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Journal of Computer and Mathematical Sciences Vol. 3, Issue 6, 31 December, 2012 Pages (557-663)

Oscillation of Solutions of Certain Fifth Order Difference Equations

B. SELVARAJ1 and S. KALEESWARI2

1Dean of Science and Humanities, Nehru Institute of Engineering and Technology,

Coimbatore, Tamil Nadu, INDIA.

2Department of Mathematics,

Nehru Institute of Engineering and Technology, Coimbatore, Tamil Nadu, INDIA.

(Received on: December 3, 2012) ABSTRACT

The objective of this paper is to study the oscillatory behavior of solutions of fifth order difference equation of the form

,

1 0

4 + =



n n n+

n

n x p x

p

a n=0,1,2,....

Examples are inserted to illustrate the results.

Keywords: fifth order, oscillation and difference equation.

AMS Subject Classification : 39A11.

1. INTRODUCTION

In this note, we consider the fifth order difference equation of the form

,

1 0

4 + =



n n n+

n

n x p x

p a

n1

n (1)

where , is the forward difference operator defined by xn = xn+1xn and

{ }

an and

{ }

pn are defined as follows:

(i)

{ }

an and

{ }

pn are real sequences and

>0

pn for infinitely many values of n.

(ii) =

= 1

1

n

n

s s

s

n a

r p as n.

We recall that a nontrivial real sequence

{ }

xn , n0 is called a solution of (1), if

{ }

xn satisfies the equation (1) for all n0. A solution

{ }

xn is said to be

(2)

oscillatory, if for every n0 >0, there exists an nn0 such that xnxn+1 0and it is called non-oscillatory otherwise. Therefore an oscillatory solution is neither eventually positive nor eventually negative.

In recent years, there has been an increasing interest in the study of oscillatory and asymptotic behavior of solutions of difference equations of the type (1) and the references cited therein.

2. MAIN RESULTS

In this section, some sufficient conditions for the oscillation of all the solutions of (1) have been proved in the following theorems.

Theorem: 1

Suppose that conditions (i) and (ii) hold and let

( ) ( )

=

= +

+

2

3 1 1

2 1

4

n

s s s

s s s

s r p s n

r p a

r

for n2 n1. (2) Then every solution

{ }

xn of equation (1) is oscillatory.

Proof:

Suppose on the contrary that

{ }

xn is a non- oscillatory solution of equation (1).

Without loss of generality, we may assume that xn>0 for nn1 .

From (1), we have

,

1 0

4 =



n n n+

n

n x p x

p

a for nn1 .

Then

{ }

xn ,

{ }

xn , 



n

n

n x

p

a ,





n

n

n x

p

2 a

, 



n

n

n x

p

3 a

are eventually non – increasing sequences.

We shall first show that 3 >0



n

n

n x

p

a .

Suppose to the contrary that

2

3 x 0,n n

p a

n n

n 



for n2 n1.

Since 



n

n

n x

p

3 a

is non-increasing, there exists a non-negative constant k1, and

2

3 n

n such that 3 x k1,

p a

n n

n 



for

n3

n and k1 >0.

Summing the last inequality from n3 to

1

n , we obtain

( )

=

=





1 1

1 3

3 3

n

n s n

n s

s s

s x k

p a

That is

( 3)

1 2

2

3 3

3 x k n n

p x a

p a

n n n n

n

n





This implies

(

3

)

1 2

2

3 3

3 x k n n

p x a

p a

n n n n

n

n





Then

−∞





n

n

n x

p

2 a

as n.

(3)

Journal of Computer and Mathematical Sciences Vol. 3, Issue 6, 31 December, 2012 Pages (557-663)

Therefore there exists an integer n4 n3

such that 2 x k2

p a

n n

n 



for nn4

and k2 >0.

Summing the last inequality from n4 to

1

n , we obtain

(

4

)

4 2

4

4 x k n n

p x a

p a

n n n n

n

n





This implies

(

4

)

4 2

4

4 x k n n

p x a

p a

n n n n

n

n





Then

−∞





n

n

n x

p

a as n.

