Oscillation of solutions of third order nonlinear neutral differential equations

where t ≥ t0 > 0, z(t) = x(t)+p(t)x(τ (t)), and α is a ratio of odd positive integers. We assume that the following conditions hold: (H1) r(t), p(t), q(t), τ (t),σ (t) ∈ C([t0,∞)), r(t), q(t) are positive and 0 ≤ p(t) ≤ p0 < ∞; (H2) limt→∞ τ (t) = limt→∞ σ (t) = ∞, σ (t) > 0, and τ (t) ≤ t; (H3) f (u) ∈ C(R) and there exists a positive constant k such that f (u)/u ≥ k for all u = 0 and γ is a ratio of odd positive integers; (H4) τ ′(t) ≥ τ0 > 0 and τ ◦ σ = σ ◦ τ . By a solution of (1.1), we mean a nontrivial function x(t) ∈ C([Tx,∞)), Tx ≥ t0, which has the properties z(t) ∈ C2([Tx,∞)), r(t)(z′′(t))α ∈ C1([Tx,∞)) and satisfies (1.1) on [Tx,∞). Our attention is restricted to those solutions x(t) of (1.1) satisfying sup{|x(t)| : t ≥ T} > 0 for all T ≥ Tx. We assume that (1.1) possesses such a solution. A solution of (1.1) is called oscillatory if it has arbitrarily large zeros on [Tx,∞); otherwise, it is termed nonoscillatory. Equation (1.1) is said to be oscillatory if all its solutions are oscillatory.


Some preliminaries
We enlist some known results which will be needed. We first present the following classes of nonoscillatory (let us say positive) solutions of (1.1): The following lemma comes directly from combining Lemma 1 and Lemma 2 in [13] with Lemma 3 and Lemma 4 in [20].

Lemma 2.1
Assume that A ≥ 0 and B ≥ 0. Then
Proof Assume that x(t) is a positive solution of Eq. (1.1) satisfying z(t) ∈ N II for t ≥ t 1 .
Going through as in the proof of Theorem 3.1, we arrive at (3.5). In the following, we consider the two cases γ > 1 and γ ≤ 1. Firstly, assume that γ > 1. Then we have (3.6).
Since z(t) is positive and decreasing, we have lim t→∞ z(t) = l ≥ 0 exists. We claim that l = 0. If not, then there exists t 3 ≥ t 2 such that z(σ (t)) > l for t ≥ t 3 . Substituting into (3.6), we get Integrating (3.30) from t 3 to t and taking into account (3.28), we have which is a contradiction. Thus l = 0 and consequently lim t→∞ x(t) = 0. In the following, we obtain the same conclusion in the case when But since τ (t) ≤ t, then we can observe that r(τ (t))(z (τ (t))) α ≥ r(t)(z (t)) α and consequently we have Integrating from t to ∞ followed by integrating from t 3 to ∞, we obtain which contradicts (3.29). Thus lim t→∞ x(t) = 0. Secondly, assume that γ ≤ 1. As in the proof of Theorem 3.1, we have (3.27). By completing the proof as in the above case of γ > 1, using (3.27) instead of (3.6), the proof is completed. Proof Let x(t) be an eventually positive solution of Eq. (1.1) satisfying z(t) ∈ N III for all t ≥ t 1 ≥ t 0 . Since z (t) < 0 and z (t) > 0, then by Lemma 2.5, there exist t 2 ≥ t 1 and a constant k 1 satisfying 0 < k 1 < 1 such that z(t) ≥ k 1 tz (t) for t ≥ t 2 , i.e., Going through as in Theorem 3.1, we arrive at (3.5). In the following, we consider the two cases γ > 1 and γ ≤ 1. Firstly, assume that γ > 1. Then we have (3.6), and using (3.32) we get But since v(t) = -r 1 α (t)z (t) is positive and increasing, then there exists a constant g 1 (3.34) Substituting into (3.33) and integrating from t 3 to t, we get But since τ (t) ≤ t, then we can conclude that r(τ (t))(z (τ (t))) α ≥ r(t)(z (t)) α . Now since from (3.35) we have Then integrating from t 4 (≥ t 3 ) to t, we get which contradicts (3.31). Secondly, assume that γ ≤ 1. As in the proof of Theorem 3.1, we arrive at (3.27), and then using (3.32) we get Going through as in the proof of the case γ > 1, using (3.36) instead of (3.33), this completes the proof.
The following results are immediate consequences of Lemma 2.4, Lemma 3.1, Theorem 3.1, and Theorem 3.2. The following results deal with the special case α ≤ 1 and γ ≥ 1 of Eq. (1.1).

Theorem 3.5 Assume that conditions (H 1 )-(H 4 ), α ≤ 1, and γ ≥ 1 hold. If there exists a positive function ρ(t)
holds for any positive constants k, M, sufficiently large t 1 ≥ t 0 , and for some t defined by (3.3) and then there exists no positive solution x(t) of Eq. (1.1) satisfying z(t) ∈ N I .

Proof
Assume that x(t) is an eventually positive solution of Eq. (1.1) satisfying z(t) ∈ N I . As in the proof of Theorem 3.1, we arrive at (3.6). Now define the function W (t) by Then W (t) > 0 for t ≥ t 1 and Since z (t) and z (t) are positive, then there exist t 2 ≥ t 1 and constant M > 0 such that z (t) ≥ M for all t ≥ t 2 . Now, from (3.38) and (2.3), we get This with (3.39) yields As we did for W , we can get But since z is increasing and τ (t) ≤ t, then This with (3.41) leads to Thus, by (3.6) and (2.4), we get Now, we consider the two cases σ (t) < t and σ (t) ≥ t. First assume that σ (t) < t. As in the proof of Theorem 3.1, we get (3.23). Substituting into (3.44), we have Secondly, assume that σ (t) ≥ t. Since z (t) > 0, it follows from (3.44) that As in the proof of Theorem 3.1, we arrive at (3.21). Then, substituting into (3.46), we have This with (3.45) yields Integrating from t 4 (> t 3 ) to t, we get This contradicts (3.37) and completes the proof.