Asymptotic behavior of higher-order neutral difference equations with general arguments
We study the asymptotic behavior of solutions of the higher-order neutral difference equation $$Δm[x(n)+cx(τ(n))]+p(n)x(σ(n))=0,N∍m≥2,n≥0,$$ where $τ (n)$ is a general retarded argument, $σ(n)$ is a general deviated argument, $c ∈ R; (p(n)) n ≥ 0$ is a sequence of real numbers, $∆$ denotes the forwa...
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| author | Chatzarakis, G. E. Khatibzadeh, H. Miliaras, G. N. Stavroulakis, I. P. Чатзаракіс, Г. Е. Хатібзадех, Г. Міліарас, Г. Н. Ставроулакіс, І. П. |
| author_facet | Chatzarakis, G. E. Khatibzadeh, H. Miliaras, G. N. Stavroulakis, I. P. Чатзаракіс, Г. Е. Хатібзадех, Г. Міліарас, Г. Н. Ставроулакіс, І. П. |
| author_sort | Chatzarakis, G. E. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
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| datestamp_date | 2020-03-18T19:15:16Z |
| description | We study the asymptotic behavior of solutions of the higher-order neutral difference equation
$$Δm[x(n)+cx(τ(n))]+p(n)x(σ(n))=0,N∍m≥2,n≥0,$$
where $τ (n)$ is a general retarded argument, $σ(n)$ is a general deviated argument, $c ∈ R; (p(n)) n ≥ 0$ is a sequence of real numbers, $∆$ denotes the forward difference operator $∆x(n) = x(n+1) - x(n)$; and $∆^j$ denotes the jth forward difference operator $∆^j (x(n) = ∆ (∆^{j-1}(x(n)))$ for $j = 2, 3,…,m$. Examples illustrating the results are also given. |
| first_indexed | 2026-03-24T02:23:16Z |
| format | Article |
| fulltext |
UDC 517.9
G. E. Chatzarakis (School Ped. and Technol. Education, Athens, Greece),
H. Khatibzadeh (Univ. Zanjan, Iran),
G. N. Miliaras (Amer. Univ. Athens, Greece),
I. P. Stavroulakis (Univ. Ioannina, Greece)
ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE
EQUATIONS WITH GENERAL ARGUMENTS
АСИМПТОТИЧНА ПОВЕДIНКА НЕЙТРАЛЬНИХ РIЗНИЦЕВИХ РIВНЯНЬ
ВИЩОГО ПОРЯДКУ IЗ ЗАГАЛЬНИМИ АРГУМЕНТАМИ
We study the asymptotic behavior of solutions of the higher-order neutral difference equation
∆m [x(n) + cx(τ(n))] + p(n)x(σ(n)) = 0, N 3 m ≥ 2, n ≥ 0,
where τ(n) is a general retarded argument, σ(n) is a general deviated argument, c ∈ R, (p(n))n≥0 is a sequence of real
numbers, ∆ denotes the forward difference operator ∆x(n) = x(n+1)−x(n), and ∆j denotes the jth forward difference
operator ∆j
(
x(n)
)
= ∆
(
∆j−1
(
x(n)
))
for j = 2, 3, . . . ,m. Examples illustrating the results are also given.
Вивчається асимптотична поведiнка розв’язкiв нейтрального рiзницевого рiвняння вищого порядку
∆m [x(n) + cx(τ(n))] + p(n)x(σ(n)) = 0, N 3 m ≥ 2, n ≥ 0,
де τ(n)— загальний аргумент iз запiзненням, σ(n) — загальний аргумент iз вiдхиленням, c ∈ R, (p(n))n≥0 —
послiдовнiсть дiйсних чисел, ∆ — оператор правої рiзницi, ∆x(n) = x(n+ 1)− x(n), та ∆j — j-й оператор правої
рiзницi, ∆j
(
x(n)
)
= ∆
(
∆j−1
(
x(n)
))
при j = 2, 3, . . . ,m. Наведено також приклади, що iлюструють отриманi
результати.
1. Introduction. Consider the mth-order neutral difference equation of the form
∆m [x(n) + cx(τ(n))] + p(n)x(σ(n)) = 0, N 3 m ≥ 2, n ≥ 0, (E)
where (p(n))n≥0 is a sequence of real numbers, c ∈ R, (τ(n))n≥0 is an increasing sequence of
integers which satisfies
τ(n) ≤ n− 1 ∀n ≥ 0 and lim
n→∞
τ(n) = +∞, (1.1)
(σ(n))n≥0 is an increasing sequence of integers such that
σ(n) ≤ n− 1 ∀n ≥ 0 and lim
n→∞
σ(n) = +∞, (1.2a)
or
σ(n) ≥ n+ 1 ∀n ≥ 0, (1.2b)
∆ denotes the forward difference operator ∆x(n) = x(n+1)−x(n), and ∆j denotes the jth forward
difference operator ∆j
(
x(n)
)
= ∆
(
∆j−1(x(n)
))
for j = 2, 3, . . . ,m.
Define
k = −min
n≥0
{τ(n), σ(n)} if σ(n) is a retarded argument.
(Clearly, k is a positive integer.)
c© G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS, 2013
430 ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 431
By a solution of (E), we mean a sequence of real numbers (x(n))n≥−k which satisfies (E)
for all n ≥ 0. It is clear that, for each choice of real numbers c−k, c−k+1, . . . , c−1, c0, there
exists a unique solution (x(n))n≥−k of (E) which satisfies the initial conditions x(−k) = c−k,
x(−k + 1) = c−k+1, . . . , x(−1) = c−1, x(0) = c0.
If σ(n) is an advanced argument, then:
By a solution of (E), we mean a sequence of real numbers (x(n))n≥0 which satisfies (E) for all
n ≥ 0.
A solution (x(n))n≥−k (or (x(n))n≥0) of (E) is called oscillatory, if the terms x(n) of the
sequence are neither eventually positive nor eventually negative. Otherwise, the solution is said to be
nonoscillatory.
In the last few decades, the asymptotic and oscillatory behavior of neutral difference equations has
been extensively studied. See, for example, [2 – 8, 10, 11, 13 – 24] and the references cited therein.
Most of these papers concern the special case where the delay (n− τ(n))n≥0 is constant, while
a small number of these papers are dealing with the general case of Eq. (E), in which the delay
(n− τ(n))n≥0 is variable. For the general theory of difference equations the reader is referred to the
monographs [1, 9, 12].
The objective in this paper is to study the asymptotic behavior of the solutions of Eq. (E). We
proceed to study the asymptotic behavior of the solutions of Eq. (E) by considering various cases on
the sign of the coefficients p(n). We examine two cases, according to whether the coefficients p(n)
are all non-negative (Case 1) or are all non-positive (Case 2). Examples illustrating the results are
also given.
2. Some preliminaries. Throughout this paper, we are going to use the following notation:
τ ◦ τ = τ2 τ ◦ τ ◦ τ = τ3, and so on. (2.1)
Let the domain of τ be the set D(τ) = Nn∗ = {n∗, n∗ + 1, n∗ + 2, . . .} , where n∗ is the smallest
natural number that τ is defined with. Then for every n > n∗ there exists a natural number m(n)
such that
x(τm(n)(n)) = x(τ(n∗)) and lim
n→∞
m(n) = +∞, (2.2)
since (m(n)) is increasing and unbounded function of n. Clearly, n∗ = τm(n)−1(n).
The following lemmas provide us with some useful tools, for establishing the main results:
Lemma 2.1. Assume that
(
z(n)
)
is a sequence of real numbers and m ∈ N. Then the following
statements hold:
(i) If
lim
n→∞
∆mz(n) > 0, (2.3)
then
(
z(n)
)
tends to +∞.
(ii) If
lim
n→∞
∆mz(n) < 0, (2.4)
then
(
z(n)
)
tends to −∞.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
432 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
(iii) If
lim
n→∞
∆mz(n) = 0 and ∆m+1z(n) ≥ 0 ∀n (2.5a)
or
lim
n→∞
∆mz(n) = 0 and ∆m+1z(n) ≤ 0 ∀n, (2.5b)
then the sequence
(
z(n)
)
is monotone and therefore its limit exists.
