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Article

Some Oscillation Results for Even-Order Differential Equations with Neutral Term

by
Maryam Al-Kandari
1,*,† and
Omar Bazighifan
2,3,†
1
Department of Mathematics, Kuwait University, P.O. Box 5969, Safat, Khaldiyah City 13060, Kuwait
2
Section of Mathematics, International Telematic University Uninettuno, CorsoVittorio Emanuele II, 39, 00186 Roma, Italy
3
Department of Mathematics, Faculty of Science, Hadhramout University, Hadhramout 50512, Yemen
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Fractal Fract. 2021, 5(4), 246; https://doi.org/10.3390/fractalfract5040246
Submission received: 3 November 2021 / Revised: 10 November 2021 / Accepted: 16 November 2021 / Published: 1 December 2021
(This article belongs to the Section General Mathematics, Analysis)

Abstract

:
The objective of this work is to study some new oscillation criteria for even-order differential equation with neutral term r x z n 1 x γ + q x y γ ζ x = 0 . By using the Riccati substitution and comparison technique, several new oscillation criteria are obtained for the studied equation. Our results generalize and improve some known results in the literature. We offer some examples to illustrate the feasibility of our conditions.

1. Introduction

In recent decades, the problem of establishing asymptotic behavior of solutions for differential equations with delay term has been a very active research area. Due to the huge advantage of neutral differential equations in describing several neutral phenomena in medicine, economics, physics, engineering, and biology, there is great academic and scientific value, both practically and theoretically, in studying neutral differential equations [1,2,3,4,5].
Many researchers have been interested in studying the qualitative and oscillatory properties of delay differential equations. Zhang et al. [6], Li and Rogovchenko [7] obtained new oscillatory conditions of higher-order neutral differential equations using some different techniques. Agarwal et al. [8] and Moaaz et al. [9] improved some criteria of oscillation for solutions of neutral differential equations. What is more, most of the articles focus on the oscillation of different order differential equations with delay terms, see [10,11,12,13,14,15].
In [16], the authors considered the oscillation of differential equation
r x z n 1 x γ + q x f y ζ x = 0 ,
where f y = y γ , and they used the integral averaging method to obtain the oscillation criteria.
Xing et al. [17] discussed the following half-linear equation
r x z n 1 x γ 1 + q x y γ 2 ζ x = 0 ,
where n is an even, they established some oscillation criteria for this equation by comparison principles.
Baculikova et al. [18] presented oscillation results by comparison principles for equation
z n 1 x γ + q x y γ ζ x = 0 .
From the above, in this paper, we are interested in studying oscillation conditions of even-order differential equation
