International Journal of Computational Engineering Research (IJCER)
Some Oscillation Properties of Third Order Linear Neutral Delay Difference Equations
A.George Maria Selvam
1, M. Paul Loganathan
2, K.R.Rajkumar
31,3
Sacred Heart College, Tirupattur – 635 601, S.India,
2
Dravidian University, Kuppam
I. INTRODUCTION
In this paper, we consider the third order linear neutral delay difference equations from
a n ( )
2x n ( ) p n x ( ) ( ( ))) n q n x ( ) ( ( )) n 0
(1) wheren N n ( )
0 n n
0,
0 1.., , n
0is a nonnegative integer, subject to the following conditions H a n p n q n
1( ), ( ), ( )
are positive sequences.0 0
0 ( ) 1, ( ) , ( ) ,lim ( ) lim ( )
n n
p n p n n n n n n
and0
1
1
( ) ( )
n
s n
R n a s
asn
0 0
2
1 ( )
( )
t nn n s n
H q t
a s
.
0
2 2
1 3
( ( )) ( 1)
lim sup ( )(1 ( ( )))
2 4 ( )
n n
s n
KM s a s
H q s p s
sa s
.We set
z n ( ) x n ( ) p n x ( ) ( ( )) n
.The oscillation theory of difference equations and their applications have received more attention in the last few decades, see [[1]-[4]], and the references cited therein. Especially the study of oscillatory behavior of second order equations of various types occupied a great deal of interest. However the study of third order difference equations have received considerably less attention even though such equations have wide applications. In [[5]- [10]] the authors investigated the oscillatory properties of solutions of third order delay difference equations and in [[11]-[15]]. Motivated by the above observations, in this paper, we investigate the oscillatory behavior of solutions of equation (1).
Let
max lim 0 ( ), ( )
x
n n
. By a solution of equation (1) we mean a real sequence x(n) which is defined for alln n
0
satisfying (1) for alln n
0. A non-trivial solution x(n) is said to be oscillatory if it is neither eventually positive or eventually negative; otherwise, it is nonoscillatory. Equation (1) is said to be oscillatory if all its solutions are oscillatory.II. MAIN RESULTS
Lemma 2.1. Let x(n) be a positive solution of equation (1) for all
n n
0suchx n ( ) 0, x n ( ) 0,
and2
x n ( ) 0
on[ , ) n
1
for somen
1 n
0. Then for each k with 0 < k < 1, there existsn
2 n
1 such that( )
2( ) ,
x n n
k n n
x n n
. (2) Proof. From the Lagrange’s Mean value theorem, we have forn n
1, for some kABSTRACT
In this paper, we establish some sufficient conditions for the oscillation of solutions of third order linear neutral delay difference equations of the form
a n ( )
2( ( ) x n p n x ( ) ( ( ))) n q n x ( ) ( ( )) n 0
.
( ) ( ( ))
( ) ( )
x n x n
x k n n
; for some k (3)such that
( ) n k n .
2x n ( ) 0
and x n ( )
is non-increasing, which implies that x k ( ) x ( ( )) n
and hence, using equation (3)
( ) ( ( )) ( ( ))( ( )) x n x n x n n n
( ) ( ( ))
1 ( ( ))
( ( )) ( ( ))
x n x n
n n
x n x n
(4)Apply Lagrange’s Mean value theorem once again for x(n) on
[ , ( )] n
1 n
forn n
1 ( ) n
. Now1 1
( ( )) ( )
( ) ( )
x n x n
x c n n
for some c such thatn
1 c ( ) n
and x c ( ) x ( ( )) n
which implies( ( )) ( ( ))( ( )
1)
x n x n n n
. Hence
( ( ))
1( ( )) ( )
x n
n n x n
For
K (0,1),
we can findn
2 n
1
( ( )) ( ( )) ( ) x n
K n x n
forn n
2 (5) From equation (4) and for alln n
2, we have( ) 1
1
( ( ))
( ( )) ( )
x n n n
x n K n
( )
1 ( ) ( )
n n
K n K n
n ( )
K n
Hence,
( ( )) ( )
( )
K n
x n
x n n
(6)
Lemma 2.2 Let x(n) be a positive solution of equation (1), then the corresponding sequence z(n) satisfies the following condition z(n) > 0,
z n 0,
2z n 0,
3z n 0
for somen
1 n
0. Then there exitsn
2 n
1 such that( )
( ) 2 , z n Mn
z n
n n
2 for each M, 0 < M < 1.Proof. We define a function H(n) for
n n
2 n
1, as2 2 2
( )
( ) ( ) ( ) ( )
2 M n n
H n n n z n z n
(7)2 2 2 2
( )
( ) ( ) ( ) ( ) ( )
2 M n n
H n z n n n z n z n
(8)By Taylor’s Theorem,
2 2 2
2 2 2
( )
( ) ( ) ( ) ( ) ( )
2
z n z n n n z n n n z n
From (8)
2 2
2 2
2 2
2 2 2 2
( ) ( )
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
2 2
n n M n n
H n z n n n z n z n n n z n z n
which implies
H n ( ) 0
andH n ( 1) H n ( ) H n ( )
2 0
for everyn n
2 from (7)2 2 2
( )
( ) ( ) ( ) 0
2 M n n
n n z n z n
which implies
( )
( ) 2
z n Mn z n
forn n
2.Theorem 2.3. Assume that
( H
1)
to( H
2)
hold, then equation (1) is oscillatory.Proof: Suppose, if possible that the equation (1) has a nonoscillatory solution. Without loss of generality suppose that x(n) is a positive solution of equation (1). We shall discuss the following cases for z(n).
