Table of Contents Author Guidelines Submit a Manuscript
Journal of Applied Mathematics

Volume 2014, Article ID 603203, 8 pages

http://dx.doi.org/10.1155/2014/603203
Research Article

Multiplicity of Positive Solutions for a Singular Second-Order Three-Point Boundary Value Problem with a Parameter

Department of Mathematics, Shijiazhuang Mechanical Engineering College, Shijiazhuang, Hebei 050003, China

Received 10 June 2014; Accepted 30 July 2014; Published 12 August 2014

Academic Editor: Alberto Cabada

Copyright © 2014 Jian Liu et al. 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

This paper is concerned with the following second-order three-point boundary value problem , , , , where , , , and is a positive parameter. First, Green’s function for the associated linear boundary value problem is constructed, and then some useful properties of Green’s function are obtained. Finally, existence, multiplicity, and nonexistence results for positive solutions are derived in terms of different values of by means of the fixed point index theory.

1. Introduction

For given positive numbers and , the existence, multiplicity, and nonexistence of positive solutions for the following boundary value problem (BVP for short) are considered, where is a positive parameter, , and may be singular at and 1.

A function is said to be a solution of BVP (1) if satisfies BVP (1). Moreover, if for any , then is said to be a positive solution of BVP (1).

Due to a wide range of applications in physics and engineering, second-order boundary value problems have been extensively investigated by numerous researchers in recent years. The study of multipoint boundary value problems was initiated by Il’in and Moiseev [1]. Gupta studied three-point boundary value problems for nonlinear ordinary differential equations in [2]. Since then, nonlinear three-point boundary value problems have been studied by many authors using the fixed point index theorem, Leray-Schauder continuation theorem, nonlinear alternative of Leray-Schauder, coincidence degree theory, and fixed point theorem in cones. For details, the readers are referred to [37] and the references therein.

In [8], positive solutions for the following three-point boundary value problem at resonance were studied. Han’s approach is to rewrite the original BVP as an equivalent one so that the Krasnosel’skii-Guo fixed point theorem can be applied and then the existence and multiplicity of positive solutions are investigated.

Then in [9], Han considered the following three-point boundary value problem: under some conditions concerning the first eigenvalue of the relevant linear operator, where is a constant and is allowed to be singular at and . The existence of positive solutions is studied by means of fixed point index theory.

Motivated by the above work, here we study the second-order three-point BVP (1). Under certain suitable conditions, the results of existence, multiplicity, and nonexistence of positive solutions for BVP (1) were established via the fixed point index theory.

We make the following assumptions:, , and ; and , where ; for and ; is nondecreasing in and for any .

The main results of the present paper are summarized as follows.

Theorem 1. Let be fulfilled and suppose that Then, there exists such that BVP (1) has at least two positive solutions for , at least one positive solution for , and no positive solution for .

The remainder of this paper is arranged as follows. Green’s function of BVP (1) and its properties are given in Section 2, and some preliminaries are also presented. The proof of Theorem 1 is given in Section 3.

2. Preliminaries

In this section we collect some preliminary results that will be used in subsequent sections.

Consider the linear boundary value problem

Lemma 2. Assume that holds. Then, for each , BVP (5) has a unique solution where

Proof. Suppose that

According to the definition and properties of Green’s function, for any , we have and thus Then by using the boundary conditions, we have Therefore For any , we have and hence By using the boundary conditions, we have Then Consequently, we can get Green’s function , and the lemma is proved.

Lemma 3. There exist a continuous function and a constant such that(i)if holds, , ;(ii)if and hold, , .

Proof. Firstly, it is obvious that for any .

Next, we will give the continuous function and the constant .

Let In the first step, we try finding the upper bounds.

We only need to show that there exists such that Case 1. .If , then and ; the conclusion is true.If , then so, for we have . Consider so, for we have .

Case 2. .If , then and ; the conclusion is true.If , then so, for we have . Consider so, for we have .

Thus, we take and then for , and accordingly .

Set and then , .

In the next step, we try finding the lower bounds.

We only need to show that there exists such that Case 1. .If , then and ; the conclusion is true.If , then so, for we have . Consider so, for we have .

Case 2. .

If , then and ; the conclusion is true.

If , then so, for we have . Consider so, for we have .

Let , where  ,   we have   for  . Thus, , Namely where .

This completes the proof of the lemma.

