Hindawi Publishing Corporation Advances in Difference Equations Volume 2010, Article ID 573281, 14 docx
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Hindawi Publishing Corporation
Advances in Difference Equations
Volume 2010, Article ID 573281, 14 pages
doi:10.1155/2010/573281
Research Article
On the Global Character of the System of Piecewise
Linear Difference Equations x
n1
|x
n
|−y
n
− 1 and
y
n1
x
n
−|y
n
|
Wirot Tikjha,
1, 2
Yongwimon L e n b u ry,
1, 2
and Evelina Giusti Lapierre
3
1
Department of Mathematics, Faculty of Science, Mahidol University, Rama 6 Road Bangkok,
10400, Thailand
2
Center of Excellence in Mathematics, PERDO Commission on Higher Education, Si A yudhya Road,
Bangkok 10400, Thailand
3
John Hazen White School of Arts and Sciences, Department of Mathematics,
Johnson and Wales University, 8 Abbott Park Place, Providence, RI 02903, USA
Correspondence should be addressed to Yongwimon Lenbury,
Received 23 June 2010; Revised 4 September 2010; Accepted 2 December 2010
Academic Editor: Donal O’Regan
Copyright q 2010 Wirot T ikjha et a l. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
We consider the system in the title where the initial condition x
0
,y
0
∈ R
2
. We show that the
system has exactly two prime period-5 solutions and a unique equilibrium point 0, −1.Wealso
show that every solution of the system is eventually one of the two prime period-5 solutions or
else the unique equilibrium point.
1. Introduction
In this paper, we consider the system of piecewise linear difference equations
x
n1
|
x
n
|
− y
n
− 1,
y
n1
x
n
−
y
n
,
n 0, 1, 2, , 1.1
where the initial condition x
0
,y
0
∈ R
2
. We show that every solution of System 1.1 is even-
tually either one of two prime period-5 solutions or else the unique equilibrium point 0, −1.
2AdvancesinDifference Equations
System 1.1 was motivated by Devaney’s Gingerbread man map 1, 2
x
n1
|
x
n
|
− x
n−1
1 1.2
or its equivalent system of piecewise linear difference equations 3, 4
x
n1
|
x
n
|
− y
n
1,
y
n1
x
n
,
n 0, 1, 2, 1.3
We believe that the methods and techniques used in this paper will be useful in
discovering the global character of solutions of similar systems, including the Gingerbread
man map.
2. The Global Behavior of the Solutions of System 1.1
System 1.1 has the equilibrium point x, y ∈ R
2
given by
x, y
0, −1
. 2.1
System 1.1 has two prime period-5 solutions,
P
1
5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
x
0
0,y
0
1
x
1
−2,y
1
−1
x
2
2,y
2
−3
x
3
4,y
3
−1
x
4
4,y
4
3
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
,
P
2
5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
x
0
0,y
0
1
7
x
1
−
8
7
,y
1
−
1
7
x
2
2
7
,y
2
−
9
7
x
3
4
7
,y
3
−1
x
4
4
7
,y
4
−
3
7
.
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
.
2.2
Advances in Difference Equations 3
Set
l
1
x, y
: x ≥ 0,y 0
,
l
2
x, y
: x 0,y≥ 0
,
l
3
x, y
: x<0,y 0
,
l
4
x, y
: x 0,y<0
,
Q
1
x, y
: x>0,y>0
,
Q
2
x, y
: x<0,y>0
,
Q
3
x, y
: x<0,y<0
,
Q
4
x, y
: x>0,y<0
.
2.3
Theorem 2.1. Let x
0
,y
0
∈ R
2
. Then there exists an integer N≥0 such that the solution
{x
n
,y
n
}
∞
nN
is eventually either the prime period-5 solution P
1
5
, the prime period-5 solution P
2
5
,or
else the unique equilibrium point 0, −1.
The proof is a direct consequence of the following lemmas.
Lemma 2.2. Suppose there exists an integer M ≥ 0 such that −1 ≤ x
M
≤ 0 and y
M
−x
M
−1.Then
x
M1
,y
M1
0, −1,andso{x
n
,y
n
}
∞
nM1
is the equilibrium solution.
Proof. Note that
x
M1
|
x
M
|
− y
M
− 1 −x
M
−
−x
M
− 1
− 1 0,
y
M1
x
M
−
y
M
x
M
−
x
M
1
−1,
2.4
and so the proof is complete.
