Volume 2010, Article ID 573281, 14 pagesdoi:10.1155/2010/573281 Research Article On the Global Character of the System of Piecewise y n1 x n − |y n | Wirot Tikjha,1, 2 Yongwimon Lenbury
Trang 1Volume 2010, Article ID 573281, 14 pages
doi:10.1155/2010/573281
Research Article
On the Global Character of the System of Piecewise
y n1 x n − |y n |
Wirot Tikjha,1, 2 Yongwimon Lenbury,1, 2
and Evelina Giusti Lapierre3
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 Ayudhya 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,scylb@mahidol.ac.th
Received 23 June 2010; Revised 4 September 2010; Accepted 2 December 2010
Academic Editor: Donal O’Regan
Copyrightq 2010 Wirot Tikjha et al 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 conditionx0, y0 ∈ R2 We show that the
system has exactly two prime period-5 solutions and a unique equilibrium point0, −1 We also
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,
where the initial conditionx0, y0 ∈ R2 We show that every solution of System1.1 is even-tually either one of two prime period-5 solutions or else the unique equilibrium point0, −1.
Trang 2System1.1 was motivated by Devaney’s Gingerbread man map 1,2
or its equivalent system of piecewise linear difference equations 3,4
xn1 |x n | − y n 1,
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
System1.1 has the equilibrium point x, y ∈ R2given by
System1.1 has two prime period-5 solutions,
P1
5
⎛
⎜
⎜
⎜
⎜
⎝
x0 0, y0 1
x1 −2, y1 −1
x2 2, y2 −3
x3 4, y3 −1
x4 4, y4 3
⎞
⎟
⎟
⎟
⎟
⎠
,
P2
5
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
x0 0, y0 1
7
x1 −8
7, y1 −1
7
x2 2
7, y2 −9
7
x3 4
7, y3 −1
x4 4
7, y4 −3
7.
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
.
2.2
Trang 3l1 x, y:x ≥ 0, y 0 ,
l2 x, y:x 0, y ≥ 0 ,
l3 x, y:x < 0, y 0 ,
l4 x, y:x 0, y < 0 ,
Q1 x, y:x > 0, y > 0 ,
Q2 x, y:x < 0, y > 0 ,
Q3 x, y:x < 0, y < 0 ,
Q4 x, y:x > 0, y < 0 .
2.3
Theorem 2.1 Let x0, y0 ∈ R2 Then there exists an integer N ≥ 0 such that the solution {x n , y n}∞nN is eventually either the prime period-5 solution P1
5, the prime period-5 solution P2
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 , yM1 0, −1, and so {x n, yn}∞nM1 is the equilibrium solution.
Proof Note that
x M1 |x M | − y M − 1 −x M − −x M − 1 − 1 0,
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, and so {x n , y n}∞
nM1 is P1
5 Proof We have
xM1 |x M | − y M − 1 x M − x M − 1 − 1 0,
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 {xn, yn}∞nM5 is P1
5.
3 If 0 ≤ y M ≤ 1/4, then y M5 8y M − 1.
Trang 4Proof We have xM 0 and y M≥ 0 Then
xM1 |x M | − y M − 1 −y M − 1 < 0, yM1 x M−y M −y M ≤ 0,
x M2 |x M1 | − y M1 − 1 2y M ≥ 0, yM2 x M1−y M1 −2y M − 1 < 0, xM3 |x M2 | − y M2 − 1 4y M ≥ 0,
y M3 x M2−y M2 −1, xM4 |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 statement1 is true
IfyM > 1/4, then yM5 x M4 − |y M4 | 1 That is, x M5, yM5 0, 1 and so
statement2 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, then y M4 −4y M − 5.
3 If y M ≤ −3/2, then {x n, yn}∞nM4 is P1
5 Proof We have x M 0 and y M < −1 Then
x M1 |x M | − y M − 1 −y M − 1 > 0,
y M1 x M−y M y M < 0, xM2 |x M1 | − y M1 − 1 −2y M − 2 > 0,
y M2 x M1−y M1 −1, xM3 |x M2 | − y M2 − 1 −2y M − 2 > 0, yM3 x M2−y M2 −2y M − 3,
x M4 |x M3 | − y M3 − 1 0,
2.7
and so statement1 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, ify 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.
Trang 5Lemma 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 P1
5.
2 If 1/4 < x M < 1, then {xn, yn}∞nM6 is P1
5.