r x z n 1 x γ + q x y γ ζ x = 0 ,
where x x 0 , n is an even and
z x = y x + m x y g x .
In this paper, we impose the following conditions:
(P1)
r C 1 [ x 0 , ) , 0 , , r x 0 , x 0 r 1 / γ s d s = ,
(P2)
m C n [ x 0 , ) , 0 , , q C [ x 0 , ) , 0 , , 0 m x < 1 ,
(P3)
0 m x m 0 < ,
(P4)
m x > 1 , ζ x g x ,
(P5)
g C n [ x 0 , ) , R , ζ C [ x 0 , ) , R , g x g 0 > 0 , ζ x > 0 , g x x , ζ x x , lim x g x = lim x ζ x = ,
(P6)
γ S = { j : j = 2 m 1 + 1 2 m 2 + 1 , m 1 , m 2 N * } .
Definition 1.
A solution of (1) is called oscillatory if it has arbitrarily large zeros on [ x 0 , ) , and otherwise is called to be non-oscillatory.
Definition 2.
Equation (1) is known as oscillatory if each solution of it is oscillatory.
Definition 3.
If the highest-order derivative of the unknown function occurs with and without delay, then the differential equation is defined as neutral.
Definition 4
([19]). Define the operator ϝ . ; l , x by
ϝ f ; l , x = l x z x , s , l f s ds ,
for x 0 1 s x and f C [ x 0 , ) , R . The function μ = μ x , s , l is defined by
z x , s , l s = μ x , s , l z x , s , l ,
such that z x , x , l = 0 , z x , l , l = 0 and z x , s , l > 0 .
It is easy to verify that ϝ · ; l , x is a linear operator and that it satisfies
ϝ f ; l , x = ϝ f μ ; l , x , for f C 1 [ x 0 , ) , R .
The purpose of this paper is to continue the previous works [17,18]. So, we extend, generalize and improve the results for (1) using Riccati transformation and comparison technique.
We need some of the following lemmas to help us prove the main results.
Lemma 1
([20]). Let y C n x 0 , , 0 , , such that y n 1 x y n x 0 for all x x 1 . If lim x y x 0 , then for every ε 0 , 1 there exists x ε x 1 such that
y x ε n 1 ! x n 1 y n 1 x f o r x x ε .
Lemma 2
([21]). Let g ( u ) = r u m u β + 1 / β , where r and m are positive constants, β S . Then, g attains its maximum value on R + at u * = β r β + 1 m β and
max u R + g = g u * = β β r β + 1 β + 1 β + 1 m β .
Lemma 3
([22]). Let γ 1 . Then
D w C w γ + 1 / γ γ γ ( γ + 1 ) γ + 1 D γ + 1 C γ , C > 0 .
Lemma 4
([23]). Let α 0 , 1 and y 1 , y 2 0 . Then,
y 1 α + y 2 α 1 2 α 1 y 1 + y 2 α i f α 1 ,
and
y 1 α + y 2 α y 1 + y 2 α i f 0 < α < 1 .
Lemma 5
([24]). Let f x C n x 0 , , 0 , . If f n 1 x f n x 0 , θ 0 , 1 and M > 0 , such that
f θ x M x n 2 f n 1 x ,
for all sufficient large x .
Lemma 6.
Let y is an eventually positive solution of (1), then there exist x x 1 x 0 , such that:
z x > 0 , z x > 0 , z n 1 x > 0 , z n x < 0 .
More precisely, z x has the following two cases for x x 1 :
I 1 z x > 0 , z x > 0 , z x > 0 , z n 1 x > 0 , z n x < 0 , I 2 z x > 0 , z ( j ) ( x ) > 0 , z ( j + 1 ) ( x ) < 0   f o r   a l l   o d d   i n t e g e r j { 1 , 2 , . . . , n 3 } , z ( n 1 ) ( x ) > 0 , z ( n ) ( x ) < 0 .
Proof. 
The proof of (4) is similar to that of ([25], Lemma 2.3), and so we omit it. Furthermore, we can conclude that case I 1 and I 2 hold. □

