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


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.

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 [110]. 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.

We consider the second-order ordinary differential equation


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

are continuous in , and


where is some positive integer,


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]).

Let with




and the constant is optimal.

Lemma 2.2 (see [12]).

Let with the boundary value conditions , then


Consider the periodic boundary value problem


Lemma 2.3.

Suppose that are -integrable periodic function, where satisfy the condition (H2), with


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


If on the contrary, (2.4) has a nonzero -periodic solution , then using (2.4), we have


where is undetermined.

Firstly, we prove that has at least one zero in . If , we may assume . Since is a -periodic solution, there exists a with . Then,


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 .

Secondly, we prove that has at least zeros on . Considering the initial value problem




is the solution of (2.8) and


where with . Since


holds under the assumptions of , there is a , such that


Now, let . By the conditions (H2), (2.11), and (2.12), we have


Since is decreasing in , we have . Therefore,


We also consider the initial value problem




is the solution of (2.16), where is the same as the previous one, and


Hence, there exists a with , such that




From (2.12) and (2.19), it follows that


By and (2.21), we have


Since is decreasing on , we have , and


We now prove that has a zero point in . If on the contrary for , then we would have the following inequalities:


In fact, from(2.4), (2.8), and (2.15), we have


with . Setting , and since


we obtain


Notice that , which implies


So, we have


Integrating from 0 to , we obtain




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 .

Thirdly, we prove that has at least zeros on . If, on the contrary, we assume that only has zeros on , we write them as




Without loss of generality, we may assume that . Since


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

Finally, we prove Lemma 2.3. Since has at least zeros on , there are two zeros and with . By Lemmas 2.1 and 2.2, we have


From , it follows that




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

3. Proof of Theorem 1.1

Firstly, we prove the existence of the solution. Consider the homotopy equation


where and . When , it holds (1.1). We assume that is the fundamental solution matrix of with . Equation (3.1) can be transformed into the integral equation


From , is a periodic solution of (3.2), then


For is invertible,


We substitute (3.4) into (3.2),


Define an operator


such that


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 .

We can rewrite (3.1) in the following form:


Let , obviously . It satisfies the following problem:


in which we have


Since , are uniformly bounded and equicontinuous, there exists continuous function , and a subsequence of (denote it again by ), such that ,  uniformly in . Using and , and are uniformly bounded. By the Hahn-Banach theorem, there exists integrable function ,, and a subsequence of (denote it again by ), such that


where denotes "weakly converges to" in . As a consequence, we have


that is,


Denote that , , then we get


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



Because for , by Leray-Schauder degree theory, we have


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

Finally, we prove the uniqueness of the equation when the condition and holds. Let and be two -periodic solutions of the problem. Denote , then is a solution of the following problem:


By Lemma 2.3, we have for .

Let . We have


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

4. An Example

Consider the system


where is a continuous function. Obviously,


satisfy Theorem 1.1, then there is a unique -periodic solution in this system.


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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.

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Correspondence to Jian Zu.

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