# Dynamics of a Higher-Order System of Difference Equations

Authors: Qi Wang; Qinqin Zhang; Qirui Li
Journal: Discrete Dynamics in Nature and Society
 (2017)

---

## Abstract

Consider the following system of difference equations:x
n
+
1
(
i
)
=
x
n
-
m
+
1
(
i
)
/
A
i
∏
j
=
0
m
-
1
x
n
-
j
(
i
+
j
+
1
)
+
α
i, x
n
+
1
(
i
+
m
)
=
x
n
+
1
(
i
), x
1
-
l
(
i
+
l
)
=
a
i
,
l, A
i
+
m
=
A
i, α
i
+
m
=
α
i, i
,
l
=
1,2
,
…
,
m; n
=
0,1
,
2
,
…
, where m is a positive integer, A
i
,
α
i, i
=
1,2
,
…
,
m, and the initial conditions a
i
,
l, i
,
l
=
1,2
,
…
,
m
, are positive real numbers. We obtain the expressions of the positive solutions of the system and then give a precise description of the convergence of the positive solutions. Finally, we give some numerical results.


---

## Body

## 1. Introduction

Difference equation or system of difference equations is a diverse field which impacts almost every branch of pure and applied mathematics. Not only does it provide us with some simple and useful mathematic models to help elucidate interesting phenomena in applications, but also it can kind of display some surprising complicated dynamics comparing with its analogue differential equations. Hence, the systems of difference equations and difference equations have attracted a lot of attention (see, e.g., the systems of difference equations [1–16] and difference equations [17–29] and the references therein). Among them, symmetric and close to symmetric systems of difference equations have attracted a considerable interest.Papaschinnopoulos and Schinas [1] studied the oscillatory behavior, the boundedness of the solutions, and the global asymptotic stability of the positive equilibrium of the system of the nonlinear difference equations:(1)
x
n
+
1
=
A
+
y
n
x
n
-
p
,
y
n
+
1
=
A
+
x
n
y
n
-
q
,
n
=
0,1
,
2
,
…
.In [2], they also investigated the boundedness, persistence, the oscillatory behavior, and the asymptotic behavior of the positive solutions of the system of difference equations:(2)
x
n
+
1
=
∑
i
=
0
k
A
i
y
n
-
i
p
i
,
y
n
+
1
=
∑
i
=
0
k
B
i
x
n
-
i
q
i
,
n
=
0,1
,
2
,
…
.Clark et al. [3, 4] investigated the global asymptotic stability of the system of difference equations:(3)
x
n
+
1
=
x
n
a
+
c
y
n
,
y
n
+
1
=
y
n
b
+
d
x
n
,
n
=
0,1
,
2
,
…
.Camouzis and Papaschinopoulos [5] studied the global asymptotic behavior of positive solutions of the system of rational difference equations:(4)
x
n
+
1
=
1
+
x
n
y
n
-
m
,
y
n
+
1
=
1
+
y
n
x
n
-
m
,
n
=
0,1
,
2
,
…
.Yang [6] studied the behavior of positive solutions of the system of difference equations:(5)
x
n
=
A
+
y
n
-
1
x
n
-
p
y
n
-
q
,
y
n
=
A
+
x
n
-
1
x
n
-
r
y
n
-
s
,
n
=
0,1
,
2
,
…
.Zhang et al. [7] studied the boundedness, the persistence, and global asymptotic stability of the positive solutions of the system of difference equations:(6)
x
n
+
1
=
A
+
y
n
-
m
x
n
,
y
n
+
1
=
A
+
x
n
-
m
y
n
,
n
=
0,1
,
2
,
…
.Yalçinkaya and Çinar [8] studied the global asymptotic stability of the system of difference equations:(7)
z
n
+
1
=
t
n
+
z
n
-
1
t
n
z
n
-
1
+
a
,
t
n
+
1
=
z
n
+
t
n
-
1
z
n
t
n
-
1
+
a
,
n
=
0,1
,
2
,
…
.Kurbanlı et al. [9] studied the behavior of the positive solutions of the following system of difference equations:(8)
x
n
+
1
=
x
n
-
1
y
n
x
n
-
1
+
1
,
y
n
+
1
=
y
n
-
1
x
n
y
n
-
1
+
1
,
n
=
0,1
,
2
,
…
.Motivated by the above studies, in this note, we consider the following system of difference equations:(9)
x
n
+
1
i
=
x
n
-
m
+
1
i
A
i
∏
j
=
0
m
-
1
x
n
-
j
i
+
j
+
1
+
α
i
,
x
n
+
1
i
+
m
=
x
n
+
1
i
,
x
1
-
l
i
+
l
=
a
i
,
l
,
A
i
+
m
=
A
i
,
α
i
+
m
=
α
i
,
i
,
l
=
1,2
,
…
,
m
;
n
=
0,1
,
2
,
…
,where m is a positive integer, A
i
,
α
i, i
=
1,2
,
…
,
m, and the initial conditions a
i
,
l, i
,
l
=
1,2
,
…
,
m, are positive real numbers. We perfect and generalize the results in related literature.


