- Changlong Yu
^{1}Email author, - Yanping Guo
^{1}and - Yude Ji
^{1}

**2009**:609143

**DOI: **10.1155/2009/609143

© Changlong Yu et al. 2009

**Received: **5 April 2009

**Accepted: **6 July 2009

**Published: **17 August 2009

## Abstract

## 1. Introduction

Multipoint boundary value problems (BVPs) for second-order differential equations in a finite interval have been studied extensively and many results for the existence of solutions, positive solutions, multiple solutions are obtained by use of the Leray-Schauder continuation theorem, Guo-Krasnosel'skii fixed point theorem, and so on; for details see [1–4] and the references therein.

where have the same signal, and are given. We first present the Green function for second-order multipoint BVPs on the half-line and then give the existence results for (1.2) using the properties of this Green function and the Leray-Schauder continuation theorem.

We use the space exists, exists with the norm , where is supremum norm on the half-line, and is absolutely integrable on with the norm .

and we suppose are the same signal in this paper and we always assume

## 2. Preliminary Results

In this section, we present some definitions and lemmas, which will be needed in the proof of the main results.

Definition 2.1 (see [15]).

Lemma 2.2.

Proof.

Therefore, the unique solution of (2.1) is which completes the proof.

*Remark of Lemma 2.2 .* Obviously
satisfies the properties of a Green function, so we call
the Green function of the corresponding homogeneous multipoint BVP of (2.1) on the half-line.

Lemma 2.3.

Proof.

Therefore, we get the result.

Lemma 2.4.

Lemma 2.5.

Proof.

Let are positive, and note , then for every , we have so that is, Because is continuous on the interval , there exists satisfying , where .

Theorem 2.6 (see [5]).

## 3. Main Results

The main result of this paper is following.

Theorem 3.1.

Lemma 3.2.

Let be an S-Carathéodory function. Then, for each is completely continuous in .

Proof.

We claim that is completely continuous in , that is, for each , is continuous in and maps a bounded subset of into a relatively compact set.

Combining (3.9) (3.13), we can see that is continuous. Let be a bounded subset; it is easy to prove that is uniformly bounded. In the same way, we can prove (3.5),(3.6), and (3.12), we can also show that is equicontinuous and equiconvergent. Thus, by Theorem 2.6, is completely continuous. The proof is completed.

Proof of Theorem 3.1.

According to Lemma 2.5, we know that for any , there exists satisfying . Hence, there are three cases as follow.

Set , which is independent of .

Set , which is independent of and is what we need.

Set and which is we need. So (1.2) has at least one solution.

## Declarations

### Acknowledgment

The Natural Science Foundation of Hebei Province (A2009000664) and the Foundation of Hebei University of Science and Technology (XL200759) are acknowledged.

## Authors’ Affiliations

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