2. Main Results

Lemma 7.
Assume that y is an eventually positive solution of (1) and
ζ 1 x ζ 0 > 0 , g x g 0 > 0 .
Then
1 ζ 0 r ζ 1 x z n 1 ζ 1 x γ + m 0 γ ζ 0 g 0 r ζ 1 g x z n 1 ζ 1 g x γ + Γ x z γ x 0 ,
where
Γ x = min q ζ 1 x , q ζ 1 g x .
Proof. 
Let y be an eventually positive solution of (1) on x 0 , . From (1), we see that
0 = 1 ζ 1 x r ζ 1 x z n 1 ζ 1 x γ + q ζ 1 x y γ x .
Thus, for all sufficiently large x, we have
0 = 1 ζ 1 x r ζ 1 x z n 1 ζ 1 x γ + m 0 γ q ζ 1 x y γ x + m 0 γ ζ 1 g x r ζ 1 g x z n 1 ζ 1 g x γ + q ζ 1 x y γ x + m 0 γ q ζ 1 g x y γ g x .
From (8) and the definition of z, we obtain
q ζ 1 x y γ x + m 0 γ q ζ 1 g x y γ g x Γ x y x + m 0 y g x γ Γ x z γ x .
Thus, by using (8) and (9), we obtain
0 1 ζ 1 x r ζ 1 x z n 1 ζ 1 x γ + m 0 γ ζ 1 g x r ζ 1 g x z n 1 ζ 1 g x γ + Γ x z γ x
From (5), we obtain
0 1 ζ 0 r ζ 1 x z n 1 ζ 1 x γ + m 0 γ ζ 0 g 0 r ζ 1 g x z n 1 ζ 1 g x γ + Γ x z γ x .
This completes the proof. □
Theorem 1.
Let θ C 1 x 0 , , R such that
lim sup x ϝ θ s Γ s Φ ^ s r ζ 1 s ζ 1 s n 2 γ + m 0 γ r ζ 1 g s g 0 ζ 1 g s n 2 γ ; l , x > 0 ,
for all M > 0 , λ 0 , 1 , where
Φ ^ s = θ s ζ 0 ( γ + 1 ) γ + 1 λ M γ μ s + θ s θ s γ + 1 ,
then (1) is oscillatory.
Proof. 
Suppose that (1) is not oscillatory. Without loss of generality, we assume that y x > 0 , y g x > 0 and y ζ x > 0 for all x x 1 . By Lemma 5, we obtain
z λ x M x n 2 z n 1 x .
Now, we define Riccati transformation by
ψ x = θ x r ζ 1 x z n 1 ζ 1 x γ z γ λ x > 0
and
ψ x = θ x r ζ 1 x z n 1 ζ 1 x γ z γ λ x + θ x r ζ 1 x z n 1 ζ 1 x γ z γ λ x γ λ θ x r ζ 1 x z n 1 ζ 1 x γ z λ x z γ + 1 λ x .
Using (12) and (13) in (14), we obtain
ψ x θ x θ x ψ x + θ x r ζ 1 x z n 1 ζ 1 x γ z γ λ x γ λ M ζ 1 x n 2 ψ x γ + 1 / γ θ x r ζ 1 x 1 / γ .
Let
ψ ˜ x = θ x r ζ 1 g x z n 1 ζ 1 g x γ z γ λ x > 0
and
ψ ˜ x θ x θ x ψ ˜ x + θ x r ζ 1 g x z n 1 ζ 1 g x γ z γ λ x γ λ M ζ 1 g x n 2 ψ ˜ x γ + 1 / γ θ x r ζ 1 g x 1 / γ .
Therefore, from (15) and (17), we obtain
1 ζ 0 ψ x + m 0 γ ζ 0 g 0 ψ ˜ x θ x ζ 0 r ζ 1 x z n 1 ζ 1 x γ z γ λ x + m 0 γ ζ 0 g 0 θ x r ζ 1 g x z n 1 ζ 1 g x γ z γ λ x + θ x ζ 0 θ x ψ x + m 0 γ ζ 0 g 0 θ x θ x ψ ˜ x 1 ζ 0 γ λ M ζ 1 x n 2 ψ x γ + 1 / γ θ x r ζ 1 x 1 / γ γ λ M m 0 γ ζ 0 g 0 ζ 1 g x n 2 ψ ˜ x γ + 1 / γ θ x r ζ 1 g x 1 / γ .
