On the Oscillation of Higher Order Impulsive Neutral Partial Differential Equations with Distributed

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International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

On the Oscillation of Higher Order Impulsive Neutral Partial Differential Equations with Distributed Deviating Arguments V. Sadhasivam1, T. Raja2, T. Kalaimani3 PG and Research Department of Mathematics, Thiruvalluvar Government Arts College, Rasipuram, Namakkal (Dt), Tamil Nadu, India.

(Affiliated to Periyar University, Salem) E-mail: ovsadha@gmail.com, trmaths19@gmail.com, kalaimaths4@gmail.com 1,2,3

Abstract — In this work, we consider a class of boundary value problems associated with even order nonlinear impulsive neutral partial functional differential equations with continuous distributed deviating arguments and damping term. Necessary and Sufficient conditions are obtained for the oscillation of solutions using impulsive differential inequalities and integral averaging scheme with Robin boundary condition. Examples are specified to point up our important results.

2010 Mathematics Subject Classification: 35B05, 35L70, 35R10, 35R12.

and the references they are cited. Consequently, it is required to study with impulse effect on higher order partial differential equations. In the monographs, J. H. Wu [22], N. Yoshida [23] provided several fundamental theories and applications of partial functional differential equations to population ecology, generic repression, climate models, viscoelastic materials, control problems, coupled oscillators, beam equations and structured population models. Strong interest on these mathematical models, we formulated this higher order problem. In this effort, we begin oscillation criteria for even order impulsive neutral partial differential equation with damping which was not formerly studied. In the consequence, the main results of this manuscript is the generalization of earlier results studied in [2,8,9,10,11,21] with additional force components along the system such as impulse, damping and distributed delay. Distributed delay is broad case of constant delay and the authors provided in the monographs [6, 7].

Keywords - Neutral partial differential equations; Oscillatio; Impulse; Distributed delay; Damping.

I. INTRODUCTION The oscillation of impulsive and non-impulsive parabolic and hyperbolic equations has been extensively studied in the literature, we refer the readers to the papers [1, 4, 8, 12, 14-19]

In this article, we consider the following impulsive system with damping of the form   a  m 1  b b     p (t ) g ( t ,  ) u ( x ,  ( t ,  )) d  (  )   q ( t ,  ) u ( x ,  ( t ,  )) d  (  )   u ( x,t)  m 1 a a  t   n   = a (t )  u ( x , t )  b j ( t )  u ( x ,  j ( t )), t  t k , ( x , t )    [0,  )  G  j =1  (i)  (i)   u ( x, tk )  u ( x, tk )   (i)   = Ik x, tk , , k = 1,2,  , i = 0,1,2,  , m  1  (i) (i)   t t    m 1      u ( x,t)   r (t ) m 1  t  t 



b

 g ( t ,  ) u ( x ,  ( t ,  )) d  (  )    





(1)



where  is a bounded domain in R N with a piecewise smooth boundary   and Equation (1) is supplemented by the following Robin boundary conditions,  (x)

u ( x, t) 

  ( x ) u ( x , t ) = 0,

where  is the outer surface normal vector to



and



is the Laplacian in the Euclidean space

( x , t )     [0,  )

 ,   C   , [0,  )  , 

2

(x)  

R

N

.

(2) 2

(x)  0

.

In the sequel, we assume that the following hypotheses ( H ) hold: (H1)

r ( t )  C  [0,  ), (0,  )  , r  ( t )  0

,

p ( t )  C ([0,  ), R )

,



t

(H2)

a ( t ), b j ( t )  PC ([0,  ), [0,  ))

continuous in

t

(H3) g ( t ,  )  C lim t  

,

j = 1,2,  , n

, where

with discontinuities of first kind only at m

([0,  )  [ a , b ], [0,  ))

 j ( t ) =  , j = 1,2,  , n

,



ds = 

 R ( t ) = exp   



t

t

r ( s )  p ( s )

0

r(s)

represents the class of functions which are

t = t k , k = 1,2, 

, and left continuous at

q ( t ,  )  C ([0,  )  [ a , b ], [0,  ))

, a, b are non-positive constants with

5

, where

R (s)

0

PC

1

a  b.

,

.

piecewise

t = t k , k = 1,2, 

 j ( t )  C ([0,  ), R )

 ds   

,

.  j (t )  t

,

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(H4)  ( t ,  ),  ( t ,  )  C ([0, 

)  [ a , b ], R )

,

 (t ,  )  t ,

with respect to t and  respectively and

 (t ,  )  t

 (t ,  ) =

liminf

t   ,   [ a , b ]

(H5)There exist a function  ( t )  C ([0, 

), [0,  ))

International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397   [ a , b ] ,  ( t ,  ) and  ( t ,  ) are nondecreasing

for

 ( t ,  ) = 

liminf

.

t   ,   [ a , b ]

satisfying  ( t )

  (t , a )

,

'

 (t ) > 0

lim  ( t ) = 

and

,

 ( ) : [ a , b ]  R

is

t  

nondecreasing and the integral is a Stieltjes integral in (1). 

(H6)

u ( x, t)

t

t = tk ,

(H7)

(i)



(i)

are piecewise continuous in t with discontinuities of first kind only at

(i)

u ( x, t k ) t

(i)



(i)

=

(i)

(i)

t

,

(i)

k = 1,2, 

i = 0,1,2,  , m  1

,

i = 0,1,2,  , m  1, k = 1,2,  ,

such that for

,

, and left continuous at

k = 1,2, 



u ( x, t k )

.

