Research ArticleExistence of Solutions for Some Nonlinear Problems withBoundary Value Conditions

Dionicio Pastor Dallos Santos

Department of Mathematics IME-USP Cidade Universitaria 05508-090 Sao Paulo SP Brazil

Correspondence should be addressed to Dionicio Pastor Dallos Santos dionicioimeuspbr

Received 7 June 2016 Revised 9 August 2016 Accepted 9 August 2016

Academic Editor Bingwen Liu

Copyright copy 2016 Dionicio Pastor Dallos Santos This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We study the existence of solutions for nonlinear boundary value problems (120593(1199061015840))1015840 = 119891(119905 119906 1199061015840) 119897(119906 1199061015840) = 0 where 119897(119906 1199061015840) = 0

denotes the boundary conditions on a compact interval [0 119879] 120593 is a homeomorphism such that 120593(0) = 0 and119891 [0 119879] times R times R rarr

R is a continuous function All the contemplated boundary value problems are reduced to finding a fixed point for one operatordefined on a space of functions and Schauder fixed point theorem or Leray-Schauder degree is used

1 Introduction

The purpose of this article is to obtain some existence resultsfor nonlinear boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119897 (119906 1199061015840) = 0

(1)

where 119897(119906 1199061015840) = 0 denotes the Dirichlet or mixed boundaryconditions on the interval [0 119879] 120593 is a bounded singularor classic homeomorphism such that 120593(0) = 0 119891 [0 119879] times

R times R rarr R is a continuous function and 119879 is a positive realnumber

Recently problem (1) in special cases when 120593 is anincreasing homeomorphism from (minus119886 119886) to R such that120593(0) = 0 and 119897(119906 119906

1015840) = 0 denotes the periodic Neumannor Dirichlet boundary conditions has been investigated byBereanu and Mawhin in [1]

In [2] the authors have studied problem (1) where 120593

R rarr (minus119886 119886) (0 lt 119886 le infin) and 119897(119906 1199061015840) = 0 denotes the

periodic boundary conditionsThey obtained the existence ofsolutions by means of the Leray-Schauder degree theory Inparticular regular periodic problems with 120593- or 119901-Laplacianon the left hand side were considered by several authors seefor example del Pino et al [3] or Yan [4]

In [5] Benevieri et al proved an existence result for theperiodic boundary value problem

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (0) = 119906 (119879)

1199061015840

(0) = 1199061015840

(119879)

(2)

assuming that 119891 [0 119879] times R119899 times R119899 rarr R119899 is a Caratheodoryfunction and 120593 R119899 rarr R119899 is a homeomorphism betweenR119899

and the open ball of R119899 with center zero and radius 1 Theyused a topological method the properties of 120593 and 119891 allowedapplying the Leray-Schauder degreeThe interest in this classof nonlinear operators 119906 997891rarr (120593(119906

1015840))1015840 is mainly due to the factthat they include the mean curvature operator

119906 997891997888rarr div(nabla119906

radic1 + |nabla119906|2

) (3)

The paper is organized as follows In Section 2 weintroduce some notations and preliminaries which will becrucial in the proofs of our results Section 3 is devoted to

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2016 Article ID 5283263 10 pageshttpdxdoiorg10115520165283263

2 Abstract and Applied Analysis

the study of existence of solutions for the Dirichlet problemswith bounded homomorphisms

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(4)

In particular Bereanu and Mawhin in [6] proved the exis-tence of at least one solution by means of the Leray-Schauderdegree

Theorem 1 (Bereanu and Mawhin) If the function 119891 satisfiesthe condition

exist119888 gt 0 such that 1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 le 119888 lt

119886

2119879

forall (119905 119909 119910) isin [0 119879] timesR timesR

(5)

the Dirichlet problem has at least one solution

The main purpose of this section is an extension of theresults obtained in the previous theorem For this we usetopological methods based upon Leray-Schauder degree [7]and more general properties of the function 119891 In Section 4we use the fixed point theorem of Schauder to show theexistence of at least one solution for boundary value problemsof the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 119906 (0) = 1199061015840

(119879)

(6)

where 120593 (minus119886 119886) rarr R (we call it singular) Of course asolution of (6) is a function 119906 [0 119879] rarr R of class 1198621 suchthat max

[0119879]|1199061015840(119905)| lt 119886 satisfying the boundary conditions

and the function 120593(1199061015840) is continuously differentiable and(120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879] In Section 5for 119906(119879) = 1199061015840(0) = 1199061015840(119879) boundary conditions and classichomeomorphisms (120593 R rarr R) we investigate the existenceof at least one solution using Leray-Schauder degree wherea solution of this problem is any function 119906 [0 119879] rarr

R of class 1198621 such that 120593(1199061015840) is continuously differentiable

which satisfies the boundary conditions and (120593(1199061015840(119905)))1015840 =

119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879] Such problems do notseem to have been studied in the literature In the presentpaper generally we follow the ideas of Bereanu and Mawhin[1 2 6 8]

2 Notation and Preliminaries

For fixed 119879 we denote the usual norm in 1198711 = 1198711([0 119879]R)

by sdot 1198711 Let 119862 = 119862([0 119879]R) denote the Banach space of

continuous functions from [0 119879] into R endowed with theuniform norm sdot

infin 1198621 = 119862

1([0 119879]R) the Banach space

of continuously differentiable functions from [0 119879] into Requipped with the usual norm 119906

1= 119906

infin+ 1199061015840

infin and 1198621

0

the closed subspace of 1198621 defined by 11986210= 119906 isin 1198621 119906(119879) =

0 = 119906(0)

We introduce the following applications

The Nemytskii operator 119873119891 1198621 rarr 119862

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (7)

The integration operator 119867 119862 rarr 1198621

119867(119906) (119905) = int119905

0

119906 (119904) 119889119904 (8)

The following continuous linear applications

119870 119862 997888rarr 1198621

119870 (119906) (119905) = minusint119879

119905

119906 (119904) 119889119904

119876 119862 997888rarr 119862

119876 (119906) (119905) =1

119879int119879

0

119906 (119904) 119889119904

119878 119862 997888rarr 119862

119878 (119906) (119905) = 119906 (119879)

119875 119862 997888rarr 119862

119875 (119906) (119905) = 119906 (0)

(9)

For 119906 isin 119862 we write

119906119898

= min[0119879]

119906

119906119872

= max[0119879]

119906

119906+= max 119906 0

119906minus= max minus119906 0

(10)

The following lemma is an adaptation of a result of[1] to the case of a homeomorphism which is not definedeverywhere We present here the demonstration for betterunderstanding of the development of our research

Lemma 2 Let 119861 = ℎ isin 119862 ℎinfin

lt 1198862 For each ℎ isin

119861 there exists a unique 119876120593

= 119876120593(ℎ) isin Im(ℎ) (where Im(ℎ)

denotes the range of ℎ) such that

int119879

0

120593minus1

(ℎ (119905) minus 119876120593(ℎ)) 119889119905 = 0 (11)

Moreover the function 119876120593

119861 rarr R is continuous and sendsbounded sets into bounded sets

Proof Let ℎ isin 119861 We define the continuous application 119866ℎ

[ℎ119898 ℎ119872] rarr R for

119866ℎ(119904) = int

119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 (12)

Abstract and Applied Analysis 3

We now show that the equation

119866ℎ(119904) = 0 (13)

has a unique solution 119876120593(ℎ) Let 119903 119904 isin [ℎ

119898 ℎ119872] be such that

int119879

0

120593minus1

(ℎ (119905) minus 119903) 119889119905 = 0

int119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 = 0

(14)

that is

int119879

0

120593minus1

(ℎ (119905) minus 119903) 119889119905 = int119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 (15)

It follows that there exists 120591 isin [0 119879] such that

120593minus1

(ℎ (120591) minus 119903) = 120593minus1

(ℎ (120591) minus 119904) (16)

Using the injectivity of 120593minus1 we deduce that 119903 = 119904 Let us

now show the existence Because120593minus1 is strictlymonotone and120593minus1(0) = 0 we have that

119866ℎ(ℎ119898) 119866ℎ(ℎ119872) le 0 (17)

It follows that there exists 119904 isin [ℎ119898 ℎ119872] such that 119866

ℎ(119904) = 0

Consequently for each ℎ isin 119861 (13) has a unique solutionThuswe define the function 119876

120593 119861 rarr R such that

int119879

0

120593minus1

(ℎ (119905) minus 119876120593(ℎ)) 119889119905 = 0 (18)

On the other hand because ℎ isin 119861 we have that10038161003816100381610038161003816119876120593(ℎ)

10038161003816100381610038161003816le ℎinfin

lt119886

2 (19)

Therefore the function119876120593sends bounded sets into bounded

setsFinally we show that119876

120593is continuous on119861 Let (ℎ

119899)119899sub 119862

be a sequence such that ℎ119899

rarr ℎ in 119862 Since the function119876120593sends bounded sets into bounded sets then (119876

120593(ℎ119899))119899

is bounded Hence (119876120593(ℎ119899))119899is relatively compact Without

loss of generality passing if necessary to a subsequence wecan assume that

lim119899rarrinfin

119876120593(ℎ119899) = (20)

where for each 119899 isin N we obtain

lim119899rarrinfin

int119879

0

120593minus1

(ℎ119899(119905) minus 119876

120593(ℎ119899)) 119889119905 = 0 (21)

Using the dominated convergence theorem we deduce that

int119879

0

120593minus1

(ℎ (119905) minus ) 119889119905 = 0 (22)

so we have that 119876120593(ℎ) = Hence the function 119876

120593is

continuous

The following extended homotopy invariance property ofthe Leray-Schauder degree can be found in [9]

Proposition 3 Let 119883 be a real Banach space 119881 sub [0 1] times 119883

be an open bounded set and 119872 be a completely continuousoperator on 119881 such that 119909 = 119872(120582 119909) for each (120582 119909) isin 120597119881Then the Leray-Schauder degree

degLS (119868 minus 119872 (120582 sdot) 119881120582 0) (23)

is well defined and independent of 120582 in [0 1] where 119881120582is the

open bounded (possibly empty) set defined by 119881120582

= 119909 isin 119883

(120582 119909) isin 119881

3 Dirichlet Problems with BoundedHomeomorphisms

In this section we are interested in Dirichlet boundary valueproblems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(24)

where 120593 R rarr (minus119886 119886) is a homeomorphism 120593(0) = 0 and119891 [0 119879]timesRtimesR rarr R is continuous In order to apply Leray-Schauder degree theory to show the existence of at least onesolution of (24) we introduce for 120582 isin [0 1] the family ofDirichlet boundary value problems

(120593 (1199061015840))1015840

= 120582119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(25)

Let

Ω = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinlt

119886

2 (26)

ClearlyΩ is an open set in [0 1]times11986210and is nonempty because

0 times 11986210

sub Ω Using Lemma 2 we can define the operator119872 Ω rarr 1198621

0by

119872(120582 119906)

= 119867(120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867(119873

119891(119906)))])

(27)

Here 120593minus1 with an abuse of notation is understood as theoperator 120593minus1 119861

119886(0) sub 119862 rarr 119862 defined by 120593minus1(V)(119905) =

120593minus1(V(119905)) It is clear that120593minus1 is continuous and sends boundedsets into bounded sets

When the boundary conditions are periodic orNeumannan operator has been considered by Bereanu andMawhin [6]

The following lemma plays a pivotal role in studying thesolutions of problem (25)

Lemma 4 The operator 119872 is well defined and continuousMoreover if (120582 119906) isin Ω is such that 119872(120582 119906) = 119906 then 119906 issolution of (25)

4 Abstract and Applied Analysis

Proof Let (120582 119906) isin Ω We show that in fact119872(120582 119906) isin 11986210 It is

clear that

(119872 (120582 119906))1015840

= 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]

(28)

where the continuity of 119872(120582 119906) and (119872(120582 119906))1015840 follows fromthe continuity of applications119867 and119873

119891

On the other hand using Lemma 2 we have

119872(120582 119906) (0) = 0 = 119872 (120582 119906) (119879) (29)

Therefore 119872(Ω) sub 11986210and 119872 is well defined The continuity

of 119872 follows by the continuity of the operators whichcompose it119872

Now suppose that (120582 119906) isin Ω is such that 119872(120582 119906) = 119906 Itfollows from (27) that

119906 (119905) = 119872 (120582 119906) (119905)

= 119867 (120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]) (119905)

(30)

for all 119905 isin [0 119879] Differentiating (30) we obtain

1199061015840

(119905) = 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))] (119905)

= 120593minus1

[120582119867 (119873119891(119906)) (119905) minus 119876

120593(120582119867 (119873

119891(119906)))]

(31)

Applying 120593 to both of its members we have that

120593 (1199061015840

(119905)) = 120582119867(119873119891(119906)) (119905) minus 119876

120593(120582119867(119873

119891(119906))) (32)

Differentiating again we deduce that

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905) (33)

for all 119905 isin [0 119879]Thus119906 satisfies problem (25)This completesthe proof

Remark 5 Note that the reciprocal of Lemma 4 is not truebecause we cannot guarantee that 120582119867(119873

119891(119906)infin

lt 1198862 forevery solution 119906 of (25)

In ourmain result we need the following lemma to obtainthe required a priori bounds for the possible fixed points of119872

Lemma 6 Assume that there exist ℎ isin 119862([0 119879]R+) and 119899 isin

1198621(RR) such that ℎ1198711 lt 1198862 119899(0) = 0

120593 (119910) 1198991015840

(119909) 119910 ge 0 (119905 119909 119910) isin [0 119879] timesR timesR (34)1003816100381610038161003816119891 (119905 119909 119910)

1003816100381610038161003816 le 119891 (119905 119909 119910) 119899 (119909) + ℎ (119905) (35)

for all (119905 119909 119910) isin [0 119879] times R times R If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(36)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Let 120582 = 0 and (120582 119906) isin Ω be such that 119872(120582 119906) = 119906Using Lemma 4 we have that 119906 is solution of (25) whichimplies that

120593 (1199061015840) = 120582119867(119873

119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))

119906 (0) = 0 = 119906 (119879)

(37)

where for all 119905 isin [0 119879] we obtain

10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119899 (119906 (119904)) 119889119904

+ int119879

0

ℎ (119904) 119889119904

(38)

On the other hand using inequality (34) we have that

minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0 (39)

Using the integration by parts formula the boundary condi-tions and the fact that 119899(0) = 0 we deduce that

int119879

0

(120593 (1199061015840

(119905)))1015840

119899 (119906 (119905)) 119889119905

= minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0

(40)

Since 120582 isin (0 1] and 119906 is solution of (25) it follows that

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119899 (119906 (119905)) 119889119905 le 0 (41)

and hence10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le ℎ1198711 (42)

