#### Abstract

By applying the least action principle and minimax methods in critical point theory, we prove the existence of periodic solutions for a class of difference systems with p-Laplacian and obtain some existence theorems.

#### 1. Introduction

Consider the following -Laplacian difference system: where is the forward difference operator defined by , , such that , , , , and is continuously differentiable in for every and -periodic in for all .

When , (1.1) reduces to the following second-order discrete Hamiltonian system:

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, and optimal control, 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 [13].

In some recent papers [418], 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. Motivated by the above papers, we consider the existence of periodic solutions for problem (1.1) by using the least action principle and minimax methods in critical point theory.

#### 2. Preliminaries

Now, we first present our main results.

Theorem 2.1. Suppose that satisfies the following conditions: (F1) there exists an integer such that for all ;(F2) there exist , and such that where for every with ,(F3)Then problem (1.1) has at least one periodic solution with period .

Theorem 2.2. Suppose that satisfies (F1) and the following conditions: (F2)' there exist , such that where for every with ;(F4)Then problem (1.1) has at least one periodic solution with period .

Theorem 2.3. Suppose that satisfies (F1), (F2), and the following condition:
(F5) Then problem (1.1) has at least one periodic solution with period .

Theorem 2.4. Suppose that satisfies (F1), (2.3), (F2)', and the following condition:
(F6) Then problem (1.1) has at least one periodic solution with period .

Remark 2.5. The lower bounds and the upper bounds of our theorems are more accurate than the existing results in the literature. Moreover, there are functions satisfying our results but not satisfying the existing results in the literature.
Let the Sobolev space be defined by
For , let , , and , then . Let As usual, let
For any , let
To prove our results, we need the following lemma.

Lemma 2.6 (see [18]). Let . If , then

#### 3. Proofs

For the sake of convenience, we denote

Proof of Theorem 2.1. From (F3), we can choose such that It follows from (F2), (2.12), and (2.13) that Hence, we have The above inequality and (3.2) imply that as . Hence, by the least action principle, problem (1.1) has at least one periodic solution with period .

Proof of Theorem 2.2. From (2.3) and (F4), we can choose a constant such that It follows from (F2)' and Lemma 2.6 that which implies that The above inequality and (3.5) imply that as . Hence, by the least action principle, problem (1.1) has at least one periodic solution with period .

Proof of Theorem 2.3. First we prove that satisfies the (PS) condition. Assume that is a (PS) sequence of ; that is, as and is bounded. By (F5), we can choose such that In a similar way to the proof of Theorem 2.1, we have Hence, we have From (2.13), we have From (3.10) and (3.11), we obtain where . Notice that implies . Hence, it follows from (3.12) that where . By the proof of Theorem 2.1, we have It follows from the boundedness of , (3.13)–(3.15) that where is a positive constant and is a constant. The above inequality and (3.8) imply that is bounded. Hence is bounded by (2.13) and (3.13). Since is finite dimensional, we conclude that satisfies (PS) condition.
In order to use the saddle point theorem ([19], Theorem 4.6), we only need to verify the following conditions:(I1), (I2).
In fact, from (F5), we have which together with (2.11) implies that
Hence, (I1) holds.
Next, for all , by (F2) and (2.12), we have which implies that for all . By Lemma 2.6, in if and only if , so from (3.20), we obtain as in ; that is, () is verified. Hence, the proof of Theorem 2.3 is complete.

Proof of Theorem 2.4. First we prove that satisfies the (PS) condition. Assume that is a (PS) sequence of ; that is, as and is bounded. By (2.3) and (F6), we can choose such that In a similar way to the proof of Theorem 2.2, we obtain Hence, we have which together with (3.11) implies that where . It follows from (3.21) that , so, we obtain where is a positive constant. By the proof of Theorem 2.2, we have
It follows from the boundedness of , (3.26), (3.27), and the above inequality that where is a positive constant and is a constant. The above inequality and (3.22) imply that is bounded. Hence, is bounded by (2.13) and (3.26).
Similar to the proof of Theorem 2.3, we only need to verify (I1) and (I2). It is easy to verify (I1) by (F6). Now, we verify that (I2) holds. For , by (F2)' and (2.12), we have Thus, we have for all . By Lemma 2.6, in if and only . So from the above inequality, we have as , that is (I2) is verified. Hence, the proof of Theorem 2.4 is complete.

#### 4. Example

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

Example 4.1. Let and 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 Moreover, we have We can choose suitable such that which shows that (F3) holds. Then from Theorem 2.1, problem (1.1) has at least one periodic solution with period .

Example 4.2. Let , then . 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 Observe that On the other hand, if we let , then we have We can choose sufficiently small such that which shows that (2.3) and () hold. Then from Theorem 2.2, problem (1.1) has at least one periodic solution with period .

Example 4.3. Let , then . 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 Observe that On the other hand, we have We can choose suitable such that which shows that (F5) holds. Then from Theorem 2.3, problem (1.1) has at least one periodic solution with period .

Example 4.4. Let , then . Let where and . It is easy to see that satisfies (F1) and where , is a positive constant and is dependent on . The above shows that (F2)' holds with