Thus there exists an integer n5 n4 such

that x k3

p a

n n

n 



for nn5 and k3 >0.

Summing the last inequality from n5 to

1

n , we obtain

(

5

)

5 3

5

5 x k n n

p x a

p a

n n n n

n

n





This implies

−∞





n

n

n x

p

a as n.

Then there exists an integer n6 n5 such

that x k4

p a

n n

n 



for nn6 and

4 >0 k

Summing the last inequality from n6 to

1

n , from

n n

n a

k p x 4

, we obtain

=

4 1

6 6

n

n

s s

s n

n a

k p x

x .

Therefore xn −∞ as n, which is a contradiction to the fact that xn is positive.

Hence 3 >0



n

n

n x

p

a .

Set

1 3

1

, 0 n n p x

a x

v r n

n n n

n

n >



=

+

(3)





+







=

+ + + + +

n n n n

n n

n n

n n

n x

p a x

r x x r p

v a 4

1 1 1 1 1 3

From equation (1), we have

n n

n n n

p p x

a

x =



+ 4 1

1 and using this in

last equation, we get

n n n

n

n n n n n n n

n r p

x x

x r r x x

p

v a 







=

+ +

+ + +

+ +

2 1

1 1

1 1 1 3

n n n n n n n

n n n

n n n

n x r p

p a x x

x r x x r p

a 











= +

+ + + +

+ +

+ + +

1 1 1 3

2 1

1

2 1 1 1 3

From equation (3), we get





= +

+ + +

+ +

1 1 3 1 2 1

1 1

n n n n

n

n x

p a x

r v

and substituting this in the last equation, we obtain









+

=

+

+ + +

+ + +

+

1 1 1 3 2 1

1 1

1

n n n n

n n n n

n n n n

n x

p a x

x x r r

r p v

r v

(4)

By using the fact that

{ }

xn is non- increasing, we get









+

+

+ + +

+ +

+

1 1 1 3 2

2 1 1

1

n n n n

n n n

n n n n

n x

p a x

x r r

r p v

r

v

(4) Consider

+

=

+

+ + + + + + + + + + +

1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1

1 1

n n n n n n n n n n n

n x

p x a p x a p x a p a

= +

+ + +

+

+ 



+

= 1 1

1 1 1

1 1

1 1

1 1

n

n s

s s s n

n

n x

p x a

p a

( )

1 , 1 1

1 1

1  +



+

+

+ x n n

p n a

n n

n n

and

= +

+ + +

+ + +

+

+ 



+

=



1 1

1 2 1 1 1 1 1

1 1

1 1 1 1

n

n s

s s s n

n n n

n

n x

p x a

p x a

p a

( )

+

+ +

1 1 2 1

1 1

1 1

n n

n x

p n a

n

Similarly,

= +

+ + +

+ + +

+

+ 



+

=



1 1

1 3 1 1 1 2 1 1 1 2 1

1 1 1 1

n

n s

s s s n

n n n

n

n x

p x a

p x a

p a

( )





+

+ +

1 1 1 3

1 n

n

n x

p n a

n

This implies

1 1

1 +

+ + n n

n x

p

a

( )





+

+ +

1 1 1 3 3

1 n

n

n x

p n a

n

which implies

+1

xn

( )





+

+ + +

+

1 1 1 3 3

1 1

1

n n n n

n x

p n a

a n p

That is,

+1

xn

( )





+

+ + +

+

1 1 1 3 3

1 1

1

n n n n

n x

p n a

a n

p

(5)

Substituting (5) in (4), we get

( )

2

1 1 3 1 1

2 2

1 3 1 1

1





+

+

+ + +

+

+ +

+

n n n n

n

n n

n n n n n

n x

p a a

x

p n n r r

r p v

r

v

Since from (3), 



= +

+ + +

+ +

1 1 3 1 2 1

1 1

n n n n

n

n x

p a x

r v

we have,

1 2 1 1

1 3 1

+ + + +

+

+ =



n n n n

n n

r x x v

p a

Consequently,

( )

