Journal of Chemical and Pharmaceutical Research, 2014, 6(7):1370-1377
Research Article
CODEN(USA) : JCPRC5
ISSN : 0975-7384
Oscillation of the systems of impulsive hyperbolic partial differential
equations
Xiaoqi Ning* and Shujun You
Department of Mathematics, Huaihua University, Huaihua, China
_____________________________________________________________________________________________
ABSTRACT
The systems of impulsive hyperbolic partial differential equations with Robin boundary value condition are investigated. Several new sufficient conditions of oscillation for such systems are established by employing impulsive differential inequalities and integration.
Key words: Impulsive; Delay; The systems of partial differential equations; Oscillation
_____________________________________________________________________________________________
INTRODUCTION
As an important study field of impulsive partial differential equation, the oscillation theory of impulsive partial differential equation has various applications therefore it has aroused great study interests in recent years, and the study about oscillation theory of impulsive partial differential equations also gradually draws people's attention [1-6]. This paper considers equations
2 2
1
1
[ ( , )
( ) (
, )]
[ ( , )
( ) (
, )]
( )
( , )
( )
(
( ), )
( , )
(
( ), ), ( , )
,
,
1, 2,
,
{1, 2,
, },
( , )
( , )
( , ),
1, 2,
,
i i i i
s
i i il i l
l m
ij j j
k m
i k i k k i k
u t x
c t u t
x
u t x
c t u t
x
t
t
a t
u t x
a t
u t
t x
p t x u t
t x
t x
R
t
t k
i
I
m
u t
x
u t
x
u t x
k
τ
τ
ρ
σ
α
=
+ =
+ −
∂
+
−
+
∂
+
−
∂
∂
=
∆
+
∆
−
−
−
∈ ×Ω
≠
=
∈
=
−
=
=
∑
∑
L
L
L
,
( , )
( , )
( , )
,
1, 2,
,
,
m
i k i k i k
k m
i
I
u t
x
u t
x
u t x
k
i
I
t
t
β
t
+ −
∈
∂
∂
∂
−
=
=
∈
∂
∂
∂
L
(1)
with boundary condition
i
( , )
( , ) ( , )
0,
,
,
,
i i k m
u t x
g t x u t x
x
t
t i
I
N
∂
+
=
∈∂Ω ≠
∈
∂
(2)where,
[0,
),
,
( , ),
ni i
2 2 1
( , )
( , )
, ( , )
,
.
n i
i m
r r
u t x
u t x
t x
G i
I
x
=∂
=
∈
∈
∂
∑
We study the oscillation problem of equation (1) with boundary condition (2), obtain sufficient condition for the oscillation of all solutions and extend the previous results by using existence condition of the eventually positive solution, combined with the impulsive differential inequality. In the following, we will use the conditions as follows:
1 2
( 1)
0
k, lim
k,
k
H
t
t
t
t
→+∞
< < < < <
L
L
= +∞
lim (
( ))
lim (
j( ))
.
t→+∞
t
−
σ
t
=
t→+∞t
−
ρ
t
= +∞
1
0
≤
α
k≤
β
k,
k
=
1, 2,
L
. ,
σ ρ
l, ,
a a
i il∈
PC
([0,
+∞
),[0,
+∞
)), ( )
c t
∈
C R R
(
+,
+);
(
2)
ij( , )
( , ),
ii( , )
0,
ii( )
min
ii( , ),
ij( )
sup |
ij( , ) |,
x G x G
H
p t x
PC G R
p t x
p t
p t x
p t
p t x
∈ ∈
∈
>
=
=
1
1,
( )
min{
( )
( )}
0,
,
.
