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 n1  |x n |−y n − 1 and y n1  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, scylb@mahidol.ac.th 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 n1  | x n | − y n − 1, y n1  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 n1  | x n | − x n−1  1 1.2 or its equivalent system of piecewise linear difference equations 3, 4 x n1  | x n | − y n  1, y n1  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 } ∞ nN 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 M1 ,y M1 0, −1,andso{x n ,y n } ∞ nM1 is the equilibrium solution. Proof. Note that x M1  | x M | − y M − 1  −x M −  −x M − 1  − 1  0, y M1  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 M1 ,y M1 0, 1,andso{x n ,y n } ∞ nM1 is P 1 5 . Proof. We have x M1  | x M | − y M − 1  x M −  x M − 1  − 1  0, y M1  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 M5  0. 2 If y M > 1/4,then{x n ,y n } ∞ nM5 is P 1 5 . 3 If 0 ≤ y M ≤ 1/4,theny M5  8y M − 1. 4AdvancesinDifference Equations Proof. We have x M  0andy M ≥ 0. Then x M1  | x M | − y M − 1  −y M − 1 < 0, y M1  x M −   y M    −y M ≤ 0, x M2  | x M1 | − y M1 − 1  2y M ≥ 0, y M2  x M1 −   y M1    −2y M − 1 < 0, x M3  | x M2 | − y M2 − 1  4y M ≥ 0, y M3  x M2 −   y M2    −1, x M4  | x M3 | − y M3 − 1  4y M ≥ 0, y M4  x M3 −   y M3    4y M − 1, x M5  | x M4 | − y M4 − 1  0, 2.6 and so statement 1 is true. If y M > 1/4, then y M5  x M4 −|y M4 |  1. That is, x M5 ,y M5 0, 1 and so statement 2 is true. If 0 ≤ y M ≤ 1/4, then y M5  x M4 −|y M4 |  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 M4  0. 2 If −3/2 <y M < −1,theny M4  −4y M − 5. 3 If y M ≤−3/2,then{x n ,y n } ∞ nM4 is P 1 5 . Proof. We have x M  0andy M < −1. Then x M1  | x M | − y M − 1  −y M − 1 > 0, y M1  x M −   y M    y M < 0, x M2  | x M1 | − y M1 − 1  −2y M − 2 > 0, y M2  x M1 −   y M1    −1, x M3  | x M2 | − y M2 − 1  −2y M − 2 > 0, y M3  x M2 −   y M2    −2y M − 3, x M4  | x M3 | − y M3 − 1  0, 2.7 and so statement 1 is true. Now if −3/2 <y M < −1, then y M3  −2y M − 3 < 0. Thus y M4  x M3 −|y M3 |  −4y M − 5, and so statement 2 is true. Lastly, if y M ≤−3/2, then y M3  −2y M − 3 ≥ 0. Thus y M4  x M3 −|y M3 |  1; that is, x M4 ,y M4 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 } ∞ nM2 is P 1 5 . 2 If 1/4 <x M < 1,then{x n ,y n } ∞ nM6 is P 1 5 . 3 If 0 ≤ x M ≤ 1/4,thenx M6  0 and y M6  8x M − 1. Proof. First consider the case x M ≥ 1andy M  0. Then x M1  | x M | − y M − 1  x M − 1 ≥ 0, y M1  x M −   y M    x M > 0, x M2  | x M1 | − y M1 − 1  −2, y M2  x M1 −   y M1    −1, 2.8 and so statement 1 is true. Next consider the case 0 ≤ x M < 1andy M  0. Then x M1  | x M | − y M − 1  x M − 1 < 0, y M1  x M −   y M    x M ≥ 0, x M2  | x M1 | − y M1 − 1  −2x M ≤ 0, y M2  x M1 −   y M1    −1, x M3  | x M2 | − y M2 − 1  2x M ≥ 0, y M3  x M2 −   y M2    −2x M − 1 < 0, x M4  | x M3 | − y M3 − 1  4x M ≥ 0, y M4  x M3 −   y M3    −1, x M5  | x M4 | − y M4 − 1  4x M ≥ 0, y M5  x M4 −   y M4    4x M − 1, x M6  | x M5 | − y M5 − 1  0. 2.9 If 1/4 <x M < 1, then y M5  4x M − 1 > 0andsoy M6  x M5 −|y M5 |  1. That is, x M6 ,y M6 0, 1 and so statement 2  is true. If 0 ≤ x M ≤ 1/4, then y M5  4x M − 1 ≤ 0. Thus y M6  x M5 −|y M5 |  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 M4  0. 