J. COMPUTATIONAL ANALYSIS AND APPLICATIONS, VOL. 15, NO.5, 852-857, 2013, COPYRIGHT 2013 EUDOXUS PRESS, LLC
Form of solutions and periodicity for systems of di¤erence equations H. El-Metwally1 and E. M. Elsayed2 . 1 Department of Mathematics, Rabigh College of Science and Art, King Abdulaziz University, P.O. Box 344, Rabigh 21911, Saudi Arabia. 2 King Abdulaziz University, Faculty of Science, Mathematics Department, P. O. Box 80203, Jeddah 21589, Saudi Arabia. 1;2 Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt. 1 E-mail:
[email protected]. 2 E-mail:
[email protected]. Abstract This paper is devoted to get the form of the solutions and the periodic nature of the following systems of rational di¤erence equations xn+1 =
yn 2 yn 2 xn
1
with initial conditions.x
2;
x
1;
1 yn
x0 ; :y
;
yn+1 =
2;
y
1;
1
xn 2 xn 2 yn
1 xn
;
y0 are real numbers.
Keywords: di¤erence equations, recursive sequence, stability, periodic solution, system of di¤erence equations. Mathematics Subject Classi…cation: 39A10. ——————————————————
1
Introduction
Di¤erence equations is a hot topic in that they are widely used to investigate equations arising in mathematical models describing real-life situations such as population biology, probability theory, and genetics. Recently, rational di¤erence equations have appealed more and more scholars for their wide applications. For details, see [1-9]. However, there are few literatures on the systems of two or three rational di¤erence equations [10–17]. In this article, we investigate the behavior of the solutions for the following systems of di¤erence equations xn+1 = with initial conditions x
2
yn 2 yn 2 xn
1 2;
x
1;
1 yn
x0 ; :y
2;
;
yn+1 =
y
1;
1 xn
;
y0 are real numbers.
yn 2 1 yn 2 xn
First system: xn+1 =
xn 2 xn 2 yn
1
; 1 yn
xn 2 1+xn 2 yn
yn+1 =
1 xn
In this section, we investigate the solutions for the following system of di¤erence equations xn+1 =
yn 2 yn 2 xn
1
1 yn
;
yn+1 =
xn 1 + xn
2
2 yn 1 xn
;
(1)
where the initial conditions are arbitrary real numbers with x 2 y 1 x0 6= 1; 6= y 2 x 1 y0 6= 1. The following theorem is devoted to the form of the solutions for System (1).
1 2
and
Theorem 1 Every solution fxn ; yn g of System (1) is periodic with period twelve and, for n = 0; 1; 2; :::; has the form x12n
2
= x
2;
x12n
1
=x
1;
x12n = x0 ; x12n+1 =
y (1 + y
x12n+2
=
y0 y 1 (1 + x 2 y 1 x0 ) ; x12n+3 = (1 + 2x 2 y 1 x0 ) ( 1 + y 2x