On the other hand since 119876120593(120582119867(119873

119891(119906))) isin

Im(120582119867(119873119891(119906))) we get

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le 2 ℎ

1198711 (43)

for all 119905 isin [0 119879] It follows that10038171003817100381710038171003817120593 (1199061015840)10038171003817100381710038171003817infin

le 2 ℎ1198711 (44)

which implies that 1199061015840infin

le 119871 where 119871 =

max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)| Using again the

boundary conditions we have that

|119906 (119905)| le int119905

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le int

119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le 119871119879

(119905 isin [0 119879])

(45)

and hence

1199061le 119871 + 119871119879 (46)

Finally if 119906 = 119872(0 119906) then 119906 = 0 so the proof is complete

Abstract and Applied Analysis 5

Let 120588 120581 isin R be such that ℎ1198711 lt 120581 lt 1198862 120588 gt 119871 + 119871119879

and consider the set

119881 = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infin

lt 120581 1199061lt 120588

(47)

Since the set 0 times 119906 isin 11986210 1199061lt 120588 sub 119881 then we deduce

that 119881 is nonempty Moreover it is clear that 119881 is open andbounded in [0 1] times 1198621

0and 119881 sub Ω On the other hand using

an argument similar to the one introduced in the proof ofLemma 6 in [6] it is not difficult to see that 119872 119881 rarr 1198621

0

is well defined and completely continuous and

119906 = 119872 (120582 119906) forall (120582 119906) isin 120597119881 (48)

31 Existence Results In this subsection we present andprove our main result

Theorem7 If119891 satisfies conditions of Lemma 6 then problem(24) has at least one solution

Proof Let 119872 be the operator given by (27) Using Proposi-tion 3 we deduce that

degLS (119868 minus 119872 (0 sdot) 1198810 0)

= degLS (119868 minus 119872 (1 sdot) 1198811 0)

(49)

where degLS(119868 minus119872(0 sdot) 1198810 0) = degLS(119868 119861120588(0) 0) = 1 Thus

there exists 119906 isin 1198811such that 119872(1 119906) = 119906 which is a solution

for (24)

Remark 8 Note that Theorem 7 is a generalization of Theo-rem 1

Corollary 9 Assume that 120593 is an increasing homomorphismLet ℎ isin 119862([0 119879]R+) be such that

ℎ1198711 lt

119886

2

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 le 119891 (119905 119909 119910) 119909 + ℎ (119905)

(50)

for all 119909 119910 isin R and 119905 isin [0 119879] If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(51)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Since 120593 is an increasing homomorphism we have that

120593 (119910) 119910 ge 0 (52)

for all 119910 isin R Using Lemma 6 with 119899(119909) = 119909 for all 119909 isin

R we can obtain the conclusion of Corollary 9 The proof isachieved

Theorem 10 If 119891 satisfies conditions of Corollary 9 thenproblem (24) has at least one solution

Let us give now an application of Theorem 10 when 119891 isunbounded

Example 11 Consider the Dirichlet problem

(120593 (1199061015840))1015840

= 119906 minus 2

119906 (0) = 119906 (119879) = 0

(53)

where 120593(119904) = 119904radic1 + 1199042It is not difficult to verify that 120593 R rarr (minus1 1) is

an increasing homeomorphism and 119891(119905 119909 119910) = 119909 minus 2 is acontinuous function such that

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 = |119909 minus 2| le (119909 minus 2) 119909 + 4

(119905 119909 119910) isin [0 119879] timesR timesR(54)

So we can choose ℎ(119905) = 4 and119879 lt 18 to see that Corollary 9holds and so usingTheorem 10we obtain that (53) has at leastone solution

4 Problems with Singular Homeomorphismsand Three-Point Boundary Conditions

In this section we study the existence of at least one solutionfor boundary value problems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 119906 (0) = 1199061015840

(119879)

(55)

where 120593 (minus119886 119886) rarr R is a homeomorphism such that 120593(0) =

0 and 119891 [0 119879] timesR timesR rarr R is a continuous functionIn order to transform problem (55) to a fixed point

problem we use a similar argument introduced in Lemma 2for ℎ isin 119862

Lemma 12 119906 isin 1198621 is a solution of (55) if and only if 119906 is a

fixed point of the operator 119872 defined on 1198621 by

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(56)

Proof If 119906 isin 1198621 is solution of (55) then

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905))

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(57)

for all 119905 isin [0 119879] Applying 119870 to both members and using thefact that 119906(0) = 119906

1015840(119879) we deduce that

120593 (1199061015840

(119905)) = 120593 (119906 (0)) + 119870 (119873119891(119906)) (119905) (58)

6 Abstract and Applied Analysis

By the inversion of 120593 in (58) we have

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) (119905) + 119888] (59)

where 119888 = 120593(119906(0)) Integrating from 0 to 119905 isin [0 119879] we havethat

119906 (119905) = 119906 (0) + 119867 (120593minus1

[119870 (119873119891(119906)) + 119888]) (119905) (60)

Because 119906(0) = 119906(119879) then

int119879

0

120593minus1

[119870 (119873119891(119906)) (119905) + 119888] 119889119905 = 0 (61)

Using an argument similar to the one introduced in Lemma 2it follows that 119888 = minus119876

120593(119870(119873119891(119906))) Hence

119906 = 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(62)

Let 119906 isin 1198621 be such that 119906 = 119872(119906) Then

119906 (119905)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]) (119905)

(63)

for all 119905 isin [0 119879] Since int119879

0120593minus1[119870(119873

119891(119906))(119905) minus

119876120593(119870(119873119891(119906)))]119889119905 = 0 therefore we have that 119906(0) = 119906(119879)

Differentiating (63) we obtain that

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))] (119905) (64)

In particular

1199061015840

(119879) = 120593minus1

(0 minus 119876120593(119870 (119873

119891(119906))))

= 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119906 (0)

(65)

Applying 120593 to both members and differentiating again wededuce that

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905)

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(66)

for all 119905 isin [0 119879] This completes the proof

Lemma 13 The operator 119872 1198621 rarr 1198621 is completelycontinuous

Proof Let Λ sub 1198621 be a bounded set Then if 119906 isin Λ thereexists a constant 120588 gt 0 such that

1199061le 120588 (67)

Next we show that119872(Λ) sub 1198621 is a compact set Let (V119899)119899be a

sequence in119872(Λ) and let (119906119899)119899be a sequence in Λ such that

V119899= 119872(119906

119899) Using (67) we have that there exists a constant

119871 gt 0 such that for all 119899 isin N10038171003817100381710038171003817119873119891(119906119899)10038171003817100381710038171003817infin

le 119871 (68)

which implies that

10038171003817100381710038171003817119870 (119873119891(119906119899)) minus 119876

120593(119870 (119873

119891(119906119899)))

10038171003817100381710038171003817infinle 2119871119879 (69)

Hence the sequence (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isbounded in 119862 Moreover for 119905 119905

1isin [0 119879] and for all 119899 isin N

we have that10038161003816100381610038161003816119870 (119873119891(119906119899)) (119905) minus 119876

120593(119870 (119873

119891(119906119899)))

minus 119870 (119873119891(119906119899)) (1199051) + 119876120593(119870 (119873

119891(119906119899)))

10038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816minus int119879

119905

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

+ int119879

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816int119905

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816le 119871

1003816100381610038161003816119905 minus 1199051

1003816100381610038161003816

(70)

which implies that (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isequicontinuous Thus by the Arzela-Ascoli theorem there isa subsequence of (119870(119873

119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899 which we

call (119870(119873119891(119906119899119895

)) minus 119876120593(119870(119873119891(119906119899119895

))))119895 which is convergent in

119862 Using the fact that 120593minus1 119862 rarr 119861119886(0) sub 119862 is continuous it

follows from

119872(119906119899119895

)1015840

= 120593minus1

[119870 (119873119891(119906119899119895

)) minus 119876120593(119870(119873

119891(119906119899119895

)))]

(71)

that the sequence (119872(119906119899119895

)1015840)119895is convergent in119862Then passing

to a subsequence if necessary we obtain that (V119899119895

)119895

=

(119872(119906119899119895

))119895is convergent in 1198621 Finally let (V

119899)119899be a sequence

in119872(Λ) Let (119911119899)119899sube 119872(Λ) be such that

lim119899rarrinfin

1003817100381710038171003817119911119899 minus V119899

10038171003817100381710038171 = 0 (72)

Let (119911119899119895

)119895be a subsequence of (119911

119899)119899such that it converges to

119911 It follows that 119911 isin 119872(Λ) and (V119899119895

)119895converge to 119911 This

concludes the proof

The next result is based on Schauderrsquos fixed point theo-rem

Theorem 14 Let 119891 [0 119879] timesRtimesR rarr R be continuous Then(55) has at least one solution

Abstract and Applied Analysis 7

Proof Let 119906 isin 1198621 Then

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(73)

where

119872(119906) (0) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (119879)

119872 (119906)1015840

(119879) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (0)

(74)

Moreover10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infin

=10038171003817100381710038171003817120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]

10038171003817100381710038171003817infinlt 119886

119872 (119906)infin

lt 119886 + 119886119879

(75)

Hence

119872 (119906)1= 119872 (119906)

infin+10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infinlt 119886 + 119886119879 + 119886

= 2119886 + 119886119879

(76)

Because the operator 119872 is completely continuous andbounded we can use Schauderrsquos fixed point theorem todeduce the existence of at least one fixed point This in turnimplies that problem (55) has at least one solution The proofis complete

5 Problems with Classic Homeomorphismsand Three-Point Boundary Conditions

We finally consider boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(77)

where 120593 R rarr R is a homeomorphism such that120593(0) = 0 and 119891 [0 119879] times R times R rarr R is a continuousfunction We remember that a solution of this problem isany function 119906 [0 119879] rarr R of class 119862

1 such that120593(1199061015840) is continuously differentiable satisfying the boundaryconditions and (120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879]

Let us consider the operator

1198721 1198621997888rarr 119862

1

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(78)

Analogously to Section 3 here 120593minus1 is understood as theoperator 120593minus1 119862 rarr 119862 defined for 120593minus1(V)(119905) = 120593minus1(V(119905)) Itis clear that 120593minus1 is continuous and sends bounded sets intobounded sets

Lemma 15 119906 isin 1198621 is a solution of (77) if and only if 119906 is a

fixed point of the operator 1198721

Proof Let 119906 isin 1198621 and we have the following equivalences

(120593 (1199061015840))1015840

= 119873119891(119906)

1199061015840

(119879) = 1199061015840

(0)

1199061015840

(0) = 119906 (119879)

lArrrArr (120593 (1199061015840))1015840

= 119873119891(119906) minus 119876 (119873

119891(119906))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 120593(1199061015840) = 119867(119873

119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906

1015840

(0))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 1199061015840= 120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))] 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))]) 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906 (119879))])

lArrrArr 119906 = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(79)

Remark 16 Note that 1199061015840(119879) = 1199061015840(0) hArr 119876(119873119891(119906)) = 0

Using an argument similar to the one introduced inLemma 13 it is easy to see that 119872

1 1198621 rarr 1198621 is completely

continuousIn order to apply Leray-Schauder degree to the operator

1198721 we introduced a family of problems depending on

parameter 120582 We remember that to each continuous function119891 [0 119879] times R times R rarr R we associate its Nemytskii operator119873119891 1198621 rarr 119862 defined by

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (80)

For 120582 isin [0 1] we consider the family of boundary valueproblems

(120593 (1199061015840))1015840

= 120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(81)

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

the study of existence of solutions for the Dirichlet problemswith bounded homomorphisms

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(4)

In particular Bereanu and Mawhin in [6] proved the exis-tence of at least one solution by means of the Leray-Schauderdegree

Theorem 1 (Bereanu and Mawhin) If the function 119891 satisfiesthe condition

exist119888 gt 0 such that 1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 le 119888 lt

119886

2119879

forall (119905 119909 119910) isin [0 119879] timesR timesR

(5)

the Dirichlet problem has at least one solution

The main purpose of this section is an extension of theresults obtained in the previous theorem For this we usetopological methods based upon Leray-Schauder degree [7]and more general properties of the function 119891 In Section 4we use the fixed point theorem of Schauder to show theexistence of at least one solution for boundary value problemsof the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 119906 (0) = 1199061015840

(119879)

(6)

where 120593 (minus119886 119886) rarr R (we call it singular) Of course asolution of (6) is a function 119906 [0 119879] rarr R of class 1198621 suchthat max

[0119879]|1199061015840(119905)| lt 119886 satisfying the boundary conditions

and the function 120593(1199061015840) is continuously differentiable and(120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879] In Section 5for 119906(119879) = 1199061015840(0) = 1199061015840(119879) boundary conditions and classichomeomorphisms (120593 R rarr R) we investigate the existenceof at least one solution using Leray-Schauder degree wherea solution of this problem is any function 119906 [0 119879] rarr

R of class 1198621 such that 120593(1199061015840) is continuously differentiable

which satisfies the boundary conditions and (120593(1199061015840(119905)))1015840 =

119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879] Such problems do notseem to have been studied in the literature In the presentpaper generally we follow the ideas of Bereanu and Mawhin[1 2 6 8]

2 Notation and Preliminaries

For fixed 119879 we denote the usual norm in 1198711 = 1198711([0 119879]R)

by sdot 1198711 Let 119862 = 119862([0 119879]R) denote the Banach space of

continuous functions from [0 119879] into R endowed with theuniform norm sdot

infin 1198621 = 119862

1([0 119879]R) the Banach space

of continuously differentiable functions from [0 119879] into Requipped with the usual norm 119906

1= 119906

infin+ 1199061015840

infin and 1198621

0

the closed subspace of 1198621 defined by 11986210= 119906 isin 1198621 119906(119879) =

0 = 119906(0)

We introduce the following applications

The Nemytskii operator 119873119891 1198621 rarr 119862

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (7)

The integration operator 119867 119862 rarr 1198621

119867(119906) (119905) = int119905

0

119906 (119904) 119889119904 (8)

The following continuous linear applications

119870 119862 997888rarr 1198621

119870 (119906) (119905) = minusint119879

119905

119906 (119904) 119889119904

119876 119862 997888rarr 119862

119876 (119906) (119905) =1

119879int119879

0

119906 (119904) 119889119904

119878 119862 997888rarr 119862

119878 (119906) (119905) = 119906 (119879)

119875 119862 997888rarr 119862

119875 (119906) (119905) = 119906 (0)

(9)

For 119906 isin 119862 we write

119906119898

= min[0119879]

119906

119906119872

= max[0119879]

119906

119906+= max 119906 0

119906minus= max minus119906 0

(10)

The following lemma is an adaptation of a result of[1] to the case of a homeomorphism which is not definedeverywhere We present here the demonstration for betterunderstanding of the development of our research

Lemma 2 Let 119861 = ℎ isin 119862 ℎinfin

lt 1198862 For each ℎ isin

119861 there exists a unique 119876120593

= 119876120593(ℎ) isin Im(ℎ) (where Im(ℎ)

denotes the range of ℎ) such that

int119879

0

120593minus1

(ℎ (119905) minus 119876120593(ℎ)) 119889119905 = 0 (11)