From (15), we obtain
1 ζ 0 ψ x + m 0 γ ζ 0 g 0 ψ ˜ x θ x Γ x + θ x ζ 0 θ x ψ x + m 0 γ ζ 0 g 0 θ x θ x ψ ˜ x 1 ζ 0 γ λ M ζ 1 x n 2 ψ x γ + 1 / γ θ x r ζ 1 x 1 / γ γ λ M m 0 γ ζ 0 g 0 ζ 1 g x n 2 ψ ˜ x γ + 1 / γ θ x r ζ 1 g x 1 / γ .
Applying ϝ . ; l , x to (19), we obtain
ϝ 1 ζ 0 ψ x + m 0 γ ζ 0 g 0 ψ ˜ x ; l , x ϝ [ θ s Γ s + θ s ζ 0 θ s ψ s + m 0 γ ζ 0 g 0 θ s θ s ψ ˜ s 1 ζ 0 γ λ M ζ 1 s n 2 ψ s γ + 1 / γ θ s r ζ 1 s 1 / γ γ λ M m 0 γ ζ 0 g 0 ζ 1 g s n 2 ψ ˜ s γ + 1 / γ θ s r ζ 1 g s 1 / γ ; l , x ] .
By (3) and (20), we see
ϝ θ s Γ s ; l , x ϝ [ 1 ζ 0 μ s + θ s θ s ψ s + m 0 γ ζ 0 g 0 μ s + θ s θ s ψ ˜ s 1 ζ 0 γ λ M ζ 1 s n 2 ψ s γ + 1 / γ θ s r ζ 1 s 1 / γ γ λ M m 0 γ ζ 0 g 0 ζ 1 g s n 2 ψ ˜ s γ + 1 / γ θ s r ζ 1 g s 1 / γ ; l , x ] .
Using Lemma 3, we set
D = 1 ζ 0 μ s + θ s θ s , C = 1 ζ 0 γ λ M ζ 1 s n 2 θ s r ζ 1 s 1 / γ and z = ψ ,
we have
1 ζ 0 μ s + θ s θ s ψ s 1 ζ 0 γ λ M ζ 1 s n 2 ψ s γ + 1 / γ θ x r ζ 1 s 1 / γ < 1 ( γ + 1 ) γ + 1 ζ 0 μ x + θ s θ s γ + 1 θ x r ζ 1 g s λ M ζ 1 s n 2 γ .
Hence, from (21) and (22), we have
ϝ θ s Γ s ; l , x ϝ [ μ s + θ s θ s γ + 1 θ s r ζ 1 s ( γ + 1 ) γ + 1 ζ 0 λ M ζ 1 s n 2 γ + μ s + θ s θ s γ + 1 m 0 γ θ s r ζ 1 g s ( γ + 1 ) γ + 1 ζ 0 g 0 λ M ζ 1 g s n 2 γ ; l , x ] .
Easily, we find
ϝ θ s Γ s ; l , x ϝ [ θ s ζ 0 ( γ + 1 ) γ + 1 λ M γ μ s + θ s θ s γ + 1 × r ζ 1 s ζ 1 s n 2 γ + m 0 γ r ζ 1 g s g 0 ζ 1 g s n 2 γ ; l , x ] .
That is,
ϝ θ s Γ s Φ ^ x r ζ 1 s ζ 1 s n 2 γ + m 0 γ r ζ 1 g s g 0 ζ 1 g s n 2 γ ; l , x 0 .
Taking the limes superior in the above inequality, we obtain
lim sup x ϝ θ s Γ s Φ ^ s r ζ 1 s ζ 1 s n 2 γ + m 0 γ r ζ 1 g s g 0 ζ 1 g s n 2 γ ; l , x 0 ,
This contradicts, which completes the proof. □
Lemma 8
([19] [Lemma 1.2]). Let ρ C 1 x 0 , , 0 , and
x 0 Ψ ^ s 2 γ γ + 1 γ + 1 r s ρ s γ + 1 ε γ s 2 γ ρ γ s d s = ,
where Ψ ^ x = θ ρ x q x 1 m ζ x γ ζ x \ x 3 γ , then z does not satisfy case I 1 .
Lemma 9.
Let θ C 1 x 0 , , R and ε 0 , 1 such that for all M > 0
lim sup x ϝ θ s Γ s Θ ^ s r ζ 1 s ζ 1 s n 2 γ + m 0 γ r ζ 1 g s g 0 ζ 1 g s n 2 γ ; l , x > 0 ,
for some ε 0 , 1 , where
Θ ^ s = θ s n 2 ! γ ε γ ζ 0 ( γ + 1 ) γ + 1 μ s + θ s θ s γ + 1 ,
then z not satisfies case I 2 .
