- Research Article
- Open Access

# Periodic Boundary Value Problems for Second-Order Functional Differential Equations

- Xiangling Fu
^{1}Email author and - Weibing Wang
^{1}

**2010**:598405

https://doi.org/10.1155/2010/598405

© X. Fu and W. Wang. 2010

**Received:**11 June 2009**Accepted:**28 January 2010**Published:**8 March 2010

## Abstract

This note considers a periodic boundary value problem for a second-order functional differential equation. We extend the concept of lower and upper solutions and obtain the existence of extreme solutions by using upper and lower solution method.

## Keywords

- Boundary Condition
- Differential Equation
- Banach Space
- Integral Equation
- Unique Solution

## 1. Introduction

Upper and lower solution method plays an important role in studying boundary value problems for nonlinear differential equations; see [1] and the references therein. Recently, many authors are devoted to extend its applications to boundary value problems of functional differential equations [2–5]. Suppose is one upper solution or lower solution of periodic boundary value problems for second-order differential equation; the condition is required. A neutral problem is that whether we can define upper and lower solution without assuming . The aim of the present paper is to discuss the following second order functional differential equation

where

In this paper, we extended the concept of lower and upper solutions for (1.1). By using the method of upper and lower solutions and monotone iterative technique, we obtained the existence of extreme solutions for the boundary value problem (1.1).

Through this paper, we assume that and

Definition 1.1.

and

Remark 1.2.

## 2. Comparison Results

We now present the main results of this section.

Lemma 2.1.

then for all .

Proof.

Suppose, to the contrary, that for some . We consider the following two cases.

Case 1.

which contradicts .

Case 2.

There exist such that and . Hence, two cases are possible.

Subcase 1.

that implies This is a contradiction.

Subcase 2.

If , then If or , then . So .

Let Then

When , same as Subcase 1, we obtain that

when , integrating the inequality (2.8) from to , we obtain

that implies This is a contradiction. The proof is complete.

Lemma 2.2.

Then for all .

Proof.

Hence by Lemma 2.1, for all , which implies that for . This completes the proof.

Lemma 2.3.

Then for all .

The proof of Lemma 2.3 is similar to that of Lemma 2.2, here we omit it.

Lemma 2.4.

Then for all .

Proof.

Suppose that for . It is easy to see that and for . From and for , we obtain that . Therefore, . It follows that , a contradiction.

Suppose that there exist such that and . Let be such that . From the first inequality of (2.16), we have

Hence, a contradiction. The proof is complete.

## 3. Linear Problem

In this section, we consider the boundary value problem

where .

Theorem 3.1.

Assume that there exist which are lower and upper solutions of (3.1) and on . Then there exists one unique solution to problem (3.1) and on .

Proof.

By Lemma 2.1, we have that for , that is, on . Similarly, one can obtain that on . Hence .

Next, we prove that if is a solution of (3.1), then .

Let . If , then on . So we have

By Lemma 2.1, we have that on .

If , then . Thus

It is easy to see that By Lemma 2.2, we have that on . Analogously, on .

If , then . Thus

It is easy to see that By Lemma 2.2, we have that on . Analogously, on .

Finally, we show that (3.1) has a solution by several steps.

Step 1.

where and are defined in (3.1). For any , we show that (3.6) has a unique solution .

It is easy to check that (3.6) is equivalent to the integral equation

Noting that , condition (1.2) implies that is a contraction mapping. There exists unique such that . Thus (3.6) has a unique solution Moreover, .

Step 2.

Step 3.

We show that there exists one such that , where is the unique solution of the problem (3.6).

Put

Using (3.18) and (3.19), one easily obtains that for any .

Put then Using Lemma 2.3, we easily obtain that on . Hence

there exists one such that that is, . Obviously, is a unique solution of the problem (3.1). This completes the proof.

## 4. Main Result

Theorem 4.1.

Let the following conditions hold.

The functions , are lower and upper solutions of (1.1), respectively, and on .

The constants in definition of upper and lower solutions satisfy

for .

Then, there exist monotone sequences with such that uniformly on , and are the minimal and the maximal solutions of (1.1), respectively, such that

on , where is any solution of (1.1) such that on .

Proof.

Theorem 3.1 implies that the problem (4.3) has a unique solution . We define an operator by , then is an operator from to .

We shall show that

(a)

(b) is nondecreasing in .

From and , we have that (a) holds. To prove (b), we show that if .

Let and then by , we have

and By Lemma 2.1, , which implies .

Define the sequence with such that for From (a) and (b), we have

on . Therefore, there exist such that , uniformly on . Clearly, are solutions of (1.1).

Finally, we prove that if is one solution of (1.1), then on . Since and , by property (b), we obtain that for . Using property (b) repeatedly, we have

for all . Passing to the limit as , we obtain This completes the proof.

## Declarations

### Acknowledgment

This work is supported by Scientific Research Fund of Hunan Provincial Education Department (09B033) and the NNSF of China (10871062).

## Authors’ Affiliations

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## Copyright

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