x12n+4
=
x
x12n+8
=
y
2;
x12n+5 =
x
1;
x12n+6 =
1 y0 )
x0 ; x12n+7 =
1 (1
+ x 2 y 1 x0 ) y0 ; x12n+9 = (1 + 2x 2 y 1 x0 ) ( 1 + y 2x 1
852
1 y0 )
2
2 x 1 y0 )
;
;
; y 2 (1 + y 2 x
1 y0 )
;
EL-METWALLY, ELSAYED: SOLVING DIFFERENCE EQUATIONS
and y12n
x 2 ; (1 + x 2 y 1 x0 ) x0 (1 + 2x 2 y 1 x0 ) x 1 (1 + y 2 x 1 y0 ); y12n+3 = ; (1 + x 2 y 1 x0 ) y 2 ( 1 + y 2 x 1 y0 ) y 1 ; y12n+5 = ; (1 + y 2 x 1 y0 ) (1 + 2x 2 y 1 x0 ) x 2 (1 + 2x 2 y 1 x0 ) y0 (1 + y 2 x 1 y0 ) ; y12n+7 = ; ( 1 + y 2 x 1 y0 ) (1 + x 2 y 1 x0 ) x0 x 1 ( 1 + y 2 x 1 y0 ); y12n+9 = : (1 + x 2 y 1 x0 )
= y
2
y12n+2
=
y12n+4
=
y12n+6
=
y12n+8
=
2;
y12n
1
=y
1;
y12n = y0 ; y12n+1 =
Proof: For n = 0 the result holds. Now suppose that n > 0 and that our assumption holds for n 1, that is x12n
14
= x
2;
x12n
10
=
y 1 (1 + x 2 y 1 x0 ) ; (1 + 2x 2 y 1 x0 )
x12n
8
=
x
x12n
4
=
y
x12n
2;
=x
13
x12n
1;
=
7
x12n
x
1;
= x0 ; x12n
12
11
x12n
9
=
y0 ( 1 + y 2x
x12n
6
=
x0 ; x12n
1 (1
+ x 2 y 1 x0 ) ; x12n (1 + 2x 2 y 1 x0 )
3
y0 ( 1 + y 2x
=
y
=
(1 + y 1 y0 )
5
=
1 y0 )
2
2 x 1 y0 )
;
;
y 2 (1 + y 2 x
1 y0 )
;
;
and y12n
14
= y
2;
y12n
y12n
10
= x
1 (1
+y y
y12n
7
=
y12n
4
= x
(1+2x 1(
13
1;
2 x 1 y0 );
1
2y
=y
1 x0 )
1+y
y12n
y12n
; y12n
6
y12n
3
y12n
11
x0 (1+2x 2 y 1 x0 ) (1+x 2 y 1 x0 ) ;
y0 (1+y ( 1+y
=
2 x 1 y0 );
=
9
= y0 ;
12
2x
1 y0 )
2x
1 y0 )
=
; y12n x0 2y
(1 + x
=
x 2 (1 + x 2 y
y12n
x
=
5
1 x0 )
8
=
;
1 x0 ) y 2 ( 1+y 2 x 1 y0 ) ; (1+y 2 x 1 y0 )
2 (1+2x (1+x 2 y
2y
1 x0 )
1 x0 )
;
:
Now it follows from System (1) that x12n
2
= =
y12n
2
= =
1
x (1+x
2y
2 (1+2x
1 x0 )
1
4 y12n 3
2 x 1 y0 )
= 1
2y
1 x0 ) x 2 y 1 x0 (1+x 2 y 1 x0 )
x12n 5 1 + x12n 5 y12n 4 x12n y 2 (1 + y
2 (1+2x 2 y 1 x0 ) (1+x 2 y 1 x0 ) x 2 (1+2x 2 y 1 x0 ) y 1 (1+x 2 y 1 x0 ) (1+2x 2 y 1 x0 ) (1+x 2 y 1 x0 )
x
y12n 5 y12n 5 x12n
=x
x0 2 y 1 x0 )
2; y
= 3
(1+y
1+ (1+y
y 2 x 1 y0 (1+y 2 x 1 y0 )
1
(1+x
y
2 x 2 x 1 y0 )
=y
1(
2 2 x 1 y0 )
1+y
2x
1 y0 ) (
1+y
y0 2x
1 y0 )
2:
Similarly, we can prove the other relations. Then the proof is so complete. Example 1. Here we consider an interesting numerical example for System (1) with the initial conditions x 2 = 0:9, x 1 = 0:4, x0 = 0:3, y 2 = 5, y 1 = 7 and y0 = 2: (See Fig.1). plot of X(n+1)=Y(n-2)/(-1-Y(n-2)X(n-1)Y(n)),Y(n+1)=X(n-2)/(1+X(n-2)Y(n-1)X(n)) 8 X(n) Y(n) 6
4
x(n),y(n)
2
0
-2
-4
-6
0
5
10
15
20
25 n
30
Figure 1.
