#### Abstract

By applying minimax methods in critical point theory, we prove the existence of periodic solutions for the following discrete Hamiltonian systems , where , , , is continuously differentiable in for every and is -periodic in is a positive integer.

#### 1. Introduction

Consider the following discrete Hamiltonian system: where is the forward difference operator defined by and is continuously differentiable in for every and is -periodic in ; is a positive integer.

Difference equations usually describe evolution of certain phenomena over the course of time. For example, if a certain population has discrete generations, the size of the th generation is a function of the th generation . In fact, difference equations provide a natural description of many discrete models in real world. Since discrete models exist in various fields of science and technology such as statistics, computer science, electrical circuit analysis, biology, neural network, optimal control, and so on, it is of practical importance to investigate the solutions of difference equations. For more details about difference equations, we refer the readers to the books [1–3].

In some recent papers [4–15], the authors studied the existence of periodic solutions and subharmonic solutions of difference equations by applying critical point theory. These papers show that the critical point theory is an effective method to the study of periodic solutions for difference equations. In 2007, Xue and Tang [11] investigated the existence of periodic solutions for (1.1) and obtained the main result.

Theorem A (see [11]). *Suppose that satisfies the following conditions: *(F1)*there exists a positive constant such that for all ;*(F2)*there are constants such that
where for every with ;*(F3)* as for all .**
Then problem (1.1) has at least one periodic solution with period .*

Let where , , , and . It is easy to see that where and is a positive constant and is dependent on . The above inequality shows that there are functions not satisfying condition (F2). If we let , , , then we have But the above equality does not satisfy (F3). This example shows that it is valuable to further improve conditions (F2) and (F3).

Before stating our main results, we first introduce some preliminaries.

#### 2. Preliminaries

Let can be equipped with the inner product by which the norm on can be reduced by where and denote the usual inner product and the usual norm in . It is easy to see that is a finite-dimensional Hilbert space and linear homeomorphic to . For any , define Obviously, and is equivalent to . Hence, there exist two positive constants , , which are independent on , such that If we define , it is easy to see that for any ,

For any , let We can compute the Fréchet derivative of (2.7) as Hence, is a critical point of on if and only if So, the critical points of are classical solutions of (1.1). The following lemmas are useful in our proof.

Lemma 2.1 (see [11]). *As a subspace of , is defined by
**
where and denote the Gauss Function. Then there hold *(i)*; *(ii)*.*

Lemma 2.2 (see [11]). *Define , , ; then one has
*

#### 3. Main Results and Proofs

Theorem 3.1. *Suppose that satisfies (F1) and the following conditions *(F2)′*there are , such that
where for every with ;*(F3)′* for all .**
Then problem (1.1) has at least one periodic solution with period .*

Theorem 3.2. *Suppose that satisfies (F1) and (F2) with . Moreover, assume the following conditions hold:
**
and *(F4)* for all .**
Then problem (1.1) has at least one periodic solution with period .*

*Remark 3.3. *It is easy to see that (F2)′ is more general than (F2) and (F3)′ is weaker than (F3). Theorem 3.2 is a new result, which completes Theorem A when .

For the sake of convenience, we denote

*Proof of Theorem 3.1. *First we prove that satisfies the (PS) condition. Suppose that a consequence is such that , where and as . Then for sufficiently large ,
From Lemma 2.1, we can write as , where . From (F3)′, we can choose such that
From (F2)′, (2.6), Hölder inequality, and Young inequality, we have
In a similar way, we have
Let . From (2.12) and (3.7), we have
It follows from (3.8) and (3.9) that
where . The fact that implies that . So it follows from (3.10) that
and so
where . It follows from the boundedness of , (2.11), (3.6), (3.11), and (3.12) that
Inequalities (3.5) and (3.13) imply that is bounded. Hence, is bounded by (3.12), and then is bounded. Since is finite dimensional, there exists a subsequence of convergent in . Thus, we conclude that (PS) condition is satisfied.

In order to use the saddle point theorem [16, Theorem 4.6], we only need to verify the following conditions: (I1);
(I2).

In fact, from (F3)′, we have
For any , since , we have
It follows from (3.14) and the above inequality that
Thus (I1) is easy to verify.

Next, for all , from (F2)′ and (2.6), we have
By (2.7), (2.12), and (3.17), we obtain
Since and , we have as in . The proof of Theorem 3.1 is complete.

*Proof of Theorem 3.2. *By (3.2) and (F4), we can choose an such that
It follows from (F2)′ with , (2.6), Hölder inequality, and Young inequality that
In a similar way, we have
From (3.8) and (3.22), we have
It follow from (3.23) that
where . The fact that implies that . So it follows from (3.24) that
and so
where . It follows from the boundedness of , (2.11), (3.21), (3.25), and (3.26) that
Inequalities (3.20) and (3.27) imply that is bounded. Hence, is bounded by (3.26), and then is bounded. Since is finite dimensional, there exists a subsequence of convergent in . Thus, we conclude that (PS) condition is satisfied.

In the following, we prove that satisfies (I1) and (I2). In fact, from (F4), we have
It follows from (3.27) and that
Thus (I1) is easy to verify.

Next, for all , from (F2)′ with and (2.6), we have
By (2.7), (2.12), and (3.30), we obtain
Since , we have as in . The proof of Theorem 3.2 is complete.

#### 4. Examples

In this section, we give two examples to illustrate our results.

*Example 4.1. *Let
where and . It is easy to see that
where and is a positive constant and is dependent on . It is easy to see that satisfies (F1). From (4.2), we can let , , and be
which shows that (F2)′ is satisfied. Moreover, if we let , then we have
If we let , then we obtain
which shows that (F3)' holds. Then from Theorem 3.1, problem (1.1) has at least one periodic solution with period .

*Example 4.2. *Let
where and . It is easy to see that satisfies (F1) and
where and is a positive constant and is dependent on . The above shows that (F2)′ holds with and
Let , then . Observe that
On the other hand, we have
We can choose sufficiently small such that and
which shows that (F4) holds. Then from Theorem 3.2, problem (1.1) has at least one periodic solution with period .

#### Acknowledgment

X. H. Tang is supported by the NNSF (no. 10771215) of China.