http://www.scirp.org/journal/am
ISSN Online: 2152-7393 ISSN Print: 2152-7385
DOI: 10.4236/am.2016.715149 September 21, 2016
Oscillation Properties of Third Order Neutral
Delay Differential Equations
*
Elmetwally M. Elabbasy
1, Osama Moaaz
1, Ebtesam Sh. Almehabresh
21Department of Mathematics, Faculty of Science, Mansoura University, Mansoura, Egypt 2Department of Mathematics, Faculty of Education, AL Asmarya Islamic University, Zliten, Libya
Abstract
Oscillation criteria are established for third-order neutral delay differential equations with deviating arguments. These criteria extend and generalize those results in the literature. Moreover, some illustrating examples are also provided to show the im-portance of our results.
Keywords
Oscillation, Third Order, Neutral Delay, Differential Equations
1. Introduction
This article is concerned with the oscillation and the asymptotic behavior of solutions of the third-order neutral delay differential equations with deviating argument of the form
( ) ( )
(
)
( )
,(
( )
,)
d 0, 0,d c
r t z t′′ α ′+
∫
q tξ xα g tξ ξ = t≥t (E) where( )
( )
b( )
,(
( )
,)
d .a
z t =x t +
∫
p tη x τ ηt η We assume that:(H) r∈C t
(
[
0,∞)
, 0,(
∞)
)
; p,τ
∈C t(
[
0,∞ ×)
[ ]
a b R, ,)
; q g, ∈C t(
[
0,∞ ×)
[ ]
c d, ,R)
,α is a quotient of odd positive integers, 0 b
( )
, d 1,ap tη η p
≤
∫
≤ < τ η( )
t, ≤t,( )
, ,g tξ ≤t limt→∞τ η
( )
t, =limt→∞g t( )
,ξ = ∞ and q t( )
,ξ >0.A function x t
( )
∈C t(
[
x,∞)
)
,tx≥t0 is called a solution of (E), if it has the properties( )
1(
[
)
)
, ,
x
z t ∈C t ∞ z t′
( )
∈C1(
[
tx,∞)
)
,( ) ( )
(
[
)
)
1,
x
r t z t′′ α∈C t ∞ and satisfies (E) on
[
tx,∞)
. We consider only those solutions x t( )
of (E) which satisfy( )
{
}
sup x t :t≥T >0 for all T ≥tx. We assume that (E) possesses such solution. A
solution of (E) is called oscillatory if it has arbitrarily large zeros on
[
tx,∞)
; otherwise,*1991 Mathematics Subject Classification: 34K10, 34K11. How to cite this paper: Elabbasy, E.M.,
Moaaz, O. and Almehabresh, E.Sh. (2016) Oscillation Properties of Third Order Neutral Delay Differential Equations. Applied Mathe- matics, 7, 1780-1788.
http://dx.doi.org/10.4236/am.2016.715149
Received: August 1, 2016 Accepted: September 18, 2016 Published: September 21, 2016 Copyright © 2016 by authors and Scientific Research Publishing Inc. This work is licensed under the Creative Commons Attribution International License (CC BY 4.0).
http://creativecommons.org/licenses/by/4.0/
it is called nonoscillatory.
In the recent years, great attention in the oscillation theory has been devoted to the os-cillatory and asymptotic properties of the third-order differential equations (see [1]-[14]). Baculikova et al. [2][3], Dzurina et al. [4] and Mihalikova et al. [11] studied the oscilla-tion of the third-order nonlinear differential equaoscilla-tion
( ) ( )
(
( )
(
( )
)
)
( )
(
(
( )
)
)
0, 0,r t x t p t x t q t f x g t t t
α
τ ′
′′
+ + = ≥
under the condition
( )
01
.
t r s
α
∞ − = ∞
∫
Li et al. [10] considered the oscillation of
( ) ( ) ( )
(
( )
(
( )
)
)
( )
(
( )
)
2 1 0, 0,
r t r t x t p t x τ t q t x g t t t
′
′′
+ + = ≥
under the assumption
( )
( )
0 0
1 1
1 and 2 .
tr s tr s
∞ − ∞ −
= ∞ < ∞
∫
∫
The aim of this paper is to discuss asymptotic behavior of solutions of class of third order neutral delay differential Equation (E) under the condition
( )
01
.
t r s
α
∞ − < ∞
∫
(1)By using Riccati transformation technique, we established sufficient conditions which insure that solution of class of third order neutral delay differential equation is oscillatory or tends to zero. The results of this study extend and generalize the previous results.
