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Article

On the Qualitative Behavior of Third-Order Differential Equations with a Neutral Term

1
Section of Mathematics, International Telematic University Uninettuno, Corso Vittorio Emanuele II, 39, 00186 Rome, Italy
2
Department of Mathematics, Faculty of Science, Hadhramout University, Hadhramout 50512, Yemen
3
Mathematical Science Department, Faculty of Science, Princess Nourah Bint Abdulrahman University, Riyadh 11546, Saudi Arabia
4
Department of Mathematics, Faculty of Science, Khon Kaen University, Khon Kaen 40002, Thailand
*
Author to whom correspondence should be addressed.
Symmetry 2021, 13(7), 1287; https://doi.org/10.3390/sym13071287
Submission received: 29 May 2021 / Revised: 14 July 2021 / Accepted: 15 July 2021 / Published: 17 July 2021

Abstract

:
In this paper, we analyze the asymptotic behavior of solutions to a class of third-order neutral differential equations. Using different methods, we obtain some new results concerning the oscillation of this type of equation. Our new results complement related contributions to the subject. The symmetry plays a important and fundamental role in the study of oscillation of solutions to these equations. An example is presented in order to clarify the main results.

1. Introduction

The importance of neutral differential equations is that they contribute to many applicationsin physics, engineering, chemistry and medicine, so third-order equations appear in computer engineering, networks, aeromechanical systems and others; see [1,2,3].
To the best of our knowledge, there is little research that has focused on the study of differential equations of third-order with neutral terms in the non-canonical case, while there are many studies discussing oscillation in the canonical case (see [4,5,6,7,8,9,10,11]).
This work is concerned with the oscillation of a class of third-order nonlinear neutral functional differential equations with delay term
m z Ψ z + p z Ψ τ z α + a b ν z , s f Ψ σ z , s d s = 0 ,
where α is the quotient of odd positive integers, and we assume that the following hypotheses are satisfied:
(C1)
m , p C z 0 , , R , ν C z 0 , × a , b , R , p z p 0 < , ν z , s 0 and
π z = z 0 m 1 / α s d s < ;
(C2)
τ C 1 z 0 , , R , σ C 1 z 0 , × a , b , R , τ z < z , τ z < τ 0 , σ z , s < z , σ has nonnegative partial derivatives and lim z τ z = , lim z σ z , s = and σ τ = τ σ ;
(C3)
f C R , R and satisfies f Ψ > k x α for all k > 0 .
By a solution of (1), we mean a function Ψ defined on [ z Ψ , ) for z Ψ z 0 , which satisfies the property m z Ψ z + p z Ψ τ z α C 1 [ z Ψ , ) , R and Ψ satisfies (1) on [ z Ψ , ) . In what follows, we assume that solutions of (1) exist on some half-line [ z Ψ , ) and satisfy the condition sup { Ψ z : z z < } > 0 for any z z Ψ .
Definition 1.
A solution of (1) is said to be oscillatory if it has arbitrarily large zeros on [ z 0 , ) . Otherwise, a solution that is not oscillatory is said to be nonoscillatory.
Definition 2.
Equation (1) is said to be oscillatory if every solution of it is oscillatory.
Lately, great attention has been devoted to the theory of oscillation in DDEs. The works [12,13,14,15] develop techniques and methods for studying the oscillations of DDEs. This development was necessarily reflected in the study of the oscillation of third-order DDEs, and this can be seen through the works [16,17,18].
The motivation for this article is to complement the results reported in [19], which discussed the oscillatory properties of a third-order equation with a late term. Therefore, we discuss their findings and results below.
Grace et al. [20] considered the equation with delay term
m z Ψ z α + ν z f Ψ σ z = 0
in canonical and noncanonical form.
Saker and Dzurina [19] obtained several criteria, which ensure that the third-order nonlinear delay differential Equation (2) with f Ψ σ z = Ψ β σ z is oscillatory or tends to zero.
Moaaz in [21] improved the results in [19] and established some criteria that ensure that all solutions of (2) are oscillatory.
Dzurina et al. [22] presented oscillation results for the general equation
m 1 z m 2 z Ψ z + p z Ψ τ z α β + ν z f Ψ σ z = 0 ,
where ϕ z is defined as in (1).
Vidhyaa et al. [23] discussed several criteria of equation
m 1 z m 2 z Ψ z + p z Ψ τ z α + ν z Ψ α σ z = 0 ,
with α = 1 and f Ψ σ z k x α σ z .
In the remainder of the article, we will adopt the following notation:
ϕ z : = Ψ z + p z Ψ τ z ν z , s : = a b ν z , s 1 p σ z , s α d s , η z , s : = min ν z , s , ν τ z , s .
Our objective of this article is to find results that ensure oscillation of (1) by using the Riccati technique and comparison method. These results extend and complement some previous results. Compared to the conditions mentioned in the previous literature, we obtain new sufficient conditions that are less restrictive.

