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# Existence of positive solutions of third-order boundary value problems with integral boundary conditions in Banach spaces

- Dan Fu
^{1}and - Wei Ding
^{1}Email author

**2013**:65

https://doi.org/10.1186/1687-1847-2013-65

© Fu and Ding; licensee Springer 2013

**Received:**13 September 2012**Accepted:**18 January 2013**Published:**20 March 2013

## Abstract

This paper deals with positive solutions of a third-order differential equation in ordered Banach spaces,

subject to the following integral boundary conditions:

where *θ* is the zero element of *E*, $g\in L[0,1]$ is nonnegative, $\phi :R\to R$ is an increasing and positive homomorphism, and $\phi (0)={\theta}_{1}$. The arguments are based upon the fixed-point principle in cone for strict set contraction operators. Meanwhile, as an application, we also give an example to illustrate our results.

## Keywords

- positive solutions
- boundary-value problem
- fixed-point principle
- cone
- measure of noncompactness

## 1 Introduction

The theory of boundary value problems is experiencing a rapid development. Many methods are used to study this kind of problems such as fixed point theorems, shooting method, iterative method with upper and lower solutions, *etc.* We refer the readers to the papers [1–14]. Among them, the fixed-point principle in cone has become an important tool used in the study of existence and multiplicity of positive solutions. Many papers that use this method have been published in recent years (see [15–20]).

Recently, scientists have noticed that the boundary conditions in many areas of applied mathematics and physics come down to integral boundary conditions. For instance, the models on chemical engineering, heat conduction, thermo-elasticity, plasma physics, and underground water flow can be reduced to the nonlocal problems with integral boundary conditions. For more information about this subject, we refer the readers to the excellent survey by Gallardo [21–23], Corduneanu [24], and Agarwal and O’Regan [25].

where $f:[a,b]\times {R}^{2}\to R$ satisfies the local Carathéodory conditions and $\mu :[a,b]\to R$ is the function of bounded variation. These criteria apply to the case where the function *f* has nonintegrable singularities in the first argument at the points *a* and *b*.

where the kernel $K(t,s)$ satisfies a continuity assumption in the ${L}^{1}$-sense and it is monotone and concave. The main method is the Krasnosel’skii fixed point theorem on a suitable cone, then the above equation has at least one positive solution.

where $J=[0,1]$, $f\in C([0,1]\times P,P)$, *θ* is the zero element of *E*, *E* is a real Banach space with the norm $\parallel x\parallel $, and $g\in L[0,1]$ is nonnegative; $\phi :R\to R$ is an increasing and positive homomorphism (see Definition 1.2) and $\phi (0)={\theta}_{1}$.

According to Definition 1.2, we know that many problems, such as the problems with *p*-Laplacian operator, three-order boundary-value problems and so on, are special cases of (1). To the best of our knowledge, there have been few results on the positive solutions for odd-order boundary-value problems (or *p*-Laplacian problems) with integral boundary conditions in Banach spaces (see [26–31]).

The plan of this paper is as follows. We introduce some notations and lemmas in the rest of this section. In Section 2, we provide some necessary backgrounds. In particular, we state some properties of the Green’s function associated with BVP (1). In Section 3, we establish the main results of the paper. Finally, one example is also included to illustrate the main results.

**Definition 1.1**Let $(E,\parallel \cdot \parallel )$ be a real Banach space. A nonempty, closed, and convex set $P\subset E$ is said to be a cone provided that the following conditions are satisfied:

- (a)
If $y\in P$ and $\lambda \ge 0$, then $\lambda y\in P$;

- (b)
If $y\in P$ and $-y\in P$, then $y=0$.

If $P\subset E$ is a cone, we denote the order induced by *P* on *E* by ≤, that is, $x\le y$ if and only if $y-x\in P$. *P* is said to be normal if there exists a positive constant *N* such that $\theta \le x\le y$ implies $\parallel x\parallel \le N\parallel y\parallel $. *N* is called the normal constant of *P*.

