VNU Journal of Science, Mathematics - Physics 24 (2008) 133-143
Oscilation and Convergence for a Neutral Difference Equation
Dinh Cong Huong*
Dept. of Math, Quy Nhon University 170 An Duong Vuong, Quynhon, Binhdinh, Vietnam
Received 24 April 2008
Abstract. The oscillation and convergence of the solutions of neutral difference equation
∆(x
n
+ δx
n−τ
) +
r

i=1
α
i
(n)F (x
n−m
i
) = 0, n = 0, 1, · · ·
are investigated, where m
i
∈ N
0
, ∀i =
1, r and F is a function mapping R to R.
Keyworks: Neutral difference equation, oscillation, nonoscillation, convergence.
1. Introduction
It is well-known that difference equation
∆(x
n

i
) = 0, (2)
for n ∈ N, n  a for some a ∈ N, where r, m
1
, m
2
, · · · , m
r
are fixed positive integers, the functions
α
i
(n) are defined on N and the function F is defined on R.
Put A = max{τ, m
1
, · · · , m
r
}. Then, by a solution of (2) we mean a function which is defined
for n  −A and sastisfies the equation (2) for n ∈ N. Clearly, if
x
n
= a
n
, n = −A, −A + 1, · · · , −1, 0
are given, then (2) has a unique solution, and it can be constructed recursively.
A nontrivial solution {x
n
}
na
of (2) is called oscillatory if for any n
1

α
i
(n)x
n−m
i
= 0, (3)
for n ∈ N, n  a for some a ∈ N, where r, m
1
, m
2
, · · · , m
r
are fixed positive integers and the
functions α
i
(n) are defined on N. It is clear that equation (3) is a particular case of (2). We shall
establish some sufficient criterias for the oscillation of solutions of the difference equation (3). First
of all we have
Theorem 1. Assume tha t
( ˜m + 1)
˜m+1
˜m
˜m
r

i=1
lim inf
n→∞
α
i

∈ N. Setting v
n
=
x
n
x
n+1
and dividing this inequality by x
n
, we obtain
1
v
n
 1 −
r

i=1
α
i
(n)
m
i

ℓ=1
v
n−ℓ
, (6)
where n  n
1
+ m, m = max

r

i=1
α
i
(n)
m
i

ℓ=1
v
n−ℓ
,
or
1
β
 1 −
r

i=1
lim inf
n→∞
α
i
(n) · β
m
i
. (7)
Since
β

i
(n)β
m
i
 1 −
r

i=1
lim inf
n→∞
α
i
(n)β
˜m
.
D.C. Huong / VNU Journal of Science, Mat hematics - Physics 24 (2008) 133-143 135
From (7) we have
lim inf
n→∞
r

i=1
α
i
(n) 
β − 1
β
˜m+1
.
But

α
i
(n)x
n−m
i
 0, n ∈ N
has no eventually negative solution. So, the proof is complete.
Corollary. Assume tha t
r

r

i=1
lim inf
n→∞
α
i
(n)

1
r
>
ˆm
ˆm
( ˆm + 1)
ˆm+1
, (8)
where δ = 0, α
i
(n)  0, n ∈ N, 1  i  r and ˆm =


r

i=1
lim inf
n→∞
α
i
(n)β
m
i

1
r
,
which is the same as
r

i=1
lim inf
n→∞
α
i
(n) · β
m
i
 r

r


n→∞
α
i
(n)

1
r
β
ˆm
.
By using the inequality (7) we have
r

r

i=1
(li m i nf
n→∞
α
i
(n))

1
r

ˆm
ˆm
( ˆm + 1)
ˆm+1
,

(ℓ)] = ∞, then z
n
< 0 and ∆z
n
 0 eventually.
Proof. (a) Since α
i
(n) ≡ 0, we have
∆z
n
= −
r

i=1
α
i
(n)x
n−m
i
< 0
eventually, so z
n
cannot be eventually identically zero. If z
n
< 0 eventually, then
z
n
 z
N
< 0, ∀n  N ∈ N.