Lemma 2.3. Suppose there exists an integer M ≥ 0 such that x
M
≥ 1 and y
M
x
M
− 1.Then
x
M1
,y
M1
0, 1,andso{x
n
,y
n
}
∞
nM1
is P
1
5
.
Proof. We have
x
M1
|
x
M
|
− y
M
− 1 x
M
−
x
M
− 1
− 1 0,
y
M1
x
M
−
y
M
x
M
−
x
M
− 1
1,
2.5
and so the proof is complete.
Lemma 2.4. Suppose there exists an integer M ≥ 0 such that x
M
0 and y
M
≥ 0. Then the following
statements are true.
1 x
M5
0.
2 If y
M
> 1/4,then{x
n
,y
n
}
∞
nM5
is P
1
5
.
3 If 0 ≤ y
M
≤ 1/4,theny
M5
8y
M
− 1.
4AdvancesinDifference Equations
Proof. We have x
M
0andy
M
≥ 0. Then
x
M1
|
x
M
|
− y
M
− 1 −y
M
− 1 < 0,
y
M1
x
M
−
y
M
−y
M
≤ 0,
x
M2
|
x
M1
|
− y
M1
− 1 2y
M
≥ 0,
y
M2
x
M1
−
y
M1
−2y
M
− 1 < 0,
x
M3
|
x
M2
|
− y
M2
− 1 4y
M
≥ 0,
y
M3
x
M2
−
y
M2
−1,
x
M4
|
x
M3
|
− y
M3
− 1 4y
M
≥ 0,
y
M4
x
M3
−
y
M3
4y
M
− 1,
x
M5
|
x
M4
|
− y
M4
− 1 0,
2.6
and so statement 1 is true.
If y
M
> 1/4, then y
M5
x
M4
−|y
M4
| 1. That is, x
M5
,y
M5
0, 1 and so
statement 2 is true.
If 0 ≤ y
M
≤ 1/4, then y
M5
x
M4
−|y
M4
| 8y
M
− 1, and so statement 3 is true.
Lemma 2.5. Suppose there exists an integer M ≥ 0 such that x
M
0 and y
M
< −1. Then the
following statements are true.
1 x
M4
0.
2 If −3/2 <y
M
< −1,theny
M4
−4y
M
− 5.
3 If y
M
≤−3/2,then{x
n
,y
n
}
∞
nM4
is P
1
5
.
Proof. We have x
M
0andy
M
< −1. Then
x
M1
|
x
M
|
− y
M
− 1 −y
M
− 1 > 0,
y
M1
x
M
−
y
M
y
M
< 0,
x
M2
|
x
M1
|
− y
M1
− 1 −2y
M
− 2 > 0,
y
M2
x
M1
−
y
M1
−1,
x
M3
|
x
M2
|
− y
M2
− 1 −2y
M
− 2 > 0,
y
M3
x
M2
−
y
M2
−2y
M
− 3,
x
M4
|
x
M3
|
− y
M3
− 1 0,
2.7
and so statement 1 is true.
Now if −3/2 <y
M
< −1, then y
M3
−2y
M
− 3 < 0. Thus y
M4
x
M3
−|y
M3
|
−4y
M
− 5, and so statement 2 is true.
Lastly, if y
M
≤−3/2, then y
M3
−2y
M
− 3 ≥ 0. Thus y
M4
x
M3
−|y
M3
| 1; that is,
x
M4
,y
M4
0, 1 and so statement 3 is true.
Advances in Difference Equations 5
Lemma 2.6. Suppose there exists an integer M ≥ 0 such that x
M
≥ 0 and y
M
0. Then the following
statements are true.
1 If x
M
≥ 1,then{x
n
,y
n
}
∞
nM2
is P
1
5
.
2 If 1/4 <x
M
< 1,then{x
n
,y
n
}
∞
nM6
is P
1
5
.
3 If 0 ≤ x
M
≤ 1/4,thenx
M6
0 and y
M6
8x
M
− 1.
Proof. First consider the case x
M
≥ 1andy
M
0. Then
x
M1
|
x
M
|
− y
M
− 1 x
M
− 1 ≥ 0,
y
M1
x
M
−
y
M
x
M
> 0,
x
M2
|
x
M1
|
− y
M1
− 1 −2,
y
M2
x
M1
−
y
M1
−1,
2.8
and so statement 1 is true.
