#### Abstract

This paper investigates the existence and nonexistence of positive solutions for a class of fourth-order nonlinear differential equation with integral boundary conditions. The associated Green's function for the fourth-order boundary value problems is first given, and the arguments are based on Krasnoselskii's fixed point theorem for operators on a cone.

#### 1. Introduction

In this paper, we consider the following fourth-order boundary value problems (BVPs) with integral boundary conditions where may be singular at and(or) , is continuous, and are nonnegative.

The existence of solutions for nonlinear higher-order nonlocal BVPs has been studied by several authors, for example, see [1–9] and the references therein. However, there are a few papers dealing with the existence of positive solutions for the fourth-order boundary value problems with integral boundary conditions, see [7–15]. For the case of , BVP (1) reduces to the following two-point BVP A great deal of research has been devoted to the existence of solutions for problem (2) by using the Leray-Schauder continuation method, topological degree, and the method of lower and upper solutions.

For the case of or , and , BVP (1) reduces to the following BVP with integral boundary conditions or By using the fixed point theorem of cone expansion and compression of norm type, Zhang and Ge [11] have obtained the existence and multiplicity of positive solutions for BVP (3) and (4).

The following assumptions will stand throughout this paper: is nonnegative, and may have singularities at and(or) ; ; are nonnegative and

The paper is organized as follows. In Section 2, we provide some necessary background material such as the Krasnoselskii’s fixed point theorem in cones. The associated Green’s function for the fourth-order boundary value problem is first given, and we also look at some properties of the Green’s function associated with the problem (1). In Section 3, the main results of problem (1) will be stated and proved. In Section 4, we give an example to illustrate how the main results can be used in practice.

#### 2. Preliminary Results

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

Lemma 1 (in [16]). *Let and be two bounded open sets in a real Banach space , such that and . Let operator be continuous, where denotes the zero element of and is a cone in . Suppose that one of the two conditions * *, and , or * *, and , is satisfied. Then, has at least one fixed point in . *

Lemma 2. *If are nonnegative, and , then for any , the BVP
**
has a unique solution which is given by
**
where
*

*Proof. *The general solution of can be written as
Now, we solve for by and , it follows that
that is,
Solving the above equations, we get
Therefore, (6) has a unique solution
The unique solution of (6) be expressed as
Therefore, the unique solution of (6) is .

Lemma 3 (in [11]). *If are nonnegative, then,**,
**, for all , for all ,** Letting , for all , one has ,* * Letting , we have , where is defined by (9). *

Lemma 4. *If are nonnegative, and , for function , we have the following properties:**,**,**,**where , .*

*Proof. * From (8) and Lemma 3, we get (1). For all , by (8) and Lemma 3, we have
For all , by (8) and Lemma 3, we have

Lemma 5. *Assume that (A1)–(A3) holds. If is a solution of the following integral equation
**
then is a solution of BVP (1), where
*

*Proof. *By using Lemma 2, the conclusion is obvious.

Lemma 6. *If (A3) holds, then one has the following three properties:**, for all ,**,**, for all ,**where
*

*Proof. *By using Lemma 3, the conclusions for (1) and the first part of (2) are obvious.

By using by (21) and Lemma 3, we have
Similarly, by (22), we get
So, by (20), (25), and (26), we concluded
By using Lemmas 3 and 4, we have

Let . Then is a Banach space with the norm , where .

For a fixed , define the cone by It's easy to prove that is a closed convex cone, and

Lemma 7. * If (A1)–(A3) hold, then , and is completely continuous. *

*Proof. *By the properties of Green's function, if then and
For all , we have by (19) and (31)
On the other hand, by (19), (33), and Lemma 3, we obtain
Similarly, by (31), (34), and Lemma 3, we obtain
So , and then .

Clearly, the operator is continuous; thus by the Arzela-Ascoli theorem, it follows that is completely continuous.

#### 3. Main Results

For convenience, we introduce the following notations:

Theorem 8. * Assume that (A1)–(A3) hold. In addition, one supposes that one of the following conditions is satisfied **, for all ,**, for all .**Then, the BVP (1) has no positive solution. *

*Proof. **Case *1. The condition () holds. Assume towards a contraction that is a positive solution of BVP (1), from the proof of Lemma 7 and (29), then for . By Lemma 6 and (29), we have
So, , which is a contradiction.*Case *2. The condition () holds. Assume towards a contraction that is a positive solution of BVP (1), from the proof of Lemma 7 and (29), then , , for . By Lemmas 4 and 6 and (30), we have
So, , which is a contradiction, and this completes the proof.

Theorem 9. *Assume that (A1)–(A3) hold. In addition, one supposes that one of the following conditions is satisfied*()*,*()*.**Then, the BVP (1) has at least one positive solution. *

* Proof. **Case *1. The condition () holds.

Considering , then there exists , such that , for , where , satisfies .

Then, for , by (19), (26), and Lemma 6, we have
which means that

Next, turning to , then there exists , such that , for , where , satisfies .

Let , for all . By (30) and Lemma 6, we have
which means that
*Case *2. The condition () holds.

Considering , then there exists , such that , for , where , satisfies .

Then, for , by (30) and Lemma 6, we have
which means that

Next, turning to , then there exists , such that , for , where , satisfies .

Let
for .

Let , for . By (26) and Lemma 6, we have
which means that

Applying Lemma 1 to (41) and (43), or (45), and (48) yields that has a fixed point with , . Thus, it follows that BVP (1) has at least one positive solution, and the theorem is proved.

Theorem 10. * Assume that (A1)–(A3) hold, as do the following two conditions:** and ,** there exists , such that ** Then, the BVP (1) has at least two positive solutions , which satisfy
*

*Proof. *We choose with .

If , then by the proof of (45), we have

If , then by the proof of (43), we have
On the other hand, by , for , we have
where

By (52), we have

Applying Lemma 1 to (50), (51), and (53) yields that has a fixed point , and a fixed point . Thus it follows that BVP (1) has at least two positive solutions and . Noticing (53), we have and . Therefore, (49) holds, and the proof is complete.

Theorem 11. *Assume that (A1)–(A3) hold, as do the following two conditions:** and ,** there exist and such that for all and . ** Then, the BVP (1) has at least two positive solutions , which satisfy
*

*Proof. *We choose with .

If , then by the proof of (41), we have

If , then by the proof of (48), we have
On the other hand, by , for , we have

Applying Lemma 1 to (55), (56), and (57) yields that has a fixed point , and a fixed point . Thus it follows that BVP (1) has at least two positive solutions and . Noticing (57), we have and . Therefore, (54) holds, and the proof is complete.

#### 4. Example

*Example 12. *Consider the following fourth-order BVP
By calculation, we obtain
About the nonexistence of positive solution, we consider BVP (58) with
where and are two positive real numbers, then
If or , by Theorem 8, BVP (58) has no positive solution. About the existence of positive solution, we consider BVP (58) with
where and are two positive real numbers, then
If and , then we have