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Existence and Uniqueness of Periodic Solution for Nonlinear Second-Order Ordinary Differential Equations
Boundary Value Problems volume 2011, Article number: 192156 (2011)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ1_HTML.gif)
Throughout this paper, we will study the existence of periodic solutions of (1.1) with the following assumptions:
are continuous in
, and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ2_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_IEq5_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ3_HTML.gif)
where is some positive integer,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ4_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ5_HTML.gif)
then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ6_HTML.gif)
and the constant is optimal.
Lemma 2.2 (see [12]).
Let with the boundary value conditions
, then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ7_HTML.gif)
Consider the periodic boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ8_HTML.gif)
Lemma 2.3.
Suppose that are
-integrable
periodic function, where
satisfy the condition (H2), with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ9_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ10_HTML.gif)
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,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ11_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ12_HTML.gif)
Obviously,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ13_HTML.gif)
is the solution of (2.8) and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ14_HTML.gif)
where with
. Since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ15_HTML.gif)
holds under the assumptions of , there is a
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ16_HTML.gif)
Now, let . By the conditions (H2), (2.11), and (2.12), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ17_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ18_HTML.gif)
Since is decreasing in
, we have
. Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ19_HTML.gif)
We also consider the initial value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ20_HTML.gif)
Clearly,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ21_HTML.gif)
is the solution of (2.16), where is the same as the previous one, and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ22_HTML.gif)
Hence, there exists a with
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ23_HTML.gif)
Then,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ24_HTML.gif)
From (2.12) and (2.19), it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ25_HTML.gif)
By and (2.21), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ26_HTML.gif)
Since is decreasing on
, we have
, and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ27_HTML.gif)
We now prove that has a zero point in
. If on the contrary
for
, then we would have the following inequalities:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ28_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ29_HTML.gif)
In fact, from(2.4), (2.8), and (2.15), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ30_HTML.gif)
with . Setting
, and since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ31_HTML.gif)
we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ32_HTML.gif)
Notice that , which implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ33_HTML.gif)
So, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ34_HTML.gif)
Integrating from 0 to , we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ35_HTML.gif)
Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ36_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ37_HTML.gif)
Obviously,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ38_HTML.gif)
Without loss of generality, we may assume that . Since
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ39_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ40_HTML.gif)
From , it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ41_HTML.gif)
Hence,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ42_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ43_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ44_HTML.gif)
From ,
is a
periodic solution of (3.2), then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ45_HTML.gif)
For is invertible,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ46_HTML.gif)
We substitute (3.4) into (3.2),
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ47_HTML.gif)
Define an operator
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ48_HTML.gif)
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ49_HTML.gif)
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:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ50_HTML.gif)
Let , obviously
. It satisfies the following problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ51_HTML.gif)
in which we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ52_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ53_HTML.gif)
where denotes "weakly converges to" in
. As a consequence, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ54_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ55_HTML.gif)
Denote that ,
, then we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ56_HTML.gif)
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
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ57_HTML.gif)
Because for
, by Leray-Schauder degree theory, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ58_HTML.gif)
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:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ59_HTML.gif)
By Lemma 2.3, we have for
.
Let . We have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ60_HTML.gif)
with . Denote
by
. So,
is the solution of the problem (1.1). The proof is complete.
4. An Example
Consider the system
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ61_HTML.gif)
where is a continuous function. Obviously,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F192156/MediaObjects/13661_2010_Article_28_Equ62_HTML.gif)
satisfy Theorem 1.1, then there is a unique -periodic solution in this system.
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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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Zu, J. Existence and Uniqueness of Periodic Solution for Nonlinear Second-Order Ordinary Differential Equations. Bound Value Probl 2011, 192156 (2011). https://doi.org/10.1155/2011/192156
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DOI: https://doi.org/10.1155/2011/192156