On a solvable of some systems of rational
di¤erence equations
M. M. El-Dessoky1;2
1
King Abdulaziz University, Faculty of Science,
Mathematics Department, P. O. Box 80203,
Jeddah 21589, Saudi Arabia.
2
Mansoura University, Faculty of Science,
Mathematics Department, Mansoura 35516, Egypt.
E-mail: dessokym@mane.edu.eg.
Abstract
In this paper, we study the existence of solutions for a class of rational
systems of di¤erence equations of order four in four-dimensional case
xn+1 =
zn+1 =
xn¡3
§1§tn zn¡1 yn¡2 xn¡3 ;
zn¡3
§1§yn xn¡1 tn¡2 zn¡3 ;
yn+1 =
tn+1 =
yn¡3
§1§xn tn¡1 zn¡2 yn¡3 ;
tn¡3
§1§zn yn¡1 xn¡2 tn¡3 ;
with the initial conditions are real numbers. Also, we study some behavior such
as the periodicity and boundedness of solutions for such systems. Finally, some
numerical examples are given to verify our theoretical results and graphed by
Matlab.
Keywords: di¤erence equations system, recursive sequences, periodic solutions.
Mathematics Subject Classi…cation: 39A10.
———————————————————
1
Introduction
The theory of discrete dynamical systems and di¤erence equations developed greatly
during the last twenty-…ve years of the twentieth century. Applications of discrete dynamical systems and di¤erence equations have experienced enormous growth in many
areas. Many applications of discrete dynamical systems and di¤erence equations have
appeared recently in the areas of biology, economics, physics, resource management,
1
and others. The theory of di¤erence equations occupies a central position in applicable analysis. There is no doubt that the theory of di¤erence equations will continue
to play an important role in mathematics as a whole. Nonlinear di¤erence equations of order greater than one are of paramount importance in applications. Such
equations also appear naturally as discrete analogues and as numerical solutions of
di¤erential and delay di¤erential equations which model various diverse phenomena
in biology, ecology, psychology, engineering, physics, probability theory, economics,
genetics, physiology and resource management. It is very interesting to investigate
the behavior of solutions of a system of higher-order rational di¤erence equations and
to discuss the local asymptotic stability of their equilibrium points [1 - 34]. There are
many papers deal with the di¤erence equations system. For example, the dynamical
behavior of positive solution for the system
xn+1 =
xn¡m+1
yn¡m+1
; yn+1 =
;
A + yn yn¡1 :::yn¡m+1
A + xn xn¡1 :::xn¡m+1
has been studied by Sroysang in [1].
In [2] El-Metwally, presented solutions of the following sixteen systems of di¤erence
equations
xn¡2 yn¡1
yn¡2 xn¡1
xn =
; yn =
:
§xn¡2 § yn¡3
§yn¡2 § xn¡3
In [3] Stević et al. obtained the solutions of the following systems of rational di¤erence
equations
xn¡k yn¡l
yn¡k xn¡l
xn =
; yn =
:
bn xn¡k + an yn¡l¡k
dn yn¡k + cn xn¡l¡k
In [4] Din has investigated the dynamics of a system of fourth-order rational di¤erenceequations
xn+1 =
®1 xn¡3
®2 yn¡3
; yn+1 =
:
¯1 + °1 yn yn¡1 xn¡2 xn¡3
¯2 + °2 xn xn¡1 yn¡2 yn¡3
Cinar [5] got the periodicity of the positive solutions of the nonlinear di¤erence equations system
1
yn
1
xn+1 = ; yn+1 =
; zn+1 =
:
zn
xn¡1 yn¡1
xn¡1
El-Dessoky et al. [6] studied the solutions of the di¤erence equation systems
xn+1 =
xn¡1
yn¡1
zn¡m
; yn+1 =
; zn+1 =
:
1 + yn xn¡1
1 + xn yn¡1
xn yn
Zkan et al. [7] investigated the periodical solutions of the following systems of third
order rational di¤erence equations
xn+1 =
yn¡2
xn¡2
xn¡2 + yn¡2
; yn+1 =
; zn+1 =
:
¡1 § yn¡2 xn¡1 yn
¡1 § xn¡2 yn¡1 xn
¡1 § xn¡2 yn¡1 xn
2
Yazlik et al. [8] studied the behaviour of solutions of the systems of di¤erence equations
xn+1 =
yn¡2 xn¡3 yn¡4
xn¡2 yn¡3 xn¡4
; yn+1 =
:
yn xn¡1 (§1 § yn¡2 xn¡3 yn¡4 )
xn yn¡1 (§1 § xn¡2 yn¡3 xn¡4 )
El-Dessoky et al. [9] investigated the form of the solution of the systems of di¤erence
equations
xn+1 =
xn¡2
yn¡2
zn¡2
; yn+1 =
; zn+1 =
;
§1 + xn¡2 zn¡1 yn
§1 + yn¡2 xn¡1 zn
§1 + zn¡2 yn¡1 xn
Also, in [10] , Kurbanli studied a three-dimensional system of rational di¤erence
equations
xn+1 =
xn¡1
yn¡1
xn
; yn+1 =
; zn+1 =
:
xn¡1 yn ¡ 1
yn¡1 xn ¡ 1
zn¡1 yn ¡ 1
To be motivated by the above studies, our aim in this paper is to investigate the
existence of solutions for the rational systems of di¤erence equations of order four in
four-dimensional case
xn¡3
yn¡3
; yn+1 =
;
§1 § tn zn¡1 yn¡2 xn¡3
§1 § xn tn¡1 zn¡2 yn¡3
zn¡3
tn¡3
=
; tn+1 =
;
§1 § yn xn¡1 tn¡2 zn¡3
§1 § zn yn¡1 xn¡2 tn¡3
xn+1 =
zn+1
where n 2 N0 and the initial conditions xi ; yi ; zi ; ti ; i = ¡3; ¡2; ¡1; 0 are arbitrary real numbers. Although we study the dynamics of theses solutions such as the
periodicity and boundedness and give some numerical examples for the systems.
2
Systems and The Expressions of Their Solutions
Here we interest to investigate the following system of some rational di¤erence equations
xn¡3
yn¡3
; yn+1 =
;
¡1 ¡ tn zn¡1 yn¡2 xn¡3
¡1 ¡ xn tn¡1 zn¡2 yn¡3
zn¡3
tn¡3
=
; tn+1 =
:
1 + yn xn¡1 tn¡2 zn¡3
1 + zn yn¡1 xn¡2 tn¡3
xn+1 =
zn+1
(1)
where n 2 N0 and the initial conditions xi ; yi ; zi ; ti ; i = ¡3; ¡2; ¡1; 0 are arbitrary
real numbers.
