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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 508247, 9 pages
http://dx.doi.org/10.1155/2013/508247
Research Article

Subharmonics with Minimal Periods for Convex Discrete Hamiltonian Systems

School of Science, Jimei University, Xiamen 361021, China

Received 19 January 2013; Accepted 24 February 2013

Academic Editor: Zhengkun Huang

Copyright © 2013 Honghua Bin. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We consider the subharmonics with minimal periods for convex discrete Hamiltonian systems. By using variational methods and dual functional, we obtain that the system has a -periodic solution for each positive integer , and solution of system has minimal period as subquadratic growth both at 0 and infinity.

1. Introduction

Consider Hamiltonian systems where , , stands for the gradient of with respect to the second variable, and is the symplectic matrix with the identity in . Moreover, is -periodic in the variable , .

Classically, solutions for systems (1) are called subharmonics. The first result concerning the subharmonics problem traced back to Birkhoff and Lewis in 1933 (refer to [1]), in which there exists a sequence of subharmonics with arbitrarily large minimal period, using perturbation techniques. More results can be found in [15], where is convex with subquadratic growth both at and infinity. Using index theory and Clarke duality, Xu and Guo [1, 5] proved that the number of solutions for systems (1) with minimal period tends towards infinity as .

For periodic and subharmonic solutions for discrete Hamiltonian systems, Guo and Yu [6, 7] obtained some existence results by means of critical point theory, where they introduced the action functional Using Clarke duality, periodic solution of convex discrete Hamiltonian systems with forcing terms has been investigated in [8]. Clarke duality was introduced in 1978 by Clarke [9], and developed by Clarke, Ekeland, and others, see [1012]. This approach is different from the direct method of variations; some scholars applied it to consider the periodic solutions, subharmonic solutions with prescribed minimal period of Hamiltonian systems; one can refer to [3, 5, 1214] and references therein. The dynamical behavior of differential and difference equations was studied by using various methods; see [1519]. We refer the reader to Agarwal [20] for a broad introduction to difference equations.

Motivated by the works of Mawhin and Willem [12] and Xu and Guo [21], we use variational methods and Clarke duality to consider the subharmonics with minimal periods for discrete Hamiltonian systems where , , with a given positive integer, and is the forward difference operator. . Moreover, hamiltonian function satisfies the following assumption:(A1) is continuous differentiable on , convex for each and for each ;(A2) there exist constants , , , such that which implies subquadratic growth both at and infinity.

Theorem 1. Assume (A1) holds. , , as uniformly in . Then there exists a -periodic solution of (3), such that , and the minimal period of tends to as .

Theorem 2. Under the assumptions (A1) and (A2), if for given integer , then the solution of (3) has minimal period .

2. Clarke Duality and Eigenvalue Problem

First we introduce a space with dimension as follows: where Equipped with inner product and norm in as where and denote the usual scalar product and corresponding norm in , respectively. It is easy to see that is a Hilbert space with dimension, which can be identified with . Then for any , it can be written as , where , , the discrete interval is denoted by , and denotes the transpose of a vector or a matrix.

Denote the subspace . Let be the direct orthogonal complement of to . Thus can be split as , and for any , it can be expressed in the form , where ,  .

Next we recall Clarke duality and some lemmas.

The Legendre transform (see [12]) of with respect to the second variable is defined by where denotes the inner product in . It follows from (A1) and (A2) that(B1) is continuous differentiable on ,(B2) for ,  ,  , one has

Associated with variational functional (2), the dual action functional is defined by Indeed, by (11), we have for any and . Therefore, can be restricted in subspace of . Moreover, in terms of Lemma  2.6 in [8] and the following lemma, the critical points of (11) correspond to the subharmonic solutions of (3).

Lemma 3 (see [8, Theorem 1]). Assume that(H1), is convex in the second variable for , (H2) there exists such that for all , , and , (H3) there exist and , such that for any , , and , (H4) for each , as .