Proof. (i) Suppose that m = 1. Then, if limn→∞∆z(n) > 0, clearly
(
z(n)
)
is eventually strictly
increasing, and therefore limn→∞ z(n) = +∞.
Assume that m > 1 and (2.3) holds. Then
(
∆m−1z(n)
)
is eventually strictly increasing. By
previous case (m = 1), we have limn→∞∆m−1z(n) = +∞.
Repeating this argument m − 1 times, we obtain limn→∞ z(n) = +∞. The proof of part (i) of
the lemma is complete.
(ii) An obvious consequence of part (i) by taking −z(n) instead of z(n).
(iii) Suppose that m = 1 and (2.5a) holds. Then (∆z(n)) is nondecreasing, and consequently
∆z(n) ≤ 0, since limn→∞∆z(n) = 0. Therefore
(
z(n)
)
is nonincreasing, which guarantees that its
limit exists.
Assume that m > 1 and (2.5a) holds. Then (∆mz(n)) is nondecreasing, and consequently
∆mz(n) ≤ 0, since limn→∞∆mz(n) = 0. Therefore
(
∆m−1z(n)
)
is nonincreasing, which guaran-
tees that its limit exists.
If limn→∞∆m−1z(n) 6= 0, by parts (i) and (ii) we have limn→∞ z(n) = ±∞.
If limn→∞∆m−1z(n) = 0, then ∆m−1z(n) ≥ 0 since
(
∆m−1z(n)
)
is nonincreasing. Therefore(
∆m−2z(n)
)
is nondecreasing, which guarantees that its limit exists.
Applying this procedure m − 2 times, we conclude that limn→∞ z(n) = ±∞ or
(
z(n)
)
is
monotone and therefore its limit exists.
In the case where (2.5b) holds, the proof is similar. The proof of part (iii) of the lemma is
complete.
Lemma 2.1 is proved.
Lemma 2.2. Assume that
(
z(n)
)
is a sequence of real numbers and N 3 m ≥ 2. Then the
following statements hold:
(i) If m is even and ∆mz(n) ≤ 0, then
(
z(n)
)
tends to ±∞ or it is nondecreasing.
(ii) If m is even and ∆mz(n) ≥ 0, then
(
z(n)
)
tends to ±∞ or it is nonincreasing.
(iii) If m is odd and ∆mz(n) ≤ 0, then
(
z(n)
)
tends to ±∞ or it is nonincreasing.
(iv) If m is odd and ∆mz(n) ≥ 0, then
(
z(n)
)
tends to ±∞ or it is nondecreasing.
Proof. (i) First we will study the case where m = 2.
Assume that ∆2z(n) ≤ 0. Then (∆z(n)) is nonincreasing, and consequently limn→∞∆z(n) =
= −∞ or limn→∞∆z(n) = A ∈ R.
If limn→∞∆z(n) = −∞, then (z(n)) is eventually nonincreasing. By part (ii) of Lemma 2.1,
we have limn→∞ z(n) = −∞.
If limn→∞∆z(n) = A < 0, then
(
z(n)
)
is eventually nonincreasing. By part (ii) of Lemma 2.1,
we have limn→∞ z(n) = −∞.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 433
If limn→∞∆z(n) = 0, then ∆z(n) ≥ 0 since (∆z(n)) is nonincreasing. Therefore (z(n)) is
nondecreasing.
If limn→∞∆z(n) = A > 0, then
(
z(n)
)
is eventually increasing. By part (i) of Lemma 2.1, we
have limn→∞ z(n) = +∞.
Now we will study the case where m is even and m > 2.
By parts (i) and (ii) of Lemma 2.1 we have that, if limn→∞∆m−1z(n) 6= 0, then limn→∞ z(n) =
= ±∞.
Suppose that limn→∞∆m−1z(n) = 0. Since ∆mz(n) ≤ 0, then ∆m−1z(n) ≥ 0. This guarantees
that
(
∆m−2z(n)
)
is nondecreasing.
If limn→∞∆m−2z(n) 6= 0, then in view of parts (i) and (ii) of Lemma 2.1, we have limn→∞ z(n) =
= ±∞.
If limn→∞∆m−2z(n) = 0, then ∆m−2z(n) ≤ 0 since
(
∆m−2z(n)
)
is nondecreasing.
Applying this procedure
m− 2
2
-times, we conclude that limn→∞ z(n) = ±∞ or ∆2z(n) ≤ 0.
This means that
(
z(n)
)
tends to ±∞ or it is nondecreasing. The proof of part (i) of the lemma is
complete.
(ii) An obvious consequence of part (i) by taking −z(n) instead of z(n).
(iii) First we will study the case where m = 3.
Assume that ∆3z(n) ≤ 0. Then
(
∆2z(n)
)
is nonincreasing, and consequently limn→∞∆2z(n) =
= −∞ or limn→∞∆2z(n) = A ∈ R.
If limn→∞∆2z(n) = −∞, then by part (ii) of Lemma 2.1 we have limn→∞ z(n) = −∞.
If limn→∞∆2z(n) = A < 0, then by part (ii) of Lemma 2.1 we have limn→∞ z(n) = −∞.
If limn→∞∆2z(n) = 0, then ∆2z(n) ≥ 0 since
(
∆2z(n)
)
is nonincreasing. By part (ii), we
conclude that
(
z(n)
)
tends to +∞ or it is nonincreasing.
If limn→∞∆2z(n) = A > 0, then by part (i) of Lemma 2.1 we have limn→∞ z(n) = +∞.
Now we will study the case where m is odd and m > 3.
By parts (i) and (ii) of Lemma 2.1 we have that, if limn→∞∆m−1z(n) 6= 0, then limn→∞ z(n) =
= ±∞.
Suppose that limn→∞∆m−1z(n) = 0. Since ∆mz(n) ≤ 0, then ∆m−1z(n) ≥ 0. This guarantees
that
(
∆m−2z(n)
)
is nondecreasing.
If limn→∞∆m−2z(n) 6= 0, then in view of parts (i) and (ii) of Lemma 2.1 we have
limn→∞ z(n) = ±∞.
If limn→∞∆m−2z(n) = 0, then ∆m−2z(n) ≤ 0 since
(
∆m−2z(n)
)
is nondecreasing.
Applying this procedure
m− 3
2
-times, we conlude that limn→∞ z(n) = ±∞ or ∆3z(n) ≤ 0.
This means that
(
z(n)
)
tends to ±∞ or it is nondecreasing. The proof of part (iii) of the lemma is
complete.
(iv) An obvious consequence of part (iii) by taking −z(n) instead of z(n).
Lemma 2.2 is proved.
Lemma 2.3. Assume that
(
x(n)
)
is a positive solution of (E). Set
z(n) := x(n) + cx(τ(n)), (2.6)
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
434 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
where (τ(n))n≥0 is an increasing sequence of integers such that (1.1) holds and c ∈ R. Then the
following statements hold:
(i) If limn→∞ z(n) = −∞ and in addition:
(ia) c < −1, then
(
x(n)
)
is unbounded;
(ib) c ≥ −1, then (E) has no positive solution.
(ii) If limn→∞ z(n) = A ∈ R− and in addition:
(iia) c < −1, then lim inf x(n) ≥ A
1 + c
and if
(
x(n)
)
has a real accumulation point greater
than
A
1 + c
, it will have infinitely many real accumulation points including
A
1 + c
;
(iib) c ≥ −1, then (E) has no positive solution.
(iii) If limn→∞ z(n) = 0 and in addition:
(iiia) c < −1, then
(
x(n)
)
tends to zero or tends to infinity or
(
x(n)
)
has infenitely many
accumulation points and lim inf x(n) = 0; furthermore, the condition z(n) > 0 guarantees that(
x(n)
)
tends to infinity;
(iiib) c = −1, then
(
x(n)
)
and (x(τ(n))) have the same set of accumulation points; furthermore,
if z(n) < 0 then
(
x(n)
)
is bounded and, if z(n) > 0 then lim inf x(n) > 0;
(iiic) c > −1, then
(
x(n)
)
tends to zero.
