Cmjv01i03p0364

J. Comp. & Math. Sci. Vol. 1(3), 364-373 (2010).

Oscillatory Properties of Fourth Order Neutral Delay Difference Equations B. SELVARAJ and I. MOHAMMED ALI JAFFER Department of Mathematics, Karunya University, Karunya Nagar, Coimbatore-641 114, Tamil Nadu, India.

Emails: selvaraj@karunya.edu and jaffermathsgac@gmail.com ABSTRACT This paper deals with oscillatory properties of solutions of Fourth Order Neutral Delay Difference Equations of the form





 cn  2  an   yn  bn yn     qn f ( yn  )  0 Examples are given to illustrate the results.

Key words : Difference Non-oscillation and Neutral delay.

equations,

Oscillation,

AMS Subject Classification: 39MA11

1 INTRODUCTION





  1 1 1      an n  no bn n no cn

We are concerned with the (H2) cn  0 , an  0 and n  no Oscillatory properties of solutions of    fourth order neutral delay difference 1 1 1     , for n  no . equation of the form

a

n  no



n

b

n  no

n

c

n  no

n



 cn  2  an   yn  bn yn     qn f (H3) yn  bn  0, qn  0 , for infinitely many

 an  yn  bn yn 

 qn f ( yn  )  0

(1.1)

values of n.

(H4)

f : R  R is

continuous and

Where the following conditions are assumed to hold.

xf ( x)  0 for all x  0 and

(H1)  ,  are non-negative constants. s.

(H5)

f ( x) K 0 x

 : N o  R is continuous for all

Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)

B. Selvaraj et al., J.Comp.&Math.Sci. Vol.1(3), 364-373 (2010)

x  0  No A non trivial solution { yn } of equation (1.1) is said to be oscillatory if for any that

n1  no there exists n  n1 such

yn yn 1  0 , otherwise the solution

is said to be non-oscillatory. The forward difference operator



is defined by

yn  yn 1  yn .

365

sufficient conditions for the oscillation of all the solution of equation (1.1).We begin with the following lemma.

Lemma 1 If

y n is an eventually positive solution

of equation (1.1) and z n  yn  bn yn  then for sufficiently large n, there are only two possible cases: Case (I) : z n  0; z n  0;  ( an zn )  0;  ( an zn )  0 2

z n years, 0; z n much  0;  (research an zn )  0;  ( an zn )  0 In recent is going in the study of oscillatory Case (II) : z  0; z  0;  ( a z )  0;  (a z )  0 n n n n n n behavior of solutions of third order 2 difference equations. zn  0; z n  0;  ( an zn )  0;  ( an zn )  0 The difference equation of the form

Proof :

  a n   bn  y n    q n f ( y n  m 1 )  hn, was studied by J.R. Greaf and E. Thandapani5 and the difference equation of the form

 3Vn  PnVn 1  0 was

studied by S.H. Saker10. Motivated by these articles, In this paper we obtain some sufficient condition for the oscillation of all the solutions of equation (1.1) using comparison method. Examples are given to illustrate the results.

Let

yn

be an eventually positive

solution of equation (1.1), then there exists n1  no such that

yn   0

and

yn   0 for n  n1 .

From the definition of clear that

zn  0

zn , is

and (cn  2 (an zn ))  0

(c  (a z ))  0 for n  n1 , thus zn , zn and

For more details on Oscillatory behavior of difference equation, we refer [1,2,4,6,7,8,9,10,11,12,13,15,16,17, 18,19,20,21,22,23].

We claim that

2 Main Results

 2 (an zn )  0

 2 ( a n z n )

are eventually of one sign.

for n  n2 for

In this section, we present some Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)

n2  n1 (1. 2)

366

B. Selvaraj et al., J.Comp.&Math.Sci. Vol.1(3), 364-373 (2010)

Suppose  Since

2

( an zn )  0 for all large n.

qn  0

and cn

 0 , its clear that

n3  n2 ,

there is an integer

we have e

 2 (an zn )  0 for sufficiently large n and completes the proof.