Moreover the function 119876120593

119861 rarr R is continuous and sendsbounded sets into bounded sets

Proof Let ℎ isin 119861 We define the continuous application 119866ℎ

[ℎ119898 ℎ119872] rarr R for

119866ℎ(119904) = int

119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 (12)

Abstract and Applied Analysis 3

We now show that the equation

119866ℎ(119904) = 0 (13)

has a unique solution 119876120593(ℎ) Let 119903 119904 isin [ℎ

119898 ℎ119872] be such that

int119879

0

120593minus1

(ℎ (119905) minus 119903) 119889119905 = 0

int119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 = 0

(14)

that is

int119879

0

120593minus1

(ℎ (119905) minus 119903) 119889119905 = int119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 (15)

It follows that there exists 120591 isin [0 119879] such that

120593minus1

(ℎ (120591) minus 119903) = 120593minus1

(ℎ (120591) minus 119904) (16)

Using the injectivity of 120593minus1 we deduce that 119903 = 119904 Let us

now show the existence Because120593minus1 is strictlymonotone and120593minus1(0) = 0 we have that

119866ℎ(ℎ119898) 119866ℎ(ℎ119872) le 0 (17)

It follows that there exists 119904 isin [ℎ119898 ℎ119872] such that 119866

ℎ(119904) = 0

Consequently for each ℎ isin 119861 (13) has a unique solutionThuswe define the function 119876

120593 119861 rarr R such that

int119879

0

120593minus1

(ℎ (119905) minus 119876120593(ℎ)) 119889119905 = 0 (18)

On the other hand because ℎ isin 119861 we have that10038161003816100381610038161003816119876120593(ℎ)

10038161003816100381610038161003816le ℎinfin

lt119886

2 (19)

Therefore the function119876120593sends bounded sets into bounded

setsFinally we show that119876

120593is continuous on119861 Let (ℎ

119899)119899sub 119862

be a sequence such that ℎ119899

rarr ℎ in 119862 Since the function119876120593sends bounded sets into bounded sets then (119876

120593(ℎ119899))119899

is bounded Hence (119876120593(ℎ119899))119899is relatively compact Without

loss of generality passing if necessary to a subsequence wecan assume that

lim119899rarrinfin

119876120593(ℎ119899) = (20)

where for each 119899 isin N we obtain

lim119899rarrinfin

int119879

0

120593minus1

(ℎ119899(119905) minus 119876

120593(ℎ119899)) 119889119905 = 0 (21)

Using the dominated convergence theorem we deduce that

int119879

0

120593minus1

(ℎ (119905) minus ) 119889119905 = 0 (22)

so we have that 119876120593(ℎ) = Hence the function 119876

120593is

continuous

The following extended homotopy invariance property ofthe Leray-Schauder degree can be found in [9]

Proposition 3 Let 119883 be a real Banach space 119881 sub [0 1] times 119883

be an open bounded set and 119872 be a completely continuousoperator on 119881 such that 119909 = 119872(120582 119909) for each (120582 119909) isin 120597119881Then the Leray-Schauder degree

degLS (119868 minus 119872 (120582 sdot) 119881120582 0) (23)

is well defined and independent of 120582 in [0 1] where 119881120582is the

open bounded (possibly empty) set defined by 119881120582

= 119909 isin 119883

(120582 119909) isin 119881

3 Dirichlet Problems with BoundedHomeomorphisms

In this section we are interested in Dirichlet boundary valueproblems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(24)

where 120593 R rarr (minus119886 119886) is a homeomorphism 120593(0) = 0 and119891 [0 119879]timesRtimesR rarr R is continuous In order to apply Leray-Schauder degree theory to show the existence of at least onesolution of (24) we introduce for 120582 isin [0 1] the family ofDirichlet boundary value problems

(120593 (1199061015840))1015840

= 120582119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(25)

Let

Ω = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinlt

119886

2 (26)

ClearlyΩ is an open set in [0 1]times11986210and is nonempty because

0 times 11986210

sub Ω Using Lemma 2 we can define the operator119872 Ω rarr 1198621

0by

119872(120582 119906)

= 119867(120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867(119873

119891(119906)))])

(27)

Here 120593minus1 with an abuse of notation is understood as theoperator 120593minus1 119861

119886(0) sub 119862 rarr 119862 defined by 120593minus1(V)(119905) =

120593minus1(V(119905)) It is clear that120593minus1 is continuous and sends boundedsets into bounded sets

When the boundary conditions are periodic orNeumannan operator has been considered by Bereanu andMawhin [6]

The following lemma plays a pivotal role in studying thesolutions of problem (25)

Lemma 4 The operator 119872 is well defined and continuousMoreover if (120582 119906) isin Ω is such that 119872(120582 119906) = 119906 then 119906 issolution of (25)

4 Abstract and Applied Analysis

Proof Let (120582 119906) isin Ω We show that in fact119872(120582 119906) isin 11986210 It is

clear that

(119872 (120582 119906))1015840

= 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]

(28)

where the continuity of 119872(120582 119906) and (119872(120582 119906))1015840 follows fromthe continuity of applications119867 and119873

119891

On the other hand using Lemma 2 we have

119872(120582 119906) (0) = 0 = 119872 (120582 119906) (119879) (29)

Therefore 119872(Ω) sub 11986210and 119872 is well defined The continuity

of 119872 follows by the continuity of the operators whichcompose it119872

Now suppose that (120582 119906) isin Ω is such that 119872(120582 119906) = 119906 Itfollows from (27) that

119906 (119905) = 119872 (120582 119906) (119905)

= 119867 (120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]) (119905)

(30)

for all 119905 isin [0 119879] Differentiating (30) we obtain

1199061015840

(119905) = 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))] (119905)

= 120593minus1

[120582119867 (119873119891(119906)) (119905) minus 119876

120593(120582119867 (119873

119891(119906)))]

(31)

Applying 120593 to both of its members we have that

120593 (1199061015840

(119905)) = 120582119867(119873119891(119906)) (119905) minus 119876

120593(120582119867(119873

119891(119906))) (32)

Differentiating again we deduce that

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905) (33)

for all 119905 isin [0 119879]Thus119906 satisfies problem (25)This completesthe proof

Remark 5 Note that the reciprocal of Lemma 4 is not truebecause we cannot guarantee that 120582119867(119873

119891(119906)infin

lt 1198862 forevery solution 119906 of (25)

In ourmain result we need the following lemma to obtainthe required a priori bounds for the possible fixed points of119872

Lemma 6 Assume that there exist ℎ isin 119862([0 119879]R+) and 119899 isin

1198621(RR) such that ℎ1198711 lt 1198862 119899(0) = 0

120593 (119910) 1198991015840

(119909) 119910 ge 0 (119905 119909 119910) isin [0 119879] timesR timesR (34)1003816100381610038161003816119891 (119905 119909 119910)

1003816100381610038161003816 le 119891 (119905 119909 119910) 119899 (119909) + ℎ (119905) (35)

for all (119905 119909 119910) isin [0 119879] times R times R If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(36)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Let 120582 = 0 and (120582 119906) isin Ω be such that 119872(120582 119906) = 119906Using Lemma 4 we have that 119906 is solution of (25) whichimplies that

120593 (1199061015840) = 120582119867(119873

119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))

119906 (0) = 0 = 119906 (119879)

(37)

where for all 119905 isin [0 119879] we obtain

10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119899 (119906 (119904)) 119889119904

+ int119879

0

ℎ (119904) 119889119904

(38)

On the other hand using inequality (34) we have that

minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0 (39)

Using the integration by parts formula the boundary condi-tions and the fact that 119899(0) = 0 we deduce that

int119879

0

(120593 (1199061015840

(119905)))1015840

119899 (119906 (119905)) 119889119905

= minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0

(40)

Since 120582 isin (0 1] and 119906 is solution of (25) it follows that

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119899 (119906 (119905)) 119889119905 le 0 (41)

and hence10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le ℎ1198711 (42)

On the other hand since 119876120593(120582119867(119873

119891(119906))) isin

Im(120582119867(119873119891(119906))) we get

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le 2 ℎ

1198711 (43)

for all 119905 isin [0 119879] It follows that10038171003817100381710038171003817120593 (1199061015840)10038171003817100381710038171003817infin

le 2 ℎ1198711 (44)

which implies that 1199061015840infin

le 119871 where 119871 =

max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)| Using again the

boundary conditions we have that

|119906 (119905)| le int119905

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le int

119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le 119871119879

(119905 isin [0 119879])

(45)

and hence

1199061le 119871 + 119871119879 (46)

Finally if 119906 = 119872(0 119906) then 119906 = 0 so the proof is complete

Abstract and Applied Analysis 5

Let 120588 120581 isin R be such that ℎ1198711 lt 120581 lt 1198862 120588 gt 119871 + 119871119879

and consider the set

119881 = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infin

lt 120581 1199061lt 120588

(47)

Since the set 0 times 119906 isin 11986210 1199061lt 120588 sub 119881 then we deduce

that 119881 is nonempty Moreover it is clear that 119881 is open andbounded in [0 1] times 1198621

0and 119881 sub Ω On the other hand using

an argument similar to the one introduced in the proof ofLemma 6 in [6] it is not difficult to see that 119872 119881 rarr 1198621

0

is well defined and completely continuous and

119906 = 119872 (120582 119906) forall (120582 119906) isin 120597119881 (48)

31 Existence Results In this subsection we present andprove our main result

Theorem7 If119891 satisfies conditions of Lemma 6 then problem(24) has at least one solution

Proof Let 119872 be the operator given by (27) Using Proposi-tion 3 we deduce that

degLS (119868 minus 119872 (0 sdot) 1198810 0)

= degLS (119868 minus 119872 (1 sdot) 1198811 0)

(49)

where degLS(119868 minus119872(0 sdot) 1198810 0) = degLS(119868 119861120588(0) 0) = 1 Thus

there exists 119906 isin 1198811such that 119872(1 119906) = 119906 which is a solution

for (24)

Remark 8 Note that Theorem 7 is a generalization of Theo-rem 1

Corollary 9 Assume that 120593 is an increasing homomorphismLet ℎ isin 119862([0 119879]R+) be such that

ℎ1198711 lt

119886

2

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 le 119891 (119905 119909 119910) 119909 + ℎ (119905)

(50)

for all 119909 119910 isin R and 119905 isin [0 119879] If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(51)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Since 120593 is an increasing homomorphism we have that

120593 (119910) 119910 ge 0 (52)

for all 119910 isin R Using Lemma 6 with 119899(119909) = 119909 for all 119909 isin

R we can obtain the conclusion of Corollary 9 The proof isachieved

Theorem 10 If 119891 satisfies conditions of Corollary 9 thenproblem (24) has at least one solution

Let us give now an application of Theorem 10 when 119891 isunbounded

Example 11 Consider the Dirichlet problem

(120593 (1199061015840))1015840

= 119906 minus 2

119906 (0) = 119906 (119879) = 0

(53)

where 120593(119904) = 119904radic1 + 1199042It is not difficult to verify that 120593 R rarr (minus1 1) is

an increasing homeomorphism and 119891(119905 119909 119910) = 119909 minus 2 is acontinuous function such that

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 = |119909 minus 2| le (119909 minus 2) 119909 + 4

(119905 119909 119910) isin [0 119879] timesR timesR(54)

So we can choose ℎ(119905) = 4 and119879 lt 18 to see that Corollary 9holds and so usingTheorem 10we obtain that (53) has at leastone solution

4 Problems with Singular Homeomorphismsand Three-Point Boundary Conditions

In this section we study the existence of at least one solutionfor boundary value problems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 119906 (0) = 1199061015840

(119879)

(55)

where 120593 (minus119886 119886) rarr R is a homeomorphism such that 120593(0) =

0 and 119891 [0 119879] timesR timesR rarr R is a continuous functionIn order to transform problem (55) to a fixed point

problem we use a similar argument introduced in Lemma 2for ℎ isin 119862

Lemma 12 119906 isin 1198621 is a solution of (55) if and only if 119906 is a

fixed point of the operator 119872 defined on 1198621 by

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(56)

Proof If 119906 isin 1198621 is solution of (55) then

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905))

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(57)

for all 119905 isin [0 119879] Applying 119870 to both members and using thefact that 119906(0) = 119906

1015840(119879) we deduce that

120593 (1199061015840

(119905)) = 120593 (119906 (0)) + 119870 (119873119891(119906)) (119905) (58)

6 Abstract and Applied Analysis

By the inversion of 120593 in (58) we have

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) (119905) + 119888] (59)

where 119888 = 120593(119906(0)) Integrating from 0 to 119905 isin [0 119879] we havethat

119906 (119905) = 119906 (0) + 119867 (120593minus1

[119870 (119873119891(119906)) + 119888]) (119905) (60)

Because 119906(0) = 119906(119879) then

int119879

0

120593minus1

[119870 (119873119891(119906)) (119905) + 119888] 119889119905 = 0 (61)

Using an argument similar to the one introduced in Lemma 2it follows that 119888 = minus119876

120593(119870(119873119891(119906))) Hence

119906 = 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(62)

Let 119906 isin 1198621 be such that 119906 = 119872(119906) Then

119906 (119905)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]) (119905)

(63)

for all 119905 isin [0 119879] Since int119879

0120593minus1[119870(119873

119891(119906))(119905) minus

119876120593(119870(119873119891(119906)))]119889119905 = 0 therefore we have that 119906(0) = 119906(119879)

Differentiating (63) we obtain that

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))] (119905) (64)

In particular

1199061015840

(119879) = 120593minus1

(0 minus 119876120593(119870 (119873

119891(119906))))

= 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119906 (0)

(65)

Applying 120593 to both members and differentiating again wededuce that

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905)

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(66)

for all 119905 isin [0 119879] This completes the proof

Lemma 13 The operator 119872 1198621 rarr 1198621 is completelycontinuous

Proof Let Λ sub 1198621 be a bounded set Then if 119906 isin Λ thereexists a constant 120588 gt 0 such that

1199061le 120588 (67)

Next we show that119872(Λ) sub 1198621 is a compact set Let (V119899)119899be a

sequence in119872(Λ) and let (119906119899)119899be a sequence in Λ such that

V119899= 119872(119906

119899) Using (67) we have that there exists a constant

119871 gt 0 such that for all 119899 isin N10038171003817100381710038171003817119873119891(119906119899)10038171003817100381710038171003817infin

le 119871 (68)

which implies that

10038171003817100381710038171003817119870 (119873119891(119906119899)) minus 119876

120593(119870 (119873

119891(119906119899)))

10038171003817100381710038171003817infinle 2119871119879 (69)

Hence the sequence (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isbounded in 119862 Moreover for 119905 119905

1isin [0 119879] and for all 119899 isin N

we have that10038161003816100381610038161003816119870 (119873119891(119906119899)) (119905) minus 119876