3
Theorem 1 Assume that fxn ; yn ; zn ; tn g are solutions of system (1). Then for n =
0; 1; 2; :::; we obtain
x4n¡3 = (¡1)n x¡3
x4n¡1 = (¡1)n x¡1
y4n¡3 = (¡1)n y¡3
y4n¡1 = (¡1)n y¡1
z4n¡3 = z¡3
z4n¡1 = z¡1
and
n¡1
Q
i=0
n¡1
Q
i=0
i=0
n¡1
Q
i=0
n¡1
Q
i=0
n¡1
Q
i=0
n¡1
Q
i=0
x4n¡2 = (¡1)n x¡2
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
;
(1+(2i+3)z¡3 t¡2 x¡1 y0 )
x4n = (¡1)n x0
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
;
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
i=0
y4n = (¡1)n y0
n¡1
Q
i=0
z4n = z0
n¡1
Q
n¡1
Q
i=0
i=0
t4n =
n¡1
Q
i=0
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
;
(1+(2i)x¡3 y¡2 z¡1 t0 )
(1+(2i+3)z¡3 t¡2 x¡1 y0 )
;
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
;
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
;
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
t4n¡2 = t¡2
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
;
(¡1+(2i+1)y¡3 z¡2 t¡1 x0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
;
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
(¡1+(2i+1)y¡3 z¡2 t¡1 x0 )
;
(¡1+(2i+2)y¡3 z¡2 t¡1 x0 )
y4n¡2 = (¡1)n y¡2
z4n¡2 = z¡2
(1+(2i)t¡3 x¡2 y¡1 z0 )
;
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
n¡1
Q
n¡1
Q
i=0
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
;
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
(1+(2i)x¡3 y¡2 z¡1 t0 )
;
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
t4n¡1 = t¡1
¡1
Q
i=0
n¡1
Q
(1+(2i)x¡3 y¡2 z¡1 t0 )
;
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
(1+(2i)z¡3 t¡2 x¡1 y0 )
;
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
t4n¡3 = t¡3
where
n¡1
Q
n¡1
Q
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
;
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
i=0
n¡1
Q (1+(2i+1)x¡3 y¡2 z¡1 t0 )
t0
;
(1+(2i+2)x¡3 y¡2 z¡1 t0 )
i=0
Ai = 1:
i=0
Proof: For n = 0 the result holds. Now suppose that n > 1 and that our assumption
holds for n ¡ 1; that is,
x4n¡7 = (¡1)n¡1 x¡3
x4n¡5 =
y4n¡7 =
y4n¡5 =
z4n¡7 =
z4n¡5 =
n¡2
Q
i=0
n¡2
Q
(1+(2i)x¡3 y¡2 z¡1 t0 )
;
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
x4n¡6 = (¡1)n¡1 x¡2
n¡2
Q
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
;
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
i=0
n¡2
Q
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
(¡1+(2i+1)y¡3 z¡2 t¡1 x0 )
(¡1)n¡1 x¡1
; x4n¡4 = (¡1)n¡1 x0
;
(1+(2i+3)z¡3 t¡2 x¡1 y0 )
(¡1+(2i+2)y¡3 z¡2 t¡1 x0 )
i=0
i=0
n¡2
n¡2
Q (¡1+(2i)y¡3 z¡2 t¡1 x0 )
Q (1+(2i+1)x¡3 y¡2 z¡1 t0 )
n¡1
(¡1)n¡1 y¡3
;
y
=
(¡1)
y
;
4n¡6
¡2
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
(1+(2i)x¡3 y¡2 z¡1 t0 )
i=0
i=0
n¡2
n¡2
Q (1+(2i+2)t¡3 x¡2 y¡1 z0 )
Q (1+(2i+3)z¡3 t¡2 x¡1 y0 )
n¡1
(¡1)n¡1 y¡1
;
y
=
(¡1)
y
;
4n¡4
0
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
i=0
i=0
n¡2
n¡2
Q (1+(2i)z¡3 t¡2 x¡1 y0 )
Q (¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
z¡3
;
z
=
z
;
4n¡6
¡2
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
i=0
i=0
n¡2
n¡2
Q (1+(2i)x¡3 y¡2 z¡1 t0 )
Q (1+(2i+1)t¡3 x¡2 y¡1 z0 )
z¡1
; z4n¡4 = z0
;
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
i=0
i=0
4
t4n¡7 = t¡3
t4n¡5 = t¡1
n¡2
Q
(1+(2i)t¡3 x¡2 y¡1 z0 )
;
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
i=0
n¡2
Q
t4n¡6 = t¡2
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
;
(¡1+(2i+1)y¡3 z¡2 t¡1 x0 )
i=0
t4n¡4 = t0
=
x4n¡7
¡1¡t4n¡4 z4n¡5 y4n¡6 x4n¡7
2
6
6
6
¡1¡6
6
6
4
=
t0
(¡1)n¡1 y¡2
Q
n¡2
(¡1)n x¡3
2
i=0
n¡1
Q
i=0
3
(1+(2i)x¡3 y¡2 z¡1 t0 )
5
(1+(2i+2)x¡3 y¡2 z¡1 t0 )
3
(1+(2i)x¡3 y¡2 z¡1 t0 )
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
n¡2
Q
i=0
(1+(2i)x¡3 y¡2 z¡1 t0 )
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
Q
n¡2
=
(1+(2i)x¡3 y¡2 z¡1 t0 )
(¡1)n¡1 x¡3
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
· i=0
¸
x¡3 y¡2 z¡1 t0
¡1¡
(1+(2n¡2)x¡3 y¡2 z¡1 t0 )
i=0
(1+(2i)x¡3 y¡2 z¡1 t0 )
;
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
y4n¡7
¡1¡x4n¡4 t4n¡5 z4n¡6 y4n¡7
2
(¡1)n¡1 y¡3
(¡1)
z¡2
n¡1
n¡2
Q
i=0
(¡1)n¡1 y¡3
2
x0
n¡2
Q
i=0
Q
n¡2
i=0
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
(¡1+(2i+1)y¡3 z¡2 t¡1 x0 )
t
(¡1+(2i+2)y¡3 z¡2 t¡1 x0 ) ¡1
Q
n¡2
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
i=0
n¡2
Q
i=0
3
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
5
(¡1+(2i+2)y¡3 z¡2 t¡1 x0 )
Q (¡1+(2i)y¡3 z¡2 t¡1 x0 )
(¡1)n¡1 y¡3
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
µ i=0
¶
¡1+(2n¡3)y¡3 z¡2 t¡1 x0
¡
¡1+(2n¡2)y¡3 z¡2 t¡1 x0
n¡2
= (¡1)n y¡3
n¡1
Q
i=0
n¡2
Q
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
(¡1)n¡1 y¡3
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
¡1¡4y¡3 z¡2 t¡1 x0
=
n¡2
Q
(1+(2i)x¡3 y¡2 z¡1 t0 )
(¡1)n¡1 x¡3
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
i=0
µ
¶
1+(2n¡1)x¡3 y¡2 z¡1 t0
¡
(1+(2n¡2)x¡3 y¡2 z¡1 t0 )
6
6
6
¡1¡6
6
6
4
=
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
:
(1+(2i+2)x¡3 y¡2 z¡1 t0 )
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
(¡1)n¡1 x¡3
(1+(2i)x¡3 y¡2 z¡1 t0 )
i=0
Q
Q
= (¡1)n x¡3
=
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
z
(1+(2i+2)x¡3 y¡2 z¡1 t0 ) ¡1
i=0
n¡2
Q
n¡2
n¡2
y4n¡3 =
i=0
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
;
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
(1+(2i)x¡3 y¡2 z¡1 t0 )
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
(1+(2i)x¡3 y¡2 z¡1 t0 )
(1+(2i+1)x¡3 y¡2 z¡1 t0 )
i=0
¡1¡4x¡3 y¡2 z¡1 t0
=
n¡2
Q
Q
n¡2
(¡1)n¡1 x¡3
i=0
n¡2
Q
i=0
We deduce from System (1) that
x4n¡3 =
n¡2
Q
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
:
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
5
=
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
(¡1+(2i+1)y¡3 z¡2 t¡1 x0 )
i=0
n¡2
Q
i=0
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
Q
n¡2
3
7
7
7
7
7
7
5
(¡1+(2i)y¡3 z¡2 t¡1 x0 )
(¡1+(2i¡1)y¡3 z¡2 t¡1 x0 )
y¡3 z¡2 t¡1 x0
¡1+
(¡1+(2n¡2)y¡3 z¡2 t¡1 x0 )
(¡1)n¡1 y¡3
i=0
7
7
7
7
7
7
5
Also, we see from Eq.(1) that
z4n¡3 =
z4n¡7
1+y4n¡4 x4n¡5 t4n¡6 z4n¡7
Q
n¡2
=
2
6
6
6
1+6
6
6
4
z¡3
(¡1)
=
Q
i=0
Q
=
z¡3
(1+(2i+3)z¡3 t¡2 x¡1 y0 )
(¡1)n¡1 x¡1
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
Q
n¡2
i=0
(1+(2i)z¡3 t¡2 x¡1 y0 )
(1+(2i+2)z¡3 t¡2 x¡1 y0 )
(1+(2i)z¡3 t¡2 x¡1 y0 )
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
i=0
1+(2n¡1)z¡3 t¡2 x¡1 y0
1+(2n¡2)z¡3 t¡2 x¡1 y0
= z¡3
Finally from Eq.(1), we see that
t4n¡3 =
2
6
6
6
1+6
6
6
4
=
t¡3
z0
n¡2
Q
n¡1
Q
i=0
i=0
(1+(2i)z¡3 t¡2 x¡1 y0 )
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
z
t
x¡1 y0
¡3 t¡2 x¡1 y0
¡3 ¡2
1+ 1+(2n¡2)z
(1+(2i)z¡3 t¡2 x¡1 y0 )
:
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
Q
i=0
(¡1)n¡1 x¡2
Q
n¡2
t¡3
2
Q
n¡2
t¡3
i=0
i=0
(1+(2i)t¡3 x¡2 y¡1 z0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
(¡1)n¡1 y¡1
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
i=0
n¡2
Q
i=0
Q
n¡2
i=0
n¡2
Q
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
t
(1+(2i+2)t¡3 x¡2 y¡1 z0 ) ¡3
3
(1+(2i)t¡3 x¡2 y¡1 z0 )
5
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
(1+(2i)t¡3 x¡2 y¡1 z0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
1+(2i¡1)t¡3 x¡2 y¡1 z0
1+(2i¡2)t¡3 x¡2 y¡1 z0
= t¡3
i=0
n¡2
Q
(1+(2i+2)t¡3 x¡2 y¡1 z0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
i=0
(1+(2i)t¡3 x¡2 y¡1 z0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
Q
n¡2
(1+(2i)t¡3 x¡2 y¡1 z0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
1+4t¡3 x¡2 y¡1 z0
=
=
z¡3
n¡2
Q
t4n¡7
1+z4n¡4 y4n¡5 x4n¡6 t4n¡7
n¡2
=
Q
n¡2
(1+(2i)z¡3 t¡2 x¡1 y0 )
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
1+z¡3 t¡2 x¡1 y0
n¡2
(1+(2i)z¡3 t¡2 x¡1 y0 )
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
3
(1+(2i+2)z¡3 t¡2 x¡1 y0 ) 7
y0
(1+(2i+3)z¡3 t¡2 x¡1 y0 ) 7
7
i=0
i=0
7
7
n¡2
n¡2
Q (1+(2i+1)z¡3 t¡2 x¡1 y0 )
Q (1+(2i)z¡3 t¡2 x¡1 y0 )
7
5
t¡2
z
(1+(2i+2)z¡3 t¡2 x¡1 y0 ) ¡3
(1+(2i+1)z¡3 t¡2 x¡1 y0 )
i=0
i=0
n¡2
Q
n¡1
n¡2
z¡3
i=0
n¡1
Q
i=0
=
t¡3
(1+(2i)t¡3 x¡2 y¡1 z0 )
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
t¡3 x¡2 y¡1 z0
1+
1+(2i+2)t¡3 x¡2 y¡1 z0
3
7
7
7
7
7
7
5
i=0
(1+(2i)t¡3 x¡2 y¡1 z0 )
:
(1+(2i+1)t¡3 x¡2 y¡1 z0 )
Similarly we can prove the other relations. This completes the proof.