2
853
35
40
45
50
EL-METWALLY, ELSAYED: SOLVING DIFFERENCE EQUATIONS
3
yn 2 1 yn 2 xn
Second system: xn+1 =
xn 2 1+xn 2 yn
; yn+1 =
1 yn
1 xn
In this section, we study the solutions for the following system of di¤erence equations xn+1 =
yn 2 yn 2 xn
1
1 yn
;
xn 2 1 + xn 2 yn
yn+1 =
1 xn
;
(2)
where n 2 N0 and the initial conditions are arbitrary real numbers: Theorem 2 Assume that fxn ; yn g be a solution for System (2). Then for n = 0; 1; 2; :::; x6n
2
= x
2
n Y1
( 1+(6i)x 2 y 1 x0 )( 1+(6i+3)x 2 y 1 x0 ) ( 1+(6i+1)x 2 y 1 x0 )( 1+(6i+4)x 2 y 1 x0 ) ;
i=0
x6n
1
= x
1
n Y1
(1+(6i+1)y (1+(6i+2)y
2x
1 y0 )(1+(6i+4)y
2x
1 y0 )
2x
1 y0 )(1+(6i+5)y
2x
1 y0 )
( 1+(6i+2)x ( 1+(6i+3)x
2y
1 x0 )(
2y
1 x0 )(
;
i=0
x6n
= x0
n Y1
1+(6i+5)x 1+(6i+6)x
2y
1 x0 )
2y
1 x0 )
;
i=0
x6n+1
=
y 2 ( 1 y 2x
1 y0 )
n Y1
(1+(6i+3)y (1+(6i+4)y
2x
1 y0 )(1+(6i+6)y
2x
1 y0 )
2x
1 y0 )(1+(6i+7)y
2x
1 y0 )
;
i=0
x6n+2
y 1 ( 1+x 2 y 1 x0 ) ( 1+2x 2 y 1 x0 )
=
n Y1
( 1+(6i+4)x ( 1+(6i+5)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+7)x 1+(6i+8)x
2y
1 x0 )
2y
1 x0 )
;
i=0
x6n+3
y0 (1+2y 2 x 1 y0 ) (1+3y 2 x 1 y0 )
=
n Y1
(1+(6i+5)y (1+(6i+6)y
2x
1 y0 )(1+(6i+8)y
2x
1 y0 )
2x
1 y0 )(1+(6i+9)y
2x
1 y0 )
;
i=0
y6n
2
= y
2
n Y1
(1+(6i)y 2 x 1 y0 )(1+(6i+3)y 2 x 1 y0 ) (1+(6i+1)y 2 x 1 y0 )(1+(6i+4)y 2 x 1 y0 ) ;
i=0
y6n
1
= y
1
n Y1
( 1+(6i+1)x ( 1+(6i+2)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+4)x 1+(6i+5)x
2y
1 x0 )
2y
1 x0 )
;
i=0
y6n
= y0
n Y1
(1+(6i+2)y (1+(6i+3)y
2x
1 y0 )(1+(6i+5)y
2x
1 y0 )
2x
1 y0 )(1+(6i+6)y
2x
1 y0 )
;
i=0
y6n+1
=
x 2 ( 1+x 2 y
1 x0 )
n Y1
( 1+(6i+3)x ( 1+(6i+4)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+6)x 1+(6i+7)x
2y
1 x0 )
2y
1 x0 )
;
i=0
y6n+2
n Y1
x 1 (1+y 2 x 1 y0 ) (1+2y 2 x 1 y0 )
=
(1+(6i+4)y (1+(6i+5)y
2x
1 y0 )(1+(6i+7)y
2x
1 y0 )
2x
1 y0 )(1+(6i+8)y
2x
1 y0 )
;
i=0
y6n+3