(i)   u ( x, tk )  (i)   PC (   [0,  )  R , R ), k = 1,2,  , i = 0,1,2,  , m  1 I k  x, tk , (i)    t  

a k ,bk

t = tk

(i)

ak



  (i) I k  x, tk ,   

(i)

(i)

, and there exist positive

u ( x, t k )   (i)  t 

u ( x, t k ) t

constants

(i)

 bk .

(i)

This work is planned as follows: In Section II, we present the definitions and lemma that will be needed. In Section III, we discuss the oscillation of the problem (1), (2). In Section IV, we present some examples to illustrated the main results.

II. PRELIMINARIES In this section, we begin with definitions and well known results which are required throughout this paper. uC

Definition 2.1. A solution u of the problem (1) is a function where   t  1 = min  0, min  inf   1 j  n  t  0

j

   ( t )  , min  inf  ( t ,  )     [ a , b ]  t  0 

(   [ t  1 ,  ), R )  C (   [ tˆ  1 ,  ), R )

m

that satisfies (1),

   tˆ  1 = min  0, min  inf  ( t ,  )   .    [ a , b ]  t  0

and

Definition 2.2. The solution u of the problem (1), (2) is said to be oscillatory in the domain G if for any positive number  there exist a point ( x 0 , t 0 )    [  ,  ) such that u ( x 0 , t 0 ) = 0 holds.

Definition 2.3. A function all t  t1 .

V (t )

is said to be eventually positive (negative) if there exists a

Lemma 2.1 [20]. Suppose that the smallest eigenvalue

0 > 0

  ( x )   ( x )

= 0

 (x)

and

 (x) > 0

as  ( x ) 

is the corresponding eigenfunction of

0 ( x   )

y (t )

(t )  0

    

in on

[ t 1 ,  )

or n  l odd for

  (0,1)

. Then  0

y (t ) y

for (n)

t1 > 0

= 0,  ( x ) = 1

(t )  0

, then there exists a

; and for t  t y , y ( t ) y

and

M > 0

such that for sufficiently large t

t y  t1

(k )

as

X

 

 = 0 (x  )

  XY   XY

 1  1

 (   1) Y  (1   ) Y

 

 0,  0,

 >1 0 <  < 1,

6

[0,  )

n2

. If

and integer

( t ) > 0, 0  k  l

y  ( t )  Mt

Lemma 2.4 [3]. If X and Y are nonnegative, then X

holds for

(3)

Lemma 2.3 [13]. Suppose that the conditions of Lemma 2.2 is satisfied, and constant

V (t ) > 0 (< 0)

of the eigenvalue problem

be a positive and n times differentiable function on

identically zero on any ray y (t ) y

0

such that

and  0

> 0,  ( x ) > 0 ( x   )

.

Lemma 2.2 [5]. Let

(n)

= 0

t1  t 0

y

( n  1)

y

; (  1)

( n  1)

(t ) .

y

(n)

(t )

is constant sign and not

l (0  l  n ) k l

(t ) y

(n)

y (t ) y

(k )

( t )  0,

, with n  l even for

( t ) > 0, l  k  n

t  t y . Then

.

there exist

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International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

where the equality holds if and only if X = Y .

For each positive solution V (t ) =

u ( x, t)



of the problem (1), (2) we combine the functions

u ( x , t )  ( x ) dx ,

 ( t )

F (t ) =

 (t )



L (t ) =

M ( ( t ))

m2

 ( t )

and

g

= 1

0



b

 (t ) p (t ) r (t )

G (t ) = g 0

r (t )

where



defined by

V (t )



b

q ( t ,  ) d  ( )

a

g (  ( t ,  ),  ) d  (  ).

a

III. OSCILLATION OF THE PROBLEM

(1) AND (2)

In this section, we establish the oscillation criteria of the problem (1), (2). The Lemma 2.1 which is very useful for establishing our main results.

Theorem 3.1. If  ( x )  0 for x    , then the necessary and sufficient condition for all solutions of the problem (1), (2) to oscillate is that all solutions of the impulsive differential equation     m 1 m 1 b b     d  d d  V (t )   V (t )  g ( t ,  )V ( ( t ,  )) d  (  )    p ( t ) g ( t ,  )V ( ( t ,  )) d  (  )    r (t )     m 1 m 1 dt  a a dt dt        n  b   q ( t ,  )V (  ( t ,  )) d  (  )   0 a ( t )V ( t )   0 b j ( t )V (  j ( t )) = 0, t  tk  a  j =1  (i)    V (t k )  (i) (i) (i) t  ak   b k , k = 1,2,  , i = 0,1,2,  , m  1 (i)   V (t k )  (i)  t 









to oscillate, where

0

is the smallest eigenvalue of (3).

Proof. (i) Sufficient Part: Suppose to the contrary that there is a non-oscillatory solution no zero in

  [ t 0 ,  )

conditions

(H 3)

and

For domain

t0  0

such that  ( t ,  )

t1 > t 0 > 0 t  t1

,

k = 1,2,  ,

 t0

t0

for

multiplying both sides of equation (1) by

for

( x , t )    [ t 0 ,  )

,

( t ,  )  [ t1 ,  )  [ a , b ]

u ( x ,  ( t ,  )) > 0

u ( x ,  j ( t )) > 0

of the problem (1), (2) which has

u ( x, t)

u ( x, t) > 0

,  (t ,  ) 

then

( x , t ,  )    [ t1 ,  )  [ a , b ] and

for

, , we attain

. Without loss of generality, we assume that

, there exists a

for

t  t0 , t  tk 

for some

(H 4 )

j = 1,2,  , n

u ( x ,  ( t ,  )) > 0

(4)

for

   



u ( x , t )  ( x ) dx 



 

b

and integrating with respect to

 j (t )  t 0

,

x

over the

 g ( t ,  ) u ( x ,  ( t ,  ))  ( x ) d  (  ) dx     

 

