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Article

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

by
Elmetwally M. Elabbasy
1,
1,2,
3,4,5,* and
Osama Moaaz
1,*
1
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt
2
Department of Mathematics, Faculty of Science, Amran University, Amran 999101, Yemen
3
Department of Mathematics and General Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia
4
Department of Medical Research, China Medical University, Taichung 40402, Taiwan
5
Department of Computer Science and Information Engineering, Asia University, Taichung 40402, Taiwan
*
Authors to whom correspondence should be addressed.
Symmetry 2020, 12(12), 1988; https://doi.org/10.3390/sym12121988
Submission received: 28 September 2020 / Revised: 25 November 2020 / Accepted: 27 November 2020 / Published: 1 December 2020
(This article belongs to the Special Issue Differential/Difference Equations and Its Application)

## Abstract

:
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.

## 1. Introduction

In this paper, we consider the third-order nonlinear damped neutral differential equation of the form
$r 2 r 1 y ′ α ′ ′ t + b t y ′ t α + q t f x σ t = 0 , for t ≥ t 0 ,$
where $y t = x t + p t x τ t$, $α$ is a ratio of positive odd integers and $f ∈ C R , R$ satisfies
$f x ≥ k x α for x ≠ 0 .$
Throughout this paper, we assume the following conditions:
(I1)
$r 1$, $r 2 ∈ C t 0 , ∞ , 0 , ∞$
$∫ t 1 ∞ r 1 − 1 / α s d s = ∞ and ∫ t 1 ∞ r 2 − 1 s d s = ∞ , t 1 ≥ t 0 , t 1 ∈ t 0 , ∞ ;$
(I2)
$p , q ∈ C t 0 , ∞ , 0 , ∞ , p t ≤ p 0 < ∞ , q$ does not vanish identically;
(I3)
$σ , τ ∈ C 1 t 0 , ∞ , R ,$$σ t < t , τ t < t , τ ′ ≥ τ 0 , σ ∘ τ = τ ∘ σ$ and $lim t → ∞ σ t = lim t → ∞ τ t = ∞ .$
A solution of (1), we mean $x ∈ C [ T x , ∞ ) , [ 0 , R )$ with $T x ≥ t 0$, which satisfies the property $y ′ , r 2 r 1 y ′ α ′ ∈ C 1 [ T x , ∞ ) , R )$ and moreover satisfies (1) on $[ T x , ∞ ) .$ We consider the nontrivial solutions of (1) existing on some half-line $[ T x , ∞ )$ and satisfying the condition $sup { x t : T ≤ t < ∞ } > 0$ for any $T ≥ T x$.
Moreover, throughout our results, we need an assumption:
(A)
There exists a nonoscillatory a solution of
$r 2 z ′ ′ t + p t r 1 t z t = 0 .$
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.
In the last three decades, a few results asymptotic behavior of oscillation of third-order have been studied in the literatures. But even-order differential equations have been deeply studied, see [8,9,10,11,12,13,14]. In the early twentieth century, it have been appeared the basic interested paper in asymptotic behavior of third-order differential equations [15]. Recently, a study has developed, especially oscillatory or nonoscillatory of solutions studying by various techniques, see [16,17,18,19,20,21,22,23].
For the sake of brevity, we define the operators
$£ 1 y t : = r 1 y ′ α , £ 2 y t : = r 2 r 1 y ′ α ′ , £ 3 y t : = r 2 r 1 y ′ α ′ ′ .$
From Equation (1) and assumption for $f x$, we obtain the inequality
$£ 3 y t + b t r 1 t £ 1 y t + k q t x α σ t ≤ 0 .$
Through this paper, we will use the following notation:
$η 1 t , t 1 : = ∫ t 1 t r 1 s − 1 α d s , η 2 t , t 1 : = ∫ t 1 t r 2 s − 1 d s ,$
$η ˜ 2 t , t 1 : = η 2 t , t 1 r 1 t 1 / α , η ^ 2 t , t 1 : = ∫ t 1 t η ˜ 2 s , t 1 d s .$
Lemma 1.
[23] Assume that $c 1 , c 2 ∈ [ 0 , ∞ )$ and $γ > 0$. Then
$c 1 + c 2 γ ≤ μ c 1 γ + c 2 γ ,$
where
$μ : = 1 i f γ ≤ 1 2 γ − 1 i f γ > 1 .$

