Bounded Oscillation Of Higher Order Neutral Differential Equations ...

130 Oscillation of Neutral Differential EquationsSince n is odd, from (H5) and (20) follows that we can apply Lemma 2. Using Lemma2, we haveHence,z (σ (t)) = z (t − (t − σ (t))) ≥[t − σ (t)]n−1z (n−1) (t) , t ≥ t 3 + σ 0 .(n − 1)![t − σ (t)]n−1z (σ (t)) ≥ z (n−1) (t) , t ≥ t 3 + σ 0 . (21)(n − 1)!Since x(t) isbounded,by(H1)itfollowsthatlim t→∞ p (t) x(τ(t)) = 0 . Then usingthis fact with lim t→∞ r (t) = 0, and by (9) there exists a t 4 ≥ t 3 such thatx (t) ≥ λz (t) , t ≥ t 4, (22)where λ is some number in (0, 1).Let t 5 ≥ max{t 4 ,t 3 + σ 0 } be such that σ (t) ≥ t 5 for all t ≥ t 5 . From (21), and (22)we havex (σ (t))[t − σ (t)] n−1 ≥ λz(n−1) (t),(n − 1)!t ≥ t 5 . (23)Using (17), and (23), then it follows from (10) that z (n) (t)+φ (t) w λcz (n−1) (t) ≤ 0, (24)where c = 1omit it.(n−1)!. The remainder of the proof is similar to that of Theorem 1 and we4 ExamplesWe consider two examples.EXAMPLE 1. Consider the equationx(t)+4e − t+π2 sin t t − π2 x + e t−π62 4+2e −56 t cos 2 3 t x 1 3 t − π 2= −4e − t 16 cos 3 t.By setting w(x) =x 1 3 , φ(t) = t − π 132e t−π6 4+2e −56 t cos 2 3 t ,wecanseethatallconditions of Theorem 1 are satisfied. Thus every bounded solution of the aboveequation is oscillatory. In fact, x(t) =e −t sin t is such a solution.EXAMPLE 2. Consider the equationx(t)+4e − t+π2 sin t t − π2 x 2= −4e − t 6 sin13 t − 2e −t cos t. + e t−2π64 − 2e −56 t sin 2 3 tx 1 3 (t − π)