 

and

. By the

j = 1,2,  , n .

     a  m 1  b   d    p (t ) u ( x , t )  ( x ) dx  g ( t ,  ) u ( x ,  ( t ,  ))  ( x ) d  (  ) dx    m 1   a dt      n b   q ( t ,  ) u ( x ,  ( t ,  ))  ( x ) d  (  ) dx = a ( t )  u ( x , t )  ( x ) dx  b j ( t )  u ( x ,  j ( t ))  ( x ) dx .   a    j =1 m 1 d  d  r (t ) m 1 dt  dt 

t0  0

( x , t ,  )    [ t1 ,  )  [ a , b ],

( x , t )    [ t1 ,  ),

 (x) > 0

,





(5)



From Green’s formula and boundary condition (2), it follows that



 u ( x , t )  ( x ) dx =







=





 u ( x,t)  ( x)   u ( x,t)  ( x )  dS     

 u ( x, t)  ( x)   u ( x, t)  ( x )  dS   0     

where dS is the surface element on Hence, we obtain



. If  ( x )

 0, x   





u ( x , t )  ( x ) dx



u ( x , t )  ( x ) dx ,



t  t1

, then from (2) we have  ( x ) 7

 0

,

u ( x, t) = 0

,

( x , t )     [0,  )

.

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International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

 u ( x, t)  ( x)   (x)  u ( x, t)    





  dS  0, t  t 1 , t  t k .  

If  ( x )  0, x    . Noting that   is piecewise smooth,  ,   C (   , [0,  we can assume that  ( x ) > 0 , x    . Then by (2) and (3), we have  u ( x, t)  ( x)   (x)  u ( x, t)     



  dS =  





)), 

2

(x)  

2

(x)  0

, without loss of generality,

   (x)  (x)    (x) u ( x, t)   ( x ) u ( x , t )  dS = 0, t  t 1 .    (x)  (x)  

Therefore, using Lemma 2.1, we obtain





and for

 u ( x , t )  ( x ) dx =   0

j = 1,2  , n









u ( x , t )  ( x ) dx , t  t 1 =   0 V ( t )

(6)

,



 u ( x ,  j ( t ))  ( x ) dx =   0



u ( x ,  j ( t ))  ( x ) dx , t  t 1 =   0V (  i ( t )).

(7)

It is easy to see that

  

b

q ( t ,  ) u ( x ,  ( t ,  ))  ( x ) d  (  ) dx

=

a



b

q (t ,  )

=



b



u ( x ,  ( t ,  ))  ( x ) dxd  (  )



a

q ( t ,  )V (  ( t ,  )) d  (  ).

(8)

a

In consideration of (5) - (8), we acquire m 1 d  d  r (t ) m 1 dt  dt 



For



  V (t )   



b a

 g ( t ,  )V ( ( t ,  )) d  (  )    a    n  b j ( t )V (  j ( t )) = 0, t  t1 , t  t k .   j =1

m 1  d g ( t ,  )V ( ( t ,  )) d  (  )    p ( t )  m 1 dt  

b

q ( t ,  )V (  ( t ,  )) d  (  )   0 a ( t )V ( t )   0 a

t  t 0 , t = t k , k = 1,2,  , i = 0,1,2,  , m  1

over the domain

, and from



(H 7 )

, we obtain

(i)

ak





V (t ) =



u ( x , t )  ( x ) dx



(i)

t 

, we have

a

(i) k



(i)

V

t 

(i)

V (t )

 tk

u ( x,

(i)

for

(i)

)

(i)

(i)

 bk .

(i)

is a positive solution of (4), which contradicts the fact that all solutions of equation (4) are oscillatory.

t  t*  0

d  d  r (t ) dt  dt  

For

t  t0

,



b

m 1 m 1

t  tk

Let



t*

 ~  V (t )   

b



a

,

k = 1,2, 

, multiplying both sides of (10) by



b a



 (x) > 0

( x , t )    [0,  )

b

(10)

we obtain

 ~ g ( t ,  )V ( ( t ,  ))  ( x ) d  (  )    a    n ~  b j ( t )  V (  j ( t ))  ( x ) = 0, t  t * , x   .   j =1

m 1  ~  g ( t ,  )V ( ( t ,  ))  ( x ) d  (  )    p ( t )  m 1 t  

a

,

 ~  V (t )   



~ ~ q ( t ,  )V (  ( t ,  ))  ( x ) d  (  )   0 a ( t )V ( t )  ( x )   0

~ ~ u ( x , t ) = V (t ) ( x )

. Without loss of generality we assume

 ~ g ( t ,  )V ( ( t ,  )) d  (  )    a    n ~  b j ( t )V (  j ( t )) = 0, t  t * , t  t k , x     j =1

m 1  ~ d g ( t ,  )V ( ( t ,  )) d  (  )    p ( t )  m 1 dt  

 ~  V (t ) ( x )   

~ V (t ) > 0

is some large number. From (4), we have

~ ~ q ( t ,  )V (  ( t ,  )) d  (  )   0 a ( t )V ( t )   0

m 1     r (t ) m 1 t  t 



, where

a

b

, integrating with respect to

(i)

u ( x, tk )

 (t k

 (x) > 0

 bk .