x0 ( 1+2x 2 y 1 x0 ) ( 1+3x 2 y 1 x0 )
=
n Y1
( 1+(6i+5)x ( 1+(6i+6)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+8)x 1+(6i+9)x
2y
1 x0 )
2y
1 x0 )
:
i=0
Proof: For n = 0 the result holds. Now suppose that n > 1 and that our assumption holds for n 1; that is x6n
8
= x
2
n Y2
( 1+(6i)x 2 y 1 x0 )( 1+(6i+3)x 2 y 1 x0 ) ( 1+(6i+1)x 2 y 1 x0 )( 1+(6i+4)x 2 y 1 x0 ) ;
i=0
x6n
7
= x
1
n Y2
(1+(6i+1)y (1+(6i+2)y
2x
1 y0 )(1+(6i+4)y
2x
1 y0 )
2x
1 y0 )(1+(6i+5)y
2x
1 y0 )
( 1+(6i+2)x ( 1+(6i+3)x
2y
1 x0 )(
2y
1 x0 )(
;
i=0
x6n
6
= x0
n Y2
1+(6i+5)x 1+(6i+6)x
2y
1 x0 )
2y
1 x0 )
;
i=0
x6n
5
=
y 2 ( 1 y 2x
1 y0 )
n Y2
(1+(6i+3)y (1+(6i+4)y
2x
1 y0 )(1+(6i+6)y
2x
1 y0 )
2x
1 y0 )(1+(6i+7)y
2x
1 y0 )
;
i=0
x6n
4
=
y 1 ( 1+x 2 y 1 x0 ) ( 1+2x 2 y 1 x0 )
n Y2
( 1+(6i+4)x ( 1+(6i+5)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+7)x 1+(6i+8)x
2y
1 x0 )
2y
1 x0 )
i=0
x6n
3
=
y0 (1+2y 2 x 1 y0 ) (1+3y 2 x 1 y0 )
n Y2
(1+(6i+5)y (1+(6i+6)y
i=0
3
854
2x
1 y0 )(1+(6i+8)y
2x
1 y0 )
2x
1 y0 )(1+(6i+9)y
2x
1 y0 )
;
;
EL-METWALLY, ELSAYED: SOLVING DIFFERENCE EQUATIONS
y6n
= y
8
2
n Y2
(1+(6i)y 2 x 1 y0 )(1+(6i+3)y 2 x 1 y0 ) (1+(6i+1)y 2 x 1 y0 )(1+(6i+4)y 2 x 1 y0 ) ;
i=0
y6n
= y
7
1
n Y2
( 1+(6i+1)x ( 1+(6i+2)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+4)x 1+(6i+5)x
2y
1 x0 )
2y
1 x0 )
;
i=0
y6n
= y0
6
n Y2
(1+(6i+2)y (1+(6i+3)y
2x
1 y0 )(1+(6i+5)y
2x
1 y0 )
2x
1 y0 )(1+(6i+6)y
2x
1 y0 )
;
i=0
y6n
x 2 ( 1+x 2 y
=
5
n Y2
1 x0 )
( 1+(6i+3)x ( 1+(6i+4)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+6)x 1+(6i+7)x
2y
1 x0 )
2y
1 x0 )
;
i=0
y6n
x 1 (1+y 2 x 1 y0 ) (1+2y 2 x 1 y0 )
=
4
n Y2
(1+(6i+4)y (1+(6i+5)y
2x
1 y0 )(1+(6i+7)y
2x
1 y0 )
2x
1 y0 )(1+(6i+8)y
2x
1 y0 )
;
i=0
y6n
x0 ( 1+2x 2 y 1 x0 ) ( 1+3x 2 y 1 x0 )
=
3
n Y2
( 1+(6i+5)x ( 1+(6i+6)x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+8)x 1+(6i+9)x
2y
1 x0 )
2y
1 x0 )