N+τn−τ
= z
N
+ x
N+τ(n−1)
< · · · < nz
N
+ x
N
.
Taking n → ∞ in the above inequality, we have x
N+τn
< 0, which is a contradiction to x
n
> 0.
(b) We have
∆z
n
= −
r

i=1
α
i
(n)x
n−m
i
< 0,
for n sufficient large. We shall prove that z
n

δ

j
x
N+(j−1)τ
, j = 1, 2, · · · .
On letting j → ∞ in the above inequality, we get x
n
→ ∞ as n → ∞. But
∆z
n
= −
r

i=1
α
i
(n)x
n−m
i
 −M
r

i=1
α
i
(n), (9)
for n sufficient large, where M > 0. Summing (9) from N to n, we obtain
z
n+1

α
i
(n) > 1, (10)
D.C. Huong / VNU Journal of Science, Mat hematics - Physics 24 (2008) 133-143 137
where −1 < δ < 0, ˜m = min
1ir
m
i
and α
i
(n)  0, α
i
(n) > α
i
(n − τ), for n sufficient large,
1  i  r. Then, (3) is oscil latory.
Proof. Assume the contrary and let {x
n
} be an eventually positive solution of (3). Let z
n
= x
n
+δx
n−τ
and w
n
= z
n
+ δz
n−τ

i
,
 −
r

i=1
α
i
(n)x
n−m
i
− δ
r

i=1
α
i
(n)x
n−τ−m
i
,
∆w
n
 −
r

i=1
α
i
(n)(x

> 0 for n sufficient large. On the other hand,
w
n
= z
n
+ δz
n−τ
 (1 + δ)z
n
,
which implies that
z
n−m
i

w
n−m
i
1 + δ
.
Hence, we obtain
∆w
n
 −
r

i=1
α
i
(n)z

solution, which is a contradiction.
Lemma 2. Assume that −1 < δ < 0 and τ > ˜m + 1, where ˜m = min
1ir
m
i
. Then, the maximum
value of f(β) =
β−1
β
˜m+2
(1 + δβ
τ
) on [1, ∞) is f (β
∗
), in which β
∗
∈ (1, (−δ)
−1/τ
) is a unique real
solution of the equation
1 + δβ
τ
+ (β − 1)[δτβ
τ
− ( ˜m + 1)(1 + δβ
τ
)] = 0.
Proof. The equation f
′
(β) = 0 is equivalent to

β→+∞
ϕ(β) = lim
β→+∞
{1 + δβ
τ
+ (β − 1)[δβ
τ
[τ − ( ˜m + 1)] − ( ˜m + 1)]} = −∞.
It implies that, ϕ is a decreasing function, starting from a positive value at β = 1, and hence (12)
has a unique real solution β
∗
∈ [1, ∞). Further, it is easy to see that β
∗
∈ (1, (−δ)
−1/τ
) and
f(β)  0, ∀β ∈ (1, (−δ)
−1/τ
), which implies that f (β
∗
) is the maximum value of f (β) on [1, ∞).
The proof is complete
Theorem 3. Assume that −1 < δ < 0; τ > ˜m + 1; α
i
(n)  0, α
i
(n) > α
i
(n − τ ),
for n suffi ci ent large, 1  i  r, ˜m = min

> 0, ∆z
n
< 0 eventually. On the other hand,
∆w
n
= ∆( z
n
+ δz
n−τ
)  −
r

i=1
α
i
(n)z
n−m
i
 0. (14)
Putting γ
n
=
z
n−1
z
n
, we have γ
n
 1 for n sufficient large. Dividing (14) by z
n

1
γ
n+1
 1 + δ

γ
n−τ+1
· · ·γ
n
− γ
n−τ+2
· · · γ
n

−
r

i=1
α
i
(n)
m
i

ℓ=0
γ
n−ℓ
. (15)
Setting lim inf
n→∞

α
i
(n) · β
m
i
+1
 1 + δβ
τ −1
(β − 1) −
1
β
= (β − 1)[
1
β
+ δβ
τ −1
],
r

i=1
lim inf
n→∞
α
i
(n) 
β − 1
β
˜m+2
(1 + δβ
τ

)
τ −m
∗
(τ − m
∗
− 1)
τ −m
∗
−1
r

i=1
lim inf
n→∞
α
i
(n) > 1, (16)
where α
i
(n)  α
i
(n − τ) for n sufficient large; δ < −1, m
∗
= ma x
1ir
m
i
, τ > m
∗
+ 1 and

+ δz
n−τ
 (1 + δ)z
n−τ
,
which is the same as
z
n−τ

1
δ + 1
w
n
.
Therefore, it follows that
∆w
n
= ∆z
n
+ δ∆z
n−τ
= −
r

i=1
α
i
(n)x
n−m
i

n
 −
r

i=1
α
i
(n)(x
n−m
i
+ δx
n−τ−m
i
),
∆w
n
 −
r

i=1
α
i
(n)z
n−m
i
 0,
so we get
0  ∆w
n
+

, we obtain
γ
n
 1 −
1
δ + 1
r

i=1
α
i
(n)
τ −m
i

ℓ=1
γ
n−m
i
+τ−ℓ
. (17)
140 D.C. Huong / VNU Journal of Science, Mathematics - Physics 24 (200 8) 133-143
Putting β = lim inf
n→∞
γ
n
, we have β  1. Taking lower limit on both sides of (17), we obtain
β  1 −
1
δ + 1