2 If −3/2 ≤ x M < −1,theny M4  −4x M − 5. 3 If x M < −3/2,then{x n ,y n } ∞ nM4 is P 1 5 . 6AdvancesinDifference Equations Proof. Let x M < −1andy M  0. Then x M1  | x M | − y M − 1  −x M − 1 > 0, y M1  x M −   y M    x M < 0, x M2  | x M1 | − y M1 − 1  −2x M − 2 > 0, y M2  x M1 −   y M1    −1, x M3  | x M2 | − y M2 − 1  −2x M − 2 > 0, y M3  x M2 −   y M2    −2x M − 3, x M4  | x M3 | − y M3 − 1  0, 2.10 and so statement 1 is true. If −3/2 ≤ x M < −1, then y M3  −2x M − 3 ≤ 0. Thus y M4  x M3 −|y M3 |  −4x M − 5, and so statement 2 is true. If x M < −3/2, then y M3  −2x M − 3 > 0andy M4  x M3 −|y M3 |  1. That is, x M4 ,y M4 0, 1 and so {x n ,y n } ∞ nM4 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 } ∞ nM is eventually the equilibrium solution. 2 If y M  1/7, then the solution {x n ,y n } ∞ nM2 is P 2 5 . 3 If y M > 1/7, then the solution {x n ,y n } ∞ nM 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 K1 . We first consider the case K  0; that is, y M ∈ 0, 1/8. By statements 1 and 3 of Lemma 2.4, x M5  0andy M5  8y M − 1. Clearly y M5 < 0, and so x M6  | x M5 | − y M5 − 1  −8y M ≤ 0, y M6  x M5 −   y M5    8y M − 1. 2.13 Advances in Difference Equations 7 Now −1 <x M6 ≤ 0andy M6  −x M6 − 1, and so by Lemma 2.2, {x n ,y n } ∞ nM7 is the equilibrium solution. Without loss of generality, we may assume K ≥ 1. For each integer n such that n ≥ 0, let Pn be the following statement: x M5n5  0, y M5n5  2 3n1 y M −  2 3n1 − 1 7  ≥ 0. 2.14 Claim 1. Pn is true for 0 ≤ n ≤ K − 1. The proof Claim 1 will be by induction on n. We will first show that P0 is true. Recall that x M  0andy M ∈ a K ,a K1  ⊂ 1/8, 1/7. Then by statements 1 and 3 of Lemma 2.4,wehavex M505  0andy M505  8y M − 1. Note that, y M505  8y M − 1  2 301 y M −  2 301 − 1 7  ≥ 0 2.15 and so P0 is true. Thus if K  1, then we have shown that for 0 ≤ n ≤ K − 1, Pn 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 Pn is true. We will show that Pn  1 is true. Since Pn is true, we know x M5n5  0,y M5n5  2 3n1 y M −  2 3n1 − 1 7  ≥ 0. 2.16 It is easy to verify that for y M ∈ 1/8, 1/7, y M5n5  2 3n1 y M −  2 3n1 − 1 7  < 1 4 . 2.17 Thus by statements 1 and 3 of Lemma 2.4, x M5n15  0, y M5n15  8  y M5n5  − 1  2 3  2 3n1 y M −  2 3n1 − 1 7  − 1  2 3n6 y M − 2 3n6 7  2 3 7 − 1  2 3n2 y M −  2 3n2 − 1 7  . 2.18 8AdvancesinDifference Equations Recall that y M ∈ a K ,a K1 2 3K − 1/7 · 2 3K , 2 3K1 − 1/7 · 2 3K1 . In particular, y M5n15  2 3n2 y M −  2 3n2 − 1 7  ≥ 2 3n2  2 3K − 1 7 · 2 3K  −  2 3n2 − 1 7   2 3n3K6 7 · 2 3K − 2 3n6 7 · 2 3K − 2 3n6 7  1 7  1 7  1 − 2 3n−K−2  ≥ 1 7  1 − 1   0, 2.19 and so Pn  1 is true. Thus the proof of the claim is complete. That is, Pn is true for 0 ≤ n ≤ K − 1. Specifically, PK − 1 is true, and so x M5K−15  0,y M5K−15  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 M5K−15 < 2 3K  2 3K3 − 1 7 · 2 3K3  −  2 3K − 1 7  . 2.21 That is, 0 ≤ y M5K−15 < 1/8, and so by case K  0, {x n ,y n } ∞ nM5K7 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 } ∞ nM 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 } ∞ nM5 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 K1 ,b K . Advances in Difference Equations 9 Note that the statement Pn which we stated and proved in the proof of statement 1 of this lemma still holds. Specifically PK − 1 is true, and so x M5K−15  0,y M5K−15  2 3K y M −  2 3K − 1 7  ≥ 0. 