Let be equipped with norm ; then is a real Banach space.

Define the cone by then is a nonempty closed subset of .

For , we write if for any . For any , let and .

Define the operator by

Lemma 4. Assume that hold; then the operator is completely continuous.

Proof. For for all , it follows from the definition of and Lemma 3 that So,

In view of Lemma 3 and (40), we have And so which shows that . By the Ascoli-Arzela theorem, it is easy to show that is completely continuous.

In view of Lemmas 2 and 3, it is easy to see that is a solution of BVP (5) if and only if is a fixed point of the operator .

The proofs of our main results are based on the fixed point index theory. The following three well-known lemmas in [10, 11] are needed in our argument.

Lemma 5. Let be a Banach space and a cone in . Assume that is a bounded open subset of . Suppose that is a completely continuous operator. If there exists such that , for all and , then the fixed point index .

Lemma 6. Let be a Banach space and a cone in . Assume that is a bounded open subset of . Suppose that is a completely continuous operator. If and , for and , then the fixed point index .

Lemma 7. Let be a Banach space and a cone in . Assume that is a bounded open subset of with . Suppose that is a completely continuous operator. If for all and , then the fixed point index .

3. Proofs of the Main Results

For convenience, we firstly introduce the following notations. and (1);; .

Lemma 8. Suppose that hold and . Then .

Proof. Let be fixed; then we can choose small enough such that . It is easy to see that By Lemma 7, it follows that From , it follows that there exists such that We may suppose that has no fixed point on . Otherwise, the proof is finished. Let for . Then . We claim that In fact, if not, there exist and such that ; then . For and , by Lemma 3 and (45), for , we have we get , which is a contradiction. Thus, (46) holds. It follows from Lemma 5 that By virtue of the additivity of the fixed point index, from (44) and (48), we have which implies that the nonlinear operator has one fixed point . Therefore, . The proof is complete.

Lemma 9. Suppose hold and . Then .

Proof. By Lemma 8, it is easy to see that . It follows from and that there exists such that for all and . Let ; by the definition of cone and Lemma 2, for , we obtain that So, . We get . This completes the proof of lemma.

Lemma 10. Suppose hold and . Then . Moreover, for any , BVP (1) has at least two positive solutions.

Proof. For any fixed , we prove that . By the definition of , there exists , such that and . Let be fixed. From the proof of Lemma 8, we see that there exist , , and such that . It is easy to see that for all . Then, by , we have

Consider now the modified BVP: where Clearly, the function is bounded for and and is continuous in . Define the operator by Then is completely continuous and all the fixed points of operator are the solutions for BVP (52). It is easy to see that there exists such that for any . From Lemma 7, we have Let We claim that if is a fixed point of operator , then . In fact, if , then From the excision property of the fixed point index and (55), we obtain that From the definition of , we know that on . Then, Hence, the nonlinear operator has at least fixed point . Then is one positive solution of BVP (1). This gives and .

We now find the second positive solution of BVP (1). By and the continuity of with respect to , there exists such that For , let We claim that is bounded in . In fact, for any , it follows from Lemma 3 and (60) that This implies . Thus is bounded in . Therefore there exists such that By Lemma 5, we get that Using a similar argument as in deriving (48), we have that where . According to the additivity of the fixed point index and by (59), (64), and (65), we have which implies that the nonlinear operator has at least one fixed point . Thus, BVP (1) has another positive solution. The proof is complete.

Lemma 11. Suppose hold and . Then .

Proof. In view of Lemma 10, it suffices to prove that . By the definition of , we can choose with such that as . By the definition of , there exists such that . We now show that is bounded. Supposing the contrary, then there exists a subsequence of (still denoted by ) such that as . It follows from that for all . Choose sufficiently large such that By , there exists such that for all and . Since as , there exists sufficiently large such that . Thus, for , we have This gives

which contradicts the choice of . Hence, is bounded. It follows from the completely continuity of that is equicontinuous; that is, for each , there is a such that where and . Then is equicontinuous. According to the Ascoli-Arzela theorem, is relatively compact. Hence, there exists a subsequence of (still denoted by ) and such that as . By , letting , we obtain that . If , using a similar argument as in deriving (69) and by , we also get a contradiction. Then , and so . This completes the proof.

Proof of Theorem 1. Theorem 1 readily follows from Lemmas 8, 9, 10, and 11.

Conflict of Interests

The authors de