 β
τ −m
∗
, ∀i =
1, r,
−
1
δ + 1
lim inf
n→∞
α
i
(n)β
τ −m
i
 −
1
δ + 1
lim inf
n→∞
α
i
(n)β
τ −m
∗
, ∀i =
1, r.
From (18) we get
−
1

∗
,
so
−
1
δ + 1
(τ − m
∗
)
τ −m
∗
(τ − m
∗
− 1)
τ −m
∗
−1
r

i=1
lim inf
n→∞
α
i
(n)  1,
which contradicts condition (16). Hence, (3) has no eventually positive solution.
Theorem 5. Suppose that
−
1
δ

ℓ=1
[

r
i=1
α
i
(ℓ)] = ∞. Then, (3) is oscilla tory.
Proof. Suppose to the contrary, and let {x
n
} be an eventually positive solution of (3). Put z
n
=
x
n
+ δx
n−τ
. By the case (b) of Lemma 1, we obtain z
n
< 0 and ∆z
n
 0. On the other hand, we
have z
n
> δx
n−τ
or x
n−τ
>
1

z
n+1
z
n
and dividing (20) by z
n
, we obtain
v
n
 1 −
1
δ
r

i=1
α
i
(n)
z
n+τ−m
i
z
n
,
or
v
n
 1 −
1
δ


i=1
lim inf
n→∞
α
i
(n) · β
τ −m
i
.
We can prove
−
1
δ
(τ − m
∗
)
τ −m
∗
(τ − m
∗
− 1)
τ −m
∗
−1
r

i=1
lim inf
n→∞

n
< 0).
Proof. Let {x
n
} be an eventually positive solution of (2). The case {x
n
} is an eventually negative
solution of (2) can be considered similarly.
(a) We have ∆z
n
= −
r

i=1
α
i
(n)F (x
n−m
i
)  0 for all large n. Thus, {z
n
} is eventually
nonincreasing.
(b) Suppose the conclusion does not hold, then since by (a) {z
n
} is nonincreasing, it follows
that eventually either z
n
≡ 0 or z
n

 (−γ)
j
x
n
for all positive integers j. Hence, x
n
→ 0 as n → ∞. It implies that {z
n
} decreases
to zero as n → ∞. This contradicts the fact that z
n
< 0.
Theorem 6. Assume that
∞

ℓ=1
r

i=1
α
i
(ℓ) = ∞, (23)
and there exists a constant η such that
−1 < η  δ  0. (24)
Suppos e further that, if |x|  c th en |F(x)|  c
1
where c and c
1
are positi ve constants. Then, every
nonoscillatory solution of (2) tends to 0 as n → ∞.

n→∞
z
n
= β. Now, suppose that β > 0. By (24), we have z
n
 x
n
. Thus,
there exists an integer n
1
 n
0
∈ N such that
β  z
n−m
i
 x
n−m
i
, ∀n  n
1
, i = 1, · · · , r.
Hence,
∆z
n
= −
r

i=1
α

n
→ −∞. This is a contradiction.
Since lim
n→∞
z
n
= 0, there exists a positive constant A such that 0 < z
n
 A and so, by (24) we
have
x
n
 −ηx
n−τ
+ A. (25)
Assume that {x
n
} is not bounded. Then, there exists a subsequence {n
k
} of N, so that lim
k→∞
x
n
k
= ∞
and x
n
k
= max
n

1
− τ and x
n
k
→ α as k → ∞. Then, from (24), we have
z
n
k
 x
n
k
+ ηx
n
k
−τ
and so
x
n
k
−τ
 −
1
η
(x
n
k
− z
n
k
).

n
} be an eventually positive solution of (2), say x
n
> 0, x
n−τ
> 0 and x
n−m
i
> 0 for
n  n
0
∈ N. By Lemma 3, {z
n
} is eventually positive and nonincreasing, so there exists lim
n→∞
z
n
.
Put lim
n→∞
z
n
= β. Summing the equation (2) from n to ∞ for n  n
0
, we obtain
z
n
= β +
∞



i=1
α
i
(ℓ)F (x
ℓ−m
i
)  z
n
− β < ∞,
which implies that x
n
→ 0 as n → ∞. The proof is similar when {x
n
} is eventually negative.
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