120593(119870 (119873

119891(119906119899)))

minus 119870 (119873119891(119906119899)) (1199051) + 119876120593(119870 (119873

119891(119906119899)))

10038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816minus int119879

119905

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

+ int119879

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816int119905

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816le 119871

1003816100381610038161003816119905 minus 1199051

1003816100381610038161003816

(70)

which implies that (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isequicontinuous Thus by the Arzela-Ascoli theorem there isa subsequence of (119870(119873

119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899 which we

call (119870(119873119891(119906119899119895

)) minus 119876120593(119870(119873119891(119906119899119895

))))119895 which is convergent in

119862 Using the fact that 120593minus1 119862 rarr 119861119886(0) sub 119862 is continuous it

follows from

119872(119906119899119895

)1015840

= 120593minus1

[119870 (119873119891(119906119899119895

)) minus 119876120593(119870(119873

119891(119906119899119895

)))]

(71)

that the sequence (119872(119906119899119895

)1015840)119895is convergent in119862Then passing

to a subsequence if necessary we obtain that (V119899119895

)119895

=

(119872(119906119899119895

))119895is convergent in 1198621 Finally let (V

119899)119899be a sequence

in119872(Λ) Let (119911119899)119899sube 119872(Λ) be such that

lim119899rarrinfin

1003817100381710038171003817119911119899 minus V119899

10038171003817100381710038171 = 0 (72)

Let (119911119899119895

)119895be a subsequence of (119911

119899)119899such that it converges to

119911 It follows that 119911 isin 119872(Λ) and (V119899119895

)119895converge to 119911 This

concludes the proof

The next result is based on Schauderrsquos fixed point theo-rem

Theorem 14 Let 119891 [0 119879] timesRtimesR rarr R be continuous Then(55) has at least one solution

Abstract and Applied Analysis 7

Proof Let 119906 isin 1198621 Then

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(73)

where

119872(119906) (0) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (119879)

119872 (119906)1015840

(119879) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (0)

(74)

Moreover10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infin

=10038171003817100381710038171003817120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]

10038171003817100381710038171003817infinlt 119886

119872 (119906)infin

lt 119886 + 119886119879

(75)

Hence

119872 (119906)1= 119872 (119906)

infin+10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infinlt 119886 + 119886119879 + 119886

= 2119886 + 119886119879

(76)

Because the operator 119872 is completely continuous andbounded we can use Schauderrsquos fixed point theorem todeduce the existence of at least one fixed point This in turnimplies that problem (55) has at least one solution The proofis complete

5 Problems with Classic Homeomorphismsand Three-Point Boundary Conditions

We finally consider boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(77)

where 120593 R rarr R is a homeomorphism such that120593(0) = 0 and 119891 [0 119879] times R times R rarr R is a continuousfunction We remember that a solution of this problem isany function 119906 [0 119879] rarr R of class 119862

1 such that120593(1199061015840) is continuously differentiable satisfying the boundaryconditions and (120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879]

Let us consider the operator

1198721 1198621997888rarr 119862

1

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(78)

Analogously to Section 3 here 120593minus1 is understood as theoperator 120593minus1 119862 rarr 119862 defined for 120593minus1(V)(119905) = 120593minus1(V(119905)) Itis clear that 120593minus1 is continuous and sends bounded sets intobounded sets

Lemma 15 119906 isin 1198621 is a solution of (77) if and only if 119906 is a

fixed point of the operator 1198721

Proof Let 119906 isin 1198621 and we have the following equivalences

(120593 (1199061015840))1015840

= 119873119891(119906)

1199061015840

(119879) = 1199061015840

(0)

1199061015840

(0) = 119906 (119879)

lArrrArr (120593 (1199061015840))1015840

= 119873119891(119906) minus 119876 (119873

119891(119906))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 120593(1199061015840) = 119867(119873

119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906

1015840

(0))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 1199061015840= 120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))] 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))]) 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906 (119879))])

lArrrArr 119906 = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(79)

Remark 16 Note that 1199061015840(119879) = 1199061015840(0) hArr 119876(119873119891(119906)) = 0

Using an argument similar to the one introduced inLemma 13 it is easy to see that 119872

1 1198621 rarr 1198621 is completely

continuousIn order to apply Leray-Schauder degree to the operator

1198721 we introduced a family of problems depending on

parameter 120582 We remember that to each continuous function119891 [0 119879] times R times R rarr R we associate its Nemytskii operator119873119891 1198621 rarr 119862 defined by

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (80)

For 120582 isin [0 1] we consider the family of boundary valueproblems

(120593 (1199061015840))1015840

= 120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(81)

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

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MathematicsJournal of

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

We now show that the equation

119866ℎ(119904) = 0 (13)

has a unique solution 119876120593(ℎ) Let 119903 119904 isin [ℎ

119898 ℎ119872] be such that

int119879

0

120593minus1

(ℎ (119905) minus 119903) 119889119905 = 0

int119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 = 0

(14)

that is

int119879

0

120593minus1

(ℎ (119905) minus 119903) 119889119905 = int119879

0

120593minus1

(ℎ (119905) minus 119904) 119889119905 (15)

It follows that there exists 120591 isin [0 119879] such that

120593minus1

(ℎ (120591) minus 119903) = 120593minus1

(ℎ (120591) minus 119904) (16)

Using the injectivity of 120593minus1 we deduce that 119903 = 119904 Let us

now show the existence Because120593minus1 is strictlymonotone and120593minus1(0) = 0 we have that

119866ℎ(ℎ119898) 119866ℎ(ℎ119872) le 0 (17)

It follows that there exists 119904 isin [ℎ119898 ℎ119872] such that 119866

ℎ(119904) = 0

Consequently for each ℎ isin 119861 (13) has a unique solutionThuswe define the function 119876

120593 119861 rarr R such that

int119879

0

120593minus1

(ℎ (119905) minus 119876120593(ℎ)) 119889119905 = 0 (18)

On the other hand because ℎ isin 119861 we have that10038161003816100381610038161003816119876120593(ℎ)

10038161003816100381610038161003816le ℎinfin

lt119886

2 (19)

Therefore the function119876120593sends bounded sets into bounded

setsFinally we show that119876

120593is continuous on119861 Let (ℎ

119899)119899sub 119862

be a sequence such that ℎ119899

rarr ℎ in 119862 Since the function119876120593sends bounded sets into bounded sets then (119876

120593(ℎ119899))119899

is bounded Hence (119876120593(ℎ119899))119899is relatively compact Without

loss of generality passing if necessary to a subsequence wecan assume that

lim119899rarrinfin

119876120593(ℎ119899) = (20)

where for each 119899 isin N we obtain

lim119899rarrinfin

int119879

0

120593minus1

(ℎ119899(119905) minus 119876

120593(ℎ119899)) 119889119905 = 0 (21)

Using the dominated convergence theorem we deduce that

int119879

0

120593minus1

(ℎ (119905) minus ) 119889119905 = 0 (22)

so we have that 119876120593(ℎ) = Hence the function 119876

120593is

continuous

The following extended homotopy invariance property ofthe Leray-Schauder degree can be found in [9]

Proposition 3 Let 119883 be a real Banach space 119881 sub [0 1] times 119883

be an open bounded set and 119872 be a completely continuousoperator on 119881 such that 119909 = 119872(120582 119909) for each (120582 119909) isin 120597119881Then the Leray-Schauder degree

degLS (119868 minus 119872 (120582 sdot) 119881120582 0) (23)

is well defined and independent of 120582 in [0 1] where 119881120582is the

open bounded (possibly empty) set defined by 119881120582

= 119909 isin 119883

(120582 119909) isin 119881

3 Dirichlet Problems with BoundedHomeomorphisms

In this section we are interested in Dirichlet boundary valueproblems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(24)

where 120593 R rarr (minus119886 119886) is a homeomorphism 120593(0) = 0 and119891 [0 119879]timesRtimesR rarr R is continuous In order to apply Leray-Schauder degree theory to show the existence of at least onesolution of (24) we introduce for 120582 isin [0 1] the family ofDirichlet boundary value problems

(120593 (1199061015840))1015840

= 120582119891 (119905 119906 1199061015840)

119906 (0) = 0 = 119906 (119879)

(25)

Let

Ω = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinlt

119886

2 (26)

ClearlyΩ is an open set in [0 1]times11986210and is nonempty because

0 times 11986210

sub Ω Using Lemma 2 we can define the operator119872 Ω rarr 1198621

0by

119872(120582 119906)

= 119867(120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867(119873

119891(119906)))])

(27)

Here 120593minus1 with an abuse of notation is understood as theoperator 120593minus1 119861

119886(0) sub 119862 rarr 119862 defined by 120593minus1(V)(119905) =

120593minus1(V(119905)) It is clear that120593minus1 is continuous and sends boundedsets into bounded sets

When the boundary conditions are periodic orNeumannan operator has been considered by Bereanu andMawhin [6]

The following lemma plays a pivotal role in studying thesolutions of problem (25)

Lemma 4 The operator 119872 is well defined and continuousMoreover if (120582 119906) isin Ω is such that 119872(120582 119906) = 119906 then 119906 issolution of (25)

4 Abstract and Applied Analysis

Proof Let (120582 119906) isin Ω We show that in fact119872(120582 119906) isin 11986210 It is

clear that

(119872 (120582 119906))1015840

= 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]

(28)

where the continuity of 119872(120582 119906) and (119872(120582 119906))1015840 follows fromthe continuity of applications119867 and119873

119891

On the other hand using Lemma 2 we have

119872(120582 119906) (0) = 0 = 119872 (120582 119906) (119879) (29)

Therefore 119872(Ω) sub 11986210and 119872 is well defined The continuity

of 119872 follows by the continuity of the operators whichcompose it119872

Now suppose that (120582 119906) isin Ω is such that 119872(120582 119906) = 119906 Itfollows from (27) that

119906 (119905) = 119872 (120582 119906) (119905)

= 119867 (120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]) (119905)

(30)

for all 119905 isin [0 119879] Differentiating (30) we obtain

1199061015840

(119905) = 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))] (119905)

= 120593minus1

[120582119867 (119873119891(119906)) (119905) minus 119876

120593(120582119867 (119873

119891(119906)))]

(31)

Applying 120593 to both of its members we have that

120593 (1199061015840

(119905)) = 120582119867(119873119891(119906)) (119905) minus 119876

120593(120582119867(119873

119891(119906))) (32)

Differentiating again we deduce that

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905) (33)

for all 119905 isin [0 119879]Thus119906 satisfies problem (25)This completesthe proof

Remark 5 Note that the reciprocal of Lemma 4 is not truebecause we cannot guarantee that 120582119867(119873

119891(119906)infin

lt 1198862 forevery solution 119906 of (25)

In ourmain result we need the following lemma to obtainthe required a priori bounds for the possible fixed points of119872

Lemma 6 Assume that there exist ℎ isin 119862([0 119879]R+) and 119899 isin

1198621(RR) such that ℎ1198711 lt 1198862 119899(0) = 0

120593 (119910) 1198991015840

(119909) 119910 ge 0 (119905 119909 119910) isin [0 119879] timesR timesR (34)1003816100381610038161003816119891 (119905 119909 119910)

1003816100381610038161003816 le 119891 (119905 119909 119910) 119899 (119909) + ℎ (119905) (35)

for all (119905 119909 119910) isin [0 119879] times R times R If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(36)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Let 120582 = 0 and (120582 119906) isin Ω be such that 119872(120582 119906) = 119906Using Lemma 4 we have that 119906 is solution of (25) whichimplies that

120593 (1199061015840) = 120582119867(119873

119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))

119906 (0) = 0 = 119906 (119879)

(37)

where for all 119905 isin [0 119879] we obtain

10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119899 (119906 (119904)) 119889119904

+ int119879

0

ℎ (119904) 119889119904

(38)

On the other hand using inequality (34) we have that

minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0 (39)

Using the integration by parts formula the boundary condi-tions and the fact that 119899(0) = 0 we deduce that

int119879

0

(120593 (1199061015840

(119905)))1015840

119899 (119906 (119905)) 119889119905

= minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0

(40)

Since 120582 isin (0 1] and 119906 is solution of (25) it follows that

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119899 (119906 (119905)) 119889119905 le 0 (41)

and hence10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le ℎ1198711 (42)

On the other hand since 119876120593(120582119867(119873

119891(119906))) isin

Im(120582119867(119873119891(119906))) we get

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le 2 ℎ

1198711 (43)

for all 119905 isin [0 119879] It follows that10038171003817100381710038171003817120593 (1199061015840)10038171003817100381710038171003817infin

le 2 ℎ1198711 (44)

which implies that 1199061015840infin

le 119871 where 119871 =

max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)| Using again the

boundary conditions we have that

|119906 (119905)| le int119905

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le int

119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le 119871119879

(119905 isin [0 119879])

(45)

and hence

1199061le 119871 + 119871119879 (46)

Finally if 119906 = 119872(0 119906) then 119906 = 0 so the proof is complete

Abstract and Applied Analysis 5

Let 120588 120581 isin R be such that ℎ1198711 lt 120581 lt 1198862 120588 gt 119871 + 119871119879

and consider the set

119881 = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infin

lt 120581 1199061lt 120588

(47)

Since the set 0 times 119906 isin 11986210 1199061lt 120588 sub 119881 then we deduce

that 119881 is nonempty Moreover it is clear that 119881 is open andbounded in [0 1] times 1198621

0and 119881 sub Ω On the other hand using

an argument similar to the one introduced in the proof ofLemma 6 in [6] it is not difficult to see that 119872 119881 rarr 1198621

0

is well defined and completely continuous and

119906 = 119872 (120582 119906) forall (120582 119906) isin 120597119881 (48)

31 Existence Results In this subsection we present andprove our main result

Theorem7 If119891 satisfies conditions of Lemma 6 then problem(24) has at least one solution

Proof Let 119872 be the operator given by (27) Using Proposi-tion 3 we deduce that

degLS (119868 minus 119872 (0 sdot) 1198810 0)

= degLS (119868 minus 119872 (1 sdot) 1198811 0)

(49)

where degLS(119868 minus119872(0 sdot) 1198810 0) = degLS(119868 119861120588(0) 0) = 1 Thus

there exists 119906 isin 1198811such that 119872(1 119906) = 119906 which is a solution

for (24)

Remark 8 Note that Theorem 7 is a generalization of Theo-rem 1

Corollary 9 Assume that 120593 is an increasing homomorphismLet ℎ isin 119862([0 119879]R+) be such that

ℎ1198711 lt

119886

2

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 le 119891 (119905 119909 119910) 119909 + ℎ (119905)

(50)

for all 119909 119910 isin R and 119905 isin [0 119879] If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(51)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Since 120593 is an increasing homomorphism we have that

120593 (119910) 119910 ge 0 (52)

for all 119910 isin R Using Lemma 6 with 119899(119909) = 119909 for all 119909 isin