Next consider the case 0 ≤ x
M
< 1andy
M
0. Then
x
M1
|
x
M
|
− y
M
− 1 x
M
− 1 < 0,
y
M1
x
M
−
y
M
x
M
≥ 0,
x
M2
|
x
M1
|
− y
M1
− 1 −2x
M
≤ 0,
y
M2
x
M1
−
y
M1
−1,
x
M3
|
x
M2
|
− y
M2
− 1 2x
M
≥ 0,
y
M3
x
M2
−
y
M2
−2x
M
− 1 < 0,
x
M4
|
x
M3
|
− y
M3
− 1 4x
M
≥ 0,
y
M4
x
M3
−
y
M3
−1,
x
M5
|
x
M4
|
− y
M4
− 1 4x
M
≥ 0,
y
M5
x
M4
−
y
M4
4x
M
− 1,
x
M6
|
x
M5
|
− y
M5
− 1 0.
2.9
If 1/4 <x
M
< 1, then y
M5
4x
M
− 1 > 0andsoy
M6
x
M5
−|y
M5
| 1. That is,
x
M6
,y
M6
0, 1 and so statement 2 is true.
If 0 ≤ x
M
≤ 1/4, then y
M5
4x
M
− 1 ≤ 0. Thus y
M6
x
M5
−|y
M5
| 8x
M
− 1, and
so statement 3 is true.
Lemma 2.7. Suppose there exists an integer M ≥ 0 such that x
M
< −1 and y
M
0. Then the
following statements are true.
1 x
M4
0.
2 If −3/2 ≤ x
M
< −1,theny
M4
−4x
M
− 5.
3 If x
M
< −3/2,then{x
n
,y
n
}
∞
nM4
is P
1
5
.
6AdvancesinDifference Equations
Proof. Let x
M
< −1andy
M
0. Then
x
M1
|
x
M
|
− y
M
− 1 −x
M
− 1 > 0,
y
M1
x
M
−
y
M
x
M
< 0,
x
M2
|
x
M1
|
− y
M1
− 1 −2x
M
− 2 > 0,
y
M2
x
M1
−
y
M1
−1,
x
M3
|
x
M2
|
− y
M2
− 1 −2x
M
− 2 > 0,
y
M3
x
M2
−
y
M2
−2x
M
− 3,
x
M4
|
x
M3
|
− y
M3
− 1 0,
2.10
and so statement 1 is true.
If −3/2 ≤ x
M
< −1, then y
M3
−2x
M
− 3 ≤ 0. Thus y
M4
x
M3
−|y
M3
| −4x
M
− 5,
and so statement 2 is true.
If x
M
< −3/2, then y
M3
−2x
M
− 3 > 0andy
M4
x
M3
−|y
M3
| 1. That is,
x
M4
,y
M4
0, 1 and so {x
n
,y
n
}
∞
nM4
is P
1
5
and the proof is complete.
We now give the proof of Theorem 2.1 when x
M
,y
M
is in l
2
{x, y : x 0,y ≥ 0}.
Lemma 2.8. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ l
2
. Then the following
statements are true.
1 If 0 ≤ y
M
< 1/7,then{x
n
,y
n
}
∞
nM
is eventually the equilibrium solution.
2 If y
M
1/7, then the solution {x
n
,y
n
}
∞
nM2
is P
2
5
.
3 If y
M
> 1/7, then the solution {x
n
,y
n
}
∞
nM
is eventually P
1
5
.
Proof. 1 We will first show that statement 1 is true. Suppose 0 ≤ y
M
< 1/7; for each n ≥ 0,
let
a
n
2
3n
− 1
7 · 2
3n
.
2.11
Observe that
0 a
0
<a
1
<a
2
< ···<
1
7
, lim
n →∞
a
n
1
7
.
2.12
Thus there exists a unique integer K ≥ 0suchthaty
M
∈ a
K
,a
K1
.
We first consider the case K 0; that is, y
M
∈ 0, 1/8. By statements 1 and 3 of
Lemma 2.4, x
M5
0andy
M5
8y
M
− 1. Clearly y
M5
< 0, and so
x
M6
|
x
M5
|
− y
M5
− 1 −8y
M
≤ 0,
y
M6
x
M5
−
y
M5
8y
M
− 1.