Lemma 2 Assume that (H1)-(H4) holds.

2

2

c n  ( a n  z n )  c n 3  ( a n 3  z nLet ) yn be 0 an eventually positive solution 3

 2 ( a n  z n )  c n3  2 ( a n 3  z n3 )  0

(1.3)

of equation (1.1) and suppose that case (I) of Lemma 1 holds. Then there exists n1  no sufficiently large such thatt

Summing (1.3) from n3 to (n  1) , we e have

 z n  

 ( a n  z n )   ( a n3  z n 3 )  c n3  ( a n3 n 1 2

)   ( a n3  z n3 )  c n3  ( a n3  z n3 ) 

s  n3

1 cs

 n  c n 2  ( a n  z n )for n  n1 a  z n  ) (2. 1)

n3

n 1

Where

n 

1

c

s  no

.

s

Proof : From case (I) of Lemma 1 and In view of (H2), we see that  ( an z n )  

 ( an zn )   as n   . Thus, there existss

n4  n3

such that

equation (1.1), we have for

an  z n  0; cn  2 ( an  z n )  0

 ( a n  z n )   ( a n4  z n4 )  0 for n  n4 Summing from

n4

to

an zn  

as

n

(n  1) , we obtaain (1.4)

n 1 2 Since   (a s zs )  (an zn )   (an1 zn1 ) s  n1

for

Dividing (1.4) by an , summing

n5

to

zn  

as

n   . This contradiction shows that

n  n1

(an zn )  (an1 zn1 ) 

 (as zs )   ( a

n 1

an,and zn )applying  (an conzn )  (n(1)

dition (H2), we obtain

and

 (cn  2 ( an zn ))  0 ,

(i.e)

from

n  n1

2

2

  (a z )   (a z ) (2.2) s

s

n

n

s  n1

Summing (2.2) from

n2 to ( n  1) , for

Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)

B. Selvaraj et al., J.Comp.&Math.Sci. Vol.1(3), 364-373 (2010)

n2  n1

a positive sequence

{}nno

such thatt

cs  2 (a s z s ) 2 n 1 a n  z n  a n2  z n2    clim  a nKz n q  (s ) as  n nsup   s s 4  cs s  n1 n  s  no  s 1 s  cs  ( a s  z s ) 2  c n n  ( a n  z n ) n 1

zn 

367

   (3.2)

This and (2.2), we get

Then every solution of equation (1.1) is oscillatory.

an zn  cn n  2 ( an zn ) (2.3)

Proof :

Since

 (cn  2 (an z n ))  0 , we

Let

get

cn   2 (an zn  )  cn  2 (an zn )

{ yn }

be a non-oscillatory

solution of equation (1.1), without loss of generality, we may assume that

yn  0, yn  0 and yn   0 for n  n1.

This and (2.3) implies that for n  n2  n1 Where  n1  no is chosen so large that n  n2  n1   sufficiently large,, Lemma 1 and Lemma 2 holds. We shall

an   z n   cn   n   2 ( an  z n  ),

consider only this case, because the proof when

2

an zn  cn n   (an zn ) and then we obtain

yn  0 is

similar..

According to Lemma 1, there are two possible cases.

an  zn   cn n   2 ( an zn ),

Case(I) :

n  n2  n1   and this leads to (2.1).