## 2. Main Results

Throughout this paper, letN and R stand for the set of natural numbers and the set of real numbers, respectively.Let{
(
x
n
(
1
)
,
x
n
(
2
)
,
…
,
x
n
(
m
)
)
}
n
=
-
m
+
1
∞ be a positive solution of (9). If we set (10)
y
n
-
m
+
1
i
=
1
x
n
-
m
+
1
i
,
i
=
1,2
,
…
,
m
;
n
∈
N
,then (9) translates into(11)
y
n
+
1
i
=
α
i
y
n
-
m
+
1
i
+
A
i
∏
j
=
0
m
-
2
y
n
-
j
i
+
j
+
1
,
y
n
+
1
i
+
m
=
y
n
+
1
i
,
y
1
-
l
i
+
l
=
b
i
,
l
,
A
i
+
m
=
A
i
,
α
i
+
m
=
α
i
,
i
,
l
=
1,2
,
…
,
m
;
n
∈
N
,where b
i
,
l
=
1
/
a
i
,
l, i
,
l
=
1,2
,
…
,
m.For convenience, in the following we will investigate (11). Set (12)
I
i
,
n
=
∏
l
=
0
m
-
1
y
n
-
l
i
+
l
,
α
=
∏
l
=
0
m
-
1
α
i
+
l
,
i
=
1,2
,
…
,
m
;
n
∈
N
,
(13)
Q
i
,
j
=
∑
l
=
0
j
-
1
∏
s
=
0
l
-
1
α
i
+
s
A
i
+
l
,
i
,
j
=
1,2
,
…
,
m
,where we appeal to the convention ∏
s
=
0
-
1
α
i
+
s
≔
1
.Combing (12) with (11), we get (14)
I
i
+
k
m
,
n
=
I
i
,
n
,
i
=
1,2
,
…
,
m
;
k
∈
N
;
n
∈
N
.By (11), (12), and (13), we get (15)
Q
i
+
k
m
,
j
=
Q
i
,
j
,
i
,
j
=
1,2
,
…
,
m
;
k
∈
N
,
(16)
Q
i
,
j
+
1
=
α
i
Q
i
+
1
,
j
+
A
i
,
i
=
1,2
,
…
,
m
;
j
=
1,2
,
…
,
m
-
1
,
(17)
Q
i
,
m
+
α
A
i
=
α
i
Q
i
+
1
,
m
+
A
i
,
i
=
1,2
,
…
,
m
.Lemma 1.
Let{
(
y
n
(
1
)
,
y
n
(
2
)
,
…
,
y
n
(
m
)
)
}
n
=
-
m
+
1
∞ be a positive solution of (11); then (18)
I
i
,
n
+
1
=
α
i
I
i
+
1
,
n
+
A
i
,
i
=
1,2
,
…
,
m
;
n
∈
N
.Proof.
From (12) we know (19)
I
i
,
n
+
1
=
∏
l
=
0
m
-
1
y
n
-
l
+
1
i
+
l
=
y
n
+
1
i
y
n
-
m
+
1
i
+
m
∏
l
=
0
m
-
1
y
n
-
l
i
+
l
+
1
=
y
n
+
1
i
y
n
-
m
+
1
i
I
i
+
1
,
n
,from (11) we obtain (20)
y
n
+
1
i
y
n
-
m
+
1
i
=
α
i
+
A
i
I
i
+
1
,
n
,and combining (19) with (20) we get the conclusion.
This completes the proof.Lemma 2.
Let{
(
y
n
(
1
)
,
y
n
(
2
)
,
…
,
y
n
(
m
)
)
}
n
=
-
m
+
1
∞ be a positive solution of (11); then (21)
I
i
,
n
+
1
m
+
j
-
1
=
α
I
i
,
n
m
+
j
-
1
+
Q
i
,
m
,
i
,
j
=
1,2
,
…
,
m
;
n
∈
N
.Proof.
Fori
,
j
=
1,2
,
…
,
m; n
∈
N, by (18), (13), and (14), we have(22)
I
i
,
n
+
1
m
+
j
-
1
=
α
i
I
i
+
1
,
n
+
1
m
+
j
-
2
+
A
i
=
α
i
α
i
+
1
I
i
+
2
,
n
+
1
m
+
j
-
3
+
A
i
+
1
+
A
i
=
∏
l
=
0
1
α
i
+
l
I
i
+
2
,
n
+
1
m
+
j
-
3
+
Q
i
,
2
=
∏
l