mii ji m m
i m
j j i
Q t
p t
p t
i
I
j
I
≤ ≤ = ≠
=
−
∑
≥
∈
∈
herePC
denotes piecewise smooth continuous function with the following properties: only discontinuous ont
=
t k
k,
=
1, 2,
L
which are discontinuity point of the first kind and left continuous ont
=
t k
k,
=
1, 2,
L
;(
H
3)
theN
in (2) is the unit vector normal to∂Ω
,pointing out ofΩ
,g t x
i( , )
is continuous nonnegative function on[0,
+∞ ×∂Ω
)
;(
H
4)
u t x
i( , )
is the solution satisfied the boundary conditions of equation (1), andu t x
i( , )
t
∂
∂
is the partial derivative ofu t x
i( , )
both are piecewise continuous function with discontinuity point of the first kind on,
1, 2,
k
t
=
t k
=
L
. Suppose that they satisfy the following equation at impulsion:
u t
i( , )
kx
u t x
i( , ),
ku t
i( , )
kx
(1
a u t x
k) ( , ),
i kk
1, 2,
,
i
I
m;
−
=
+= +
=
∈
L
(3)( , )
( , )
( , )
( , )
,
(1
)
,
1, 2,
,
.
i k i k i k i k
k m
u t
x
u t x
u t
x
u t x
k
i
I
t
t
t
β
t
− +
∂
=
∂
∂
= +
∂
=
∈
∂
∂
∂
∂
L
(4)1. THE MAIN RESULT
Theorem 1. Suppose condition
( 1) (
H
−
H
4)
satisfied, if second-order impulsive differential inequalities0
( )
( )
( ) (
( ))[1
(
( ))]
0,
,
,
1, 2,
,
( )
(1
) ( ),
1, 2,
,
( )
(1
)
( ),
1, 2,
,
k
k k k
k k k
w t
w t
Q t w t
t
c t
t
t
t t
t k
w t
w t
k
w t
w t
k
σ
σ
α
β
+ +
′′
+
′
+
−
−
−
≤
≥
≠
=
= +
=
′
≤ +
′
=
L
L
L
(5)
doesn’t have eventually positive solutions, then every non zero solution of problems (1)-(2) is oscillation in domain
G
.Proof. Suppose systems (1)-(2) have a nonzero and non-oscillation solution
T
1 2
( , )
( ( , ),
( , ),
,
m( , )) ,
u t x
=
u t x u t x
L
u t x
let us assume whent
≥ ≥
t
00
,we have|
u t x
i( , ) | 0,
>
i
=
1, 2,
L
,
m
. Setsgn ( , ),
( , )
( , ),
i
u t x z t x
i i iu t x
iThen
z t x
i( , )
>
0, ( , ) [ ,
t x
∈
t
0+∞ ×Ω
)
.
From condition( 1)
H
we can see there isT
1>
t
0, whent
≥
T
1 we havez t x
i( , )
>
0,
z t
i(
−
ρ
l( ), )
t x
>
0,
z t
i(
−
σ
( ), )
t x
>
0,
( , ) [ ,
t x
∈
T
1+∞ ×Ω =
)
,
i
1, 2,
L
, ,
m l
=
1, 2,
L
, .
s
Whent
≠
t
k时, we integrate both sides of equation (1) onΩ
2 2
1
1 1
( , )
( )
(
, )
( , )
( )
(
, )
( )
( , )
( )
(
( ), )
( , )
(
( ), )
,
,
,
1, 2,
,
(
i i)
(
i i)
s
i i il i l
l m
ij j m
j
d
d
u t x dx c t
u t
x dx
u t x dx c t
u t
x dx
dt
dt
a t
u t x dx
a t
u t
t x dx
p t x u t
t x dx t
T i
I
k
τ
τ
ρ
σ
Ω Ω Ω Ω
Ω = Ω
Ω =
+
−
+
+
−
=
∆
+
∆
−
+
−
−
≥
∈
=
∫
∫
∫
∫
∑
∫
∫
∑∫
L
(6)
then
(
) (
)
2 2
1
1 1
( , )
( )
(
, )
( , )
( )
(
, )
( )
( , )
( )
(
( ), )
( , ) (
( ), )
,
,
,
1, 2,
.
