research article on positive periodic solutions to...

Research ArticleOn Positive Periodic Solutions to Nonlinear Fifth-OrderDifferential Equations with Six Parameters

Yunhai Wang1 and Fanglei Wang2

1College of Mechanical Engineering Guizhou Institute of Technology Guiyang 550003 China2College of Science Hohai University Nanjing 210098 China

Correspondence should be addressed to Yunhai Wang wangyhnuaa126com

Received 6 November 2015 Revised 4 February 2016 Accepted 15 February 2016

Academic Editor Juan Martinez-Moreno

Copyright copy 2016 Y Wang and F WangThis is an open access article distributed under theCreativeCommonsAttribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study the existence and multiplicity of positive periodic solutions to the nonlinear differential equation 119906(5)(119905) + 119896119906(4)(119905)minus120573119906(3)minus 12058511990610158401015840(119905) + 120572119906

1015840(119905) + 120596119906(119905) = 120582ℎ(119905)119891(119906) in 0 le 119905 le 1 119906119894(0) = 119906119894(1) 119894 = 0 1 2 3 4 where 119896 120572 120596 120582 gt 0 120573 120585 isin 119877 ℎ isin 119862(119877 119877)

is a 1-periodic function The proof is based on the Krasnoselskii fixed point theorem

1 Introduction

Fifth-order boundary value problems (BVPs) are knownto arise in the mathematical modeling of viscoelastic flowand other branches of mathematical physical and engineer-ing sciences see [1ndash4] So the existence multiplicity andnonexistence of solutions to the fifth-order BVPs are noticedby many authors For example Agarwal and Odda [5 6]gave some theorems which list conditions for the existenceand uniqueness of solutions to the fifth-order BVPs by thetopological methods Gamel and Lv [7 8] focused theirattention on the study of numerical solution to the fifth-orderBVPs

On the other hand all kinds of topological methods suchas the method of upper and lower solutions degree theoryand some fixed point theorems in cones have been widelyapplied to study the singular and regular periodic boundaryvalue problems see [9ndash19]

Inspired by the above references we establish the exis-tence and multiplicity results of the nonlinear differentialequation in this paper

119906(5)(119905) + 119896119906

(4)(119905) minus 120573119906

(3)minus 12058511990610158401015840(119905) + 120572119906

1015840(119905) + 120596119906 (119905)

= 120582ℎ (119905) 119891 (119906) in 0 le 119905 le 1

119906119894(0) = 119906

119894(1) 119894 = 0 1 2 3 4

(1)

where 119896 120572 120596 120582 gt 0 120573 120585 isin 119877 and ℎ isin 119862(119877 119877) is a 1-periodicfunction

Both of regular and singular cases are considered Theproof is based on the Krasnoselskii fixed point theorem in acone

Lemma 1 (see [20]) Let 119864 be a Banach space and let 119870 sub 119864be a cone in 119864 Assume Ω1 and Ω2 are open subsets of 119864 with0 isin Ω1 Ω1 sub Ω2 and let 119879 119870 cap (Ω2 Ω1) rarr 119870 be acompletely continuous operator such that either

(i) 119879119906 le 119906 119906 isin 119870 cap 120597Ω1 and 119879119906 ge 119906 119906 isin119870 cap 120597Ω2 or

(ii) 119879119906 ge 119906 119906 isin 119870 cap 120597Ω1 and 119879119906 le 119906 119906 isin119870 cap 120597Ω2

Then 119879 has a fixed point in 119870 cap (Ω2 Ω1)

This paper is organized as follows in Section 2 somepreliminaries are given in Section 3 we give themain resultsin Section 4 we give an example to illustrate ourmain results

2 Preliminaries

If 119896 120572 120596 120573 120585 satisfy the assumption

(H1) 119896 gt 0 120573 gt minus2120587 0 lt 120572 lt ((12)120573 + 21205872)2 1205721205874 +1205731205872+ 1 gt 0 120585 = 119896120573 120596 = 119896120572

Hindawi Publishing CorporationJournal of Function SpacesVolume 2016 Article ID 8567679 5 pageshttpdxdoiorg10115520168567679

