- Research Article
- Open Access

# Triple Positive Solutions for a Type of Second-Order Singular Boundary Problems

- Jiemei Li
^{1, 2}Email author and - Jinxiang Wang
^{1}

**2010**:376471

https://doi.org/10.1155/2010/376471

© Jiemei Li and Jinxiang Wang. 2010

**Received:**7 April 2010**Accepted:**26 August 2010**Published:**1 September 2010

## Abstract

This paper deals with the existence of triple positive solutions for a type of second-order singular boundary problems with general differential operators. By using the Leggett-Williams fixed point theorem, we establish an existence criterion for at least three positive solutions with suitable growth conditions imposed on the nonlinear term.

## Keywords

- Banach Space
- Integral Equation
- Integral Operator
- Fixed Point Theorem
- Continuous Operator

## 1. Introduction

It is easy to see that and may be singular at and/or

When or and , the two kinds of singular boundary value problems have been discussed extensively in the literature; see [1–10] and the references therein. Hence, the problem that we consider is more general and is different from those in previous work.

Furthermore, we will see in the later that the presence of brings us three main difficulties:

(1) the Green's function cannot be explicitly expressed;

(2) the equivalence between BVP (1.1) and its associated integral equation has to be proved;

(3) the compactness of associated integral operator has to be verified.

We will overcome the above mentioned difficulties in Section 2. Also, although the Leggett-William fixed point theorem is used extensively in the study of triple positive differential equations, the method has not been used to study this type of second-order singular boundary value problem with general differential operators. We are concerned with solving these problems in this paper.

To state our main tool used in this paper, we give some definitions and notations.

Theorem 1.1 (Leggett-Williams fixed point theorem).

Let be completely continuous, and let be a nonnegative continuous concave functional on such that for all . Suppose that there exist such that

(i) and , for ;

(ii) , for ;

(iii) , for with .

Then has at least three fixed points in satisfying and .

Remark 1.2.

We note the existence of triple positive solutions of other kind of boundary value problems; see He and Ge [11], Zhao et al. [12], Zhang and Liu [13], Graef et al. [14], and the references therein.

The rest of the paper is organized as follows. In Section 2, we overcome the above-mentioned difficulties in this work. The main results are formulated and proved in Section 3. Finally, an example is presented to demonstrate the application of the main theorems in Section 4.

## 2. Preliminaries and Lemmas

Throughout this paper, we assume the following:

(H1) ;

(H2) , and ;

(H3) is continuous and does not vanish identically on any subinterval of , and ;

(H4) is continuous.

Lemma 2.1.

Suppose that (H1) and (H2) hold. Then

has a unique solution and ;

has a unique solution and .

Proof.

We only prove (i). (ii) can be treated in the same way.

which implies that is a solution of integral equation (2.11).

Conversely, if is a solution of (2.11) with , by reversing the above argument we could deduce that the function is a solution of (2.1) and satisfy and . Therefore, to prove that (2.1) has a unique solution, , and is equivalent to prove that (2.11) has a unique solution .

Since , has a unique fixed point by Banach contraction principle. That is, (2.11) has a unique solution .

Remark 2.2.

Lemma 2.1 generalizes Theorem of [1], where .

Lemma 2.3.

Suppose that (H1) and (H2) hold. Then

(i) is nondecreasing in ;

(ii) is nonincreasing in .

Proof.

We only prove (i). (ii) can be treated in the same way.

which is a contradiction!

Remark 2.4.

then, , which is a contradiction!

The other inequality can be treated in the same manner.

Lemma 2.5.

Proof.

Lemma 2.6.

Moreover, on .

Proof.

This together with Remark 2.4 implies that the right side of (2.31) is well defined.

Lemma 2.7.

Proof.

Then Lemma 2.3 guarantees that , and Lemma 2.7 guarantees that (2.43) holds.

Remark 2.8.

Lemma 2.9.

Let (H1)–(H4) hold. Then is a completely continuous operator.

Proof.

From (H3) and (H4) and Lemma 2.6, it is easy to see that , and is continuous by the Lebesgue s dominated convergence theorem.

Let be any bounded set. Then (H3) and (H4) imply that is a bounded set in .

From this fact, it is easy to verify that is equicontinuous. Therefore, by the Arzela-Ascoli theorem, is a completely continuous operator.

## 3. Main Result

We notice that, for each , , and also that by Lemma 2.6, is a solution of (1.1) if and only if is a fixed point of the operator .

Theorem 3.1.

Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:

(S1) , for ;

(S2) , for ;

(S3) , for .

Then the boundary value problem (1.1) has at least three positive solutions in satisfying , and .

Proof.

Hence, . In the same way, if , then . Therefore, condition (ii) of Leggett-williams fixed-point theorem holds.

To sum up, all the conditions of Leggett-williams fixed-point theorem are satisfied. Therefore, has at least three fixed points, that is, problem (1.1) has at least three positive solutions in satisfying and .

Theorem 3.2.

Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:

(A1) , for ;

(A2) , for .

Then the boundary value problem (1.1) has at least positive solutions.

Proof.

When , it follows from condition (A1) that , which means that has at least one fixed point by the Schauder fixed-point Theorem. When , it is clear that Theorem 3.1 holds (with ). Then we can obtain at least three positive solutions , and satisfying and with . Following this way, we finish the proof by the induction method.

## 4. Example

The other inequalities in (4.4) can be proved by the same method.

## Declarations

### Acknowledgment

The first author was partially supported by NNSF of China (10901075).

## Authors’ Affiliations

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