2.13
Advances in Difference Equations 7
Now −1 <x
M6
≤ 0andy
M6
−x
M6
− 1, and so by Lemma 2.2, {x
n
,y
n
}
∞
nM7
is the
equilibrium solution.
Without loss of generality, we may assume K ≥ 1.
For each integer n such that n ≥ 0, let Pn be the following statement:
x
M5n5
0,
y
M5n5
2
3n1
y
M
−
2
3n1
− 1
7
≥ 0.
2.14
Claim 1. Pn is true for 0 ≤ n ≤ K − 1.
The proof Claim 1 will be by induction on n. We will first show that P0 is true.
Recall that x
M
0andy
M
∈ a
K
,a
K1
⊂ 1/8, 1/7. Then by statements 1 and 3
of Lemma 2.4,wehavex
M505
0andy
M505
8y
M
− 1.
Note that,
y
M505
8y
M
− 1 2
301
y
M
−
2
301
− 1
7
≥ 0 2.15
and so P0 is true. Thus if K 1, then we have shown that for 0 ≤ n ≤ K − 1, Pn is true. It
remains to consider the case K ≥ 2. So assume that K ≥ 2. Let n be an integer such that
0 ≤ n ≤ K − 2 and suppose Pn is true. We will show that Pn 1 is true.
Since Pn is true, we know
x
M5n5
0,y
M5n5
2
3n1
y
M
−
2
3n1
− 1
7
≥ 0. 2.16
It is easy to verify that for y
M
∈ 1/8, 1/7,
y
M5n5
2
3n1
y
M
−
2
3n1
− 1
7
<
1
4
. 2.17
Thus by statements 1 and 3 of Lemma 2.4,
x
M5n15
0,
y
M5n15
8
y
M5n5
− 1
2
3
2
3n1
y
M
−
2
3n1
− 1
7
− 1
2
3n6
y
M
−
2
3n6
7
2
3
7
− 1
2
3n2
y
M
−
2
3n2
− 1
7
.
2.18
8AdvancesinDifference Equations
Recall that y
M
∈ a
K
,a
K1
2
3K
− 1/7 · 2
3K
, 2
3K1
− 1/7 · 2
3K1
.
In particular,
y
M5n15
2
3n2
y
M
−
2
3n2
− 1
7
≥ 2
3n2
2
3K
− 1
7 · 2
3K
−
2
3n2
− 1
7
2
3n3K6
7 · 2
3K
−
2
3n6
7 · 2
3K
−
2
3n6
7
1
7
1
7
1 − 2
3n−K−2
≥
1
7
1 − 1
0,
2.19
and so Pn 1 is true. Thus the proof of the claim is complete. That is, Pn is true for 0 ≤
n ≤ K − 1. Specifically, PK − 1 is true, and so
x
M5K−15
0,y
M5K−15
2
3K
y
M
−
2
3K
− 1
7
≥ 0. 2.20
In particular,
2
3K
2
3K
− 1
7 · 2
3K
−
2
3K
− 1
7
≤ y
M5K−15
< 2
3K
2
3K3
− 1
7 · 2
3K3
−
2
3K
− 1
7
. 2.21
That is, 0 ≤ y
M5K−15
< 1/8, and so by case K 0, {x
n
,y
n
}
∞
nM5K7
is the equilibrium
solution, and the proof of statement 1 is complete.
2 We will next show that statement 2 is true. Suppose x
M
,y
M
0, 1/7 .Note
that 0, 1/7 ∈ P
2
5
. Thus the solution {x
n
,y
n
}
∞
nM
is P
2
5
.
3 Finally, we will show that statement 3 is true. Suppose y
M
> 1/7.
First consider y
M
> 1/4. By statement 2 of Lemma 2.4,thesolution{x
n
,y
n
}
∞
nM5
is P
1
5
.
Next consider the case y
M
∈ 1/7, 1/4.Foreachn ≥ 1, let
b
n
2
3n−1
3
7 · 2
3n−1
.
2.22
Observe that
1
4
b
1
>b
2
>b
3
> ···>
1
7
, lim
n →∞
b
n
1
7
.
2.23
Thus there exists a unique integer K ≥ 1suchthaty
M
∈ b
K1
,b
K
.