:
i=0
It follows from System (2) that x6n
2
=
1
=
0
y6n 5 y6n 5 x6n
4 y6n 3
x 2 ( 1+x 2 y
1 x0 ) i=0
x
1 x0 )
y 1 ( 1+x 2 y 1 x0 ) ( 1+2x 2 y 1 x0 ) x0 ( 1+2x 2 y 1 x0 ) ( 1+3x 2 y 1 x0 )
x 2 ( 1+x 2 y
=
nQ2
x 2 ( 1+x 2 y
1
B B B B B B B B B B B @
2(
=
nQ2
1 x0 ) i=0
; ( 1+(6i+3)x ( 1+(6i+4)x
nQ2
i=0 nQ2
i=0 nQ2 i=0
2y
1 x0 )(
2y
1 x0 )(
( 1+(6i+3)x ( 1+(6i+4)x
2y
1+(6i+6)x 1+(6i+7)x
1 x0 )
1+(6i+6)x 2 y 1 x0 )( 1+(6i+7)x
( 1+(6i+4)x ( 1+(6i+5)x
2y
1 x0 )(
2y
1 x0 )(
( 1+(6i+5)x ( 1+(6i+6)x
2y
1 x0 )(
2y
1 x0 )(
nQ2
( 1+(6i+3)x ( 1+(6i+4)x (1 (6n 3)x 2 y 1 x0 x i=0
1 x0 ) 1 x0 )
1 x0 )(
2y
1 x0 )(
2y
1 x0 )(
2y
2y
1 x0 )
2y
1 x0 )
1+(6i+7)x 1+(6i+8)x
2y
1 x0 )
2y
1 x0 )
1+(6i+8)x 1+(6i+9)x
2y
1 x0 )
2y
1 x0 )
( 1+(6i+3)x 2 y 1 x0 )( 1+(6i+6)x ( 1+(6i+4)x 2 y 1 x0 )( 1+(6i+7)x x 2 y 1 x0 1 ( 1+(6n 3)x y x ) 2 1 0
1+(6n 3)x 2 y ( 1+x 2 y 1 x0 )
2y 2y
2y
1 x0 )
2y
1 x0 )
1+(6i+6)x 1+(6i+7)x
2y
1 x0 )
2y
1 x0 )
1 x0 )
1 C C C C C C C C C C C A
:
Then we see that x6n
2
=x
2
n Y1
( 1+(6i)x 2 y 1 x0 )( 1+(6i+3)x 2 y 1 x0 ) ( 1+(6i+1)x 2 y 1 x0 )( 1+(6i+4)x 2 y 1 x0 ) :
i=0
Again we see from System (2) that y6n
1
x6n 4 1 + x6n 4 y6n
=
3 x6n 2
y 1 ( 1+x 2 y 1 x0 ) ( 1+2x 2 y 1 x0 )
( 1+(6i+4)x 2 y 1 x0 )( 1+(6i+7)x 2 y ( 1+(6i+5)x 2 y 1 x0 )( 1+(6i+8)x 2 y i=0 n 2 y 1 ( 1+x 2 y 1 x0 ) Q ( 1+(6i+4)x 2 y 1 x0 )( 1+(6i+7)x ( 1+2x 2 y 1 x0 ) ( 1+(6i+5)x 2 y 1 x0 )( 1+(6i+8)x i=0 nQ2 x0 ( 1+2x 2 y 1 x0 ) ( 1+(6i+5)x 2 y 1 x0 )( 1+(6i+8)x ( 1+3x 2 y 1 x0 ) ( 1+(6i+6)x 2 y 1 x0 )( 1+(6i+9)x i=0 nQ1 ( 1+(6i)x 2 y 1 x0 )( 1+(6i+3)x 2 y 1 x0 ) x 2 ( 1+(6i+1)x 2 y 1 x0 )( 1+(6i+4)x 2 y 1 x0 ) i=0
0
=
y
B B B 1+B B B @
1 ( 1+x 2 y 1 x0 ) ( 1+2x 2 y 1 x0 )
=
nQ2 i=0
1+ y
1(
1+x
2y
( 1+(6i+4)x ( 1+(6i+5)x
Then 1
=y
1
n Y1
2y
1 x0 )(
2y
1 x0 )(
x 2 y 1 x0 ( 1+(6n 2)x 2 y
1 x0 )( 1+(6n 2)x ( 1+2x 2 y 1 x0 )
=
y6n
nQ2
2y
1 x0 )
1+(6i+7)x 1+(6i+8)x
