• Research Article
• Open Access

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

Journal of Inequalities and Applications20102010:598405

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

• Accepted: 28 January 2010
• Published:

## 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 [25]. 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

(1.1)

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

(1.2)
(1.3)

Definition 1.1.

Functions and are called lower solution and upper solution of the boundary value problem (1.1), respectively if
(1.4)
(1.5)
where
(1.6)
(1.7)

and

Remark 1.2.

Clearly, and . For example,
(1.8)

## 2. Comparison Results

We now present the main results of this section.

Lemma 2.1.

Assume that satisfies
(2.1)

then for all .

Proof.

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

Case 1.

on . It is easy to obtain that on . Thus from . Consequently, we obtain that
(2.2)

Case 2.

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

Subcase 1.

. There exists a such that
(2.3)
Let Then
(2.4)
Integrating the above inequality from to , we obtain
(2.5)
and then integrate from to to obtain
(2.6)

that implies This is a contradiction.

Subcase 2.

. There exists a such that
(2.7)

If , then If or , then . So .

Let Then

(2.8)

When , same as Subcase 1, we obtain that

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

(2.9)
and then integrate from to to obtain
(2.10)

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

Lemma 2.2.

Assume that satisfies
(2.11)

Then for all .

Proof.

Put
(2.12)
then and for all and
(2.13)
We have
(2.14)

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

Lemma 2.3.

Assume that satisfies
(2.15)

Then for all .

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

Lemma 2.4.

Assume that satisfies
(2.16)

Then for all .

Proof.

Suppose that for some . Then from boundary conditions, we have that there exists a such that
(2.17)

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

(2.18)
Integrating the above inequality from to , we obtain
(2.19)
and then integrate from to to obtain
(2.20)

Hence, a contradiction. The proof is complete.

## 3. Linear Problem

In this section, we consider the boundary value problem

(3.1)

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.

We first show that the solution of (3.1) is unique. Let and be solutions of (3.1) and set . Thus
(3.2)

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

(3.3)

By Lemma 2.1, we have that on .

If , then . Thus

(3.4)

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

If , then . Thus

(3.5)

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.

Consider the equation
(3.6)

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

(3.7)
where
(3.8)
Define a mapping by
(3.9)
Obviously is a Banach space with norm . For any , we have
(3.10)

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.

We show that for any , and are continuous in , where is a unique solution of the problem (3.6). Let be the solution of
(3.11)
Then
(3.12)
From (3.12), we have that
(3.13)
Hence
(3.14)
Since , exists for any and . From (3.12) and (3.14), we have
(3.15)

Step 3.

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

Put

(3.16)
(3.17)
then for any and
(3.18)
(3.19)

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

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

(3.20)
(3.21)
Define a function
(3.22)
where is the unique solution of the problem (3.6). Since is continuous and
(3.23)

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

(4.1)

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

(4.2)

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

Proof.

Let . For any , we consider the equation
(4.3)

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

(4.4)

and By Lemma 2.1, , which implies .

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

(4.5)

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

(4.6)

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

(1)
Department of Mathematics, Hunan University of Science and Technology, Xiangtan, Hunan, 411201, China

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