2. Auxiliary Results

We now present some lemmas that will be used later.
Lemma 1.
Assume Ψ z is nonoscillatory solution of (1). Then, ϕ z > 0 and there are three possible cases of ϕ z :
N 1 ϕ z > 0 , ϕ z > 0 ; N 2 ϕ z > 0 , ϕ z < 0 ; N 3 ϕ z < 0 , ϕ z > 0 .
Lemma 2.
(Thandapani [15]) Assume that b 1 , b 2 [ 0 , ) and γ > 0 . Then
b 1 + b 1 γ μ b 1 γ + b 2 γ ,
where
(i) 
μ = 1 when γ 1 ;
(ii) 
μ = 2 γ 1 when γ > 1 .
Lemma 3.
(Zhang [18]) Let g v = C v D v α + 1 / α where C , D > 0 and leg α be a ratios of two odd positive integers. Then g attains its maximum value on R at v * = α C / α + 1 D such that
max v R g v = g v * = α α α + 1 α + 1 C α + 1 D α .
Lemma 4.
Let
Ψ be a positive solution of ( 1 ) .
Then
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α + k μ a b η z , s ϕ α σ z , s d s 0 .
Furthermore, if ϕ z > 0 and p z 0 , 1 . Then
m z ϕ z α + k y α σ z , a ν z , s 0 .
and
ϕ σ z c ´ σ z ,
where c ˜ : = c k .
Proof. 
Let Ψ z be a positive solution of (1). Then, there exists z 1 z 0 such that Ψ σ z , s > 0 and Ψ τ z > 0 for all z z 1 z 0 . From hypothesis (C3), it follows from (1) that
m z ϕ z α + k a b ν z , s Ψ α σ z , s d s 0 .
By Lemma 2, we obtain
ϕ α σ z , s μ Ψ α σ z , s + p 0 α Ψ α σ τ z , s .
In view of (C2) and inequality (6), we see
p 0 α τ z m τ z ϕ τ z α + k p 0 α a b ν τ z , s Ψ α σ τ z , s d s 0 .
Hence
p 0 α τ 0 m τ z ϕ τ z α + k p 0 α a b ν τ z , s Ψ α σ τ z , s d s 0 .
Using (6) with above inequality, and taking into account (7), we have
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α k a b ν z , s Ψ α σ z , s + p 0 α ν τ z , s Ψ α σ τ z , s d s .
It follows that
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α k a b O ^ z , s Ψ α σ z , s + p 0 α Ψ α σ τ z , s d s .
Thus,
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α + k μ a b η z , s ϕ α σ z , s d s 0 .
Now, assume that Ψ z is a positive solution of (1) and ϕ z > 0 ; we find
Ψ z = ϕ z p z Ψ τ z ϕ z 1 p z ,
and
Ψ α σ z , s ϕ α σ z , s 1 p σ z , s α .
Combining (6) and (8), we have
m z ϕ z α + k a b ν z , s ϕ α σ z , s 1 p σ z , s α d s 0 .
By the fact that σ z , s is nondecreasing with respect to s, we have
ϕ α σ z , s > ϕ α σ z , a .
From (9), we get (4).
Furthermore, it is easy to see that
ϕ z ϕ z 1 for c [ z 1 , ) .
Integrating from σ z to z 1 , we get
ϕ σ z c σ z z 1 .
Hence, for any κ ( 0 , 1 ) and z z 2 , we see that
ϕ σ z c ˜ σ z , where c ˜ = c κ .
The proof is complete. □
Lemma 5.
Let Ψ z be a positive solution of (1) and ϕ z satisfying case N 3 . If
z 0 a b η z , s d s d u = ,
or
z 0 1 m ζ z a b η u , s d s d u 1 α d ζ = ,
then lim z ϕ z = 0 .
Proof. 
Assume that Ψ z is a positive solution of (1); there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 z 0 . Since ϕ z > 0 and ϕ z < 0 , there is λ 0 such that lim z ϕ z = λ . Assume that λ > 0 . Integrating (3) from z 2 to z, we have