**Definition 1.2**A projection $\phi :R\to R$ is called an increasing and positive homomorphism if the following conditions are satisfied:

- (1)
If $x\le y$, then $\phi (x)\le \phi (y)$ for all $x,y\in R$;

- (2)
*φ*is a continuous bijection and its inverse is also continuous; - (3)
$\phi (xy)=\phi (x)\phi (y)$ for all $x,y\in {R}_{+}=[0,+\mathrm{\infty})$.

- (4)
$\phi (xy)=\phi (x)\phi (y)$ for all $x,y\in R$, where $R=(-\mathrm{\infty},+\mathrm{\infty})$.

**Definition 1.3** Let *E* be a real Banach space and $P\subset E$ be a cone in *E*. If ${P}^{\ast}=\{\psi \in {E}^{\ast}|\psi (x)\ge 0,\mathrm{\forall}x\in P\}$, then ${P}^{\ast}$ is a dual cone of the cone *P*.

**Definition 1.4** Let *S* be a bounded set in a real Banach space *E*. Let $\alpha (S)$ = inf{$\delta >0:S$ be expressed as the union of a finite number of sets such that the diameter of each set does not exceed *δ*, *i.e.*, $S={\bigcup}_{i=1}^{m}{S}_{i}$ with $diam({S}_{i})\le \delta $, $i=1,2,\dots ,m$}. Clearly, $0\le \alpha (S)<\mathrm{\infty}$. $\alpha (S)$ is called Kuratowski’s measure of noncompactness.

**Definition 1.5** Let *E* be an ordered Banach space, *D* be a bounded set of *E*. The operator $A:D\to E$ is said to be a *k*-set contraction if $A:D\to E$ is continuous and bounded, and there is a constant $k\ge 0$ such that $\alpha (A(S))\le k\alpha (S)$ for any bounded $S\subset D$; a *k*-set contraction with $k<1$ is called a strict set contraction.

*E*can be found in [15–18]. For a bounded set

*C*in the Banach space

*E*, we denote by $\alpha (C)$ the Kuratowski’s measure of noncompactness. In the following, we denote by $\alpha (\cdot )$, ${\alpha}_{C}(\cdot )$ the Kuratowski’s measure of noncompactness of a bounded subset in

*E*and in $C(J,E)$, respectively. And we set

where $x\in P$, *β* denotes 0 or ∞, $\psi \in {P}^{\ast}$, $\parallel \psi \parallel =1$, and $h(t,x)={\phi}^{-1}({\int}_{0}^{t}f(s,x(s))\phantom{\rule{0.2em}{0ex}}ds)$.

**Lemma 1.1** [1]

*If* $H\in C(J,E)$ *is bounded and equicontinuous*, *then* ${\alpha}_{C}(H)=\alpha (H(J))={max}_{t\in J}\alpha (H(t))$, *where* $H(J)=\{x(t):t\in J,x\in H\}$, $H(t)=\{x(t):x\in H\}$.

**Lemma 1.2** [1]

*Let*

*D*

*be a bounded set of*

*E*;

*if*

*f*

*is uniformly continuous and bounded from*$J\times S$

*into*

*E*,

*then*

*where* ${\eta}_{l}$ *is a nonnegative constant*.

**Lemma 1.3**

*Let*

*K*

*be a cone of the Banach space*

*E*

*and*${K}_{r}=\{x\in K:\parallel x\parallel \le r\}$, ${K}_{r,{r}^{\prime}}=\{x\in K,r\le \parallel x\parallel \le {r}^{\prime}\}$

*with*${r}^{\prime}>r>0$.

*Suppose that*$A:{K}_{{r}^{\prime}}\to K$

*is a strict set contraction such that one of the following two conditions is satisfied*:

- (a)
$\parallel Ax\parallel \ge \parallel x\parallel $, $\mathrm{\forall}x\in K$, $\parallel x\parallel =r$; $\parallel Ax\parallel \le \parallel x\parallel $, $\mathrm{\forall}x\in K$, $\parallel x\parallel ={r}^{\prime}$.

- (b)
$\parallel Ax\parallel \le \parallel x\parallel $, $\mathrm{\forall}x\in K$, $\parallel x\parallel =r$; $\parallel Ax\parallel \ge \parallel x\parallel $, $\mathrm{\forall}x\in K$, $\parallel x\parallel ={r}^{\prime}$.