Lemma 1. If xi ; yi ; zi ; ti ; i = ¡3; ¡2; ¡1; 0 arbitrary real numbers and let
fxn ; yn ; zn ; tn g are solutions of system (1) then the following statements are true:(i) If x¡3 = 0; then we have x4n¡3 = 0 and y4n¡2 = (¡1)n y¡2 ; z4n¡1 = z¡1 ; t4n =
t0 :
(ii) If x¡2 = 0; then we have x4n¡2 = 0 and y4n¡1 = (¡1)n y¡1 ; z4n = z0 ; t4n¡3 =
t¡3 :
(iii) If x¡1 = 0; then we have x4n¡1 = 0 and y4n = (¡1)n y0 ; z4n¡3 = z¡3 ; t4n¡2 =
t¡2 :
6
(iv) If x0 = 0; then we have x4n = 0 and y4n¡3 = (¡1)n y¡3 ; z4n¡2 = z¡2 ; t4n¡1 =
t¡1 :
(v) If y¡3 = 0; then we have y4n¡3 = 0 and x4n = (¡1)n x0 ; z4n¡2 = z¡2 ; t4n¡1 =
t¡1 :
(vi) If y¡2 = 0; then we have y4n¡2 = 0 and x4n¡3 = (¡1)n x¡3 ; z4n¡1 = z¡1 ; t4n =
t0 :
(vii) If y¡1 = 0; then we have y4n¡1 = 0 and x4n¡2 = (¡1)n x¡2 ; z4n = z0 ; t4n¡3 =
t¡3 :
(viii) If y0 = 0; then we have y4n = 0 and x4n¡1 = (¡1)n x¡1 ; z4n¡3 = z¡3 ; t4n¡2 =
t¡2 :
(ix) If z¡3 = 0; then we have z4n¡3 = 0 and x4n¡1 = (¡1)n x¡1 ; y4n = (¡1)n y0 ; t4n¡2 =
t¡2 :
(x) If z¡2 = 0; then we have z4n¡2 = 0 and x4n = (¡1)n x0 ; y4n¡3 = (¡1)n y¡3 ; t4n¡1 =
t¡1 :
(xi) If z¡1 = 0; then we have z4n¡1 = 0 and x4n¡3 = (¡1)n x¡3 ; y4n¡2 =
(¡1)n y¡2 ; t4n = t0 .
(xii) If z0 = 0; then we have z4n = 0 and x4n¡2 = (¡1)n x¡2 ; y4n¡1 = (¡1)n y¡1 ; t4n¡3 =
t¡3 .
(xiii) If t¡3 = 0; then we have t4n¡3 = 0 and x4n¡2 = (¡1)n x¡2 ; y4n¡1 = (¡1)n y¡1 ; z4n =
z0 :
(ivx) If t¡2 = 0; then we have t4n¡2 = 0 and x4n¡1 = (¡1)n x¡1 ; y4n = (¡1)n y0 ; z4n¡3 =
z¡3 :
(vx) If t¡1 = 0; then we have t4n¡1 = 0 and x4n = (¡1)n x0 ; y4n¡3 = (¡1)n y¡3 ; z4n¡2 =
z¡2 .
(vxi) If t0 = 0; then we have t4n = 0 and x4n¡3 = (¡1)n x¡3 ; y4n¡2 = (¡1)n y¡2 ; z4n¡1 =
z¡1 .
Proof: The proof follows directly from the expressions of the solutions of system (1).
Theorem 2 Assume that fxn ; yn ; zn ; tn g are solutions of the system
xn+1 =
zn+1 =
xn¡3
;
1+tn zn¡1 yn¡2 xn¡3
zn¡3
;
1¡yn xn¡1 tn¡2 zn¡3
yn+1 =
tn+1 =
yn¡3
;
1¡xn tn¡1 zn¡2 yn¡3
tn¡3
;
1+zn yn¡1 xn¡2 tn¡3
(2)
with x¡3 y¡2 z¡1 t0 6= §1; t¡3 x¡2 y¡1 z0 6= ¡1; t¡3 x¡2 y¡1 z0 6= ¡ 12 ; z¡3 t¡2 x¡1 y0 6= §1;
y¡3 z¡2 t¡1 x0 6= 1; y¡3 z¡2 t¡1 x0 6= 12 ; takes the form
x4n¡3 =
x4n¡1 =
y4n¡3 =
y4n¡1 =
x¡3
;
(1+x¡3 y¡2 z¡1 t0 )n
x¡1
;
(1+z¡3 t¡2 x¡1 y0 )n
y¡3
;
(1¡y¡3 z¡2 t¡1 x0 )n
x4n¡2 =
x¡2 (1+t¡3 x¡2 y¡1 z0 )n
;
(1+2t¡3 x¡2 y¡1 z0 )n
x4n = x0 (1 ¡ y¡3 z¡2 t¡1 x0 )n ;
y4n¡2 = y¡2 (1 + x¡3 y¡2 z¡1 t0 )n ;
y¡1 (1+2t¡3 x¡2 y¡1 z0 )n
;
(1+t¡3 x¡2 y¡1 z0 )n
7
y4n = y0 (1 + z¡3 t¡2 x¡1 y0 )n ;
z4n¡3 =
z4n¡1 =
t4n¡3 =
t4n¡1 =
z¡3
;
(1¡z¡3 t¡2 x¡1 y0 )n
z¡1
;
(1¡x¡3 y¡2 z¡1 t0 )n
t¡3
;
(1+t¡3 x¡2 y¡1 z0 )n
z¡2 (¡1+y¡3 z¡2 t¡1 x0 )n
;
(¡1+2y¡3 z¡2 t¡1 x0 )n
n
z0 (1 + t¡3 x¡2 y¡1 z0 ) ;
z4n¡2 =
z4n =
t4n¡2 = t¡2 (1 ¡ z¡3 t¡2 x¡1 y0 )n ;
t¡1 (¡1+2y¡3 z¡2 t¡1 x0 )n
;
(¡1+y¡3 z¡2 t¡1 x0 )n
t4n = t0 (1 ¡ x¡3 y¡2 z¡1 t0 )n :
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption
holds for n ¡ 1. That is
x4n¡7 =
x4n¡5 =
y4n¡7 =
y4n¡5 =
z4n¡7 =
z4n¡5 =
t4n¡7 =
t4n¡5 =
x¡3
;
(1+x¡3 y¡2 z¡1 t0 )n¡1
x¡1
;
(1+z¡3 t¡2 x¡1 y0 )n¡1
y¡3
;
(1¡y¡3 z¡2 t¡1 x0 )n¡1
x4n¡6 =
x4n¡4 = x0 (1 ¡ y¡3 z¡2 t¡1 x0 )n¡1 ;
y4n¡6 = y¡2 (1 + x¡3 y¡2 z¡1 t0 )n¡1 ;
y¡1 (1+2t¡3 x¡2 y¡1 z0 )n¡1
;
(1+t¡3 x¡2 y¡1 z0 )n¡1
z¡3
;
(1¡z¡3 t¡2 x¡1 y0 )n¡1
z¡1
;
(1¡x¡3 y¡2 z¡1 t0 )n¡1
t¡3
;
(1+t¡3 x¡2 y¡1 z0 )n¡1
x¡2 (1+t¡3 x¡2 y¡1 z0 )n¡1
;
(1+2t¡3 x¡2 y¡1 z0 )n¡1
y4n¡4 = y0 (1 + z¡3 t¡2 x¡1 y0 )n¡1 ;
z4n¡6 =
z4n¡4 =
z¡2 (¡1+y¡3 z¡2 t¡1 x0 )n¡1
;
(¡1+2y¡3 z¡2 t¡1 x0 )n¡1
z0 (1 + t¡3 x¡2 y¡1 z0 )n¡1
;
t4n¡6 = t¡2 (1 ¡ z¡3 t¡2 x¡1 y0 )n ;
t¡1 (¡1+2y¡3 z¡2 t¡1 x0 )n¡1
;
(¡1+y¡3 z¡2 t¡1 x0 )n¡1
t4n¡4 = t0 (1 ¡ x¡3 y¡2 z¡1 t0 )n¡1 :
It follows from system (2) that
x4n¡3 =
=
y4n¡2 =
=
z4n¡1 =
=
t4n =
=
x4n¡7
1+t4n¡4 z4n¡5 y4n¡6 x4n¡7
x¡3
(1+x¡3 y¡2 z¡1 t0 )n¡1
[1+x¡3 y¡2 z¡1 t0 ]
=
=
x¡3
:
