm our theoretical results stated in Theorem 3.3.


References

[1] I. Boglaev, “Approximate solution of a non-linear boundary value problem with a small parameter forthe highest-order differential,” USSR Computational Mathematics and Mathematical Physics, vol. 24, no.6, pp. 30–35, 1984.

[2] I. Boglaev, “Numerical method for quasi-linear parabolic equation with boundary layer,” USSRComputational Mathematics and Mathematical Physics, vol. 30, pp. 716–726, 1990.

[3] J. J. H. Miller, E. O’Riordan, and G. I. Shishkin, Fitted Numerical Methods for Singular PerturbationProblems: Error Estimates in the Maximum Norm for Linear Problems in One and Two Dimensions, WorldScientific, Singapore, 1996.

[4] I. Boglaev and M. Hardy, “Uniform convergence of a weighted average scheme for a nonlinearreaction-diffusion problem,” Journal of Computational and Applied Mathematics, vol. 200, no. 2, pp. 705–721, 2007.

[5] J. M. Ortega and W. C. Rheinboldt, Iterative Solution of Nonlinear Equations in Several Variables, AcademicPress, New York, NY, USA, 1970.

[6] A. A. Samarskii, The Theory of Difference Schemes, vol. 240 of Monographs and Textbooks in Pure and AppliedMathematics, Marcel Dekker, New York, NY, USA, 2001.

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