3 If 0 ≤ x M ≤ 1/4, then x M6 0 and y M6 8x M − 1.
Proof First consider the case x M ≥ 1 and y 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, yM2 x M1−y M1 −1,
2.8
and so statement1 is true
Next consider the case 0≤ x M < 1 and yM 0 Then
x M1 |x M | − y M − 1 x M − 1 < 0, yM1 x M−y M x M ≥ 0, xM2 |x M1 | − y M1 − 1 −2x M ≤ 0, yM2 x M1−y M1 −1, xM3 |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, yM4 x M3−y M3 −1, xM5 |x M4 | − y M4 − 1 4x M ≥ 0, yM5 x M4−y M4 4x M − 1, xM6 |x M5 | − y M5 − 1 0.
2.9
If 1/4 < x M < 1, then y M5 4x M − 1 > 0 and so y M6 x M5 − |y M5| 1 That is,
x M6 , yM6 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 statement3 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, then yM4 −4x M − 5.
3 If x M < −3/2, then {x n , y n}∞
nM4 is P1
5.
Trang 6Proof Let xM < −1 and yM 0 Then
xM1 |x M | − y M − 1 −x M − 1 > 0, yM1 x M−y M x M < 0,
x M2 |x M1 | − y M1 − 1 −2x M − 2 > 0, yM2 x M1−y M1 −1, xM3 |x M2 | − y M2 − 1 −2x M − 2 > 0,
y M3 x M2−y M2 −2x M − 3, xM4 |x M3 | − y M3 − 1 0,
2.10
and so statement1 is true
If−3/2 ≤ x M < −1, then yM3 −2x M − 3 ≤ 0 Thus y M4 x M3 − |y M3 | −4x M− 5, and so statement2 is true
Ifx M < −3/2, then y M3 −2x M − 3 > 0 and y M4 x M3 − |y M3| 1 That is,
x M4 , yM4 0, 1 and so {x n, yn}∞nM4isP1
5 and the proof is complete
We now give the proof ofTheorem 2.1whenx M, yM is in l2 {x, y : x 0, y ≥ 0}.
Lemma 2.8 Suppose there exists an integer M ≥ 0 such that x M , y M ∈ l2 Then the following statements are true.
1 If 0 ≤ y M < 1/7, then {xn, yn}∞nM is eventually the equilibrium solution.
2 If y M 1/7, then the solution {x n, yn}∞nM2 is P2
5.
3 If y M > 1/7, then the solution {xn, yn}∞nM is eventually P1
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 23n− 1
Observe that
0 a0< a1 < a2 < · · · <1
7, lim
n → ∞ an 1
Thus there exists a unique integerK ≥ 0 such that y M ∈ a K , a K1
We first consider the caseK 0; that is, yM ∈ 0, 1/8 By statements 1 and 3 of
Lemma 2.4,xM5 0 and y M5 8y M − 1 Clearly y M5 < 0, and so
xM6 |x M5 | − y M5 − 1 −8y M ≤ 0,
Trang 7Now−1 < x M6 ≤ 0 and y M6 −x M6 − 1, and so by Lemma 2.2,{x n, yn}∞nM7 is the equilibrium solution
Without loss of generality, we may assumeK ≥ 1.
For each integern such that n ≥ 0, let Pn be the following statement:
x M5n5 0,
y M5n5 23n1y M− 23n1− 1
7
Claim 1 Pn is true for 0 ≤ n ≤ K − 1.
The proofClaim 1will be by induction onn We will first show that P0 is true.
Recall thatxM 0 and y M ∈ a K, aK1 ⊂ 1/8, 1/7 Then by statements 1 and 3
ofLemma 2.4, we havex M505 0 and y M505 8y M− 1
Note that,
y M505 8y M− 1 2301y M− 2301− 1
7
and soP0 is true Thus if K 1, then we have shown that for 0 ≤ n ≤ K − 1, Pn is true It
remains to consider the caseK ≥ 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.
SincePn is true, we know
xM5n5 0, yM5n5 23n1yM− 23n1− 1
7
It is easy to verify that foryM ∈ 1/8, 1/7,
y M5n5 23n1y M− 23n1− 1
7
< 1
Thus by statements1 and 3 ofLemma 2.4,
x M5n15 0,
y M5n15 8y M5n5− 1
23
23n1y M− 23n1− 1
7
− 1
23n6 yM−23n6
7 23
7 − 1
23n2yM− 23n2− 1
7
.
2.18
Trang 8Recall thatyM ∈ a K, aK1 23K − 1/7 · 23K , 23K1− 1/7 · 23K1.
