#### Abstract

The authors study the existence and uniqueness of a set with -periodic solutions for a class of second-order differential equations by using Mawhin's continuation theorem and some analysis methods, and then a unique homoclinic orbit is obtained as a limit point of the above set of -periodic solutions.

#### 1. Introduction

In this paper, we study the existence and uniqueness of homoclinic solutions for the following nonlinear second-order differential equations: where , , and are all in .

As usual we say that a nonzero solution of (1.1) is homoclinic (to 0) if and as .

Equation (1.1) is important in the applied sciences such as nonlinear vibration of masses, see [1–3] and the references therein. But most of the authors in those papers are interested in the study of problems of periodic solutions. Recently, the existence of homoclinic solutions for some second-order ordinary differential equation (system) has been extensively studied by using critical point theory, see [4–13] and the references therein. For example, in [9], by using the Mountain Pass theorem, Lv et al. discussed the existence of homoclinic solutions for the following second-order Hamiltonian systems: and in [13], the authors by means of variational method studied the problem of homoclinic solutions for the forced pendulum equation without the first derivative term. But, as far as we know, there were few papers studying the existence of homoclinic solution for the equation such as (1.1). This is due to the fact that (1.1) contains the first derivative term . This implies that the differential equation is not the Euler Lagrange equation associated with some functional . So the method of critical point theory (or variational method) in [4–13] cannot be applied directly. Although paper [13] discussed the existence of homoclinic solutions for the following equation containing the first derivative term: the term containing the first derivative is only linear with respect to .

In order to investigate the homoclinic solutions to (1.1), firstly, we study the existence of -periodic solutions to the following equation for each : where is a -periodic function such that is a given constant, and is a constant independent of . Then a homoclinic solution to (1.1) is obtained as a limit point of the set , where is an arbitrary -periodic solution to (1.4) for each .

The significance of present paper is that we not only investigate the existence of homoclinic solution to (1.1), but also study the uniqueness of the homoclinic solution and, the existence of -periodic solutions to (1.4) is obtained by using Mawhin’s continuation theorem [14], not by using the methods of critical point theory, which is quite different from the approaches of [4–13, 15]. Furthermore, the method to obtain the homoclinic solution to (1.1) is also different from the corresponding ones of [15].

#### 2. Main Lemmas

For each , let , , the norms of and are defined by and , respectively, then and are all Banach spaces. Furthermore for , , where .

Lemma 2.1 (see [12]). *Let and , then for every , the following inequality holds:
**
where are constants. *

Lemma 2.2 (see [12]). *Let , then the following inequality holds:
**
where and are constants with and .*

In order to use Mawhin’s continuation theorem for investigating the existence of -periodic solutions to (1.4), we give some definitions associated with Mawhin’s continuation theorem.

*Definition 2.3 (see [14]). *Let and be two Banach spaces with norms and , respectively. A linear operator
is said to be a Fredholm opeartor with index zero provided that (1) is a closed subset of ; (2).

If is a Fredholm operator with index zero, then and . Let and be the continuous projectors. Clearly, , thus the restriction is invertible. Denote by the inverse of .

*Definition 2.4 (see [14]). *Let and be two Banach spaces with norms and , respectively, and the operator
is a Fredholm operator with index zero, is an open bounded set with . A continuous operator is said to be -compact in , provided that (1) is a relative compact set of ; (2) is a bounded set of .

Lemma 2.5 (see [14]). *Suppose that and are two Banach spaces, and is a Fredholm operator with index zero. Furthermore, is an open bounded subset and is -compact on . If all the following conditions hold:*(1)*, for all ;*(2)*, for all ;*(3)*,*

where is an isomorphism. Then equation has a solution on .

Lemma 2.6. *Assume that there are positive constants , , , , and with , such that the following conditions hold.*(A1)*, and .*(A2)*(A3)** with for all .*

Then for every , (1.4) possesses a -periodic solution.

*Remark 2.7. *From (1.5), we see
which together with assumption (A1) yields that and are two constants independent of .

Similarly, we have that and are two constants independent of .

