Bounded Oscillation Of Higher Order Neutral Differential Equations ...

A. Zein and T. Abu-Kaff 129Setting u (t) =λcz (n−1) (t),and integrating (14) divided by w (u (t)) from t 7 to t, weobtain t u(t7 )dsλc φ (v) dv ≤t 7 u(t) w (s) . (15)Since u (t) < 0, u (t) is decreasing. And since u (t) > 0, it follows that lim t→∞ u (t) =L ≥ 0. If L = 0, then by (15) we must have ∞φ (t) dt < ∞, (16)t 7which is contrary to (8). In the case when L = 0, letting t →∞in (15) and using (7),we again obtain (16). The proof is complete.In the next theorem besides conditions (H1)-(H4) we further assume that:(H5) 0 0fort>0 is continuous and nondecreasing on R + with:|y||f (t, x, y)| ≥ φ(t)w[t − σ (t)] n−1 (17)andIf n is odd and ±α0dx< ∞ for every α > 0. (18)w (x) ∞φ(t)dt = ∞, (19)then every bounded solution x(t) of equation (1) is oscillatory.PROOF. Assume that equation (1) has a bounded non-oscillatory solution x(t).Without loss of generality, we may assume that x(t) is eventually positive (the proofis similar when x(t) is eventually negative). That is, let x(t) > 0, x(τ(t)) > 0andx(σ(t)) > 0fort ≥ t 0 ≥ 0. Set z (t) as in (9). Then from (1) and (9) we have (10). From(10) z (n) (t) < 0. It follows that z (i) (t) (i =0, 1, ..., n − 1) is strictly monotonic and ofconstant sign eventually. Since p (t) andr (t) are oscillating functions, lim t→∞ p (t) =lim t→∞ r (t) =0, and x(t) is bounded, there exists a t 1 ≥ t 0 such that z(t) > 0fort ≥ t 1 . Since x (t) is bounded, and by using lim t→∞ p (t) = lim t→∞ r (t) =0,thenitfollows from (9) that there is a t 2 ≥ t 1 ,suchthatz (t) is also bounded for t ≥ t 2 .Now, by applying Lemma 1, there exist a t 3 ≥ t 2 and an integer l with n − l oddsuch that (2), and (3) are satisfied by z (t) fort ≥ t 3 . Since n is odd and z (t) isboundedthen l =0(otherwisez (t) is not bounded). Hence from relations (2), and (3) we have(−1) k z (k) (t) > 0, k =0, 1, ..., n − 1. (20)

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)