(ii) Necessary Part: Suppose that equation (4) has a non-oscillatory solution ~ V (t ) > 0

(9)

)

(i)

V (t k )

t

Therefore

b

, multiplying both sides of the equation (1) by

t

According to





 x

  V (t )   

 ~  V (t ) ( x )   



w(x) =  0 w(x)

,

b



. By Lemma 2.1, we have

8

x

. Then (11) implies

(11)

Integrated Intelligent Research (IIR)

m 1     r (t ) m 1 t  t 





For

b

~  u ( x, t)   

a

m 1   ~ g ( t ,  ) u ( x ,  ( t ,  )) d  (  )    p ( t )  m 1 t  

a



~ b j ( t )  u ( x ,  j ( t )),

j =1

(i)



(i)



(i)

ak

(i)

(i)

t

, multiplying both sides of equation (4) by



 (x) > 0

b

(12)

, we have

(i)

 ~ ~  ~ (i)  V (t k ) ( x )  V (t k ) ( x )  b k V (t k ) ( x ) (i) (i) t t

(i)

t ~ ~ u ( x , t ) = V (t ) ( x )

( x , t )    [0,  )

,

(i)

(i)

  (i)  ~ ~ ~ u ( x , tk )  u ( x , tk )  bk u ( x , tk ) (i) (i) t t

(i)

(i)     (i) ~ ~ u ( x , tk ) = I k  x , tk , u ( x , tk )  (i) (i)   t t   ~ ~ shows that u ( x , t ) = V ( t )  ( x ), ( x , t )    [ t * ,  )



 ~ g ( t ,  ) u ( x ,  ( t ,  )) d  (  )    a     t  t* , x   .  

~  u (x, t)   

n

ak

which

b

~ ~ q ( t ,  ) u ( x ,  ( t ,  )) d  (  ) = a ( t )  u ( x , t ) 

t  t 0 , t = t k , k = 1,2, 

since



International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

(i)

 (x)

w ( x) 

  ( x ) w ( x ) = 0,

, satisfies (1). From Lemma 2.1, we get

x  

which implies

Hence

~ u ( x, t)

~   ( x ) u ( x , t ) = 0, ( x , t )     [0,  ).  ~ ~ u ( x , t ) = v ( t )  ( x ) > 0 is a non-oscillatory solution of the

 (x)

(13) problem (1), (2) which is a contradiction. The proof is complete. 

Remark 3.1. Theorem 3.1 shows that the oscillation of problem (1), (2) is equivalent to the oscillation of the impulsive differential equation (4).

Theorem 3.2. If  ( x )

 0

r ( t ) Z

x  

for



and the impulsive differential inequality

'

( m  1)

(t )

p (t ) Z

( m  1)

(t )  g 0



b

        

q ( t ,  ) Z ( ( t )) d  (  )  0

a

 (i)

ak

(i)



Z (tk )

t





(i)

(i)

(i)

Z (tk )

t

 b k , t = t k , k = 1,2,  , i = 0,1,2,  , m  1

(i)

(14)

has no eventually positive solution, then every solution of the problem (1), (2) is oscillatory in G .

Proof. Suppose to the contrary that there is a non-oscillatory solution u ( x , t ) of the problem (1), (2) which has no zero in   [ t 0 ,  )

for some

that there exists a

. Without loss of generality, we assume that

such that  ( t ,  ) 

u ( x ,  ( t ,  )) > 0

then and

t0  0

t1 > t 0

u ( x ,  j ( t )) > 0

t  t0 , t  tk

For

d  d  r (t ) dt  dt 



for

b

m 1

,

k = 1,2,  ,



b

a

 (t ,  )  t 0

j = 1,2,  , n . As



b

u ( x, t) > 0

,

( x , t )    [ t 0 ,  )

( t ,  )  [ t 1 ,  )  [ a , b ]

and

 j (t )  t 0

u ( x ,  ( t ,  )) > 0

for

,

t0  0

,

 g ( t ,  )V ( ( t ,  )) d  (  )   a   n  b j ( t )V (  j ( t ))  0, t  t 1 , t  t k  j =1 

m 1 d   g ( t ,  )V ( ( t ,  )) d  (  )    p ( t )  V (t )  m 1 dt   

g ( t ,  )V ( ( t ,  )) d  (  )



( m  1)



'

(t )

 p (t ) Z

( m  1)

(t ) 

b



. Equation (15), can be written as



b

q ( t ,  )V (  ( t ,  )) d  (  )  0,

a

9

t  tk .

for

t  t1

,

( x , t ,  )    [ t 1 ,  )  [ a , b ]

in the proof of the Theorem 3.1.

a

r ( t ) Z

. By assumption

j = 1,2,  , n

we obtain equation (9). By Lemma 2.1, from (9) we have

q ( t ,  )V (  ( t ,  )) d  (  ) =   0 a ( t )V ( t )   0

Z (t ) = V (t ) 

for

( x , t ,  )    [ t 1 ,  )  [ a , b ],

for

a

Set

,

( x , t )    [ t 1 ,  ),

  V (t )  

m 1

t0

(16)

(15)

Integrated Intelligent Research (IIR)

We have

We can claim that r (T ) Z

( m  1)

From

p ( t ) Z

( m  1)

(t )

for t > t 1 . Hence

< 0

t  t1

for

(t ) > 0



'

( m  1)

. In fact, in

( m  2)

( t )  r ( t )( m  1) Z

 r ( t )  p ( t ) R ( t ) = R ( t )   r (t ) 

, we have

(H 1 )

( m  1)

r (t ) Z

r ( t ) Z

and

r (t ) Z

r (t ) Z

( m  1)

( m  1)

(t )  0

(t )

for

is a decreasing from in the interval t  t1

, then there exists a

T  t1

[ t 1 ,  )

.

such that

. This implies that

(T ) < 0

r ( t ) 

t  t1

, for

Z (t ) > 0

International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

( t )  0.

