Non Oscillation of Second Order Neutral Delay Difference Equations

Non Oscillation of Second Order Neutral Delay Difference Equations

B. SELVARAJ and G.GOMATHI JAWAHAR Department of Mathematics

Karunya University, Karunya Nagar, Coimbatore – 641 114 Tamil Nadu, India.

Email : [email protected].

Email:[email protected]

ABSTRACTS

In this paper we deal with the non oscillatory behavior of all solutions of the second order neutral delay difference equations . Example is provided to illustrate the main results.

Key words: Non oscillation, Difference equations , Neutral Delay.

Subject classification : 2000 mathematics 39 A11.

INTRODUCTION

Con side r th e ne utra l de lay difference equation of the form,

(1:1) Where k,l are non negative integers, q

(n=

0,1,2….) are real numbers, and  denotes the forward difference operator defined by

 y

= y

-y

In addition to the above, we assume the following.

H

: q

>0, eventually positive . H

: y

> y

By a solution of (1.1) we mean a real sequence {y

} which satisfies the equation (1.1) for all n  N(no). A solution {y

} of (1.1) is said to be oscillatory if the terms {y

} of the sequence are not eventually positive, or not eventually negative. Otherwise it is called non oscillatory. Equation (1.1) is said to be oscillatory if all its solutions are oscillatory.

In the past few years, there has

been an increasing interest in the study of oscillatory and non oscillatory behavior of difference equations. For example see [2,3,4,5,6,7,8,9,10,11].

Following this trend, in this paper we obtain some sufficient conditions for the nonoscillation of all solutions of equation (1.1). An example is provided to illustrate the main results.

MAIN RESULTS

In this section, we present some sufficient condition for the non oscillation of all solutions of equation (1.1)

Theorem 2:1

Assume that = 

and K {1,2….}, 1 {0,1,2….}

Then every non oscillatory solution of equation (1.1) tends to zero as n tends to .

Proof

Let {y

} be a nonoscillatory solution of equation (1.1). Let us assume that {y

} is eventually positive.

Thus, there exists an integer n

>1 such that q

>0 and y

> 0 for n > n

.

Let z

= y

+y

(1.2) Then from (1.1), 

z

= q

y

  

z

> 0

Hence there are 2 possibilities

z

< 0 ,  z

> 0

Case I

z

<0 and z

>0 , for n>n

from (1.1),



z

= q

y

(z

- z

) = q

y

z

= z

- q

y

==> z

= z

- z

- q

y

Hence

 z

-  z

– q

y

+ q

y

– y

+ y

+y

– y

= z

– z

– q

y

– (z

-z

-q

y

) – q

y

+q

y

– y

+ y

+y

– y

=0 (1.3)

Suppose that lim sup q

=  <

 , for some n > n

Then from (1.3), takin n  we have,

 z

 z

 (y

– y

) – y

+ y

+ y

-y

< 0 for ne > n

(1.4)

Now we want to prove that lim z

= o n



Suppose lim z

= a > 0, then n



obviously lim y

= a>o.

n



Summing both sides of (1.4) from n

to N, we have

z

– z

– (z

– z

) N

< - a  

 as N

 n

Which implies that z

 as n

 , a

contradiction.

Hence lim z

= 0 n



Which implies that lim y

= 0 n



Hence every non oscillatory solution of equation (1.1) tends to Zero as n

 .

Case II

 z

>o and z

> o for n > n

Suppose that lim inf q

= 

<  for some n



n > n

.

Substituting in (1.3), we have

 z

 z



(y

y

) – y

- y

+ y

y

- y

> 0 for n > n

(1.5) Now we want to prove that

lim z

= o n



Suppose lim z

= a

> 0, then n



obviously lim y

= a

> 0.

n



Summing both sides of (1.5) from n

to N, we have

Z

- Z

- (Z

- Z

) N

> -a

 

 as N

 n

Which implies that Z

 as n

 , a contradiction.

Hence lim Z

= 0 n



Which implies that lim y

= 0 n



Hence every non oscillatory solution of (1.1) tends to Zero as n

 .



Theorem 2.2 Assume that  q

=  . n= k

Let q

= min (q

, q

) when  z

< 0 and q

= max (q

, q

) When  z

> 0.

Then every non oscillatory solution of equation (1.1) tends to zero as n tends to

 .

Proof

Let {y

} be a eventually positive solution of (1.1).

Let z

= y

+y

from (1.1), 

z

= q

y

Hence 

z

> 0

Hence there are 2 possibilities.

 z

< 0 ,  z

>0

Case I

Suppose  z

< 0.

Hence z

is eventually positive and decreasing

So lim z

= M > 0 exists and is finite.

n



since 

z

- q

y

=0, we have

* *



z

- q

y

= 0 (1.6)

Subtracting ( 1.6) from (1.1) ,



z

- q

y

- ( 

z

- q

y

) =0



z

- 

z

- q

y

+ q

y

=0 z

-2 z

+ z

- z

+ 2 z

- z

- q

y

+ q

y

= 0 (1.7)

Since q

= min ( q

, q

), we have z

- 2 z

+ z

- z

+ 2 z

- z

+ q

(y

- y

) < 0, For some n > n

.

Hence

z

- 2 z

+ z

- z

+ 2 z

- z

< - q

( y

- y

) (1.8)

Summing both sides from n

to N, we have N

 ( z

- 2 z

+ z

- z

n

+2 z

- z

) < -  q

( y N

- y

)



as N

 . n

Which implies that z

  as n

 , which is a contradiction.

Hence lim z

= 0 n



Therefore lim y

=0.

n



Case II

Suppose  z

> 0

Hence z

is eventually positive and increasing sequence.

So, lim z

= M > 0 exists and is finite.

n



Here q

= max (q

, q

)

Hence equation (1.7) becomes,

z

- 2 z

+ z

-z

+ 2z

- z

> - q

(y

- y

) For some n > n

.

Taking summation from n6 to N, we have N

 (z

- 2 z

+ z

n

- z

+ 2 z

- z

) > -  q

(y

- y

)



 as N



Hence lim z

 as n

 . n



Which is a contradiction, since

lim z

= M > 0 and is finite.

n



Therefore lim Z

= 0 n



Hence lim y

=0 n



Therefore every non oscillatory solution of (1.1) tends to zero as n

 .

*

*

*

*

*

EXAMPLES

1. Consider the difference equation



(y

+y

) - 5 y

= 0 for n > 3 (3.1) 32

Here k= 2:, l=3:, q

= 5 /32

It is easy to see that all conditions of theorem (2.1), (2.2) are satisfied. Hence all non oscillatory solutions of (3.1) goes to zero as n

 . One such solution is 1/2

.

2. Consider the difference equation



(y

+y

) - 4n+8

y

= 0 for n>4 (3.2) n

-5n+4

Here k=2; l=0; q

= 4n+8 n

-5n+4

It is easy to see that all conditions of theorem (2.1), (2.2) are satisfied. Hence all non oscillatory solutions of (3.2) goes to zero as n

 . one such solution is

1

 n

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