In this case, we define the function

The proof is complete.

by

Theorem 1 : Assume that (H1)-(H4) holds and n 1



1  1   s  n2 as t  s  u t cu

lim sup  n







q    i

i u



(3.1)

Furthermore, assume that there exists

zn  0

for

n  n1  no

wn

cn  2 (an zn ) wn  n For n  n1 (3.3) zn  Then by equation (1.1) and Lemma 2, we have

wn   Kn qn 

n  wn 1  n n1 n1

Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)

wn 1 n

368

K n qn 

K n qn 

n n1

B. Selvaraj et al., J.Comp.&Math.Sci. Vol.1(3), 364-373 (2010)

 wn 1  n ( wn 1 ) 2 n1an

obtain, (3 .4)

cn  2 (an zn )  cn1  2 ( an1 zn1 )  K n 1

cn  ( an z2 n )  cn  ( an zn )  K  qs ys   0 (n ) an   n  n an  wn   Kn qn   wn  s  n1 n 4n1 n  n 1 n  2n 1  2 From Lemma 1, since ncn ( an z n )  0 2 and decreasing, we have e    cn a( an zn )  0

n an n  n  wn 1   4n1 n  2n1  n 1an

n 1 n 

 n 

 

(3.5)



 c n1  2 ( a n1  z n1 )  K  q i y i    0 i  n1

This implies that

and hence

1  ( an zn )  K cn 2

 ( n ) 2 an  wn    K n qn   4n 1 n   (3.6) Summing (3.6), we have for

n  n2 ,

n 1  (s ) 2 as   wn  wn2    Ks qs   4s 1 s   s  n2 

Letting have

n

(3.7)   , inview of (3.2), wee

wn  

a contradiction.



q y i

i 

0

i  n1

Summing again from

n too , we havee



1  (an zn )  K   qi yi   0 u  n cu i u Summing from

n

to o

, we havee

   1  an zn  K     qi yi  t  n1  u t cu i u

Summing from

 0 

n1 to ( n  1) , we have ve

n 1

z n  0

Case(II) :

This implies that

for

n  n1  no

yn

is positive and

decreasing function. Summing equation (1.1) from

n1 too (n  1) , ( n  n1 ) , wee

 1    1     qi yi    0  s  n1 as t  s  u t cu i u 

zn  zn1  K 

Hence, using the fact that

yn

is decre-

asing, we get

yn  bn yn   yn1  bn1 yn1   Kyn1

Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)

qs y s

0



B. Selvaraj et al., J.Comp.&Math.Sci. Vol.1(3), 364-373 (2010)

n 1







 1  1    qi   0  s  n1 as t  s  u t cu i u 

 yn1  bn1 yn1   Kyn1 

369

so large that Lemma 1 and Lemma 2 holds. We shall consider only this case, because the proof when y n

 0 is similar..

According to Lemma 1, there are two possible cases.

This implies n 1

1    1   bn      qi   K s  n1 as t  s  u t cu i u  Which contradicts (3.1). Hence the theorem was proved.

Theorem 2 :

If the case(I) holds,then by defining again wn by (3.3) as in the Theorem 1, we have wn  0 and (3.6) holds. From (3.6) we have for n  n1

 (s ) 2 as   (n  s)  Ks qs  4  s  n1 s 1 s   n 1

r

Let all the assumption of Theorem 1 holds except the condition (3.2), which changed to n 1

n 1  r     (n  s) ws s  n1  

 (n  s)  K s qs   s  n1 n 1   1 r

lim sup n 

n

r

 (n  s)

s  no

r

 Ks qs  

Since n 1

  r

 ( n  s ) w

s

2

 (s ) as   n  s  Ks qs   r 4   ( n  s )   s 1` s   ws  r

n 1  r     (n  s) ws s n1 

(3.9)

n 1

 r  (n  s ) r 1 ws  wn (n  n ) r

s  n1

s  n1

(3. 8)

(n  s ) r ws  wn1 (n  n1 ) r

(3.10)

We get Then every solution

yn

of equation

1 nr

(1.1) is oscillatory.