=
0
1
α
i
+
l
α
i
+
2
I
i
+
3
,
n
+
1
m
+
j
-
4
+
A
i
+
2
+
Q
i
,
2
=
∏
l
=
0
2
α
i
+
l
I
i
+
3
,
n
+
1
m
+
j
-
4
+
Q
i
,
3
=
⋯
=
∏
l
=
0
m
-
1
α
i
+
l
I
i
+
m
,
n
m
+
j
-
1
+
Q
i
,
m
=
α
I
i
,
n
m
+
j
-
1
+
Q
i
,
m
.Hence, (21) holds.Lemma 3.
Let{
(
y
n
(
1
)
,
y
n
(
2
)
,
…
,
y
n
(
m
)
)
}
n
=
-
m
+
1
∞ be a positive solution of (11); then (23)
I
i
,
n
m
+
j
-
1
=
α
n
I
i
,
j
-
1
+
∑
l
=
0
n
-
1
α
l
Q
i
,
m
,
i
,
j
=
1,2
,
…
,
m
;
n
∈
N
,where we appeal to the convention ∑
l
=
0
-
1
α
l
≔
0
.Proof.
We will prove the conclusion by induction. Fori
,
j
=
1,2
,
…
,
m, n
=
0, it is obvious that (23) holds. For i
,
j
=
1,2
,
…
,
m, n
=
1, from Lemma 2, we know that (23) holds.
Suppose that (23) holds for n
=
k, then for n
=
k
+
1, by Lemma 2 we have(24)
I
i
,
k
+
1
m
+
j
-
1
=
α
I
i
,
k
m
+
j
-
1
+
Q
i
,
m
=
α
k
+
1
I
i
,
j
-
1
+
α
∑
l
=
0
k
-
1
α
l
Q
i
,
m
+
Q
i
,
m
=
α
k
+
1
I
i
,
j
-
1
+
∑
l
=
0
k
α
l
Q
i
,
m
.Hence, (23) holds for n
=
k
+
1, from which we get the conclusion.In the following, set(25)
r
n
=
0
,
w
h
e
n
n
+
1
mod
⁡
m
=
0
;
n
+
1
mod
⁡
m
-
m
,
w
h
e
n
n
+
1
mod
⁡
m
≠
0
,
p
n
=
n
+
1
m
-
1
,
w
h
e
n
n
+
1
mod
⁡
m
=
0
;
n
+
1
m
,
w
h
e
n
n
+
1
mod
⁡
m
≠
0
,where ⌊
·
⌋ is floor function.Lemma 4.
Let{
(
y
n
(
1
)
,
y
n
(
2
)
,
…
,
y
n
(
m
)
)
}
n
=
-
m
+
1
∞ be a positive solution of (11); then (26)
y
n
+
1
i
=
y
r
n
i
∏
l
=
0
p
n
α
i
+
A
i
α
l
I
i
+
1
,
m
+
r
n
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
,
i
=
1,2
,
…
,
m
;
n
∈
N
.Proof.
In fact, fori
=
1,2
,
…
,
m; n
∈
N by (11) and Lemma 3 we have(27)
y
n
+
1
i
=
y
n
-
m
+
1
i
α
i
+
A
i
I
i
+
1
,
n
=
y
n
-
2
m
+
1
i
α
i
+
A
i
I
i
+
1
,
n
-
m
α
i
+
A
i
I
i
+
1
,
n
=
⋯
=
y
r
n
i
∏
l
=
0
p
n
α
i
+
A
i
I
i
+
1
,
n
-
l
m
=
y
r
n
i
∏
l
=
0
p
n
α
i
+
A
i
α
l
I
i
+
1
,
m
+
r
n
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
.Hence, (26) holds.In the following, set(28)
∑
l
=
0
∞
ln
⁡
α
i
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
≔
η
i
,
m
-
j
,
i
,
j
=
1,2
,
…
,
m
.It is obvious that η
i
,
m
-
j
∈
{
-
∞
}
∪
R
∪
{
+
∞
}
.Lemma 5.
Fori
=
1,2
,
…
,
m, the following statements are true.(1)
Suppose thatα
=
1, α
i
=
1 or α
≥
1, α
i
>
1 or α
