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R E S E A R C H

Open Access

Existence of concave symmetric positive

solutions for a three-point boundary value

problems

Ummahan Akcan

1

and Nuket A Hamal

2*

*Correspondence: nuket.aykut@ege.edu.tr 2Department of Mathematics, Ege

University, Bornova, Izmir, 35100, Turkey

Full list of author information is available at the end of the article

Abstract

In this paper, we investigate the existence of triple concave symmetric positive solutions for the nonlinear boundary value problems with integral boundary conditions. The proof is based upon the Avery and Peterson fixed point theorem. An example which supports our theoretical result is also indicated.

MSC: 34B10; 39B18; 39A10

Keywords: three-point boundary value problem; symmetric positive solution; integral boundary condition; fixed point theorem

1 Introduction

Consider the boundary value problem (BVP)

u(x) +f(x,u(x),u(x)) = , x∈(, ),

u() =u() =αηu(s)ds, (.)

wheref: (, )×[,∞)×R→[,∞) is continuous andα,η∈(, ). A functionuC[, ] is said to be symmetric on [, ] if

u(x) =u( –x), x∈[, ].

By a symmetric positive solution of BVP (.) we mean a symmetric functionuC[, ] such thatu(x)≥ forx∈(, ) andu(x) satisfies BVP (.).

Recently, many authors have focused on the existence of symmetric positive solutions for ordinary differential equation boundary value problems; for example, see [–] and the references therein. However, multi-point boundary value problems included the most recent works [–, –] and boundary value problems with integral boundary conditions for ordinary differential equations have been studied by many authors; one may refer to [, –]. Motivated by the works mentioned above, we aim to investigate existence results for concave symmetric positive solutions of BVP (.) by applying the fixed point theorem of Avery and Peterson.

The organization of this paper is as follows. Section  of this paper contains some pre-liminary lemmas. In Section , by applying the Avery and Peterson fixed point theorem,

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we obtain concave symmetric positive solutions for BVP (.). In Section , an example will be presented to illustrate the applicability of our result.

Throughout this paper, we always assume that the following assumption is satisfied:

(H) fC((, )×[, +∞)×R, [, +∞)),f(x,u,v) =f( –x,u, –v)forx∈(,], and

f(x,u,v)≥for all(x,u,v)∈(, )×[, +∞)×R.

2 Preliminaries

In this section, we present several lemmas that will be used in the proof of our result.

Lemma . Let hC[, ]andαη= ,then the BVP

u(x) +h(x) = ,  <x< , (.)

u() =u() =α η

u(s)ds, (.)

has a solution

u(x) =

H(s) +G(x,s)h(s)ds, (.)

where

V(s) =

(ηs), sη,

, ηs, H(s) =

αη

( –αη)( –s) –

α

( –αη)V(s), (.)

G(x,s) =

s( –x), ≤sx≤,

x( –s), ≤xs≤. (.)

Proof Suppose thatuC[, ] is a solution of problem (.) and (.). Then we have u(x) = –h(x).

Forx∈[, ], by integration from  to , we have

u(x) =u() –

x

h(s)ds.

Forx∈[, ], by integration again from  to , we have

u(x) =u()x

x

τ

h(s)ds

.

That is,

u(x) =u() +u()x

x

(xs)h(s)ds, (.)

therefore,

u() =u() +u() –

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From condition (.), we have

u() =

( –s)h(s)ds.

Integrating (.) from  toη, whereη∈(, ), we have

η

u(s)ds=u()η+u()η

 –

η

τ

(τs)h(s)ds

=u()η+u()η

 –  

η

(ηs)h(s)ds,

and fromu() =αηu(s)ds, we have

u() = αη

( –αη)u

() – α

( –αη)

η

(ηs)h(s)ds,

therefore, (.) and (.) have a unique solution

u(x) = αη

( –αη)

( –s)h(s)dsα ( –αη)

η

(ηs)h(s)ds

+x

( –s)h(s)ds

x

(xs)h(s)ds.

From (.) and (.), we obtain

u(x) =

H(s) +G(x,s)h(s)ds.

The proof is complete.

The functionsHandGhave the following properties.

Lemma . Ifα,η∈(, ),then we have H(s)≥ for s∈[, ].

Proof From the definition ofH(s),s∈(, ), andα,η∈(, ), we haveH(s)≥. Lemma . G( –x,  –s) =G(x,s), ≤G(x,s)≤G(s,s)for x,s∈[, ].

Proof From the definition ofG(x,s), we getG( –x,  –s) =G(x,s) and ≤G(x,s)≤G(s,s)

forx,s∈[, ].

Lemma . If f(x,u(x),u(x))∈C((, )×[, +∞)×R, [, +∞))and we letα,η∈(, ), then the unique solution u of BVP(.)satisfies u(x)≥for x∈[, ].

