http://dx.doi.org/10.4236/am.2014.519281
On a Max-Type Difference System
Decun Zhang, Xibao Li, Liying Wang, Shiwei Cui
Institute of Systems Science and Mathematics, Naval Aeronautical and Astronautical University, Yantai, China
Email: [email protected], [email protected]
Received 22 August 2014; revised 25 September 2014; accepted 6 October 2014
Copyright © 2014 by authors and Scientific Research Publishing Inc.
This work is licensed under the Creative Commons Attribution International License (CC BY).
http://creativecommons.org/licenses/by/4.0/
Abstract
In this paper, we show that every well-defined solution of the max-type system of difference
equa-tions
1 max , 1
n n
n
x y
x
+ = −
β ,
1 max , 1
n n
n
y x
y
+ = +
β ,
0
n∈ is eventually periodic with period four.
Keywords
Periodic Solution, Max-Type Difference System
1. Introduction
Max-type difference equations and max-type difference systems have been wisely applied in biology, computer science and automatic control systems and so on. There has been great interest in studying these equations in recent years.
For example, Briden et al. [1] investigated the periodicity character of the solution of the max-type difference equation
1 0
1 1
max , n ,
n
n n
A
x n
x x
+
−
= ∈
Xiao Qian et al. [2] showed that the solution of the max-type difference equation
1 max , 1 , 0
n n
n
x x n
x β
+ −
= ∈
is periodic with period two.
1 1
0
1 1
max ,
,
max ,
n n
n
n n
n
x x
y
n
y y
x
β
β
+ −
+ −
=
∈
=
is periodic with period two.
In addition, E. M. Elasyed, Stevo Stević and others investigated some periodic max-type difference equations and periodic max-type difference systems in [4]-[7].
In this paper we show that every solution of the following max-type difference system
1 1
1 1
max ,
, 0,1,
max ,
n n
n
n n
n
x y
x
n
y x
y
β
β
+ −
+ −
=
=
=
(1)
where the initial conditions x−1, , , x0 y−1 y0 are arbitrary non-zero real numbers and n∈, is periodic with
period four.
Remark 1. Note that if β =0, then System (1) becomes xn+1= yn−1, xn−1=yn+1, from which it follows that 4
n n
x =x+ , yn =yn+4 and every solution is periodic with period four.
2. Some Lemmas
Lemma 1 Assume that
{
}
1 ,
n n n
x y ∞=− is a solution of System (1) and there exists a k0∈ −
{
1, 0,1, 2,}
such that0 0 2, 0 1 0 3, 0 0 2, 0 1 0 3
k k k k k k k k
x =y + x + =y + y =x + y + =x + (2) Then every solution is periodic with period four.
Proof Frist, we will prove that
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
4 1 4 1 2 4 2 3 4 3
4 1 4 1 2 4 2 3 4 3
, , , ;
, , , .
k k m k k m k k m k k m k k m k k m k k m k k m
x x x x x x x x
y y y y y y y y
+ + + + + + + + + +
+ + + + + + + + + +
= = = =
= = = = (3)
where m∈, from which the lemma follows.
Now, we use the method of induction. For m=1, Equation (3) becomes the following equations
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
4 1 5 2 6 3 7
4 1 5 2 6 3 7
, , , ;
, , , .
k k k k k k k k k k k k k k k k
x x x x x x x x
y y y y y y y y
+ + + + + + +
+ + + + + + +
= = = =
= = = = (4)
By System (1) and Equation (2), we obtain that
0 0 0 0 0
0 0
0 0 0 0 0
0 0
0 0 0
0 0
4 2 2
3 1
4 2 2
3 1
5 3 1
4 2
max , max , ,
max , max , ,
max , max ,
k k k k k
k k
k k k k k
k k
k k k
k k
x y x y x
x y
y x y x y
y x
x y x y
x y
β β
β β
β β
+ + +
+ +
+ + +
+ +
+ + +
+ +
= = = =
= = = =
= = =
0 0
0 0 0 0 0
0 0
3 1
5 3 1 3 1
4 2
,
max , max , ,
k k
k k k k k
k k
x
y x y x y
y x
β β
+ +
+ + + + +
+ +
=
= = = =
0 0 0 0
0 0
0 0 0 0
0 0
0 0 0 0
0 0
0
6 4 2
5 1
6 4 2
5 1
7 5 1 3
6 2
max , max , ,
max , max , ,
max , max , ,
k k k k
k k
k k k k
k k
k k k k
k k
k
x y y x
x x
y x x y
y y
x y y x
x x y β β β β β β + + + + + + + + + + + + + + + + = = = = = = = = =
0 0 0
0 0
7 5 1 3
6 2
max , k max , k k .
k k
x x y
y y β β + + + + + + = = =
From which, Equation (4) holds.