m z ϕ z α + p 0 α τ 0 m ϕ τ z α m z 1 ϕ z 1 α + p 0 α τ 0 m τ z 1 ϕ τ z 1 α k μ z 2 z a b η u , s ϕ α σ u , s d s d u m z 1 ϕ z 1 α + p 0 α τ 0 m τ z 1 ϕ τ z 1 α k μ λ α z 2 z a b η u , s d s d u .
Which contradicts (10), hence λ = 0 . The proof is complete. □
Lemma 6.
Let Ψ z be a positive solution of (1). If
z 0 a b η u , s σ α u , s d s d u = ,
then case N 1 is impossible.
Proof. 
Let Ψ z be a positive solution of (1). Then, there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 z 0 . On the contrary, assume that ϕ z satisfies case N 1 . Integrating (3) from z 2 to z and using (5), we get
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α m z 2 ϕ z 2 α + p 0 α τ 0 m τ z 2 ϕ τ z 2 α c ˜ α k μ z 1 z a b η u , s σ α u , s d s d u ,
which is a contradiction. □
Lemma 7.
Let ϕ z be a positive increasing solution of (1). If
z 0 1 m 1 / α z z 0 z a b η u , s σ α u , s d s d u 1 / α d z = ,
then ϕ satisfies case N 2 .
Proof. 
Let Ψ z be a positive solution of (1). Then, there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 z 0 . Since ϕ is increasing, ϕ satisfies either case N 1 or case N 2 . In view of π z < and (14), we see that (12) holds. By Lemma 6, ϕ z satisfies case N 2 . The proof is complete. □
Lemma 8.
Let ϕ z be a positive increasing solution of (1). If ϕ satisfies (14), then:
(a) 
ϕ z t y z , ϕ z / z is decreasing, eventually, and lim z ϕ z / z = ϕ = 0 ,
(b) 
ϕ z π z m 1 α z ϕ z and ϕ z / π z is increasing eventually.
Proof. 
Let Ψ z be a positive solution of (1). Then, there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 z 0 . Using the fact that ϕ z is decreasing, we see that there exists constant λ 0 such that lim z ϕ z = λ 0 . We claim that λ = 0 . As the proof of Lemma 6; we have (13). Take into account m z ϕ z α 0 and ϕ z < 0 , we have
m z ϕ z α 1 + p 0 α τ 0 c ˜ α k μ z 1 z a b η u , s σ α u , s d s d u .
It follows that
m z ϕ z α c ˜ α τ 0 τ 0 + p 0 α k μ z 1 z a b η u , s σ α u , s d s d u
and
ϕ z c ˜ 1 m z k τ 0 μ τ 0 + p 0 α 1 / α z 1 z a b η u , s σ α u , s d s d u 1 / α .
Integrating from z 2 to z, we get
ϕ z ϕ z 2 c ˜ k τ 0 μ τ 0 + p 0 α 1 α z 2 z z 2 u a b η u , s σ α u , s d s d u 1 α m 1 α ζ d ζ .
Which in view of (14) contradicts the positivity of ϕ z . Thus, λ = 0 . By Hospital’s rule, we see that
lim z ϕ z z = ϕ z = 0 .
Thus
ϕ z 1 ϕ z z 1 > 0 .
So
ϕ z > ϕ z z 1 ,
for z z 2 . Hence, by monotonicity of ϕ z , one can obtain that
ϕ z = ϕ z 1 + z 1 z ϕ s d s ϕ z 1 + ϕ z z z 1 .
By (15), we have
ϕ z z = t y ϕ z 2 < 0 .
Now, it is easy to see that
ϕ z u 1 m 1 / α s m 1 / α s ϕ s d s m 1 / α z ϕ z π z .
Thus,
ϕ z π z = m 1 / α z ϕ z π z + ϕ z m 1 / α z π 2 z 0 .
The proof is complete. □