(¡1+x¡3 y¡2 z¡1 t0 )n
y4n¡6
1¡x4n¡3 t4n¡4 z4n¡5 y4n¡6
n¡1
y·
¡2 (1+x¡3 y¡2 z¡1 t0 ) ¸
x¡3 y¡2 z¡1 t0
1¡
(1+x¡3 y¡2 z¡1 t0 )
=
=
z4n¡5
1¡y4n¡2 x4n¡3 t4n¡4 z4n¡5
z¡1
(1¡x¡3 y¡2 z¡1 t0 )n¡1
[1¡x¡3 y¡2 z¡1 t0 ]
=
=
"
y¡2 (1+x¡3 y¡2 z¡1 t0 )n¡1
#
x¡3 t0 (1¡x¡3 y¡2 z¡1 t0 )n¡1 z¡1 y¡2 (1+x¡3 y¡2 z¡1 t0 )n¡1
1¡
n
(1+x¡3 y¡2 z¡1 t0 )
(1¡x¡3 y¡2 z¡1 t0 )n¡1
n¡1
y¡2
·(1+x¡3 y¡2 z¡1 t0 ) ¸
1
(1+x¡3 y¡2 z¡1 t0 )
= y¡2 (1 + x¡3 y¡2 z¡1 t0 )n :
z¡1
(1¡x
y
z¡1 t0 )n¡1
¡3
¡2
"
#
n
x¡3 y¡2 (1+x¡3 y¡2 z¡1 t0 ) z¡1 t0 (1¡x¡3 y¡2 z¡1 t0 )n¡1
1¡
n
(1+x¡3 y¡2 z¡1 t0 )
(1¡x¡3 y¡2 z¡1 t0 )n¡1
z¡1
:
(¡1+x¡3 y¡2 z¡1 t0 )n
t4n¡4
1+z4n¡1 y4n¡2 x4n¡3 t4n¡4
t·0 (1¡x¡3 y¡2 z¡1 t0 )n¡1¸
x¡3 y¡2 z¡1 t0
1+
(1¡x¡3 y¡2 z¡1 t0 )
x¡3
(1+x
y
z¡1 t0 )n¡1
¡3
¡2
"
#
n¡1
z¡1 t0 (1¡x¡3 y¡2 z¡1 t0 )
x¡3 y¡2 (1+x¡3 y¡2 z¡1 t0 )n¡1
1+
(1¡x¡3 y¡2 z¡1 t0 )n¡1
(1+x¡3 y¡2 z¡1 t0 )n¡1
=
=
"
t0 (1¡x¡3 y¡2 z¡1 t0 )n¡1
#
z¡1 y¡2 (1+x¡3 y¡2 z¡1 t0 )n x¡3 t0 (1¡x¡3 y¡2 z¡1 t0 )n¡1
1+
(1¡x¡3 y¡2 z¡1 t0 )n
(1+x¡3 y¡2 z¡1 t0 )n
n¡1
t0 (1¡x
· ¡3 y¡2 z¡1 t0 ) ¸
1
1¡x¡3 y¡2 z¡1 t0
8
= t0 (1 ¡ x¡3 y¡2 z¡1 t0 )n :
Also, we can prove the other relations similarly. The proof is complete.
Theorem 3 Let fxn ; yn ; zn ; tn g are solutions of the system
xn¡3
yn¡3
; yn+1 = ¡1¡xn tn¡1
;
¡1+tn zn¡1 yn¡2 xn¡3
zn¡2 yn¡3
zn¡3
tn¡3
; tn+1 = 1+zn yn¡1
;
1+yn xn¡1 tn¡2 zn¡3
xn¡2 tn¡3
xn+1 =
zn+1 =
(3)
with the initial values are arbitrary real numbers satis…es x¡3 y¡2 z¡1 t0 6= 1; x¡3 y¡2 z¡1 t0 6=
1
; z¡3 t¡2 x¡1 y0 6= ¡1; z¡3 t¡2 x¡1 y0 6= ¡ 12 ; t¡3 x¡2 y¡1 z0 6= §1; y¡3 z¡2 t¡1 x0 6= §1.
2
Then the solution are given by the following formulae for n = 0; 1; 2; :::;
x4n¡3 =
x4n¡1 =
y4n¡3 =
y4n¡1 =
z4n¡3 =
z4n¡1 =
t4n¡3 =
t4n¡1 =
x¡3
;
(¡1+x¡3 y¡2 z¡1 t0 )n
x4n¡2 = (¡1)n x¡2 (1 + t¡3 x¡2 y¡1 z0 )n ;
(¡1)n x¡1 (1+2z¡3 t¡2 x¡1 y0 )n
;
(1+z¡3 t¡2 x¡1 y0 )n
x4n = x0 (¡1 + y¡3 z¡2 t¡1 x0 )n ;
(¡1)n y¡3
(¡1)n y¡2 (¡1+x¡3 y¡2 z¡1 t0 )n
;
n ; y4n¡2 =
(1+y¡3 z¡2 t¡1 x0 )
(¡1+2x¡3 y¡2 z¡1 t0 )n
y¡1
; y4n = (¡1)n y0 (1 + z¡3 t¡2 x¡1 y0 )n
(¡1+t¡3 x¡2 y¡1 z0 )n
z¡3
; z4n¡2 = z¡2 (1 + y¡3 z¡2 t¡1 x0 )n ;
(1+z¡3 t¡2 x¡1 y0 )n
(¡1)n z¡1 (1¡2x¡3 y¡2 z¡1 t0 )n
;
(¡1+x¡3 y¡2 z¡1 t0 )n
t¡3
;
(1+t¡3 x¡2 y¡1 z0 )n
t4n¡2 =
(¡1)n t¡1
;
(¡1+y¡3 z¡2 t¡1 x0 )n
;
z4n = (¡1)n z0 (¡1 + t¡3 x¡2 y¡1 z0 )n ;
t¡2 (1+z¡3 t¡2 x¡1 y0 )n
;
(1+2z¡3 t¡2 x¡1 y0 )n
t4n = (¡1)n t0 (¡1 + x¡3 y¡2 z¡1 t0 )n :
Proof: As the proof of Theorem 2.
Here for con…rming the results of this section, we consider an interesting numerical
examples of the systems (1) - (2).
Example 1. We consider the system (1) with the initial conditions x¡3 = 0:6; x¡2 =
3; x¡1 = 0:9; x0 = 1:3; y¡3 = 2; y¡2 = 1:3; y¡1 = ¡0:5; y0 = 0:1; z¡3 = 1:1; z¡2 =
0:6; z¡1 = ¡0:7; z0 = 1:5; t¡3 = ¡2; t¡2 = 0:9; t¡1 = ¡3 and t0 = 0:8. (See Fig. 1).
8
x(n)
y(n)
z(n)
t(n)
6
x(n),y(n),z(n),t(n)
4
2
0
-2
-4
-6
-8
0
5
10
15
20
25
n
30
35
40
45
Figure 1. Plot the behavior of the solution of system (1).
9
50
Example 2. See Figure (2) for an example for the system (2) with the initial values
x¡3 = 0:46; x¡2 = 0:23; x¡1 = 0:29; x0 = 1:16; y¡3 = 0:2; y¡2 = 1:3; y¡1 = ¡0:5;
y0 = 0:61; z¡3 = 0:21; z¡2 = 0:26; z¡1 = 0:27; z0 = 1:85; t¡3 = 0:2; t¡2 =
0:09; t¡1 = 0:28 and t0 = 0:58.
3
x(n)
y(n)
z(n)
t(n)
2.5
x(n),y(n),z(n),t(n)
2
1.5
1
0.5
0
-0.5
0
5
10
15
20
n
25
30
35
40
Figure 2. Plot the behavior of the solution of system (2).
3
Systems have a Periodic Solutions:
In this section, we investigate the solutions and periodic nature of the solutions of
the following system of four nonlinear di¤erence equations
xn+1 =
zn+1 =
xn¡3
;
¡1¡tn zn¡1 yn¡2 xn¡3
zn¡3
;
¡1¡yn xn¡1 tn¡2 zn¡3
yn+1 =
tn+1 =
yn¡3
;
¡1¡xn tn¡1 zn¡2 yn¡3
tn¡3
;
¡1¡zn yn¡1 xn¡2 tn¡3
(4)
where n 2 N0 and the initial conditions are arbitrary real numbers.