*Then* *A* *has a fixed point* $x\in {K}_{r,{r}^{\prime}}$.

## 2 Preliminaries

To establish the existence of positive solutions in ${C}^{3}(J,P)$ of (1), let us list the following assumptions:

(H_{0}) $f\in C(J\times P,P)$, and for any $l>0$, *f* is uniformly continuous on $J\times {P}_{l}$. Further suppose that $g\in L[0,1]$ is nonnegative, $\sigma ={\int}_{0}^{1}sg(s)\phantom{\rule{0.2em}{0ex}}ds$, and $\sigma \in [0,1)$, $\gamma =\frac{1+{\int}_{0}^{1}(1-s)g(s)\phantom{\rule{0.2em}{0ex}}ds}{1-\sigma}$, $h(s,x(s))={\phi}^{-1}({\int}_{0}^{s}f(t,x(t))\phantom{\rule{0.2em}{0ex}}dt)$, where ${P}_{l}=\{x\in P,\parallel x\parallel \le l\}$.

_{1}) There exists a nonnegative constant ${\eta}_{l}$ with $\gamma {\eta}_{l}<1$ such that

Evidently, $(C(J,E),{\parallel \cdot \parallel}_{c})$ is a Banach space, and the norm is defined as ${\parallel x\parallel}_{C}={max}_{t\in J}\parallel x(t)\parallel $.

In the following, we construct a cone $K=\{x\in Q:x(t)\ge \delta x(v),t\in {J}_{\delta},v\in [0,1]\}$, where $Q=\{x\in {C}^{3}(J,P):x(t)\ge \theta ,t\in J\}$, and let ${B}_{l}=\{x\in C(J,P):{\parallel x\parallel}_{c}\le l\}$, $l>0$. It is easy to see that *K* is a cone of ${C}^{3}(J,E)$ and ${K}_{r,{r}^{\prime}}=\{x\in K:r\le \parallel x\parallel \le {r}^{\prime}\}\subset K$, $K\subset Q$.

In our main results, we will make use of the following lemmas.

**Lemma 2.1**

*Assume that*(H

_{0})

*is satisfied*.

*Then*$x(t)$

*is a solution of problem*(1)

*if and only if*$x\in K$

*is a solution of the integral equation*

*Here*,

*we define an operator*

*A*

*by*

*That is*, *x* *is a fixed point of the operator* *A* *in* *K*.

**Lemma 2.2** *If condition* (H_{0}) *is satisfied*, *then the operator* *A* *defined by* (5) *is a continuous operator*.

*Proof* It can be verified easily by the definition of $(Ax)(t)$, we omit it here. □

**Lemma 2.3** *For* $t,s\in [0,1]$, $0\le G(t,s)\le \frac{1}{4}$.

*Proof* It is obvious that $G(t,s)\ge 0$ for any $s,t\in [0,1]$. When $0\le s\le t\le 0$, $G(t,s)=s(1-t)\le t(1-t)=-{(t-\frac{1}{2})}^{2}+\frac{1}{4}$, thus when $t=\frac{1}{2}$, ${max}_{0\le t,s\le 1}G(t,s)=\frac{1}{4}$, therefore $G(t,s)\le \frac{1}{4}$, which implies the proof is complete. □

**Lemma 2.4**

*Assume that*(H

_{0})

*holds*,

*choose*$\delta \in (0,\frac{1}{2})$

*and let*${J}_{\delta}=[\delta ,1-\delta ]$,

*then for all*$v,s\in [0,1]$,

*where* *γ* *is as defined in* (H_{0}).

*Proof* First, we prove that $G(t,s)\ge \delta G(v,s)$. Obviously, for $t\in {J}_{\delta}$, $v,s\in \{0,1\}$, $G(t,s)\ge \delta G(v,s)$ hold. And for $v,s\in (0,1)$, we have the following four cases.

To sum up, we get that $G(t,s)\ge \delta G(v,s)$.

So, we complete the proof. □

*Proof of Lemma 2.1*Necessity. First, we suppose that

*x*is a solution of equation (1). By taking the integral of (1) on $[0,t]$, we have

*t*, we have

which implies $x\in K$. To sum up, we know that *x* is a solution of the integral equation (4) in *K*.

