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.

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

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.

The following results will be used later.

Lemma 2.1 (see [12]).

Let with

then

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 .

Proof.

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

Obviously,

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

Clearly,

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

Hence, there exists a with , such that

Then,

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

Therefore,

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

Obviously,

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

Hence,

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

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.

Denote

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.

Consider the system

where is a continuous function. Obviously,

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

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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