2. Main Results
In this section, we will establish some new oscillation criteria for solutions of (E). Theorem 2.1. Assume that conditions (1) and (H) are satisfied. If for some function
[
)
(
)
(
)
1
0, , 0, ,
C t
ρ
∈ ∞ ∞ for all sufficiently large t1≥t0 and for t3≥t2≥t1, one has( ) ( )(
) ( )
(
)
( ) ( )
(
)
( )
31 1
1
lim sup 1 d ,
1
t t t
r s s
s q s p G s s
s
α α
α α
ρ ρ
ρ α
+ ∗
+ →∞
′
− − = ∞
+
∫
(2)where
( )
( )
( )
( )
( )
( )
2 1
1
1 ,
1
d d
, , d ,
d
g s c v
t t
d c s
t
r u u v
G s q s q s
r u u
α α
α
ξ ξ −
∗ −
= =
∫
∫
∫
∫
(3)
( )
( )
01 1
, d d d d .
d
t v r u u cq s s u v
α
ξ ξ
∞ ∞ ∞
= ∞
∫ ∫
∫ ∫
(4)
If
( ) ( )(
)
(
( )
)
(
)
(
( ) ( )
)
2 1 1
1 1
lim sup 1 , d
1
t t
t s q s p g s c t s r s s
α α
α α
δ
α δ
∗
+ →∞
− − − = ∞
+
∫
(5)where
( )
( )
1
1 d ,
t
t s
rα s
δ =
∫
∞(6)
then every solution x t
( )
of (E) is either oscillatory or converges to zero as t→ ∞.Proof. Assume that x t
( )
is a positive solution of (E). Based on the condition (1),there exist three possible cases
(1) z t
( )
>0,z t′( )
>0,z t′′( )
>0,(
r t( ) ( )
z t′′ α)
′≤0;(2) z t
( )
>0,z t′( )
<0,z t′′( )
>0,(
r t( ) ( )
z t′′ α)
′≤0;(3) z t
( )
>0,z t′( )
>0,z t′′( )
<0,(
r t( ) ( )
z t′′ α)
′≤0,for t≥t1, t1 is large enough. We consider each of three cases separately. Suppose first
that z t
( )
has the property (1). We define the function ω( )
t by( )
( )
( ) ( )
(
)
( )
(
)
.r t z t
t t
z t
α
α
ω =ρ ′′
′ (7)
Then, ω
( )
t >0 for t≥t1. Using z t′( )
>0, we have( )
( )
( )
(
( )
)
( )
( )
(
( )
)
( )
(
( )
)
( )
(
) ( )
, , d
, , d
, , d
1 .
b a b a
b a
x t z t p t x t
z t p t z t
z t z t b p t
p z t
η τ η η
η τ η η
τ η η
= −
≥ −
≥ −
≥ −
∫
∫
∫
(8)Since
( )
(
( ) ( )
)
( )
(
( ) ( )
)
( )
1 1
1
1
1 1
1
d d ,
t t
t t
r s z s
z t s r t z t s
r s r s
α α
α α
α α
′′
′ ≥
∫
≥ ′′ ∫
we have that
( )
( )
11 0.