R we can obtain the conclusion of Corollary 9 The proof isachieved

Theorem 10 If 119891 satisfies conditions of Corollary 9 thenproblem (24) has at least one solution

Let us give now an application of Theorem 10 when 119891 isunbounded

Example 11 Consider the Dirichlet problem

(120593 (1199061015840))1015840

= 119906 minus 2

119906 (0) = 119906 (119879) = 0

(53)

where 120593(119904) = 119904radic1 + 1199042It is not difficult to verify that 120593 R rarr (minus1 1) is

an increasing homeomorphism and 119891(119905 119909 119910) = 119909 minus 2 is acontinuous function such that

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 = |119909 minus 2| le (119909 minus 2) 119909 + 4

(119905 119909 119910) isin [0 119879] timesR timesR(54)

So we can choose ℎ(119905) = 4 and119879 lt 18 to see that Corollary 9holds and so usingTheorem 10we obtain that (53) has at leastone solution

4 Problems with Singular Homeomorphismsand Three-Point Boundary Conditions

In this section we study the existence of at least one solutionfor boundary value problems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 119906 (0) = 1199061015840

(119879)

(55)

where 120593 (minus119886 119886) rarr R is a homeomorphism such that 120593(0) =

0 and 119891 [0 119879] timesR timesR rarr R is a continuous functionIn order to transform problem (55) to a fixed point

problem we use a similar argument introduced in Lemma 2for ℎ isin 119862

Lemma 12 119906 isin 1198621 is a solution of (55) if and only if 119906 is a

fixed point of the operator 119872 defined on 1198621 by

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(56)

Proof If 119906 isin 1198621 is solution of (55) then

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905))

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(57)

for all 119905 isin [0 119879] Applying 119870 to both members and using thefact that 119906(0) = 119906

1015840(119879) we deduce that

120593 (1199061015840

(119905)) = 120593 (119906 (0)) + 119870 (119873119891(119906)) (119905) (58)

6 Abstract and Applied Analysis

By the inversion of 120593 in (58) we have

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) (119905) + 119888] (59)

where 119888 = 120593(119906(0)) Integrating from 0 to 119905 isin [0 119879] we havethat

119906 (119905) = 119906 (0) + 119867 (120593minus1

[119870 (119873119891(119906)) + 119888]) (119905) (60)

Because 119906(0) = 119906(119879) then

int119879

0

120593minus1

[119870 (119873119891(119906)) (119905) + 119888] 119889119905 = 0 (61)

Using an argument similar to the one introduced in Lemma 2it follows that 119888 = minus119876

120593(119870(119873119891(119906))) Hence

119906 = 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(62)

Let 119906 isin 1198621 be such that 119906 = 119872(119906) Then

119906 (119905)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]) (119905)

(63)

for all 119905 isin [0 119879] Since int119879

0120593minus1[119870(119873

119891(119906))(119905) minus

119876120593(119870(119873119891(119906)))]119889119905 = 0 therefore we have that 119906(0) = 119906(119879)

Differentiating (63) we obtain that

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))] (119905) (64)

In particular

1199061015840

(119879) = 120593minus1

(0 minus 119876120593(119870 (119873

119891(119906))))

= 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119906 (0)

(65)

Applying 120593 to both members and differentiating again wededuce that

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905)

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(66)

for all 119905 isin [0 119879] This completes the proof

Lemma 13 The operator 119872 1198621 rarr 1198621 is completelycontinuous

Proof Let Λ sub 1198621 be a bounded set Then if 119906 isin Λ thereexists a constant 120588 gt 0 such that

1199061le 120588 (67)

Next we show that119872(Λ) sub 1198621 is a compact set Let (V119899)119899be a

sequence in119872(Λ) and let (119906119899)119899be a sequence in Λ such that

V119899= 119872(119906

119899) Using (67) we have that there exists a constant

119871 gt 0 such that for all 119899 isin N10038171003817100381710038171003817119873119891(119906119899)10038171003817100381710038171003817infin

le 119871 (68)

which implies that

10038171003817100381710038171003817119870 (119873119891(119906119899)) minus 119876

120593(119870 (119873

119891(119906119899)))

10038171003817100381710038171003817infinle 2119871119879 (69)

Hence the sequence (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isbounded in 119862 Moreover for 119905 119905

1isin [0 119879] and for all 119899 isin N

we have that10038161003816100381610038161003816119870 (119873119891(119906119899)) (119905) minus 119876

120593(119870 (119873

119891(119906119899)))

minus 119870 (119873119891(119906119899)) (1199051) + 119876120593(119870 (119873

119891(119906119899)))

10038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816minus int119879

119905

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

+ int119879

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816int119905

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816le 119871

1003816100381610038161003816119905 minus 1199051

1003816100381610038161003816

(70)

which implies that (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isequicontinuous Thus by the Arzela-Ascoli theorem there isa subsequence of (119870(119873

119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899 which we

call (119870(119873119891(119906119899119895

)) minus 119876120593(119870(119873119891(119906119899119895

))))119895 which is convergent in

119862 Using the fact that 120593minus1 119862 rarr 119861119886(0) sub 119862 is continuous it

follows from

119872(119906119899119895

)1015840

= 120593minus1

[119870 (119873119891(119906119899119895

)) minus 119876120593(119870(119873

119891(119906119899119895

)))]

(71)

that the sequence (119872(119906119899119895

)1015840)119895is convergent in119862Then passing

to a subsequence if necessary we obtain that (V119899119895

)119895

=

(119872(119906119899119895

))119895is convergent in 1198621 Finally let (V

119899)119899be a sequence

in119872(Λ) Let (119911119899)119899sube 119872(Λ) be such that

lim119899rarrinfin

1003817100381710038171003817119911119899 minus V119899

10038171003817100381710038171 = 0 (72)

Let (119911119899119895

)119895be a subsequence of (119911

119899)119899such that it converges to

119911 It follows that 119911 isin 119872(Λ) and (V119899119895

)119895converge to 119911 This

concludes the proof

The next result is based on Schauderrsquos fixed point theo-rem

Theorem 14 Let 119891 [0 119879] timesRtimesR rarr R be continuous Then(55) has at least one solution

Abstract and Applied Analysis 7

Proof Let 119906 isin 1198621 Then

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(73)

where

119872(119906) (0) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (119879)

119872 (119906)1015840

(119879) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (0)

(74)

Moreover10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infin

=10038171003817100381710038171003817120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]

10038171003817100381710038171003817infinlt 119886

119872 (119906)infin

lt 119886 + 119886119879

(75)

Hence

119872 (119906)1= 119872 (119906)

infin+10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infinlt 119886 + 119886119879 + 119886

= 2119886 + 119886119879

(76)

Because the operator 119872 is completely continuous andbounded we can use Schauderrsquos fixed point theorem todeduce the existence of at least one fixed point This in turnimplies that problem (55) has at least one solution The proofis complete

5 Problems with Classic Homeomorphismsand Three-Point Boundary Conditions

We finally consider boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(77)

where 120593 R rarr R is a homeomorphism such that120593(0) = 0 and 119891 [0 119879] times R times R rarr R is a continuousfunction We remember that a solution of this problem isany function 119906 [0 119879] rarr R of class 119862

1 such that120593(1199061015840) is continuously differentiable satisfying the boundaryconditions and (120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879]

Let us consider the operator

1198721 1198621997888rarr 119862

1

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(78)

Analogously to Section 3 here 120593minus1 is understood as theoperator 120593minus1 119862 rarr 119862 defined for 120593minus1(V)(119905) = 120593minus1(V(119905)) Itis clear that 120593minus1 is continuous and sends bounded sets intobounded sets

Lemma 15 119906 isin 1198621 is a solution of (77) if and only if 119906 is a

fixed point of the operator 1198721

Proof Let 119906 isin 1198621 and we have the following equivalences

(120593 (1199061015840))1015840

= 119873119891(119906)

1199061015840

(119879) = 1199061015840

(0)

1199061015840

(0) = 119906 (119879)

lArrrArr (120593 (1199061015840))1015840

= 119873119891(119906) minus 119876 (119873

119891(119906))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 120593(1199061015840) = 119867(119873

119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906

1015840

(0))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 1199061015840= 120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))] 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))]) 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906 (119879))])

lArrrArr 119906 = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(79)

Remark 16 Note that 1199061015840(119879) = 1199061015840(0) hArr 119876(119873119891(119906)) = 0

Using an argument similar to the one introduced inLemma 13 it is easy to see that 119872

1 1198621 rarr 1198621 is completely

continuousIn order to apply Leray-Schauder degree to the operator

1198721 we introduced a family of problems depending on

parameter 120582 We remember that to each continuous function119891 [0 119879] times R times R rarr R we associate its Nemytskii operator119873119891 1198621 rarr 119862 defined by

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (80)

For 120582 isin [0 1] we consider the family of boundary valueproblems

(120593 (1199061015840))1015840

= 120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(81)

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

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Mathematical Problems in Engineering

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

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Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

Proof Let (120582 119906) isin Ω We show that in fact119872(120582 119906) isin 11986210 It is

clear that

(119872 (120582 119906))1015840

= 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]

(28)

where the continuity of 119872(120582 119906) and (119872(120582 119906))1015840 follows fromthe continuity of applications119867 and119873

119891

On the other hand using Lemma 2 we have

119872(120582 119906) (0) = 0 = 119872 (120582 119906) (119879) (29)

Therefore 119872(Ω) sub 11986210and 119872 is well defined The continuity

of 119872 follows by the continuity of the operators whichcompose it119872

Now suppose that (120582 119906) isin Ω is such that 119872(120582 119906) = 119906 Itfollows from (27) that

119906 (119905) = 119872 (120582 119906) (119905)

= 119867 (120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))]) (119905)

(30)

for all 119905 isin [0 119879] Differentiating (30) we obtain

1199061015840

(119905) = 120593minus1

[120582119867 (119873119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))] (119905)

= 120593minus1

[120582119867 (119873119891(119906)) (119905) minus 119876

120593(120582119867 (119873

119891(119906)))]

(31)

Applying 120593 to both of its members we have that

120593 (1199061015840

(119905)) = 120582119867(119873119891(119906)) (119905) minus 119876

120593(120582119867(119873

119891(119906))) (32)

Differentiating again we deduce that

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905) (33)

for all 119905 isin [0 119879]Thus119906 satisfies problem (25)This completesthe proof

Remark 5 Note that the reciprocal of Lemma 4 is not truebecause we cannot guarantee that 120582119867(119873

119891(119906)infin

lt 1198862 forevery solution 119906 of (25)

In ourmain result we need the following lemma to obtainthe required a priori bounds for the possible fixed points of119872

Lemma 6 Assume that there exist ℎ isin 119862([0 119879]R+) and 119899 isin

1198621(RR) such that ℎ1198711 lt 1198862 119899(0) = 0

120593 (119910) 1198991015840

(119909) 119910 ge 0 (119905 119909 119910) isin [0 119879] timesR timesR (34)1003816100381610038161003816119891 (119905 119909 119910)

1003816100381610038161003816 le 119891 (119905 119909 119910) 119899 (119909) + ℎ (119905) (35)

for all (119905 119909 119910) isin [0 119879] times R times R If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(36)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Let 120582 = 0 and (120582 119906) isin Ω be such that 119872(120582 119906) = 119906Using Lemma 4 we have that 119906 is solution of (25) whichimplies that

120593 (1199061015840) = 120582119867(119873

119891(119906)) minus 119876

120593(120582119867 (119873

119891(119906)))

119906 (0) = 0 = 119906 (119879)

(37)

where for all 119905 isin [0 119879] we obtain

10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119899 (119906 (119904)) 119889119904

+ int119879

0

ℎ (119904) 119889119904

(38)

On the other hand using inequality (34) we have that

minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0 (39)

Using the integration by parts formula the boundary condi-tions and the fact that 119899(0) = 0 we deduce that

int119879

0

(120593 (1199061015840

(119905)))1015840

119899 (119906 (119905)) 119889119905

= minusint119879

0

120593 (1199061015840

(119905)) 1198991015840

(119906 (119905)) 1199061015840

(119905) 119889119905 le 0

(40)

Since 120582 isin (0 1] and 119906 is solution of (25) it follows that

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119899 (119906 (119905)) 119889119905 le 0 (41)

and hence10038161003816100381610038161003816120582119867 (119873

119891(119906)) (119905)

10038161003816100381610038161003816le ℎ1198711 (42)

On the other hand since 119876120593(120582119867(119873

119891(119906))) isin

Im(120582119867(119873119891(119906))) we get

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le 2 ℎ

1198711 (43)

for all 119905 isin [0 119879] It follows that10038171003817100381710038171003817120593 (1199061015840)10038171003817100381710038171003817infin

le 2 ℎ1198711 (44)

which implies that 1199061015840infin

le 119871 where 119871 =

max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)| Using again the

boundary conditions we have that

|119906 (119905)| le int119905

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le int

119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 le 119871119879

(119905 isin [0 119879])

(45)

and hence

1199061le 119871 + 119871119879 (46)

Finally if 119906 = 119872(0 119906) then 119906 = 0 so the proof is complete

Abstract and Applied Analysis 5

Let 120588 120581 isin R be such that ℎ1198711 lt 120581 lt 1198862 120588 gt 119871 + 119871119879

and consider the set

119881 = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infin

lt 120581 1199061lt 120588

(47)

Since the set 0 times 119906 isin 11986210 1199061lt 120588 sub 119881 then we deduce

that 119881 is nonempty Moreover it is clear that 119881 is open andbounded in [0 1] times 1198621

0and 119881 sub Ω On the other hand using

an argument similar to the one introduced in the proof ofLemma 6 in [6] it is not difficult to see that 119872 119881 rarr 1198621

0

is well defined and completely continuous and

119906 = 119872 (120582 119906) forall (120582 119906) isin 120597119881 (48)

31 Existence Results In this subsection we present andprove our main result

Theorem7 If119891 satisfies conditions of Lemma 6 then problem(24) has at least one solution

Proof Let 119872 be the operator given by (27) Using Proposi-tion 3 we deduce that

degLS (119868 minus 119872 (0 sdot) 1198810 0)

= degLS (119868 minus 119872 (1 sdot) 1198811 0)

(49)

where degLS(119868 minus119872(0 sdot) 1198810 0) = degLS(119868 119861120588(0) 0) = 1 Thus

there exists 119906 isin 1198811such that 119872(1 119906) = 119906 which is a solution

for (24)

Remark 8 Note that Theorem 7 is a generalization of Theo-rem 1

Corollary 9 Assume that 120593 is an increasing homomorphismLet ℎ isin 119862([0 119879]R+) be such that

ℎ1198711 lt

119886

2

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 le 119891 (119905 119909 119910) 119909 + ℎ (119905)

(50)

for all 119909 119910 isin R and 119905 isin [0 119879] If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(51)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Since 120593 is an increasing homomorphism we have that