i i i i
s
i i il i l
l m
i
ij j m
j j
d
d
z t x dx
c t
z t
x dx
z t x dx c t
z t
x dx
dt
dt
a t
z t x dx
a t
z t
t x dx
p t x z t
t x dx t
T i
I
k
τ
τ
ρ
δ
σ
δ
Ω Ω Ω Ω
Ω = Ω
Ω =
+
−
+
+
−
=
∆
+
∆
−
+
−
−
≥
∈
=
∫
∫
∫
∫
∑
∫
∫
∑ ∫
L
(7)
It can be derived from Green Theorem and the boundary condition (2) that
( , )
( , )
i( , ) ( , )
0,
i i i
z t x
z t x dx
ds
g t x z t x ds
N
Ω ∂Ω ∂Ω
∂
∆
=
= −
≤
∂
∫
∫
∫
t
≥
T i
1,
∈
I
m,
k
=
1, 2,
L
,
(8)
1
(
( ), )
(
( ), )
(
( ), ) (
( ), )
0,
,
,
1, 2,
,
i l i l
i l i l
m
z t
t x
z t
t x dx
ds
N
g t
t x z t
t x ds
t
T i
I
k
ρ
ρ
ρ
ρ
Ω ∂Ω
∂Ω
∂
−
∆
−
=
∂
= −
−
−
≤
≥
∈
=
∫
∫
∫
L
(9)
here
ds
denotes the area element on∂Ω
. Combining (7),(8),(9) and condition(
H
2)
, we can work out the form
(
) (
)
2 2
1,
1
( , )
( )
(
, )
( , )
( )
(
, )
( )
(
( ), )
( )
(
( ), )
,
,
,
1, 2,
.
i i i i
m
ii i ij j
j j i m
d
d
z t x dx
c t
z t
x dx
z t x dx c t
z t
x dx
dt
dt
p t
z t
t x dx
p t
z t
t x dx
t
T i
I
k
τ
τ
σ
σ
Ω Ω Ω Ω
Ω = ≠ Ω
+
−
+
+
−
≤ −
−
+
−
≥
∈
=
∫
∫
∫
∫
∑
∫
∫
L
(10)
Set
1
( )
( , )
,
,
,
i i m
v t
z t x dx
t
T i
I
Ω
=
∫
≥
∈
By form (10), we have
1,
1
[ ( )
( ) (
)]
[ ( )
( ) (
)]
[
( ) (
( ))
( ) (
( ))]
0,
,
,
1, 2,
.
i i i i
m
ii i ij j
j j i k
v t
c t v t
v t
c t v t
p t v t
t
p t v t
t
t
T t
t k
τ
τ
σ
σ
′′ ′
= ≠
+
−
+
+
−
+
+
−
−
−
≤
≥
≠
=
∑
L
Set
1 1
( )
( ),
,
m i i
V t
v t
t
T
==
∑
≥
Combining this form and form (11), we have
1 ,
1
1
[
( ) (
( ))
( ) (
( ))]
[ ( )
( ) (
)]
[ ( )
( ) (
)]
0,
,
1, 2,
m m
ii i ij j
i j j i
V t
c t V t
V t
c t V t
t
T
p t v t
t
p t v t
t
k
σ
σ
τ
′′τ
′= = ≠
+
−
+
+
−
≤
≥
=
+
+
∑
−
−
∑
−
L
(12)
and hence
11 1 1
1, 1
22 2 2
1, 2
1,
1
[ ( )
( ) (
)]
[ ( )
( ) (
)]
( ) (
( ))
( ) (
( ))
( ) (
( ))
(
( ))
( )
(
( ))
( ) (
( ))
0,
,
1, 2,
.
{[
]
[
]
[
]}
m
j j j j
m
j j j j
m
mm m mj j
j j m
V t
c t V t
V t
c t V t
p
t v t
t
p
t v t
t
p
t v t
t
p v t
t
p
t v t
t
p
t v t
t
t
T k
τ
τ
σ
σ
σ
σ
σ
σ
′′ ′
= ≠
= ≠
= ≠
+
−
+
+
−
+
+
−
−
−
+
+
−
−
−
+
+ +
−
−
−
≤
≥
=
∑
∑
∑
L
L
furthermore we have
11 1 1 22 2 2
1, 1 1, 2
1 1,
[ ( )
( ) (
)]
[ ( )
( ) (
)]
( )
( )
(
( ))
( )
( )
(
( ))
( )
( )
(
( ))
0,
,
1, 2,
,
{[
]
[
]
[
]
}
m m
j j
j j j j
m
mm jm m
j j m
V t
c t V t
V t
c t V t
p
t
p
t v t
t
p
t
p
t v t
t
p
t
p
t v t
t
t
T k
τ
τ
σ
σ
σ
′′ ′
= ≠ = ≠
= ≠
+
−
+
+
−
+
−
−
+
−
−
+
+ +
−
−
≤
≥
=
+
∑
∑
∑
L
L
then
1 1
1, 1
[ ( )
( ) (
)]
[ ( )
( ) (
)]
min
{
( )
( )
}
(
( ))
0,
,
1, 2,
,
m m
ii ji i
i m
j j i i
V t
c t V t
V t
c t V t
p t
p t
v t
t
t
T k
τ
τ
σ
′′ ′
≤ ≤ = ≠ =
+
−
+
+
−
+
−
−
≤
≥
+
=
∑
∑
L
Consequently, we have
1
[ ( )
( ) (
)]
[ ( )
( ) (
)]
( ) (
( ))
0,
,
1, 2,
.