(iv) If limn→∞ z(n) = A ∈ R+ and in addition:
(iva) c ≤ −1, then
(
x(n)
)
tends to infinity;
(ivb) −1 < c ≤ 0, then limn→∞ x(n) =
A
1 + c
;
(ivc) c > 0, then
(
x(n)
)
is bounded.
(v) If limn→∞ z(n) = +∞ and in addition:
(va) c ≤ 0, then
(
x(n)
)
tends to infinity;
(vb) c > 0, then
(
x(n)
)
is unbounded.
Proof. (i) Assume that limn→∞ z(n) = −∞ and c < −1. By (2.6) we have
lim
n→∞
[x(n) + cx(τ(n))] = −∞,
which guarantees that (x(τ(n))) tends to infinity, and therefore
(
x(n)
)
is unbounded.
Assume that limn→∞ z(n) = −∞ and c ≥ −1.
If c = −1, we have limn→∞ [x(n)− x(τ(n))] = −∞, which guarantees that
(
x(n)
)
is un-
bounded. On the other hand x(n)− x(τ(n)) < 0 eventually, or
x(n) < x(τ(n)) < x(τ2(n)) < . . . < x(τm(n`)(n)) = x(τ(nλ)),
where, in view of (2.2), nλ = τm(n`)−1(n). This means that
(
x(n)
)
has an upper bound, which
contradicts
(
x(n)
)
is unbounded. Therefore, Eq. (E) has no positive solution.
If −1 < c < 0, we have limn→∞ [x(n) + cx(τ(n))] = −∞, which guarantees that
(
x(n)
)
is
unbounded. On the other hand x(n) + cx(τ(n)) < 0 eventually, or
x(n) < −cx(τ(n)) < (−c)2x(τ2(n)) < . . . < (−c)m(nµ)x(τ(ns))→ 0 as n→∞,
where, in view of (2.2), ns = τm(nµ)−1(n). Thus limn→∞ x(n) = 0 which contradicts
(
x(n)
)
is
unbounded. Therefore, Eq. (E) has no positive solution.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 435
If c ≥ 0, we have limn→∞ [x(n) + cx(τ(n))] = −∞, which contradicts z(n) ≥ x(n) > 0.
Therefore, Eq. (E) has no positive solution. The proof of part (i) of the lemma is complete.
(ii) Assume that limn→∞ z(n) = A ∈ R− and c < −1. Then for every ε > 0 with 0 < ε < −A,
there exists n1 ≥ n∗ such that z(n) < A+ ε ∀n ≥ n1, or
x(n) < −cx(τ(n)) +A+ ε <
< −c
[
−cx(τ2(n)) +A+ ε
]
+A+ ε =
= (−c)2 x(τ2(n))− c (A+ ε) +A+ ε < . . .
. . . < (−c)m(nλ) x(τ(nt))−
A+ ε
1 + c
[
(−c)m(nλ) − 1
]
=
= (−c)m(nλ)
[
x(τ(nt))−
A+ ε
1 + c
]
+
A+ ε
1 + c
∀n ≥ n1
where, in view of (2.2), nt = τm(nλ)−1(n). Since x(n) > 0, clearly x(τ(nt)) ≥
A+ ε
1 + c
, or eventually
x(τ(n)) ≥ A+ ε
1 + c
.
On the other hand, since limn→∞ z(n) = A, we have z(n) > A− ε, or
x(n) > −cx(τ(n)) +A− ε > −cA+ ε
1 + c
+A− ε =
A− (1 + 2c)ε
1 + c
.
The last inequality guarantees that lim inf x(n) ≥ A
1 + c
> 0.
It is clear that
A
1 + c
could be an accumulation point of
(
x(n)
)
. Let L >
A
1 + c
be an accumula-
tion point of
(
x(n)
)
. Then there exists a subsequence (x(θ(n))) of
(
x(n)
)
such that limn→∞ x(θ(n)) =
= L. Taking into account that limn→∞ z(θ(n)) = A, we obtain limn→∞ x(τ(θ(n))) =
A
c
+
L
−c
.
In view of this, we have
lim
n→∞
[
x(τ(θ(n))) + cx(τ2(θ(n)))
]
= A,
or
lim
n→∞
x(τ2(θ(n))) =
A
c
+
A
(−c)2
+
L
(−c)2
.
Following the above procedure, we can construct a sequence (bn)n≥1 of accumulation points with
bn =
A
c
+
A
(−c)2
+
A
(−c)3
+ . . .+
A
(−c)n
+
L
(−c)n
, n ≥ 1.
Notice that this sequence of accumulation points converges to
A
1 + c
.
Assume that limn→∞ z(n) = A ∈ R− and c ≥ −1.
If c = −1, we have limn→∞ [x(n)− x(τ(n))] = A. Then for every ε > 0 with 0 < ε < −A
there exists n2 > n∗ such that x(n)− x(τ(n)) < A+ ε ∀n ≥ n2, or
x(n) < x(τ(n)) +A+ ε < x(τ2(n)) + 2 (A+ ε) < . . .
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
436 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
. . . < x(τm(nρ)(n)) +m(nρ) (A+ ε) =
= x(τ(nη)) +m(nρ) (A+ ε) ∀n ≥ n2,
where, in view of (2.2), nη = τm(nρ)−1(n). This inequality, for sufficiently large n guarantees that
x(n) < 0 which contradicts x(n) > 0. Therefore, Eq. (E) has no positive solution.
If −1 < c < 0, we have limn→∞ [x(n) + cx(τ(n))] = A < 0, which, for sufficiently large n,
means that x(n) + cx(τ(n)) < 0, or
x(n) < −cx(τ(n)) < (−c)2 x(τ2(n)) < . . .
. . . < (−c)m(n`) x(τm(n`)(n))→ 0 as n→∞.
Thus limn→∞ x(n) = 0, and consequently limn→∞ z(n) = 0 which contradicts limn→∞ z(n) =
= A < 0. Therefore, Eq. (E) has no positive solution.
If c ≥ 0, clearly z(n) > 0 which contradicts limn→∞ z(n) = A < 0. Therefore, Eq. (E) has no
positive solution. The proof of part (ii) of the lemma is complete.
(iii) Assume that limn→∞ z(n) = 0.
If c < −1, we have limn→∞ [x(n) + cx(τ(n)] = 0, which means that limn→∞ x(n) = 0 or
limn→∞ x(n) = +∞ or (x(n)) has infinitely many accumulation points. Indeed, in the case where
(x(n)) does not tend to zero or to infinity, let L0 > 0 an accumulation point of (x(n)). Then there
exists a subsequence (x(θ(n))) of (x(n)) such that limn→∞ x(θ(n)) = L0. Taking into account that
limn→∞ z(θ(n)) = 0, we obtain limn→∞ x(τ(θ(n))) =
L0
−c
.
In view of this, we have
lim
n→∞
[
x(τ(θ(n))) + cx(τ2(θ(n)))
]
= 0,
or
lim
n→∞
x(τ2(θ(n))) =
L0
(−c)2
.
Based on the above procedure, we can construct a sequence (λn)n≥1 of accumulation points with
λn =
L0
(−c)n
, n ≥ 1.
Notice that this sequence of accumulation points converges to zero, and therefore lim inf x(n) = 0.
Furthermore, the condition z(n) > 0 guarantees that
(
x(n)
)
tends to infinity. Indeed, x(n) +
+ cx(τ(n)) > 0, or
x(n) > −cx(τ(n)) > . . . > (−c)m(n)x(τ(n∗))→ +∞ as n→∞,
which means that
(
x(n)
)
tends to infinity.
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ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 437
If c = −1 then limn→∞ [x(n)− x(τ(n))] = 0, which means that
(
x(n)
)
and (x(τ(n))) have the
same set of accumulation points.