2 Journal of Function Spaces

then (1) can be rewritten as

119906(5)(119905) + 119896119906

(4)(119905) minus 120573119906

(3)minus 119896120573119906

10158401015840(119905) + 120572119906

1015840(119905)

+ 119896120572119906 (119905) = 120582ℎ (119905) 119891 (119906)

(2)

Furthermore we have

(119906(4)(119905) minus 120573119906

(2)(119905) + 120572119906 (119905))

1015840

+ 119896 (119906(4)(119905) minus 120573119906

10158401015840(119905) + 120572119906 (119905)) = 120582ℎ (119905) 119891 (119906)

(3)

From Lemma 3 in [12] the fourth-order linear problem

119906(4)(119905) minus 120573119906

10158401015840(119905) + 120572119906 (119905) = 0 in 0 le 119905 le 1

119906119894(0) = 119906

119894(1) 119894 = 0 1 2

119906(3)(0) minus 119906

(3)(1) = 1

(4)

has a unique continuous solution (119905) and (119905) gt 0 0 le 119905 le 1Let

1198981 = min0le119905le1 (119905)

1198721 = max0le119905le1 (119905)

(5)

For given constants 120572 and 120573 theminimum1198981 andmaximum1198721 can be computed explicitly Let119862[0 1] denote the Banachspace 119906 isin 119862(119877 119877) 119906(119905) = 119906(119905+1)with themaximumnorm119906 = max119905isin[01]|119906(119905)|

Lemma 2 (see [12]) Assume that (H1) holds For 120593(119905) isin119862[0 1] the linear boundary value problem

119906(4)(119905) minus 120573119906

10158401015840(119905) + 120572119906 (119905) = 120593 (119905) 119894119899 0 le 119905 le 1

119906119894(0) = 119906

119894(1) 119894 = 0 1 2 3

(6)

has a unique solution 119906(119905) which is given by expression

119906 (119905) = int

1

0

1198661 (119905 119904) 120593 (119904) 119889119904(7)

where

1198661 (119905 119904) =

(119905 minus 119904) 0 le 119904 le 119905 le 1

(1 + 119905 minus 119904) 0 le 119905 le 119904 le 1

(8)

Lemma 3 (see [13]) For any 119896 gt 0 and 120595(119905) isin 119862[0 1] thelinear boundary value problem

V1015840 (119905) + 119896V (119905) = 120595 (119905) 119894119899 0 le 119905 le 1

V (0) = V (1)(9)

has a unique solution V(119905) which can be expressed as

V (119905) = int1

0

1198662 (119905 119904) 120593 (119904) 119889119904(10)

where

1198662 (119905 119904) =

119890minus119896(119905minus119904)

1 minus 119890minus119896 0 le 119904 le 119905 le 1

119890minus119896(1+119905minus119904)


(11)

satisfying the estimates

1198982 =119890minus119896

1 minus 119890minus119896le 1198662 (119905 119904) le

1

1 minus 119890minus119896= 1198722

(12)

From Lemmas 2 and 3 it follows that 119906(119905) is the solutionof (1) if and only if 119906(119905) satisfies

119906 (119905) = 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904(13)

We define an operator 119879 on 119862[0 1] by

119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

(14)

Let 119870 be a cone in 119862[0 1] defined as

119870 = 119906 (119905) isin 119862 [0 1] 119906 (119905) ge 0 119905 isin [0 1] 119906 (119905)

ge 120579 119906

(15)

where 120579 = 1198981119898211987211198722 For 119903 gt 0 let Ω119903 = 119906 isin 119862[0 1] 119906 lt 119903 Note that 120597Ω119903 = 119906 isin 119862[0 1] 119906 = 119903

Lemma4 Assume that (H1) holds In addition we suppose thefollowing conditions hold

(H2) ℎ(119905) isin 119862[0 1] and int10ℎ(119905)119889119905 gt 0

(H3) 119891 [0infin) rarr [0infin) is continuous(H4) 119891(119906) gt 0 for 119906 gt 0

Then 119879(119870) sub 119870 and 119879 119870 rarr 119870 is completely continuous