Then system (3) has at least one periodic solution , such that minimizes the dual action .

The following lemmas will be useful in our proofs, where Lemma 4 can be proved by means of Euler formula , and Lemma 5 is a Hölder inequality.

Lemma 4. For any , .

Lemma 5. For any , , , one has , where , and .

Lemma 6 (see [12, proposition 2.2]). Let be of and convex functional, , where , , , , . Then , where .

In order to know the form of , we consider eigenvalue problem where , , , . We can rewrite (12) as the following form: Denoting then problem (12) is equivalent to Letting be the solution of (15), for some , we have and . Consider the polynomial and condition ; it follows that

In the following we denote by , , , and . By , it follows that Thus

Let Obviously, and satisfy (15). Moreover , , , , , .

For , subspace is defined bywhere denotes the greatest-integer function and Therefore, Moreover, for any , we may express as where .

Since , we consider eigenvalue problem where . The second order difference equation (24) has complexity solution for , where . Moreover, ; that is, ,  .

By the previous, it follows Lemma 7.

Lemma 7. For any , one has , and , where Moreover, if , then .

3. Proofs of Main Results

Lemma 8. Consider

Proof. Letting , then . By Lemmas 5 and 7, we have

Lemma 9. If there exist , and , such that for all and , then each solution of (3) satisfies the inequalities

Proof. Let be the solution of (3). By Lemma 6, we have Obviously, by (3), and it follows that ; that is, By means of Lemma 8, we have which gives first conclusion.
Now, in view of (28); therefore by convex and Lemma 8, we have which gives the second conclusion. The proof is completed.

Proof of Theorem 1. Let . By assumption in Theorem 1, there exists , such that , for and . Moreover, there exist , such that Thus, by convex of , for all with , we have Therefore there exist and , such that Combining the previous argument, by Lemma 3, the system (3) has a -periodic solution such that minimizes the dual action It follows that and .
We next prove that as .
Suppose not, there exist and a subsequence such that In terms of (3), it follows that for some , and , . Consequently, by , we have where and
By (36), if , we have , and . Letting and , in terms of (12), associated with is given by which belongs to , andMoreover, by Lemma 4 we have Thus . Combining (39), we have , which is impossible as large. So the claim is valid.
It remains only to prove that the minimal period of tends to as .
If not, there exists and a sequence such that the minimal period of satisfies . By assumption in Theorem 1, there exists and such that By (36) and Lemma 9 with replaced by , we get
Write , where . Inequality (46) implies that By Lemma 7 and (45), it follows that which implies that is bounded, therefore is bounded; a contradiction with the second claim . This completes the proof.

Proof of Theorem 2. Under the assumptions (A1) and (A2), all conditions in Theorem 1 are satisfied. Then, for each integer , there exists a -periodic solution of (3) such that minimizes the dual action
If the critical point of dual action functional has minimal period , where , then by Lemma 7 with replaced by , we have the following estimate: By Lemma 5 and the previous inequality, we have where for . It follows from assumption (B2) that thus One can obtain the previous inequality by minimizing in (53) with respect to , and the minimum is attained at .
On the other hand, let where , . Then , and Taking , where , by Lemma 4, it follows that where and Therefore, taking , by eigenvalue problem (24) and (B2), it follows that
Let equal the right-hand side of (59) where . It is easy to see that the absolute minimum of is attained at and given by . Hence, by (19), let where .
If is even, then . Set For , we have Combining (54), (59), and (62), we have By , and , it follows that For integer , , , , we have , .
If is even, then . By assumption we have , which implies that or . If , then . So we have .
If is odd, then . By assumption , we have , so . This completes the proof.

Acknowledgments

This research is supported by the National Natural Science Foundation of China under Grants (11101187), NCETFJ (JA11144), the Excellent Youth Foundation of Fujian Province (2012J06001), and the Foundation of Education of Fujian Province (JA09152).

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