Nonexistence of Homoclinic Solutions for a Class of Discrete Hamiltonian Systems
نویسنده
چکیده
and Applied Analysis 3 From (A0), (17), (19), (20), and the first equation of system (6), we have x (n+1) = n ∑ τ=−∞ β (τ) y (τ) μ−2 y (τ) n ∏ s=τ [1−α (s)] −1 , ∀n∈Z, (21) x (n+1) =− +∞ ∑ τ=n+1 β (τ) y (τ) μ−2 y (τ) τ−1 ∏ s=n+1 [1−α (s)] , ∀n∈Z. (22) Combining (19) with (21), one has |x (n + 1)| ν = n ∑ τ=−∞ β (τ) y (τ) μ−2 y (τ) n ∏ s=τ [1 − α (s)] −1 ν ≤ ζ (n) n ∑ τ=−∞ β (τ) y (τ) μ , ∀n ∈ Z. (23) Similarly, it follows from (20) and (22) that |x (n + 1)| ν = +∞ ∑ τ=n+1 β (τ) y (τ) μ−2 y (τ) τ−1 ∏ s=n+1 [1 − α (s)] ν ≤ η (n) +∞ ∑ τ=n+1 β (τ) y (τ) μ , ∀n ∈ Z. (24) Combining (23) with (24), one has |x (n + 1)| ν ≤ ζ (n) η (n) ζ (n) + η (n) +∞ ∑ τ=−∞ β (τ) y (τ) μ , ∀n ∈ Z. (25) Now, it follows from (16), (18), and (25) that +∞ ∑ n=−∞ γ + (n) |x (n + 1)| ν ≤ [ +∞ ∑ n=−∞ ζ (n) η (n) ζ (n) + η (n) γ + (n)] × +∞ ∑ n=−∞ β (n) y (n) μ < +∞. (26)
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