Theorem 4 Assume that x¡3 y¡2 z¡1 t0 6= ¡1; t¡3 x¡2 y¡1 z0 6= ¡1; z¡3 t¡2 x¡1 y0 6= ¡1;
y¡3 z¡2 t¡1 x0 6= ¡1 and x¡3 y¡2 z¡1 t0 6= ¡2; t¡3 x¡2 y¡1 z0 6= ¡2; z¡3 t¡2 x¡1 y0 6= ¡2;
y¡3 z¡2 t¡1 x0 6= ¡2; then all solutions of the system (4) are unbounded and given by
the expressions
x4n¡3 =
x4n¡1 =
y4n¡3 =
y4n¡1 =
z4n¡3 =
z4n¡1 =
t4n¡3 =
t4n¡1 =
x¡3
;
(¡1¡x¡3 y¡2 z¡1 t0 )n
x¡1
;
(¡1¡z¡3 t¡2 x¡1 y0 )n
y¡3
;
(¡1¡y¡3 z¡2 t¡1 x0 )n
y¡1
;
(¡1¡t¡3 x¡2 y¡1 z0 )n
z¡3
;
(¡1¡z¡3 t¡2 x¡1 y0 )n
z¡1
;
(¡1¡x¡3 y¡2 z¡1 t0 )n
t¡3
;
(¡1¡t¡3 x¡2 y¡1 z0 )n
t¡1
;
(¡1¡y¡3 z¡2 t¡1 x0 )n
x4n¡2 = x¡2 (¡1 ¡ t¡3 x¡2 y¡1 z0 )n ;
x4n = x0 (¡1 ¡ y¡3 z¡2 t¡1 x0 )n ;
y4n¡2 = y¡2 (¡1 ¡ x¡3 y¡2 z¡1 t0 )n ;
y4n = y0 (¡1 ¡ z¡3 t¡2 x¡1 y0 )n ;
z4n¡2 = z¡2 (¡1 ¡ y¡3 z¡2 t¡1 x0 )n ;
z4n = z0 (¡1 ¡ t¡3 x¡2 y¡1 z0 )n ;
t4n¡2 = t¡2 (¡1 ¡ z¡3 t¡2 x¡1 y0 )n ;
t4n = t0 (¡1 ¡ x¡3 y¡2 z¡1 t0 )n :
10
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption
holds for n ¡ 1. That is
x4n¡7 =
x4n¡5 =
y4n¡7 =
y4n¡5 =
z4n¡7 =
z4n¡5 =
t4n¡7 =
t4n¡5 =
x¡3
;
(¡1¡x¡3 y¡2 z¡1 t0 )n¡1
x¡1
;
(¡1¡z¡3 t¡2 x¡1 y0 )n¡1
y¡3
;
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
y¡1
;
(¡1¡t¡3 x¡2 y¡1 z0 )n¡1
z¡3
;
(¡1¡z¡3 t¡2 x¡1 y0 )n¡1
z¡1
;
(¡1¡x¡3 y¡2 z¡1 t0 )n¡1
t¡3
;
(¡1¡t¡3 x¡2 y¡1 z0 )n¡1
t¡1
;
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
x4n¡6 = x¡2 (¡1 ¡ t¡3 x¡2 y¡1 z0 )n¡1 ;
x4n¡4 = x0 (¡1 ¡ y¡3 z¡2 t¡1 x0 )n¡1 ;
y4n¡6 = y¡2 (¡1 ¡ x¡3 y¡2 z¡1 t0 )n¡1 ;
y4n¡4 = y0 (¡1 ¡ z¡3 t¡2 x¡1 y0 )n¡1 ;
z4n¡6 = z¡2 (¡1 ¡ y¡3 z¡2 t¡1 x0 )n¡1 ;
z4n¡4 = z0 (¡1 ¡ t¡3 x¡2 y¡1 z0 )n¡1 ;
t4n¡6 = t¡2 (¡1 ¡ z¡3 t¡2 x¡1 y0 )n¡1 ;
t4n¡4 = t0 (¡1 ¡ x¡3 y¡2 z¡1 t0 )n¡1 :
It follows from system (4) that
x4n =
=
y4n¡3 =
=
z4n¡2 =
=
t4n¡1 =
=
x4n¡4
¡1¡t4n¡1 z4n¡2 y4n¡3 x4n¡4
n¡1
x
·0 (¡1¡y¡3 z¡2 t¡1 x0 ) ¸
y¡3 z¡2 t¡1 x0
¡1¡
¡1¡y¡3 z¡2 t¡1 x0
=
y4n¡7
¡1¡x4n¡4 t4n¡5 z4n¡6 y4n¡7
y¡3
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
(¡1¡y¡3 z¡2 t¡1 x0 )
=
z4n¡6
¡1¡y4n¡3 x4n¡4 t4n¡5 z4n¡6
n¡1
z·
¡2 (¡1¡y¡3 z¡2 t¡1 x0 ) ¸
y¡3 x0 z¡2 t¡1
¡1¡
¡1¡y¡3 z¡2 t¡1 x0
t4n¡5
¡1¡z4n¡2 y4n¡3 x4n¡4 t4n¡5
t¡1
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
(¡1¡y¡3 z¡2 t¡1 x0 )
=
=
x0 (¡1¡y¡3 z¡2 t¡1 x0 )n¡1
#
t¡1 z¡2 (¡1¡y¡3 z¡2 t¡1 x0 )n y¡3 x0 (¡1¡y¡3 z¡2 t¡1 x0 )n¡1
¡1¡
n
n
(¡1¡y¡3 z¡2 t¡1 x0 )
(¡1¡y¡3 z¡2 t¡1 x0 )
"
n¡1
x0 (¡1¡y
¡3 z¡2 t¡1 x0 ) ¸
·
1
¡1¡y¡3 z¡2 t¡1 x0
=
= x0 (¡1 ¡ y¡3 z¡2 t¡1 x0 )n ;
y¡3
n¡1
(¡1¡y
z
¡3
¡2 t¡1 x0 )
"
#
n¡1
t¡1 x0 (¡1¡y¡3 z¡2 t¡1 x0 )
y¡3 z¡2 (¡1¡y¡3 z¡2 t¡1 x0 )n¡1
¡1¡
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
y¡3
;
(¡1¡y¡3 z¡2 t¡1 x0 )n
=
=
=
z¡2 (¡1¡y¡3 z¡2 t¡1 x0 )n¡1
#
y¡3 x0 (¡1¡y¡3 z¡2 t¡1 x0 )n¡1 t¡1 z¡2 (¡1¡y¡3 z¡2 t¡1 x0 )n¡1
¡1¡
n
(¡1¡y¡3 z¡2 t¡1 x0 )
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
"
n¡1
z¡2·(¡1¡y¡3 z¡2 t¡1 x0 )¸
1
¡1¡y¡3 z¡2 t¡1 x0
= z¡2 (¡1 ¡ y¡3 z¡2 t¡1 x0 )n ;
t¡1
n¡1
(¡1¡y
z
¡3
¡2 t¡1 x0 )
"
#
n
y¡3 z¡2 (¡1¡y¡3 z¡2 t¡1 x0 ) t¡1 x0 (¡1¡y¡3 z¡2 t¡1 x0 )n¡1
¡1¡
(¡1¡y¡3 z¡2 t¡1 x0 )n
(¡1¡y¡3 z¡2 t¡1 x0 )n¡1
t¡1
:
(¡1¡y¡3 z¡2 t¡1 x0 )n
Also, we can prove the other relations similarly. The proof is complete.
Theorem 5 If the sequences fxn ; yn ; zn ; tn g are solutions of di¤erence equation system (4) such that x¡3 y¡2 z¡1 t0 = t¡3 x¡2 y¡1 z0 = z¡3 t¡2 x¡1 y0 = y¡3 z¡2 t¡1 x0 = ¡2;
11
then all solutions of the system are periodic with period four and takes the form
x4n¡3
y4n¡3
z4n¡3
t4n¡3
=
=
=
=
x¡3 ;
y¡3 ;
z¡3 ;
t¡3 ;
x4n¡2 = x¡2 ; x4n¡1 = x¡1 ; x4n = x0 ;
y4n¡2 = y¡2 ; y4n¡1 = y¡1 ; y4n = y0 ;
z4n¡2 = z¡2 ; z4n¡1 = z¡1 ; z4n = z0 ;
t4n¡2 = t¡2 ; t4n¡1 = t¡1 ; t4n = t0 :
Or
fxn g
fyn g
fzn g
ftn g
=
=
=
=
fx¡3 ;
fy¡3 ;
fz¡3 ;
ft¡3 ;
x¡2 ; x¡1 ; x0 ; x¡3 ; x¡2 ; :::g ;
y¡2 ; y¡1 ; y0 ; y¡3 ; y¡2 ; :::g ;
z¡2 ; z¡1 ; z0 ; z¡3 ; z¡2 ; :::g ;
t¡2 ; t¡1 ; t0 ; t¡3 ; t¡2 ; :::g :
Proof: The proof follows from the previous Theorem and will be omitted.
Example 3. We put the initial conditions as follows x¡3 = 0:6; x¡2 = 3; x¡1 = 0:9;
x0 = 1:3; y¡3 = 0:22; y¡2 = 1:3; y¡1 = ¡0:5; y0 = 0:1; z¡3 = 1:1; z¡2 = 0:6; z¡1 =
0:7; z0 = 1:5; t¡3 = 0:02; t¡2 = 0:09; t¡1 = ¡0:3 and t0 = 0:8; for the di¤erence
system (4), see Fig. 3.
6
x(n)
y(n)
z(n)
t(n)
5
4
x(n),y(n),z(n),t(n)
3
2
1
0
-1
-2
-3
-4
0
2
4
6
8
10
n
12
14
16
18
20
Figure 3. Draw of the behavior of the solution of system (4).