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
>
0; then η
i
,
m
-
j
=
+
∞.
(2)
Suppose thatα
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
=
0 or α
>
1, α
i
=
1; then η
i
,
m
-
j
∈
R.
(3)
Suppose thatα
≥
1, α
i
<
1 or α
<
1, α
i
-
1
Q
i
,
m
+
1
-
α
A
i
<
0; then η
i
,
m
-
j
=
-
∞.Proof.
(1) Case  1.α
=
1, α
i
=
1. Note that (29)
ln
⁡
1
+
A
i
I
i
+
1
,
j
-
1
+
l
Q
i
+
1
,
m
~
A
i
I
i
+
1
,
j
-
1
+
l
Q
i
+
1
,
m
,
a
s
l
⟶
∞
,
∑
l
=
0
∞
A
i
I
i
+
1
,
j
-
1
+
l
Q
i
+
1
,
m
=
+
∞
.It follows that η
i
,
m
-
j
=
+
∞
.
Case  2.α
≥
1, α
i
>
1. Note that (30)
l
i
m
l
→
+
∞
ln
⁡
α
i
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
=
ln
⁡
α
i
>
0
.Hence, η
i
,
m
-
j
=
+
∞
.
Case  3.α
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
>
0. By (16) and (17), we have (31)
α
i
+
A
i
1
-
α
Q
i
+
1
,
m
=
α
i
Q
i
+
1
,
m
+
A
i
-
α
A
i
Q
i
+
1
,
m
=
Q
i
,
m
Q
i
+
1
,
m
.
(32)
Q
i
,
m
-
Q
i
+
1
,
m
=
∑
l
=
0
m
-
1
∏
p
=
0
l
-
1
α
i
+
p
A
i
+
l
-
∑
l
=
0
m
-
1
∏
p
=
0
l
-
1
α
i
+
p
+
1
A
i
+
l
+
1
=
1
-
α
α
i
A
i
+
α
i
-
1
∑
l
=
0
m
-
2
∏
p
=
0
l
-
1
α
i
+
p
+
1
A
i
+
l
+
1
=
1
-
α
α
i
A
i
+
α
i
-
1
Q
i
+
1
,
m
-
1
=
1
α
i
α
i
-
1
Q
i
,
m
+
1
-
α
A
i
,That is, (33)
Q
i
,
m
-
Q
i
+
1
,
m
=
1
α
i
α
i
-
1
Q
i
,
m
+
1
-
α
A
i
.
Whenα
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
>
0, combining (31) with (33) we get (34)
l
i
m
l
→
+
∞
ln
⁡
α
i
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
=
ln
⁡
α
i
+
A
i
1
-
α
Q
i
+
1
,
m
>
0
.Hence, η
i
,
m
-
j
=
+
∞
.
(2) Case  1.α
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
=
0. In this case, by (33) we know Q
i
,
m
=
Q
i
+
1
,
m and(35)
α
i
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
=
1
+
α
i
-
1
α
l
1
-
α
I
i
+
1
,
j
-
1
+
Q
i
+
1
,
m
1
-
α
α
l
I
i
+
1
,
j
-
1
+
1
-
α
l
Q
i
+
1
,
m
.Hence, (36)
ln
⁡
α
i
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
~
α
i
-
1
α
l
1
-
α
I
i
+
1
,
j
-
1
-
Q
i
+
1
,
m
1
-
α
α
l
I
i
+
1
,
j
-
1
+
1
-
α
l
Q
i
+
1
,
m
as
l
⟶
∞
.Note that the series ∑
l
=
0
∞
‍
(
α
i
-
1
)
α
l
(
1
-
α
)
I
i
+
1
,
j
-
1
-
Q
i
+
1
,
m
/
(
1
-
α
)
α
l
I
i
+
1
,
j
-
1
+
(
1
-
α
l
)
Q
i
+
1
,
m is convergent, and we have η
i
,
m
-
j
∈
R
.
