Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 6 Issue 2(2014), Pages 23-31.
OSCILLATION OF A CLASS OF SECOND ORDER NEUTRAL
DIFFERENCE EQUATIONS WITH DELAYS
(COMMUNICATED BY CHUANZHI BAI)
VILDAN KUTAY, HUSEYIN BEREKETOGLU
Abstract. By using Riccati transformation techniques we will establish some
oscillation criteria for the second order neutral delay difference equation
∆ [p(n) (∆ (x(n) + q(n)x(n − τ ) + h(n)x(n − σ)))γ ]
+f (n, x(n − σ)) = g(n, x(n − σ), x(n − τ )), n ≥ 0.
Moreover, in some special cases, we show that our conditions can be reduced
to those given in [[10],[15]]. Finally, some examples are given to illustrate our
results.
1. Introduction
In this paper, we consider the neutral delay difference equation
∆ [p(n) (∆ (x(n) + q(n)x(n − τ ) + h(n)x(n − σ)))γ ]
+ f (n, x(n − σ)) = g(n, x(n − σ), x(n − τ )), n ≥ 0
(1.1)
where ∆ is forward difference operator defined by ∆x(n) = x(n + 1) − x(n), γ > 0
is a quotient of odd positive integers, {p(n)} is a positive real sequence, {q(n)}
and {h(n)} are nonnegative real sequences, τ and σ are fixed nonnegative integers;
f (n, x) and g(n, x, y) are defined for all n ∈ N = {0, 1, 2, ...} and x, y ∈ R.
Neutral difference equations can be applied in several fields such as bifurcation
analysis, population dynamics, stability theory, the dynamics of delayed network
systems and others. In recent years, the oscillatory behaviour of neutral delay
difference equations has been investigated by many authors [See: [1]-[21]].
In [12], some new oscillation results have been found for the second order nonlinear neutral delay difference equation
∆(a(n) (∆ (x(n) + p(n)x(n − τ )))γ ) + f (n, x(n − σ)) = 0,
n = 0, 1, 2, ...
(1.2)
Here, we aim to have similar oscillation results for Eq. (1.1) which is more
general than Eq. (1.2).
2010 Mathematics Subject Classification. 39A10, 39A21.
Key words and phrases. Oscillation; Riccati transformation; Neutral delay difference equation.
c
2014
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted March 23, 2014. Published Jun 1, 2014.
23
24
V. KUTAY, H. BEREKETOGLU
Definition 1.1. By a solution of (1.1), we mean a nontrivial sequence {x(n)}
which is defined for n ≥ −L, where L = max {τ, σ} and satisfies equation (1.1) for
n = 0, 1, 2, ...
Definition 1.2. A solution {x(n)} of (1.1) is said to be oscillatory if for every n1 >
0 there exists an n ≥ n1 such that x(n)x(n + 1) ≤ 0, otherwise it is nonoscillatory.
If all solutions of Eq. (1.1) are oscillatory, then Eq. (1.1) is said to be oscillatory.
Before given the main results, we note that the initial value problem which
consists of Eq. (1.1) and the initial function x(n) = ϕ(n) has a unique solution,
where ϕ(n) is defined for n = −L, ..., −1, 0.
2. Main results
Theorem 2.1. Let γ > 0. Suppose that ∆p(n) ≥ 0 and the following conditions
hold:
γ1
∞
P
1
= ∞, 0 ≤ q(n) + h(n) < 1,
(H1)
p(n)
n=n0
(H2) There exist two real sequences {r(n)} and {s(n)} such that r(n) − s(n) ≥ 0
and
n
X
lim inf
(r(i) − s(i)) ≥ 0 for all large n0 ,
(2.1)
n→∞
i=n0
f (n, u) ≥ r(n)uγ and g(n, u, v) ≤ s(n)uγ for u 6= 0.
n+σ
P
Furthermore, we assume that
Q(i) > 0 and
(2.2)
i=n+1
∞
X

Q(n) 
n=n0
n+σ
X
1
! 1+σ
Q(i)

(σ + 1) − σ  = ∞,
(2.3)
i=n+1
where
γ
(1 − q(n − σ) − h(n − σ))
Q(n) = (r(n) − s(n))
p(n − σ)
Then every solution of equation (1.1) oscillates.
n−σ
2
γ
.