Example 4. Figure (4) shows the periodicity behavior of the solution of the di¤erence
system (4) with the initial conditions x¡3 = 2; x¡2 = ¡0:5; x¡1 = 1; x0 = 4;
y¡3 = ¡5; y¡2 = 10; y¡1 = 2; y0 = 0:1; z¡3 = 15; z¡2 = 0:6; z¡1 = ¡0:7; z0 = 1;
t¡3 = 2; t¡2 = ¡4=3; t¡1 = 1=6 and t0 = 1=7.
12
15
x(n)
y(n)
z(n)
t(n)
x(n),y(n),z(n),t(n)
10
5
0
-5
0
5
10
15
n
20
25
30
Figure 4. Plot the periodicity of the solution of system (4).
The following theorems can be proved similarly.
4
Other Systems:
In this section, we get the solutions of the following systems of the di¤erence equations
xn+1 =
zn+1 =
xn+1 =
zn+1 =
xn+1 =
zn+1 =
xn+1 =
zn+1 =
xn+1 =
zn+1 =
xn¡3
;
1¡tn zn¡1 yn¡2 xn¡3
zn¡3
;
1¡yn xn¡1 tn¡2 zn¡3
yn+1 =
xn¡3
;
¡1+tn zn¡1 yn¡2 xn¡3
zn¡3
;
¡1¡yn xn¡1 tn¡2 zn¡3
yn+1 =
xn¡3
;
1+tn zn¡1 yn¡2 xn¡3
zn¡3
;
1+yn xn¡1 tn¡2 zn¡3
yn+1 =
xn¡3
;
1¡tn zn¡1 yn¡2 xn¡3
zn¡3
;
1+yn xn¡1 tn¡2 zn¡3
yn+1 =
tn+1 =
tn+1 =
tn+1 =
tn+1 =
yn¡3
;
1¡xn tn¡1 zn¡2 yn¡3
tn¡3
:
1¡zn yn¡1 xn¡2 tn¡3
yn¡3
;
¡1+xn tn¡1 zn¡2 yn¡3
tn¡3
:
¡1¡zn yn¡1 xn¡2 tn¡3
yn¡3
;
1¡xn tn¡1 zn¡2 yn¡3
tn¡3
:
1¡zn yn¡1 xn¡2 tn¡3
yn¡3
;
1+xn tn¡1 zn¡2 yn¡3
tn¡3
:
1¡zn yn¡1 xn¡2 tn¡3
xn¡3
yn¡3
; yn+1 = ¡1+xn tn¡1
;
¡1+tn zn¡1 yn¡2 xn¡3
zn¡2 yn¡3
zn¡3
tn¡3
; tn+1 = 1+zn yn¡1
:
1¡yn xn¡1 tn¡2 zn¡3
xn¡2 tn¡3
where n 2 N0 and the initial conditions are arbitrary real numbers.
13
(5)
(6)
(7)
(8)
(9)
Theorem 6 If fxn ; yn ; zn ; tn g are solutions of di¤erence equation system (5). Then
for n = 0; 1; 2; :::;
n¡1
Q
x4n¡3 = x¡3
x4n¡1 = x¡1
y4n¡3 = y¡3
y4n¡1 = y¡1
z4n¡3 = z¡3
i=0
n¡1
Q
i=0
n¡1
Q
i=0
n¡1
Q
i=0
n¡1
Q
z4n¡1 = z¡1
t4n¡3 = t¡3
t4n¡1 = t¡1
where
¡1
Q
i=0
n¡1
Q
i=0
n¡1
Q
i=0
n¡1
Q
i=0
(¡1+(4i)x¡3 y¡2 z¡1 t0 )
;
(¡1+(4i+1)x¡3 y¡2 z¡1 t0 )
x4n¡2 = x¡2
x4n = x0
(¡1+(4i)y¡3 z¡2 t¡1 x0 )
;
(¡1+(4i+1)y¡3 z¡2 t¡1 x0 )
y4n¡2 = y¡2
n¡1
Q
y4n = y0
(¡1+(4i)z¡3 t¡2 x¡1 y0 )
;
(¡1+(4i+1)z¡3 t¡2 x¡1 y0 )
z4n¡2 = z¡2
i=0
(¡1+(4i)t¡3 x¡2 y¡1 z0 )
;
(¡1+(4i+1)t¡3 x¡2 y¡1 z0 )
t4n¡2 = t¡2
t4n = t0
n¡1
Q
i=0
n¡1
Q
z4n = z0
i=0
(¡1+(4i+1)y¡3 z¡2 t¡1 x0 )
;
(¡1+(4i+2)y¡3 z¡2 t¡1 x0 )
(¡1+(4i+3)t¡3 x¡2 y¡1 z0 )
;
(¡1+(4i+4)t¡3 x¡2 y¡1 z0 )
i=0
n¡1
Q
n¡1
Q
(¡1+(4i+1)x¡3 y¡2 z¡1 t0 )
;
(¡1+(4i+2)x¡3 y¡2 z¡1 t0 )
(¡1+(4i+3)z¡3 t¡2 x¡1 y0 )
;
(¡1+(4i+4)z¡3 t¡2 x¡1 y0 )
i=0
(¡1+(4i+2)x¡3 y¡2 z¡1 t0 )
;
(¡1+(4i+3)x¡3 y¡2 z¡1 t0 )
(¡1+(4i+1)t¡3 x¡2 y¡1 z0 )
;
(¡1+(4i+2)t¡3 x¡2 y¡1 z0 )
(¡1+(4i+3)y¡3 z¡2 t¡1 x0 )
;
(¡1+(4i+4)y¡3 z¡2 t¡1 x0 )
i=0
n¡1
Q
(¡1+(4i+2)t¡3 x¡2 y¡1 z0 )
;
(¡1+(4i+3)t¡3 x¡2 y¡1 z0 )
(¡1+(4i+2)y¡3 z¡2 t¡1 x0 )
;
(¡1+(4i+3)y¡3 z¡2 t¡1 x0 )
i=0
n¡1
Q
(¡1+(4i+2)z¡3 t¡2 x¡1 y0 )
;
(¡1+(4i+3)z¡3 t¡2 x¡1 y0 )
n¡1
Q
i=0
(¡1+(4i+1)z¡3 t¡2 x¡1 y0 )
;
(¡1+(4i+2)z¡3 t¡2 x¡1 y0 )
(¡1+(4i+3)x¡3 y¡2 z¡1 t0 )
:
(¡1+(4i+4)x¡3 y¡2 z¡1 t0 )
Ai = 1:
i=0
Lemma 2. If xi ; yi ; zi ; ti ; i = ¡3; ¡2; ¡1; 0 arbitrary real numbers and let
fxn ; yn ; zn ; tn g are solutions of system (1) then the following statements are true:(i) If x¡3 = 0; then we have x4n¡3 = 0 and y4n¡2 = y¡2 ; z4n¡1 = z¡1 ; t4n = t0 :
(ii) If x¡2 = 0; then we have x4n¡2 = 0 and y4n¡1 = y¡1 ; z4n = z0 ; t4n¡3 = t¡3 :
(iii) If x¡1 = 0; then we have x4n¡1 = 0 and y4n = y0 ; z4n¡3 = z¡3 ; t4n¡2 = t¡2 :
(iv) If x0 = 0; then we have x4n = 0 and y4n¡3 = y¡3 ; z4n¡2 = z¡2 ; t4n¡1 = t¡1 :
(v) If y¡3 = 0; then we have y4n¡3 = 0 and x4n = x0 ; z4n¡2 = z¡2 ; t4n¡1 = t¡1 :
(vi) If y¡2 = 0; then we have y4n¡2 = 0 and x4n¡3 = x¡3 ; z4n¡1 = z¡1 ; t4n = t0 :
(vii) If y¡1 = 0; then we have y4n¡1 = 0 and x4n¡2 = x¡2 ; z4n = z0 ; t4n¡3 = t¡3 :
(viii) If y0 = 0; then we have y4n = 0 and x4n¡1 = x¡1 ; z4n¡3 = z¡3 ; t4n¡2 = t¡2 :
(ix) If z¡3 = 0; then we have z4n¡3 = 0 and x4n¡1 = x¡1 ; y4n = y0 ; t4n¡2 = t¡2 :
(x) If z¡2 = 0; then we have z4n¡2 = 0 and x4n = x0 ; y4n¡3 = y¡3 ; t4n¡1 = t¡1 :
(xi) If z¡1 = 0; then we have z4n¡1 = 0 and x4n¡3 = x¡3 ; y4n¡2 = y¡2 ; t4n = t0 .
(xii) If z0 = 0; then we have z4n = 0 and x4n¡2 = x¡2 ; y4n¡1 = y¡1 ; t4n¡3 = t¡3 .
(xiii) If t¡3 = 0; then we have t4n¡3 = 0 and x4n¡2 = x¡2 ; y4n¡1 = y¡1 ; z4n = z0 :
(ivx) If t¡2 = 0; then we have t4n¡2 = 0 and x4n¡1 = x¡1 ; y4n = y0 ; z4n¡3 = z¡3 :
14
(vx) If t¡1 = 0; then we have t4n¡1 = 0 and x4n = x0 ; y4n¡3 = y¡3 ; z4n¡2 = z¡2 .
(vxi) If t0 = 0; then we have t4n = 0 and x4n¡3 = x¡3 ; y4n¡2 = y¡2 ; z4n¡1 = z¡1 .