2y
1 x0 )
2y
1 x0 )
1 x0 ) 1 x0 ) 2y
1 x0 )
2y
1 x0 )
2y
1 x0 )
2y
1 x0 )
1 C C C C C C A
1 x0 )
nQ2 i=0
( 1+(6i+4)x ( 1+(6i+5)x
1 + (6n
2)x
( 1+(6i+1)x ( 1+(6i+2)x
2y
1 x0 )(
2y
1 x0 )(
2 y 1 x0
+x
2y
1 x0 )(
2y
1 x0 )(
1+(6i+7)x 1+(6i+8)x
2y
1 x0 )
2y
1 x0 )
2 y 1 x0
1+(6i+4)x 1+(6i+5)x
2y
1 x0 )
2y
1 x0 )
:
i=0
Similarly we can prove the other relations. This completes the proof. Lemma 1. Let fxn ; yn g be a solution for System (2) with the initial conditions x x 1 , x0 , y 2 , y 1 , y0 2 R, then the following statements are true: 4
855
2,
EL-METWALLY, ELSAYED: SOLVING DIFFERENCE EQUATIONS
(i) If x 2 = 0; y 1 6= 0; x0 6= 0; then we have x6n 2 = y6n+1 = x0 ; x6n+2 = y 1 ; y6n 1 = y 1 ; y6n+3 = x0 : (ii) If x 1 = 0; y 2 6= 0; y0 6= 0; then we have x6n 1 = y6n+2 = 0 y 2 ; x6n+3 = y0 ; y6n 2 = y 2 ; y6n = y0 : (iii) If x0 = 0; y 1 6= 0; x 2 6= 0; then we have x6n = y6n+3 = 0 x 2 ; x6n+2 = y 1 ; y6n 1 = y 1 ; y6n+1 = x 2 : (iv) If y 2 = 0; x 1 6= 0; y0 6= 0; then we have y6n 2 = x6n+1 = 0 x 1 ; x6n+3 = y0 ; y6n = y0 ; y6n+2 = x 1 : (v) If y 1 = 0; x0 6= 0; x 2 6= 0; then we have y6n 1 = x6n+2 = 0 x 2 ; x6n = x0 ; y6n+1 = x 2 ; y6n+3 = x0 : (vi) If y0 = 0; y 2 6= 0; x 1 6= 0; then we have y6n = x6n+3 = 0 x 1 ; x6n+1 = y 2 ; y6n 2 = y 2 ; y6n+2 = x 1 :
0 and x6n = and x6n+1 = and x6n
2
=
and x6n
1
=
and x6n
2
=
and x6n
1
=
Proof: The proof follows by direct substitutions in the obtained form of the solutions for System (2) in Theorem 2. Example 2. Figure (2) shows the behavior of the solution for System (2) with the initial conditions x 2 = 5, x 1 = 0:4, x0 = 0:13, y 2 = 0:3, y 1 = 0:9 and y0 = 2. plot of X(n+1)=Y(n-2)/(-1-Y(n-2)X(n-1)Y(n)),Y(n+1)=X(n-2)/(-1+X(n-2)Y(n-1)X(n)) 5 x(n) y(n) 4
3
2
x(n),y(n)
1
0
-1
-2
-3
-4
0
10
20
30
40
50
60
70
n
Figure 2. The following theorems can be treated similarly to the previous results.