V t
c t V t
V t
c t V t
Q t V t
t
t
T k
τ
′′τ
′σ
+
−
+
+
−
+
−
≤
≥
=
L
(13)Set
1
( )
( )
( ) (
)
( )
0,
,
w t
=
V t
+
c t V t
− ≥
τ
V t
>
t
≥
T
then form (13) can be rewrite as
w t
′′
( )
+
w t
′
( )
+
Q t V t
( ) (
−
σ
( ))
t
≤
0,
t
≥
T k
1,
=
1, 2,
L
,
(14) Thus, we have1
[
e w t
t′
( )]
′+
e Q t V t
t( ) (
−
σ
( ))
t
≤
0,
t
≥
T k
,
=
1, 2,
L
.
From the above discussion , we can see that
( )
0,
(
( ))
0, [
t( )]
0,
k,
1, 2,
,
w t
>
V t
−
σ
t
>
e w t
′
′≤
t
≠
t k
=
L
So
e w t
t′
( )
is a monotone decreasing function in( ,
t t
k k+1]
,w t
′
( )
is also a monotone decreasing function in1
( ,
t t
k k+]
, it meansw t
′′ ≤
( )
0,
t
≠
t k
k,
=
1, 2,
L
.
( , )
( , )
( , )
(1
) ( , ),
i k(1
)
i k,
i k k i k k
u t
x
u t x
u t
x
u t x
t
t
α
+β
+
= +
∂
= +
∂
∂
∂
Hence
( , )
( , )
sgn ( , )
sgn ( , ), sgn
i ksgn
i k,
i k i k
u t
x
u t x
u t
x
u t x
t
t
+
+
=
∂
=
∂
∂
∂
then
1
1 1
1
1
( ( , )
( , ))
sgn( ( , )) ( , ) sgn( ( , )) ( , )
sgn( ( , ))[ ( , )
( , )]
sgn( ( , )
( )
( )
[ ( )
( )
)
(
]
m
i k i k i
m
i k i k i k i k i
m
i k m
k k
i k i k i
m
i k k i k i
i k i k i
z t
x
z t
x dx
u t
x u t
x
u t
x u t
x dx
u t x
u t
x
u t
x d
V t
V t
x
v
u
u
t
t
v
t
t
x
α
+ −
Ω =
+ + − −
+ −
Ω =
+ −
Ω =
Ω =
+ −
=
=
−
=
−
=
−
=
−
=
−
∑
∑
∑∫
∑
∫
∑∫
∫
, )
( ),
k k
x dx
V t
α
=
furthermore ,we have( )
( )
( )
( ) (
) [ ( )
( ) (
)]
( )
( )
( )[ (
)
(
)]
[ ( )
( ) (
)]
( ),
k k k k k k k k
k k k k k
k k k k k k
w t
w t
V t
c t V t
V t
c t V t
V t
V t
c t
V t
V t
V t
c t V t
t
τ
τ
τ
τ
α
τ
α ω
+ − + + + − − −
+ − + −
−
=
+
− −
+
−
=
−
+
− −
−
=
+
−
=
Thus, we have
w t
( )
kw t
( )
kα
kw t
( ).
k+
−
−=
Note that( )
k( ),
kw t
−=
w t
we can also havew t
( )
k+= +
(1
α
k) ( ).