Proof. Suppose that {x(n)} is an eventually positive solution of (1.1) such that
x(n − L) > 0 for all n ≥ n0 > 0. Define
z(n) = x(n) + q(n)x(n − τ ) + h(n)x(n − σ).
(2.4)
This function is positive, that is, z(n) > 0 for n ≥ n0 . From (1.1) and (2.2),
γ
∆ [p(n) (∆z(n)) ] ≤ − (r(n) − s(n)) (x (n − σ))γ ≤ 0, n ≥ n0 .
γ
(2.5)
Hence {p(n) (∆z(n)) } is an eventually nonincreasing sequence. Firstly, we claim
that ∆z(n) ≥ 0 for n ≥ n0 . Otherwise, there would be an integer n1 ≥ n0 such that
γ
γ
p(n1 ) (∆z(n1 )) = α < 0 and then p(n) (∆z(n)) ≤ α for n ≥ n1 , i.e.,
γ1
α
∆z(n) ≤
.
p(n)
Summing both sides of the last inequality from n1 to n − 1, we get
n−1 γ1
1 X
1
γ
z(n) ≤ z(n1 ) + α
→ −∞ as n → ∞.
p(i)
i=n1
OSCILLATION OF A CLASS OF SECOND ORDER NEUTRAL DIFFERENCE EQUATIONS25
But, this contradicts with z(n) > 0 for n ≥ n0 . So, ∆z(n) ≥ 0 for n ≥ n0 . From
this fact and (2.4), it follows (1 − q(n) − h(n)) z(n) ≤ x(n). Therefore,
(1 − q(n − σ) − h(n − σ)) z(n − σ) ≤ x(n − σ), n ≥ n1 = n0 + σ.
(2.6)
From (1.1), (2.2) and (2.6), we have
γ
γ
∆ [p(n) (∆z(n)) ] + (r(n) − s(n)) (1 − q(n − σ) − h(n − σ)) (z(n − σ))γ ≤ 0 (2.7)
for n ≥ n1 .
Secondly, we claim that ∆2 z(n) ≤ 0 for n ≥ n0 . Otherwise, there would be an
integer n1 ≥ n0 such that ∆2 z(n) > 0, i.e., ∆z(n + 1) > ∆z(n). Since ∆p(n) ≥ 0,
p(n + 1)(∆z(n + 1))γ > p(n + 1)(∆z(n))γ ≥ p(n)(∆z(n))γ
and this is a contradiction. So, ∆2 z(n) ≤ 0 and then {∆z(n)} is a nonincreasing
sequence. Therefore
z(n) − z(n1 ) =
n−1
X
∆z(k) ≥ (n − n1 ) ∆z(n)
k=n1
n
∆z(n) for n ≥ n1 ≥ 2n0 + 1. Then
2
n−σ
z(n − σ) ≥
∆z(n − σ), n ≥ n2 = n1 + σ.
2
Then (2.7) and (2.8) imply that
and z(n) ≥
γ
(2.8)
γ
∆ [p(n) (∆z(n)) ] + (r(n) −
s(n)) (1
γ− q(n − σ) − h(n − σ))
n−σ
γ
×
(∆z(n − σ)) ≤ 0, n ≥ n2 .
2
(2.9)
γ
Let y(n) = p(n) (∆z(n)) , hence y(n) > 0 and
∆y(n) + Q(n)y(n − σ) ≤ 0, n ≥ n2 .
(2.10)
Define
∆y(n)
.
(2.11)
y(n)
Since {y(n)} is nonincreasing sequence, we have that 0 ≤ λ(n) < 1 for large n.