4
Third system: xn+1 =
yn 2 1 yn 2 xn
1 yn
; yn+1 =
xn 2 1 xn 2 yn
1 xn
In this section, we obtain the form of the solutions for the following system of di¤erence equations xn 2 yn 2 ; yn+1 = ; (3) xn+1 = 1 yn 2 xn 1 yn 1 xn 2 yn 1 xn where the initial conditions are arbitrary real numbers such that x y 2 x 1 y0 6= 1; 6= 12 :
2 y 1 x0
6=
1 and
Theorem 3 Every solution fxn ; yn g for System (3) is periodic with period twelve and has the form 8 9 y 2 y0 (1+2y 2 x 1 y0 ) > < x 2 ; x 1 ; x0 ; (1+y 2 x 1 y0 ) ; y 1 (1 x 2 y 1 x0 ); (1+y 2 x 1 y0 ) ; > = x 1 x 2 (1+x 2 y 1 x0 ) x0 (1 x 2 y 1 x0 ) y 2 (1+2y 2 x 1 y0 ) fxn g = ; ; ; ; ; (1 x 2 y 1 x0 ) (1+2y 2 x 1 y0 ) (1+x 2 y 1 x0 ) (1+y 2 x 1 y0 ) > > : ; y0 y 1 (1 + x 2 y 1 x0 ); (1+y 2 x 1 y0 ) ; x 2 ; x 1 ; x0 ; :::
and
fyn g =
5
(
y
x 2 2 ; y 1 ; y0 ; (1 x 2 y x 2 (1 x 2 y 1 x0 ) ;
1 (1+y 2 x 1 y0 ) ; x (1+2y ; (1+x x20y 1 x0 ) ; y 2 ; y 1 ; 1 x0 ) 2 x 1 y0 ) x 1 (1+y 2 x 1 y0 ) x0 (1+2y 2 x 1 y0 ) ; (1+x 2 y 1 x0 ) ; y 2 ; y 1 ; y0 ; :::
Fourth system: xn+1 =
yn 2 1 yn 2 xn
1 yn
; yn+1 =
y0 ;
xn 2 1 xn 2 yn
)
:
1 xn
We get, in this section, the form of the solutions for the following system of di¤erence equations xn 2 yn 2 ; yn+1 = ; (4) xn+1 = 1 yn 2 xn 1 yn 1 xn 2 yn 1 xn where n 2 N0 and the initial conditions are arbitrary real numbers such that x 1 and y 2 x 1 y0 = 6 1: 1
Theorem 4 Let fxn ; yn gn=
2
is a solution of System (4) then 5
856
2 y 1 x0
6=
EL-METWALLY, ELSAYED: SOLVING DIFFERENCE EQUATIONS
1
2
is a periodic solution with period six i.e. xn+6 = xn ; yn+6 = yn for
1
2
has the following form
1. fxn ; yn gn= all n 2. 2. fxn ; yn gn= x6n
2
= x
x6n+2
=
y6n
= y
2;
y
x6n
1 (1
1
+x
=x
1;
x6n = x0 ;
2 y 1 x0 );
x6n+3 =
x6n+1 = (1 + y
y0 2x
y (1 + y 1 y0 )
2
2 x 1 y0 )
;
;
and 2
y6n+2
=
2;
x
y6n 1 (1
1
+y
=y
1;
y6n = y0 ;
2 x 1 y0 );
y6n+3 =
y6n+1 = (1 + x
x0 2y
x (1 + x 1 x0 )
2
2 y 1 x0 )
;
:
Lemma 2. All solutions for System (4) are periodic of period three i¤ y 2 x 1 y0 = 2, y 2 = x 2 , y 1 = x 1 and y0 = x0 ; and has the form f:::; x 2 ; x 1 ; x0 ; x 2 ; x 1 ; x0 ; :::g.
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