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.