We study periodic solutions for nonlinear secondorder ordinary differential problem . By constructing upper and lower boundaries and using LeraySchauder degree theory, we present a result about the existence and uniqueness of a periodic solution for secondorder 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 secondorder differential equations have been obtained by combining the classical method of lower and upper solutions and the method of alternative problems (The LyapunovSchmidt method) as discussed by many authors [1–10]. In [11], the author gives a simple method to discuss the existence and uniqueness of nonlinear twopoint boundary value problems. In this paper, we will extend this method to the periodic problem.
We consider the secondorder 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
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 (H_{2}), 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 (H_{2}), (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.
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 HahnBanach 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 LeraySchauder 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.
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.
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