120593 (119910) 119910 ge 0 (52)

for all 119910 isin R Using Lemma 6 with 119899(119909) = 119909 for all 119909 isin

R we can obtain the conclusion of Corollary 9 The proof isachieved

Theorem 10 If 119891 satisfies conditions of Corollary 9 thenproblem (24) has at least one solution

Let us give now an application of Theorem 10 when 119891 isunbounded

Example 11 Consider the Dirichlet problem

(120593 (1199061015840))1015840

= 119906 minus 2

119906 (0) = 119906 (119879) = 0

(53)

where 120593(119904) = 119904radic1 + 1199042It is not difficult to verify that 120593 R rarr (minus1 1) is

an increasing homeomorphism and 119891(119905 119909 119910) = 119909 minus 2 is acontinuous function such that

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 = |119909 minus 2| le (119909 minus 2) 119909 + 4

(119905 119909 119910) isin [0 119879] timesR timesR(54)

So we can choose ℎ(119905) = 4 and119879 lt 18 to see that Corollary 9holds and so usingTheorem 10we obtain that (53) has at leastone solution

4 Problems with Singular Homeomorphismsand Three-Point Boundary Conditions

In this section we study the existence of at least one solutionfor boundary value problems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 119906 (0) = 1199061015840

(119879)

(55)

where 120593 (minus119886 119886) rarr R is a homeomorphism such that 120593(0) =

0 and 119891 [0 119879] timesR timesR rarr R is a continuous functionIn order to transform problem (55) to a fixed point

problem we use a similar argument introduced in Lemma 2for ℎ isin 119862

Lemma 12 119906 isin 1198621 is a solution of (55) if and only if 119906 is a

fixed point of the operator 119872 defined on 1198621 by

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(56)

Proof If 119906 isin 1198621 is solution of (55) then

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905))

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(57)

for all 119905 isin [0 119879] Applying 119870 to both members and using thefact that 119906(0) = 119906

1015840(119879) we deduce that

120593 (1199061015840

(119905)) = 120593 (119906 (0)) + 119870 (119873119891(119906)) (119905) (58)

6 Abstract and Applied Analysis

By the inversion of 120593 in (58) we have

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) (119905) + 119888] (59)

where 119888 = 120593(119906(0)) Integrating from 0 to 119905 isin [0 119879] we havethat

119906 (119905) = 119906 (0) + 119867 (120593minus1

[119870 (119873119891(119906)) + 119888]) (119905) (60)

Because 119906(0) = 119906(119879) then

int119879

0

120593minus1

[119870 (119873119891(119906)) (119905) + 119888] 119889119905 = 0 (61)

Using an argument similar to the one introduced in Lemma 2it follows that 119888 = minus119876

120593(119870(119873119891(119906))) Hence

119906 = 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(62)

Let 119906 isin 1198621 be such that 119906 = 119872(119906) Then

119906 (119905)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]) (119905)

(63)

for all 119905 isin [0 119879] Since int119879

0120593minus1[119870(119873

119891(119906))(119905) minus

119876120593(119870(119873119891(119906)))]119889119905 = 0 therefore we have that 119906(0) = 119906(119879)

Differentiating (63) we obtain that

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))] (119905) (64)

In particular

1199061015840

(119879) = 120593minus1

(0 minus 119876120593(119870 (119873

119891(119906))))

= 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119906 (0)

(65)

Applying 120593 to both members and differentiating again wededuce that

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905)

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(66)

for all 119905 isin [0 119879] This completes the proof

Lemma 13 The operator 119872 1198621 rarr 1198621 is completelycontinuous

Proof Let Λ sub 1198621 be a bounded set Then if 119906 isin Λ thereexists a constant 120588 gt 0 such that

1199061le 120588 (67)

Next we show that119872(Λ) sub 1198621 is a compact set Let (V119899)119899be a

sequence in119872(Λ) and let (119906119899)119899be a sequence in Λ such that

V119899= 119872(119906

119899) Using (67) we have that there exists a constant

119871 gt 0 such that for all 119899 isin N10038171003817100381710038171003817119873119891(119906119899)10038171003817100381710038171003817infin

le 119871 (68)

which implies that

10038171003817100381710038171003817119870 (119873119891(119906119899)) minus 119876

120593(119870 (119873

119891(119906119899)))

10038171003817100381710038171003817infinle 2119871119879 (69)

Hence the sequence (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isbounded in 119862 Moreover for 119905 119905

1isin [0 119879] and for all 119899 isin N

we have that10038161003816100381610038161003816119870 (119873119891(119906119899)) (119905) minus 119876

120593(119870 (119873

119891(119906119899)))

minus 119870 (119873119891(119906119899)) (1199051) + 119876120593(119870 (119873

119891(119906119899)))

10038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816minus int119879

119905

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

+ int119879

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816int119905

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816le 119871

1003816100381610038161003816119905 minus 1199051

1003816100381610038161003816

(70)

which implies that (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isequicontinuous Thus by the Arzela-Ascoli theorem there isa subsequence of (119870(119873

119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899 which we

call (119870(119873119891(119906119899119895

)) minus 119876120593(119870(119873119891(119906119899119895

))))119895 which is convergent in

119862 Using the fact that 120593minus1 119862 rarr 119861119886(0) sub 119862 is continuous it

follows from

119872(119906119899119895

)1015840

= 120593minus1

[119870 (119873119891(119906119899119895

)) minus 119876120593(119870(119873

119891(119906119899119895

)))]

(71)

that the sequence (119872(119906119899119895

)1015840)119895is convergent in119862Then passing

to a subsequence if necessary we obtain that (V119899119895

)119895

=

(119872(119906119899119895

))119895is convergent in 1198621 Finally let (V

119899)119899be a sequence

in119872(Λ) Let (119911119899)119899sube 119872(Λ) be such that

lim119899rarrinfin

1003817100381710038171003817119911119899 minus V119899

10038171003817100381710038171 = 0 (72)

Let (119911119899119895

)119895be a subsequence of (119911

119899)119899such that it converges to

119911 It follows that 119911 isin 119872(Λ) and (V119899119895

)119895converge to 119911 This

concludes the proof

The next result is based on Schauderrsquos fixed point theo-rem

Theorem 14 Let 119891 [0 119879] timesRtimesR rarr R be continuous Then(55) has at least one solution

Abstract and Applied Analysis 7

Proof Let 119906 isin 1198621 Then

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(73)

where

119872(119906) (0) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (119879)

119872 (119906)1015840

(119879) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (0)

(74)

Moreover10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infin

=10038171003817100381710038171003817120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]

10038171003817100381710038171003817infinlt 119886

119872 (119906)infin

lt 119886 + 119886119879

(75)

Hence

119872 (119906)1= 119872 (119906)

infin+10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infinlt 119886 + 119886119879 + 119886

= 2119886 + 119886119879

(76)

Because the operator 119872 is completely continuous andbounded we can use Schauderrsquos fixed point theorem todeduce the existence of at least one fixed point This in turnimplies that problem (55) has at least one solution The proofis complete

5 Problems with Classic Homeomorphismsand Three-Point Boundary Conditions

We finally consider boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(77)

where 120593 R rarr R is a homeomorphism such that120593(0) = 0 and 119891 [0 119879] times R times R rarr R is a continuousfunction We remember that a solution of this problem isany function 119906 [0 119879] rarr R of class 119862

1 such that120593(1199061015840) is continuously differentiable satisfying the boundaryconditions and (120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879]

Let us consider the operator

1198721 1198621997888rarr 119862

1

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(78)

Analogously to Section 3 here 120593minus1 is understood as theoperator 120593minus1 119862 rarr 119862 defined for 120593minus1(V)(119905) = 120593minus1(V(119905)) Itis clear that 120593minus1 is continuous and sends bounded sets intobounded sets

Lemma 15 119906 isin 1198621 is a solution of (77) if and only if 119906 is a

fixed point of the operator 1198721

Proof Let 119906 isin 1198621 and we have the following equivalences

(120593 (1199061015840))1015840

= 119873119891(119906)

1199061015840

(119879) = 1199061015840

(0)

1199061015840

(0) = 119906 (119879)

lArrrArr (120593 (1199061015840))1015840

= 119873119891(119906) minus 119876 (119873

119891(119906))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 120593(1199061015840) = 119867(119873

119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906

1015840

(0))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 1199061015840= 120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))] 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))]) 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906 (119879))])

lArrrArr 119906 = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(79)

Remark 16 Note that 1199061015840(119879) = 1199061015840(0) hArr 119876(119873119891(119906)) = 0

Using an argument similar to the one introduced inLemma 13 it is easy to see that 119872

1 1198621 rarr 1198621 is completely

continuousIn order to apply Leray-Schauder degree to the operator

1198721 we introduced a family of problems depending on

parameter 120582 We remember that to each continuous function119891 [0 119879] times R times R rarr R we associate its Nemytskii operator119873119891 1198621 rarr 119862 defined by

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (80)

For 120582 isin [0 1] we consider the family of boundary valueproblems

(120593 (1199061015840))1015840

= 120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(81)

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

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Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

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Decision SciencesAdvances in

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Let 120588 120581 isin R be such that ℎ1198711 lt 120581 lt 1198862 120588 gt 119871 + 119871119879

and consider the set

119881 = (120582 119906) isin [0 1] times 1198621

010038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infin

lt 120581 1199061lt 120588

(47)

Since the set 0 times 119906 isin 11986210 1199061lt 120588 sub 119881 then we deduce

that 119881 is nonempty Moreover it is clear that 119881 is open andbounded in [0 1] times 1198621

0and 119881 sub Ω On the other hand using

an argument similar to the one introduced in the proof ofLemma 6 in [6] it is not difficult to see that 119872 119881 rarr 1198621

0

is well defined and completely continuous and

119906 = 119872 (120582 119906) forall (120582 119906) isin 120597119881 (48)

31 Existence Results In this subsection we present andprove our main result

Theorem7 If119891 satisfies conditions of Lemma 6 then problem(24) has at least one solution

Proof Let 119872 be the operator given by (27) Using Proposi-tion 3 we deduce that

degLS (119868 minus 119872 (0 sdot) 1198810 0)

= degLS (119868 minus 119872 (1 sdot) 1198811 0)

(49)

where degLS(119868 minus119872(0 sdot) 1198810 0) = degLS(119868 119861120588(0) 0) = 1 Thus

there exists 119906 isin 1198811such that 119872(1 119906) = 119906 which is a solution

for (24)

Remark 8 Note that Theorem 7 is a generalization of Theo-rem 1

Corollary 9 Assume that 120593 is an increasing homomorphismLet ℎ isin 119862([0 119879]R+) be such that

ℎ1198711 lt

119886

2

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 le 119891 (119905 119909 119910) 119909 + ℎ (119905)

(50)

for all 119909 119910 isin R and 119905 isin [0 119879] If (120582 119906) isin Ω is such that119872(120582 119906) = 119906 then

10038171003817100381710038171003817120582119867 (119873

119891(119906))

10038171003817100381710038171003817infinle ℎ1198711

10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

le 119871

1199061le 119871 + 119871119879

(51)

where 119871 = max|120593minus1(minus2ℎ1198711)| |120593minus1(2ℎ

1198711)|

Proof Since 120593 is an increasing homomorphism we have that

120593 (119910) 119910 ge 0 (52)

for all 119910 isin R Using Lemma 6 with 119899(119909) = 119909 for all 119909 isin

R we can obtain the conclusion of Corollary 9 The proof isachieved

Theorem 10 If 119891 satisfies conditions of Corollary 9 thenproblem (24) has at least one solution

Let us give now an application of Theorem 10 when 119891 isunbounded

Example 11 Consider the Dirichlet problem

(120593 (1199061015840))1015840

= 119906 minus 2

119906 (0) = 119906 (119879) = 0

(53)

where 120593(119904) = 119904radic1 + 1199042It is not difficult to verify that 120593 R rarr (minus1 1) is

an increasing homeomorphism and 119891(119905 119909 119910) = 119909 minus 2 is acontinuous function such that

1003816100381610038161003816119891 (119905 119909 119910)1003816100381610038161003816 = |119909 minus 2| le (119909 minus 2) 119909 + 4

(119905 119909 119910) isin [0 119879] timesR timesR(54)

So we can choose ℎ(119905) = 4 and119879 lt 18 to see that Corollary 9holds and so usingTheorem 10we obtain that (53) has at leastone solution

4 Problems with Singular Homeomorphismsand Three-Point Boundary Conditions

In this section we study the existence of at least one solutionfor boundary value problems of the type

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 119906 (0) = 1199061015840

(119879)

(55)

where 120593 (minus119886 119886) rarr R is a homeomorphism such that 120593(0) =

0 and 119891 [0 119879] timesR timesR rarr R is a continuous functionIn order to transform problem (55) to a fixed point

problem we use a similar argument introduced in Lemma 2for ℎ isin 119862

Lemma 12 119906 isin 1198621 is a solution of (55) if and only if 119906 is a

fixed point of the operator 119872 defined on 1198621 by

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(56)

Proof If 119906 isin 1198621 is solution of (55) then

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905))

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(57)

for all 119905 isin [0 119879] Applying 119870 to both members and using thefact that 119906(0) = 119906

1015840(119879) we deduce that

120593 (1199061015840

(119905)) = 120593 (119906 (0)) + 119870 (119873119891(119906)) (119905) (58)

6 Abstract and Applied Analysis

By the inversion of 120593 in (58) we have

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) (119905) + 119888] (59)

where 119888 = 120593(119906(0)) Integrating from 0 to 119905 isin [0 119879] we havethat

119906 (119905) = 119906 (0) + 119867 (120593minus1

[119870 (119873119891(119906)) + 119888]) (119905) (60)

Because 119906(0) = 119906(119879) then

int119879

0

120593minus1

[119870 (119873119891(119906)) (119905) + 119888] 119889119905 = 0 (61)

Using an argument similar to the one introduced in Lemma 2it follows that 119888 = minus119876

120593(119870(119873119891(119906))) Hence

119906 = 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(62)

Let 119906 isin 1198621 be such that 119906 = 119872(119906) Then

119906 (119905)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]) (119905)

(63)

for all 119905 isin [0 119879] Since int119879

0120593minus1[119870(119873

119891(119906))(119905) minus

119876120593(119870(119873119891(119906)))]119889119905 = 0 therefore we have that 119906(0) = 119906(119879)

Differentiating (63) we obtain that

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))] (119905) (64)

In particular

1199061015840

(119879) = 120593minus1

(0 minus 119876120593(119870 (119873

119891(119906))))

= 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119906 (0)

(65)

Applying 120593 to both members and differentiating again wededuce that

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905)

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(66)

for all 119905 isin [0 119879] This completes the proof

Lemma 13 The operator 119872 1198621 rarr 1198621 is completelycontinuous

Proof Let Λ sub 1198621 be a bounded set Then if 119906 isin Λ thereexists a constant 120588 gt 0 such that