Proof For any 119906 isin 119870 on one hand we have

119879 (119906) (119905) = 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

(16)

On the other hand we have

119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

(17)

Journal of Function Spaces 3

Thus

119879 (119906) (119905) ge

11989811198982

11987211198722

119879119906 = 120579 119879119906 (18)

Therefore 119879(119870) sub 119870 Using the Ascoli-Arzela theorem(Chapter 2 [21]) it is easy to show that 119879 119870 rarr 119870 is compactand continuous

From Lemma 4 and [18] it follows that

Lemma 5 Assume that (H1) and (H2) hold In addition wesuppose the following conditions hold

(H5) lim119906rarr0+119891(119906) = infin(H6) 119891 119877+ 0 rarr (0infin) is continuous

Then 119879(119870 0) sub 119870 and 119879 119870 0 rarr 119870 is completelycontinuous

3 Main Results

For convenience we give the notations

1198910 = lim119906rarr0+

119891 (119906)

119906

119891infin = lim119906rarr+infin

119891 (119906)

119906

(19)

Theorem 6 (regular case) Assume that (H1)ndash(H4) hold

(a) If 1198910 = 0 and 119891infin = infin then for all 120582 gt 0 (1) has apositive solution

(b) If 1198910 = infin and 119891infin = 0 then for all 120582 gt 0 (1) has apositive solution

(c) If 1198910 = 0 or 119891infin = 0 then there exists a 1205820 gt 0 suchthat for all 120582 gt 1205820 (1) has a positive solution

(d) If 1198910 = infin or 119891infin = infin then there exists a 1205820 gt 0 suchthat for all 0 lt 120582 lt 1205820 (1) has a positive solution

(e) If 1198910 = 119891infin = 0 then there exists a 1205820 gt 0 such thatfor all 120582 gt 1205820 (1) has two positive solutions

(f) If 1198910 = 119891infin = infin then there exists a 1205820 gt 0 such thatfor all 0 lt 120582 lt 1205820 (1) has two positive solutions

Proof (a) Since 1198910 = 0 there exists a 119903 gt 0 such that 119891(119906) le120598119906 for any 119906 isin [0 119903] where 120598 satisfies 12059812058211987211198722 int

1

0ℎ(119905)119889119905 lt 1

Then for any 119906 isin 120597Ω119903 cap 119870 we have

119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722 int

1

0

ℎ (120591) 120598119906 (120591) 119889120591

le 12059812058211987211198722 119906int

1

0

ℎ (120591) 119889120591 lt 119906

(20)

Since 119891infin = infin there exists a 119877 gt 0 such that 119891(119906) ge 120578119906for any 119906 ge 119877 where 120578 satisfies 12057812058212057911989811198982 int

1

0ℎ(119905)119889119905 gt 1 Take

119877 = max119877120579 2119903 + 1 Then for any 119906 isin 120597Ω119877 cap 119870 we have119906 ge 120579119906 = 120579119877 gt 119877 and

119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982 int

1

0

ℎ (120591) 120578119906 (120591) 119889120591

ge 12057812058211989811198982120579 119906int

1

0

ℎ (120591) 119889120591 gt 119906

(21)

By the above inequalities and Lemma 1119879 has a fixed pointin 119870 cap (Ω119877 Ω119903)

(b) Since1198910 = infin there exists a 119903 gt 0 such that119891(119906) ge 120578119906for any 119906 isin [0 119903] where 120578 satisfies 12057812058212057911989811198982 int

1

0ℎ(119905)119889119905 gt 1


119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982 int

1

0

ℎ (120591) 120578119906 (120591) 119889120591

ge 12057812058211989811198982120579 119906int

1

0

ℎ (120591) 119889120591 gt 119906

(22)

Since 119891infin = 0 there exists a 119877 gt 0 such that 119891(119906) le 120598119906for any 119906 ge 119877 where 120598 satisfies 12059812058211987211198722 int

1

0ℎ(119905)119889119905 lt 1

Take 119877 = max119877120579 2119877 + 1 Then for any 119906 isin 120597Ω119877 cap 119870we have 119906 ge 120579119906 = 120579119877 gt 119877 and