In particular,
y M5n15 23n2y M− 23n2− 1
7
≥ 23n2 23K− 1
7· 23K
− 23n2− 1
7
23n3K6
7· 23K − 23n6
7· 23K −23n6
7
1 7
1− 23n−K−2
≥ 1
71 − 1
0,
2.19
and soPn 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 23K y M− 23K− 1
7
In particular,
23K 2
3K− 1
7· 23K
− 23K− 1 7
≤ y M5K−15 < 23K 23K3− 1
7· 23K3
− 23K− 1 7
That is, 0 ≤ y M5K−15 < 1/8, and so by case K 0, {xn, yn}∞nM5K7 is the equilibrium solution, and the proof of statement1 is complete
2 We will next show that statement 2 is true Suppose x M , y M 0, 1/7 Note
that0, 1/7 ∈ P2
5 Thus the solution{x n, yn}∞nMisP2
5
3 Finally, we will show that statement 3 is true Suppose y M > 1/7.
First consideryM > 1/4 By statement 2 ofLemma 2.4, the solution{x n, yn}∞nM5
isP1
5
Next consider the caseyM ∈ 1/7, 1/4 For each n ≥ 1, let
bn 23n−1 3
Observe that
1
4 b1 > b2> b3> · · · > 1
7, lim
n → ∞ bn 1
Thus there exists a unique integerK ≥ 1 such that y M ∈ b K1 , b K
Trang 9Note that the statementPn which we stated and proved in the proof of statement
1 of this lemma still holds Specifically PK − 1 is true, and so
xM5K−15 0, yM5K−15 23K yM− 23K− 1
7
Recall that fory M ∈ b K1 , b K
In particular,
yM5K 23K yM− 23K− 1
7
> 23K 23K2 3
7· 23K2
− 23K− 1 7
1
By statement2 ofLemma 2.4, the solution{x n , y n}∞nM5K5isP1
5
We now give the proof ofTheorem 2.1whenx M, yM is in l4 {x, y : x 0, y < 0}.
Lemma 2.9 Suppose there exists an integer M ≥ 0 such that x M, yM ∈ l4 Then the following statements are true.
1 If −9/7 < y M < 0, then {xn, yn}∞nM is eventually the equilibrium solution.
2 If y M −9/7, then the solution {x n , y n}∞
nM1 is P2
5.
3 If y M < −9/7, then the solution {xn, yn}∞nM is eventually P1
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
xM1 |x M | − y M − 1 −y M − 1 ≤ 0,
In particular,−1 < x M1 ≤ 0 and y M1 −x M1 − 1, and so byLemma 2.2,{x n , y n}∞
nM2is the equilibrium solution
Case 2 Suppose −5/4 ≤ y M < −1 By statements 1 and 2 ofLemma 2.5,xM4 0 and
y M4 −4y M− 5 Then
xM5 |x M4 | − y M4 − 1 4y M 4 < 0,
Thus−1 ≤ x M5 < 0 and yM5 −x M5− 1, and so byLemma 2.2,{x n, yn}∞nM6is the equilibrium solution
Case 3 Suppose −9/7 < y M < −5/4 By statements 1 and 2 of Lemma 2.5,xM4 0 and yM4 −4y M − 5 Note that 0 < y M4 < 1/7 and so by statement 1 ofLemma 2.8,
{x n , y n}∞
nM4is eventually equilibrium solution
Trang 102 We will next show that statement 2 is true Suppose y M −9/7 By direct
calcu-lations we havex M1 , y M1 2/7, −9/7 So the solution {x n , y n}∞
nM1isP2
5
3 Finally, we will show that statement 3 is true Suppose x M 0 and y M < −9/7 Case 1 Suppose −3/2 < y M < −9/7 By statements 1 and 2 ofLemma 2.5, we havexM4
0 andy M4 −4y M − 5 Note that 1/7 < y M4 < 1 and so by statement 3 ofLemma 2.8, the solution{x n, yn}∞nM4is eventuallyP1
5
Case 2 Suppose yM ≤ −3/2 By statement 3 ofLemma 2.5, the solution{x n, yn}∞nM4is
P1
5
We now give the proof ofTheorem 2.1whenx M , y M is in l1 {x, y : x ≥ 0, y 0}.
Lemma 2.10 Suppose there exists an integer M ≥ 0 such that x M , y M ∈ l1 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, yn}∞nM3 is P2
5.