   

(17)

R ( t ) > 0, R  ( t )  0

and

for

t  t1

. We have multiply

R (t )

on both sides of the

r (t )

equation (17), we have R (t )

r ( t ) 

p ( t ) Z

( m  1)

R (t )

(t ) 

r (t )

r ( t )( m  1) Z

( m  2)

(t )  0

r (t )

R ( t ) Z

( m  1)



'

(t )

 0.

(18)

From (18), we have ( m  1)

R (t ) Z

( m  1)

( t )  R (T ) Z

( T )  0,

t  T.

Thus t

Z

( m  1)

T

Z Z

Since



t

( m  2)

(H 1 )

R (T )

T

R (s) ( m  1)

 R (T ) Z

T

( m  2)

t



( s ) ds 

( m  2)

(t )  Z

Z

( m  1)

t

1

T

R (s)



(T )

( T ) ds ,

(T )  R (T ) Z

ds ,

( m  1)

(T )

t T

t T t

1

T

R (s)



t  T.

ds ,

, we see that the right side tends to negative infinity. Thus

negative. This contradicts the face that t 2  t1

Lemma 2.2, there exits a (i)

Z (  1)

By choosing

lim

Z

( m  2)

( t ) = 

, which implies

Z (t )

is eventually

t  

( i  1)

(i)

Z

i =1,

Z (t ) > 0

and a odd number l ,

( t ) > 0,

0  i  l,

( t ) > 0,

l  i  m  1.

we have

. At the same time, we can further prove

Z  (t ) > 0

0  l  m 1

Z (t )  x (t ) > 0

, since

Z ( ( t ,  ))  Z ( ( t ,  )   ( t ,  ))  x ( ( t ,  )   ( t ,  ))

,

, and for

Z ( t )  0

t  t2

Z

( m  1)

( t )  0, t  t 2

. Furthermore, from

, we have

, we have

,

and thus

r ( t ) Z

( m  1)



'

(t )

 p (t ) Z

( m  1)

(t ) 

 q ( t ,  ) Z (  ( t ,  ))  1  a 



b



b

a

 g (  ( t ,  ),  ) d  (  )  d  (  )  0 . 

From equation (16), we get

r ( t ) Z

( m  1)



'

(t )

 p (t ) Z

( m  1)

(t )  g 0



b

q ( t ,  ) Z (  ( t ,  )) d  (  )  0.

(19)

a

From

(H 4)

and

t  t2.

Then (19) can be written as

r ( t ) Z For

( m  1)

 '

(t )

p (t ) Z

t  t 0 , t = t k , k = 1,2, 

(i)

ak



(i)



(i)

Z (t )

Theorem 3.3

(t )  g 0



b

 ( t )   ( t ,  )  t . Thus Z ( ( t ))  Z ( ( t , a ))

q ( t ,  ) Z ( ( t ) ) d  (  )  0.

for

(20)

a

from (4) we have 

(i)

Z ( x, t k ) t

Therefore

( m  1)

  [ a , b ] and

Z ( x, t k ) t



Z ( ( t ,  ))  Z ( ( t , a )) > 0,

, we obtain

(H 5)

(i)

 bk .

(i)

is an eventually positive solution of (14). This contradicts the hypothesis and completes the proof.  Assume that

d

 (t , a )

exists and  ( x )

 0

for

x  

dt

 ( t )  C  ([0,  ), (0,  ))

which is nondecreasing with respect to t , such that

10

. If for

t0 > 0

and there exists a function

Integrated Intelligent Research (IIR)

 b ( m  1)  k  a (0) < s k

   

 ( s )   ( s )G ( s ) 

F





t

where

 t t 0 k

0

International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

1

 ( s ) ds =  ,

2

( s ) ( s )

(21)

, then every solution of the boundary value problem (1), (2) is oscillatory

in G .

4 L(s)

Proof. To prove the solutions of (1), (2) are oscillatory in G , from Theorem 3.2, it is enough to prove that the impulsive differential inequality (14) has no eventually positive solution. Suppose that Z (t ) > 0 is a solution of the inequality (14). Define

then

W (t )  0

for

( m  1)

r (t ) Z

W (t ) =  (t )

(t )

,

Z ( ( t ))

t  t0

t  t0 ,

, and 

W ( t ) 

From

Z

(m )

 (t )   p (t ) Z

 ( t )

( m  1)

(t )  g 0



W (t ) 

 (t )

(22)



b

a

 q ( t ,  ) Z ( ( t )) d  (  )  

Z ( ( t ))





 (t ) r (t ) Z

( m  1)

Z

2



( t ) Z  ( ( t ))   ( t )

( ( t ))

.

, according to Lemma 2.3, we obtain

(t )  0

Z  ( ( t ))  M ( ( t ))

m2

Z

( m  1)

(23)

( t ).

Thus W  ( t )  F ( t )W ( t )  G ( t )  ( t ) 

L (t )

2

W

 (t )

(t )

( m  1)

bk



W (t k ) 

W ( t k ).

(0)

ak

Define  b ( m  1)  k  a (0) < t k



U (t ) =

t t 0 k

   

1

W ( t ).