Advances in Difference Equations 9
Note that the statement Pn which we stated and proved in the proof of statement
1 of this lemma still holds. Specifically PK − 1 is true, and so
x
M5K−15
0,y
M5K−15
2
3K
y
M
−
2
3K
− 1
7
≥ 0. 2.24
Recall that for y
M
∈ b
K1
,b
K
.
In particular,
y
M5K
2
3K
y
M
−
2
3K
− 1
7
> 2
3K
2
3K2
3
7 · 2
3K2
−
2
3K
− 1
7
1
4
. 2.25
By statement 2 of Lemma 2.4,thesolution{x
n
,y
n
}
∞
nM5K5
is P
1
5
.
We now give the proof of Theorem 2.1 when x
M
,y
M
is in l
4
{x, y : x 0,y <0}.
Lemma 2.9. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ l
4
. Then the following
statements are true.
1 If −9/7 <y
M
< 0,then{x
n
,y
n
}
∞
nM
is eventually the equilibrium solution.
2 If y
M
−9/7, then the solution {x
n
,y
n
}
∞
nM1
is P
2
5
.
3 If y
M
< −9/7, then the solution {x
n
,y
n
}
∞
nM
is eventually P
1
5
.
Proof. 1 We will first show that statement 1 is true. So suppose −9/7 <y
M
< 0.
Case 1. Suppose −1 ≤ y
M
< 0. Then
x
M1
|
x
M
|
− y
M
− 1 −y
M
− 1 ≤ 0,
y
M1
x
M
−
y
M
y
M
.
2.26
In particular, −1 <x
M1
≤ 0andy
M1
−x
M1
− 1, and so by Lemma 2.2, {x
n
,y
n
}
∞
nM2
is
the equilibrium solution.
Case 2. Suppose −5/4 ≤ y
M
< −1. By statements 1 and 2 of Lemma 2.5, x
M4
0and
y
M4
−4y
M
− 5. Then
x
M5
|
x
M4
|
− y
M4
− 1 4y
M
4 < 0,
y
M5
x
M4
−
y
M4
−4y
M
− 5.
2.27
Thus −1 ≤ x
M5
< 0andy
M5
−x
M5
− 1, and so by Lemma 2.2, {x
n
,y
n
}
∞
nM6
is
the equilibrium solution.
Case 3. Suppose −9/7 <y
M
< −5/4. By statements 1 and 2 of Lemma 2.5, x
M4
0
and y
M4
−4y
M
− 5. Note that 0 <y
M4
< 1/7 and so by statement 1 of Lemma 2.8,
{x
n
,y
n
}
∞
nM4
is eventually equilibrium solution.
10 Advances in Difference Equations
2 We will next show that statement 2 is true. Suppose y
M
−9/7. By direct calcu-
lations we have x
M1
,y
M1
2/7, −9/7.Sothesolution{x
n
,y
n
}
∞
nM1
is P
2
5
.
3 Finally, we will show that statement 3 is true. Suppose x
M
0andy
M
< −9/7.
Case 1. Suppose −3/2 <y
M
< −9/7. By statements 1 and 2 of Lemma 2.5,wehavex
M4
0andy
M4
−4y
M
− 5. Note that 1/7 <y
M4
< 1 and so by statement 3 of Lemma 2.8,the
solution {x
n
,y
n
}
∞
nM4
is eventually P
1
5
.
Case 2. Suppose y
M
≤−3/2. By statement 3 of Lemma 2.5,thesolution{x
n
,y
n
}
∞
nM4
is
P
1
5
.
We now give the proof of Theorem 2.1 when x
M
,y
M
is in l
1
{x, y : x ≥ 0,y 0}.
Lemma 2.10. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ l
1
. Then the following
statements are true.
1 If 0 ≤ x
M
< 1/7,then{x
n
,y
n
}
∞
nM
is eventually the equilibrium solution.
2 If x
M
1/7, then the solution {x
n
,y
n
}
∞
nM3
is P
2
5
.
3 If x
M
> 1/7, then the solution {x
n
,y
n
}
∞
nM
is eventually P
1
5
.
Proof. 1 We will first show that statement 1 is true. So suppose 0 ≤ x
M
< 1/7andy
M
0.
