- Research Article
- Open Access

# Existence and Uniqueness of Periodic Solution for Nonlinear Second-Order Ordinary Differential Equations

- Jian Zu
^{1}Email author

**Received:**22 May 2010**Accepted:**6 March 2011**Published:**15 March 2011

## Abstract

We study periodic solutions for nonlinear second-order ordinary differential problem . By constructing upper and lower boundaries and using Leray-Schauder degree theory, we present a result about the existence and uniqueness of a periodic solution for second-order ordinary differential equations with some assumption.

## Keywords

- Ordinary Differential Equation
- Periodic Solution
- Degree Theory
- Periodic Problem
- Important Branch

## 1. Introduction

The study on periodic solutions for ordinary differential equations is a very important branch in the differential equation theory. Many results about the existence of periodic solutions for second-order differential equations have been obtained by combining the classical method of lower and upper solutions and the method of alternative problems (The Lyapunov-Schmidt method) as discussed by many authors [1–10]. In [11], the author gives a simple method to discuss the existence and uniqueness of nonlinear two-point boundary value problems. In this paper, we will extend this method to the periodic problem.

Throughout this paper, we will study the existence of periodic solutions of (1.1) with the following assumptions:

The following is our main result.

Theorem 1.1.

Assume that and hold, then (1.1) has a unique periodic solution.

## 2. Basic Lemmas

The following results will be used later.

Lemma 2.1 (see [12]).

and the constant is optimal.

Lemma 2.2 (see [12]).

Lemma 2.3.

then (2.4) has only the trivial -periodic solution .

Proof.

where is undetermined.

we could get a contradiction.

Without loss of generality, we may assume that ; then there exists a sufficiently small such that . Since is a continuous function, there must exist a with .

which implies (2.24). By a similar argument, we have (2.25). Therefore, , a contradiction, which shows that has at least one zero in , with .

We let . If , then from a similar argument, there is a , such that and so on. So, we obtain that has at least zeros on .

we obtain , which contradicts . Therefore, has at least zeros on .

which implies for . Also . Therefore, for , a contradiction. The proof is complete.

## 3. Proof of Theorem 1.1

Clearly, is a completely continuous operator in .

There exists , such that every possible periodic solution satisfies ( denote the usual normal in . If not, there exists and the solution with .

which also satisfy the condition . Notice that and are integrable on , so satisfies Lemma 2.3. Hence, we have , which contradicts . Therefore, is bounded.

So, we conclude that has at least one fixed point in , that is, (1.1) has at least one solution.

By Lemma 2.3, we have for .

with . Denote by . So, is the solution of the problem (1.1). The proof is complete.

## 4. An Example

## Declarations

### Acknowledgments

The author expresses sincere thanks to Professor Yong Li for useful discussion. He would like to thank the reviewers for helpful comments on an earlier draft of this paper.

## Authors’ Affiliations

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## Copyright

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