d
t t
z t
r α s s
′
−
′
≤
∫
(9)
( )
( )
( )
( )
( )
( )
( )
( )
2 1 1 2 1 1 1 2 1 1 1 d d dd d ,
d t s t t s t t s t t t t z t
z t z t r u u s
r u u
z t
r u u s
r u u
α α α α − − − − ′ = + ′ ≥
∫
∫
∫
∫ ∫
∫
(10)for t≥ >t2 t1. Differentiating (7), we obtain
( )
( ) ( ) ( )
(
)
( )
(
)
( )
( ) ( )
(
)
( )
(
)
( ) ( ) ( )
(
(
( )
)
)
1 1 .r t z t
r t z t r t z t
t t t t
z t z t z t
α
α α
α α α
ω
ρ
ρ
αρ
+ + ′ ′′ ′′ ′′ ′ = ′ + − ′ ′ ′
It follows from (E), (7) and (8) that
( )
( )
( ) (
)
(
( )
)
( )
(
)
( ) ( )
( )
(
( ) ( )
( )
)
1 1 ,, d 1 ,
d c
z g t c t t
t t q t p t
t
z t t r t
α α α α α α ρ ω
ω ρ ξ ξ ω α
ρ ρ + ′ ′ ≤ − − + − ′
∫
that is( )
( )
( ) (
)
(
( )
)
( )
(
)
(
)
( )
(
)
(
)
( )
(
)
( )
( ) ( )
(
( ) ( )
)
1 1 , ,, d 1
,
,
d c
z g t c z g t c
t t q t p
z t z g t c
t t t
t r t
α α α α α α α α
ω ρ ξ ξ
ρ ω α ω
ρ ρ + ′ ′ ≤ − − ′ ′ ′ + −
∫
which follows from (9) and (10) that
( )
( ) ( ) (
)
( )( )
( )
( )
( ) ( )
(
( ) ( )
)
( )
2 1 1 1 , 1 1 1 d d, d 1
d
.
g t c s
t t
d
c t
t
r u u s
t t q t p
r u u
t
t t
t
t r t
α α α α α α α
ω ρ ξ ξ
ρ ω α ω
ρ ρ − − + ′ ≤ − − ′ + −
∫
∫
∫
∫
Hence, we have
( )
( ) ( ) (
)
( )( )
( )
(
)
( ) ( )
(
)
( )
2 1 1 1 , 1 1 1 d d, d 1
d
1
. 1
g t c s
t t
d
c t
t
r u u s
t t q t p
r u u
r t t
t α α α α α α α
ω ρ ξ ξ
ρ ρ α − − + + ′ ≤ − − ′ + +
∫
∫
∫
∫
Integrating the last inequality from t3
(
>t2)
to t, we get( )
( ) ( )(
) ( )
(
)
( ) ( )
(
)
( )
3 1 3 1 11 d ,
1
t t
r s s
t s q s p G s s
s α α α α ρ ω ρ ρ α + ∗ + ′ ≥ − − +
which contradicts (2). Assume now that z t
( )
has the property (2). Using the similarproof ([1], Lemma 2), we can get limt→∞x t
( )
=0 due to condition (4). Thirdly, as-sume that z t
( )
has the property (3). From(
r t( ) ( )
z t′′ α)
′<0, r t( ) ( )
z t′′ α is de-creasing. Thus we get( ) ( )
( ) ( )
, 1.r s z′′ s α ≤r t z t′′ α s≥ ≥t t
Dividing the above inequality by r s
( )
and integrating it from t to l, we obtain( )
( )
1( ) ( )
l 1( )
d .t
z l′ ≤z t′ +rα t z t′′
∫
r−α s sLetting l→ ∞, we get
( )
1( ) ( )
1( )
0 d ,
t
z t′ rα t z t′′ ∞r−α s s
≤ +
∫
that is
( ) ( )
( )
( )
1
1
d 1.
t
r t z t
r s s
z t
α
α − ∞
′′
− ≤
′
∫
(12)
Define function ψ by
( )
( ) ( )
(
)
( )
(
)
, 1.r t z t
t t t
z t
α
α
ψ = ′′ ≥
′
(13)
Then ψ
( )
t <0 for t≥t1. Hence, by (12) and (13), we obtain( )
t α1( )
t 1δ ψ
− ≤ . (14)
Differentiating (13), we get
( )
(
( ) ( )
)
( )
(
)
( ) ( )
(
(
( )
)
)
1 1.
r t z t z t
t r t
z t z t
α α
α α
ψ
α
+
+
′ ′′
′′
′ = −
′ ′
Using z t′
( )
>0, we have (8). From (E) and (8), we have( )
( ) (
)
(
( )
)
( )
(
)
( )
( )
( )
1
,
, d 1 .