3. Main Results

Theorem 1.
If
z 0 1 m ξ z 0 a b η u , s d s d u 1 α d ξ = ,
then (1) satisfies case N 3 .
Proof. 
Assume that Ψ z is a positive solution of (1). Then, there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 . Suppose to the contrary that ϕ z satisfies case N 1 or N 2 . Since ϕ is increasing, it follows that
ϕ z ϕ z 1 = ϱ for z z 2 .
Hence,
ϕ σ z , s ϕ z 1 , s = ϱ .
By m z ϕ z α 0 , we have
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α m z ϕ z α 1 + p 0 α τ 0 .
In (3), we get
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α k μ a b η z , s ϕ α σ z , s d s .
Integrating (19) from z 2 to z and using (17), we obtain
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α m z 1 ϕ z 1 α + p 0 α τ 0 m τ z 1 ϕ τ z 1 α ϱ α k μ z 2 z a b η u , s d s d u .
First, assume that ϕ z satisfies case N 1 . We note that
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α > 0 .
Using the fact π z < together with (16) yields that z 2 z a b η u , s d s d u = , and this contradicts the positivity of m z ϕ z α .
If ϕ z satisfies case N 2 , using (18) in (20) becomes
m z ϕ z α 1 + p 0 α τ 0 ϱ α k μ z 2 z a b η u , s d s d u ,
that is,
ϕ z ϱ k τ 0 μ τ 0 + p 0 α 1 α 1 m z z 2 z a b η u , s d s d u 1 α .
Integrating from z 2 to z, we have
ϕ z ϕ z 2 ϱ k τ 0 μ τ 0 + p 0 α 1 α z 2 z 1 m ξ z 2 z a b η u , s d s d u 1 α d ξ .
This contradicts the positivity of ϕ z . The proof of the lemma is complete. □
Theorem 2.
If
lim inf z σ z z 1 m ζ z 2 u a b η u , s σ α u , s d s d u 1 α d ζ > μ τ 0 + p 0 α k τ 0 e α 1 α ,
then (1) satisfies case N 3 .
Proof. 
Let Ψ z be a positive solution of (1). Then, there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 . Suppose to the contrary that ϕ satisfies case N 1 or N 2 . In view of (21), (14) holds. Hence, by Lemma 8, ϕ z satisfies case N 2 and properties (a) and (b). This implies that
ϕ σ z σ z ϕ σ z .
Combining the above inequality along with (3), we get
m ϕ z α + p 0 α τ 0 m ϕ τ z α k μ a b η z , s σ α z , s ϕ σ z , s α d s k μ ϕ σ z , b α a b η z , s σ α z , s d s .
Integrating from z 2 to z and using (18), we have
m z ϕ z α 1 + p 0 α τ 0 k μ ϕ σ z , b α z 2 z a b η u , s σ α u , s d s d u .
Using the fact that ϕ z < 0 , we see that
ϕ z k τ 0 μ τ 0 + p 0 α 1 α ϕ σ z 1 m z z 2 z a b η u , s σ α u , s d s d u 1 α .
Now, set χ z = ϕ z ; we obtain
χ z + k τ 0 μ τ 0 + p 0 α 1 α 1 m z z 2 z a b η u , s σ α u , s d s d u 1 α χ σ z 0 .
In view of ([13], Theorem 1), we see that, the associated delay Equation (23) has positive solution, which is a contradiction. The proof is complete. □
Theorem 3.
Assume that (14) hold. If
lim sup z π α z z 2 z a b η u , s σ α u , s d s d u > μ τ 0 + p 0 α k τ 0 ,
then (1) satisfies case N 3 .
Proof. 