Furthermore, if z(n) < 0, then x(n)− x(τ(n)) < 0, or
x(n) < x(τ(n)) < . . . < x(τ(n∗))
which means that
(
x(n)
)
is bounded.
If z(n) > 0, then x(n)− x(τ(n)) > 0, or
x(n) > x(τ(n)) > . . . > x(τ(n∗))
which means that lim inf x(n) > 0.
Suppose that c > −1.
If −1 < c < 0, then for every ε > 0 there exists n3 > n∗ such that x(n) + cx(τ(n)) < ε
∀n ≥ n3. Thus
x(n) < −cx(τ(n)) + ε < −c
[
−cx(τ2(n)) + ε
]
+ ε < . . .
. . . < (−c)m(n`) x(τm(n`)(n)) + ε− cε+ . . .+ (−c)m(n`)−1 ε ∀n ≥ n3.
As n→∞, clearly m(n`)→∞, and therefore
lim
n→∞
x(n) ≤ lim
m(n`)→∞
[
ε− cε+ . . .+ (−c)m(n`)−1 ε
]
=
ε
1 + c
.
Since ε is an arbitrarily small, real positive number and as in addition, x(n) > 0, the last inequality
guarantees that limn→∞ x(n) = 0.
If c ≥ 0, clearly z(n) > 0. Taking into account that limn→∞ z(n) = 0, it is obvious that
limn→∞ x(n) = 0. The proof of part (iii) of the lemma is complete.
(iv) Assume that limn→∞ z(n) = A ∈ R+.
If c < −1, then limn→∞ [x(n) + cx(τ(n))] = A > 0, which means that eventually x(n) +
+ cx(τ(n)) > 0. Thus
x(n) > −cx(τ(n)) > (−c)2 x(τ2(n)) . . . > (−c)m(nµ) x(τ(ns))→ +∞ as n→∞
which guarantees that (x(n)) tends to infinity.
If c = −1, then limn→∞ [x(n)− x(τ(n))] = A > 0. Thus, for every ε > 0 with 0 < ε < A,
there exists n4 > n∗ such that x(n)− x(τ(n)) > A− ε ∀n ≥ n4, or
x(n) > x(τ(n)) +A− ε > x(τ2(n)) + 2 (A− ε) > . . .
. . . > x(τm(nµ)(n)) +m(nµ) (A− ε) =
= x(τ(ns)) +m(nµ) (A− ε)→ +∞ as n→∞,
which means that (x(n)) tends to infinity.
Suppose that c > −1.
If −1 < c < 0, then for every ε > 0, there exists n5 > n∗ such that |z(n)−A| < ε ∀n ≥ n5.
Thus
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438 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
−cx(τ(n)) +A− ε < x(n) < −cx(τ(n)) +A+ ε
or
−c
[
−cx(τ2(n)) +A− ε
]
+A− ε < x(n) < −c
[
−cx(τ2(n)) +A+ ε
]
+A+ ε.
Based on the above procedure we get
x(n) < (−c)m(nµ) x(τ(ns))−
A+ ε
1 + c
[
(−c)m(nµ) − 1
]
and
x(n) > (−c)m(nµ) x(τ(ns))−
A− ε
1 + c
[
(−c)m(nµ) − 1
]
or
x(n) < (−c)m(nµ)
[
x(τ(ns))−
A+ ε
1 + c
]
+
A+ ε
1 + c
and
x(n) > (−c)m(nµ)
[
x(τ(ns))−
A− ε
1 + c
]
+
A− ε
1 + c
.
Therefore as n→∞ (clearly m(nµ)→∞) we obtain
A− ε
1 + c
≤ lim inf x(n) ≤ lim supx(n) ≤ A+ ε
1 + c
.
Since ε is an arbitrary real positive number, the last inequality gives limn→∞ x(n) =
A
1 + c
.
If c = 0, then z(n) = x(n) and therefore limn→∞ x(n) = A.
If c > 0, then limn→∞ [x(n) + cx(τ(n))] = A > 0, which guarantees that (x(n)) is bounded.
The proof of part (iv) of the lemma is complete.
(v) Assume that limn→∞ z(n) = +∞.
If c ≤ 0, then limn→∞ [x(n) + cx(τ(n))] = +∞, which guarantees that
(
x(n)
)
tends to infinity.
If c > 0, then limn→∞ [x(n) + cx(τ(n))] = +∞, which guarantees that
(
x(n)
)
is unbounded.
The proof of part (v) of the lemma is complete.
Lemma 2.3 is proved.
3. Main results. Throughout this section, we are going to use the following remarks:
Remark 3.1. Assume that the coefficients p(i) are always nonnegative or nonpositive,∑∞
i=0
p(i) = ±∞ and lim inf x(n) > 0. Then
∑∞
i=0
p(i)x(σ(i)) = ±∞, respectively.
Remark 3.2. Assume that the coefficients p(i) are always nonnegative or nonpositive,∑∞
i=0
p(i) = ±∞ and
∑∞
i=0
p(i)x(σ(i)) ∈ R. Then lim inf x(n) = 0.
The asymptotic behavior of the solutions of the neutral type difference equation (E) is described
in the following two cases:
Case 1. p(n) ≥ 0.
Theorem 3.1. Assume that p(n) ≥ 0 ∀n ≥ 0 and
∑∞
i=0
p(i) = +∞. Then for Eq. (E) the
following statements hold:
(i) If c < −1, then every nonoscillatory solution
(
x(n)
)
:
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ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 439
(ia) has no real non-zero limit, if m is even;
(ib) is unbounded, if m is odd.
(ii) If c = −1, then:
(iia) every nonoscillatory solution
(
x(n)
)
is bounded and lim inf x(n) = 0, if m is even;
(iib) every non-zero solution
(
x(n)
)
oscillates, if m is odd.
(iii) If −1 < c < 0, then every nonoscillatory solution
(
x(n)
)
tends to zero.
(iv) If c = 0 then:
(iva) every non-zero solution
(
x(n)
)
oscillates, if m is even;
(ivb) every nonoscillatory solution
(
x(n)
)
tends to zero, if m is odd.
(v) If 0 < c < 1, then every nonoscillatory solution
(
x(n)
)
:
(va) is unbounded but lim inf x(n) = 0, if m is even;
(vb) tends to zero or it is unbounded but lim inf x(n) = 0, if m is odd.
(vi) If c ≥ 1, then every nonoscillatory solution
(
x(n)
)
:
(via) cannot tend to zero but lim inf x(n) = 0, if m is even;
(vib) tends to zero or lim inf x(n) = 0, if m is odd.
Proof. Assume that a solution (x(n))n≥−k of (E) is nonoscillatory. Then it is either eventually
positive or eventually negative. As (−x(n))n≥−k is also a solution of (E), we may restrict ourselves
only to the case where x(n) > 0 for all large n. Let n0 ≥ −k be an integer such that x(n) > 0 for
all n ≥ n0 ≥ n∗. Then, there exists n1 ≥ n0 such that x(τ(n)) > 0, x(σ(n)) > 0 ∀n ≥ n1.
In view of (2.6), Eq.(E) becomes ∆mz(n) + p(n)x(σ(n)) = 0, or
∆mz(n) = −p(n)x(σ(n)). (3.1)
Therefore, for sufficiently large n and since p(n) ≥ 0, we have ∆mz(n) ≤ 0.
Summing up (3.1) from n1 to n, n ≥ n1 we obtain
∆m−1z(n+ 1) = ∆m−1z(n1)−
n∑
i=n1
p(i)x(σ(i)). (3.2)
(i) c < −1.
Assume that m is even. Since ∆mz(n) ≤ 0, by part (i) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing.
If limn→∞ z(n) = −∞, then in view of part (ia) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. Then
∑∞
i=n1
p(i)x(σ(i)) = +∞. By (3.2),
(
∆m−1z(n+ 1)
)
tends to −∞, which, in view
of part (ii) of Lemma 2.1, guarantees that
(
z(n)
)
tends to −∞. This contradicts our assumption.