2 1 2

1 1

1 3 1 1

1

+ + +

+ +

+

+

n n n

n n

n n n n n

n r

v a

p n n r r

r p v

r

v (6)

(5)

Journal of Computer and Mathematical Sciences Vol. 3, Issue 6, 31 December, 2012 Pages (557-663)

Consider

( )

( ) ( ) ( )

(

1

)

3

1 1 2

3 1 1

1 2

2 1 2

1 1

1 3 1 1

1

4

4 r p n n

a r n

n p r

a r

r v a

p n n r r

r p v

r v

n n

n n n

n

n n

n n n

n n

n n n n n n

+

+

+ + +

+

+ + +

+ +

+

(adding and subtracting last two terms) That is

( ) ( )

( )

( )

2

1 1

1 1

1 1

1 1

1 1

3 1 1

1 2

2 4

+

+ + +

+ + +

+ +

n n

n n

n n

n n n

n n

n n n

n n

p r

a n n n n v r

a p r r

n n n n

n n p r

a p r

r v

This implies that

( )

1

(

1

)

3

1 2

4r p n n a p r

r v

n n

n n n

n

n

+

<

+ +

That is

( ) ( )

<

+ +

3 1 1

1 2

4r p n n a p r

r v

n n

n n n

n n

Summing the last inequality from n2 to n1 , we have

( ) ( )

= +

+

1 3

1 1

1 2

2

2 4

n

n

s s s

s s s

s n

n r p s n

a p r

r v

v

In view of (2), we see that

vn −∞ as n

which is a contradiction to the fact that vn is positive and hence the proof.

Next, we introduce a double sequence

{

H

( )

m,n

}

by means of which we will study the oscillation criteria of equation (1).

Consider a double sequence

{

H

( )

m,n /mn0

}

such that

(6)

(i) H

(

m,m

)

=0 for m0

(ii) H

( )

m,n >0 for m>n0

(iii) 2H

( )

m,n =h

( ) ( )

m,n H m,n for m>n0 where 2H

( )

m,n =H

(

m,n+1

)

H

( )

m,n 0 for mn0 Theorem: 2

Suppose that conditions (i) and (ii) hold. If there exists a double sequence H

( )

m,n such

that

( )

( ) ( ) ( )

,

( )

, ,

( )

7

, 4 , sup 1 lim

1 2

1 3

1 1

2 1 1 2

2

=





×

= + +

+ +

n

n

s s

s s

s s s s

s n

s n r H

s r n n h

s p r

r p a

r s n H

n n H

for n2 n1. Then every solution

{ }

xn of equation (1) is oscillatory.

Proof:

Proceeding as in the proof of theorem 1 with the assumption that equation (1) has a

non-oscillatory solution, say xn>0 for nn1 and by taking vn as in equation (3), we get

>0

vn and equation (6) holds and so we obtain from the equation (6), for nn2,

( )

2 1 2

1 1

1 3 1 1

1

+ + +

+ +

+

+

n n n

n n

n n n n n

n r

v a

p n n r r

r v v

p r

This implies

( ) ( ) ( )

( ) ( )

2 1 2

1 1

1 3 1 1

1 1 1

1 1

2

2 2

2

,

, ,

,

+ + +

+

=

+

+

=

=

=

+

s s s

s s

n

n s

s s s n

n s s n

n s n

n s

s s

r v a

p n s s r n H

r r s v

n H v

s n H p

r s n H

Using summing by parts, we get

( ) ( ( ) ) ( )

( ) ( ) ( )

2 1 2

1 1

1 3 1 1

1 1 1

1 1

2 1

2 2

2 2 2

, ,

, ,

,

+ + +

+ + =

+

=

+

=

=

+

s s s

s s

n

n s s

s s n

n s

s n

n s n s n n

n s

s s

r v a

p n s s r n r H

r s v

n H

v s n H v

s n H p

r s n H

References

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