Oscillation Criteria for a Class of Third-Order Damped Neutral Differential Equations

: In this paper, we study the asymptotic and oscillatory properties of a certain class of third-order neutral delay differential equations with middle term. We obtain new characterizations of oscillation of the third-order neutral equation in terms of oscillation of a related, well-studied, second-order linear equation without damping. An Example is provided to illustrate the main results.

Moreover, throughout our results, we need an assumption: (A) There exists a nonoscillatory a solution of A solution x of (1) is said to be oscillatory if it has arbitrarily large zeros and otherwise, it is called nonoscillatory. Equation (1) is said to be oscillatory if all its solutions are oscillatory.
Recently, it is easy to notice the growing interest in studying the qualitative properties of solutions to fractional/functional differential and difference equations, see [1][2][3][4]. The third-order differential equations have an important applications in many problems for instance, economy, physics, biology and population dynamics, see [5][6][7]. Although importance of those kind of equations in applications they had been realized very early.
For the sake of brevity, we define the operators From Equation (1) and assumption for f (x), we obtain the inequality Through this paper, we will use the following notation: (r 2 (s)) −1 ds, Lemma 1. [23] Assume that c 1 , c 2 ∈ [0, ∞) and γ > 0. Then

Results and Proofs
Lemma 2. Assume that (A) holds. If x is a nonoscillatory solution of (1), then there are two possible classes for y : Proof. Let x be a positive solution of (1). Then there exists t 1 ≥ t 0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0. By (3), it is easy to see that where z (t) = −£ 1 y (t) . Let u(t) > 0 be a solution of (2) for t ≥ t 1 ≥ t 0 . Assume that z > 0 is oscillatory. Hence z has consecutive zeros at a and b (t This contradiction completes the proof. Lemma 3. Assume x is a nonoscillatory solution of (1) with y ∈ N 1 . Then and y (t) ≥ η 2 (t, t 1 ) (£ 2 y (t)) 1/α .
Proof. Let x be a positive solution and y ∈ N 1 be a solution of (1). Then there exists t 1 ≥ t 0 such that x (σ (t)) > 0 and x (g (t)) > 0. From (1), we see that £ 3 y (t) ≤ 0. Thus That is, Now, integrating (7) from t 1 to t, we get Thus, the proof is complete.
Lemma 4. Assume x is nonoscillatory solution of (1) with y ∈ N 2 . Then The proof of the lemma is complete.

Lemma 5.
Assume that x is a positive a solution and y ∈ N 1 and p (t) ∈ (0, 1) . Then where Proof. Let x be a positive a solution of (1) and y ∈ N 1 . Then there exists t 1 ≥ t 0 such that x (σ (t)) > 0 and x (g (t)) > 0. The corresponding y(t) satisfies That is Combining (3) and (10), we have The proof of the lemma completed.
Lemma 6. Assume x is a positive a solution of (1) and y ∈ N 2 . Then Proof. Let x be a positive a solution of (1) and y ∈ N 2 . Then there exists t 1 ≥ t 0 such that x (t) > 0, x (τ (t)) > 0 and x (σ (t)) > 0. From Lemma 1, we obtain Now, from (3) and (I 3 ), we have Combining (3) along with (13), we get By virtue of (12) and using £ 3 y (t) ≤ 0, we have The proof of the lemma completed.
Proof. Let x be a positive a solution of (1) and y ∈ N 1 . Then there exists t 1 ≥ t 0 such that x (σ (t)) > 0 and x (g (t)) > 0. By (1), we see that £ 3 y (t) ≤ 0 for t 2 ≥ t 1 , that is £ 2 y (t) > 0 otherwise lim t→∞ £ 1 y (t) = −∞, a contradiction. Define a positive function by Using (7), we have Also by (6), it is easy to see that Now, by differentiating (15), we get Using (15) and (9), we obtain It follows from (15) and (5) that From (16), we get By (17), we have Applying the inequality Thus, Integrating (20) from t 2 to t, we obtain The proof of the lemma is complete.
We present the following theorem.

Theorem 2. Assume (A) holds. If every a solution of the first-order equation
or £ 2 y (t) is oscillatory, then N 1 = ∅.
In view of [24](Theorem 1), we see that the first-order delay differential Equation (21) has a positive a solution, a contradiction. Then, the proof is complete. Proof. Let x be a positive a solution of (1) and y ∈ N 1 , there exists t 1 ≥ t 0 such that x (σ (t)) > 0 and x (g (t)) > 0. Set χ = £ 1 y (t) in (9), we see that In view of [19] (Lemma 2.6), (2) has positive a solution, a contradiction. Then, the proof is complete. Now, we can extend Theorem 2 to where

Theorem 4.
If every solution of the first-order equation or y is oscillatory, then N 1 = ∅.
Proof. Let x be a positive solution of (1) and y ∈ N 1 . Then there exists t 1 ≥ t 0 such that x (σ (t)) > 0 and x (g (t)) > 0. Now, we can easily extend Lemma 5 to the equation Set χ = £ 1 y (t), we see that In view of [19] (Lemma 2.6), (25) has a positive solution, a contradiction. Then, the proof is complete.
Theorem 5. Assume (A) holds and α ≥ 1, and there exists a function h ∈ C 1 (I, R) such that holds with where c is positive constant, then N 2 = ∅ or £ 2 y (t) is oscillatory.
By Integrating (28) from h(t) ≥ t 1 to t, we get which contradicts (26). The proof is complete.
holds with Q (t) defined as in Theorem 5, then N 2 = ∅ or £ 2 y (t) is oscillatory.
Proof. Let x be a positive solution of (1) and y ∈ N 2 . Then there exists t 1 ≥ t 0 such that x (σ (t)) > 0 and x (g (t)) > 0. As in Theorem 5 ,we obtain (28) and Integrating (28) from u to t, we get Integrating from h (t) to t, we have which contradicts (30). The proof is complete.
If (31) hold, then it is clear that all conditions of Theorem 5 are satisfied, and hence every solution of (1), or y (t), is oscillatory.