By statement 3 of Lemma 2.6, x
M6
0andy
M6
8x
M
− 1. In particular, −1 <y
M6
< 1/7
and so by statement 1 of Lemma 2.8 and statement 1 of Lemma 2.9, {x
n
,y
n
}
∞
nM6
is
eventually the equilibrium solution.
2 We will next show that statement 2 is true. Suppose x
M
1/7. By direct
calculations we have x
M3
,y
M3
2/7, −9/7. Thus the solution {x
n
,y
n
}
∞
nM3
is P
2
5
.
3 Finally, we will show statement 3 is true.
First consider the case 1/7 <x
M
≤ 1/4. By statement 3 of Lemma 2.6, x
M6
0and
y
M6
8x
M
− 1. Now, 1/7 <y
M6
≤ 1 and so by statement 3 of Lemma 2.8,thesolution
{x
n
,y
n
}
∞
nM6
is eventually P
1
5
.
Next consider the case x
M
> 1/4. Then by statements 1 and 2 of Lemma 2.6,if
x
M
≥ 1then{x
n
,y
n
}
∞
nM2
is P
1
5
,andif1/4 <x
M
< 1then{x
n
,y
n
}
∞
nM6
is P
1
5
.
We next give the proof of Theorem 2.1 when x
M
,y
M
is in l
3
{x, y : x<0,y 0}.
Lemma 2.11. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ l
3
. Then the following
statements are true.
1 If −9/7 <x
M
< 0,then{x
n
,y
n
}
∞
nM
is eventually the equilibrium solution.
2 If x
M
−9/7, then the solution {x
n
,y
n
}
∞
nM1
is P
2
5
.
3 If x
M
< −9/7, then the solution {x
n
,y
n
}
∞
nM
is eventually P
1
5
.
Proof. 1 We will first prove statement 1 is true. Suppose −9/7 <x
M
< 0.
First consider the case −1 ≤ x
M
< 0. Then
x
M1
|
x
M
|
− y
M
− 1 −x
M
− 1,
y
M1
x
M
−
y
M
x
M
.
2.28
Advances in Difference Equations 11
In particular, −1 <x
M1
≤ 0andy
M1
−x
M
− 1andsobyLemma 2.2, {x
n
,y
n
}
∞
nM2
is the
equilibrium solution.
Next consider the case −9/7 <x
M
< −1. By statements 1 and 2 of Lemma 2.7,
x
M4
0andy
M4
−4x
M
− 5. In particular, −1 <y
M4
< 1/7 and so by statement 1 of
Lemma 2.8 and statement 1 of Lemma 2.9, {x
n
,y
n
}
∞
nM4
is eventually the equilibrium
solution.
2 We will next show that statement 2 is true. Suppose x
M
−9/7. By direct
calculations, we have x
M1
,y
M1
2/7, −9/7.Thatis,{x
n
,y
n
}
∞
nM1
is P
2
5
.
3 Lastly, we will show that statement 3 is true. Suppose x
M
< −9/7.
First consider the case −3/2 ≤ x
M
< −9/7. By statements 1 and 2 of Lemma 2.7,
x
M4
0andy
M4
−4x
M
− 5. In particular, 1/7 <y
M4
≤ 1 and so by statement 3 of
Lemma 2.8,thesolution{x
n
,y
n
}
∞
nM4
is eventually P
1
5
.
Next consider the case x
M
< −3/2. By statement 3 of Lemma 2.7,thesolution
{x
n
,y
n
}
∞
nM4
is P
1
5
.
We next give the proof of Theorem 2.1 when x
M
,y
M
is in Q
1
{x, y : x>0,y> 0}.
Lemma 2.12. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ Q
1
. Then the following
statements are true.
1 If y
M
≤ x
M
− 1, then the solution {x
n
,y
n
}
∞
nM2
is P
1
5
.
2 If y
M
>x
M
− 1, then there exists an integer N such that x
MN
,y
MN
∈ l
2
∪ l
4
.
Proof. Suppose x
M
> 0andy
M
> 0.
Then
x
M1
|
x
M
|
− y
M
− 1 x
M
− y
M
− 1,
y
M1
x
M
−
y
M
x
M
− y
M
.