119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722 int

1

0

ℎ (120591) 120598119906 (120591) 119889120591

le 12059812058211987211198722 119906int

1

0

ℎ (120591) 119889120591 lt 119906

(23)



(c) For any 120588 gt 0 let119898(120588) = min119891(119906) 119906 ge 0 120579120588 le 119906 le120588 Then there exists a 1205820 gt 0 such that

119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982119898(120588)int

1

0

ℎ (120591) 119889120591 gt 119906

(24)

for any 119906 isin 120597Ω120588 cap 119870 120582 gt 1205820If 1198910 = 0 from the proof of (a) it follows that there exists

a 119903 lt 120588 such that

119879 (119906) lt 119906 for 119906 isin 120597Ω119903 cap 119870 (25)

If119891infin = 0 from the proof of (b) it follows that there existsa 119877 gt 0 such that

119879 (119906) lt 119906 for 119906 isin 120597Ω119877 cap 119870 (26)

By Lemma 1 119879 has a fixed point in 119870 cap (Ω120588 Ω119903) or 119870 cap(Ω119877 Ω120588)

(d) For any 120588 gt 0 let119872(120588) = max119891(119906) 119906 ge 0 120579120588 le119906 le 120588Then there exists a 1205820 gt 0 such that

119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722119872(120588)int

1

0

ℎ (120591) 119889120591 lt 119906

(27)

for any 119906 isin 120597Ω119877 cap 119870 0 lt 120582 lt 1205820From the part proof of (a) and (b) the results of (d) follow

(e) From the proof of (c) it is easy to obtain the result(f) From the proof of (d) it is easy to obtain the result

Theorem7 (singular case) Assume that (H1) (H2) (H5) and(H6) hold

(A) There exists a 1205820 gt 0 such that (1) has a positivesolution for 0 lt 120582 lt 1205820

(B) If lim119906rarrinfin(119891(119906)119906) = 0 then for any 120582 gt 0 (1) has apositive solution

(C) If lim119906rarrinfin(119891(119906)119906) = infin then (1) has two positivesolutions for all sufficiently small 120582 gt 0

Proof (A) On one hand from the proof of (d) inTheorem 6it is clear to see that there exists a 1205820 gt 0 such that

119879119906 lt 119906 for 119906 isin 120597Ω120588 0 lt 120582 lt 1205820 On the other handsince lim119906rarr0+119891(119906) = infin there exists a 119903 gt 0 with 119903 lt 120588such that 119891(119906) ge 120583119906 for 0 lt 119906 le 119903 where 120583 satisfies12058212058312057911989811198982 int

1

0ℎ(119905)119889119905 gt 1 For any 119906 isin 120597Ω119903 cap 119870 we have

119879119906 gt 119906 Therefore by Lemma 1 119879 has a fixed point in119870 cap (Ω120588 Ω119903)

Finally from (A) and (b) result (B) follows From (A) and(d) result (C) follows

4 Examples

Example 1 Consider the fourth-order nonlinear problems

119906(5)(119905) + 2119906

(4)(119905) minus 5119906

(3)minus 10119906

10158401015840(119905) + 4119906

1015840(119905) + 8119906 (119905)

= 120582sin2 (2120587119905) 119891 (119906)

119906119894(0) = 119906

119894(1) 119894 = 0 1 2 3 4

(28)

where

119896 = 2

120572 = 5

120596 = 10

120573 = 4

120585 = 8

ℎ (119905) = sin2 (2120587119905)

119891 (119906) =

119906120583 0 lt 119906 le 1

119906] 119906 gt 1

(29)

First via some simple computations it is easy to see that(H1) and (H2) hold Second if 120583 ] gt 0 then (H3) and (H4)obviously hold Now we illustrate the cases of Theorem 6 viathe discussions of 120583 and ]

(a) If 120583 gt 1 and ] gt 1 then for all 120582 gt 0 (28) has apositive solution

(b) If 0 lt 120583 lt 1 and 0 lt ] lt 1 then for all 120582 gt 0 (28)has a positive solution