3 If x M > 1/7, then the solution {x n , y n}∞
nM is eventually P1
5.
Proof 1 We will first show that statement 1 is true So suppose 0 ≤ x M < 1/7 and yM 0
By statement3 ofLemma 2.6,x M6 0 and y M6 8x M − 1 In particular, −1 < y M6 < 1/7
and so by statement 1 ofLemma 2.8 and statement1 of Lemma 2.9, {x n, yn}∞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 havex M3 , yM3 2/7, −9/7 Thus the solution {x n, yn}∞nM3isP2
5
3 Finally, we will show statement 3 is true
First consider the case 1/7 < x M ≤ 1/4 By statement 3 ofLemma 2.6,x M6 0 and
yM6 8x M − 1 Now, 1/7 < y M6 ≤ 1 and so by statement 3 ofLemma 2.8, the solution
{x n, yn}∞nM6is eventuallyP1
5 Next consider the casex M > 1/4 Then by statements 1 and 2 of Lemma 2.6, if
xM ≥ 1 then {x n, yn}∞nM2isP1
5, and if 1/4 < xM < 1 then {xn, yn}∞nM6isP1
5
We next give the proof ofTheorem 2.1whenx M, yM is in l3 {x, y : x < 0, y 0}.
Lemma 2.11 Suppose there exists an integer M ≥ 0 such that x M, yM ∈ l3 Then the following statements are true.
1 If −9/7 < x M < 0, then {xn, yn}∞nM is eventually the equilibrium solution.
2 If x M −9/7, then the solution {x n , y n}∞nM1 is P2
5.
3 If x M < −9/7, then the solution {xn, yn}∞nM is eventually P1
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
xM1 |x M | − y M − 1 −x M − 1,
Trang 11In particular,−1 < x M1 ≤ 0 and y M1 −x M− 1 and so byLemma 2.2,{x n, yn}∞nM2is the equilibrium solution
Next consider the case−9/7 < x M < −1 By statements 1 and 2 ofLemma 2.7,
xM4 0 and y M4 −4x M − 5 In particular, −1 < y M4 < 1/7 and so by statement 1 of
Lemma 2.8and 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 havex M1 , y M1 2/7, −9/7 That is, {x n , y n}∞
nM1isP2
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 ofLemma 2.7,
x M4 0 and y M4 −4x M − 5 In particular, 1/7 < y M4 ≤ 1 and so by statement 3 of
Lemma 2.8, the solution{x n , y n}∞nM4is eventuallyP1
5 Next consider the case xM < −3/2 By statement 3 of Lemma 2.7, the solution
{x n , y n}∞
nM4isP1
5
We next give the proof ofTheorem 2.1whenx M , y M is in Q1 {x, y : x > 0, y > 0}.
Lemma 2.12 Suppose there exists an integer M ≥ 0 such that x M, yM ∈ Q1 Then the following statements are true.
1 If y M ≤ x M − 1, then the solution {x n , y n}∞
nM2 is P1
5.
2 If y M > x M − 1, then there exists an integer N such that x MN , y MN ∈ l2∪ l4 Proof Suppose xM > 0 and yM > 0.
Then
x M1 |x M | − y M − 1 x M − y M − 1,
Case 1 Suppose y M ≤ x M −1 Then, in particular, x M1 x M −y M −1 ≥ 0 and y M1 x M −y M >
0 Thus
xM2 |x M1 | − y M1 − 1 −2,
and so statement1 is true
Case 2 Suppose yM > xM − 1 Then, in particular, x M1 x M − y M − 1 < 0.
Subcase 1 Suppose xM − y M < 0.
ThenyM1 x M − y M < 0 It follows by a straight forward computation, which will be
omitted, thatx M5 0 Hence x M5 , y M5 ∈ l2∪ l4
Subcase 2 Suppose x M − y M≥ 0
ThenyM1 x M − y M ≥ 0 It follows by a straight forward computation, which will
be omitted, thatx M6 0 Hence x M6 , y M6 ∈ l2∪ l4, and the proof is complete
We next give the proof ofTheorem 2.1whenx M , y M is in Q3 {x, y : x < 0, y < 0}.
...remains to consider the caseK ≥ So assume that K ≥ Let n be an integer such that
0≤ n ≤ K − and suppose Pn is true We will show that Pn 1 is true.
SincePn... 5
Lemma 2.6 Suppose there exists an integer M ≥ such that x M ≥ and y M Then the following statements are true.... 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 ≥ such that −1 ≤ x