1199061le 120588 (67)

Next we show that119872(Λ) sub 1198621 is a compact set Let (V119899)119899be a

sequence in119872(Λ) and let (119906119899)119899be a sequence in Λ such that

V119899= 119872(119906

119899) Using (67) we have that there exists a constant

119871 gt 0 such that for all 119899 isin N10038171003817100381710038171003817119873119891(119906119899)10038171003817100381710038171003817infin

le 119871 (68)

which implies that

10038171003817100381710038171003817119870 (119873119891(119906119899)) minus 119876

120593(119870 (119873

119891(119906119899)))

10038171003817100381710038171003817infinle 2119871119879 (69)

Hence the sequence (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isbounded in 119862 Moreover for 119905 119905

1isin [0 119879] and for all 119899 isin N

we have that10038161003816100381610038161003816119870 (119873119891(119906119899)) (119905) minus 119876

120593(119870 (119873

119891(119906119899)))

minus 119870 (119873119891(119906119899)) (1199051) + 119876120593(119870 (119873

119891(119906119899)))

10038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816minus int119879

119905

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

+ int119879

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816int119905

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816le 119871

1003816100381610038161003816119905 minus 1199051

1003816100381610038161003816

(70)

which implies that (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isequicontinuous Thus by the Arzela-Ascoli theorem there isa subsequence of (119870(119873

119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899 which we

call (119870(119873119891(119906119899119895

)) minus 119876120593(119870(119873119891(119906119899119895

))))119895 which is convergent in

119862 Using the fact that 120593minus1 119862 rarr 119861119886(0) sub 119862 is continuous it

follows from

119872(119906119899119895

)1015840

= 120593minus1

[119870 (119873119891(119906119899119895

)) minus 119876120593(119870(119873

119891(119906119899119895

)))]

(71)

that the sequence (119872(119906119899119895

)1015840)119895is convergent in119862Then passing

to a subsequence if necessary we obtain that (V119899119895

)119895

=

(119872(119906119899119895

))119895is convergent in 1198621 Finally let (V

119899)119899be a sequence

in119872(Λ) Let (119911119899)119899sube 119872(Λ) be such that

lim119899rarrinfin

1003817100381710038171003817119911119899 minus V119899

10038171003817100381710038171 = 0 (72)

Let (119911119899119895

)119895be a subsequence of (119911

119899)119899such that it converges to

119911 It follows that 119911 isin 119872(Λ) and (V119899119895

)119895converge to 119911 This

concludes the proof

The next result is based on Schauderrsquos fixed point theo-rem

Theorem 14 Let 119891 [0 119879] timesRtimesR rarr R be continuous Then(55) has at least one solution

Abstract and Applied Analysis 7

Proof Let 119906 isin 1198621 Then

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(73)

where

119872(119906) (0) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (119879)

119872 (119906)1015840

(119879) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (0)

(74)

Moreover10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infin

=10038171003817100381710038171003817120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]

10038171003817100381710038171003817infinlt 119886

119872 (119906)infin

lt 119886 + 119886119879

(75)

Hence

119872 (119906)1= 119872 (119906)

infin+10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infinlt 119886 + 119886119879 + 119886

= 2119886 + 119886119879

(76)

Because the operator 119872 is completely continuous andbounded we can use Schauderrsquos fixed point theorem todeduce the existence of at least one fixed point This in turnimplies that problem (55) has at least one solution The proofis complete

5 Problems with Classic Homeomorphismsand Three-Point Boundary Conditions

We finally consider boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(77)

where 120593 R rarr R is a homeomorphism such that120593(0) = 0 and 119891 [0 119879] times R times R rarr R is a continuousfunction We remember that a solution of this problem isany function 119906 [0 119879] rarr R of class 119862

1 such that120593(1199061015840) is continuously differentiable satisfying the boundaryconditions and (120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879]

Let us consider the operator

1198721 1198621997888rarr 119862

1

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(78)

Analogously to Section 3 here 120593minus1 is understood as theoperator 120593minus1 119862 rarr 119862 defined for 120593minus1(V)(119905) = 120593minus1(V(119905)) Itis clear that 120593minus1 is continuous and sends bounded sets intobounded sets

Lemma 15 119906 isin 1198621 is a solution of (77) if and only if 119906 is a

fixed point of the operator 1198721

Proof Let 119906 isin 1198621 and we have the following equivalences

(120593 (1199061015840))1015840

= 119873119891(119906)

1199061015840

(119879) = 1199061015840

(0)

1199061015840

(0) = 119906 (119879)

lArrrArr (120593 (1199061015840))1015840

= 119873119891(119906) minus 119876 (119873

119891(119906))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 120593(1199061015840) = 119867(119873

119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906

1015840

(0))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 1199061015840= 120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))] 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))]) 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906 (119879))])

lArrrArr 119906 = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(79)

Remark 16 Note that 1199061015840(119879) = 1199061015840(0) hArr 119876(119873119891(119906)) = 0

Using an argument similar to the one introduced inLemma 13 it is easy to see that 119872

1 1198621 rarr 1198621 is completely

continuousIn order to apply Leray-Schauder degree to the operator

1198721 we introduced a family of problems depending on

parameter 120582 We remember that to each continuous function119891 [0 119879] times R times R rarr R we associate its Nemytskii operator119873119891 1198621 rarr 119862 defined by

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (80)

For 120582 isin [0 1] we consider the family of boundary valueproblems

(120593 (1199061015840))1015840

= 120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(81)

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

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Differential EquationsInternational Journal of

Volume 2014

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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

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Mathematical PhysicsAdvances in

Complex AnalysisJournal of

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OptimizationJournal of

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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

By the inversion of 120593 in (58) we have

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) (119905) + 119888] (59)

where 119888 = 120593(119906(0)) Integrating from 0 to 119905 isin [0 119879] we havethat

119906 (119905) = 119906 (0) + 119867 (120593minus1

[119870 (119873119891(119906)) + 119888]) (119905) (60)

Because 119906(0) = 119906(119879) then

int119879

0

120593minus1

[119870 (119873119891(119906)) (119905) + 119888] 119889119905 = 0 (61)

Using an argument similar to the one introduced in Lemma 2it follows that 119888 = minus119876

120593(119870(119873119891(119906))) Hence

119906 = 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(62)

Let 119906 isin 1198621 be such that 119906 = 119872(119906) Then

119906 (119905)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]) (119905)

(63)

for all 119905 isin [0 119879] Since int119879

0120593minus1[119870(119873

119891(119906))(119905) minus

119876120593(119870(119873119891(119906)))]119889119905 = 0 therefore we have that 119906(0) = 119906(119879)

Differentiating (63) we obtain that

1199061015840

(119905) = 120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))] (119905) (64)

In particular

1199061015840

(119879) = 120593minus1

(0 minus 119876120593(119870 (119873

119891(119906))))

= 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119906 (0)

(65)

Applying 120593 to both members and differentiating again wededuce that

(120593 (1199061015840

(119905)))1015840

= 119873119891(119906) (119905)

119906 (0) = 119906 (119879)

119906 (0) = 1199061015840

(119879)

(66)

for all 119905 isin [0 119879] This completes the proof

Lemma 13 The operator 119872 1198621 rarr 1198621 is completelycontinuous

Proof Let Λ sub 1198621 be a bounded set Then if 119906 isin Λ thereexists a constant 120588 gt 0 such that

1199061le 120588 (67)

Next we show that119872(Λ) sub 1198621 is a compact set Let (V119899)119899be a

sequence in119872(Λ) and let (119906119899)119899be a sequence in Λ such that

V119899= 119872(119906

119899) Using (67) we have that there exists a constant

119871 gt 0 such that for all 119899 isin N10038171003817100381710038171003817119873119891(119906119899)10038171003817100381710038171003817infin

le 119871 (68)

which implies that

10038171003817100381710038171003817119870 (119873119891(119906119899)) minus 119876

120593(119870 (119873

119891(119906119899)))

10038171003817100381710038171003817infinle 2119871119879 (69)

Hence the sequence (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isbounded in 119862 Moreover for 119905 119905

1isin [0 119879] and for all 119899 isin N

we have that10038161003816100381610038161003816119870 (119873119891(119906119899)) (119905) minus 119876

120593(119870 (119873

119891(119906119899)))

minus 119870 (119873119891(119906119899)) (1199051) + 119876120593(119870 (119873

119891(119906119899)))

10038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816minus int119879

119905

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

+ int119879

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816

le

100381610038161003816100381610038161003816100381610038161003816int119905

1199051

119891 (119904 119906119899(119904) 1199061015840

119899(119904)) 119889119904

100381610038161003816100381610038161003816100381610038161003816le 119871

1003816100381610038161003816119905 minus 1199051

1003816100381610038161003816

(70)

which implies that (119870(119873119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899

isequicontinuous Thus by the Arzela-Ascoli theorem there isa subsequence of (119870(119873

119891(119906119899)) minus 119876

120593(119870(119873119891(119906119899))))119899 which we

call (119870(119873119891(119906119899119895

)) minus 119876120593(119870(119873119891(119906119899119895

))))119895 which is convergent in

119862 Using the fact that 120593minus1 119862 rarr 119861119886(0) sub 119862 is continuous it

follows from

119872(119906119899119895

)1015840

= 120593minus1

[119870 (119873119891(119906119899119895

)) minus 119876120593(119870(119873

119891(119906119899119895

)))]

(71)

that the sequence (119872(119906119899119895

)1015840)119895is convergent in119862Then passing

to a subsequence if necessary we obtain that (V119899119895

)119895

=

(119872(119906119899119895

))119895is convergent in 1198621 Finally let (V

119899)119899be a sequence

in119872(Λ) Let (119911119899)119899sube 119872(Λ) be such that

lim119899rarrinfin

1003817100381710038171003817119911119899 minus V119899

10038171003817100381710038171 = 0 (72)

Let (119911119899119895

)119895be a subsequence of (119911

119899)119899such that it converges to

119911 It follows that 119911 isin 119872(Λ) and (V119899119895

)119895converge to 119911 This

concludes the proof

The next result is based on Schauderrsquos fixed point theo-rem

Theorem 14 Let 119891 [0 119879] timesRtimesR rarr R be continuous Then(55) has at least one solution

Abstract and Applied Analysis 7

Proof Let 119906 isin 1198621 Then

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(73)

where

119872(119906) (0) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (119879)

119872 (119906)1015840

(119879) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (0)

(74)

Moreover10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infin

=10038171003817100381710038171003817120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]

10038171003817100381710038171003817infinlt 119886

119872 (119906)infin

lt 119886 + 119886119879

(75)

Hence

119872 (119906)1= 119872 (119906)

infin+10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infinlt 119886 + 119886119879 + 119886

= 2119886 + 119886119879

(76)

Because the operator 119872 is completely continuous andbounded we can use Schauderrsquos fixed point theorem todeduce the existence of at least one fixed point This in turnimplies that problem (55) has at least one solution The proofis complete

5 Problems with Classic Homeomorphismsand Three-Point Boundary Conditions

We finally consider boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(77)

where 120593 R rarr R is a homeomorphism such that120593(0) = 0 and 119891 [0 119879] times R times R rarr R is a continuousfunction We remember that a solution of this problem isany function 119906 [0 119879] rarr R of class 119862

1 such that120593(1199061015840) is continuously differentiable satisfying the boundaryconditions and (120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879]

Let us consider the operator

1198721 1198621997888rarr 119862

1

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(78)

Analogously to Section 3 here 120593minus1 is understood as theoperator 120593minus1 119862 rarr 119862 defined for 120593minus1(V)(119905) = 120593minus1(V(119905)) Itis clear that 120593minus1 is continuous and sends bounded sets intobounded sets

Lemma 15 119906 isin 1198621 is a solution of (77) if and only if 119906 is a

fixed point of the operator 1198721

Proof Let 119906 isin 1198621 and we have the following equivalences

(120593 (1199061015840))1015840

= 119873119891(119906)

1199061015840

(119879) = 1199061015840

(0)

1199061015840

(0) = 119906 (119879)

lArrrArr (120593 (1199061015840))1015840

= 119873119891(119906) minus 119876 (119873

119891(119906))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 120593(1199061015840) = 119867(119873

119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906

1015840

(0))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 1199061015840= 120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))] 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))]) 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906 (119879))])

lArrrArr 119906 = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(79)

Remark 16 Note that 1199061015840(119879) = 1199061015840(0) hArr 119876(119873119891(119906)) = 0

Using an argument similar to the one introduced inLemma 13 it is easy to see that 119872

1 1198621 rarr 1198621 is completely

continuousIn order to apply Leray-Schauder degree to the operator

1198721 we introduced a family of problems depending on

parameter 120582 We remember that to each continuous function119891 [0 119879] times R times R rarr R we associate its Nemytskii operator119873119891 1198621 rarr 119862 defined by

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (80)

For 120582 isin [0 1] we consider the family of boundary valueproblems

(120593 (1199061015840))1015840

= 120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(81)

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

Proof Let 119906 isin 1198621 Then

119872(119906)

= 120593minus1

(minus119876120593(119870 (119873

119891(119906))))

+ 119867(120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))])

(73)

where

119872(119906) (0) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (119879)

119872 (119906)1015840

(119879) = 120593minus1

(minus119876120593(119870 (119873

119891(119906)))) = 119872 (119906) (0)

(74)

Moreover10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infin

=10038171003817100381710038171003817120593minus1

[119870 (119873119891(119906)) minus 119876

120593(119870 (119873

119891(119906)))]

10038171003817100381710038171003817infinlt 119886

119872 (119906)infin

lt 119886 + 119886119879

(75)

Hence

119872 (119906)1= 119872 (119906)

infin+10038171003817100381710038171003817119872 (119906)

101584010038171003817100381710038171003817infinlt 119886 + 119886119879 + 119886

= 2119886 + 119886119879

(76)

Because the operator 119872 is completely continuous andbounded we can use Schauderrsquos fixed point theorem todeduce the existence of at least one fixed point This in turnimplies that problem (55) has at least one solution The proofis complete

5 Problems with Classic Homeomorphismsand Three-Point Boundary Conditions

We finally consider boundary value problems of the form

(120593 (1199061015840))1015840

= 119891 (119905 119906 1199061015840)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(77)

where 120593 R rarr R is a homeomorphism such that120593(0) = 0 and 119891 [0 119879] times R times R rarr R is a continuousfunction We remember that a solution of this problem isany function 119906 [0 119879] rarr R of class 119862

1 such that120593(1199061015840) is continuously differentiable satisfying the boundaryconditions and (120593(1199061015840(119905)))1015840 = 119891(119905 119906(119905) 1199061015840(119905)) for all 119905 isin [0 119879]

Let us consider the operator

1198721 1198621997888rarr 119862

1

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(78)