(c) If 120583 gt 1 or 0 lt ] lt 1 then there exists a 1205820 gt 0 suchthat for all 120582 gt 1205820 (28) has a positive solution

(d) If 0 lt 120583 lt 1 or ] gt 1 then there exists a 1205820 gt 0 suchthat for all 0 lt 120582 lt 1205820 (28) has a positive solution

(e) If 120583 gt 1 and 0 lt ] lt 1 then there exists a 1205820 gt 0 suchthat for all 120582 gt 1205820 (28) has two positive solutions

(f) If 0 lt 120583 lt 1 and ] gt 1 then there exists a 1205820 gt 0such that for all 0 lt 120582 lt 1205820 (28) has two positivesolutions


Finally if 120583 lt 0 ] ge 0 then (H5) and (H6) hold Nowwe illustrate the cases of Theorem 7 via the discussions of 120583and ]

(A) If 120583 lt 0 then there exists a 1205820 gt 0 such that (28) hasa solution for 0 lt 120582 lt 1205820

(B) If 120583 lt 0 0 le ] lt 1 then for any 120582 gt 0 (28) has apositive solution

(C) If 120583 lt 0 ] gt 1 then (28) has two positive solutionsfor all sufficient small 120582 gt 0

Disclosure

The first author is supported by Doctoral Scientific ResearchFoundation of Guizhou Province The Joint Science andTechnology FundQKH[2014 7366] The second author issupported by by NSF of China (no 11501165) NSF of JiangsuProvince (no BK20130825) and the Fundamental ResearchFunds for the Central Universities

Conflict of Interests

The authors declare that they have no competing interests

Authorsrsquo Contribution

Both authors contributed to each part of this work equallyand read and approved the final paper

References

[1] A R Davies A Karageoghis and T N Phillips ldquoSpectralGlarkien methods for the primary two-point boundary-valueproblems in modeling viscelastic flowsrdquo International Journalfor Numerical Methods in Engineering vol 26 no 3 pp 647ndash662 1988

[2] D J Fyfe ldquoLinear dependence relations connecting equalinterval Nth degree splines and their derivativesrdquo Journal of theInstitute ofMathematics and Its Applications vol 7 pp 398ndash4061971

[3] A Karageoghis T N Phillips and A R Davies ldquoSpectralcollocationmethods for the primary two-point boundary-valueproblems in modeling viscelastic flowsrdquo International Journalfor NumericalMethods in Engineering vol 26 pp 805ndash813 1998

[4] G Liu ldquoNew research directions in singular perturbationtheory artificial parameter approach and inverse-perturbationtechniquerdquo in Proceeding of the 7th Conference of the ModernMathematics and Mechanics Shanghai China September 1997

[5] R P Agarwal Boundary Value Problems for High OrdinaryDifferential Equations World Scientific Singapore 1986

[6] S N Odda ldquoExistence solution for 5th order differentialequations under some conditionsrdquo Applied Mathematics vol 1no 4 pp 279ndash282 2010

[7] M El-Gamel ldquoSinc and the numerical solution of fifth-orderboundary value problemsrdquo Applied Mathematics and Computa-tion vol 187 no 2 pp 1417ndash1433 2007

[8] X Lv and M Cui ldquoAn efficient computational method forlinear fifth-order two-point boundary value problemsrdquo Journalof Computational and Applied Mathematics vol 234 no 5 pp1551ndash1558 2010

[9] J Chu P J Torres andM Zhang ldquoPeriodic solutions of secondorder non-autonomous singular dynamical systemsrdquo Journal ofDifferential Equations vol 239 no 1 pp 196ndash212 2007

[10] C DeCoster and P Habets ldquoUpper and lower solutions in thetheory of ODE boundary value problems classical and recentresultsrdquo inNonlinear Analysis and Boundary Value Problems forOrdinary Differential Equations CISM-ICMS F Zanolin Edvol 371 pp 1ndash78 Springer New York NY USA 1996

[11] P Jebelean and J Mawhin ldquoPeriodic solutions of forced dissipa-tive p-Lienard equations with singularitiesrdquo Vietnam Journal ofMathematics vol 32 pp 97ndash103 2004