Theorem 7 The form of the solutions of system (6) are given by the following formulae:
x4n¡3 =
x4n¡1 =
y4n¡3 =
y4n¡1 =
z4n¡3 =
z4n¡1 =
t4n¡3 =
t4n¡1 =
x¡3
;
(¡1+x¡3 y¡2 z¡1 t0 )n
x¡1
;
(¡1+z¡3 t¡2 x¡1 y0 )n
y¡3
;
(¡1+y¡3 z¡2 t¡1 x0 )n
(¡1)n x¡2 (1+t¡3 x¡2 y¡1 z0 )n
;
(1+2t¡3 x¡2 y¡1 z0 )n
n
x0 (¡1 + y¡3 z¡2 t¡1 x0 ) ;
x4n¡2 =
x4n =
y4n¡2 = y¡2 (¡1 + x¡3 y¡2 z¡1 t0 )n ;
(¡1)n y¡1 (1+2t¡3 x¡2 y¡1 z0 )n
;
(1+t¡3 x¡2 y¡1 z0 )n
y4n = y0 (¡1 + z¡3 t¡2 x¡1 y0 )n ;
(¡1)n z¡2 (¡1+y¡3 z¡2 t¡1 x0 )n
;
(¡1+2y¡3 z¡2 t¡1 x0 )n
(¡1)n z¡3
;
(1+z¡3 t¡2 x¡1 y0 )n
z4n¡2 =
(¡1)n t¡3
;
(1+t¡3 x¡2 y¡1 z0 )n
t4n¡2 = (¡1)n t¡2 (¡1 + z¡3 t¡2 x¡1 y0 )n ;
(¡1)n z¡1
;
(1+x¡3 y¡2 z¡1 t0 )n
z4n = (¡1)n z0 (1 + t¡3 x¡2 y¡1 z0 )n ;
(¡1)n t¡1 (¡1+2y¡3 z¡2 t¡1 x0 )n
;
(¡1+y¡3 z¡2 t¡1 x0 )n
t4n = (¡1)n t0 (1 + x¡3 y¡2 z¡1 t0 )n ;
where x¡3 y¡2 z¡1 t0 6= §1; z¡3 t¡2 x¡1 y0 6= §1; t¡3 x¡2 y¡1 z0 6= ¡1; t¡3 x¡2 y¡1 z0 6= ¡ 12 ,
y¡3 z¡2 t¡1 x0 6= 1, y¡3 z¡2 t¡1 x0 6= 12 .
Theorem 8 Let fxn ; yn ; zn ; tn g are solutions of di¤erence equation system (7) with
x¡3 y¡2 z¡1 t0 6= ¡1; y¡3 z¡2 t¡1 x0 6= 1; t¡3 x¡2 y¡1 z0 6= 1; z¡3 t¡2 x¡1 y0 6= ¡1; then for
n = 0; 1; 2; :::;
x4n¡3 =
x4n¡1 =
y4n¡3 =
y4n¡1 =
z4n¡3 =
z4n¡1 =
t4n¡3 =
t4n¡1 =
x¡3
;
(1+x¡3 y¡2 z¡1 t0 )n
x¡1
;
(1+z¡3 t¡2 x¡1 y0 )n
x4n¡2 = (¡1)n x¡2 (¡1 + t¡3 x¡2 y¡1 z0 )n ;
x4n = (¡1)n x0 (¡1 + y¡3 z¡2 t¡1 x0 )n ;
(¡1)n y¡3
; y4n¡2 = y¡2 (1 + x¡3 y¡2 z¡1 t0 )n ;
(¡1+y¡3 z¡2 t¡1 x0 )n
(¡1)n y¡1
; y4n = y0 (1 + z¡3 t¡2 x¡1 y0 )n ;
(¡1+t¡3 x¡2 y¡1 z0 )n
z¡3
; z4n¡2 = (¡1)n z¡2 (¡1 + y¡3 z¡2 t¡1 x0 )n
(1+z¡3 t¡2 x¡1 y0 )n
z¡1
; z4n = (¡1)n z0 (¡1 + t¡3 x¡2 y¡1 z0 )n ;
(1+x¡3 y¡2 z¡1 t0 )n
(¡1)n t¡3
; t4n¡2 = t¡2 (1 + z¡3 t¡2 x¡1 y0 )n ;
(¡1+t¡3 x¡2 y¡1 z0 )n
(¡1)n t¡1
;
(¡1+y¡3 z¡2 t¡1 x0 )n
;
t4n = t0 (1 + x¡3 y¡2 z¡1 t0 )n :
Theorem 9 Suppose that the initial conditions of the system (8) are arbitrary real
numbers satis…es x¡3 y¡2 z¡1 t0 6= §1; y¡3 z¡2 t¡1 x0 6= ¡1; y¡3 z¡2 t¡1 x0 6= ¡ 12 ; t¡3 x¡2 y¡1 z0 6=
1; t¡3 x¡2 y¡1 z0 6= 12 ; z¡3 t¡2 x¡1 y0 6= §1; and if fxn ; yn ; zn ; tn g are solutions of system
15
(8). Then for n = 0; 1; 2; :::;
x4n¡3 =
x4n¡1 =
y4n¡3 =
y4n¡1 =
z4n¡3 =
z4n¡1 =
t4n¡3 =
t4n¡1 =
(¡1)n x¡3
x¡2 (¡1+t¡3 x¡2 y¡1 z0 )n
;
n ; x4n¡2 =
(¡1+x¡3 y¡2 z¡1 t0 )
(¡1+2t¡3 x¡2 y¡1 z0 )n
n
(¡1) x¡1
; x4n = x0 (1 + y¡3 z¡2 t¡1 x0 )n ;
(¡1+z¡3 t¡2 x¡1 y0 )n
y¡3
; y4n¡2 = (¡1)n y¡2 (¡1 + x¡3 y¡2 z¡1 t0 )n
(1+y¡3 z¡2 t¡1 x0 )n
n
y¡1 (¡1+2t¡3 x¡2 y¡1 z0 )
(¡1+t¡3 x¡2 y¡1 z0 )n
z¡3
;
(1+z¡3 t¡2 x¡1 y0 )n
z¡1
;
(1+x¡3 y¡2 z¡1 t0 )n
(¡1)n t¡3
(¡1+t¡3 x¡2 y¡1 z0 )n
;
; y4n = (¡1)n y0 (¡1 + z¡3 t¡2 x¡1 y0 )n ;
z¡2 (1+y¡3 z¡2 t¡1 x0 )n
;
(1+2y¡3 z¡2 t¡1 x0 )n
n
(¡1) z0 (¡1 + t¡3 x¡2 y¡1 z0 )n ;
z4n¡2 =
z4n =
; t4n¡2 = t¡2 (1 + z¡3 t¡2 x¡1 y0 )n ;
t¡1 (1+2y¡3 z¡2 t¡1 x0 )n
;
(1+y¡3 z¡2 t¡1 x0 )n
t4n = t0 (1 + x¡3 y¡2 z¡1 t0 )n :
Theorem 10 Assume that fxn ; yn ; zn ; tn g are solutions of the system (9), with x¡3 y¡2 z¡1 t0 6=
1; z¡3 t¡2 x¡1 y0 6= 1; y¡3 z¡2 t¡1 x0 6= 1; t¡3 x¡2 y¡1 z0 6= ¡1; x¡3 y¡2 z¡1 t0 6= 2; z¡3 t¡2 x¡1 y0 6=
2; y¡3 z¡2 t¡1 x0 6= 2; t¡3 x¡2 y¡1 z0 6= ¡2. Then for n = 0; 1; 2; :::;
x4n¡3 =
x4n¡1 =
y4n¡3 =
y4n¡1 =
z4n¡3 =
z4n¡1 =
t4n¡3 =
t4n¡1 =
x¡3
; x4n¡2 = x¡2 (¡1 ¡ t¡3 x¡2 y¡1 z0 )n ;
(¡1+x¡3 y¡2 z¡1 t0 )n
x¡1
; x4n = x0 (¡1 + y¡3 z¡2 t¡1 x0 )n ;
(¡1+z¡3 t¡2 x¡1 y0 )n
y¡3
; y4n¡2 = y¡2 (¡1 + x¡3 y¡2 z¡1 t0 )n ;
(¡1+y¡3 z¡2 t¡1 x0 )n
y¡1
; y4n = y0 (¡1 + z¡3 t¡2 x¡1 y0 )n ;
(¡1¡t¡3 x¡2 y¡1 z0 )n
z¡3
; z4n¡2 = z¡2 (1 ¡ y¡3 z¡2 t¡1 x0 )n ;
(1¡z¡3 t¡2 x¡1 y0 )n
z¡1
; z4n = z0 (1 + t¡3 x¡2 y¡1 z0 )n ;
(1¡x¡3 y¡2 z¡1 t0 )n
t¡3
; t4n¡2 = t¡2 (1 ¡ z¡3 t¡2 x¡1 y0 )n ;
(1+t¡3 x¡2 y¡1 z0 )n
t¡1
;
(1¡y¡3 z¡2 t¡1 x0 )n
t4n = t0 (1 ¡ x¡3 y¡2 z¡1 t0 )n :
Theorem 11 If the sequences fxn ; yn ; zn ; tn g are solutions of di¤erence equation system (9) such that x¡3 y¡2 z¡1 t0 = z¡3 t¡2 x¡1 y0 = y¡3 z¡2 t¡1 x0 = 2; t¡3 x¡2 y¡1 z0 = ¡2;
then fxn ; yn g are periodic with period four and fzn ; tn g are periodic with period eight
and takes the form
x4n¡3
y4n¡3
z4n¡3
t4n¡3
=
=
=
=
x¡3 ; x4n¡2 = x¡2 ; x4n¡1 = x¡1 ; x4n = x0 ;
y¡3 ; y4n¡2 = y¡2 ; y4n¡1 = y¡1 ; y4n = y0 ;
(¡1)n z¡3 ; z4n¡2 = (¡1)n z¡2 ; z4n¡1 = (¡1)n z¡1 ; z4n = (¡1)n z0 ;
(¡1)n t¡3 ; t4n¡2 = (¡1)n t¡2 ; t4n¡1 = (¡1)n t¡1 ; t4n = (¡1)n t0 :
Example 5. Figure (5) shows the periodicity behavior of the solution of the di¤erence
system (9) with the initial conditions x¡3 = 2; x¡2 = ¡0:5; x¡1 = 1; x0 = 4;
16
y¡3 = ¡5; y¡2 = 3; y¡1 = ¡2; y0 = 0:1; z¡3 = 5; z¡2 = 0:6; z¡1 = ¡0:1; z0 = 1;
t¡3 = ¡2; t¡2 = 4; t¡1 = ¡1=6 and t0 = ¡10=3.