## 2. Results and Proofs

Lemma 2.
Assume that (A) holds. If x is a nonoscillatory solution of (1), then there are two possible classes for $y :$
$N 1 = y t : £ 1 y t > 0 , £ 2 y t > 0 ; N 2 = y t : £ 1 y t < 0 , £ 2 y t > 0 .$
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
$r 2 z ′ t ′ t + b t r 1 t z t > 0 ,$
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 $b ( t 1 < a < b )$ such that $z ′ ( a ) ≥ 0 , z ′ ( b ) ≤ 0$ for $t ∈ ( a , b )$. This implies that
$0 < ∫ a b r 2 z ′ t ′ t + b t r 1 t z t u t d t = r 2 t z ′ t u t a b − ∫ a b r 2 t z ′ t u ′ t d t + ∫ a b b t r 1 t z t u t d t = r 2 t z ′ t u t a b − r 2 t z t u ′ t | a b + ∫ a b r 2 t u ′ t ′ z t d t + ∫ a b b t r 1 t z t u t d t = r 2 t z ′ t u t a b + ∫ a b r 2 t u ′ t ′ + b t r 1 t u t z t d t = r 2 t z ′ t u t a b ≤ 0 .$
This contradiction completes the proof. □
Lemma 3.
Assume x is a nonoscillatory solution of (1) with $y ∈ N 1 .$ Then
$£ 1 y t ≥ η 2 t , t 1 £ 2 y t$
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
$£ 1 y t = ∫ t 1 t £ 1 y s ′ d s + £ 1 y t 1 ≥ ∫ t 1 t 1 r 2 s £ 2 y s d s ≥ £ 2 y t ∫ t 1 t 1 r 2 s d s = η 2 t , t 1 £ 2 y t .$
That is,
$y ′ t ≥ η 2 t , t 1 r 1 t 1 / α £ 2 y t 1 / α = η ˜ 2 t , t 1 £ 2 y t 1 / α .$
Now, integrating (7) from $t 1$ to t, we get
$y t ≥ ∫ t 1 t η ˜ 2 s , t 1 £ 2 y s 1 / α d s ≥ £ 2 y t 1 / α ∫ t 1 t η ˜ 2 s , t 1 d s = η ^ 2 t , t 1 £ 2 y t 1 / α .$
Thus, the proof is complete. □
Lemma 4.
Assume x is nonoscillatory solution of (1) with $y ∈ N 2$. Then
$y u ≥ − £ 1 y v 1 / α η 1 v , u , f o r v ≥ u ≥ t .$
Proof.
Since $£ 1 y t$ is nondecreasing, for $v ≥ u ≥ t ,$ we have
$y u = y v − ∫ u v 1 r 1 1 α ϑ £ 1 y ϑ 1 / α d ϑ ≥ − £ 1 y v 1 / α ∫ u v 1 r 1 1 α ϑ d ϑ = − £ 1 y v 1 / α η 1 v , u ,$
i.e.,
$y u ≥ − £ 1 y v 1 / α η 1 v , u .$
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
$£ 3 y t + b t r 1 t £ 1 y t + k Q t y α σ t ≤ 0 ,$
where
$Q t = q t 1 − p σ t α .$
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
$x t = y t − p t x τ t ≥ y t − p t y τ t ≥ y t − p t y t = y t 1 − p t .$
That is
$x α σ t ≥ y α σ t 1 − p σ t α .$
Combining (3) and (10), we have
$£ 3 y t + b t r 1 t £ 1 y t + k Q t y α σ t ≤ 0 .$
The proof of the lemma completed. □
Lemma 6.
Assume x is a positive a solution of (1) and $y ∈ N 2$. Then
$1 + p 0 α τ 0 £ 3 y t + b t r 1 t £ 1 y t + p 0 α τ 0 1 α b τ t r 1 τ t £ 1 y τ t + k μ q ˜ t y α σ t ≤ 0 ,$
where $q ˜ t : = min q t , q τ t .$
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
$y α t ≤ μ x α t + p 0 α x α σ t .$
Now, from (3) and (I$3$), we have
$0 ≥ p 0 α τ 0 £ 3 y τ t + p 0 α b τ t τ 0 1 α r 1 τ t £ 1 y τ t + p 0 α k q τ t x α τ σ t .$