[12] Y Li ldquoPositive solutions of fourth-order periodic boundaryvalue problemsrdquo Nonlinear Analysis Theory Methods amp Appli-cations vol 54 no 6 pp 1069ndash1078 2003

[13] D OrsquoRegan and H Wang ldquoPositive periodic solutions ofsystems of first order ordinary differential equationsrdquo Results inMathematics vol 48 no 3-4 pp 310ndash325 2005

[14] P J Torres ldquoExistence of one-signed periodic solutions of somesecond-order differential equations via a Krasnoselskii fixedpoint theoremrdquo Journal of Differential Equations vol 190 no2 pp 643ndash662 2003

[15] FWang andYAn ldquoExistence andmultiplicity results of positivedoubly periodic solutions for nonlinear telegraph systemrdquoJournal of Mathematical Analysis and Applications vol 349 no1 pp 30ndash42 2009

[16] F Wang W Li and Y An ldquoNonnegative doubly periodicsolutions for nonlinear telegraph systemwith twin-parametersrdquoApplied Mathematics and Computation vol 214 no 1 pp 310ndash317 2009

[17] F Wang ldquoDoubly periodic solutions of a coupled nonlineartelegraph system with weak singularitiesrdquo Nonlinear AnalysisReal World Applications vol 12 no 1 pp 254ndash261 2011

[18] H Wang ldquoPositive periodic solutions of singular systems witha parameterrdquo Journal of Differential Equations vol 249 no 12pp 2986ndash3002 2010

[19] Q Yao ldquoExistence multiplicity and infinite solvability of pos-itive solutions to a nonlinear fourth-order periodic boundaryvalue problemrdquoNonlinear AnalysisTheory Methods amp Applica-tions vol 63 no 2 pp 237ndash246 2005

[20] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press New York NY USA 1988

[21] R F Brown A Topological Introduction to Nonlinear AnalysisSpringer 2014

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then (1) can be rewritten as

119906(5)(119905) + 119896119906

(4)(119905) minus 120573119906

(3)minus 119896120573119906

10158401015840(119905) + 120572119906

1015840(119905)

+ 119896120572119906 (119905) = 120582ℎ (119905) 119891 (119906)

(2)

Furthermore we have

(119906(4)(119905) minus 120573119906

(2)(119905) + 120572119906 (119905))

1015840

+ 119896 (119906(4)(119905) minus 120573119906

10158401015840(119905) + 120572119906 (119905)) = 120582ℎ (119905) 119891 (119906)

(3)

From Lemma 3 in [12] the fourth-order linear problem

119906(4)(119905) minus 120573119906

10158401015840(119905) + 120572119906 (119905) = 0 in 0 le 119905 le 1

119906119894(0) = 119906

119894(1) 119894 = 0 1 2

119906(3)(0) minus 119906

(3)(1) = 1

(4)

has a unique continuous solution (119905) and (119905) gt 0 0 le 119905 le 1Let

1198981 = min0le119905le1 (119905)

1198721 = max0le119905le1 (119905)

(5)

For given constants 120572 and 120573 theminimum1198981 andmaximum1198721 can be computed explicitly Let119862[0 1] denote the Banachspace 119906 isin 119862(119877 119877) 119906(119905) = 119906(119905+1)with themaximumnorm119906 = max119905isin[01]|119906(119905)|

Lemma 2 (see [12]) Assume that (H1) holds For 120593(119905) isin119862[0 1] the linear boundary value problem

119906(4)(119905) minus 120573119906

10158401015840(119905) + 120572119906 (119905) = 120593 (119905) 119894119899 0 le 119905 le 1

119906119894(0) = 119906

119894(1) 119894 = 0 1 2 3

(6)

has a unique solution 119906(119905) which is given by expression

119906 (119905) = int

1

0

1198661 (119905 119904) 120593 (119904) 119889119904(7)

where

1198661 (119905 119904) =

(119905 minus 119904) 0 le 119904 le 119905 le 1

(1 + 119905 minus 119904) 0 le 119905 le 119904 le 1

(8)