Case  2.α
>
1, α
i
=
1. Since (37)
ln
⁡
1
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
~
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
,
as
l
⟶
∞
.The series ∑
l
=
0
∞
‍
A
i
/
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
‍
α
s
Q
i
+
1
,
m is convergent, and we know that η
i
,
m
-
j
∈
R
.
(3) Case  1.α
≥
1, α
i
<
1. Note that (38)
l
i
m
l
→
+
∞
ln
⁡
α
i
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
=
ln
⁡
α
i
<
0
.Hence, η
i
,
m
-
j
=
-
∞
.
Case  2.α
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
<
0. Combining (31) with (33) we get (39)
l
i
m
l
→
+
∞
ln
⁡
α
i
+
A
i
α
l
I
i
+
1
,
j
-
1
+
∑
s
=
0
l
-
1
α
s
Q
i
+
1
,
m
=
ln
⁡
α
i
+
A
i
1
-
α
Q
i
+
1
,
m
<
0
.Hence, η
i
,
m
-
j
=
-
∞.Theorem 6.
Let{
(
y
n
(
1
)
,
y
n
(
2
)
,
…
,
y
n
(
m
)
)
}
n
=
-
m
+
1
∞ be a positive solution of (11). The following statements are true.(1)
Suppose thatα
=
1, α
i
=
1 or α
≥
1, α
i
>
1 or α
<
1, α
i
-
1
Q
i
,
m
+
1
-
α
A
i
>
0; then l
i
m
n
→
∞
y
n
+
1
(
i
)
=
+
∞, i
=
1,2
,
…
,
m.
(2)
Suppose thatα
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
=
0 or α
>
1, α
i
=
1; then l
i
m
k
→
∞
y
k
m
-
j
+
1
(
i
)
=
y
-
j
+
1
(
i
)
e
x
p
(
η
i
,
j
-
1
), i
,
j
=
1,2
,
…
,
m.
(3)
Suppose thatα
≥
1, α
i
<
1 or α
<
1, α
i
-
1
Q
i
,
m
+
1
-
α
A
i
<
0; then l
i
m
n
→
∞
y
n
+
1
(
i
)
=
0, i
=
1,2
,
…
,
m.Proof.
By Lemma4 and (28) we know (40)
l
i
m
k
→
∞
ln
⁡
y
k
m
-
j
+
1
i
=
ln
⁡
y
-
j
+
1
i
+
η
i
,
j
-
1
.The conclusion follows by Lemma 5.Theorem 7.
Let{
(
x
n
(
1
)
,
x
n
(
2
)
,
…
,
x
n
(
m
)
)
}
n
=
-
m
+
1
∞ be a positive solution of (9). The following statements are true.(1)
Suppose thatα
=
1, α
i
=
1 or α
≥
1, α
i
>
1 or α
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
>
0; then l
i
m
n
→
∞
x
n
+
1
(
i
)
=
0, i
=
1,2
,
…
,
m.
(2)
Suppose thatα
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
=
0 or α
>
1, α
i
=
1; then l
i
m
k
→
∞
x
k
m
-
j
+
1
(
i
)
=
x
-
j
+
1
(
i
)
e
x
p
(
-
η
i
,
j
-
1
), i
,
j
=
1,2
,
…
,
m.
(3)
Suppose thatα
≥
1, α
i
<
1 or α
<
1, (
α
i
-
1
)
Q
i
,
m
+
(
1
-
α
)
A
i
<
0; then l
i
m
n
→
∞
x
n
+
1
(
i
)
=
+
∞, i
=
1,2
,
…
,
m.Proof.
The proof follows by Theorem6 and (10).