Suppose to the contrary that ϕ satisfies case N 1 or N 2 . We see that (12) holds due to π z < (this mean that lim z π z = 0 ) and condition (24). Hence, by Lemma 8, ϕ z satisfies case N 2 in addition to properties (a) and (b). As in the proof of Theorem 2 with the fact that m z ϕ z α is nonincreasing and integrating (22) from z 2 to z, we obtain
m z ϕ z α 1 + p 0 α τ 0 k μ π α z m z ϕ z α z 2 z a b η u , s σ α u , s d s d u .
That is,
1 + p 0 α τ 0 π α z z 2 z a b η u , s σ α u , s d s d u .
This contradicts (24). The proof is complete. □
Theorem 4.
Assume that (14) holds. If there exists a nondecreasing function ρ C 1 z 0 , , 0 , and σ z > 0 , such that
lim sup z z z k τ 0 μ τ 0 + p 0 α 1 α ρ ζ σ ζ m 1 α ζ z 2 u a b η u , s σ α u , s d s d u ρ 2 ζ 4 p ζ σ ζ d ζ = ,
for any z z 0 , , then (1) satisfies case N 3 .
Proof. 
Let Ψ z be a positive solution of (1). Then there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 . Suppose to the contrary that ϕ satisfies case N 1 or N 2 . By Lemma 8, ϕ z satisfies case N 2 and the properties (a) and (b). Define the function w z by
w z : = ρ z ϕ z ϕ σ z .
Then w z > 0 , and
w z = ρ z ρ z w z + ρ z ϕ z ϕ σ z ρ z ϕ z ϕ σ z σ z ϕ 2 σ z .
Using the fact ϕ z is decreasing, we have
w z ρ z w z ρ z + ρ z ϕ z ϕ σ z ρ z ϕ z 2 σ z ϕ 2 σ z = ρ z w z ρ z + ρ z ϕ z ϕ σ z σ z ρ z ρ z ϕ z ϕ σ z 2 .
By (26), we obtain
w z ρ z w z ρ z + ρ z ϕ z ϕ σ z σ z ρ z w 2 z .
Integrating (3) from z 2 to z and ϕ σ z / σ z < 0 , we have
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α m z 2 ϕ z 2 α + p 0 α τ 0 m z 2 ϕ τ z 2 α + k μ z 2 z a b η u , s ϕ α σ u , s d s d u m z 2 ϕ z 2 α + p 0 α τ 0 m z 2 ϕ τ z 2 α + k μ ϕ σ z σ z α z 2 z a b η u , s σ α u , s d s d u .
Since lim z ϕ z / z = 0 , there exists z 3 > z 2 such that
m z 2 ϕ z 2 α + p 0 α τ 0 m z 2 ϕ τ z 2 α k μ ϕ σ s σ s α z 2 z a b η u , s σ α u , s d s d u > 0 .
Combining the above inequality with (28) implies
m z ϕ z α + p 0 α τ 0 m τ z ϕ τ z α m ϕ z 2 α + p 0 α τ 0 m ϕ τ z 2 α + k μ ϕ σ z σ z α z 2 z a b η u , s σ α u , s d s d u k μ ϕ σ z σ z α z 2 z a b η u , s σ α u , s d s d u k μ ϕ σ z σ z α z 2 z a b O ^ u , s σ α u , s d s d u .
Using (18), we have
m z ϕ z α 1 + p 0 α τ 0 k μ ϕ σ z σ z α z 2 z a b η u , s σ α u , s d s d u ,
that is,
ϕ z ϕ σ z k τ 0 μ τ 0 + p 0 α 1 α 1 σ z m 1 α z z 2 z a b η u , s σ α u , s d s d u .
Substituting (29) in (27), yields that
w z ρ z ρ z w z ρ z σ s k τ 0 μ τ 0 + p 0 α 1 α 1 m 1 α z z 2 z a b η u , s σ α u , s d s d u σ z ρ z w 2 z = σ z ρ z w z ρ z 2 σ z 2 + ρ 2 z 4 p z σ z ρ z σ s k τ 0 μ τ 0 + p 0 α 1 α 1 m 1 α z z 2 z a b η u , s σ α u , s d s d u .
Hence,
w z ρ z σ s k τ 0 μ τ 0 + p 0 α 1 α 1 m 1 α z z 2 z a b η u , s σ α u , s d s d u + ρ   2 z 4 p z σ z .
Integrating from z 3 to z, we have