Lemma 3 (see [13]) For any 119896 gt 0 and 120595(119905) isin 119862[0 1] thelinear boundary value problem

V1015840 (119905) + 119896V (119905) = 120595 (119905) 119894119899 0 le 119905 le 1

V (0) = V (1)(9)

has a unique solution V(119905) which can be expressed as

V (119905) = int1

0

1198662 (119905 119904) 120593 (119904) 119889119904(10)

where

1198662 (119905 119904) =

119890minus119896(119905minus119904)


119890minus119896(1+119905minus119904)


(11)

satisfying the estimates

1198982 =119890minus119896

1 minus 119890minus119896le 1198662 (119905 119904) le

1

1 minus 119890minus119896= 1198722

(12)

From Lemmas 2 and 3 it follows that 119906(119905) is the solutionof (1) if and only if 119906(119905) satisfies

119906 (119905) = 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904(13)

We define an operator 119879 on 119862[0 1] by

119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

(14)

Let 119870 be a cone in 119862[0 1] defined as

119870 = 119906 (119905) isin 119862 [0 1] 119906 (119905) ge 0 119905 isin [0 1] 119906 (119905)

ge 120579 119906

(15)

where 120579 = 1198981119898211987211198722 For 119903 gt 0 let Ω119903 = 119906 isin 119862[0 1] 119906 lt 119903 Note that 120597Ω119903 = 119906 isin 119862[0 1] 119906 = 119903

Lemma4 Assume that (H1) holds In addition we suppose thefollowing conditions hold

(H2) ℎ(119905) isin 119862[0 1] and int10ℎ(119905)119889119905 gt 0

(H3) 119891 [0infin) rarr [0infin) is continuous(H4) 119891(119906) gt 0 for 119906 gt 0

Then 119879(119870) sub 119870 and 119879 119870 rarr 119870 is completely continuous

Proof For any 119906 isin 119870 on one hand we have

119879 (119906) (119905) = 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

(16)

On the other hand we have

119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

(17)


Thus

119879 (119906) (119905) ge

11989811198982

11987211198722

119879119906 = 120579 119879119906 (18)






3 Main Results


1198910 = lim119906rarr0+

119891 (119906)

119906


119891 (119906)

119906

(19)









1

0ℎ(119905)119889119905 lt 1


119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722 int

1

0

ℎ (120591) 120598119906 (120591) 119889120591

le 12059812058211987211198722 119906int

1

0

ℎ (120591) 119889120591 lt 119906

(20)


1

0ℎ(119905)119889119905 gt 1 Take


119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982 int

1

0

ℎ (120591) 120578119906 (120591) 119889120591

ge 12057812058211989811198982120579 119906int

1

0

ℎ (120591) 119889120591 gt 119906

(21)



1

0ℎ(119905)119889119905 gt 1


119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982 int

1

0

ℎ (120591) 120578119906 (120591) 119889120591

ge 12057812058211989811198982120579 119906int

1

0

ℎ (120591) 119889120591 gt 119906

(22)


1

0ℎ(119905)119889119905 lt 1


119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722 int

1

0

ℎ (120591) 120598119906 (120591) 119889120591

le 12059812058211987211198722 119906int

1

0

ℎ (120591) 119889120591 lt 119906

(23)




119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982119898(120588)int

1

0

ℎ (120591) 119889120591 gt 119906

(24)


a 119903 lt 120588 such that

119879 (119906) lt 119906 for 119906 isin 120597Ω119903 cap 119870 (25)


119879 (119906) lt 119906 for 119906 isin 120597Ω119877 cap 119870 (26)



119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722119872(120588)int

1

0

ℎ (120591) 119889120591 lt 119906

(27)









1




4 Examples


119906(5)(119905) + 2119906

(4)(119905) minus 5119906

(3)minus 10119906

10158401015840(119905) + 4119906

1015840(119905) + 8119906 (119905)

= 120582sin2 (2120587119905) 119891 (119906)

119906119894(0) = 119906

119894(1) 119894 = 0 1 2 3 4

(28)

where

119896 = 2

120572 = 5

120596 = 10

120573 = 4

120585 = 8

ℎ (119905) = sin2 (2120587119905)