*x*be as in (4). Taking the derivative of (4), it implies that

hold, which implies $x(t)$ is a solution of (1). The proof is complete. □

**Lemma 2.5** *Suppose that* (H_{0}) *and* (H_{1}) *hold*. *Then*, *for each* $l>0$, *A* *is a strict set contraction on* $Q\cap {B}_{l}$, *i*.*e*., *there exists a constant* $0\le {k}_{l}<1$ *such that* ${\alpha}_{C}(A(S))\le {k}_{l}{\alpha}_{C}(S)$ *for any* $S\subset Q\cap {B}_{l}$.

*Proof*By Lemmas 2.1 and 2.3, we know that $A:Q\to Q$ is continuous and bounded. Now, let

*S*be a bounded set in

*Q*. Then by (H

_{1}), we get that

where ${k}_{l}=\gamma {\eta}_{l}$, $0\le {k}_{l}<1$. The proof is complete. □

## 3 Main results

*f*which allow us to apply Lemma 1.3 to establish the existence of positive solutions of (1). At the beginning, we introduce the notation

**Theorem 3.1** *Suppose* (H_{0}) *and* (H_{1}) *hold and* *P* *is normal*. *If* $\gamma {h}^{0}<1<\mathrm{\Lambda}{(\psi h)}_{\mathrm{\infty}}$, *then problem* (1) *has at least one positive solution*.

*Proof* *A* is defined as (5). Considering $\gamma {h}^{0}<1$, there exists ${\overline{r}}_{1}$ such that $\parallel h(t,x)\parallel \le ({h}^{0}+{\epsilon}_{1})\parallel x\parallel $ for $t\in J$, $x\in K$, $\parallel x\parallel \le {\overline{r}}_{1}$, where ${\epsilon}_{1}>0$ satisfies $\gamma ({h}^{0}+{\epsilon}_{1})\le 1$.

*i.e.*, for $x\in K$, ${\parallel x\parallel}_{C}={r}_{1}$, ${\parallel Ax\parallel}_{C}\le {\parallel x\parallel}_{C}$ holds.

Next, turning to $1<\mathrm{\Lambda}{(\psi h)}_{\mathrm{\infty}}$, there exists ${\overline{r}}_{2}>0$ such that $\psi (h(t,x(t)))>({(\psi h)}_{\mathrm{\infty}}-{\epsilon}_{2})\parallel x\parallel $, $t\in J$, $x\in P$, $\parallel x\parallel \ge {\overline{r}}_{2}$, where ${\epsilon}_{2}>0$ satisfies $({(\psi h)}_{\mathrm{\infty}}-{\epsilon}_{2})\parallel x\parallel \mathrm{\Lambda}\ge 1$.

*i.e.*, for $x\in K$, ${\parallel x\parallel}_{C}={r}_{2}$, ${\parallel Ax\parallel}_{C}\ge {\parallel x\parallel}_{C}$ holds.

Lemma 1.3 yields that *A* has at least one fixed point ${x}^{\ast}\in {\overline{K}}_{{r}_{1},{r}_{2}}$, ${r}_{1}\le \parallel {x}^{\ast}\parallel \le {r}_{2}$ and ${x}^{\ast}(t)\ge \delta \parallel {x}^{\ast}\parallel >0$, $t\in {J}_{\delta}$. Thus, BVP (1) has at least one positive solution ${x}^{\ast}$. The proof is complete. □

**Remark 3.1**If $\phi =I$, where

*I*denotes a unit operator, then the differential equation can change into the general differential equation

This system has been studied in Ref. [1]. By Theorem 3.1, we can easily obtain the main result (Theorem 3.1 in Ref. [1]).

**Corollary 3.1** *Assume that* (H_{0}) *holds and* *P* *is normal*. *If* $\frac{1}{2}\gamma {f}^{0}<1<\mathrm{\Lambda}{(\psi f)}_{\mathrm{\infty}}$, *then problem* (14) *has at least one positive solution*.