2.29
Case 1. Suppose y
M
≤ x
M
−1. Then, in particular, x
M1
x
M
−y
M
−1 ≥ 0andy
M1
x
M
−y
M
>
0. Thus
x
M2
|
x
M1
|
− y
M1
− 1 −2,
y
M2
x
M1
−
y
M1
−1,
2.30
and so statement 1 is true.
Case 2. Suppose y
M
>x
M
− 1. Then, in particular, x
M1
x
M
− y
M
− 1 < 0.
Subcase 1. Suppose x
M
− y
M
< 0.
Then y
M1
x
M
− y
M
< 0. It follows by a straight forward computation, which will be
omitted, that x
M5
0. Hence x
M5
,y
M5
∈ l
2
∪ l
4
.
Subcase 2. Suppose x
M
− y
M
≥ 0.
Then y
M1
x
M
− y
M
≥ 0. It follows by a straight forward computation, which will
be omitted, that x
M6
0. Hence x
M6
,y
M6
∈ l
2
∪ l
4
, and the proof is complete.
We next give the proof of Theorem 2 .1 when x
M
,y
M
is in Q
3
{x, y : x<0,y<0}.
12 Advances in Difference Equations
Lemma 2.13. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ Q
3
. Then the following
statements are true.
1 If y
M
≥−x
M
− 1, then the solution {x
n
,y
n
}
∞
nM2
is the equilibrium solution.
2 If y
M
< −x
M
− 1,thenx
M4
,y
M4
∈ l
2
∪ l
4
.
Proof. By assumption, we have x
M
< 0andy
M
< 0.
If y
M
≥−x
M
− 1, then
x
M1
|
x
M
|
− y
M
− 1 −x
M
− y
M
− 1 ≤ 0,
y
M1
x
M
−
y
M
x
M
y
M
< 0,
x
M2
|
x
M1
|
− y
M1
− 1 0,
y
M2
x
M1
−
y
M1
−1.
2.31
Hence {x
n
,y
n
}
∞
nM2
is the equilibrium solution and statement 1 is true.
If y
M
< −x
M
− 1, then it follows by a straight forward computation, which will be
omitted, that x
M4
0. Thus x
M4
,y
M4
∈ l
2
∪ l
4
and statement 2 is true.
We next give the proof of Theorem 2 .1 when x
M
,y
M
is in Q
2
{x, y : x<0,y>0}.
Lemma 2.14. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ Q
2
. Then the following
statements are true.
1 If y
M
≥−x
M
− 1,thenx
M1
,y
M1
∈ Q
3
∪ l
4
.
2 If y
M
≤−x
M
− 3/2,thenx
M3
,y
M3
∈ Q
1
∪ l
1
.
3 If y
M
< −x
M
− 1, y
M
> −x
M
− 3/2 and x
M
≤−5/4 ,thenx
M4
,y
M4
∈ Q
1
∪ l
1
.
4 If y
M
< −x
M
−1, y
M
> −x
M
−3/2, x
M
> −5/4 and y
M
≤ x
M
5/4,thenx
M5
,y
M5
∈
Q
3
∪ l
4
.
5 If y
M
< −x
M
−1, y
M
> −x
M
−3/2, x
M
> −5/4 and y
M
>x
M
5/4,thenx
M6
,y
M6
∈
Q
3
∪ l
4
.
Proof. Now x
M
< 0andy
M
> 0.
1 If y
M
≥−x
M
− 1, then
x
M1
−x
M
− y
M
− 1 ≤ 0,
y
M1
x
M
− y
M
< 0.
2.32
Thus x
M1
,y
M1
∈ Q
3
∪ l
4
.
2 If y
M
≤−x
M
− 3/2, then x
M1
−x
M
− y
M
− 1 > 0. It follows by a straight forward
computation, which will be omitted, that
x
M3
−2x
M
2y
M
− 2 > 0,
y
M3
−2x
M
− 2y
M
− 3 ≥ 0.
2.33
Hence x
M3
,y
M3
∈ Q
1
∪ l
1
.
Advances in Difference Equations 13
3 If y
M
< −x
M
− 1, y
M
> −x
M
− 3/2, and x
M
≤−5/4, then x
M1
−x
M
− y
M
− 1 > 0.
It follows by a straight forward computation, which will be omitted, that
x
M4
4y
M
> 0,
y
M4
−4x
M
− 5 ≥ 0.
2.34
Thus x
4
,y
4
∈ Q
1
∪ l
1
.