d c
z g t c z t
t q t p r t
z t z t
α α
α
α
ψ ξ ξ α
+
′′
′ ≤ − − − ′
′
∫
(15)In view of (3), we see that
( )
( )(
1)
.z t ≥z t′ t−t (16)
Hence,
( )
(
1)
0, z t
t t
′
≤
−
which implies that
( )
(
)
( )
( )
(
)
(
1)
1, ,
.
z g t c g t c t
z t t t
− ≥
By (13) and (15)-(17), we get
( )
( ) (
, d 1)
(
( )
, 1)
1( )
1( )
.d c
t q t p g t c t r t t
α α
α α α
ψ′ ≤ −
∫
ξ ξ − − −α − ψ +Multiplying the last inequality by δ
( )
t and integrating from t2( )
>t1 to t, weob-tain
( ) ( )
( ) ( )
( ) ( )(
)
(
( )
)
( ) ( )
( )
( )
( )
22 2
2 2 1
1 1
0 1 , d
d d ,
t t
t t
t t
t t t t s q s p g s c t s
s s s
s s
r s r s
α α
α α
α
ψ δ ψ δ δ
αψ δ ψ
∗
+
≥ − + − −
+ −
∫
∫
∫
which follows that
( ) ( )
( ) ( )(
)
(
( )
)
(
)
(
( ) ( )
)
2
2 2 1 1
1 1
1 1 , d ,
1
t t
t t s q s p g s c t s
s r s
α α
α α
ψ δ δ
α δ
∗
+
+ ≥ − − −
+
∫
which contradicts (5). This completes the proof.
3. Examples
The following examples illustrate applications of our result in this paper.
Example 3.1. For t≥1 and λ>0, considerthe third-order differential equation
( )
(
)
4
1 1
3
1 5
0 0
3 2
d d 0.
2
t
t x t p x x t
t λ
η ξ ξ ξ
′
′′
+ + − =
∫
∫
(18)
Let ρ
( )
t =1, α=1, a=0, b=1, c=0, d=1,( )
4 3,
r t =t p t n
( )
, =p1 suchthat 1
1 0
0≤
∫
pdη≤ <p 1,( )
, , 2 t tτ η = q t
( )
,ξ =2λξ t35, g t( )
,ξ = −t ξ. Note that,( )
( )
0
1 4 1
3 3
1
d d 3 , 3 ,
tr s s s s t t
α δ
− − −
∞ ∞
= = < ∞ =
∫
∫
and
4 1 3
5
1 0
3 2
d d d d .
vu u s u v
s
λξ ξ
−∞ ∞ ∞
= ∞
∫ ∫
∫ ∫
Furthermore
( ) ( )(
) ( )
(
)
( ) ( )
(
)
( )
(
)
3
3
1 1
2 1
5 3 3
1 3
1 1
3 3
1
1
lim sup 1 d
1
9 6
lim sup 1 d ,
6
t t t
t t t
r s s
s q s p G s s
s
s t s
p s s
s t
α α
α α
ρ ρ
ρ α
β λ
+ ∗
+ →∞
− −
→∞ − −
′
− −
+
− +
= − = ∞
−
∫
∫
such that q∗
( )
s , G s( )
are defined as in (3) and1 2
3 3
1 2 2
6t t 9t
( ) ( )(
)
(
( )
)
(
)
(
( ) ( )
)
(
) (
)
2 2 1 1 2 1 1 1 1lim sup 1 , d
1
1
lim sup 3 1 d .
12 t t t t t t
s q s p g t c t s
s r s
p s s t s s
α α α α δ α δ λ ∗ + →∞ − − →∞ − − − + = − − − = ∞
∫
∫
Using our result, every solution of (18) is either oscillatory or converges to zero as
t→ ∞ if λ>1 36 1
(
−p)
.Example 3.2. For t≥1 and µ>0, consider the third-order differential equation
( )
3
2 1
5 2 3
1 0
2
d d 0.
