CUBO A Mathematical Journal Vol.14, No¯ 02, (111–152). June 2012 On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| E.A. Grove Department of Mathematics, University of Rhode Island, Kingston, Rhode Island, 02881-0816, USA email: grove@math.uri.edu and E. Lapierre 2 Department of Mathematics, Johnson and Wales University, Providence, Rhode Island 02903, USA. email: elapierre@jwu.edu and W. Tikjha Faculty of Science and Technology, Pibulsongkram Rajabhat University, Muang District, Phitsanuloke, 65000, Thailand email: wirot tik@yahoo.com ABSTRACT In this paper we consider the system of piecewise linear difference equations in the title, where the initial conditions x0 and y0 are real numbers. We show that there exists a unique equilibrium solution and exactly two prime period-3 solutions, and that except for the unique equilibrium solution, every solution of the system is eventually one of the two prime period-3 solutions. RESUMEN En este art́ıculo consideramos el sistema de ecuaciones en diferencia lineales por partes indicado en el t́ıtulo, donde las condiciones iniciales x0 e y0 son números reales. De- mostramos que existe una única solución de equilibrio y exactamente dos soluciones de peŕıodo 3-primo, y que exceptuando la solución única de equilibrio, toda solución del sistema es eventualmente una de las dos soluciones de periodo 3-primo. Keywords and Phrases: Periodic solution; systems of piecewise linear difference equations 2010 AMS Mathematics Subject Classification: 39A10, 65Q10. 2 Corresponding author. 112 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) 1 Introduction In this paper we consider the system of piecewise linear difference equations        xn+1 = |xn| − yn − 1 , n = 0, 1, . . . yn+1 = xn + |yn| (1.1) where the initial conditions x0 and y0 are arbitrary real numbers. We show that every solution of System(1.1) is either (from the beginning) the unique equilibrium point (x̄, ȳ) = ( − 2 5 , − 1 5 ) or else is eventually one of the following period-3 cycles: P 1 3 =          x0 = 0 , y0 = −1 x1 = 0 , y1 = 1 x2 = −2 , y2 = 1          or P2 3 =            x0 = 0 , y0 = − 1 3 x1 = − 2 3 , y1 = 1 3 x2 = − 2 3 , y2 = − 1 3            . This study of System(1.1) was motivated by Devaney’s celebrated Gingerbreadman map        xn+1 = |xn| − yn + 1 , n = 0, 1, . . . . yn+1 = xn See Ref. [1, 2, 3, 4]. We believe that the methods and techniques used in this paper will be useful in discovering the global behavior of similar piecewise linear systems of the form        xn+1 = |xn| + ayn + b , n = 0, 1, 2... yn+1 = xn + c |yn| + d For another system of this form see [5]. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 113 2 The Global Behavior Of The Solutions Of System(1.1) Set l1 = {(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}. Theorem 1. Let {(xn, yn)} ∞ n=0 be a solution of System(1.1) with (x0, y0) ∈ R 2. Then either {(xn, yn)} ∞ n=0 is the unique equilibrium (x̄, ȳ), or else there exists a non-negative integer N ≥ 0 such that the solution {(xn, yn)} ∞ n=N of System(1.1) is either the prime period-3 cycle P 1 3 or the prime period-3 cycle P2 3 . The proof of Theorem 1 is a direct consequence of the following lemmas. Lemma 2. Suppose there exists a non-negative integer N ≥ 0 such that yN = −xN − 1 and yN ≥ 0. Then (xN+1, yN+1) = (0, −1), and so {(xn, yn)} ∞ n=N+1 is the period-3 cycle P 1 3 . Proof. Note that xN = −yN − 1 ≤ −1, and so xN+1 = |xN| − yN − 1 = −xN − (−xN − 1) − 1 = 0 yN+1 = xN + |yN| = xN + (−xN − 1) = −1. The proof is complete. Lemma 3. Suppose there exists a non-negative integer N ≥ 0 such that (xN, yN) ∈ l2. Then {(xn, yn)} ∞ n=N+2 is the period-3 cycle P 1 3 . 