Analogously to Section 3 here 120593minus1 is understood as theoperator 120593minus1 119862 rarr 119862 defined for 120593minus1(V)(119905) = 120593minus1(V(119905)) Itis clear that 120593minus1 is continuous and sends bounded sets intobounded sets

Lemma 15 119906 isin 1198621 is a solution of (77) if and only if 119906 is a

fixed point of the operator 1198721

Proof Let 119906 isin 1198621 and we have the following equivalences

(120593 (1199061015840))1015840

= 119873119891(119906)

1199061015840

(119879) = 1199061015840

(0)

1199061015840

(0) = 119906 (119879)

lArrrArr (120593 (1199061015840))1015840

= 119873119891(119906) minus 119876 (119873

119891(119906))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 120593(1199061015840) = 119867(119873

119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906

1015840

(0))

119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 1199061015840= 120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))] 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (1199061015840

(0))]) 119876 (119873119891(119906)) = 0 119906

1015840

(0) = 119906 (119879)

lArrrArr 119906 = 119906 (119879) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119906 (119879))])

lArrrArr 119906 = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[119867 (119873119891(119906) minus 119876 (119873

119891(119906))) + 120593 (119878 (119906))])

(79)

Remark 16 Note that 1199061015840(119879) = 1199061015840(0) hArr 119876(119873119891(119906)) = 0

Using an argument similar to the one introduced inLemma 13 it is easy to see that 119872

1 1198621 rarr 1198621 is completely

continuousIn order to apply Leray-Schauder degree to the operator

1198721 we introduced a family of problems depending on

parameter 120582 We remember that to each continuous function119891 [0 119879] times R times R rarr R we associate its Nemytskii operator119873119891 1198621 rarr 119862 defined by

119873119891(119906) (119905) = 119891 (119905 119906 (119905) 119906

1015840

(119905)) (80)

For 120582 isin [0 1] we consider the family of boundary valueproblems

(120593 (1199061015840))1015840

= 120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(81)

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

Notice that (81) coincides with (77) for 120582 = 1 So for each120582 isin [0 1] the operator associated with (81) by Lemma 15 isthe operator119872(120582 sdot) where119872 is defined on [0 1] times 1198621 by

119872(120582 119906) = 119878 (119906) + 119876 (119873119891(119906))

+ 119870 (120593minus1

[120582119867 (119873119891(119906) minus 119876 (119873

119891(119906)))

+ 120593 (119878 (119906))])

(82)

Using the same arguments as in the proof of Lemma 13we show that the operator 119872 is completely continuousMoreover using the same reasoning as above system (81) (seeLemma 15) is equivalent to the problem

119906 = 119872(120582 119906) (83)

51 Existence Results In this subsection we present andprove our main results These results are inspired by worksof Bereanu and Mawhin [6] and Manasevich and Mawhin[10] We denote by degB the Brouwer degree and by degLS theLeray-Schauder degree and define the mapping119866 R2 rarr R2

by

119866 R2997888rarr R

2

(119886 119887) 997891997888rarr (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879)

(84)

Theorem 17 Assume thatΩ is an open bounded set in1198621 suchthat the following conditions hold

(1) For each 120582 isin (0 1) problem

(120593 (1199061015840))1015840

= 120582119873119891(119906)

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(85)

has no solution on 120597Ω(2) The equation

119866 (119886 119887) = (0 0) (86)

has no solution on 120597Ω cap R2 where we consider thenatural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 119862

1(3) The Brouwer degree

deg119861(119866Ω capR

2 0) = 0 (87)

Then problem (77) has a solution

Proof Let 120582 isin (0 1] If 119906 is a solution of (85) then119876(119873119891(119906)) = 0 hence 119906 is a solution of problem (81) On the

other hand for 120582 isin (0 1] if 119906 is a solution of (81) and because

119876(120582119873119891(119906) + (1 minus 120582)119876 (119873

119891(119906))) = 119876 (119873

119891(119906)) (88)

we have 119876(119873119891(119906)) = 0 then 119906 is a solution of (85) It follows

that for 120582 isin (0 1] problems (81) and (85) have the samesolutions We assume that for 120582 = 1 (81) does not have asolution on 120597Ω since otherwise we are done with proof Itfollows that (81) has no solutions for (120582 119906) isin (0 1] times 120597Ω If120582 = 0 then (81) is equivalent to the problem

(120593 (1199061015840))1015840

= 119876 (119873119891(119906))

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(89)

and thus if 119906 is a solution of (89) we must have

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 (90)

Moreover 119906 is a function of the form 119906(119905) = 119886+119887119905 119886 = 119887minus119887119879Thus by (90)

int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 = 0 (91)

which together with hypothesis (2) implies that 119906 = 119887minus119887119879+

119905119887 notin 120597Ω Thus we have proved that (81) has no solution in 120597Ω

for all 120582 isin [0 1] Then we have that for each 120582 isin [0 1] theLeray-Schauder degree degLS(119868minus119872(120582 sdot) Ω 0) is well definedand by the homotopy invariance one has

degLS (119868 minus 119872 (0 sdot) Ω 0) = degLS (119868 minus 119872 (1 sdot) Ω 0) (92)

On the other hand we have that

degLS (119868 minus 119872 (0 sdot) Ω 0)

= degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

(93)

But the range of the mapping

119906 997891997888rarr 119878 (119906) + 119876 (119873119891(119906)) + 119870 (119878 (119906)) (94)

is contained in the subspace of related functions isomorphicto R2 Thus using a reduction property of Leray-Schauderdegree [7 11]

degLS (119868 minus (119878 + 119876119873119891+ 119870119878) Ω 0)

= degB (119868 minus (119878 + 119876119873119891+ 119870119878)

10038161003816100381610038161003816 ΩcapR2 Ω capR

2 0)

= degB (119866Ω capR2 0) = 0

(95)

Then degLS(119868 minus 119872(1 sdot) Ω 0) = 0 where 119868 denotes the unitoperator Hence there exists 119906 isin Ω such that 119872

1(119906) = 119906

which is a solution for (77)

The following result gives a priori bounds for the possiblesolutions of (85) and adapts a technique introduced by WardJr [12]

Theorem 18 Assume that 119891 satisfies the following conditions

(1) There exists 119888 isin 119862 such that

119891 (119905 119909 119910) ge 119888 (119905) (96)

for all (119905 119909 119910) isin [0 119879] timesR timesR

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

(2) There exists1198721lt 1198722such that for all 119906 isin 1198621

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119898

ge 1198722

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0 if 1199061015840119872

le 1198721

(97)

If (120582 119906) isin (0 1) times 1198621 is such that 119906 is solution of (85) then

1199061lt 119903 (2 + 119879) (98)

where

119903

= max 10038161003816100381610038161003816120593minus1

(119871 + 21003817100381710038171003817119888minus10038171003817100381710038171198711)

1003816100381610038161003816100381610038161003816100381610038161003816120593minus1

(minus119871 minus 21003817100381710038171003817119888minus10038171003817100381710038171198711)

10038161003816100381610038161003816

119871 = max 1003816100381610038161003816120593 (119872

2)1003816100381610038161003816

1003816100381610038161003816120593 (1198721)1003816100381610038161003816

(99)

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) Then for all 119905 isin [0 119879]

(120593 (1199061015840

(119905)))1015840

= 120582119873119891(119906) (119905)

1199061015840

(0) = 1199061015840

(119879) = 119906 (119879)

int119879

0

119891 (119905 119906 (119905) 1199061015840

(119905)) 119889119905 = 0

(100)

Using hypothesis (2) we have that

1199061015840

119898lt 1198722

1199061015840

119872gt 1198721

(101)

It follows that there exists 120596 isin [0 119879] such that 1198721lt 1199061015840(120596) lt

1198722and

int119905

120596

(120593 (1199061015840

(119904)))1015840

119889119904 = 120582int119905

120596

119873119891(119906) (119904) 119889119904 (102)

which implies that

10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816le

10038161003816100381610038161003816120593 (1199061015840

(120596))10038161003816100381610038161003816

+ int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

(103)

where

int119879

0

10038161003816100381610038161003816119891 (119904 119906 (119904) 119906

1015840

(119904))10038161003816100381610038161003816119889119904

le int119879

0

119891 (119904 119906 (119904) 1199061015840

(119904)) 119889119904 + 2int119879

0

119888minus

(119904) 119889119904

(104)

Hence10038161003816100381610038161003816120593 (1199061015840

(119905))10038161003816100381610038161003816lt 119871 + 2

1003817100381710038171003817119888minus10038171003817100381710038171198711 (105)

where 119871 = max|120593(1198722)| |120593(119872

1)| and 119905 isin [0 119879] It follows

that10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 (106)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)|

Because 119906 isin 1198621 is such that 1199061015840(0) = 119906(119879) we have that

|119906 (119905)| le |119906 (119879)| + int119879

0

100381610038161003816100381610038161199061015840

(119904)10038161003816100381610038161003816119889119904 lt 119903 + 119903119879

(119905 isin [0 119879])

(107)

and hence 1199061= 119906infin

+ 1199061015840infin

lt 119903 + 119903119879 + 119903 = 119903(2 + 119879) Thisproves the theorem

Now we show the existence of at least one solution forproblem (77) by means of Leray-Schauder degree

Theorem 19 Let 119891 be continuous and let it satisfy conditions(1) and (2) ofTheorem 18 Assume that the following conditionshold for some 120588 ge 119903(2 + 119879)

(1) The equation

119866 (119886 119887) = (0 0) (108)

has no solution on 120597119861120588(0) capR2 where we consider the

natural identification (119886 119887) asymp 119886 + 119887119905 ofR2 with relatedfunctions in 1198621

(2) The Brouwer degree

deg119861(119866 119861120588(0) capR

2 0) = 0 (109)

Then problem (77) has a solution

Proof Let (120582 119906) isin (0 1) times 1198621 be such that 119906 is a solution of(85) UsingTheorem 18 we have

1199061= 119906infin

+10038171003817100381710038171003817119906101584010038171003817100381710038171003817infin

lt 119903 + 119903119879 + 119903 = 119903 (2 + 119879) (110)

where 119903 = max|120593minus1(119871 + 2119888minus1198711)| |120593minus1(minus119871 minus 2119888minus

1198711)| Thus

the conditions of Theorem 17 are satisfied with Ω = 119861120588(0)

where 119861120588(0) is the open ball in 1198621 center 0 and radius 120588 This

concludes the proof

Let us give now an application of Theorem 19

Example 20 Let us consider the problem

((1199061015840)3

)1015840

=1198901199061015840

2minus 1

119906 (119879) = 1199061015840

(0) = 1199061015840

(119879)

(111)

Let 1198721= minus1 and 119872

2= 1 If we suppose that 1199061015840

119898ge 1198722and

1199061015840119872

le 1198721 then

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 ge (

1198901198722

2minus 1)119879 gt 0

int119879

0

(1198901199061015840(119905)

2minus 1)119889119905 le (

1198901198721

2minus 1)119879 lt 0

(112)

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

On the other hand if we choose 120588 ge (1 + 2119879)13

(2 + 119879) and119888(119905) = minus1 for all 119905 isin [0 119879] we have that the equation

119866 (119886 119887) = (119886119879 + 1198871198792minus 119887119879

minus1

119879int119879

0

119891 (119905 119886 + 119887119905 119887) 119889119905 119887 minus 119886 minus 119887119879) = (0 0)

= (119886119879 + 1198871198792minus 119887119879 minus

1

119879int119879

0

(119890119887

2minus 1)119889119905 119887 minus 119886

minus 119887119879) = (0 0) = (119886119879 + 1198871198792minus 119887119879 minus

119890119887

2+ 1 119887 minus 119886

minus 119887119879) = (0 0)

(113)

has no solution on 120597119861120588(0)capR2Thenwehave that theBrouwer

degree degB(119866 119861120588(0) cap R2 (0 0)) is well defined and by the

properties of that degree we have that

degB (119866 119861120588(0) capR

2 (0 0)) = sum

119909isin119866minus1(00)

sgn 119869119866(119909)

= 0

(114)

where (0 0) is a regular value of 119866 and 119869119866(119909) = det1198661015840(119909) is

the Jacobian of 119866 at 119909 So using Theorem 19 we obtain thatthe boundary value problem (111) has at least one solution

Competing Interests

The author declares that there is no conflict of interestsregarding the publication of this article

Acknowledgments

This researchwas supported byCAPES andCNPqBrazilTheauthor would like to thankDr Pierluigi Benevieri for his kindadvice and for the constructive revision of this paper

References

[1] C Bereanu and J Mawhin ldquoExistence and multiplicity resultsfor some nonlinear problemswith singular120593-laplacianrdquo Journalof Differential Equations vol 243 no 2 pp 536ndash557 2007

[2] C Bereanu and J Mawhin ldquoPeriodic solutions of nonlinearperturbations of 120593rdquo Nonlinear Analysis Theory Methods ampApplications vol 68 no 6 pp 1668ndash1681 2008

[3] M A del Pino R F Manasevich and A E Murua ldquoExistenceandmultiplicity of solutionswith prescribed period for a secondorder quasilinear ODErdquo Nonlinear Analysis Theory Methodsamp Applications vol 18 no 1 pp 79ndash92 1992

[4] P Yan ldquoNonresonance for one-dimensional p-Laplacian withregular restoringrdquo Journal of Mathematical Analysis and Appli-cations vol 285 no 1 pp 141ndash154 2003

[5] P Benevieri J M do O and E S de Medeiros ldquoPeriodicsolutions for nonlinear systems with mean curvature-like oper-atorsrdquo Nonlinear Analysis Theory Methods amp Applications vol65 no 7 pp 1462ndash1475 2006

[6] C Bereanu and J Mawhin ldquoBoundary-value problems withnon-surjective 120593- laplacian and one-sided bounded nonlinear-ityrdquo Advances Differential Equations vol 11 no 1 pp 35ndash602006

[7] J Mawhin Topological Degree Methods in Nonlinear BoundaryValue Problems vol 40 of CBMS Series American Mathemati-cal Society Providence RI USA 1979

[8] C Bereanu and J Mawhin ldquoBoundary value problems for somenonlinear systems with singular 120601-laplacianrdquo Journal of FixedPoint Theory and Applications vol 4 no 1 pp 57ndash75 2008

[9] JMawhin ldquoLeray-Schauder degree a half century of extensionsand applicationsrdquo Topological Methods in Nonlinear Analysisvol 14 no 2 pp 195ndash228 1999

[10] RManasevich and JMawhin ldquoPeriodic solutions for nonlinearsystems with p-Laplacian-like operatorsrdquo Journal of DifferentialEquations vol 145 no 2 pp 367ndash393 1998

[11] K Deimling Nonlinear Functional Analysis Springer BerlinGermany 1985

[12] J R Ward Jr ldquoAsymptotic conditions for periodic solutionsof ordinary differential equationsrdquo Proceedings of the AmericanMathematical Society vol 81 no 3 pp 415ndash420 1981

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of