2.24 Recall that for y M ∈ b K1 ,b K . In particular, y M5K  2 3K y M −  2 3K − 1 7  > 2 3K  2 3K2  3 7 · 2 3K2  −  2 3K − 1 7   1 4 . 2.25 By statement 2 of Lemma 2.4,thesolution{x n ,y n } ∞ nM5K5 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 } ∞ nM is eventually the equilibrium solution. 2 If y M  −9/7, then the solution {x n ,y n } ∞ nM1 is P 2 5 . 3 If y M < −9/7, then the solution {x n ,y n } ∞ nM 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 M1  | x M | − y M − 1  −y M − 1 ≤ 0, y M1  x M −   y M    y M . 2.26 In particular, −1 <x M1 ≤ 0andy M1  −x M1 − 1, and so by Lemma 2.2, {x n ,y n } ∞ nM2 is the equilibrium solution. Case 2. Suppose −5/4 ≤ y M < −1. By statements 1 and 2 of Lemma 2.5, x M4  0and y M4  −4y M − 5. Then x M5  | x M4 | − y M4 − 1  4y M  4 < 0, y M5  x M4 −   y M4    −4y M − 5. 2.27 Thus −1 ≤ x M5 < 0andy M5  −x M5 − 1, and so by Lemma 2.2, {x n ,y n } ∞ nM6 is the equilibrium solution. Case 3. Suppose −9/7 <y M < −5/4. By statements 1 and 2 of Lemma 2.5, x M4  0 and y M4  −4y M − 5. Note that 0 <y M4 < 1/7 and so by statement 1 of Lemma 2.8, {x n ,y n } ∞ nM4 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 M1 ,y M1 2/7, −9/7.Sothesolution{x n ,y n } ∞ nM1 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 M4  0andy M4  −4y M − 5. Note that 1/7 <y M4 < 1 and so by statement 3 of Lemma 2.8,the solution {x n ,y n } ∞ nM4 is eventually P 1 5 . Case 2. Suppose y M ≤−3/2. By statement 3 of Lemma 2.5,thesolution{x n ,y n } ∞ nM4 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 } ∞ nM is eventually the equilibrium solution. 2 If x M  1/7, then the solution {x n ,y n } ∞ nM3 is P 2 5 . 3 If x M > 1/7, then the solution {x n ,y n } ∞ nM 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 M6  0andy M6  8x M − 1. In particular, −1 <y M6 < 1/7 and so by statement 1 of Lemma 2.8 and statement 1 of Lemma 2.9, {x n ,y n } ∞ nM6 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 M3 ,y M3 2/7, −9/7. Thus the solution {x n ,y n } ∞ nM3 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 M6  0and y M6  8x M − 1. Now, 1/7 <y M6 ≤ 1 and so by statement 3 of Lemma 2.8,thesolution {x n ,y n } ∞ nM6 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 } ∞ nM2 is P 1 5 ,andif1/4 <x M < 1then{x n ,y n } ∞ nM6 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 } ∞ nM is eventually the equilibrium solution. 2 If x M  −9/7, then the solution {x n ,y n } ∞ nM1 is P 2 5 . 3 If x M < −9/7, then the solution {x n ,y n } ∞ nM 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 M1  | x M | − y M − 1  −x M − 1, y M1  x M −   y M    x M . 2.28 [...]... 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... omitted, that xM 6 0 Hence xM 6 , yM 6 ∈ l2 ∪ l4 , and the proof is complete We next give the proof of Theorem 2.1 when xM , yM is in Q3 { x, y : x < 0, y < 0} 12 Advances in Difference Equations Lemma 2.13 Suppose there exists an integer M ≥ 0 such that xM , yM ∈ Q3 Then the following statements are true 1 If yM ≥ −xM − 1, then the solution { xn , yn }∞ M n 2 If yM < −xM − 1, then xM 4 , yM 2 is the equilibrium... hence xM > 0 and yM < 0 Thus xM 1 yM |xM | − yM − 1 1 xM − yM xM − yM − 1, xM yM 2.39 14 Advances in Difference Equations We have xM 1 , yM 1 ∈ Q4 , and thus xM |xM 1 | − yM 2 yM xM 2 1 −1 1 − yM −2yM − 2, 2xM − 1 1 2.40 We also have x2 , y2 ∈ Q4 , and hence xM |xM 2 | − yM 3 yM xM 3 Finally, we have xM 3 , yM xM In particular, xM 4 − yM −1 4 −2xM − 2yM − 2, 2xM − 2yM − 3 2 2.41 ∈ Q4 , and so |xM 3... 