(i)
z n ( ) 0, z n ( ) 0,
2z n ( ) 0,
3z n ( ) 0,
(ii)
z n ( ) 0, z n ( ) 0,
2z n ( ) 0,
3z n ( ) 0,
Case 1.
z n ( ) 0, z n ( ) 0,
2z n ( ) 0,
3z n ( ) 0,
Since z(n) > 0and
z n ( ) 0,
then there exists finite limitslim ( )
n
z n k
. We shall prove that k = 0.Assume that k > 0. Then for any
0
, we havek z n ( ) k
. Let(1 )
0 k p
p
, we have( ) ( ) ( ( )) k x n k p n x n
.( ) ( ) ( )
x n k p k m k
x n ( ) mz n ( )
When
( )
( )
k p k
m k
. Now from the equation (1) we have
a n ( )
2x n ( ) p n x ( ) ( ( ) n q n x ( ) ( ( )) n a n ( )
2z n ( ) q n mz ( ) ( ( )) s
Summing the above inequality from n to ∞ we get,
( )
2( ) ( ) ( ( ))
t n s n
a t z t m q s z s
( )
2( ) ( ) ( ( ))
s n
a n z n m
q s z s
Using the fact that
z ( ( )) n k
we obtain,( )
2( ) ( )
s n
a n z n mk q s
which implies2
1
( ( )) ( )
( )
s nz n mk q s
a n
. Summing from n to ∞ we have,2
1
( ) ( )
s n s n
( )
t sz s mk q t
a s
1
( ) ( )
s n
( )
t sz n mk q t
a s
Summing the last inequality
n
1 to
1
1
( ) 1 ( )
n n s n
( )
t sz n mk q t
a s
This contradicts
H
2 . Thus k = 0. Moreover, the inequality,0 x n ( ) z n ( )
implieslim ( ) 0
n
x n
.Case 2.
z n ( ) 0, z n ( ) 0,
2z n ( ) 0,
3z n ( ) 0,
We have x(n) = z(n) - p(n)
x ( ( )) n
, we obtain further( ( )) ( ( )) ( ( )) ( ( ) ) x n z n p n x n
z ( ( )) n p ( ( )) ( ( )) n x n
(1 p ( ( ))) ( ( )) n z n
. From equation (1) we have, a n ( )
2z n ( ) q n x ( ) ( ( )) n
a n ( )
2z n ( ) q n ( )(1 p ( ( ))) ( ( )) n z n
2
1
( ) ( )
( ) ,
( ) a n z n
w n n n n
z n
(9)2 2
( 1) ( 1) ( ) ( )
( ) ( 1) ( )
a n z n a n z n
w n n
z n z n
2 2 2
( 1) ( ( ) ( )) ( ) ( ) ( )
1 ( 1) ( ) ( 1)
w n a n z n a n z n z n
n n z n z n z n
2 2 2
2
( 1) ( ( ) ( )) ( ( 1))
1 ( ) ( ) ( ( 1))
w n a n z n z n
n a n
n z n z n
2
2 2
( 1) ( )(1 ( ( ))) ( ( )) ( )
( 1)
1 ( ) ( 1) ( 1)
w n nq n p n z n na n
n z n n a n w n
(10)Also from Lemma (2.1) with x(n) =
z n ( )
( ( )) ( ) , ( ) ( )
x n K n
n n
x n n
( ( )) ( ) ( )
z n K n
z n n
(11)1 ( ) 1
( ) ( ( ))
K n
z n n z n
for ( ) n n
1n
2.By Lemma (2.2)
( ( )) ( ) ( ( ))
( ) ( ( ))
z n K n z n
z n n z n
( ) ( ) 2
K n M n
n
(12)( ( )) ( ( ))
2( ) 2
z n KM n
z n n
Using (11) and (12) in (10)
2
2 2 2
( ) ( 1) ( )
( ) ( )(1 ( ( ))) ( 1)
2 1 ( 1) ( 1)
KM n w n na n
w n q n p n w n
n n a n
(13)Using the inequality
2
2
1
, 0
4
Vx Ux V U
U
And put
1
2( )
2( 1), , ,
1 ( 1) ( 1)
x w n V U na n
n n a n
we have
2 2
2 2
( 1) ( ) ( 1)
( 1)
( 1) ( 1) ( 1) 4 ( )
w n na n a n
n n a n w n na n
(14)From equation (13)
2 2
( ( )) ( 1)
( ) ( )(1 ( ( )))
2 4 ( )
KM n a n
w n nq n p n
n na n
(15)Summing the last inequality from
n
2 ton 1
we obtain2
2 2
1
2
( ( )) ( 1)
( )(1 ( ( ))) ( )
2 4 ( )
n
n n
KM s a s
q s p s w n
sa s
Taking lim sup in the above inequality, we obtain contradiction with
H
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