Proof : n 1 Proceeding as in rthe proof of n ( n  s)that Qs equation  wn  Theorem 1, rwe assume n s  n1  (1.1) has non-oscillatory solution, say



yn  0 , yn   0

and

yn   0

n  n1 . Where n  n1  no

r

n 1

 n  n1  ( n  s ) Q  w  s n1    r n   n s  n1 r

r

n  r   r n  n

n 1

 (n  s )

ws

s n1

for

is chosen

r 1

Where Qs  Ks qs 

(s )2 as  4s 1 s 

Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)

(3.11)

n 1

 (n  s )

s n1

r

370

1 nr

B. Selvaraj et al., J.Comp.&Math.Sci. Vol.1(3), 364-373 (2010) n 1



r

(n  s) Q s  wn

( n  s ) Q s  w n1

{}nno and

H such that

n

n 1  1 lim sup H ( n , s )  K s qs  n  H ( n, no ) s no 

s  n1

 n  n1    n  

a sequence

r

(3.12)

 s 1as  Q 2 (n, s )  H n s  Ks qs   (n1, o ) 4   s   1 r sup r  (n  s ) Qs  wn1 Then lim n   h (n, s) n s  n1 Q (n, s)   s Where

H ( n, s )

Which contradicts the condition (3.8).

Then every solution If the case (II) holds, we come back to the proof of the second part of Theorem 1 and hence the proof is omitted. This completes the proof.

Next, we present some new oscillation results for equation (1.1), we introduce a double sequence {H ( m, n) / m  n  0} such that

H (m, n)  0 for m  0 (ii ) H ( m, n )  0 for m  n  0 and (i )

(iii )  2 H (m, n)  h(m, n) H (m, n) ; mn0

Theorem 3 : Assume that (H1)-(H4) holds.

Furthermore, assume that there exists

(3.14)

yn of equation (1.1)

is oscillatory.

Proof : Let

yn be a non-oscillatory solution

of equation (1.1). Let us first assume

yn

that

is eventually positive and that

yn  0 , yn   0 and yn   0 for n  n1 . The case where yn is eventually negative with similarly and is omitted.As in the proof of Lemma 1, there are two possible cases.

Let the case (I) hold. Again defining wn as in (3.3), we obtain (3.4).

Where  2 H (m, n)  H ( m, n  1)  H ( m, n)

for m  n  0 .

s 1

(3.13)

Let

us denot e

Rs 

s 

s  s 1

 s  s 1as 

Then from (3.4), we get

Journal of Computer and Mathematical Sciences Vol. 1 Issue 3, 30 April, 2010 Pages (274-402)

and

4

s

B. Selvaraj et al., J.Comp.&Math.Sci. Vol.1(3), 364-373 (2010)

n 1

371

n 1

 H (n, s) K q   H (n, s)[w s s

s

s  n1

  s ws 1  Rs ( ws ) 2 ]

s  n1

n 1 n s s  n1

 [ H (n, s ) w ]

  { 2 H (n, s ) ws 1  H (n, s )[ s ws 1  Rs ( ws )2 ]} s  n1

n 1





 H ( n, n1 ) wn1   [ H ( n, s ) h( n, s )  H ( n, s ) s ws 1  H (n, s ) Rs w2 s 1 ] s  n1

2

 1 Q(n, s )  n 1 Q 2 (n, s) H (n, s)  H (n, n1 ) wn1   H (n, s)  Rs ws 1    2 4 Rs R s  n1  s n1 s   n 1

It follows that, n 1  1 Q 2 ( n, s )   H (n, s)  Ks qs  4 R   wn1 H ( n, n1 ) s  n1 s  

This contradicts (3.13).

(3.15)

satisfies all the conditions of Theorem

If the case (II) holds, we come back to the proof of the second part of Theorem 1 and hence the proof is omitted. This completes the proof.

1 and Theorem 2 for

h ( n, s ) 

H (n, s )  n  s ,

1 and  n  1. Hence all ns

the solutions of equation (E1) are 3 Examples :

oscillatory. In fact

Example 1 :



one such a solution of equation (E1).



 (n  1)2  n  yn  (n  2) yn3   

n yn  (n  2) yn3 80

88

80

{ yn }  {n( 1) n } is

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2. R.P. Agarwal, Martin Bohner, Said 148) (9 yn3  5 y 3n3 )  0 R. Grace, Donal O’Regan, Discrete (E1)



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