( m  1)

In fact,

W (t )

is continuous on each interval

(t k , t k 1 ] ,

and in consideration of



W (t k ) 

bk

(0)

W (t k )

ak

t  t0



,

U (tk ) =

t  t0

and for all





t

0

t

k

U (t )



U ( t ) 

t



 b ( m  1)  k  a (0) k  t t t 0 j k 1

which implies that



   

1 

W (tk ) 

 b ( m  1)  k  a (0) k t t <t  0 j k



   

1

W (tk ) = U (tk )

,

U (tk ) =

=

 b ( m  1)  k  a (0) k t t t  0 j k

0

t

k

1

 b ( m  1)  k  a (0) k t t <t  0 j k





W (tk ) 

   

1

W (tk ) = U (tk )

is continuous on [ t 0 ,  ) .

 b ( m  1)  k  a (0) k < t

 b ( m  1)  k  a (0) k < t

   

   

1

 U 2 (t ) L (t )     (t ) 

 W  ( t )  W 

2

(t )

 t

0

t

k

 b ( m  1)  k  a (0) k < t

   

1

G ( t )  ( t )  F ( t )U ( t )

  W ( t ) F ( t )  G ( t )  ( t )   0.  (t )  L (t )

That is



U ( t )   t

Taking

X =

0

 t t 0 k

t

k

 b ( m  1)  k  a (0) k < t

 b ( m  1)  k  a (0) k < t

 L (t )  U   (t ) 

 L (t )  U ( t ),   (t ) 

2



( t )  F ( t )U ( t )  t

Y =

F (t ) 2

 t

0

t

k

0

t

k

 b ( m  1)  k  a (0) k < t

 b ( m  1)  k  a (0) k < t

from Lemma 2.4, we have 11

   

1

   

1

G ( t )  ( t ).

 (t ) L (t )

,

(24)

. It follows that for

Integrated Intelligent Research (IIR)



F ( t )U ( t ) 

t t 0 k

International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

 b ( m  1)  k (0)  < t a k

 L (t )  U   (t ) 

 b ( m  1)  k (0)  < t a k

   

2

F

(t ) 

2

( t ) ( t )



4 L (t )

t t 0 k

 b ( m  1)  k (0)  < t a k

   

1

.

Thus



U ( t )  

t t 0 k

1

2  F ( t ) ( t )   G ( t ) ( t )  .

4 L (t )  

(25)

Integrating both sides from t 0 to t , we have t

U (t )  U (t 0 ) 

  t

t  

Letting

0 t t 0 k

 b ( m  1)  k (0)  < s a k

   

1

2  F ( s ) ( s )   G ( s ) ( s )   ds .

4 L(s)  

lim U ( t ) = 

, from (21), we have

, which leads to a contradiction with

U (t )  0

. This complete the proof.

t  

 Theorem 3.4. Assume that

d

exists and  ( x )

 (t , a )

 0

for

x  

. Moreover, suppose that there exist functions

 (t )

and

dt

 ( s )  C  ([0,  ), (0,  ))

(H8) (H9)

H ( t , t ) = 0, t  t 0 ; ' t (t ,

H

(H10)



 s

is nondecreasing with respect to

H (t , s )  0

t

, and the functions

H ( t , s ), h ( t , s )  C  ( D , R )

, in

, such that

H ( t , s ) > 0, t > s  t 0

' s

s )  0,

 (t )

in which

D = {( t , s ) | t  s  t 0 > 0}

which

,

,

[ H ( t , s )  ( s )]  H ( t , s )  ( s ) F ( s ) = h ( t , s )

.

If limsup

H (t , t 0 )

t  

where

t

1

  t

0 t t 0 k

 b ( m  1)  k (0)  < s a k

   

 ( s ) ds   ,

2

h (t , s )

1

 ( s )  G ( s ) ( s ) H ( t , s )  ( s ) 

1

 (s)

4 L ( s ) H (t , s )  ( s )

(26)

, then every solution of the boundary value problem (1), (2) is oscillatory

in G .

Proof. Assume that the boundary value problem (1), (2) has a non-oscillatory solution that u ( x , t ) Theorem

> 0, ( x , t )    [0,  )

. The case for we have

3.3,

( x , t )    [ t1 ,  ), ( x , t ,  )    [ t1 ,  )  [ a , b ]



U ( t )   t

0

t

k

 b ( m  1)  k  a (0) k < t

 L (t )  U   (t ) 

t

T

  T

t

t



0

  T

t t 0 k

t

can be considered in the same method. Proceeding as the proof of u ( x ,  ( t ,  )) > 0, u ( x ,  ( t ,  )) > 0, u ( x ,  j ( t )) > 0, j = 1,2,  , n for

u ( x, t) < 0



( t )  F ( t )U ( t )  t

t

 

k

   < s

for

( m  1) bk (0) ak

 b ( m  1)  k (0)  < s a k

. Without loss of generality, assume

, and

H (t , s )  ( s )

multiplying the above inequality by

 U  ( s ) H ( t , s )  ( s ) ds

2

u ( x, t)

   

0

t

t  s T

 L(s)  U   (s) 

2

k

 b ( m  1)  k  a (0) k < t

   

1

G ( t ) ( t )

, and integrating from

( s ) H ( t , s )  ( s ) ds 



t

T

to t , we have

F ( s )U ( s ) H ( t , s )  ( s ) ds

T

1

G ( s )  ( s ) H ( t , s )  ( s ) ds .

Thus, we have t

  T

t

0

t

k

 b ( m  1)  k  a (0) k < s

   

1

G ( s )  ( s ) H ( t , s )  ( s ) ds  U ( T ) H ( t , T )  ( T ) 



t

h ( t , s )U ( s ) ds

T

t



  T

t t 0 k

 b ( m  1)  k (0)  < s a k

 L(s)  U   (s) 

2

12

( s ) H ( t , s )  ( s ) ds .