Combining (3) along with (13), we get
$0 ≥ £ 3 y t + p 0 α τ 0 £ 3 y τ t + b t r 1 t £ 1 y t + p 0 α τ 0 1 α b τ t r 1 τ t £ 1 y τ t + k q ˜ t x α σ t + p 0 α x α τ σ t .$
By virtue of (12) and using $£ 3 y t ≤ 0$, we have
$0 ≥ 1 + p 0 α τ 0 £ 3 y t + b t r 1 t £ 1 y t + p 0 α τ 0 1 α b τ t r 1 τ t £ 1 y τ t + k μ q ˜ t y α σ t .$
The proof of the lemma completed. □
Theorem 1.
Assume (A) holds, $α ≥ 1$ and $σ ′ t > 0 .$ If there exist a function $δ t ∈ C 1 t 0 , ∞ , 0 , ∞ ,$ for all sufficiently large $t 1 ≥ t 0$, there is a $t 2 ≥ t 1$ such that
$lim sup t → ∞ ∫ t 1 t k μ Q t δ s − δ ′ t r 1 t − δ t b t 2 4 α δ t r 1 t 2 σ ′ t η ˜ 2 σ t , t 1 η ^ 2 σ t , t 1 α − 1 d s = ∞ ,$
then, $N 1 = ∅ .$
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 $£ 3 y t ≤ 0$
$£ 1 y t = ∫ t 2 t £ 1 y s ′ d s + £ 1 y t 2 ≤ ∫ t 2 t 1 r 2 s £ 2 y s d s + £ 1 y t 2 ≤ £ 2 y t ∫ t 2 t 1 r 2 s d s = £ 1 y t 2 + η 2 t , t 2 £ 2 y t ,$
for $t 2 ≥ t 1 ,$ that is $£ 2 y t > 0$ otherwise $lim t → ∞ £ 1 y t = − ∞ ,$ a contradiction. Define a positive function by
$w t = δ t £ 2 y t y α σ t .$
Using (7), we have
$y ′ σ t ≥ η 2 σ t , t 1 r 1 σ t 1 / α £ 2 y σ t 1 / α ≥ η 2 σ t , t 1 r 1 σ t 1 / α £ 2 y t 1 / α ,$
hence,
$y ′ σ t y σ t ≥ η 2 σ t , t 1 r 1 σ t δ t 1 / α δ 1 / α t £ 2 y t 1 / α y σ t = η 2 σ t , t 1 δ t r 1 σ t 1 / α w 1 / α t .$
Also by (6), it is easy to see that
$w t = δ t £ 2 y t y α σ t ≤ δ t £ 2 y σ t y α σ t ≤ δ t η ^ 2 σ t , t 1 α ,$
hence
$w t 1 / α − 1 ≤ δ t 1 / α − 1 η ^ 2 σ t , t 1 1 − α .$
Now, by differentiating (15), we get
$w ′ t = δ ′ t £ 2 y t y α σ t + δ t £ 3 y t y α σ t − α δ t y α − 1 σ t y ′ σ t σ ′ t £ 2 y t y 2 α σ t .$
Using (15) and (9), we obtain
$w ′ t = δ ′ t δ t w t + £ 3 y t £ 2 y t w t − α σ ′ t y ′ σ t y σ t w t ≤ δ ′ t δ t w t − b t r 1 t £ 1 y t − k Q t y α σ t £ 2 y t w t − α σ ′ t y ′ σ t y σ t w t .$
It follows from (15) and (5) that
$w ′ t ≤ δ ′ t δ t w t − b t r 1 t η 2 t , t 1 w t − k Q t δ t − α σ ′ t y ′ σ t y σ t w t ≤ δ ′ t δ t − b t r 1 t η 2 t , t 1 w t − k Q t δ t − α σ ′ t y ′ σ t y σ t w t .$
From (16), we get
$w ′ t ≤ δ ′ t δ t − b t r 1 t η 2 t , t 1 w t − k Q t δ t − α σ ′ t η 2 σ t , t 1 δ t r 1 σ t 1 / α w 1 / α − 1 t w 2 t .$
By (17), we have
$w ′ t ≤ − k Q t δ t + δ ′ t δ t − b t r 1 t η 2 t , t 1 w t − α σ ′ t δ t η ˜ 2 σ t , t 1 η ^ 2 σ t , t 1 α − 1 w 2 .$
Applying the inequality
$A u − B u 2 ≤ A 2 4 B ,$
with
$A = δ ′ t δ t − b t r 1 t η 2 t , t 1 , B = α σ ′ t δ t η ˜ 2 σ t , t 1 η ^ 2 σ t , t 1 α − 1 .$
Thus,
$w ′ t ≤ − k μ Q t δ t + δ ′ t r 1 t − δ t b t 2 δ t 2 r 1 t 2 × δ t 4 α σ ′ t η ˜ 2 σ t , t 1 η ^ 2 σ t , t 1 α − 1 ,$
that is