Assume Equation (3) holds for 1≤ ≤m m0, and by using System (1) and Equation (2), we obtain that
( )
( )
( )
0 0 0 0 0
0 0
0 0 0
0 0 0 0 0
0 0
0 0 0
0
0 0
0 0
4 2 2 4
4 1
4 3 3
4 2 2 4
4 1
4 3 3
4 1 4 1
4 4
max , max , ,
max , max , ,
max ,
k m k k k
k m
k m k
k m k k k
k m
k m k
k k m
k m
x y y x x
x x
y x x y y
y y x y x β β β β β + + + + + + + + + + + + + + + + + + + + + + + + = = = = = = = = = ( ) ( )
0 0 0 0
0
0 0 0 0 0
0 0
0 0 0
0 0
0 0
0 0 0
3 3 5 1
4
4 3 3 5 1
1 4 1
4 4 4
4 4 2 4 1
4 5
max , ,
max , max , ,
max , max
m k k k
k
k m k k k
k m
k m k
k m k m
k m k
y x x
x
y x x y y
y y x y x x β β β β β + + + + + + + + + + + + + + + + + + + + + + + + = = = = = = = = = ( ) ( )
0 0 0
0 0 0 0 0
0 0
0 0 0
0 0 0 0
0 0
0 0 0
4 6 2
5
4 4 4 6 2
2 4 1
4 5 5
4 5 5 7
3 4 1
4 6 6
, ,
max , max , ,
max , max ,
k k k
k m k k k
k m
k m k
k m k k
k m
k m k
y x x
y x x y y
y y
x y y x
x x β β β β + + + + + + + + + + + + + + + + + + + + + + + + = = = = = = = = = ( ) 0
0 0 0 0 0
0 0
0 0 0
3
4 5 5 7 3
3 4 1
4 6 6
,
max , max , .
k
k m k k k
k m
k m k
x
y x x y y
y y β β + + + + + + + + + + + + = = = = =
So we complete the proof.
Lemma 2 Assume that β >0. Then every solution of System (1) is positive if initial conditions satisfy one of the following conditions x−1>0 or x0>0 or y−1>0 or y0 >0.
Proof Without loss of generality, we assume that x−1>0 and from System (1) we have
1 1 2 0
0 1
3 1 4 2
2 3
max , 0, max , 0;
max , 0, max , 0.
y x y x
y y
x y x y
x x
β β
β β
−
= > = >
= > = >
By using the method of induction, we have
, 0, 3 n n
x y > n≥
0 , 0, n n
x y > n≥n
The proof is completed.
Lemma 3 Assume that β >0. Then every solution of System (1) with positive initial conditions is periodic with period four.
Proof By System (1), we obtain that
1 1 1 1
0 0
max , 0, max , 0
x y y x
x y
β β
− −
= > = >
Let p=max
{
y x0, 0}
, 0 1max ,
q y
y β
−
=
, 1 0
max ,
s x
x β
−
=
and there are four cases which need to be
discussed.
Case 1. 1 1
0 0 ,
y x
x y
β β
− −
≥ ≥ . We have
{
}
{
}
1 1 1 1
0 0 0 0
2 0 0 0 2 0 0 0
1 1
3 1 1 3 1
2 0 0 2 0 0
max , , max , ;
max , max , , max , max , ;
max , max , , max , max ,
x y y x
x x y y
x y x y p y x y x p
x y
x y y y x
x p y y y p x x
β β β β
β β
β β β β β β β β
− −
= = = =
= = = = = =
= = = = = = =
{
}
{
}
1
4 2 0 2 4 2 0 2
3 3
;
max , max , , max , max , .
x
x y y p p y y x x p p x
x y
β β
=
= = = = = = = =
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four. Moreover, we have
4 1 4 3 1 4 1 4 3 1
0 0
4 2 4 4 2 4 2 4 4 2
, ;
, ,
n n n n
n n n n
x y x y x y
x y
x y x p y x y p
β β
+ + + +
+ + + +
= = = = = =
= = = = = =
where n∈0, and the solution has the following form
{ }
{ }
1 0 1
0 0 0 0
1 0 1
0 0 0 0
, , , , , , , , , , ,
, , , , , , , , , , .