From (2.11), we obtain
y(n + 1)
= 1 − λ(n)
y(n)
and
n−1
Y
y(n − σ)
−1
=
(1 − λ(i)) .
y(n)
i=n−σ
λ(n) = −
Using (2.10) and (2.11), then
λ(n) ≥ Q(n)
n−1
Y
(1 − λ(i))
i=n−σ
Set b(n) =
n+σ
P
−1
!−σ
n−1
1 X
≥ Q(n) 1 −
λ(i)
.
σ i=n−σ
(2.12)
Q(i). By (2.12),
i=n+1
n−1
X
1
b(n)
λ(i)
λ(n) ≥ Q(n) 1 −
σb(n)
i=n−σ
!−σ
.
(2.13)
26
V. KUTAY, H. BEREKETOGLU
From (2.13) and the inequality
1
−σ
1+σ (σ + 1) − σ
r
1
1 − rx
≥x+
,
σ
r
for r > 0 and x <
σ
,
r
we obtain that
"
#
n−1
1
1 X
1 1+σ
(b(n))
λ(n) ≥ Q(n)
λ(i) +
(σ + 1) − σ .
b(n) i=n−σ
b(n)
(2.14)
Rearranging (2.14), we conclude that

n−1
X
λ(n)b(n) − Q(n)
λ(i) ≥ Q(n) 
i=n−σ
n+σ
X
1
! 1+σ

(σ + 1) − σ 
Q(i)
i=n+1
and for N > n2
N
X
λ(n)b(n)−
n=n2
N
X
Q(n)
n=n2
n−1
X
N
X
λ(i) ≥

Q(n) 
n=n2
i=n−σ
n+σ
X
1
! 1+σ
Q(i)

(σ + 1) − σ  .
i=n+1
(2.15)
If we change the bounds of summation, we have
N
P
Q(n)
n=n2
n−1
P
λ(i) ≥
i=n−σ
N −σ−1
i+σ
P
P
i=n2
λ(i)Q(n)
=
N −σ−1
P
=
i=n2
N −σ−1
P
n=i+1
λ(i)
i+σ
P
Q(n)
λ(n)
n=i+1
n+σ
P
Q(i).
n=n2
i=n+1
(2.16)
Using (2.15) and (2.16), we can deduce that


1
! 1+σ
N
n+σ
N
n+σ
X
X
X
X
(σ + 1) − σ  .
λ(n)
Q(i) ≥
Q(n) 
Q(i)
n=N −σ
i=n+1
n=n2
(2.17)
i=n+1
Because of {y(n)} is positive and nonincreasing,
n+σ
X
Q(i) ≤ 1.
(2.18)
i=n+1
So, from (2.3), (2.17) and (2.18), it follows that


1
! 1+σ
N
N
n+σ
X
X
X
λ(n) ≥
Q(i)
(σ + 1) − σ  → ∞ as n → ∞.
Q(n) 
n=N −σ
n=n2
i=n+1
From the definition of λ(n), we get
N
X
n=N −σ
λ(n) =
N
X
n=N −σ
1−
y(n + 1)
y(n)
< σ + 1.
And this contradicts with (2.17). It is noted that even in the case of {x(n)} is an
eventually negative solution of (1.1), we get similar contradiction by substitution
y(n) = −x(n) into Eq. (1.1). So, every solution of (1.1) oscillates.
OSCILLATION OF A CLASS OF SECOND ORDER NEUTRAL DIFFERENCE EQUATIONS27
Theorem 2.2. Let γ ≥ 1. Suppose that ∆p(n) ≥ 0, (H1) and (H2) hold. Moreover,
assume that there exists a positive sequence {ρ(n)} such that
"
#
n
2
X
p(k − σ) (∆ρ(k))
lim sup
ρ(k)M (k) −
=∞
(2.19)
γ−1
n→∞
4γ k−σ
ρ(k)
k=0
2
where M (n) = (r(n) − s(n))(1 − q(n − σ) − h(n − σ))γ . Then every solution of Eq.
(1.1) oscillates.