w t
k (15) Similarly, we can prove( )
k( )
k k( ).
kw t
′
+−
w t
′
−≤
β
w t
′
note that1
1
1 1
(
( )
( )
[ ( )
( )]
, )
( , )
( , )
( , )
( , )
( , )
sgn(
)
sgn(
)
( , )
( , )
(
sgn(
)
m
i k i k i
m
i k i k i
m
k k i k i
k i k i
m
i k k i
k i
i k
i
z t
x
z t
x
dx
t
t
u t
x
u t
x
u t
x
u t
x
dx
t
t
t
t
u t x
u t
x
u t
t
t
V t
V t
v t
v t
+ −
Ω =
+ + − −
Ω =
+ − + −
+ −
Ω =
=
∂
∂
=
∂
−
∂
′
−
′
=
′
′
∂
∂
∂
∂
=
∂
∂
−
∂
∂
∂
∂
∂
=
−
−
∂
∂
∑∫
∑∫
∑∫
∑
1
1
, )
( , )
( , )
sgn(
)
( )
( ),
m
i k i k k i
m
k i k k k i
x
dx
t
u t x
u t x
dx
t
t
v t
V t
β
β
β
Ω =
=
∂
∂
∂
=
∂
∂
′
′
=
=
∑∫
then
( )
( )
[ ( )
( ) (
)]
[ ( )
( ) (
)]
( ) [ ( ) (
)]
( ) [ ( ) (
)]
( )
( )
( ) (
)
( )
(
)
( ) (
)
( )
(
)
[
k k k k k k k k
k k k k k k
k k k k k k
k k k k
w t
w t
V t
c t V t
V t
c t V t
V t
c t V t
V t
c t V t
V t
V t
c t V t
c t V t
c t V t
c t V t
τ
τ
τ
τ
τ
τ
τ
τ
+ − + + + ′ − − − ′
+ + + ′ − − − ′
+ − + + + +
− − − −
′
−
′
=
+
−
−
+
−
′
′
=
+
−
−
−
−
′
′
′
′
=
−
+
− +
−
′
′
−
− −
−
′
=
( )
( )]
( )[ (
)
(
)]
( )[
(
)
(
)]
( )
( )
(
)
( )
(
)
[
( )
( ) (
)
( )
(
)]
[ ( )
( ) (
)]
( ),
k k k k k
k k k
k k k k k k k k
k k k k k k
k k k k k k
V t
V t
c t
V t
V t
c t
V t
V t
V t
c t
V t
c t
V t
V t
c t V t
c t V t
V t
c t V t
w t
τ
τ
τ
τ
β
α
τ
β
τ
β
τ
τ
β
τ
β
+ − + −
+ −
′
′
′
−
+
− −
−
′
′
+
− −
−
′
′
′
=
+
− +
−
′
′
′
≤
+
− +
−
=
+
−
′
=
thus, we have
( )
k( )
k k( ).
kw t
′
+−
w t
′
−≤
β
w t
′
furthermore , becausew t
′
( )
k−=
w t
′
( ),
k we have
w t
′
( )
k+≤ +
(1
β
k)
w t
′
( ).
k (16) By using (15),(16),we can work out1
( )
0,
,
k.
w t
′ ≥
t
≥
T t
≠
t
In fact, if the above form is wrong then
∃ ≥
T
2T
1,
such thatw T
′
( )
2<
0,
let us assumew T
′
( )
2= −
c c
, (
>
0).
By( )
0,
k,
1, 2,
,
w t
′′ ≤
t
≠
t k
=
L
w t
′
( )
k+≤ +
(1
β
k)
w t
′
( ),
kt
=
t k
k,
=
1, 2,
L
,
together with lemma 5.1.3 in [6], we have
2 2
2 2
( )
( )
(1
)
(1
),
.