n n
n n
x x x p p p p
x y x y
y y y p p p p
y x y x
β β β β
β β β β
∞
− =−
∞
− =−
=
=
Case 2. 1 1
0 0 ,
y x
x y
β β
− −
< ≥ . We have
{
}
1 1 1 1 1
0 0 0
2 0 0 2 0 0 0
1 1 1
3 1 1 3 1 1
2 0 0 2
max , , max , ;
max , max , , max , max , ;
max , max , , max , max ,
x y y y x
x y y
x y y q y x y x p
x y y
x y y y x y
x q y y y p
β β β
β β β
β β β β β β
− − −
−
−
= = = =
= = = = = =
= = = = = =
{
}
1 1
4 2 0 2 4 2 2
3 3 1
;
max , max , , max , max , .
y x
x y y p p y y x q q x
x y y
β β β
−
−
= =
= = = = = = = =
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four. Moreover, we have
4 1 4 3 1 1 4 1 4 3 1 0
4 2 4 4 2 4 2 4 4 2
, ;
, ,
n n n n
n n n n
x y x y y x y
y
x y x q y x y p
β
+ + − + +
+ + + +
= = = = = =
= = = = = =
where n∈0, and the solution has the following form
{ }
{ }
1 0 1 1
1
0 0
1 0 1 1
1
0 0
, , , , , , , , , , ,
, , , , , , , , , , .
n n
n n
x x x y q p y q p
y y
y y y p y q p y q
y y
β β
β β
∞
− − −
=−
∞
− − −
=−
=
=
Case 3. 1 1
0 0 ,
y x
x y
β β
− −
≥ < . We have
{
}
1 1 1 1 1
0 0 0
2 0 0 0 2 0 0
1 1 1
3 1 1 1 1 3 1
2 2 0
max , , max , ;
max , max , , max , max , ;
max , max , , max , max ,
x y y x x
x x y
x y x y p y x x s
x y x
x y x x y y x
x p y s x
β β β
β β β
β β β β β
− − −
−
− −
= = = =
= = = = = =
= = = = = =
{
}
1 0
4 2 2 4 2 0 2
3 1 3
;
max , max , , max , max , .
x x
x y s s y y x x p p x
x x y
β
β β β
−
= =
= = = = = = = =
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four. Moreover, we have
4 1 4 3 1 4 1 4 3 1 1 0
4 2 4 4 2 4 2 4 4 2
, ;
, ,
n n n n
n n n n
x y x y x y x
x
x y x p y x y s
β
+ + + + −
+ + + +
= = = = = =
= = = = = =
where n∈0, and the solution has the following form
{ }
{ }
1 0 1 1
1
0 0
1 0 1 1
1
0 0
, , , , , , , , , , ,
, , , , , , , , , , .
n n
n n
x x x p x s p x s
x x
y y y x s p x s p
x x
β β
β β
∞
− − −
=−
∞
− − −
=−
=
=
Case 4. 1 1
0 0 ,
y x
x y
β β
− −
< < . We have
1 1 1 1 1 1
0 0
2 0 0 2 0 0
1 1 1 1
3 1 1 1 1 3 1
2 2
max , , max , ;
max , max , , max , max , ;
max , max , , max , max
x y y y x x
x y
x y y q y x x s
x y y x
x y x x y y x
x q y
β β
β β β β
β β β
− − − −
− −
− −
= = = =
= = = = = =
= = = = = =
1 1 1
4 2 2 4 2 2
3 1 3 1
, ;
max , max , , max , max , .
y y x
s
x y s s y y x q q x
x x y y
β
β β β β
− −
− −
= =
= = = = = = = =
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four. Moreover, we have
4 1 4 3 1 1 4 1 4 3 1 1
4 2 4 4 2 4 2 4 4 2
, ;
, ,
n n n n
n n n n
x y x y y x y x
x y x q y x y s
+ + − + + −
+ + + +
= = = = = =
= = = = = =
where n∈0, and the solution has the following form
{ }
{
}
{ }
{
}
1 0 1 1 1 1 1
1 0 1 1 1 1 1
, , , , , , , , , , , , , , , , , , , , , . n n
n n
x x x y q x s y q x s
y y y x s y q x s y q
∞
− − − − −
=− ∞
− − − − −
=− = =
So we complete the proof.
Lemma 4 Assume that β >0. Then every solution of System (1) with negative initial conditions is periodic with period four.
Proof Since x−1, , , x0 y−1 y0<0 and β >0, by induction we have xn<0, 0yn< . If we use the change ,
n n n n
z = −x w = −y and System (1) can be rewritten as follows
1 1
1 1
min , ,
min , .
n n
n
n n
n
z w
z
w z
w
β
β
+ −
+ −
=
=
(5)
where zn, 0, 1, 0,1,wn > n= − .