Proof. Suppose that, {x(n)} is an eventually positive solution of (1.1), i.e., x(n) > 0
and x(n − σ) > 0 for all n ≥ n0 . Following similar steps in the proof of Theorem
2.1, we have (2.7). Define the sequence {w(n)} as
γ
w(n) = ρ(n)
p(n) (∆z(n))
.
(z(n − σ))γ
(2.20)
So, w(n) > 0 and we have
γ
ρ(n)
∆ (p(n) (∆z(n)) )
∆w(n) = p(n + 1) (∆z(n + 1)) ∆
. (2.21)
+
ρ(n)
(z(n − σ))γ
(z(n − σ))γ
γ
Using (2.7) and (2.21), we obtain
∆ρ(n)
w(n + 1)
ρ(n + 1)
γ
p(n + 1) (∆z(n + 1)) ∆ ((z(n − σ))γ )
− ρ(n)
.
(z(n + 1 − σ))γ (z(n − σ))γ
∆w(n) ≤
−ρ(n)M (n) +
(2.22)
Because of (2.7) and ∆z(n) ≥ 0, we get
γ
p(n−σ)∆ (z(n − σ)) ≥ p(n+1) (∆z(n + 1))
γ
and z(n+1−σ) ≥ z(n−σ). (2.23)
From (2.22) and (2.23), we have
∆w(n) ≤
∆ρ(n)
w(n + 1)
ρ(n + 1)
γ
p(n + 1) (∆z(n + 1)) ∆ (z γ (n − σ))
−ρ(n)M (n) +
− ρ(n)
(z γ (n + 1 − σ))
2
.
By using the inequality
xγ − y γ ≥ γy γ−1 (x − y) for all x 6= y > 0 and γ ≥ 1,
we obtain
∆ρ(n)
w(n + 1)
ρ(n + 1)
γ−1
γ
p(n + 1)γ (z(n − σ))
∆ (z(n − σ)) (∆z(n + 1))
∆w(n) ≤ −ρ(n)M (n) +
− ρ(n)
(z γ (n + 1 − σ))
2
.
(2.24)
Then from (2.8), (2.23) and (2.24), it follows that for n ≥ n2
γ−1
n−σ
∆ρ(n)
w(n + 1) − γ
∆w(n) ≤ −ρ(n)M (n) +
ρ(n + 1)
2
2
2
2γ
ρ(n)
(p(n + 1)) (ρ(n + 1)) (∆z(n + 1))
×
.
2
2
(ρ(n + 1)) p(n − σ)
(z γ (n + 1 − σ))
(2.25)
28
V. KUTAY, H. BEREKETOGLU
So, we have
∆w(n) ≤
=
<
∆ρ(n)
−ρ(n)M (n) +
w(n + 1)
ρ(n
+ 1)
γ−1
ρ(n)
n−σ
w2 (n + 1)
−γ
2
2
(ρ(n + 1)) p(n − σ)
p(n − σ)(∆ρ(n))2
−ρ(n)M (n) +
γ−1
4γ n−σ
ρ(n)
2
2
q
p
γ−1
ρ(n)
γ n−σ
p(n
−
σ)∆ρ(n)
2

p
w(n + 1) − q
−
γ−1
ρ(n + 1) p(n − σ)
ρ(n)
2 γ n−σ
2
"
#
p(n − σ)(∆ρ(n))2
− ρ(n)M (n) −
.
γ−1
ρ(n)
4γ n−σ
2
(2.26)
Summing both sides of (2.26) from n2 to n, it follows that
"
#
n
X
p(k − σ)(∆ρ(k))2
−w(n2 ) < w(n + 1) − w(n2 ) < −
ρ(k)M (k) −
.
γ−1
4γ k−σ
ρ(k)
k=n2
2
Then we have
n
X
k=n2
"
p(k − σ)(∆ρ(k))2
ρ(k)M (k) −
γ−1
4γ k−σ
ρ(k)
2
#
<c
which is contrary to (2.19); where c > 0 is a finite constant. On the other hand, the
proof of the case of {x(n)} not to be eventually negative is similar to the previous
part. Hence, every solution of (1.1) oscillates.
