Journal of Function Spaces
Volume 2015 (2015), Article ID 928105, 8 pages
http://dx.doi.org/10.1155/2015/928105

Research Article

Existence of Positive Solutions for a Fourth-Order -Point Boundary Value Problem

Liu Yang and Chunfang Shen

Department of Mathematics, Hefei Normal University, Hefei, Anhui 230061, China

Received 3 August 2014; Accepted 5 February 2015

Academic Editor: Jesús G. Falset

Copyright © 2015 Liu Yang and Chunfang Shen. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract.

By using Krasnosel’skii’s fixed point theorem and the fixed point index theorem in the special function space, we obtain some sufficient conditions for the existence of positive solutions of fourth-order boundary value problem with multipoint boundary conditions. Applications of our results to some special problems are also discussed.

1. Introduction

In this paper, we study the existence of positive solutions of the boundary value problem consisting of the nonlinear fourth-order differential equation

and the multipoint boundary condition

or

where , , , and , , and there exist such that .

It is well known that boundary value problems of fourth-order ordinary differential equations are used to describe a large number of physical, biological, and chemical phenomena. For an example, (1) is often used to describe the deformation of an elastic beam under a variety of boundary conditions [1–3]. For this reason, there is a wide literature that deals with the existence and the multiplicity of solutions for fourth-order boundary value problems; see [4–19].

Krasnoselskiis theorem in a cone and fixed point index theorem have often been used to study the existence and multiplicity of positive solutions of nonlocal boundary value problems over the last several years. As to fourth-order Bvps, by using Krasnoselskiis fixed point theorem, Graef et al. [11] established the existence results of positive solutions for the nonlinear fourth-order ordinary differential equation

with the boundary conditions

However, the aforementioned papers mainly considered the two or three point boundary conditions. There are very few works on the multipoint boundary value problem for fourth-order ordinary differential equations. As to -point boundary value problem ((1),  (2)) or ((1),  (3)), there is no existence results of solution or positive solution. The goal of this paper is to fill this gap. In this paper, by constructing an available integral operator and combining fixed point theorem and fixed point index theorem, we establish some sufficient conditions of the existence of positive solutions for problems ((1),  (2)) and ((1),  (3)).

The rest of this paper is organized as follows. In Section 2, we given some preliminaries and lemmas for use later. Section 3 is devoted to the existence and multiplicity of positive solutions of problem ((1),  (2)). In Section 4, we establish the existence results of positive solutions for problem ((1),  (3)). In Section 5, we give some examples to demonstrate the main results of this paper.

2. Preliminaries

Definition 1. Let be a real Banach space over the real numbers. A nonempty convex closed set is said to be a cone provided that

(i), for all , ,(ii) implies .

Definition 2. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
Let be the Banach space endowed with the norm . We denote

Lemma 3 (see [20]). Let be a Banach space and be a cone. Assume and are open bounded subsets of with , and let

be a completely continuous operator such that

or

then, has a fixed point in .

Lemma 4 (see [20]). Let be a Banach space and let be a cone. For , define and assume that is a completely continuous operator such that for . Then,

3. Existence Results of Problems (1) and (2)

We begin with the fourth-order -point boundary value problem

Lemma 5. Denoting , , , and , problem (11),  (12) has the unique solution

where

for , .

Proof. Let be Green’s function of problem with boundary condition (12). We can suppose

where , , are unknown coefficients. Considering the properties of Green’s function and boundary condition (12), we have

A straightforward calculation shows that

These give the explicit expression of Green’s function. Then, we have

Lemma 6. One can see that , .

Proof. For , ,

Then, , , , which induces that is decreasing on . By a simple computation, we see

This ensures that , .

Lemma 7. If , and is the solution of problem (11),  (12), then

where .

Proof. Since , , then is decreasing on . Considering , we have , . Thus, is decreasing on . Considering this together with the boundary condition , we conclude that . Then, is concave on . Taking into account that , we get that

From the concavity of , we have

Multiplying both sides with and considering the boundary condition, we have

Problem (1),  (2) has a solution if and only if solves the operator equation

Denote the cone

Since is a cone, by Lemma 7 we have that . For the convenience, we denote

Theorem 8. Problem (1),  (2) has at least one positive solution if

(1), or(2), .