k k
k k
T t t T t t
w t
w T
β
c
β
t
T
< < < <
′
≤
′
∏
+
= −
∏
+
≥
Note that
w t
( )
k+= +
(1
α
k) ( ),
w t
k then we have2
2 2
2
2 2
2
2
( )
( )
(1
)
(1
)
(1
)
1
(1
)
( )
,
1
k k k
k k
t
k T k k
T t t s t t T t s t
k
k T
T t t T t s k
w t
w T
c
ds
w T
c
ds
α
α
β
β
α
α
< < < < < <
< < < <
≤
+
−
+
+
+
=
+
−
+
∏
∫
∏
∏
∏
∫
∏
note
0
<
α
k≤
β
k in( 1)
H
, therefor we havew t
( )
≤
0,
t
≥
T
2,
according the above form , this contradicts tow t
( )
>
0,
t
≥
T
1. Byw t
( )
=
V t
( )
+
c t V t
( ) (
−
τ
),
t
≥
T t
1,
≠
t k
k,
=
1, 2,
L
. together with form (14) we have1
( )
( )
( )[ (
( ))
(
( )) (
( )
)]
0,
,
k,
1, 2,
.
w t
w t
Q t w t
t
c t
t V t
t
t
T t
t k
σ
σ
σ
τ
′′
+
′
+
−
−
−
−
−
≤
≥
≠
=
L
(17)Since
( )
( ),
( )
0,
k,
1, 2,
,
1
( )
( )
( ) (
( ))[1
(
( ))]
0,
,
k,
1, 2,
.
w t
′′
+
w t
′
+
Q t w t
−
σ
t
−
c t
−
σ
t
≤
t
≥
T t
≠
t k
=
L
(18)therefor combining form (18) and form (15),(16) we can see that
w t
( )
is a eventually positive solution of differential inequality (5),this contradicts to condition of theorem 1,and we complete the proof of theorem1. Theorem 2. Suppose condition( 1) (
H
−
H
4)
satisfied, further more suppose(i)
0
≤
σ
( )
t
≤
σ
, 0
≤
ρ
j( )
t
≤
ρ
,
j
=
1, 2,
L
, ;
m
(ii) if1 1
,
,
k k
k k
α
β
∞ ∞
= =
< +∞
< +∞
∑
∑
and for each are sufficiently largeT
, we have{
}
(1
)
(1
)
( )[1
(
( ))]
lim inf
,
(1
)
(1
)
k k
k k
t
k k
T T
t t s t t t
k k
T
T t t t
Q s
c s
s
dsd
d
η
η η
η η
α
β
σ
η
β
α η
< < < < →+∞
< < < <
+
+
−
−
−
= −∞
+
+
∏
∏
∫
∫
∏
∏
∫
Then every nonzero solution of problems (1)-(2) is oscillation in domain G
Proof. By theorem 1 we only need to prove the nonexistence of the eventually positive solution of the second order
impulsive differential inequality (5). Suppose
w t
( )
is a eventually positive solution of the impulsive differential inequality (5), then there exists a real numberT
≥
max{ , },
σ ρ
andw T
(
−
σ
)
≥
0,
in form (5), we notice that( )
0
w t
′ ≥
andw t
(
−
σ
( ))
t
≥
0,
t
≥
T t
,
≠
t k
k,
=
1, 2,
L
,
hence we have( )
(
) ( )[1
(
( ))]
0.