Corollary 2.3. If q(n) = h(n) = g(n, x(n − σ), x(n − τ )) = 0,
f (n, x(n − σ)) = η(n)(x(n − σ))γ , then the condition (2.19) is reduced to
"
#
n
2
X
p(k − σ) (∆ρ(k))
lim sup
ρ(k)η(k) −
=∞
γ−1
n→∞
4γ k−σ
ρ(k)
k=0
2
which is the same as that in (Corollary 2.1, [10] ).
Corollary 2.4. In the special case of Eq. (1.1)
∆2 x(n) + µ(n)x(n − σ) = 0,
n ≥ 0,
the condition (2.19) is reduced to
lim sup
n→∞
n
X
k=0
"
1
ρ(k) µ(k) −
4
∆ρ(k)
ρ(k)
2 #
=∞
which is the same as that in (Corollary 1, [15] ).
Theorem 2.5. Let 0 < γ < 1. Suppose that q(n) = h(n) = 0, ∆p(n) ≥ 0 and
γ1
∞ X
1
<∞
(2.27)
p(n)
n=n
0
OSCILLATION OF A CLASS OF SECOND ORDER NEUTRAL DIFFERENCE EQUATIONS29
holds. Moreover, we assume that there exists a positive sequence {ρ(n)} such that
(2.19) holds and also there is a positive sequence {δ(n)} such that
∞
X
∆δ(n) ≥ 0,
δ(n + 1) [r(n) − s(n)] = ∞ and
n=n0
! γ1
n−1
X
1
=∞
δ(i + 1) [r(i) − s(i)]
p(n)δ(n) i=n
∞
X
n=n0
(2.28)
0
for some n0 > 0. Then every solution of equation
γ
∆ [p(n) (∆x(n)) ] + f (n, x(n − σ)) = g(n, x(n − σ), x(n − τ ))
(2.29)
oscillates or converges to zero.
Proof. Assume that {x(n)} is an eventually positive solution of (2.29), i.e., x(n) > 0
and x(n − σ) > 0 for all n ≥ n0 . From Eq. (2.29),
γ
∆ [p(n) (∆x(n)) ] ≤ 0, n ≥ n0 .
(2.30)
So, {p(n)∆x(n)} is an eventually nonincreasing sequence. From (2.1), the nonincreasing sequence {p(n)∆x(n)} is either eventually positive or eventually negative.
Hence there exist two possible cases of ∆x(n) : ∆x(n) < 0, ∆x(n) > 0.
In case of ∆x(n) < 0 for n ≥ n1 > n0 , it is lim x(n) = a ≥ 0. We now claim that
n→∞
a = 0. If not, then (x(n − σ))γ → aγ > 0 as n → ∞. Therefore there exists n2 ≥ n1
such that (x(n − σ))γ ≥ aγ . Hence from (2.2), we have
γ
∆ [p(n) (∆x(n)) ] ≤ − (r(n) − s(n)) aγ .
γ
Let us define ν(n) = δ(n)p(n) (∆x(n)) for n ≥ n2 . Then we can write
γ
∆ν(n) ≤ −aγ δ(n + 1) (r(n) − s(n)) + p(n) (∆x(n)) ∆δ(n).
Summing both sides of the last inequality from n2 to n − 1, we find
ν(n) ≤ ν(n2 ) − aγ
n−1
X
δ(i + 1) (r(i) − s(i)) +
i=n2
n−1
X
γ
p(i) (∆x(i)) ∆δ(i).
i=n2
From (2.28), we get
n−1
X
ν(n) ≤ ν(n2 ) − aγ
δ(i + 1) (r(i) − s(i)) .
i=n2
Due to (2.28), since
∞
P
δ(n + 1) [r(n) − s(n)] = ∞, it is possible to take an integer n3 sufficiently large
n=n0
such that for all n ≥ n3
ν(n) ≤ −
n−1
aγ X
δ(i + 1) (r(i) − s(i)) .
