#### Abstract

We study the existence of periodic solutions of the second-order differential equation , where are two constants satisfying , , is a constant satisfying , are continuous, and is -periodic. When the limits exist and are finite, we give some sufficient conditions for the existence of -periodic solutions of the given equation.

#### 1. Introduction

In this paper, we are concerned with the existence of periodic solutions of the second-order differential equation with an asymmetric nonlinearity and a deviating argument: where are two constants satisfying , , is a constant satisfying , are continuous, and is -periodic.

In recent years, the periodic problem of the second-order differential equation with a deviating argument has been widely studied because of its background in applied sciences (see [1–6] and the references cited therein).

In case when and , (1) becomes Assume that the limits exist and are finite. Lazer and Leach [7] proved that (2) has one -periodic solution provided that the function is of constant sign.

In case when and satisfy the equation , , (1) becomes Equation (4) was first introduced by Fučík [8]. Lately, the periodic problem of (4) was widely studied in the literature (see [9–13] and the references cited therein). To deal with the existence of periodic solutions of (4), Dancer [9] introduced a -periodic function where is a -periodic function defined by Obviously, is a periodic solution of the equation satisfying the initial value , . It was proved in [9] that (4) has at least one -periodic solution provided that has a constant sign in .

In the present paper, we will deal with the periodic solutions of (1) under condition . Owing to the appearance of the asymmetric nonlinearity , the methods in [4, 5] are no longer valid. To overcome this difficulty, we embed (1) into an operator equation with the form instead of as in [4, 5]. We first prove a continuation lemma and then apply this continuation lemma to prove the existence of periodic solution of (1).

Let us denote Obviously, we have We obtain the following result.

Theorem 1. *Assume that condition holds and . Then (1) has at least one -periodic solution provided that either
**
or
**
holds, where the function is defined by , .*

*Remark 2. *In the case when , we can obtain the similar sufficient conditions. For brevity, we omit the detailed description.

*Remark 3. *Obviously, if = or = , , then the first inequality of Theorem 1 reduces to the condition as in [8]; namely,

Throughout this paper, we always use to denote the real number set. For a multivariate function depending on , the notation always means that, for , holds uniformly with respect to other variables, whereas (or ) always means that (or ) is bounded for large enough. For any continuous -periodic function , we always set .

#### 2. Preliminary Lemmas

We now embed (1) into a family of equations with one parameter , where is continuous and satisfies the sign condition as follows:

Lemma 4. *Suppose that there exist two positive constants and such that, for any -periodic solution of (12), the following conditions hold:
**
Then (1) has at least one -periodic solution.*

*Proof. *We follow an argument in [14] to prove Lemma 4. At first, we introduce some notations. Let and be two Banach spaces defined by
with the norms
Define a linear operator by
where , and a nonlinear operator ,
It is easy to see that
It follows that is a Fredholm mapping of index zero.

Let us define two continuous projectors and by setting
Set . Then is an algebraic isomorphism, and we define by
Clearly, we have that, for any ,
For any open bounded set , we can prove by standard arguments that and are relatively compact on the closure . Therefore, is -compact on .

It is noted that (12), together with the -periodic boundary condition, is equivalent to the operator equation
Let be the open bounded set defined by
From (14), we have
Since is a Fredholm operator with index zero and is -compact on , we get from the homotopic invariance of the coincidence degree that
Next, we will compute . To this end, we introduce an auxiliary operator defined by
Clearly, is -compact on and
Now, we will prove that
Obviously, it follows from (25) and (28) that
On the other hand, if is a solution of , then satisfies the equation as follows:
Multiplying both sides of (31) by and integrating over , we get
If , then we infer from (13) and (32) that for every . Furthermore, for every , where is a constant. Consequently, we have , and then .

From the homotopic invariance of the coincidence degree, we have
In the following, we will compute . To this end, we use the equality [15] as follows:
which holds provided that the following conditions are satisfied,
In what follows, we will prove that conditions (35) and (36) are satisfied. In fact, if is a solution of , then satisfies the equation as follows:
Using the same method as before, we can get . This is a contradiction. To check condition (36), we notice that if , then with . Hence, we have that, for ,
Finally, we can easily calculate the Brouwer degree and obtain
Therefore, we have
Consequently, the equation
has at least one -periodic solution. Equivalently, (1) has at least one -periodic solution.

*Remark 5. *In (12), if satisfies the following condition,
then the conclusion of Lemma 4 still holds. This claim can be proved by using the same method as the one used for proving Lemma 4. In fact, we only need to modify the term in the auxiliary operator to the term .

#### 3. Periodic Solutions of Duffing Equation with a Deviating Argument

At first, we choose a continuous function satisfying where are constants. Moreover, satisfies condition (13).

Considering the equivalent system of (12), Let be any (possible) -periodic solution of (12). Write . Then, is a -periodic solution of system (44).

In what follows, we will introduce a transformation. To this end, let us denote by a solution of equation satisfying the initial condition , . Obviously, is -periodic. The derivative of will be denoted by . It is easy to check that the following properties are satisfied:(1), .(2), .(3), .Let us define a mapping as follows: where .

Under the transformation , if , , then the -periodic solution of system (44) can be expressed in the form satisfying the equations as follows: Let us denote . From now on, we always assume that is bounded. From the first equation of (46) we get that Therefore, we have Furthermore, we get From the second equation of (46), we have As a result, Substituting (51) in (47), we obtain that, for , Consequently, Similarly, substituting (51) in the second equality of (44), we get that, for , Therefore, we have Write Recalling that and , we have the following estimates.

Lemma 6. *Assume that condition holds. Then, for ,
*

*Proof. *We only give the proof for the case . The other cases can be treated similarly. Since is -periodic, it follows from the expression of that
From the dominated convergent theorem, we have that, for ,

Lemma 7. *Assume that condition holds. Then, for ,
*

*Proof. *We also only give the proof for the case . The other cases can be treated similarly. Since is -periodic, it follows from the expression of and the dominated convergent theorem that, for ,

Lemma 8. *Assume that condition (43) holds. Then, for ,
*

*Proof. *From the expression of and the dominated convergent theorem we have that, for ,
Similarly, we have that, for ,

*Proof of Theorem 1. *We proceed to prove Theorem 1 in two different cases.

Assume that the first inequality of Theorem 1 holds. Without loss of generality, we assume
Let us set
We now choose a function satisfying (43) and (13). Moreover, satisfy
Then we infer from Lemmas 7 and 8 that, for ,
Since , , and , there exists a constant such that, for and ,
From (68) and (69) we have that, for ,
Consequently, there exists a constant such that if is a -periodic solution of system (46), then , . Furthermore, there exist constants and such that if is a -periodic solution of (12), then
From Lemma 4, we know that (1) has at least one -periodic solution.

We assume that the second inequality of Theorem 1 holds. Without loss of generality, we assume
Let us set
Similarly, we choose a continuous function satisfying (43) and (13). Moreover, satisfy
Then we infer from Lemmas 7 and 8 that, for ,
Since , , and , there exists a constant such that, for and and ,
From (75) and (76) we have that, for sufficiently large ,
Consequently, there exists a constant such that if is a -periodic solution of system (46), then , . Furthermore, there exist constants and such that if is a periodic solution of (12), then
From Lemma 4 we know that (1) has at least one periodic solution.

#### Acknowledgments

This research is supported by Research Fund for the Doctoral Program of Higher Education of China, no. 11AA0013, Beijing Natural Science Foundation (Existence and multiplicity of periodic solutions in nonlinear oscillations), no. 1112006, and the Grant of Beijing Education Committee Key Project, no. KZ201310028031.