Abstract
By applying minimax methods in critical point theory, we prove the existence of periodic solutions for the following discrete Hamiltonian systems Δ2u(t − 1)+∇F(t, u(t)) = 0, where t ∈ ℤ, u ∈ ℝN, F : ℤ × ℝN → ℝ, F(t, x) is continuously differentiable in x for every t ∈ ℤ and is T‐periodic in t; T is a positive integer.
Highlights
Consider the following discrete Hamiltonian system: Δ2u t − 1 ∇F t, u t 0, t ∈ Z, 1.1 where Δ is the forward difference operator defined by Δu t u t 1 − u t, Δ2u t Δ Δu t, t ∈ Z, u ∈ RN, F : Z × RN → R, and F t, x is continuously differentiable in x for every t ∈ Z and is T -periodic in t; T is a positive integer.Difference equations usually describe evolution of certain phenomena over the course of time
By applying minimax methods in critical point theory, we prove the existence of periodic solutions for the following discrete Hamiltonian systems Δ2u t − 1 ∇F t, u t 0, where t ∈ Z, u ∈ RN, F : Z × RN → R, F t, x is continuously differentiable in x for every t ∈ Z and is T -periodic in t; T is a positive integer
Difference equations provide a natural description of many discrete models in real world
Summary
Consider the following discrete Hamiltonian system: Δ2u t − 1 ∇F t, u t 0, t ∈ Z, 1.1 where Δ is the forward difference operator defined by Δu t u t 1 − u t , Δ2u t Δ Δu t , t ∈ Z, u ∈ RN, F : Z × RN → R, and F t, x is continuously differentiable in x for every t ∈ Z and is T -periodic in t; T is a positive integer. 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. 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. Suppose that F satisfies the following conditions: F1 there exists a positive constant T such that F t T, x F t, x for all t, x ∈ Z × RN; F2 there are constants L1 > 0, L2 > 0, and 0 ≤ α < 1 such that. 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
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