Proof From the definition ofu(x), Lemma ., Lemma ., andf(x,u(x),u(x))∈C((, )×

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LetE=C[, ]. ThenEis Banach space with the norm u =max{ u

∞, u ∞}, where

u ∞=maxx∈[,]|u(x)|. We define the conePEby

P= uE:u(x)≥ is concave andu(x) =u( –x),x∈[, ]. Define the operatorT:PEas follows:

(Tu)(x) =

H(s) +G(x,s)fs,u(s),u(s)ds forx∈[, ], (.)

whereG(x,s) andH(s) are given by (.) and (.). Clearly,uis the solution of BVP (.) if and only ifuis a fixed point of the operatorT.

Lemma . Let <α< and <η< .For uP,there exists a real number M> such that

max

≤x≤u(x)≤Mmax≤x≤u

(x),

where M=(–αη).

Proof For anyuP, we have

max

≤x≤

u(x)=u

 

, max

≤x≤

u(x)=u(),

the concavity ofuimplies that

u()≥u(  ) –u()

 

,

then

u

 

u()

 +u(). (.)

Now we divide the proof into two cases.

Case . If  <η, then the concavity ofuimplies that

η

u(s)dsηu(η),

whereηu(η) is the area of rectangle, then

u()≤αηu(η)≤αηu

 

.

So

u()≤αηu

 

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then from (.) and (.), we have

and because the graph ofuis concave, we have

η

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Case . If  <η, then using concavity and positivity ofu, we get u(η) –u()

η

u() –u()  

.

So

η– 

η u()≥u

 

– 

ηu(η), (.)

then from (.) and (.), we have

u()≥ αη

αηαη+ u

 

,

that is,

min

≤x≤u(x)≥

αη

αηαη+  u ∞.

Case . If 

≤η< , then using concavity and positivity ofu, we get

η

u(s)ds≥(u( 

) –u())  

 +

u()  .

So

u()≥ α  –αu

 

,

that is,

min

≤x≤

u(x)≥ α

 –α u ∞,

whereN=max{αηαηαη+,

α

–α}, and forα,η∈(, ), one can easily see thatN∈(, ). These

equations complete the proof.

3 Existence of triple concave symmetric positive solutions for BVP (1.1)

In this section, we will apply the Avery-Peterson fixed point theorem to establish the ex-istence of at least three concave symmetric positive solutions of BVP (.).

Letα,γ,θ,ψbe maps onPwithαa nonnegative continuous concave functional;γ,θ

nonnegative continuous convex functionals, andψ a nonnegative continuous functional. Then for positive numbersa,b,c,dwe define the following subsets ofP:

P(γ,d) = uP|γ(u) <d,

P(α,γ,b,d) = uP(γ,d)|α(u)≥b,

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Now we state the Avery-Peterson fixed point theorem.

Theorem .([, , , ]) Let P be cone in Banach space E.Letγ andθbe nonnegative continuous convex functionals on P,αbe a nonnegative continuous concave functional on P,andψbe a nonnegative continuous functional on P leading to

ψ(λu)≤λψ(u) for all≤λ≤ and

α(u)≤ψ(u), u(u) for all uP(γ,d)with M,d positive numbers. Suppose T:PP is completely continuous and there exist positive numbers a,b,c with a<b such that

(S): uP(α,θ,γ,b,c,d)|α(u) >b

=∅ and

α(Tu) >b for uP(α,θ,γ,b,c,d);

(S): α(Tu) >b for uP(α,γ,b,d)withθ(Tu) >c;

(S):  /∈R(ψ,γ,a,d) and ψ(Tu) <a for uR(ψ,γ,a,d)withψ(u) =a. Then T has at least three fixed points u,u,uP(γ,d)such that

γ(ui)≤d for i= , , ; ψ(u) <a;

ψ(u) >a withα(u) <b; α(u) >b.

Lemma . Assume that(H)is satisfied and letαη∈(, ).Then the operator T is com-pletely continuous.

Proof For anyuP, from the expression ofTu, we know

(Tu)(x) +f(x,u(x),u(x)) = , x∈(, ), (Tu)() = (Tu)() =αη(Tu)(s)ds.

Clearly,Tuis concave. From the definition ofTu, Lemma ., and Lemma . we see that Tuis nonnegative on [, ]. We now show thatTuis symmetric about 

. From Lemma . and (H), forx∈[, ], we have

(Tu)( –x) =

H(s) +G( –x,s)fs,u(s),u(s)ds

=

H(s)fs,u(s),u(s)ds+

G( –x,s)fs,u(s),u(s)ds

=

H(s)fs,u(s),u(s)ds

G( –x,  –s)f –s,u( –s),u( –s)ds

=

H(s)fs,u(s),u(s)ds+

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= pact. Suppose thatDPis a bounded set. Then there existsrsuch that

D= uP| ur.

These equations imply that the operatorT is uniformly bounded. Now we show thatTu is equi-continuous.

We separate these three conditions:

Case (i). ≤xx≤; Case (ii). 

≤x≤x≤;

Case (iii). ≤x≤x≤.