## 3. Numerical Results

In this section, we give some numerical simulations to illustrate our results. Consider the following system of difference equations:(41)
x
n
+
1
i
=
x
n
-
2
i
A
i
∏
j
=
0
2
x
n
-
j
i
+
j
+
1
+
α
i
,
A
i
+
3
=
A
i
,
α
i
+
3
=
α
i
,
i
=
1,2
,
3
;
n
=
3,4
,
5
,
…
.For convenience, setΘ
=
(
α
1
,
α
2
,
α
3
), Ξ
=
(
A
1
,
A
2
,
A
3
) and Λ
=
(
x
1
(
1
)
,
x
1
(
2
)
,
x
1
(
3
)
,
x
2
(
1
)
,
x
2
(
2
), x
2
(
3
)
,
x
3
(
1
)
,
x
3
(
2
)
,
x
3
(
3
)
).Example 1.
In (41), we take Θ
=
(
1.5,1.3,1.8
), Ξ
=
(
3,4
,
2
), Λ
=
(
1,4
,
7,3
,
6,9
,
2,5
,
8
)
. From Table 1 and Figure 1(a) we see that(42)
l
i
m
n
→
∞
x
n
+
1
1
=
0
,
l
i
m
n
→
∞
x
n
+
1
2
=
0
,
l
i
m
n
→
∞
x
n
+
1
3
=
0
.Table 1
n
95
96
97
98
99
100
x
n
(
1
)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
x
n
(
2
)
0.0006
0.0009
0.0000
0.0005
0.0007
0.0000
x
n
(
3
)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000Figure 1
α
>
1
.
(a)
(b)
(c)
(d)Example 2.
In (41), we take Θ
=
(
1.2,1
,
1.1
), Ξ
=
(
2,5
,
3
), Λ
=
(
0.5,0.3,0.9,0.7,1
,
0.2,0.9,0.7,0.3
)
. From Table 2 and Figure 1(b) we see that(43)
L
i
m
n
→
∞
x
n
+
1
1
=
0
,
l
i
m
k
→
∞
x
3
k
+
j
2
=
x
j
2
exp
⁡
-
η
2,3
-
j
,
j
=
1,2
,
3
,
l
i
m
n
→
∞
x
n
+
1
3
=
0
.Table 2
n
95
96
97
98
99
100
x
n
(
1
)
0.0020
0.0022
0.0011
0.0017
0.0018
0.0009
x
n
(
2
)
0.2396
0.4081
0.1478
0.2396
0.4081
0.1478
x
n
(
3
)
0.0073
0.0118
0.0085
0.0066
0.0108
0.0077Example 3.
In (41), we take Θ
=
(
1,1
,
2
), Ξ
=
(
3,4
,
6
), Λ
=
(
1,4
,
5.5,2
,
5,8
,
3,6
,
7
)
. From Table 3 and Figure 1(c) we see that(44)
l
i
m
k
→
∞
x
3
k
+
j
1
=
x
j
1
e
x
p
-
η
1,3
-
j
,
l
i
m
k
→
∞
x
3
k
+
j
2
=
x
j
2
e
x
p
-
η
2,3
-
j
,
j
=
1,2
,
3
,
l
i
m
n
→
∞
x
n
+
1
3
=
0
.Table 3
n
95
96
97
98
99
100
x
n
(
1
)
0.9344
2.0000
0.0052
0.9344
2.0000
0.0052
x
n
(
2
)
2.4481
3.8536
0.0132
2.4481
3.8536
0.0132
x
n
(
3
)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000Example 4.
In (41), we take Θ
=
(
1,0.9,1.2
), Ξ
=
(
4,0.4,0.3
), Λ
=
(
1,0.5,0.6,0.8,0.8,0.3,0.6,0.2,0.5
)
. From Table 4 and Figure 1(d) we see that(45)
l
i
m
k
→
∞
x
3
k
+
j
1
=
x
j
1
e
x
p
-
η
1,3
-
j
,
j
=
1,2
,
3
,
l
i
m
n
→
∞
x
n