w z w z 3 z 3 z k τ 0 μ τ 0 + p 0 α 1 α ρ ζ σ ζ m 1 α ζ z 2 u a b η u , s σ α u , s d s d u ρ   2 ζ 4 p ζ σ ζ d ζ ,
which is a contradiction. The proof is complete. □
Theorem 4 can be used in a wide range of applications by choosing ρ z = 1 π ; we conclude the following corollary.
Corollary 1.
Assume that (14) holds. If there exists a nondecreasing function ρ C 1 z 0 , , 0 , and σ z > 0 such that
lim sup z z z k τ 0 μ τ 0 + p 0 α 1 α π ζ σ ζ m 1 α ζ z 2 u a b η z , s σ α z , s d s d u m 2 / α ζ 4 σ ζ π 3 ζ d ζ = ,
for any z z 0 , , then (1) satisfies case N 3 .
Theorem 5.
Assume that (14) hold. If there exists a nondecreasing function δ C 1 z 0 , , 0 , , such that
lim sup z z 2 z δ u k a b η u , s d s μ δ u α + 1 α + 1 α + 1 σ u α π α u δ α u d u > δ z π α z σ α z ,
then (1) satisfies case N 3 .
Proof. 
Let Ψ z be a positive solution of (1); there exists z 1 z 0 such that Ψ τ z > 0 and Ψ σ z > 0 for all z z 1 . Suppose to the contrary that ϕ satisfies case N 1 or N 2 . By Lemma 8, ϕ z satisfies case N 2 and the properties (a) and (b).
Define the function w z by
w z : = δ z m ϕ z α ϕ α σ z + 1 π α σ α z .
From Lemma 8, it is easy to see that
ϕ σ z σ z ϕ σ z σ z ϕ z σ z π z m 1 α z ϕ z .
That is, w z > 0 and
δ z m ϕ z α ϕ α σ z δ z π α σ α z .
Using (19) and the fact ϕ z is decreasing, we have
m z ϕ z α k μ ϕ α σ z a b η z , s d s ,
hence,
m z ϕ z α ϕ α σ z k μ a b η z , s d s .
That is,
w z = δ z δ z w z + δ z m z ϕ z α ϕ α σ z α δ z m z ϕ z α ϕ σ z σ z ϕ α + 1 σ z + α δ z π z σ z α + 1 σ z m 1 α z σ z π z δ z δ z w z δ z k μ a b η z , s d s α σ z ϕ z δ 1 α z m 1 α z ϕ z w z δ z π z σ z 1 α + 1 + α δ z π z σ z α + 1 σ z m 1 α z σ z π z .
In view of property (b) in Lemma 8, we find
w z δ z δ z w z δ z k a b η z , s d s μ α σ z π z δ 1 α z w z δ z π α z σ α z 1 α + 1 + α δ z π z σ z α + 1 σ z m 1 α z σ z π z .
Set
A : = δ z δ z , B : = α σ z π z δ 1 α z , C : = δ z π α z σ α z .
Using Lemma 3, we obtain
w z = δ z k a b η z , s d s μ + δ z π α z σ α z + 1 α + 1 α + 1 δ z α + 1 σ z α π α z δ α z + α δ z π z σ z α + 1 σ z m 1 α z σ z π z .
It is clear that
δ z π α z σ α z = δ z π α z σ α z + α δ z π z σ z α + 1 σ z m 1 α z σ z π z .
In (35), we get
w z = δ z k a b η z , s d s μ + δ z π α z σ α z + 1 α + 1 α + 1 δ z α + 1 σ z α π α z δ α z .
Integrating the above inequality from z 2 to z yields
z 2 z δ u k a b η z , s d s μ δ u α + 1 α + 1 α + 1 σ u α π α u δ α u d u + δ z π α z σ α z δ z 2 π α z 2 σ α z 2 w z 2 w z .
From (32), we are led to
z 2 z δ u k a b η z , s d s μ δ u α + 1 α + 1 α + 1 σ u α π α u δ α u d u + δ z π α z σ α z δ z 2 π α z 2 σ α z 2 δ z 2 m ϕ z 2 α ϕ α σ z 2 δ z m ϕ z α ϕ α σ z .
By (34), (36) becomes
z 2 z δ u k a b η z , s d s μ δ u α + 1 α + 1 α + 1 σ u α π α u δ α u d u δ z π α σ α z .
The proof is complete. □