* Proof. *Set , , , , where , and
Clearly, , , which implies that is a closed subset of , and . So is a Fredholm operator with index zero. Let
be defined respectively by and let
Then has a unique continuous pseudo-inverse on defined by , where
For each open bounded set , from the above formula, it is easy to see that the mapper is -compact on .*Step 1.* For each , let , that is,
We will show that is bounded in . Suppose that , then
Multiplying both sides of (2.12) by and integrating on the interval , we have from assumption (A2) that
By using hólder inequality, we get
which together with the conclusion of Remark 2.7 shows
Clearly, is a constant independent of and .

Multiplying both sides of (2.12) by and integrating on the interval , we have
It follows from (2.16) and assumption (A3) that
which implies
By using Lemma 2.2, we have
Clearly, is a constant independent of and .

On the other hand, multiplying both sides of (2.12) by and integrating on the interval , we have
It follows from assumption (A2) that
which together with (2.19) and results in
that is,
Therefore
where is a constant independent of and . By using Lemma 2.2 again, we get
Obviously, is a constant independent of and . Therefore, if , then by (2.19) we see that
Clearly, is a constant independent of and ; that is, is uniformly bounded for all and .*Step 2.* From assumptions (A2) and (A3), we see that there must be a constant such that and . Set , where . We will show that , for all .

In fact, by assumption (A2), we see that , and if , then or . So
where . This implies that , for all .*Step 3.* Set , , we will show .

Let , for all , when , we have and
So for all , , and then
Therefore, by Lemma 2.5, (1.4) has a -periodic solution .

*Remark 2.8. *Suppose that all the conditions in Lemma 2.6 hold. We see that for each , (1.4) has a -periodic solution . This implies that
Furthermore, as same as the proof of step 1 in Lemma 2.6 with replacing by , we have
where and are two positive constants independent of .

Lemma 2.9 (see [12]). *Let be the -periodic solution for (1.4) and satisfies (2.30) and (2.31) for all . Then there exists a function such that for each interval , there is a subsequence of with uniformly on . *

#### 3. Main Results

Theorem 3.1. *Suppose that assumptions (A1), (A2), and (A3) in Lemma 2.6 hold. Then (1.1) has a unique homoclinic solution.*

* Proof. *Since assumptions (A1), (A2), onsisting of Kuratowski operations we used following principles and (A3) in Lemma 2.6 hold, by using Lemma 2.6, we see that (1.4) has a -periodic solution satisfying (2.30) and (2.31) for each . It follows from Lemma 2.9 that there exists a such that for each interval , there is a subsequence of satisfying uniformly on . Below, we will show that is just a unique homoclinic solution to (1.1).*Step 1.* We show that is a solution of (1.1).

In view of being a -periodic solution to (1.4), we have
Take such that , there exists such that for all
Integrating (3.2) from to , we have
Since Lemma 2.9 shows that uniformly on and uniformly on as , let in (3.3), we get
In view of and are arbitrary, (3.4) shows that is a solution of (1.1).*Step 2.* We prove that , as .

Obviously, for every , there exists such that for all , we have
It follows that
and then
which yields
By using Lemma 2.1, as ,
So we have , as .*Step 3.* We will show that
From the Remark 2.8 and Lemma 2.9, we have
which together with (1.1) implies that
where and . If , as , then there exist a and a sequence such that
From this, we have for
It follows that
which contradicts (3.6), and so (3.10) holds.*Step 4.* Finally, we will prove that (1.1) possesses a unique homoclinic solution. In order to do it, let , where and are two arbitrary homoclinic solutions of (1.1). Then
We will show that
If (3.17) does not hold, then there must be a such that
or

If , then from (3.16), we see that there is a constant such that and for . Let such that , then
that is,
So and , and then from (1.1), we see
By using the condition (A3), we have that
which contradicts to (3.20). This contradiction implies that (3.18) does not hold. Similarly, we can prove that (3.19) does not hold, either. So .

As an application, we consider the following example:
where are constants and, . Corresponding to (1.1), we can chose and such that assumptions (A2) and (A3) hold. Furthermore, by the direct calculation, we can easily obtain that
This implies that assumption (A1) also holds. So by applying Theorem 3.1, we know that (3.25) possesses a unique homoclinic solution.

#### Acknowledgments

The authors are very grateful to the referee for her/his careful reading of the original paper and for her/his valuable suggestions for improving this paper. This work was sponsored by the key NNSF of China (no. 11271197) and Science Foundation of NUIST (no. 20090202; 2012r101).