w t
′′ +
w T
−
σ
Q t
−
c t
−
σ
t
≤
(19) combining (19) andw t
′
( )
k+≤ +
(1
β
k)
w t
′
( ),
k in (5), we can use lemma 5.1.3 in [6] to get( )
( )
(1
)
(1
)
{
(
) ( )[1
(
( ))]
}
,
k
k
k T t t t
k T
s t t
w t
w T
w T
Q s
c s
s
ds
β
β
σ
σ
< <
< <
′
≤
′
+
+
+
+
−
−
−
−
∏
∏
∫
(20)and together with
w t
( )
k+= +
(1
α
k) ( )
w t
k in (5) we have( )
( )
(1
)
(1
)
( )
(1
)
(1
)
(
) ( )[1
(
( ))]
( )
(1
)
( )
(1
)
(1
)
(
)
(1
)
(1
)
{
[
] }
k k k
k
k k k
k k
t
k T k k
T t t t t T t
k T
s t
t
k T k k
T t t t t T t
t
k k
T T
t t s t
w t
w T
w T
w T
Q s
c s
s
ds d
w T
w T
d
w T
η η
η
η
η η
η η
α
α
β
β
σ
σ
η
α
α
β η
σ
α
β
< < < < < <
< <
< < < < < <
< < < <
′
≤
+
+
+
+
+
+
+
−
−
−
−
′
=
+
+
+
+
+
+
−
+
+
∏
∫
∏
∏
∏
∫
∏
∫
∏
∏
∏
∏
∫
∫
η[
−
Q s
( )[1
−
c s
(
−
σ
( ))]
s
]
dsd
η
(21)
1 1 1 1
1
1 1 1 1
1
1 1
1
1 1 1
1 1
(1
)
(1
)
(1
)
(1
)
(1
)
(1
)
(1
)(
)
(1
)
(1
)(
)
(1
)(
)
k k
j m
j m
t
k k
T
T t t t
j
m m m m
t t t
k k k k
T t t
j
k k k j k
j
m m m
k k k j j
j
k k k j
m
k m m k
d
d
d
d
t
T
t
t
t
t
η η
β
α η
α η
β
α η
β η
α
β
α
β
+ +
< < < < −
=
= = = + =
−
+ =
= = = +
+ =
+
+
≤
+
+
+
+
+
+
=
+
− +
+
+
−
+
+
−
< +∞
∏
∏
∫
∑
∏
∏
∏
∏
∫
∫
∫
∑
∏
∏
∏
∏
.
(22)
For
t
>
T
, the inequality (21) with both sides divided by the left side of (22), we obtain(1
) ( )
( )
( )
(1
)
(1
)
(1
)
(1
)
(
)
(1
)
(1
)
( ) 1
(
( ))
,
(1
)
(1
)
{
[
]}
k
k k k k
k k
k k
k T t t
t t
k k k k
T T
T t t t T t t t
t
k k
T T
t t s t t
k k
T
T t t t
T
t
T
d
d
T
Q s
c s
s
dsd
d
η η η η
η
η η
η η
α ω
ω
ω
β
α η
β
α η
ω
σ
α
β
σ
η
β
α η
< <
< < < < < < < <
< < < <
< < < <
+
′
≤
+
+
+
+
+
+
−
+
+
−
−
−
+
+
+
∏
∏
∏
∏
∏
∫
∫
∏
∏
∫
∫
∏
∏
∫
Notes
(1
)
(1
)
,
,
k k
t
k k
T
T t t t
d
t
η η
β
α η
< < < <
+
+
→ +∞
→ +∞
∏
∏
∫
We get
( )
lim inf
.
(1
)
(1
)
k k
t t
k k
T
T t t t
w t
d
η η
β
α η
→+∞
< < < <
= −∞
+
+
∏
∏
∫
(23)According to condition (ii) in theorem 2, on the other hand by
w t
( )
>
0,
t
≥
T
,
we have( )
lim inf
0,
(1
)
(1
)
k k
t t
k k
T
T t t t
w t
d
η η
β
α η
→+∞
< < < <
≥
+
+
∏
∏
∫
This form contradicts to (23), hence impulsive differential equation (5) doesn’t have eventually positive solution, and we complete the proof of theorem 2.
REFERENCES
[1] L.P.Luo, Z.G.Ouyang. Oscillation of systems of a class of impulsive delay neutral parabolic equations. Acta
Analysis Functionalis Applicata, 8(3): 257-264, 2006.
[2] Q.Zhao,W.A.Liu. Oscillation of solution of a class of impulsive delay neutral hyperbolic equations. Journal of
Mathematics, 26(5): 563-568, 2006.
[3] Zhang Y T, Luo Q. On the forced oscillation of solutions for systems of impulsive neutral parabolic differential equations with several delays. J. of Math(PRC), 26(3): 272-276, 2006.
[4] L.P.Luo,Z.G.Ouyang . Oscillation of the system of impulsive neutral felay parabolic partial differential equations. Acta Mathematicae Applicatae Sinica, 30(5): 822-830, 2007.
[5] L.P.Luo. Oscillation of impulsive delay neutral hyperbolic equations with nonlinear diffusion coefficient.
Progress in Nature Science, 18(3): 341-344, 2008.