Proof. 
Proceeding as in the proof of Theorem 1. From Lemma 1, we obtain
z x ε n 2 ! x n 2 z n 1 x .
We define Riccati transformation by
ς x = θ x r ζ 1 x z n 1 ζ 1 x γ z γ x > 0
and
ς x θ x θ x ς x + θ x r ζ 1 x z n 1 ζ 1 x γ z γ λ x γ ε n 2 ! ζ 1 x n 2 ς x γ + 1 / γ θ x r ζ 1 x 1 / γ .
Similarly, define
ς ˜ x = θ x r ζ 1 g x z n 1 ζ 1 g x γ z γ x > 0
and
ς ˜ x θ x θ x ς ˜ x + θ x r ζ 1 g x z n 1 ζ 1 g x γ z γ x γ ε n 2 ! ζ 1 g x n 2 ς ˜ x γ + 1 / γ θ x r ζ 1 g x 1 / γ .
Thus, we find
lim sup x ϝ θ s Γ s Θ ^ s r ζ 1 s ζ 1 s n 2 γ + m 0 γ r ζ 1 g s g 0 ζ 1 g s n 2 γ ; l , x 0 ,
This contradicts our assumption (25), which completes the proof. □
Theorem 2.
Let (24) and (25) hold. Then Equation (1) is oscillatory.
Theorem 3.
Let γ > 0 , P 1 , P 3 , P 5 hold and
x 1 C x dx = ,
where C x = min q x , q g x , then (1) is oscillatory.
Proof. 
Proceeding as in the proof of Theorem 1. By γ > 0 , we need to divide into two situations to discuss, that is γ 1 and 0 < γ < 1 .
When γ 1 is satisfied, owing to Lemma 6, we find (4) holds. According to (1), we see
r x z n 1 x γ = q x y γ ζ x < 0 , x x 1 .
Thus, r x z n 1 x γ is not increasing for x x 1 . Let φ x = r x z n 1 x γ , Ψ x = y γ ζ x . From (2) and (32), we obtain
φ x + q x Ψ x + m 0 γ g x φ g x + m 0 γ q g x Ψ g x 0 ,
which leads to
φ x + m 0 γ g 0 φ g x + C x Ψ x + + m 0 γ Ψ g x 0 .
According to Lemma 4 and P 3 , we have
φ x + m 0 γ g 0 φ g x + 1 2 γ 1 C x z γ ζ x 0 .
Integrating (34) from x 1 to x, we obtain
1 2 γ 1 x 1 x C s z γ ζ s ds φ x 1 φ x + m 0 γ g 0 φ g x 1 φ g x .
By z x > 0 , we obtain z ζ x > α > 0 . By virtue of P 1 , (4) and (32), we know that φ x > 0 , φ x < 0 , and so φ x is bounded. Thus, the right of (35) is bounded, contrary to (31).
If 0 < γ < 1 , the argument is analogous to that in the above discussion, so is omitted. This completes the proof. □
Corollary 1.
Let γ > 0 , P 1 , P 3 , P 5 and (31) hold. If the following inequality
φ x + m 0 γ g 0 φ g x + 1 2 γ 1 C x r ζ x Ω 1 γ ζ x φ ζ x 0 ,
has no eventually positive solution, where
Ω 1 x = δ x n 1 n 1 ! ,
then (1) is oscillatory.
Corollary 2.
Let γ > 0 , P 1 , P 3 , P 5 and (31) hold, and
φ x + m 0 γ g 0 φ g x + C x r ζ x Ω 1 γ ζ x φ ζ x 0 ,