Proof. Firstly, we consider case . By , we choose satisfying . Then, there exists such that

Define ; then, for , we have

On the other hand, by , choose satisfying . Then, there exists such that

Let and let . Then, for , we see . Then,

Then, by Lemma 3, has a fixed point , which implies problem (1),  (2) has at least one positive solution .
Next, the case , is considered. For , there exists such that

where satisfies . Denote ; then, for , we have

Considering , there exists such that

where satisfies . We distinguish two cases to discuss.
Case 1. Suppose that is bounded; then, there exists satisfying . Taking and , then for , we have

Case 2. Suppose that is unbounded. Since , then there certainly exists such that

For , where , we have

So, in either case, we can choose such that , for . Then, Lemma 3 implies that has a fixed point. Consequently, problem (1),  (2) has at least one positive solution.

Theorem 9. Problem (1),  (2) has at least one positive solution in one of the following cases:

(3), , and , ;(4), , and , ;(5), , and ;(6), , and ;(7), , and ;(8), , and .

Proof. We consider the case , firstly. Since , for satisfying , there exists such that

Define ; then, for , we see

Considering , there exists and satisfies such that

Let and let ; then, for , we get ; then,

Then, by Lemma 3, problem (1),  (2) has at least one positive solution. As for other cases , the proof is considerably analogous with the case and is omitted here.

Theorem 10. Assume following conditions are satisfied:

;there exist such that for ;

then, problem (1),  (2) has at least two positive solutions satisfying

Proof. Considering , similarly with Theorem 8, we claim there exists such that

Since , similarly with above, there exists such that

For , ,

Hence, .
Therefore,

Thus, there exists at least two positive solutions such that

Theorem 11. Assume the following conditions are satisfied:

;there exists such that for ;

then, problem (1),  (2) has at least two positive solutions satisfying

Proof. Considering , , we claim there exists , such that

For , ,

Hence, . Therefore,

Thus, there exists at least two positive solutions such that

4. Existence Results of Problem (1),  (3)

Lemma 12. Denoting , , and , Green’s function of problem

is

for .

Proof. For , , we suppose

Considering the properties of Green’s function together with the boundary condition (54), we have

Hence,

The proof of Lemma 12 is completed.

Lemma 13. One can see that , .

Proof. For , ,

Then, , , , which implies that is increasing on . The fact that

ensures that , .

Lemma 14. If , , and is the solution of problem (53),  (54), then

where .

Proof. It follows from the same methods as Lemma 7 that is concave on . Taking into account that , one see that is increasing on and

From the concavity of , we have

Multiplying both sides with and considering the boundary condition, we have

Thus,

Problem (1),  (3) has a solution if and only if solves the operator equation

Denote the cone

It is obvious that . We denote

Theorem 15. Problem (1),  (3) has at least one positive solution if

(1), ,(2), .

Theorem 16. Problem (1),  (3) has at least one positive solution in one of the following cases:

(3), , and , ;(4), , and , ;(5), , and ;(6), , and ;(7), , and ;(8), , and .

Theorem 17. Assume following conditions are satisfied:

;there exist such that for ;

then, problem (1),  (3) has at least two positive solutions , satisfying

Theorem 18. Assume following conditions are satisfied:

;there exists such that for ;

then, problem (1),  (3) has at least two positive solutions satisfying

5. Examples

In this section, we given some examples to illustrate the main results established in this paper.

Example 1. Consider the fourth-order boundary value problem

where . Then, problems (71) have at least one positive solution. In fact, when , and ; when , and . From Theorem 8, there exists at least one positive solution of problem (71).

Example 2. Consider the fourth-order boundary value problem

We see that

It is easy to check that . Taking , we see

From Theorem 18, there exist at least two positive solutions of problem (72) such that

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work is sponsored by the NSFC (11201109), Anhui Provincial Natural Science Foundation (1408085QA07), and the Higher School Natural Science Project of Anhui Province (KJ2014A200).

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