$w ′ t ≤ − k μ Q t δ t + δ ′ t r 1 t − δ t b t 2 4 α δ t r 1 t 2 σ ′ t η ˜ 2 σ t , t 1 η ^ 2 σ t , t 1 α − 1 .$
Integrating (20) from $t 2$ to t, we obtain
$∫ t 2 t k μ Q t δ s − δ ′ t r 1 t − δ t b t 2 4 α δ t r 1 t 2 σ ′ t η ˜ 2 σ t , t 1 η ^ 2 σ t , t 1 α − 1 d s ≤ w t 2 .$
The proof of the lemma is complete. □
Now, let
$P 1 t = b t r 1 t η 2 t , t 1 α P 2 t = k Q t η ^ 2 t , t 1 α$
and
$μ t = exp ∫ t 1 t P 1 s d s .$
We present the following theorem.
Theorem 2.
Assume (A) holds. If every a solution of the first-order equation
$z ′ t + P 2 t z σ t = 0 .$
or $£ 2 y t$ is oscillatory, then $N 1 = ∅ .$
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 $x g t > 0 .$ As in Theorem 1, we have $£ 2 y t > 0 .$ Hence by (9), we obtain
$0 ≥ £ 3 y t + b t r 1 t £ 1 y t + k Q t y α σ t .$
Using (5) and (6), we have
$0 ≥ £ 3 y t + b t r 1 t η 2 t , t 1 £ 2 y t + k Q t η ^ 2 σ t , t 1 α £ 2 y σ t .$
Now, set $ϖ t = £ 2 y t$, we get
$ϖ ′ t + P 1 t ϖ t + P 2 t ϖ σ t ≤ 0 .$
Multiplying (22) by $μ t$, we have
$μ ϖ ′ t + μ t P 2 t ϖ σ t ≤ 0 .$
Now, setting the positive function $z = μ ϖ$ and taking into account $μ t$ is increasing function, we obtain
$z ′ t + μ t μ σ t P 2 t z σ t ≤ 0 .$
That is
$z ′ t + P 2 t z σ t ≤ 0 .$
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. □
Corollary 1.
Assume (A) holds. If
$lim inf t → ∞ ∫ σ t t P 2 t d s > 1 e ,$
then $N 1 = ∅ .$
Theorem 3.
Assume (2) is a oscillatory, then $N 1 = ∅ .$
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
$r 2 χ ′ ′ t + b t r 1 t χ t ≤ 0 .$
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
$r 2 r 1 y ′ α ′ ′ t + b t y ′ h t α + q t f x σ t = 0 ,$
where $h ∈ C 1 t 0 , ∞ , R$ is such that $σ t ≤ h t ≤ t$ and $h ′ t ≥ 0 .$
Theorem 4.
If every solution of the first-order equation
$r 2 χ ′ ′ t + b t r 1 h t χ h t = 0 ,$
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
$£ 3 y t + b t r 1 h t £ 1 y h t + k Q t y α σ t ≤ 0 .$
Set $χ = £ 1 y t$, we see that
$r 2 χ ′ ′ t + b t r 1 h t χ h t ≤ 0 .$
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 $σ t < h t < t , h ′ t ≥ 0$. If
$lim sup t → ∞ ∫ h ( t ) t Q ˜ s η 2 h t , h s d s > 1$
holds with
$Q ˜ t = τ 0 τ 0 + p 0 α c k μ q ˜ t η 1 h t , σ t − p 0 α τ 0 1 α b τ t r 1 τ t − b t r 1 t for all t ≥ t 1 ,$
where c is positive constant, 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 $x g t > 0$. We consider $£ 2 y t ≤ 0$ cannot hold for all large t, by a double integration of
$y ′ t = £ 1 y t r 1 t 1 / α ≤ £ 1 y t 2 r 1 t 1 / α$
that is $y ′ < 0$, a contradiction. Thus $£ 2 y t ≥ 0 .$ By using (8) with $u =$ $σ t$ and $v = h t$, we get $( σ t < h t < τ t )$