5
x(n)
y(n)
z(n)
t(n)
4
3
x(n),y(n),z(n),t(n)
2
1
0
-1
-2
-3
-4
-5
0
2
4
6
8
10
n
12
14
16
18
20
Figure 5. Plot the periodicity of the solution of system (9).
Acknowledgements
This article was funded by the Deanship of Scienti…c Research (DSR), King Abdulaziz University, Jeddah. The authors, therefore, acknowledge with thanks DSR
technical and …nancial support.
References
[1] B. Sroysang, Dynamics of a system of rational higher-order di¤erence equation,
Discrete Dyn. Nat. Soc., 2013, (2013), Article ID 179401, 5 pages.
[2] H. El-Metwally, Solutions form for some rational systems of di¤erence equations,
Discrete Dyn. Nat. Soc., 2013, (2013), Article ID 903593, 10 pages.
[3] Stevo Stević, Josef Diblík, Bratislav Iriµcanin and Zden¼ek Šmarda, On a solvable
system of rational di¤erence equations, J. Di¤erence Equ. Appl., 20, (2014),
811–825.
[4] Q. Din, On a system of fourth-order rational di¤erence equations, Acta Univ.
Apulensis, 39, (2014), 137-150.
[5] C. Cinar and I. Yalçinkaya, On the positive solutions of the di¤erence equation
system xn+1 = 1=zn ; yn+1 = yn =xn¡1 yn¡1 ; zn+1 = 1=xn¡1 ; J. Inst. Math. Comp.
Sci., 18, (2005), 91-93.
[6] M. M. El-Dessoky, M. Mansour and E. M. Elsayed, Solutions of some rational
systems of di¤erence equations, Utilitas Mathematica 92, (2013), 329-336.
17
[7] O. Zkan and A. S. Kurbanli, On a system of di¤erence equation, Discrete Dyn.
Nat. Soc., 2013, (2013) Article ID 970316, 7 pages.
[8] Y. Yazlik, D. T. Tollu, N. Taskara, On the Behaviour of Solutions for Some
Systems of Di¤erence Equations, J. Comp. Anal. Appl., 18 (1), (2015), 166-178.
[9] M. M. El-Dessoky and E. M. Elsayed, On a solution of system of three fractional
di¤erence equations, J. Comp. Anal. Appl., 19(4), (2015), 760-769.
[10] A. S. Kurbanli, On the behavior of solutions of the system of rational di¤erence equations: xn+1 = xn¡1 =xn¡1 yn ¡ 1; yn+1 = yn¡1 =yn¡1 xn ¡ 1; and zn+1 =
xn =zn¡1 yn ¡ 1; Discrete Dyn. Nat. Soc., 2011, (2011), Article ID 932362, 12
pages.
[11] Lin-Xia Hu and Xiu-Mei Jia, Global Asymptotic Stability of a Rational System,
Abstr. Appl. Anal., 2014, (2014), Article ID 286375, 6 pages.
[12] C. Wang, Shu. Wang and W. Wang, Global asymptotic stability of equilibrium
point for a family of rational di¤erence equations, Appl. Math. Lett., 24 (5),
(2011), 714-718.
[13] T. F. Ibrahim and N. Touafek, On a third order rational di¤erence equation
with variable coe¢cients, Dyn. Contin. Discrete Impuls. Syst. Ser. B Appl. Algorithms, 20, (2013) 251-264.
[14] E. M. Elsayed and H. A. El-Metwally, On the solutions of some nonlinear systems
of di¤erence equations, Adv. Di¤er. Equ. 2013, 2013: 16.
[15] Q. Din, T. F. Ibrahim and K. A. Khan, Behavior of a Competitive System of
Second-Order Di¤erence Equations, The Scienti…c World Journal, 2014, (2014),
Article ID 283982, 9 pages.
[16] A. Q. Khan, M. N. Qureshiand Q. Din, Global dynamics of some systems of
higher-order rational di¤erence equations, Adv. Di¤er. Equ. 2013, 2013: 354.
[17] A. Q. Khan, M. N. Qureshi, Global dynamics of some systems of rational di¤erence equations, J. Egypt. Math. Soc., 24 (1), (2016), 30-36.
[18] A. Kurbanli, C. Cinar and M. Erdo¼gan, On the behavior of solutions of the
xn¡1
yn¡1
system of rational di¤erence equations xn+1 = xn¡1
; yn+1 = yn¡1
; zn+1 =
yn ¡1
xn ¡1
xn
. Appl. Math., 2, (2011), 1031-1038.
zn¡1 yn
[19] Hui-li Ma and Hui Feng, On Positive Solutions for the Rational Di¤erence EquaByn
tion Systems xn+1 = xnAy 2 and yn+1 = xn¡1
, International Scholarly Research
yn¡1
n
Notices, 2014, (2014), Article ID 857480, 4 pages.
18
[20] M. M. El-Dessoky, On the solutions and periodicity for some nonlinear systems
of di¤erence equations, J. Nonlinear Sci. Appl., 9 (5), (2016), 2190-2207.
[21] Miron B. Bekker, Martin J. Bohner, Hristo D. Voulov, Asymptotic behavior of
solutions of a rational system of di¤erence equations, J. Nonlinear Sci. Appl., 7,
(2014), 379-382.
[22] I. Yalcinkaya and C. Cinar, Global asymptotic stability of two nonlinear di¤erence equations, Fasciculi Mathematici, 43, (2010), 171–180.
[23] L. Keying, Z. Zhongjian, L. Xiaorui and Li Peng, More on three-dimensional
systems of rational di¤erence equations, Discrete Dyn. Nat. Soc., 2011, (2011),
Article ID 178483, 9 pages.
[24] Q. Din, Global behavior of a plant-herbivore model, Adv. Di¤er. Equ. 2015,
2015: 119 DOI 10.1186/s13662-015-0458-y.
[25] E. M. E. Zayed and M. A. El-Moneam, On the global attractivity of two nonlinear
di¤erence equations, J. Math. Sci., 177, (2011), 487-499.
[26] Q. Din, Asymptotic behavior of an anti-competitive system of second-order difference equations, J. Egypt. Math. Soc., 24 (1), (2016), 37-43.
[27] N. Touafek and E. M. Elsayed, On a second order rational systems of di¤erence
equations, Hokkaido Mathematical Journal, 44, (2015) 29–45.
[28] Q. Din, On a system of rational di¤erence equations, Demonstratio Math., XLVII
(2) (2014), 324-335.
[29] Stevo Stević, Boundedness and persistence of some cyclic-type systems of di¤erence equations, Appl. Math. Lett., 56, (2016), 78-85.
[30] R. Abu-Saris, C. Cinar, I. Yalcinkaya, On the asymptotic stability of xn+1 =
a+xn xn¡k
, Comput. Math. Appl., 56, (2008) 1172–1175.
xn +xn¡k
[31] Y. Halim, N. Touafek, Y. Yazlik, Dynamic behavior of a second-order nonlinear
rational di¤erence equation, Turkish J. Math., 39, (2015), 1004-1018
[32] R. Abo-Zeid, Attractivity of two nonlinear third order di¤erence equations, J.
Egypt. Math. Soc., 21, (2013), 241–247.
[33] M. M. El-Dessoky, E. M. Elsayed and M. Alghamdi, Solutions and periodicity
for some systems of fourth order rational di¤erence equations, J. Comp. Anal.
Appl., 18(1), (2015), 179-194.
[34] Qianhong Zhang, Jingzhong Liu and Zhenguo Luo, Dynamical Behavior of a
System of Third-Order Rational Di¤erence Equation, Discrete Dyn. Nat. Soc.,
2015, (2015), Article ID 530453, 6 pages.
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