4. Corollaries of the Main Theorems

Corollary 2.
Assume that (16) holds. Then (1) is almost oscillatory.
Corollary 3.
Assume that (21) and (10) or (11) hold. Then (1) is almost oscillatory.
Corollary 4.
Assume that (14), (24), and either (10) or (11) hold. Then (1) is almost oscillatory.
Corollary 5.
Assume that (14) holds and if there exists a nondecreasing function ρ C 1 z 0 , , 0 , and σ z > 0 , such that (25) and either (10) or (11) hold. Then (1) is almost oscillatory.
Corollary 6.
Assume that (14) holds and if there exists a nondecreasing function δ C 1 z 0 , , 0 , , such that (31) and either (10) or (11) hold. Then (1) is almost oscillatory.
Example 1.
Consider the equation
z 2 Ψ z + p 0 Ψ ϵ t α + ν 0 z ϕ α λ t = 0 , z 1 , λ 0 , 1 ,
where m = z 2 , ν z = ν 0 z > 0 and σ z = λ t , τ z = ϵ t . Set δ z = π z τ z = ϵ . We conclude that Theorem 3 fails.
By Theorem 2, the same conclusion holds for (37) if
ν 0 λ ln 1 λ > τ 0 + p 0 τ 0 k e .
By Theorem 3, the same conclusion holds for (37) if
ν 0 λ > μ τ 0 + p 0 α k τ 0 .
Furthermore, Corollary 1 if
ν 0 > 1 4 τ 0 + p 0 α λ τ 0 .
That is, since Lemma 5 is satisfied, and we see that (37) is almost oscillatory if conditions (38)–(40) hold.

5. Conclusions

In this article, we are interested in studying the asymptotic behavior of third-order neutral differential equations. We found more than one criterion to check the oscillation by Riccati method. Our results are an extension and complement to some results published in the literature. We supported the results obtained in this paper with examples. An interesting problem for further research could be to study the problem of oscillation when
z 0 m 1 / p 1 s d s = .

Author Contributions

Conceptualization, O.B., F.M. and K.N.; methodology, O.B., F.M. and K.N.; investigation, O.B., F.M. and K.N.; resources, O.B., F.M. and K.N.; data curation, O.B., F.M. and K.N.; writing—original draft preparation, O.B., F.M. and K.N.; writing—review and editing, O.B., F.M. and K.N.; supervision, O.B., F.M. and K.N.; project administration, O.B., F.M. and K.N.; funding acquisition, O.B., F.M. and K.N. All authors read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

The authors thank the reviewers for their useful comments, which led to the improvement of the content of the paper. This research was funded by the Deanship of Scientific Research at Princess Nourah bint Abdulrahman University through the Fast-track Research Funding Program.

Conflicts of Interest

The authors declare no conflict of interest.

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Bazighifan, O.; Mofarreh, F.; Nonlaopon, K. On the Qualitative Behavior of Third-Order Differential Equations with a Neutral Term. Symmetry 2021, 13, 1287. https://doi.org/10.3390/sym13071287

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Bazighifan O, Mofarreh F, Nonlaopon K. On the Qualitative Behavior of Third-Order Differential Equations with a Neutral Term. Symmetry. 2021; 13(7):1287. https://doi.org/10.3390/sym13071287

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Bazighifan, Omar, Fatemah Mofarreh, and Kamsing Nonlaopon. 2021. "On the Qualitative Behavior of Third-Order Differential Equations with a Neutral Term" Symmetry 13, no. 7: 1287. https://doi.org/10.3390/sym13071287

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