Asymptotic Dichotomy in a Class of Fourth-Order Nonlinear Delay Differential Equations with Damping

and Applied Analysis 3 here g t ≤ t∗ ≤ t.Hence x l ( g t ) ≥ x n−1 t∗ n − l − 1 ! t − g t n−l−1 ≥ x n−1 t n − l − 1 ! t − g t n−l−1. 1.7 Combining 1.7 and 1.5 , we see that x′ ( g t ) ≥ g l−1 t 2l−1 l − 1 ! n − l − 1 ! n−1 t t − g t n−l−1 1.8 for all large t. Lemma 1.2. Suppose the linear third-order differential equation z′′′ p t z 0, t ∈ I 1.9 has an eventually positive increasing solution on I and y is a nonoscillatory solution of 1.1 on I. Then there exists t0 ∈ a,∞ such that y t y′ t > 0 or y t y′ t < 0 for t ≥ t0. Proof. Without loss of generality, we may assume that y t > 0 and y g t > 0 for t ≥ a. Then x t −y′ t is a solution of the third-order delay differential equation x′′′ p t x q t f ( y ( g t )) , t ≥ a. 1.10 We claim that, all solutions of 1.10 are nonoscillatory so that y′ t is eventually positive or eventually negative . To see this, let z be the solution of 1.9 such that z t > 0 and z′ t > 0 for t ≥ a. Then from Lemma 1.1, we have z′′ t > 0 for t greater than or equal to a positive number t1. Suppose to the contrary that x is an oscillatory solution of 1.10 . We assert that s x′z − xz′ oscillates. Indeed, since x is oscillatory, for any T > a, there are b and c such that T ≤ b < c, x b x c 0, x′ b ≥ 0 and x′ c ≤ 0. Hence s b x′ b z b ≥ 0 and s c x′ c z c ≤ 0. If s b 0 or s c 0, then either b or c is a zero of s in T,∞ . Otherwise, s b > 0 and s c < 0 so that there is d ∈ b, c such that s d 0. These show that s has arbitrarily large zeros. Now that s is oscillatory, either s t ≤ 0 for all t ∈ t1,∞ or else there are strictly increasing and divergent sequences {bn} and {cn} such that s t > 0 for t ∈ bn, cn ⊂ t1,∞ , s bn s cn 0, s′ bn ≥ 0 and s′ cn ≤ 0. Suppose the former case holds. Then there are b, c ⊂ t1,∞ such that s t ≤ 0 for t ∈ b, c and s′ b ≥ 0 and s′ c ≤ 0. From 1.9 and 1.10 , we have zq t f ( y ( g t )) x′′′z − z′′′x s′′ − z ′ z s′ z′′ z s, 1.11 or, ( s′ z )′ q t f ( y ( g t )) − z ′′s z2 . 1.12 4 Abstract and Applied Analysis Hence we have 0 ≥ ∫ c b ( s′ t z t )′ dt ∫ c b { q t f ( y ( g t )) − z ′′ t s t z2 t } dt > 0, 1.13 which is a contradiction. Suppose the latter case holds. From 1.9 and 1.10 , we have x′′′z′ − z′′′x′ p t xz′ − x′z z′q t fyg t , 1.14