Now, we will prove that every solution of System (5) with positive initial conditions is periodic with period four.
Let p′ =min
{
z w0, 0}
, 0 1min ,
q w
w β
−
′ =
, 1 0
min ,
s z
z β
−
′ =
. Similar to the proof of Lemma (3), there are
four cases which need to be discussed.
Case 1. 1 1
0 0 ,
w z
z w
β β
− −
< < . We obtain that
{ }
{ }
1 0 1
0 0 0 0
1 0 1
0 0 0 0
, , , , , , , , , , ,
, , , , , , , , , , .
n n
n n
z z z p p p p
z w z w
w w w p p p p
w z w z
β β β β
β β β β
∞
− =−
∞
− =−
′ ′ ′ ′
=
′ ′ ′ ′
=
Case 2. 1 1
0 0 ,
w z
z w
β β
− −
> < . We obtain that
{ }
{ }
1 0 1 1
1
0 0
1 0 1 1
1
0 0
, , , , , , , , , , ,
, , , , , , , , , , .
n n
n n
z z z w q p w q p
w w
w w w p w q p w q
w w
β β
β β
∞
− − −
=−
∞
− − −
=−
′ ′ ′ ′
=
′ ′ ′ ′
=
Case 3. 1 1
0 0 ,
w z
z w
β β
− −
< > . We obtain that
{ }
{ }
1 0 1 1
1
0 0
1 0 1 1
1
0 0
, , , , , , , , , , ,
, , , , , , , , , , .
n n
n n
z z z p z s p z s
z z
w w w z s p z s p
z z
β β
β β
∞
− − −
=−
∞
− − −
=−
′ ′ ′ ′
=
′ ′ ′ ′
=
Case 4. 1 1
0 0 ,
w z
z w
β β
− −
> > . We obtain that
{ }
{
}
{ }
{
}
1 0 1 1 1 1 1
1 0 1 1 1 1 1
, , , , , , , , , , ,
, , , , , , , , , , .
n n
n n
z z z w q z s w q z s
w w w z s w q z s w q
∞
− − − − −
=− ∞
− − − − −
=−
′ ′ ′ ′
=
′ ′ ′ ′
=
So we complete the proof.
Lemma 5 Assume that β <0. Then every solution of System (1) is periodic with period four if initial conditions satisfy one of the following conditions
( )
( )
( )
11 00 1 00 1( )
1 01 01 0 0 1, , , 0; , , , 0; , 0, , 0; , 0, , 0.
a x x y y b x x y y
c x y x y d x y x y
− − − −
− − − −
> <
> < < >
Proof (a) If x−1, , , x0 y−1 y0>0, by using System (1), we know that there is only one case which needs to be discussed. That is
1 1
0 0
,
y x
x y
β β
− −
< <
Then we have
1 1 1 1 1 1
0 0
2 0 0 2 0 0
1 1
3 1 1 1 3 1 1 1
2 2
4 2 2 0
3
max , 0, max , 0;
max , 0, max , 0;
max , 0, max , 0;
max , 0,
x y y y x x
x y
x y y y x x
x y
x y y x y x x y
x y
x y y x y
x
β β
β β
β β
β
− − − −
− −
= = > = = >
= = > = = >
= = = > = = = >
= = = >
4 3 2 2 0
max ,x x y 0.
y β
= = = >
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four.
The proof of case
( )( )( )
b c d is similar to the proof of case( )
a , so we omit it. Then, the proof is completed.Lemma 6 Assume that β <0. Then every solution of System (1) is periodic with period four if initial conditions satisfy one of the following conditions
( )
( )
( )
( )
( )
( )
( )
( )
1 0 1 0 0 1 1 0
1 1 0 0 0 1 0 1
1 0 1 0 0 1 1 0
1 1 0 0 0 1 0 1 0, , , 0; 0, , , 0; 0, , , 0; 0, , , 0; 0, , , 0; 0, , , 0;
0, , , 0; 0, , , 0.
e x x y y f x x y y
j y x x y h y x x y
i x x y y j x x y y
k y x x y l y x x y
− − − −
− − − −
− − − −
− − − −
< > < >
< > < >
> < > <
< > > <
Proof
( )
e If x−1<0, , , x0 y−1 y0>0, by using System (1), we know that 1 0y x
β −
< so there are two cases
which need to be discussed. That is
( )
1( )
10 0
1 , 2
e x e x
y y
β β
− −
≥ <
Case 1.