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In addition

(Tu)(x) – (Tu)(x)=

x

x

fs,u(s),u(s)dsM|xx|.

So, we see that Tuis equi-continuous. By applying the Arzela-Ascoli theorem, we can guarantee thatT(D) is relatively compact, which meansT is compact. Then we find that T is completely continuous. This completes the proof.

For convenience, we denote

L=

H(s) +G

 ,s

ds,

N=max

αη

αηαη+ ,

α

 –α

,

M=  ( –αη),

δ=

H(s)ds.

Theorem . Suppose(H)holds and letα,η∈(, ).Moreover,there exist nonnegative numbers <a<b≤dδsuch that

(B) f(x,u,v)≤d for(x,u,v)∈[, ]×[,Md]×[–d,d]; (B) f(x,u,v) >

b

δ for(x,u,v)∈[, ]×

b, b N

×[–d,d];

(B) f(x,u,v) < a

L for(x,u,v)∈[, ]×[,a]×[–d,d],

then BVP(.)has at least three concave symmetric positive solutions u,u,usuch that

max

≤x≤

ui(x)≤d for i= , , ; max

≤x≤

u(x)<a,

max

≤x≤u(x)>a withmin≤x≤u(x)<b, min≤x≤u(x)>b.

Proof BVP (.) has a solution u=u(x) if and only if u solves the operator equation u=T(u). So we need to verify that operatorT satisfies the Avery-Peterson fixed point theorem, which will prove the existence of at least three fixed points ofT.

Complete continuity ofTis clear from Lemma .. Define the nonnegative functionals

α,θ,γ, andψby

γ(u) = max

≤x≤

u(x), ψ(u) =θ(u) = max

≤x≤

u(x), α(u) = min

≤x≤

u(x).

Then in the coneP,θandγare convex asαis concave. It is well known thatψ(λu)≤λψ(u) for all ≤λ≤ andα(u)≤ψ(u). Moreover, from Lemma ., u(u). Now, we will prove the main theorem in four steps.

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If uP(γ,d), thenγ(u) =max≤x≤|u(x)| ≤d. Lemma . yields max≤x≤|u(x)| ≤

This shows that condition (S) of Theorem . is satisfied.

Step . We will show that condition (S) of Theorem . is satisfied. TakeuP(α,γ,b,d)

Step . We will show that condition (S) of Theorem . is also satisfied. Obviously,

ψ() =  <a, and we have  /∈R(ψ,γ,a,d). Assume thatuR(ψ,γ,a,d) withψ(u) =a. Then, from condition (B), we have

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It proves that condition (S) holds. All conditions of Theorem . are satisfied and we assert that BVP (.) has at least three concave symmetric positive solutionsu,u,uP such that

max

≤x≤u

i(x)≤d fori= , , ; maxxu(x)<a,

max

≤x≤

u(x)>a with min ≤x≤

u(x)<b, min ≤x≤

u(x)>b.

Therefore, our proof is complete.

4 Example

Example . We consider the following three-point second-order BVP with integral boundary conditions:

u(x) +f(x,u(x),u(x)) = , x∈(, ), u() =u() = 

   u(s)ds,

(.)

where

fx,u,u=

⎧ ⎨ ⎩

x( –x) +u+

|u|

,, ≤u≤, x( –x) + , +

|u|

,, u≥.

We can see from (.) thatα=,η=. ThenM=,N=,L=,,δ=, . We choose a= ,b= ,d= ,. Clearly  <a<b≤. Moreover,f satisfies (H).

(B) f(x,u,v)≤ 

+ , +

,

, < d= ,

for (x,u,v)∈[, ]×

,, 

×[–,, ,];

(B) f(x,u,v) > , > b

δ = ,. for (x,u,v)∈[, ]×[, ]×[–,, ,];

(B) f(x,u,v) <  +  +

, , <

a L≈.

for (x,u,v)∈[, ]×[, ]×[–,, ,].

So, by Theorem . we find that BVP (.) has at least three concave symmetric positive solutionsu,u,usuch that

max

≤x≤

ui(x)≤, fori= , , ; max

≤x≤

u(x)< ,

max

≤x≤u(x)>  with min≤x≤u(x)< , min≤x≤u(x)> .

Competing interests

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Authors’ contributions

All authors contributed equally to the manuscript and typed, read, and approved the final manuscript.

Author details

1Department of Mathematics, Faculty of Science, Anadolu University, Eskisehir, 26470, Turkey.2Department of Mathematics, Ege University, Bornova, Izmir, 35100, Turkey.

Received: 10 September 2014 Accepted: 17 November 2014 Published:09 Dec 2014

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problem. Nonlinear Anal.71, 9-18 (2009)

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5. Pang, H, Tong, Y: Symmetric positive solutions to a second-order boundary value problem with integral boundary conditions. Bound. Value Probl.2013, 150 (2013). doi:10.1186/1687-2770-2013-150

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References

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