In this paper some sufficient
were obtained with the assumption of e the following conditions.
condition for the oscillation of first order
neutral delay difference equation were
H1:
{ pn } is an positive sequence.
obtained. .
Neutral Delay Difference Equation, Oscillation, Nonoscillation, Eventually
H2: f is a continuous function such that
uf (u) ≥ 0 .
H3: If there exists a function w such that
positive.
w(u)>0, for u>0 and
f (uv) ≤ w(u) f (v) .
H4: If there exists a function φ such that
In this paper some sufficient
φ(u) is increasing and
uφ(u) > 0, for
u ≠ 0
condition for the oscillation of first order
& φ(u + v) ≤
f (u) f (v) .
neutral delay difference equation of the form
∆ (an xn − pn xn−k ) + qn f ( xn−l ) = 0, n ∈ N (n0 )
(1.1.1)
and
∆ ( xn + pn xn−k ) + qn f ( xn−l ) = 0, n ∈ N (n0 )
(1.1.2)
In this section, I obtain some sufficient condition for the oscillatory solutions of the equation (1.1.1) and (1.1.2)
Assume that
pn ≤ 1
Hence
zn → −∞ ,
n → ∞
an−k
and xn be an eventually positive solution of
as n → ∞.
.Which contradicts the fact that
the equation (1.1.1) and
zn is eventually positive. Hence the proof.
yn = (an xn − pn xn−k )
. Then eventually yn >0.
pn
Assume that pn, qn > 0 and
an−k
≤ 1.
Let us consider xn > 0, xn-l>0, xn-k >0 for
lim
inf q∗ (1+ pn−l λn−k ) > 0,
some n > n1.
If,
n→∞ n
q∗ = qn
qn−k
From the equation (1.1.1),
where
a
n
n−l
,then every solution of
∆yn
= −qn f (xn−l ) < 0.
Hence yn is a
equation (1.1.1) is an oscillatory solution.
decreasing function.
Suppose yn is not eventually positive, then eventually yn < 0.
Hence there exists n2 > n1 and M>0, such
Let us assume the contradiction that equation (1.1.1) has an non oscillatory solution. Let us consider xn is eventually
that
Let,
Then,
yn < −M .
zn = an xn > 0.
zn = yn + pn xn−k .
positive.
Let us consider xn > 0, xn-l>0, xn-k >0 for some n > n1.
By theorem 1.2.1, yn is eventually positive.
zn < −M +
pn an−k
.
zn−k
Also we have
∆yn
= −qn f (xn−l )
n
n
λ ≥ q∗ +
p λ q
∗
n−l n−k n
q
yn = an xn − pn xn−k
n−k
∗
pn−l λn−k
∆yn ≤ −qn xn−l
(1.2.1)
Hence
limn→∞ inf qn (1 +
qn−k
) ≤ λn ,
∆yn
≤ −qn
yn−l + pn−l xn−k −l
a
Therefore,
lim
n→∞
n
inf q∗ (1 + pn−l λn−k ) ≤ 0.
qn−k
n−l
∆yn
= − qn yn−l
an−l
qn pn−l xn−k −l
an−l
Which contradicts the given condition of the theorem. Hence every solution of the
From the equation (1.2.1),
equation (1.1.1) is an oscillatory solution.
n
∆y ≤ − qn yn−l
+ qn pn−l ∆yn−k
an−l
an−l
qn−k
an = 1, forn = 1,2,3....
Hence yn satisfies the inequality,
Suppose that
Then
n
∆y +
qn y
n−l
qn
pn−l
∆yn−k
≤ 0.
∆ ( xn + pn xn−k ) + qn f ( xn−l ) = 0, n ∈ N (n0 )
is oscillatory if there exists a function λ
an−l
an−l
qn−k
such that
0 ≤ λn ≤ 1
for
n ≥ n0
and the
Let
y
=
λ
− ∆yn n
n
, then
differenc inequality
∆ zn + Qnφ(zn−l +k ) ≤ 0,
λ y ≥ qn yn−l
qn pn−l ∆yn−k
(1.2.2)
n n a
n−l
an−l
qn−k
has oscillatory solution where,
q ∗ = qn
(1 −λ )q
a
n
n−l
, Hence we have,
Qn = min λn qn ,
n−k
wpn−l
n−l
Suppose to the contrary that there is a non oscillatory solution xn. Assume that ,
m
∑ yn
n=n0
m
> ∑Qn
n=n0
f ( x
n−l ) +
m
∑ Qn
n=n0 +k
f ( p
n−l
xn−k −l )
xn > 0,
m
∑ yn
n=n0
m
> ∑
n=n0
Qn { f ( x
n−l
) + f ( p
n−l
xn−k −l )}
For all n>n0. Let
m m
∑ yn > ∑Qn {φ( xn−l + pn−l xn−k −l )}
yn = xn + pn xn−k
n=n0
n=n0 +k
.
∆( yn ) = −qn f (xn−l ) < 0
m
∑ yn >
n=n0
m
∑Qn {yn−l }
n=n0 +k
Also yn+1<yn, yn is decreasing function.,
Let
m
zn = ∑Qnφ{yn−l } > 0.
Hence
yn+1 + qn f ( xn−l ) = yn
n0 +k
yn > qn f ( xn−l ), n ≥ n0.
Then
∆zn = zn+1 − zn
Taking summation from n0 to m , m>n0,
∆zn =
m
∑(Qn+1φ{yn+1−l } − Qnφ( yn−l ))
n=n0 +k
m m
∑ yn > ∑
qn f ( xn−l )
∆zn
= Qm+1φ( y
m+1−l
0
) − φn +k
φ( y
n0 +k −l )
n=n0
n=n0
∆zn > −Qnφ( yn−l )
m m
∑ yn > ∑((λn qn f ( xn−l ) + (1 − λn )qn f ( xn−l ))
∆zn > −Qnφ(zn−l +k )
n=n0
m
n=n0
m m
∆zn + Qnφ(zn−l +k ) > 0.
This condition holds
∑ yn > ∑Qn f ( xn−l ) +
∑ (1 − λn−k )qn−k f ( xn−k −l )
n=n0
n=n0
n=n0 +k
when zn
is eventually positive solution. This
m m m
is a contradiction to the equation (1.2.2).
∑ yn > ∑Qn f ( xn−l ) +
∑ Qn w( pn−l ) f ( xn−k −l )
n=n0
n=n0
n=n0 +k
Hence the proof compltes. Similarly we prove that, when xn is eventually negative.
Consider the first order neutral delay difference equation
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∆ (nxn
Here
− xn−1
) + (2n + 3)x
n−2
3 = 0, n > 0
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an = n, k = 1,l = 2, pn = −1, qn = (2n + 3)
All the conditions of the theorem 1.2.2 are satisfied. Hence all its solutions are oscillatory. One such solution is (-1) n.
Consider the first order neutral delay difference equation
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∆ (xn
Here
− 1
n −1
xn−1 ) +
2n + 3
(n − 2)3
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an = 1, k = 1,l = 2, pn
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= 2n + 3 (n − 2)3
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Hence all the conditions of the theorem 1.2.2 are satisfied.
Hence all its solutions are oscillatory. One such solution is n(-1)n.