Therefore limn→∞ z(n) = +∞ is false.
If
(
z(n)
)
is nondecreasing, clearly its limit exists.
Suppose that limn→∞ z(n) = A < 0. Then in view of part (iia) of Lemma 2.3 we have that
lim inf x(n) ≥ A
1 + c
. Therefore for every ε > 0 with ε <
A
1 + c
, there exists n2 such that
x(n) >
A
1 + c
− ε ∀n ≥ n2. (3.3)
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440 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
Thus, for every n3 with σ(n3) ≥ n2, by (3.2) and (3.3) we obtain
∆m−1z(n+ 1) < ∆m−1z(n3)−
(
A
1 + c
− ε
) n∑
i=n3
p(i)→ −∞ as n→∞,
which guarantees that limn→∞∆m−1z(n+ 1) = −∞. As in previous case, we conclude that (z(n))
tends to −∞, which contradicts
(
z(n)
)
is nondecreasing. Therefore limn→∞ z(n) = A < 0 is false.
Suppose that limn→∞ z(n) = 0. Then in view of part (iiia) of Lemma 2.3 we have that
(
x(n)
)
either tends to zero or
(
x(n)
)
has infenitely many accumulation points and lim inf x(n) = 0 or(
x(n)
)
tends to infinity. But, if limn→∞ x(n) = +∞, then as in previous case we have that
limn→∞ z(n) = −∞ which contradicts our assumption.
Suppose that limn→∞ z(n) = A > 0. Since (z(n)) is nondecreasing, we have z(n) > 0 eventu-
ally. Therefore
x(n) > −cx(τ(n)) > . . . > (−c)m(nµ) x(τ(ns))→∞ as n→∞, (3.4)
which guarantees that
(
x(n)
)
tends to infinity. As in previous case, we have limn→∞∆m−1z(n +
+ 1) = −∞, which means that
(
z(n)
)
tends to −∞. This contradicts our assumption. Therefore
limn→∞ z(n) = A > 0 is false.
In other words,
(
x(n)
)
cannot have a non-zero real limit. Indeed, assume that limn→∞ x(n) =
= ` > 0. Then limn→∞ z(n) = (1 + c)` < 0, which contradicts ”limn→∞ z(n) = A < 0 is
false”.
Assume that m is odd. Since ∆mz(n) ≤ 0, by part (iii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing.
If limn→∞ z(n) = −∞, then in view of part (ia) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. Then
∑∞
i=n1
p(i)x(σ(i)) = +∞. By (3.2),
(
∆m−1z(n+ 1)
)
tends to −∞, and therefore
(z(n)) tends to −∞, which contradicts our assumption. Therefore limn→∞ z(n) = +∞ is false.
If
(
z(n)
)
is nonincreasing, clearly its limit exists.
Suppose that limn→∞ z(n) = A < 0. As in previous case, this is false.
Suppose that limn→∞ z(n) = A ≥ 0. Since (z(n)) is nonincreasing, we have z(n) ≥ 0 even-
tually. Therefore (3.4) holds, and consequently
(
x(n)
)
tends to infinity. Thus, as in previous case,
we are led to a contradiction. Therefore limn→∞ z(n) = A ≥ 0 is false. The proof of part (i) of the
theorem is complete.
(ii) c = −1.
Assume that m is even. Since ∆mz(n) ≤ 0, by part (i) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
Suppose that limn→∞ z(n) = A > 0. Since (z(n)) is nondecreasing, we have z(n) > 0 eventu-
ally. Therefore
x(n) > x(τ(n)) > . . . > x(τ(ns)), (3.5)
which guarantees that
∑n
i=n1
p(i)x(σ(i)) = +∞.
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ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 441
Hence, by (3.2) we have limn→∞∆m−1z(n+ 1) = −∞, which as in part (i) gives
(
z(n)
)
tends
to −∞. This leads to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = 0, then z(n) ≤ 0 since (z(n)) is nondecreasing. Then in view of part
(iiib) of Lemma 2.3 we have that
(
x(n)
)
is bounded. Furthermore, in view of Remark 3.2 we
have lim inf x(n) = 0. Indeed, if lim inf x(n) > 0 then, by Remark 3.1, we conclude that∑n
i=n1
p(i)x(σ(i)) = +∞. This guarantees that
(
z(n)
)
tends to −∞, which contradicts our as-
sumption.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. This guarantees that limn→∞∆m−1z(n + 1) = −∞, which as in part (i) gives that
(
z(n)
)
tends to −∞. This leads a contradiction. Therefore limn→∞ z(n) = +∞ is false.
Assume that m is odd. Since ∆mz(n) ≤ 0, by part (iii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = A ≥ 0, then z(n) > 0 eventually since
(
z(n)
)
is nonincreasing. Thus
(3.5) holds and consequently limn→∞ z(n) = −∞. This contradicts our assumption. Therefore
limn→∞ z(n) = A ≥ 0 is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. This guarantees that limn→∞∆m−1z(n + 1) = −∞, which as in part (i) gives that
(
z(n)
)
tends to −∞. This leads to a contradiction. Therefore limn→∞ z(n) = +∞ is false.
Consequently,
(
x(n)
)
oscillates. The proof of part (ii) of the theorem is complete.
(iii) −1 < c < 0.
Assume that m is even. Since ∆mz(n) ≤ 0, by part (i) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
=
A
1 + c
> 0. This guarantees that
∑n
i=n1
p(i)x(σ(i)) = +∞. Thus, as in previous parts, we are
led to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = 0, then by part (iiic) we have that
(
x(n)
)
tends to zero.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. Thus, as in previous parts, we are led to a contradiction. Therefore limn→∞ z(n) = +∞ is
false.
Assume that m is odd. Since ∆mz(n) ≤ 0, by part (iii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) of Lemma 2.3 we have that
(
x(n)
)
tends to zero.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
=
A
1 + c
> 0. As in case where m is even, we are led to a contradiction. Therefore limn→∞ z(n) =
= A > 0 is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. Thus, as in previous cases, we are led to a contradiction. Therefore limn→∞ z(n) = +∞ is
false. The proof of part (iii) of the theorem is complete.
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442 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
(iv) c = 0.
Assume that m is even. Since ∆mz(n) ≤ 0, by part (i) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
= A > 0. This guarantees that
∑n
i=n1
p(i)x(σ(i)) = +∞. Hence, as in previous cases, we are led
to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = 0, then
(
z(n)
)
≤ 0 since
(
z(n)
)
is nondecreasing. This contradicts z(n) =
= x(n) > 0. Therefore limn→∞ z(n) = 0 is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. This guarantees that
∑n
i=n1
p(i)x(σ(i)) = +∞. Hence, as in previous cases, we are led to
a contradiction. Therefore limn→∞ z(n) = +∞ is false. Consequently,
(
x(n)
)
oscillates.
Assume that m is odd. Since ∆mz(n) ≤ 0, by part (iii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) of Lemma 2.3 we have that
(
x(n)
)
tends to zero.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
= A > 0. As in case where m is even, we are led to a contradiction. Therefore limn→∞ z(n) = A >
> 0 is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. This guarantees that
∑+∞
i=n1
p(i)x(σ(i)) = +∞. Hence, as in previous cases, we are led
to a contradiction. Therefore limn→∞ z(n) = +∞ is false. The proof of part (iv) of the theorem is
complete.
(v) 0 < c < 1.
Assume that m is even. Since ∆mz(n) ≤ 0, by part (i) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then
(
z(n)
)
≤ 0 since
(
z(n)
)
is nondecreasing. This contradicts z(n) =
= x(n) + cx(τ(n)) > 0. Therefore limn→∞ z(n) = 0 is false.