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 1 2 eventually either the prime period-5 solution P5 , the prime period-5 solution P5 , or else the unique equilibrium point 0, −1 The proofs involve careful consideration.. .Advances in Difference Equations 11 In particular, −1 < xM 1 ≤ 0 and yM 1 −xM − 1 and so by Lemma 2.2, { xn , yn }∞ M 2 is the n equilibrium solution Next consider the case −9/7 < xM < −1 By statements 1 and 2 of Lemma 2.7, −4xM − 5 In particular, −1 < yM 4 < 1/7 and so by statement 1 of xM 4 0 and yM 4 Lemma 2.8 and... yN / Q4 , then the ∈ proof of Theorem 2.1 is complete Finally, we consider the case where the initial condition xM , yM ∈ Q4 { x, y : x > 0, y < 0} Lemma 2.15 Suppose there exists an integer M ≥ 0 such that xM , yM ∈ Q4 Then there exists a ∈ positive integer N ≤ 4 such that xM N , yM N / Q4 Proof Without loss of generality, it suffices to consider the case where xM n , yM n ∈ Q4 for 0 ≤ n ≤ 3 2.38... 0} Lemma 2.12 Suppose there exists an integer M ≥ 0 such that xM , yM ∈ Q1 Then the following statements are true 1 If yM ≤ xM − 1, then the solution { xn , yn }∞ M n 2 1 is P5 2 If yM > xM − 1, then there exists an integer N such that xM N , yM N ∈ l2 ∪ l4 Proof Suppose xM > 0 and yM > 0 Then xM 1 yM |xM | − yM − 1 xM − yM − 1, Case 1 Suppose yM ≤ xM −1 Then, in particular, xM 0 Thus xM |xM 1 | −... < −9/7 First consider the case −3/2 ≤ xM < −9/7 By statements 1 and 2 of Lemma 2.7, −4xM − 5 In particular, 1/7 < yM 4 ≤ 1 and so by statement 3 of xM 4 0 and yM 4 1 Lemma 2.8, the solution { xn , yn }∞ M 4 is eventually P5 n Next consider the case xM < −3/2 By statement 3 of Lemma 2.7, the solution 1 { xn , yn }∞ M 4 is P5 n We next give the proof of Theorem 2.1 when xM , yM is in Q1 { x, y : x... computation, which will be omitted, that xM 4 0 Thus xM 4 , yM 4 ∈ l2 ∪ l4 and statement 2 is true We next give the proof of Theorem 2.1 when xM , yM is in Q2 { x, y : x < 0, y > 0} Lemma 2 .14 Suppose there exists an integer M ≥ 0 such that xM , yM ∈ Q2 Then the following statements are true 1 If yM ≥ −xM − 1, then xM 1 , yM ∈ Q3 ∪ l4 1 2 If yM ≤ −xM − 3/2, then xM 3 , yM 3 ∈ Q1 ∪ l1 3 If yM < −xM − 1, yM... − 3/2, then xM 1 −xM − yM − 1 > 0 It follows by a straight forward computation, which will be omitted, that xM 3 −2xM yM Hence xM 3 , yM 3 3 −2xM − 2yM − 3 ≥ 0 ∈ Q1 ∪ l1 2yM − 2 > 0, 2.33 Advances in Difference Equations 13 3 If yM < −xM − 1, yM > −xM − 3/2, and xM ≤ −5/4, then xM 1 −xM − yM − 1 > 0 It follows by a straight forward computation, which will be omitted, that xM yM 4yM > 0, 4 2.34 −4xM . 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. equilibrium point. 1. Introduction In this paper, we consider the system of piecewise linear difference equations x n1  | x n | − y n − 1, y n1  x n −   y n   , n  0, 1, 2, , 1.1 where the initial. point 0, −1. 2AdvancesinDifference Equations System 1.1 was motivated by Devaney’s Gingerbread man map 1, 2 x n1  | x n | − x n−1  1 1.2 or its equivalent system of piecewise linear difference