has no eventually positive solution, then (1) is oscillatory.
Theorem 4.
Let n 4 is an even and P 1 , P 2 , P 5 hold. If there exist ξ , σ C 1 [ x 0 , ) , 0 ,
such that
x 1 ξ x E x K x dx =
and
x 1 σ x x s x n 4 E * s n 4 ! r 1 / γ s ds σ x 2 4 σ x ζ x dx = ,
where
E x = q x 1 m ζ x γ , E * x = x E s ds 1 / γ
and
K x = r x ξ x η + 1 η + 1 η + 1 ξ x ζ x Ω 2 ζ x γ , Ω 2 x = δ x n 2 n 2 ! ,
then (1) is oscillatory.
Proof. 
Proceeding as in the proof of Theorem 1. By Lemma 5, z satisfies case I 1 or case I 2 .
Assume that case I 1 hold. Then, lim x z x . From that and Lemma 1, we achieve
z x Ω 2 x z n 1 x .
By ζ x x and the fact that z n 1 x is not increasing, we obtain
z ζ x z n 1 x z ζ x z n 1 ζ x Ω 2 ζ x .
Owing to z x > 0 and (2), we obtain
y ζ x 1 m ζ x z ζ x .
Let
η x = ξ x r x z n 1 x γ z γ ζ x > 0 ,
we set
d x = ξ x ξ x , e x = γ ζ x Ω 2 ζ x ξ x r x 1 / γ .
Then,
η x ξ x E x + d x η x e x η η + 1 / η .
By Lemma 2, we obtain
d x η x e x η η + 1 / η K x .
Thus,
η x ξ x E x + K x .
This yields
x 1 ξ x E x K x dx η x 1 ,
which contradicts (38).
For the case I 2 , according to (1) and (41), we achieve
r x z n 1 x γ + E x z γ ζ x 0 .
Integrating (43) from x to , from z x > 0 and P 5 , we find
z n 1 x + z ζ x E * x r 1 / γ x 0 .
Integrating (44) from x to , we see
z n 2 x + z ζ x x E * s r 1 / γ s ds 0 .
Continuously, if we integrate (45) from x to for all ( n 4 ) times we find
z x + z ζ x x s x n 4 E * s n 4 ! r 1 / γ s ds 0 .
Let
ν x = σ x z x z ζ x > 0 .
Since z x is decreasing and ζ x x , according to Lemma 2, we find
ν x σ x x s x n 4 E * s n 4 ! r 1 / γ s ds + σ x 2 4 σ x ζ x .
This implies that
x 1 σ x x s x n 4 E * s n 4 ! r 1 / γ s ds σ x 2 4 σ x ζ x ν x 1 .
This contradicts our assumption (39), which completes the proof. □
Corollary 3.
Let n 4 is an even and P 1 , P 4 , P 5 hold. If there exist ξ , σ C 1 [ x 0 , ) , 0 ,
and ε 0 , 1 such that
x 1 q x ξ x M ζ x K * x dx =
and
x 1 σ x x s x n 4 M * s n 4 ! r 1 / γ s ds σ x 2 4 σ x ζ x R x dx = ,
where
M x   = 1 m g 1 x 1 g 1 g 1 x n 1 / ε g 1 x n 1 / ε m g 1 g 1 x , M * x   = x q s M γ ζ s ds 1 / γ
and
K * x   = r x ξ x η + 1 η + 1 η + 1 ξ x ζ x R x Ω 2 g 1 ζ x γ , R x   = g 1 ζ x ,
then (1) is oscillatory.