$y σ t ≥ η 1 h t , σ t − £ 1 y h t 1 α .$
Substituting (27) into (11), we obtain
$0 ≥ 1 + p 0 α τ 0 £ 3 y t + b t r 1 t £ 1 y t + p 0 α τ 0 1 α b τ t r 1 τ t £ 1 y τ t + k μ q ˜ t y α σ t .$
Since $h t < τ t < t ,$ we find
$0 ≥ 1 + p 0 α τ 0 £ 3 y t + £ 1 y h t b t r 1 t + p 0 α τ 0 1 α b τ t r 1 τ t + k μ q ˜ t η 1 h t , σ t − £ 1 y h t 1 α .$
Set $θ t = − £ 1 y t$
$0 ≤ 1 + p 0 α τ 0 r 2 t θ ′ t ′ + θ h t b t r 1 t + p 0 α τ 0 1 α b τ t r 1 τ t − c k μ q ˜ t η 1 h t , σ t θ h t 1 α − 1 θ h t .$
Taking into account that is $θ ′ t ≤ 0$ and $α ≥ 1 ,$ there exists positive constant c such that $θ h t 1 α − 1 ≥ c$. Thus
$r 2 t θ ′ t ′ ≥ τ 0 τ 0 + p 0 α c k μ q ˜ t η 1 h t , σ t − p 0 α τ 0 1 α b τ t r 1 τ t − b t r 1 t θ h t .$
This implies
$r 2 t θ ′ t ′ ≥ Q ˜ t θ h t .$
From (28) we see that $r 2 θ ′$ is increasing, we get
$θ t = θ t 2 + ∫ t 2 t r 2 s θ ′ s r 2 s d s > θ t 2 + r 2 t 2 θ ′ t 2 ∫ t 2 t 1 r 2 s d s = θ t 2 + r 2 t 2 θ ′ t 2 η 2 t , t 1 .$
Thus, $θ ′ t 2 < 0$ otherwise we imply $lim t → 0 θ t = ∞$ a contradiction to the boundedness of $θ$. So for $t 2 ≥ t ,$ we have
$θ > 0 θ ′ < 0 r 2 θ ′ ′ > 0 .$
Therefor, for $v ≥ u ≥ t 1 ,$ we find
$θ u > θ u − θ v = − ∫ u v θ ′ s d s = − ∫ u v r 2 s θ ′ s r 2 s d s .$
Since $r 2 θ ′$ is increasing
$θ u > − r 2 v θ ′ v ∫ u v 1 r 2 s d s = − η 2 v , u r 2 v θ ′ v .$
In (29), setting $u = h s$ and $v = h t ,$ we have
$θ h s > − η 2 h t , h s r 2 h t θ ′ h t .$
By Integrating (28) from $h ( t ) ≥ t 1$ to t, we get
$− r 2 h ( t ) θ ′ h ( t ) > ∫ h ( t ) t Q ˜ s θ h s d s > − r 2 h t θ ′ h t ∫ h ( t ) t Q ˜ s η 2 h t , h s r 2 h t θ ′ h t .$
Thus,
$∫ h ( t ) t Q ˜ s η 2 h t , h s d s < 1 ,$
which contradicts (26). The proof is complete. □
Theorem 6.
Assume (A) holds and $α ≥ 1$, there is function $h ∈ C 1 ( I , R )$ such that $σ t < h t < t , h ′ t ≥ 0$. If
$lim sup t → ∞ ∫ h t t 1 r 2 u ∫ u t Q ˜ s d s d u > 1$
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
$θ > 0 θ ′ < 0 r 2 θ ′ ′ > 0 .$
Integrating (28) from u to t, we get
$− r 2 u θ ′ u > ∫ u t Q ˜ s θ h s d s ≥ θ h t ∫ u t Q ˜ s d s ,$
that is
$− θ ′ u > 1 r 2 u θ h t ∫ u t Q ˜ s d s .$
Integrating from $h t$ to t, we have
$θ h t > θ h t ∫ h t t 1 r 2 u ∫ u t Q ˜ s d s d u .$
Thus
$∫ h t t 1 r 2 u ∫ u t Q ˜ s d s d u < 1 ,$
which contradicts (30). The proof is complete. □
Note that the conditions (14), (23) and (25) eliminate solutions from the class $N 1$, while conditions (26) and (30) eliminate solutions from the class $N 2$. By combining condition eliminate solutions from the class $N 1$ and condition eliminate solutions from the class $N 2$, we ensure that the solutions of (1) are oscillatory. Therefore, we get the following theorem.