(27)

Integrated Intelligent Research (IIR)

Put



X =

t t 0 k

 b ( m  1)  k (0)  < s a k

 L(s) 1  H ( t , s )  ( s ) U ( s ), Y = h (t , s )   (s) 2 

from Lemma 2.4, we attain for



h ( t , s )U ( s ) 

t t 0 k

International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

t  T  t0

 b ( m  1)  k (0)  < s a k

 t t 0 k

 b ( m  1)  k (0)  < s a k

   

1

 (s) L ( s ) H (t , s )  ( s )

,

that

 L(s)  H ( t , s )  ( s )U   (s) 

2

(s) 

h (t , s )

1

2

 (s)

4 L ( s ) H (t , s )  ( s )

 t t 0 k

 b ( m  1)  k (0)  < s a k

   

1

(28)

.

In addition, from (27) and (28), we have t

  T

t

0

t

k

 b ( m  1)  k  a (0) k < s

   

1

2

G ( s )  ( s ) H ( t , s )  ( s ) ds 

1 4

t

h (t , s )  ( s )

T

L ( s ) H (t , s )  ( s )



 U ( T ) H ( t , T )  ( T )  H ( t , t 0 )  ( T )U ( T ),

 t

0

t

k

 b ( m  1)  k  a (0) k < s

   

1

ds

t  T  t0 .

(29)

Thus t

1 H (t , t0 )

  t

0t

0

t

k

 b ( m  1)  k  a (0) k < s

T

 



t

0 t t 0 k

   

1

 b ( m  1)  k (0)  < s a k

2  h (t , s )  ( s )  1  G ( s ) ( s ) H ( t , s )  ( s )   ds 4 L ( s ) H (t , s )  ( s )    

   

1

G ( s )  ( s )  ( s ) ds   ( T )U ( T ).

Letting t   , we have limsup

H (t , t0 )

t  

T



t

1

  t

0 t t 0 k

  t

0t

t

0

k

 b ( m  1)  k (0)  < s a k

   

 b ( m  1)  k  a (0) k < s

   

1

2  h (t , s )  ( s )  1  G ( s ) ( s ) H ( t , s )  ( s )   ds 4 L ( s ) H (t , s )  ( s )    

1

G ( s )  ( s )  ( s ) ds   ( T )U ( T )

<  ,



which leads to a contradiction with (26). The proof is complete.

Remark 3.2. In Theorem 3.4, by choosing  ( s )

=  (s)  1

, we have the following corollary.

Corollary 3.1 Assume that the conditions of Theorem 3.4 hold, and limsup t  

t

1 H (t , t0 )

  t

0t

t

0

k

 b ( m  1)  k  a (0) k < s

   

1

2   h (t , s ) 1 G ( s ) H (t , s )   ds =  , 4 L ( s ) H (t , s )    

then every solution of the boundary value problem (1), (2) is oscillatory in

G

.

Remark 3.3 From Theorem 3.4 and Corollary 3.1, we can attain variety oscillatory criteria by different choices of the weighted function H ( t , s ) . For example, choosing H ( t , s ) = ( t  s )   1 , t  s  t 0 , in which  > 2 is an integer, then h (t , s ) = (t  s )

 1

 1   F (s)  t  s 

Corollary 3.2 If there exists a

,

d

t  s  t0

 (t , a )

. From Corollary 3.1, we have

and an integer

 > 2

such that

dt

limsup t  

t

1  1

(t  t0 )

  t

0t

0

t

k

 b ( m  1)  k  a (0) k < s

   

1

(t  s )

 1

2  1  1   G ( s )   F ( s )    ds =  , 4 L(s)  t  s   

then every solution of the boundary value problem (1), (2) is oscillatory in G .

Now

we

consider

h ( t , s ) = [ R ( t )  R ( s )]



H ( t , s ) = [ R ( t )  R ( s )]

    F (s)  R ( t )  R ( s )  



,

t  s  t0

,

where

R (t ) =



t

.

13

t

0

1 r(s)

ds

and

lim t  

R ( t ) = 

,

then

Integrated Intelligent Research (IIR)

d

Corollary 3.3 If there exists

International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

 (t , a )

and an integer

 > 2

such that

dt t

1

limsup

[ R ( t )  R ( t 0 )]

t  

 



t

0t

0

t

k

2     1   [ R ( t )  R ( s )] G ( s )   F ( s )   ds =  ,  4 L ( s )  R (t )  R ( s )    <s 



then every solution of the boundary value problem (1), (2) is oscillatory in G .

IV. EXAMPLE In this part, we present couple of examples to point up our results established in Section III. Example 4.1. Consider the following impulsive neutral partial differential equation of the form  u ( x , t  2 ) d      /2  /2     /4 4 4 3 3   u ( x , t  2 ) d  = u (x, t)  u (x, t  ), t  t k , k = 1,2,  ,  3   /2 3 4 2  (i) (i)  k      u ( x, tk ) = u ( x , t k ), u ( x, tk ) = u ( x , t k ), i = 1,2,3,4,5, k = 1,2,  (i) (i) k 1  t t  5   3   t  4 t 5 

  /4

 2  u (x, t)   3 



  1  5 u ( x , t  2 ) d         5    2  t

  /4

 2  u ( x, t)   3 





for

( x , t )  (0,  )  [0,  )

, with the boundary condition

u (0, t ) = u (  , t ) = 0,

Here

 = (0,  )

,

,

t  tk .

