Oscillation Criteria for Some New Generalized Emden-Fowler Dynamic Equations on Time Scales

نویسندگان

  • Haidong Liu
  • Puchen Liu
  • Patricia J. Y. Wong
چکیده

and Applied Analysis 3 Theorem 4. Assume that (H 1 )–(H 6 ) and (7) hold. If there exists a function ξ(t) ∈ C rd(T , (0,∞)) such that for any positive numberM, lim t→∞ ∫ t t0 (ξ (s) p (s) − Q (s)) Δs = ∞, (12) where p (s) = q (s) [1 − p (δ (s))] β , Q (s) = αM(R (σ (s))) α−β r (δ (s)) ((ξ Δ (s)) + ) α+1 (α + 1) α+1 βξ (s) (δ (s)) α , (ξ Δ (s)) + := max {ξΔ (s) , 0} , (13) then (1) is oscillatory. Proof. Suppose that (1) has a nonoscillatory solution x(t), then there exists T 0 ⩾ t 0 such that x(t) ̸ = 0 for all t ⩾ T 0 . Without loss of generality, we assume that x(t) > 0, x(τ(t)) > 0 and x(δ(t)) > 0 for t ⩾ T 0 , because a similar analysis holds for x(t) < 0, x(τ(t)) < 0 and x(δ(t)) < 0. Then the following are deduced from (1), (H 3 ), and (H 6 ): Z (t) ⩾ x (t) > 0 for t ⩾ T0, (r (t) 󵄨󵄨󵄨󵄨 Z Δ (t) 󵄨󵄨󵄨󵄨 α−1 Z Δ (t)) Δ ⩽ 0, t ⩾ T 0 . (14) Therefore r(t)|Z(t)|Z(t) is a nonincreasing function and Z(t) is eventually of one sign. We claim that Z Δ (t) > 0 or ZΔ (t) = 0, t ⩾ T0. (15) Otherwise, if there exists a t 1 ⩾ T 0 such that Z(t) < 0 for t ⩾ t 1 , then from (14), for some positive constant K, we have −r (t) (−Z Δ (t)) α ⩽ −K, t ⩾ t 1 , (16) that is, −Z Δ (t) ⩾ ( K r (t) ) 1/α , t ⩾ t 1 , (17) integrating the above inequality from t 1 to t, we have Z (t) ⩽ Z (t1) − K 1/α (R (t) − R (t1)) . (18) Letting t → ∞, from (7), we get lim t→∞ Z(t) = −∞, which contradicts (14). Thus, we have proved (15). We choose some T 1 ⩾ T 0 such that δ(t) ⩾ T 0 for t ⩾ T 1 . Therefore from (14), (15), and the fact δ(t) ⩽ σ(t), we have that r (σ (t)) (Z Δ (σ (t))) α ⩽ r (δ (t)) (Z Δ (δ (t))) α , t ⩾ T 1 , (19) which follows that Z Δ (δ (t)) ⩾ Z Δ (σ (t)) ( r (σ (t)) r (δ (t)) ) 1/α , t ⩾ T 1 . (20) On the other hand, from (1), (H 6 ), and (15), we have (r (t) (Z Δ (t)) α ) Δ + q (t) (Z (δ (t)) − p (δ (t)) x (τ (δ (t)))) β ⩽ 0, t ⩾ T 1 . (21) Noticing (15) and the fact Z(t) ⩾ x(t), we get (r (t) (Z Δ (t)) α ) Δ + p (t) Z β (δ (t)) ⩽ 0, t ⩾ T1, (22) where p(t) = q(t)[1 − p(δ(t))]. Define w (t) = ξ (t) r (t) (Z Δ (t)) α Z (δ (t)) , for t ⩾ T 1 . (23) Obviously, w(t) > 0. By (22), (23) and the product rule and the quotient rule, we obtain w Δ (t) = ξ (t) Z (δ (t)) (r (t) (Z Δ (t)) α ) Δ + r (σ (t)) (Z Δ (σ (t))) α × ξ (t) Z β (δ (t)) − ξ (t) (Z β (δ (t))) Δ Z (δ (t)) Z (δ (σ (t))) ⩽ − ξ (t) p (t) + ξ (t) ξ (σ (t)) w (σ (t)) − r (σ (t)) (Z Δ (σ (t))) α ξ (t) (Z β (δ (t))) Δ Z (δ (t)) Z (δ (σ (t))) . (24) Now we consider the following two cases. Case 1. Let β ⩾ 1. By (15), Lemmas 1 and 2, we have

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تاریخ انتشار 2014