Theorem 7.
Assume that (A) holdsand there is a function $h ∈ C 1 ( I , R )$ such that $σ t < h t < t$ and $h ′ t ≥ 0$. Let one of the following statements are true:
(a)
$α ≤ 1 ,$$σ ′ t > 0$ and there exists a function $δ t ∈ C 1 t 0 , ∞ , 0 , ∞$ such that (14) and (26) hold;
(b)
$α ≤ 1 ,$$σ ′ t > 0$ and there exists a function $δ t ∈ C 1 t 0 , ∞ , 0 , ∞$ such that (14) and (30) hold;
(c)
$α ≥ 1 ,$ (26) and (23) hold;
(d)
$α ≥ 1 ,$ (30) and (23) hold.
Then every solution of (1), or $£ 2 y t$, is oscillatory.
Theorem 8.
Assume that (A) holds and there is a function $h ∈ C 1 ( I , R )$ such that $σ t < h t < t$ and $h ′ t ≥ 0$. Let one of the following statements are true:
(a)
$α ≥ 1$, (2) is oscillatory and (26) holds;
(b)
$α ≥ 1$, (2) is oscillatory and (30) holds;
(c)
$α ≥ 1$, (25) is oscillatory and (26) holds;
(d)
$α ≥ 1$, (25) is oscillatory and (30) holds.
Then every solution of (1), or $y ′ t$, is oscillatory.
Example 1.
Consider the damped neutral differential equation
$y ‴ t + 1 2 y ′ t + 1 2 x t − 3 π 2 = 0 ,$
where $y t = x t + p 0 x τ 0 , α = 1 .$ Let $h t = t − π ,$ we see that
$σ t < h t < t , h ′ t ≥ 0 ,$
also
$Q ˜ t = τ 0 τ 0 + p 0 α c k μ 1 2 η 1 t − π , t − 3 π 2 − p 0 α τ 0 1 α b τ t r 1 τ t − 1 ,$
where
$η 1 t − π , t − 3 π 2 = 3 π 2 − π .$
By Theorem 6, condition (30) becomes
$c τ 0 π + 2 p 0 + 4 τ 0 π 2 8 τ 0 + p 0 > 1 .$
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.

## Author Contributions

Formal analysis, E.M.E. and T.A.; Investigation, B.Q. and O.M.; Supervision, E.M.E. and O.M.; Writing—original draft, B.Q.; Writing—review and editing, E.M.E., B.Q., T.A. and O.M. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

## Acknowledgments

The authors thank the reviewers for their useful comments, which led to the improvement of the content of the paper.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Elabbasy, E.M.; Qaraad, B.; Abdeljawad, T.; Moaaz, O. Oscillation Criteria for a Class of Third-Order Damped Neutral Differential Equations. Symmetry 2020, 12, 1988. https://doi.org/10.3390/sym12121988

AMA Style

Elabbasy EM, Qaraad B, Abdeljawad T, Moaaz O. Oscillation Criteria for a Class of Third-Order Damped Neutral Differential Equations. Symmetry. 2020; 12(12):1988. https://doi.org/10.3390/sym12121988

Chicago/Turabian Style

Elabbasy, Elmetwally M., Belgees Qaraad, Thabet Abdeljawad, and Osama Moaaz. 2020. "Oscillation Criteria for a Class of Third-Order Damped Neutral Differential Equations" Symmetry 12, no. 12: 1988. https://doi.org/10.3390/sym12121988

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