1 3 3 2
t
t x t x t x t
t
η η η µ −ξ ξ ξ
′ ′′ + + + − = +
∫
∫
(19)Let ρ
( )
t =1, α =3, a=1, b=2, c=0, d =1, r t( )
=t5,( )
, 1p t n t
η =
+ such
that 2
(
(
)
)
13
0 1 d 1,
4 t
η η
≤
∫
+ ≤ < τ η( ) (
t, = +t η)
2 ,( )
, 2 2 , 3 q tξ = µt−ξ( )
, .2
g t ξ = −t ξ
Note that,
( )
( )
1 5 2
3 3
1 0
3 3
d d , ,
2 2
tr s s s s t t
α δ
− − −
∞ ∞
= = < ∞ =
∫
∫
and 1 3 1 5 2 1 0 2d d d d . 3
v u u µξ ξs s u v
∞ ∞ − ∞ = ∞
∫ ∫
∫ ∫
Furthermore( ) ( )(
) ( )
(
)
( ) ( )
(
)
( )
( ) ( )
3 3 1 1 3 1 2 3 3 2 14 3 2 2
3 3
1 1
lim sup 1 d
1
9 3
lim supp d ,
3 4 t t t t t t
r s s
s q s p G s s
s
s t s
s s s t α α α α ρ ρ ρ α β µ + ∗ + →∞ − − →∞ − − ′ − − + − + = = ∞ −
∫
∫
such that q∗
( )
s , G s( )
are defined as in (3) and2 1
3 3
1 2 2
3t t 9 ,t
β = − −
( ) ( )(
)
(
( )
)
(
)
( ) ( )
(
)
(
)
2 2 1 1 8 3 13 3 1 1 1lim sup 1 , d
1
1
lim sup d .
128 864 t t t t t t
s q s p g t c t s
s r s
s s t s s
α α α α δ α δ µ ∗ + →∞ − − →∞ − − − + = − − = ∞
∫
∫
Using our result, every solution of (19) is either oscillatory or converges to zero as
t→ ∞ if λ>0 for some 1 0, 1 . 12
t ∈
( )
(
)
( )
5
2 1
4
7
1 0
4
1 3
d d 0.
3
t x t x t x t
t γ
η η ξ ξ
′
′′
+ − + =
∫
∫
(20)
Let ρ
( )
t =1, α =1, a=1, b=2, c=0, d=1,( )
5 4,
r t =t
( )
, 1 3p t n = such that
1 0
1 1
0 d 1,
3 η 3
≤
∫
≤ < τ η( )
t, = −t η,( )
7 4
, 3 ,
q t ξ = γξ t g t
( )
,ξ =t. Note that,( )
( )
1 5 1
4 4
1 0
d d 4 , 4 ,
tr s s s s t t
α δ
− − −
∞ ∞
= = < ∞ =
∫
∫
and
5 1 4
7
1 0
4 3
d d d d .
2
vu u s u v
s
γξ ξ
−∞ ∞ ∞
= ∞
∫ ∫
∫ ∫
Furthermore
( ) ( )(
) ( )
(
)
( ) ( )
(
)
( )
33
1 1
3 1
7 4 4
1 4
1 1
4 4
1
1
lim sup 1 d
1
4 3
lim sup d ,
3
t t t
t t t
r s s
s q s p G s s
s
s t s
s s
s t
α α
α α
ρ ρ
ρ α
β γ
+ ∗
+ →∞
− −
→∞ − −
′
− −
+
− +
= = ∞
−
∫
∫
such that q∗
( )
s , G s( )
are defined as in (3) and1 3
4 4
1 2 2
3t t 4t
β = − − ,
( ) ( )(
)
(
( )
)
(
)
(
( ) ( )
)
(
)
2
2
1 1
2 1
1
1 1
lim sup 1 , d
1
1
lim sup 4 d .
16
t t t
t t t
s q s p g t c t s
s r s
s s t s s
α α
α α
δ
α δ
γ ∗
+ →∞
− −
→∞
− − −
+
= − − = ∞
∫
∫
Using our result, every solution of (20) is either oscillatory or converges to zero as
t→ ∞ if 1. 64
λ>
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http://dx.doi.org/10.1016/j.mcm.2010.02.011
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