Suppose that limn→∞ z(n) = A > 0. By part (ivc) of Lemma 2.3 we have that
(
x(n)
)
is bounded. This guarantees that
∑+∞
i=n1
p(i)x(σ(i)) < +∞. Indeed, if
∑+∞
i=n1
p(i)x(σ(i)) =
= +∞, then by (3.2) and part (ii) of Lemma 2.1 we have that limn→∞ z(n) = −∞ which con-
tradicts limn→∞ z(n) = A > 0. By Remark 3.2 we have that lim inf x(n) = 0. Then there ex-
ists a subsequence (x(θ(n))) of
(
x(n)
)
such that limn→∞ x(θ(n)) = 0. Taking into account that
limn→∞ z(θ(n)) = A, we obtain limn→∞ x(τ(θ(n))) =
A
c
.
In view of this, we have
lim
n→∞
[
x(τ(θ(n))) + cx(τ2(θ(n)))
]
= A,
or
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ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 443
lim
n→∞
x(τ2(θ(n))) =
A− A
c
c
< 0,
which is impossible. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. If
∑+∞
i=n1
p(i)x(σ(i)) = +∞, then by (3.2)
(
∆m−1z(n+ 1)
)
tends to −∞. By part (ii)
of Lemma 2.1 we conclude that
(
z(n)
)
tends to −∞, which contradicts limn→∞ z(n) = +∞.
Therefore
∑+∞
i=n1
p(i)x(σ(i)) < +∞. By Remark 3.2, we conclude that lim inf x(n) = 0.
Assume that m is odd. Since ∆mz(n) ≤ 0, by part (iii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) we have that
(
x(n)
)
tends to zero.
Suppose that limn→∞ z(n) = A > 0. By part (ivc) of Lemma 2.3 we have that
(
x(n)
)
is
bounded. Thus, as in case where m is even, we are led to a contradiction. Therefore limn→∞ z(n) =
= A > 0 is false.
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. Thus, as in case where m is even, we have lim inf x(n) = 0. The proof of part (v) of the
theorem is complete.
(vi) c ≥ 1.
Assume that m is even. Since ∆mz(n) ≤ 0, by part (i) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then
(
z(n)
)
≤ 0 since
(
z(n)
)
is nondecreasing. This contradicts z(n) =
= x(n) + cx(τ(n)) > 0. Therefore limn→∞ z(n) = 0 is false.
Suppose that limn→∞ z(n) = A > 0. By part (ivc) of Lemma 2.3 we have that
(
x(n)
)
is
bounded. Therefore, as in previous part, we have that
∑+∞
i=n1
p(i)x(σ(i)) < +∞. By Remark 3.2
we have that lim inf x(n) = 0.
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. If
∑+∞
i=n1
p(i)x(σ(i)) = +∞, then by (3.2)
(
∆m−1z(n+ 1)
)
tends to −∞. By part (ii) of
Lemma 2.1 we conclude that
(
z(n)
)
tends to −∞, which contradicts limn→∞ z(n) = +∞. There-
fore
∑+∞
i=n1
p(i)x(σ(i)) < +∞. By Remark 3.2 we conclude that lim inf x(n) = 0. In other words,(
x(n)
)
cannot tend to zero but lim inf x(n) = 0.
Assume that m is odd. Since ∆mz(n) ≤ 0, by part (iii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) we have that
(
x(n)
)
tends to zero.
Suppose that limn→∞ z(n) = A > 0. Then by part (ivc) of Lemma 2.3 we have that
(
x(n)
)
is
bounded. Thus we have
∑+∞
i=n1
p(i)x(σ(i)) < +∞. By Remark 3.2, lim inf x(n) = 0.
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444 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. As in previous case, lim inf x(n) = 0. The proof of part (vi) of the theorem is complete.
Theorem 3.1 is proved.
Case 2. p(n) ≤ 0.
Theorem 3.2. Assume that p(n) ≤ 0 ∀n ≥ 0 and
∑∞
i=0
p(i) = −∞. Then for Eq. (E) the
following statements hold:
(i) If c < −1, then every nonoscillatory solution
(
x(n)
)
:
(ia) is unbounded but lim inf x(n) = 0 or tends to infinity, if m is even;
(ib) has no a non-zero real limit but lim inf x(n) = 0, if m is odd.
(ii) If c = −1, then every nonoscillatory solution
(
x(n)
)
:
(iia) tends to infinity, if m is even;
(iib) tends to infinity or it is bounded and lim inf x(n) = 0, if m is odd.
(iii) If −1 < c < 0, then every nonoscillatory solution
(
x(n)
)
tends to zero or tends to infinity.
(iv) If c = 0, then every nonoscillatory solution
(
x(n)
)
:
(iva) tends to zero or tends to infinity, if m is even;
(ivb) tends to infinity, if m is odd.
(v) If 0 < c < 1, then every nonoscillatory solution
(
x(n)
)
:
(va) is unbounded or tends to zero, if m is even;
(vb) is unbounded, if m is odd.
(vi) If c ≥ 1, then every nonoscillatory solution
(
x(n)
)
:
(via) is unbounded or is bounded and lim inf x(n) = 0 or tends to zero, if m is even;
(vib) cannot tend to zero and it is unbounded or it is bounded with lim inf x(n) = 0, if m is odd.
Proof. Assume that the solution (x(n))n≥−k of (E) is nonoscillatory. Then it is either eventually
positive or eventually negative. As (−x(n))n≥−k is also a solution of (E), we may restrict ourselves
only to the case where x(n) > 0 for all large n. Let n0 ≥ −k be an integer such that x(n) > 0 for
all n ≥ n0 ≥ n∗. Then, there exists n1 ≥ n0 such that x(τ(n)) > 0, x(σ(n)) > 0 ∀n ≥ n1.
In view of (2.6), Eq.(E) becomes ∆mz(n)+p(n)x(σ(n)) = 0, or (3.1). Therefore, for sufficiently
large n and since p(n) ≤ 0, we have ∆mz(n) ≥ 0.
Summing up (3.1) from n1 to n, n ≥ n1 we obtain (3.2).
(i) c < −1.
Assume that m is even. Since ∆mz(n) ≥ 0, by part (ii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing.
If limn→∞ z(n) = −∞, then in view of part (ia) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. If
∑n
i=n1
p(i)x(σ(i)) = −∞, then by (3.2)
(
∆m−1z(n+ 1)
)
tends to +∞. By part (i)
of Lemma 2.1 we conclude that
(
z(n)
)
tends to +∞, which contradicts limn→∞ z(n) = −∞.
Therefore
∑n
i=n1
p(i)x(σ(i)) > −∞. By Remark 3.2, we conclude that lim inf x(n) = 0.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity.
If
(
z(n)
)
is nonincreasing, clearly its limit exists.
Suppose that limn→∞ z(n) = A < 0. Then in view of part (iia) of Lemma 2.3 we have that
lim inf x(n) ≥ A
1 + c
. Therefore for every ε > 0 with ε <
A
1 + c
, there exists n2 such that (3.3)
holds. Thus, for every n3 with σ(n3) ≥ n2, by (3.2) and (3.3) we obtain
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ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 445
∆m−1z(n+ 1) > ∆m−1z(n3)−
(
A
1 + c
− ε
)∑n
i=n3
p(i)→ +∞ as n→∞,
which guarantees that limn→∞∆m−1z(n + 1) = +∞. By part (i) of Lemma 2.1 we conclude that
(z(n)) tends to +∞, which contradicts our assumption. Therefore limn→∞ z(n) = A < 0 is false.
Suppose that limn→∞ z(n) = A ≥ 0. Then since (z(n)) is nonincreasing, we have z(n) > 0
eventually. Therefore (3.4) holds, which guarantees that
(
x(n)
)
tends to infinity. Thus, as in previous
case, we have that limn→∞∆m−1z(n + 1) = ∞ which means that
(
z(n)
)
tends to +∞. This
contradicts limn→∞ z(n) = A ≥ 0. Therefore limn→∞ z(n) = A ≥ 0 is false.
Assume that m is odd. Since ∆mz(n) ≥ 0, by part (iv) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing.
If limn→∞ z(n) = −∞, then in view of part (ia) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. Also, as in case where m is even, lim inf x(n) = 0 is satisfied.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity.
