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

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.

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.

We consider the second-order ordinary differential equation

(1.1)

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

are continuous in , and

(1.2)

(1.3)

where is some positive integer,

(1.4)

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

(2.1)

then

(2.2)

and the constant is optimal.

Lemma 2.2 (see [12]).

Let with the boundary value conditions , then

(2.3)

Consider the periodic boundary value problem

(2.4)

Lemma 2.3.

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

(2.5)

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

Proof.

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

(2.6)

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,

(2.7)

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

(2.8)

Obviously,

(2.9)

is the solution of (2.8) and

(2.10)

where with . Since

(2.11)

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

(2.12)

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

(2.13)

(2.14)

Since is decreasing in , we have . Therefore,

(2.15)

We also consider the initial value problem

(2.16)

Clearly,

(2.17)

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

(2.18)

Hence, there exists a with , such that

(2.19)

Then,

(2.20)

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

(2.21)

By and (2.21), we have

(2.22)

Since is decreasing on , we have , and

(2.23)

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

(2.24)

(2.25)

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

(2.26)

with . Setting , and since

(2.27)

we obtain

(2.28)

Notice that , which implies

(2.29)

So, we have

(2.30)

Integrating from 0 to , we obtain

(2.31)

Therefore,

(2.32)

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

(2.33)

Obviously,

(2.34)

Without loss of generality, we may assume that . Since

(2.35)

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

(2.36)

From , it follows that

(2.37)

Hence,

(2.38)

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

(3.1)

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

(3.2)

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

(3.3)

For is invertible,

(3.4)

We substitute (3.4) into (3.2),

(3.5)

Define an operator

(3.6)

such that

(3.7)

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:

(3.8)

Let , obviously . It satisfies the following problem:

(3.9)

in which we have

(3.10)

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

(3.11)

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

(3.12)

that is,

(3.13)

Denote that , , then we get

(3.14)

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

Denote

(3.15)

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

(3.16)

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:

(3.17)

By Lemma 2.3, we have for .

Let . We have

(3.18)

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

4. An Example

Consider the system

(4.1)

where is a continuous function. Obviously,

(4.2)

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