4 If y
M
< −x
M
− 1, y
M
> −x
M
− 3/2, x
M
> −5/4, and y
M
≤ x
M
5/4, then x
M1
−x
M
− y
M
− 1 > 0. It follows by a straight forward computation, which will be
omitted, that
x
M5
4x
M
4y
M
4 < 0,
y
M5
−4x
M
4y
M
− 5 ≤ 0.
2.35
Thus x
M5
,y
M5
∈ Q
3
∪ l
4
.
5 Finally, suppose that y
M
< −x
M
−1, y
M
> −x
M
−3/2, x
M
> −5/4, and y
M
>x
M
5/4.
Then x
M1
−x
M
− y
M
− 1 > 0. It follows by a straight forward computation, which
will be omitted, that
x
M5
4x
M
4y
M
4 < 0,
y
M5
−4x
M
4y
M
− 5 > 0.
2.36
Note that
y
M5
−4x
M
4y
M
− 5 > −4x
M
− 4y
M
− 5 −x
M5
− 1 2.37
and so by the first statement of this Lemma, x
M6
,y
M6
∈ Q
3
∪ l
4
.
Thus we see that if there exists an integer N ≥ 0suchthatx
N
,y
N
/
∈ Q
4
, then the
proof of Theorem 2.1 is complete. Finally, we consider the case where the initial condition
x
M
,y
M
∈ Q
4
{x, y : x>0,y<0}.
Lemma 2.15. Suppose there exists an integer M ≥ 0 such that x
M
,y
M
∈ Q
4
. Then there exists a
positive integer N ≤ 4 such that x
MN
,y
MN
/
∈ Q
4
.
Proof. Without loss of generality, it suffices to consider the case where
x
Mn
,y
Mn
∈ Q
4
for 0 ≤ n ≤ 3. 2.38
Now x
M
,y
M
∈ Q
4
, and hence x
M
> 0andy
M
< 0.
Thus
x
M1
|
x
M
|
− y
M
− 1 x
M
− y
M
− 1,
y
M1
x
M
−
y
M
x
M
y
M
.
2.39
14 Advances in Difference Equations
We have x
M1
,y
M1
∈ Q
4
,andthus
x
M2
|
x
M1
|
− y
M1
− 1 −2y
M
− 2,
y
M2
x
M1
−
y
M1
2x
M
− 1.
2.40
We also have x
2
,y
2
∈ Q
4
, and hence
x
M3
|
x
M2
|
− y
M2
− 1 −2x
M
− 2y
M
− 2,
y
M3
x
M2
−
y
M2
2x
M
− 2y
M
− 3.
2.41
Finally, we have x
M3
,y
M3
∈ Q
4
,andso
x
M4
|
x
M3
|
− y
M3
− 1 −4x
M
< 0,
y
M4
x
M3
−
y
M3
−4y
M
− 5.
2.42
In particular, x
M4
< 0 and hence x
M4
,y
M4
/
∈ Q
4
.
3. Conclusion
We have presented the complete results concerning the global character of the solutions to
System 1.1. We divided the real plane into 8 sections and utilized mathematical induction,
proof by iteration, and direct computations to show that every solution of System 1.1 is
eventually either the prime period-5 solution P
1
5
, the prime period-5 solution P
2
5
,orelsethe
unique equilibrium point 0, −1. The proofs involve careful consideration of the various cases
and subcases.
Acknowledgments
The a uthors would like to express their gratitude to the Strategic Scholarships Fellowships
Frontier Research Networks, the Office of the Commission on Higher Education, and
National Center for Genetic Engineering and Biotechnology.
References
1 R. L. Devaney, “A piecewise linear model for the zones of instability of an area-preserving map,”
Physica D, vol. 10, no. 3, pp. 387–393, 1984.
2 H. O. Peitgen and D. Saupe, Eds., The Science of Fractal Images, Springer, New York, NY, USA, 1991.
3 E. A. Grove and G. Ladas, Periodicities in Nonlinear Difference Equations, Chapman & Hall/CRC, Boca
Raton, Fla, USA, 2005.
4 M. R. S. Kulenovi
´
candO.Merino,Discrete Dynamical Systems and Difference Equations with Mathematica,
Chapman & Hall/CRC, Boca Raton, Fla, USA, 2002.