If
(
z(n)
)
is nondecreasing, clearly its limit exists.
Suppose that limn→∞ z(n) = A < 0. As in previous case, this is false.
Suppose that limn→∞ z(n) = 0. Then in view of part (iiia) of Lemma 2.3 we have that
(
x(n)
)
either tends to zero or
(
x(n)
)
has infenitely many accumulation points or
(
x(n)
)
tends to infin-
ity. But, if limn→∞ x(n) = +∞, then as in previous case we have limn→∞ z(n) = +∞, which
contradicts our assumption.
Suppose that limn→∞ z(n) = A > 0. Thus (3.4) holds, and as in previous case we are led to a
contradiction. Therefore limn→∞ z(n) = A > 0 is false.
In other words,
(
x(n)
)
cannot have a non-zero real limit. Indeed, assume that limn→∞ x(n) =
= ` > 0. Then limn→∞ z(n) = (1 + c)` < 0, which contradicts “limn→∞ z(n) = A < 0 is false”.
The proof of part (i) of the theorem is complete.
(ii) c = −1.
Assume that m is even. Since ∆mz(n) ≥ 0, by part (ii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = A ≥ 0, then z(n) ≥ 0 eventually since
(
z(n)
)
is nonincreasing. Thus (3.5)
holds which means that
(
x(n)
)
has a lower bound, i.e., lim inf x(n) > 0. By Remark 3.1 and
relation (3.2) we conclude that limn→∞∆m−1z(n) > 0. Therefore, by part (i) of Lemma 2.1 we
have that limn→∞ z(n) = +∞, which contradicts our assumption. Therefore limn→∞ z(n) = A ≥ 0
is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity.
Assume that m is odd. Since ∆mz(n) ≥ 0, by part (iv) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
Suppose that limn→∞ z(n) = A > 0. Since (z(n)) is nondecreasing, we have z(n) > 0 eventu-
ally. Therefore (3.5) holds, and therefore as in case m is even, we are led to a contradiction. Therefore
limn→∞ z(n) = A > 0 is false.
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446 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
If limn→∞ z(n) = 0, then z(n) ≤ 0 since (z(n)) is nondecreasing. Then in view of part (iiib)
of Lemma 2.3 we have that
(
x(n)
)
is bounded. Furthermore, in view of Remark 3.2 we have
lim inf x(n) = 0.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. The proof of part (ii) of the theorem is complete.
(iii) −1 < c < 0.
Assume that m is even. Since ∆mz(n) ≥ 0, by part (iii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) we have that
(
x(n)
)
tends to zero.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
=
A
1 + c
> 0. This guarantees that
∑n
i=n1
p(i)x(σ(i)) = −∞. Thus, as in previous case, we are led
to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity.
Assume that m is odd. Since ∆mz(n) ≥ 0, by part (iv) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
=
A
1 + c
> 0. This guarantees that
∑n
i=n1
p(i)x(σ(i)) = −∞. Thus, as in case where m is even,
we are led to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = 0, then by part (iiic) we have that
(
x(n)
)
tends to zero.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. The proof of part (iii) of the theorem is complete.
(iv) c = 0.
Assume that m is even. Since ∆mz(n) ≥ 0, by part (ii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) of Lemma 2.3 we have that
(
x(n)
)
tends to zero.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
= A > 0. This guarantees that
∑n
i=n1
p(i)x(σ(i)) = −∞. Hence, as in previous cases, we are led
to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity.
Assume that m is odd. Since ∆mz(n) ≥ 0, by part (iv) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
Suppose that limn→∞ z(n) = A > 0. By part (ivb) of Lemma 2.3 we have that limn→∞ x(n) =
= A > 0. This guarantees that
∑n
i=n1
p(i)x(σ(i)) = +∞. Hence, as in previous cases, we are led
to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
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ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 447
If limn→∞ z(n) = 0, then
(
z(n)
)
≤ 0 since
(
z(n)
)
is nondecreasing. This contradicts z(n) =
= x(n) > 0. Therefore limn→∞ z(n) = 0 is false.
If limn→∞ z(n) = +∞, then in view of part (va) of Lemma 2.3 we have that
(
x(n)
)
tends to
infinity. The proof of part (iv) of the theorem is complete.
(v) 0 < c < 1.
Assume that m is even. Since ∆mz(n) ≥ 0, by part (ii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) of Lemma 2.3 we have that
(
x(n)
)
tends to zero.
If limn→∞ z(n) = A > 0, then
∑n
i=n1
p(i)x(σ(i)) > −∞. By Remark 3.2 we have that
lim inf x(n) = 0. Then there exists a subsequence (x(θ(n))) of (x(n)) such that limn→∞ x(θ(n)) =
= 0. By a similar procedure as in part (v) of Theorem 3.1 we are led to a contradiction. Therefore
limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
unbounded.
Assume that m is odd. Since ∆mz(n) ≥ 0, by part (iv) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then z(n) ≤ 0 since (z(n)) is nondecreasing. This contradicts z(n) =
= x(n) + cx(τ(n)) > 0. Therefore limn→∞ z(n) = 0 is false.
Suppose that limn→∞ z(n) = A > 0. By part (ivc) of Lemma 2.3 we have that
(
x(n)
)
is
bounded. This guarantees that
∑n
i=n1
p(i)x(σ(i)) > −∞. As in case where m is even, we are led
to a contradiction. Therefore limn→∞ z(n) = A > 0 is false.
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. The proof of part (v) of the theorem is complete.
(vi) c ≥ 1.
Assume that m is even. Since ∆mz(n) ≥ 0, by part (ii) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nonincreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then by part (iiic) we have that
(
x(n)
)
tends to zero.
Suppose that limn→∞ z(n) = A > 0. By part (ivc) of Lemma 2.3 we have that
(
x(n)
)
is
bounded. This guarantees that
∑n
i=n1
p(i)x(σ(i)) > −∞. By Remark 3.2, lim inf x(n) = 0.
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded.
Assume that m is odd. Since ∆mz(n) ≥ 0, by part (iv) of Lemma 2.2 we have that
(
z(n)
)
tends
to ±∞ or it is nondecreasing. By parts (ib) and (iib) of Lemma 2.3, the cases limn→∞ z(n) = −∞
and limn→∞ z(n) = A < 0 are not valid.
If limn→∞ z(n) = 0, then
(
z(n)
)
≤ 0 since
(
z(n)
)
is nondecreasing. This contradicts z(n) =
= x(n) + cx(τ(n)) > 0. Therefore limn→∞ z(n) = 0 is false.
Suppose that limn→∞ z(n) = A > 0. By part (ivc) of Lemma 2.3 we have that
(
x(n)
)
is bounded. This guarantees that
∑n
i=n1
p(i)x(σ(i)) > −∞. By Remark 3.2 we have that
lim inf x(n) = 0.
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448 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
If limn→∞ z(n) = +∞, then in view of part (vb) of Lemma 2.3 we have that
(
x(n)
)
is un-
bounded. The proof of part (vi) of the theorem is complete.
Theorem 3.2 is proved.
4. Examples. In this section we present some examples to illustrate the main results.
Example 4.1. Consider the difference equation
∆2(x(n)− 2x(n− 1)) + p(n)x(n2 + 1) = 0, n ≥ 2,
where p(n) =
(2n+ 10)
(
n2 + 1
)
n (n+ 1) (n− 1) (n+ 2)
> 0 ∀n ≥ 2.
Here m is even and
∑∞
i=2
p(i) = +∞. It is easy to see that all conditions of part (ia) of the
Theorem 3.1 are satisfied, and hence every nonoscillatory solution (x(n)) of the above equation has
no a real non-zero limit. In fact (x(n)) =
(
1
n
)
is one such solution, since it satisfies the above
equation for all n ≥ 2 and limn→∞ x(n) = 0.
Example 4.2. Consider the difference equation
∆3(x(n)− x(n− 1)) + 16x(n2 + 2) = 0, n ≥ 1.