**Remark 3.2**If $\phi (x)={\mathrm{\Phi}}_{p}(x)={|x|}^{p-2}x$ for some $p>1$, where ${\mathrm{\Phi}}_{P}^{-1}={\mathrm{\Phi}}_{q}$, then (1) can be written as a BVP with a

*p*-Laplace operator,

First, we reduce (H_{0}) as ${({\mathrm{H}}_{0})}^{\ast}$.

${({\mathrm{H}}_{0})}^{\ast}$ $f\in C(J\times P,P)$, and for any $l>0$, *f* is uniformly continuous on $J\times {P}_{l}$. Further suppose that $g\in L[0,1]$ is nonnegative, $\sigma ={\int}_{0}^{1}sg(s)\phantom{\rule{0.2em}{0ex}}ds$, and $\sigma \in [0,1)$, $\gamma =\frac{1+{\int}_{0}^{1}(1-s)g(s)\phantom{\rule{0.2em}{0ex}}ds}{1-\sigma}$, ${h}_{1}(s,x(s))={\mathrm{\Phi}}_{q}({\int}_{0}^{s}f(t,x(t))\phantom{\rule{0.2em}{0ex}}dt)$, where ${P}_{l}=\{x\in P,\parallel x\parallel \le l\}$.

Then we can get the following similar conclusion.

**Corollary 3.2** *Suppose* ${({\mathrm{H}}_{0})}^{\ast}$ *and* (H_{1}) *hold and* *P* *is normal*. *If* $\gamma {h}_{1}^{0}<1<\mathrm{\Lambda}{(\psi {h}_{1})}_{\mathrm{\infty}}$, *then problem* (15) *has at least one positive solution*.

## 4 Example

Next, we will give an example to illustrate our results.

**Example 4.1**

Let $E={R}^{n}=\{x=({x}_{1},{x}_{2},\dots ,{x}_{n}):{x}_{i}\in R,i=1,2,\dots ,n\}$ with the norm $\parallel x\parallel ={max}_{1\le i\le n}|{x}_{i}|$, and $P=\{x=({x}_{1},{x}_{2},\dots ,{x}_{n}):{x}_{i}\ge 0,i=1,2,\dots ,n\}$. Then we have the following result.

**Theorem 4.1** *For any* $t\in J$, *system* (16) *has at least one positive solution* $x(t)$.

*Proof* First, according to (16), (17), we have $g=1$ and ${f}_{i}(t,{x}_{i})=8t{x}_{i}^{3}$. Then, when ${x}_{i}={t}^{2}$, we have $f(t,x)=8{t}^{7}$. Moreover, we can get that ${P}^{\ast}=P$. Choose $\psi =(1,1,\dots ,1)$, then it is clear that (H_{0}) and (H_{1}) are satisfied.

Choose $\delta =\frac{1}{4}\in (0,\frac{1}{2})$, then $\mathrm{\Lambda}=\delta {\int}_{\delta}^{1-\delta}H(\frac{1}{2},s)\phantom{\rule{0.2em}{0ex}}ds=\frac{1}{4}{\int}_{\frac{1}{4}}^{\frac{3}{4}}H(\frac{1}{2},s)\phantom{\rule{0.2em}{0ex}}ds=\frac{1}{16}$.

Then $h(s,x(s))={\phi}^{-1}({\int}_{0}^{s}f(t,x(t))\phantom{\rule{0.2em}{0ex}}dt)=\sqrt{{\int}_{0}^{s}f(t,x(t))\phantom{\rule{0.2em}{0ex}}dt}={t}^{4}={x}^{2}$.

thus we have $\gamma {h}^{0}=0<1$.

which implies that $\mathrm{\Lambda}{(\psi h)}_{\mathrm{\infty}}\to \mathrm{\infty}>1$. So, all the conditions in Theorem 3.1 are satisfied, then the conclusion follows, and the proof is complete. □

## Declarations

### Acknowledgements

This work was supported by the NNSF of China under Grant (No. 11271261), Natural Science Foundation of Shanghai (No. 12ZR1421600), Shanghai municipal education commission (No. 10YZ74), and Shanghai Normal University Leading Academic Discipline project (No. DZW912).

## Authors’ Affiliations

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