(0)

m = 6,

 (t ,  ) =  (t ,  ) = t  2

ak

a (t ) =

. Since

t0 = 1

,

k

(0)

4

,

(0)

ak

t  

  t

(i)

0 t t < s 0 k

=

ds =





1

bk

k



,

3

k

<s

(i)

, ak

,

4

= bk

G (s) =





3

k k 1

= 1

ds =

3

i = 1,2,3,4,5

,

,

L(s) =

4s

t 1

k

 

 n=0

r (t ) =

3

,

g (t ,  ) =

2

2

p (t ) = 

,

,  ( )

. Then hypotheses

(H1)  (H 7 )

= 

1

,

q (t ,  ) =

2

3

[ a , b ] = [   /2,   /4]

M = 1 ,  (t ) = t

,

4

k

<s

k 1

t

ds 

  2

 t 1 1< t

k

<s

k k 1

t

ds 

  t

,

  (t ) = 1 ,

k

<s

hold, moreover k

3

 2 1< t

k 1

ds  

n

n 1

=  .

Thus, limsup t  

1 ( t  1)

2

    

t

  1

1< t

k

<s

k 1 k

2     1 1 2   3 2    (t  s )       ds 2 4  3  9 3( t  s )   18  4 s    (t  s ) 

   =  . 

Therefore all the conditions of the Corollary 3.2 are satisfied. Therefore, every solution of equation (30) - (31) is oscillatory in G . In fact u ( x , t ) = sin x cos t is such a solution.

Example 4.2 Consider the following impulsive neutral partial differential equation of the form  u ( x, t   )d      0 3   2 2  u ( x , t   ) d  = t  u ( x , t )  ( t  3)  u ( x , t  ), t  t k , k = 1,2,  ,  2  2  (i) (i)  k      u ( x, tk ) = u ( x , t k ), u ( x, tk ) = u ( x , t k ), i = 1,2,3, k = 1,2,  (i) (i) k 1  t t  3   2  t 3 t  t 

 1  u (x,t)   2 



3   u ( x , t   ) d      2 t   3  t  0

 1  u (x,t)   2 



for

( x , t )  (0,  )  [0,  )



0

(32)

, with the boundary condition

u x ( 0 , t )  u (0, t ) = u x (  , t )  u (  , t ) = 0,

t  tk .

14

(33)

,

3

4

3

18

1< t

,

j =1

2





,

2

1

=

,

4

 1 (t ) = t 

6

 1< t

(i)

k 1

b1 (t ) =

g0 = 1

tk = 2 ,

t

lim

(31)

= bk

3

 = 3

(30)

Integrated Intelligent Research (IIR)

Here

 = (0,  )

,

 (t ,  ) =  (t ,  ) = t  

 = 3

. Since

(0)

m = 4,

t0 = 1

,

,

k

tk = 2 ,

limsup t  

(0)

ak

a (t ) = t

g

1 ( t  1)

2

2

0

    

International Journal of Computing Algorithm Volume: 06 Issue: 01 June 2017 Page No.5-15 ISSN: 2278-2397

= bk

,

k

=

b1 (t ) = t

= 1



,

2

3

  1

1< t

k

<s

, ak

,

G (s) =

2 t

(i)

k 1

3

k



,

= 1

i = 1,2,3

,

r (t ) = t

2

g (t ,  ) =

,

1

,

p (t ) =  2 t

,

q (t ,  ) =

3



,

j =1

,

[ a , b ] = [   ,0]

,  ( )

= 

,

3

,

2

2

 1 (t ) = t 

2 k 1

(i)

= bk

M = 1

,

 (t ) = t

,

  (t ) = 1 ,

2

2

,

L(s) = 1 .

4

2   3 3 2 (t  s )    2 4  

Then hypotheses

(H1)  (H 7 )

 1   1 2        ds 2 2  s (t  s )   (t  s ) s  

hold, thus,

   =  . 

Therefore all the conditions of the Corollary 3.2 are satisfied. Therefore, every solution of equation (32) - (33) is oscillatory in G .

In fact u ( x , t ) = e  x sin t is such a solution. continuous distributed deviating arguments, Open Acknowledgment Access Library Journal, 1(2014), 1-8.

The authors wish to express their sincere thanks to the referees [13] Ch.G. Philos, A new criterion for the oscillatory and for valuable comments and suggestions. asymptotic behavior of delay differential equations, Bull.

Acad. Pol. Sci. Ser. Sci. Math., 39(1981),61-64.

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[3] G.H. Hardy, J.E. Littlewood and G. Polya, Inequalities, [16] V. Sadhasivam, T. Raja and T. Kalaimani, Oscillation of nonlinear impulsive neutral functional hyperbolic Cambridge University Press, Cambridge, UK, 1988. equations with damping, International Journal of [4] P. Hartman and A. Wintner, On a comparison theorem for Pure and Applied Mathematics, 106(8)(2016), 187-197. self-adjoint partial differential equations of elliptic type, [17] V. Sadhasivam, T. Raja and T. Kalaimani, Oscillation of Proc. Amer. Math.

Soc., 6(1955), 862-865. impulsive neutral hyperbolic equations with continuous [5] I.T. Kiguradze, On the oscillation of solutions of the distributed deviating arguments, Global Journal of Pure m d u n and Applied Mathematics, 12(3)(2016), 163-167.  a ( t ) u sgnu = 0 , Math. Sb., 65(1964), 172equation m dt [18] S. Tanaka and N. Yoshida, Forced oscillation of certain 187 (in Russian). hyperbolic equations with continuous distributed deviating [6] G.S. Ladde, V. Lakshmikantham and B.G. Zhang, arguments, Ann. Polon. Math., 85(2005), 37-54.

Oscillation Theory of Differential Equations with [19] E. Thandapani and R. Savithri, On oscillation of a neutral Deviating Arguments, Marcel Dekker, Inc, New York, partial functional differential equations, Bull. Inst. Math.

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[9] W.N. Li and L. Debnath, Oscillation of higher-order [22] J.H. Wu, Theory and Applications of Partial Functional neutral partial functional differential equations, Appl.

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