Here m is odd and
∑∞
i=1
p(i) = +∞. All conditions of part (iib) of the Theorem 3.1 are
satisfied, and hence every non-zero solution (x(n)) oscillates. In fact (x(n)) = ((−1)n) is one such
solution, since it satisfies the above equation for all n ≥ 1 and oscillates.
Example 4.3. Consider the difference equation
∆2
(
x(n)− 1
4
x(n− 1)
)
+ p(n)x(n− 2) = 0, n ≥ 41,
where
p(n) =
5
2
· n
3 − 57n2 + 722n− 2320
n(n− 1)(n+ 1)(n+ 2)
> 0 ∀n ≥ 41.
Clearly
∑∞
i=41
p(i) = +∞. All conditions of part (iii) of the Theorem 3.1 are satisfied, and
hence every nonoscillatory solution (x(n)) tends to zero. In fact (x(n)) =
(
10n
n!
)
is one such
solution, since it satisfies the above equation for all n ≥ 41 and limn→∞ x(n) = 0.
Example 4.4. Consider the difference equation
∆3
(
x(n) + x(n− 1)
)
+ p(n)x(n+ 3) = 0, n ≥ 4,
where p(n) =
3n− 11
n+ 3
> 0 ∀n ≥ 4.
Here m is odd and
∑∞
i=4
p(i) = +∞. All conditions of part (vib) of the Theorem 3.1 are
satisfied, and hence every nonoscillatory solution (x(n)) tends to zero or lim inf x(n) = 0. In
fact (x(n)) =
( n
2n
)
is one such solution, since it satisfies the above equation for all n ≥ 4 and
limn→∞ x(n) = 0.
Example 4.5. Consider the difference equation
∆2 (x(n)− 2x(n− 2)) +
(
− 1
16
)
x(n+ 3) = 0, n ≥ 3.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
ASYMPTOTIC BEHAVIOR OF HIGHER-ORDER NEUTRAL DIFFERENCE EQUATIONS . . . 449
Here m is even and
∑∞
i=3
p(i) = −∞. All conditions of part (ia) of the Theorem 3.2 are
satisfied, and hence every nonoscillatory solution (x(n)) is unbounded but lim inf x(n) = 0 or tends
to infinity. In fact (x(n)) = (2n) is one such solution, since it satisfies the above equation for all
n ≥ 3 and limn→∞ x(n) = +∞.
Example 4.6. Consider the difference equation
∆2(x(n)− x(n− 2)) +
(
−3
8
)
x(n+ 1) = 0, n ≥ 2.
Here m is even and
∑∞
i=2
p(i) = −∞. All conditions of part (iia) of the Theorem 3.2 are
satisfied, and hence every nonoscillatory solution (x(n)) tends to infinity. In fact (x(n)) = (2n) is
one such solution, since it satisfies the above equation for all n ≥ 2 and limn→∞ x(n) = +∞.
Example 4.7. Consider the difference equation
∆3(x(n)− 1
2
x(n− 1)) +
(
− 4
37
)
x(n− 4) = 0, n ≥ 4.
Clearly
∑∞
i=4
p(i) = −∞. All conditions of part (iii) of the Theorem 3.2 are satisfied, and hence
every nonoscillatory solution (x(n)) tends to zero or tends to infinity. In fact (x(n)) = (3−n) is one
such solution, since satisfies the above equation for all n ≥ 4 and limn→∞ x(n) = 0.
Example 4.8. Consider the difference equation
∆2
(
x(n) + x(n− 1)
)
+ p(n)x(n3 + 2) = 0, n ≥ 10,
where
p(n) =
n3 + 2
ln (n3 + 2)
ln
n+1
√
n+ 1 n
√
n
n+2
√
n+ 2 n−1
√
n− 1
< 0 ∀n ≥ 10.
Here m is even and
∑∞
i=4
p(i) = −∞. All conditions of part (via) of the Theorem 3.2 are satis-
fied, and hence every nonoscillatory solution (x(n)) is unbounded or is bounded and lim inf x(n) = 0
or tends to zero. In fact (x(n)) =
(
lnn
n
)
is one such solution, since it satisfies the above equation
for all n ≥ 10 and limn→∞ x(n) = 0.
Remark 4.1. Similarly, one can construct examples to illustrate the other parts of Theorems 3.1
and 3.2.
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450 G. E. CHATZARAKIS, H. KHATIBZADEH, G. N. MILIARAS, I. P. STAVROULAKIS
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Received 26.01.13
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 3
|
| id | umjimathkievua-article-2429 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:23:16Z |
| publishDate | 2013 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/cd/24fb44ecb13f6c344a1740ab58797ccd.pdf |
| spelling | umjimathkievua-article-24292020-03-18T19:15:16Z Asymptotic behavior of higher-order neutral difference equations with general arguments Асимптотична поведiнка нейтральних рiзницевих рiвнянь вищого порядку iз загальними аргументами Chatzarakis, G. E. Khatibzadeh, H. Miliaras, G. N. Stavroulakis, I. P. Чатзаракіс, Г. Е. Хатібзадех, Г. Міліарас, Г. Н. Ставроулакіс, І. П. We study the asymptotic behavior of solutions of the higher-order neutral difference equation $$Δm[x(n)+cx(τ(n))]+p(n)x(σ(n))=0,N∍m≥2,n≥0,$$ where $τ (n)$ is a general retarded argument, $σ(n)$ is a general deviated argument, $c ∈ R; (p(n)) n ≥ 0$ is a sequence of real numbers, $∆$ denotes the forward difference operator $∆x(n) = x(n+1) - x(n)$; and $∆^j$ denotes the jth forward difference operator $∆^j (x(n) = ∆ (∆^{j-1}(x(n)))$ for $j = 2, 3,…,m$. Examples illustrating the results are also given. Вивчається асимптотична поведiнка розв’язкiв нейтрального рiзницевого рiвняння вищого порядку $$Δm[x(n) + cx(τ(n))] + p(n)x(σ(n)) = 0, N ∍ m ≥ 2, n≥0,$$ де $τ(n)$— загальний аргумент iз запiзненням, $σ(n)$ — загальний аргумент iз вiдхиленням, $c ∈ R, (p(n)) n ≥ 0$ — послiдовнiсть дiйсних чисел, $∆$ — оператор правої рiзницi, $∆x(n) = x(n + 1) − x(n)$, та $∆^j$ — $j$-й оператор правої рiзницi, $∆^j (x(n) = ∆ (∆^{j-1}(x(n)))$ при $j = 2, 3,…,m$.. Наведено також приклади, що iлюструють отриманi результати. Institute of Mathematics, NAS of Ukraine 2013-03-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2429 Ukrains’kyi Matematychnyi Zhurnal; Vol. 65 No. 3 (2013); 430-450 Український математичний журнал; Том 65 № 3 (2013); 430-450 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2429/1625 https://umj.imath.kiev.ua/index.php/umj/article/view/2429/1626 Copyright (c) 2013 Chatzarakis G. E.; Khatibzadeh H.; Miliaras G. N.; Stavroulakis I. P. |
| spellingShingle | Chatzarakis, G. E. Khatibzadeh, H. Miliaras, G. N. Stavroulakis, I. P. Чатзаракіс, Г. Е. Хатібзадех, Г. Міліарас, Г. Н. Ставроулакіс, І. П. Asymptotic behavior of higher-order neutral difference equations with general arguments |
| title | Asymptotic behavior of higher-order neutral difference equations with general arguments |
| title_alt | Асимптотична поведiнка нейтральних рiзницевих рiвнянь вищого порядку iз загальними аргументами |
| title_full | Asymptotic behavior of higher-order neutral difference equations with general arguments |
| title_fullStr | Asymptotic behavior of higher-order neutral difference equations with general arguments |
| title_full_unstemmed | Asymptotic behavior of higher-order neutral difference equations with general arguments |
| title_short | Asymptotic behavior of higher-order neutral difference equations with general arguments |
| title_sort | asymptotic behavior of higher-order neutral difference equations with general arguments |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2429 |
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