( )
10 1
e x
y
β −
{
}
1 1 1 1 1
0 0 0
2 0 0 0 2 0 0 0
1 1 1
3 1 1 3 1
2 0 0 0 2
max , 0, max , 0;
max , max , 0, max , max , 0;
max , max , 0, max ,
x y y y x
x y y
x y y y y x y x p
x y y
x y y y x
x y y y y
β β β
β β β
β β β β β
− − −
−
= = > = = <
= = = > = = = >
= = = = < =
{
}
1 1 1
4 2 0 2 4 2 0 0 2
3 3 1
max , 0;
max , max , 0, max , max , 0.
y y x
p
x y y p p y y x y y x
x y y
β
β β β
− −
−
= = = >
= = = = > = = = = >
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four.
Case 2.
( )
10 2
e x
y
β −
< . We have
1 1 1 1 1 1
0 0
2 0 0 0 2 0 0
1 1 1 1
3 1 1 1 1 3
2 0 2
max , 0, max , 0;
max , max , 0, max , max , 0;
max , max , 0, max ,
x y y y x x
x y
x y y y y x x s
x y y x
x y x x y y
x y y
β β
β β β β
β β β
− − − −
− −
− −
= = > = = <
= = = > = = = >
= = = = < =
1 1 1 1
4 2 2 4 2 0 0 2
3 1 3 1
max , 0;
max , max , 0, max , max , 0.
x y y x
s
x y s s y y x y y x
x x y y
β
β β β β
− −
− −
= = = >
= = = = > = = = = >
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four.
The proof of case
( ) ( )
f − l is similar to the proof of case( )
e , so we omit it. Then, the proof is completed.Lemma 7 Assume that β <0. Then every solution of System (1) is periodic with period four if initial conditions satisfy one of the following conditions
( )
m x , 0, , 0;−1 x0> y−1 y0<( )
n x , 0, , −1 x0< y−1 y0>0; 0 ,( )
x−1 y−1>0, , 0;x0 y0<( )
p x , −1 y−1<0, , x0 y0>0.Proof (m) If x−1, 0, x0> y−1, 0y0< , by using System (1), we know that there are four cases which need to
be discussed. That is
( )
1 1( )
1 1( )
1 1( )
1 10 0 0 0 0 0 0 0
1 , ; 2 , ; 3 , ; 4 , .
m y x m y x m y x m y x
x y x y x y x y
β β β β β β β β
− − − − − − − −
≥ ≥ < ≥ ≥ < < <
Case 1.
( )
1 10 0 1 ,
m y x
x y
β β
− −
≥ ≥ . We have
{
}
{
}
1 1 1 1
0 0 0 0
2 0 0 0 0 2 0 0 0 0
1 1
3 1 1 3 1
2 0 0 0 2 0
max , 0, max , 0;
max , max , 0, max , max , 0;
max , max , 0, max , max
x y y x
x x y y
x y x y x y x y x x
x y
x y y y x
x x y y y x
β β β β
β β
β β β β β β
− −
= = < = = >
= = = > = = = >
= = = = > = =
{
}
{
}
1
0 0
4 2 0 0 0 2 4 2 0 0 0 2
3 3
, 0;
max , max , 0, max , max , 0.
x
x x
x y y x x y y x x x x x
x y
β β
β β
= = <
= = = = > = = = = >
Hence, x1 =y3, , , x2 =y4 y1 =x3 y2 =x4. And by Lemma 1, we have that the solution is periodic with period four.
The proof of case
( )( )( )
m2 m3 m4 is similar to the proof of case( )( )( )
b c d in Lemma 3, so we omit it and case( )
m is completed.Similarly, the proof of case
( )( )( )
n o p is similar to the proof of case( )
m , so we omit it.3. Main Results
By using Lemma 2 and Lemma 3, we obtain the following result.
Theorem 1 Assume that β >0. Then every solution of System (1) is periodic with period four if initial conditions satisfies one of the following conditions x−1 >0 or x0>0 or y−1 >0 or y0 >0.
By using Theorem 1 and Lemma 4, we obtain the following result.
Theorem 2 Assume that β >0. Then every well-defined solution of System (1) is periodic with period four. By using Lemma 5, Lemma 6 and Lemma 7, we obtain the following result.
Theorem 3 Assume that β <0. Then every well-defined solution of System (1) is periodic with period four. By using Theorem 2 and Theorem 3, we obtain the following result.
Theorem 4 Assume that β∈ℜ. Then every well-defined solution of System (1) is periodic with period four.
Acknowledgements
We thank the Editor and the referee for their comments. Research supported by Distinguished Expert Founda-tion and Youth Science FoundaFounda-tion of Naval Aeronautical and Astronautical University.
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