2 i=n
2
Summing both sides of this inequality from n3 to n, we get
x(n + 1) ≤ x(n3 ) −
aγ
2
γ1 X
n
s=n3
! γ1
s−1
X
1
δ(i + 1) [r(i) − s(i)]
.
p(i)δ(i) i=n
2
30
V. KUTAY, H. BEREKETOGLU
Condition (2.28) implies that {x(n)} is eventually negative. So, we have a contradiction. Thus {x(n)} converges to zero.
On the other hand, the case of ∆x(n) > 0 leads us to a similar contradiction.
Therefore, x(n) is not eventually positive. Similarly, if we assume that a solution
of (2.29) {x(n)} is eventually negative, then {x(n)} → 0 or there is a contradiction.
Hence, the proof is complete.
Example 2.6. Consider the second order neutral delay difference equation
" 79 #
1
1
∆ n ∆ x(n) + 2 x(n − 4) + 3 x(n − 3)
n
n
9
5 7n
3−6n
(2.31)
+
+
+ 2
(x (n − 3)) 7
4
2
n +4
!
23
1
(x(n − 3)) 7
(x(n − 4))4
= 2
, n ≥ 4.
n
1 + (x(n − 3))2 1 + (x(n − 4))4
Since
f (n, x(n − σ)) = 54 +
7n
2
3−6n
n2 +4
9
(x(n − 3)) 7 ≥
5
4
7n
2
9
(x(n − 3)) 7 and
!
23
(x(n − 3)) 7
(x(n − 4))4
9
1
g(n, x(n − σ), x(n − τ )) = n2
≤ n12 (x(n − 3)) 7 ,
2
4
1 + (x(n − 3)) 1 + (x(n − 4))
all the hypotheses of Theorem 2.1 are satisfied. Thus every solution of equation
(2.31) oscillates.
+
+
Example 2.7. Consider the second order neutral delay difference equation
h
i
1
∆ (n + 1)2 (∆(x(n)) 9
n2 + 5
7
1
+ 6(n + 1) 3 + 2
(x(n − 2)) 9
(2.32)
n + 4n + 6
!
19
4
1
(x(n − 2)) 9
(x(n − 3))
=
, n ≥ 2.
8 1 + (x(n − 2))2 1 + (x(n − 3))4
It follows that 2
7
1
7
1
+5
f (n, x(n − σ)) = 6(n + 1) 3 + n2n+4n+6
(x(n − 2)) 9 ≥ 6(n + 1) 3 (x(n − 2)) 9
and
!
19
(x(n − 3))4
1
(x(n − 2)) 9
1
1
≤ (x(n − 2)) 9 .
g(n, x(n − σ), x(n − τ )) =
2
4
8 1 + (x(n − 2)) 1 + (x(n − 3))
8
If we take δ(n) = n, then all the hypotheses of Theorem 2.5 are satisfied. Thus
every solution of equation (2.32) oscillates or converges to zero.
Acknowledgments. We wish to thank the referee for the positive and helpful
criticisms.
References
[1] R. P. Agarwal, Difference Equations and Inequalities Theory Methods and Applications,
Marcel Dekker, New York 2000.
[2] R. P. Agarwal and P. J. Y. Wong, Advanced Topics in Difference Equations, Kluwer Academic
Publishers, Drodrecht 1997.
[3] S. Chen, and L. H. Erbe, Riccati techniques and discrete oscillation, J. Math. Appl. 142
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OSCILLATION OF A CLASS OF SECOND ORDER NEUTRAL DIFFERENCE EQUATIONS31
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VILDAN KUTAY
Department of Mathematics, Faculty of Science, Ankara University, 06100 Ankara,
TURKEY
E-mail address: vkutay@ankara.edu.tr
HUSEYIN BEREKETOGLU
Department of Mathematics, Faculty of Science, Ankara University, 06100 Ankara,
TURKEY
E-mail address: bereket@science.ankara.edu.tr