3. Examples

Example 1.
Consider the equation
y x + 2 y x 5 π + q 0 y x π   = 0 .
We note that r x   = 1 ,   m x   = 2 ,   g x   = x 5 π , ζ x   = x π , ζ 1 x   = x + π and q x   = Γ x   = q 0 0 . Thus, if we choose z x , s , l   =   x s s l , then it is easy to see that
μ x , s , l   = x s s l x s s l
and
Φ ^ s = θ s ζ 0 ( γ + 1 ) γ + 1 λ M γ μ s + θ s θ s γ + 1 = 1 4 λ M x s s l x s s l 2 .
Thus,
lim sup x ϝ θ s Γ s Φ ^ s r ζ 1 s ζ 1 s n 2 γ + m 0 γ r ζ 1 g s g 0 ζ 1 g s n 2 γ ; l , x = lim sup x ϝ q 0 3 4 λ M x s s l x s s l 2 ; l , x > 0 .
By Theorem 1, equation (50) is oscillatory.
Example 2.
Consider the equation
y x + m 0 y x 5 π 4 + q 0 y x π = 0 ,
where q 0 > 0 . Let r x = 1 , m x = m 0 , g x = x 5 π , ζ x = x π , ζ 1 s = x + π and q x = Γ x = q 0 , then we find
x 0 r 1 / γ s d s = .
If we set z x   =   x s s l , thus, by Theorem 2, Equation (51) is oscillatory.
Example 3.
Consider the equation
e x y x + 1 2 y x 3 1 / 5 + e x y 1 / 5 x 4 = 0 , x 1 .
Let n = 4 , γ = 1 / 5 , r x   =   q x   =   e x ,   m x   =   1 / 2 ,   g x   =   x / 3 , ζ x   =   x / 4 , then it is easy to see that
C x   =   q g x   =   e x / 3
and
1 e s / 3 d s = .
By Theorem 3, Equation (52) is oscillatory.
Example 4.
Let the equation
x y x + 1 2 y x 2 + q 0 x 3 y 9 x 10 = 0 , x 1 ,
where q 0 1 . Let n = 4 , γ = 1 , r x = x , q x = q 0 / x 3 ,   m x = 1 / 2 ,   g x = x / 2 , ζ x = 9 x / 10 , we set ξ x = x 2 , σ x = x , δ = 9 3 1 / 9 3 , then it is easy to see that
E x   = q 0 2 x 3 , K x = 2000 9 3 δ ,
then
x 1 q 0 2 2000 9 3 δ x 1 d x if q 0 4000 9 3 1 5.5 ,
also, it is easy to see that
E x   = q 0 4 x 2 ,
then
x 1 q 0 8 5 18 x 1 d x if q 0 20 9 2.2
By Theorem 4, Equation (53) is oscillatory if q 0 4000 9 3 1 5.5 .

4. Conclusions

In this work, by using comparison method and Riccati technique, the oscillation criteria of (1) is discussed. The criteria presented criteria extend and complement some of the published papers. Additionally, we give some examples to prove the effectiveness of the criteria we obtained. In the future, we will continue this work by studying the oscillation conditions of the following equation:
r x z n 1 x p 1 + q x y p 1 μ x   =   0 , x x 0 ,
where z x   =   y x + m x y g x .

Author Contributions

Conceptualization, M.A.-K., O.B.; Data duration, M.A.-K., O.B.; Formal analysis, M.A.-K., O.B.; Investigation, M.A.-K., O.B.; Methodology, M.A.-K., O.B. 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.

Conflicts of Interest

The authors declare no conflict of interest.

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Al-Kandari, M.; Bazighifan, O. Some Oscillation Results for Even-Order Differential Equations with Neutral Term. Fractal Fract. 2021, 5, 246. https://doi.org/10.3390/fractalfract5040246

AMA Style

Al-Kandari M, Bazighifan O. Some Oscillation Results for Even-Order Differential Equations with Neutral Term. Fractal and Fractional. 2021; 5(4):246. https://doi.org/10.3390/fractalfract5040246

Chicago/Turabian Style

Al-Kandari, Maryam, and Omar Bazighifan. 2021. "Some Oscillation Results for Even-Order Differential Equations with Neutral Term" Fractal and Fractional 5, no. 4: 246. https://doi.org/10.3390/fractalfract5040246

APA Style

Al-Kandari, M., & Bazighifan, O. (2021). Some Oscillation Results for Even-Order Differential Equations with Neutral Term. Fractal and Fractional, 5(4), 246. https://doi.org/10.3390/fractalfract5040246

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