+
1
2
=
+
∞
,
l
i
m
n
→
∞
x
n
+
1
3
=
0
.Table 4
n
195
196
197
198
199
200
x
n
(
1
)
0.0551
0.2586
0.0801
0.0551
0.2586
0.0801
x
n
(
2
)
150.6878
359.6003
566.9402
167.4252
399.5398
629.9081
x
n
(
3
)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000Example 5.
In (41), we take Θ
=
(
10
/
9,0.8,9
/
8
), Ξ
=
(
5,1
,
7
), Λ
=
(
7,0.4,10,2
,
0.9,5
,
5,0.6,9
)
. From Table 5 and Figure 2(a) we see that(46)
l
i
m
k
→
∞
x
3
k
+
j
1
=
x
j
1
e
x
p
-
η
1,3
-
j
,
j
=
1,2
,
3
,
l
i
m
n
→
∞
x
n
+
1
2
=
+
∞
,
l
i
m
n
→
∞
x
n
+
1
3
=
0
.Table 5
n
95
96
97
98
99
100
x
n
(
1
)
0.0035
0.0315
0.0006
0.0031
0.0280
0.0005
x
n
(
2
)
570.3772
418.8940
35.3935
710.9386
522.1435
44.1177
x
n
(
3
)
0.0112
0.0233
0.0001
0.0098
0.0204
0.0001Figure 2
α
=
1
.
(a)
(b)Example 6.
In (41), we take Θ
=
(
1,0.8,1.25
), Ξ
=
(
2,0.1,4
), Λ
=
(
20,0.8,30,15,0.5,18,10,0.2,25
)
. From Table 6 and Figure 2(b) we see that(47)
l
i
m
n
→
∞
x
n
+
1
1
=
0
,
l
i
m
n
→
∞
x
n
+
1
2
=
+
∞
,
l
i
m
n
→
∞
x
n
+
1
3
=
0
.Table 6
n
95
96
97
98
99
100
x
n
(
1
)
0.1805
1.8961
0.0333
0.1783
1.8737
0.0329
x
n
(
2
)
460.0534
187.1326
24.3153
574.7237
233.7778
30.3762
x
n
(
3
)
0.0010
0.0014
0.0000
0.0008
0.0011
0.0000Example 7.
In (41), we take Θ
=
(
1.1,0.8,1
), Ξ
=
(
2,3
,
1
), Λ
=
(
20,50,20,80,58,18,10,60,16
)
. From Table 7 and Figure 3(a) we see that(48)
l
i
m
n
→
∞
x
n
+
1
1
=
0
,
l
i
m
n
→
∞
x
n
+
1
2
=
+
∞
,
l
i
m
n
→
∞
x
n
+
1
3
=
0
.Table 7
n
75
76
77
78
79
80
x
n
(
1
)
0.1590
0.0000
0.9893
0.1387
0.0000
0.8631
x
n
(
2
)
422.5695
0.0031
195.9576
491.4949
0.0036
227.9814
x
n
(
3
)
6.6679
0.0007
5.4747
6.5355
0.0007
5.3663Figure 3
α
<
1
.
(a)
(b)Example 8.
In (41), we take Θ
=
(
2,0.5,0.4
), Ξ
=
(
1,6
,
2
), Λ
=
(
0.3,3
,
1,0.7,5
,
1.2,0.5,2
,
1.4
)
. From Table 8 and Figure 3(b) we see that(49)
l
i
m
n
→
∞
x
n
+
1
1
=
0
,
l
i
m
k
→
∞
x
3
k
+
j
2
=
x
j
2
e
x
p
-
η
2,3
-
j
,
j
=
1,2
,
3
,
l
i
m
n
→
∞
x
n
+
1
3
=
+
∞
.Table 8
n
45
46
47
48
49
50
x
n
(
1
)
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
x
n
(
2
)
0.6839
0.1089
1.0899
0.6839
0.1089
1.0899
x
n
(
3
)
33410
4620
33100
69610
9620
68960



---

*Source: 4312640-2017-01-19.xml*