119891 (119906) =

119906120583 0 lt 119906 le 1

119906] 119906 gt 1

(29)













Disclosure






References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










Thus

119879 (119906) (119905) ge

11989811198982

11987211198722

119879119906 = 120579 119879119906 (18)






3 Main Results


1198910 = lim119906rarr0+

119891 (119906)

119906


119891 (119906)

119906

(19)









1

0ℎ(119905)119889119905 lt 1


119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722 int

1

0

ℎ (120591) 120598119906 (120591) 119889120591

le 12059812058211987211198722 119906int

1

0

ℎ (120591) 119889120591 lt 119906

(20)


1

0ℎ(119905)119889119905 gt 1 Take


119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982 int

1

0

ℎ (120591) 120578119906 (120591) 119889120591

ge 12057812058211989811198982120579 119906int

1

0

ℎ (120591) 119889120591 gt 119906

(21)



1

0ℎ(119905)119889119905 gt 1


119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982 int

1

0

ℎ (120591) 120578119906 (120591) 119889120591

ge 12057812058211989811198982120579 119906int

1

0

ℎ (120591) 119889120591 gt 119906

(22)


1

0ℎ(119905)119889119905 lt 1


119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722 int

1

0

ℎ (120591) 120598119906 (120591) 119889120591

le 12059812058211987211198722 119906int

1

0

ℎ (120591) 119889120591 lt 119906

(23)




119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982119898(120588)int

1

0

ℎ (120591) 119889120591 gt 119906

(24)


a 119903 lt 120588 such that

119879 (119906) lt 119906 for 119906 isin 120597Ω119903 cap 119870 (25)


119879 (119906) lt 119906 for 119906 isin 120597Ω119877 cap 119870 (26)



119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722119872(120588)int

1

0

ℎ (120591) 119889120591 lt 119906

(27)









1




4 Examples


119906(5)(119905) + 2119906

(4)(119905) minus 5119906

(3)minus 10119906

10158401015840(119905) + 4119906

1015840(119905) + 8119906 (119905)

= 120582sin2 (2120587119905) 119891 (119906)

119906119894(0) = 119906

119894(1) 119894 = 0 1 2 3 4

(28)

where

119896 = 2

120572 = 5

120596 = 10

120573 = 4

120585 = 8

ℎ (119905) = sin2 (2120587119905)

119891 (119906) =

119906120583 0 lt 119906 le 1

119906] 119906 gt 1

(29)













Disclosure






References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119879 (119906) ge 119879 (119906) (119905)

= 120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

ge 12058211989811198982 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

ge 12058211989811198982119898(120588)int

1

0

ℎ (120591) 119889120591 gt 119906

(24)


a 119903 lt 120588 such that

119879 (119906) lt 119906 for 119906 isin 120597Ω119903 cap 119870 (25)


119879 (119906) lt 119906 for 119906 isin 120597Ω119877 cap 119870 (26)



119879 (119906)

= max119905isin[01]

120582∬

1

0

1198661 (119905 119904) 1198662 (119904 120591) ℎ (120591) 119891 (119906 (120591)) 119889120591 119889119904

le 12058211987211198722 int

1

0

ℎ (120591) 119891 (119906 (120591)) 119889120591

le 12058211987211198722119872(120588)int

1

0

ℎ (120591) 119889120591 lt 119906

(27)









1




4 Examples


119906(5)(119905) + 2119906

(4)(119905) minus 5119906

(3)minus 10119906

10158401015840(119905) + 4119906

1015840(119905) + 8119906 (119905)

= 120582sin2 (2120587119905) 119891 (119906)

119906119894(0) = 119906

119894(1) 119894 = 0 1 2 3 4

(28)

where

119896 = 2

120572 = 5

120596 = 10

120573 = 4

120585 = 8

ℎ (119905) = sin2 (2120587119905)

119891 (119906) =

119906120583 0 lt 119906 le 1

119906] 119906 gt 1

(29)













Disclosure






References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra














Disclosure






References





























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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