114 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Proof. We have xN+1 = |xN| − yN − 1 = 0 − yN − 1 = −yN − 1 < 0 yN+1 = xN + |yN| = 0 + yN = yN ≥ 0 and so it follows by Lemma 2 that {(xn, yn)} ∞ n=N+2 is the period-3 cycle P 1 3 . Lemma 4. Suppose there exists a non-negative integer N ≥ 0 such that xN = 0 and yN < −1. Then (1) xN+3 = 2yN + 2 < 0. (2) If − 3 2 ≤ yN < −1, then yN+3 = −2yN − 3 ≤ 0. (3) If yN < − 3 2 , then {(xn, yn)} ∞ n=N+4 is the period-3 cycle P 1 3 . Proof. We have xN+1 = |xN| − yN − 1 = −yN − 1 > 0 yN+1 = xN + |yN| = −yN > 0 xN+2 = |xN+1| − yN+1 − 1 = −2 yN+2 = xN+1 + |yN+1| = −2yN − 1 > 0 xN+3 = |xN+2| − yN+2 − 1 = 2yN + 2 < 0 yN+3 = xN+2 + |yN+2| = −2yN − 3. If − 3 2 ≤ yN < −1, then yN+3 = −2yN − 3 ≤ 0. If yN < − 3 2 , then yN+3 = −2yN − 3 > 0 and so by Lemma 2 {(xn, yn)} ∞ n=N+4 is the period-3 cycle P 1 3 . The proof is complete. Lemma 5. Suppose there exists a non-negative integer N ≥ 0 such that xN = 0 and −1 < yN ≤ 0. Then (1) If − 1 4 < yN ≤ 0, then {(xn, yn)} ∞ n=N+5 is the period-3 cycle P 1 3 . (2) If − 1 2 < yN ≤ − 1 4 , then xN+5 = 8yN + 2, yN+5 = −8yN − 3, and xN+6 = 0. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 115 (3) If −1 < yN ≤ − 1 2 , then {(xn, yn)} ∞ n=N+6 is the period-3 cycle P 1 3 . Proof. We have xN+1 = |xN| − yN − 1 = −yN − 1 < 0 yN+1 = xN + |yN| = −yN ≥ 0 xN+2 = |xN+1| − yN+1 − 1 = 2yN ≤ 0 yN+2 = xN+1 + |yN+1| = −2yN − 1 xN+3 = |xN+2| − yN+2 − 1 = 0. If − 1 4 < yN ≤ 0, then yN+2 < 0 and yN+3 = xN+2 + |yN+2| = 4yN +1 > 0. It follows by Lemma 3 that {(xn, yn)} ∞ n=N+5 is the period-3 cycle P 1 3 , and so Statement 1 is true. If − 1 2 < yN ≤ − 1 4 , then yN+2 < 0 and yN+3 = xN+2 + |yN+2| = 4yN + 1 ≤ 0 xN+4 = |xN+3| − yN+3 − 1 = −4yN − 2 < 0 yN+4 = xN+3 + |yN+3| = −4yN − 1 ≥ 0 xN+5 = |xN+4| − yN+4 − 1 = 8yN + 2 ≤ 0 yN+5 = xN+4 + |yN+4| = −8yN − 3 xN+6 = |xN+5| − yN+5 − 1 = 0 and so Statement 2 is true. If −1 < yN ≤ − 1 2 , then yN+6 = xN+5 + |yN+5| = −1 and so {(xn, yn)} ∞ n=N+6 is the period-3 cycle P1 3 . The proof is complete. Lemma 6. Suppose there exists a non-negative integer N ≥ 0 such that (xN, yN) ∈ l4. Then the following five statements are true: 116 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) (1) Suppose − 1 3 < yN ≤ 0. Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 . (2) Suppose yN = − 1 3 . Then {(xn, yn)} ∞ n=N is the period-3 cycle P 2 3 . (3) Suppose − 4 3 < yN < − 1 3 . Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 . (4) Suppose yN = − 4 3 . Then {(xn, yn)} ∞ n=N+3 is the period-3 cycle P 2 3 . (5) Suppose yN < − 4 3 . Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 . Proof. We have xN = 0 and yN ≤ 0. (1) Suppose − 1 3 < yN ≤ 0. Note that by Statement 1 of Lemma 5, that if − 1 4 < yN ≤ 0, then {(xn, yn)} ∞ n=N+5 is the period-3 cycle P 1 3 . So suppose − 1 3 < yN ≤ − 1 4 . For each integer n ≥ 1, let an = −22n + 1 3 · 22n . Observe that − 1 4 = a1 > a2 > a3 > . . . > − 1 3 and lim n→∞ an = − 1 3 . Thus there exists a unique integer K ≥ 1 such that yN ∈ (aK+1, aK]. We first consider the case K = 1; that is, yN ∈ ( − 5 16 , −1 4 ] . It follows from Statement 2 of Lemma 5 that xN+5 = 8yN + 2 ≤ 0, yN+5 = −8yN − 3 < 0, and xN+6 = 0. Thus yN+6 = xN+5 + |yN+5| = 16yN + 5 > 0, and so by Lemma 3 we have {(xn, yn)} ∞ n=N+8 is the period-3 cycle P 1 3 . Hence without loss of generality, we may assume K ≥ 2. For each integer m ≥ 1, let P(m) be the following statement: xN+3m+3 = 0 yN+3m+3 = 2 2m+2yN + 22m+2 − 1 3 ≤ 0. Claim: P(m) is true for 1 ≤ m ≤ K − 1. The proof of the Claim will be by induction on m. We shall first show that P(1) is true. Recall that xN = 0 and yN ∈ (aK+1, aK] ⊂ ( −1 3 , − 5 16 ] , and so by Statement 2 of Lemma 5 CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 117 we have xN+5 = 8yN + 2 < 0 and yN+5 = −8yN − 3 < 0. Then xN+3(1)+3 = 0 yN+3(1)+3 = 16yN + 5 = 2 2(1)+2yN + 22(1)+2 − 1 3 ≤ 0 and so P(1) is true. Thus if K = 2, then we have shown that for 1 ≤ m ≤ K − 1, P(m) is true. It remains to consider the case K ≥ 3. So assume that K ≥ 3. Let m be an inte- ger such that 1 ≤ m ≤ K−2, and suppose P(m) is true. We shall show that P(m+1) is true. Since P(m) is true we know xN+3m+3 = 0 yN+3m+3 = 2 2m+2yN + 22m+2 − 1 3 ≤ 0. Recall that yN ∈ (aK+1, aK] = ( −22(K+1) + 1 3 · 22(K+1) , −22K + 1 3 · 22K ] . Then xN+3m+4 = |xN+3m+3| − yN+3m+3 − 1 = −2 2m+2yN − ( 22m+2 − 1 3 ) − 1. Note that xN+3m+4 = −yN+3m+3 − 1. In particular, xN+3m+4 = −2 2m+2yN − ( 22m+2 − 1 3 ) − 1 < −22m+2 ( −22(K+1) + 1 3 · 22(K+1) ) − ( 22m+2 − 1 3 ) − 1 = 22m+2K+4 3 · 22K+2 − 22m+2 3 · 22K+2 − 22m+2 3 + 1 3 − 1 = − 22m−2K 3 − 2 3 < 0 and yN+3m+4 = xN+3m+3 + |yN+3m+3| = 0 + |yN+3m+3| = −yN+3m+3 ≥ 0. 118 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Thus xN+3m+5 = |xN+3m+4| − yN+3m+4 − 1 = yN+3m+3 + 1 − (−yN+3m+3) − 1 = 2yN+3m+3 ≤ 0 and yN+3m+5 = xN+3m+4 + |yN+3m+4| = −yN+3m+3 − 1 + (−yN+3m+3) = −2yN+3m+3 − 1. In particular, yN+3m+5 = −2 ( 22m+2yN + 22m+2 − 1 3 ) − 1 < −2 [ 22m+2 ( −22(K+1) + 1 3 · 22(K+1) ) + 22m+2 − 1 3 ] − 1 = 22m+2K+5 3 · 22K+2 − 22m+3 3 · 22K+2 − 22m+3 3 + 2 3 − 1 = − 22m−2K+1 3 − 1 3 < 0. Finally, xN+3(m+1)+3 = xN+3m+6 = |xN+3m+5| − yN+3m+5 − 1 = −2yN+3m+3 − (−2yN+3m+3 − 1) − 1 = 0 CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 119 and yN+3(m+1)+3 = yN+3m+6 = xN+3m+5 + |yN+3m+5| = 2yN+3m+3 + 2yN+3m+3 + 1 = 4yN+3m+3 + 1 = 22 ( 22m+2yN + 22m+2 − 1 3 ) + 1 = 22m+4yN + 22m+4 − 4 3 + 1 = 22(m+1)+2yN + 22(m+1)+2 − 1 3 . In particular, yN+3(m+1)+3 ≤ 2 2(m+1)+2 ( −22K + 1 3 · 22K ) + 22(m+1)+2 − 1 3 = − 22m+2K+4 3 · 22K + 22m+4 3 · 22K + 22m+4 3 − 1 3 = − 1 3 ( 1 − 22m−2K+4 ) ≤ 0 and so P(m + 1) is true. Thus the proof of the Claim is complete. That is, P(m) is true for 1 ≤ m ≤ K − 1. Specifically, P(K − 1) is true, and so xN+3(K−1)+3 = xN+3K = 0 yN+3(K−1)+3 = yN+3K = 2 2KyN + 22K − 1 3 < 0. Note that 22K ( −22K+2 + 1 3 · 22K+2 ) + 22K − 1 3 < yN+3K ≤ 2 2K ( −22K + 1 3 · 22K ) + 22K − 1 3 . So as 120 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) 22K ( −22K+2 + 1 3 · 22K+2 ) + 22K − 1 3 = −24K+2 3 · 22K+2 + 22K 3 · 22K+2 + 22K 3 − 1 3 = 1 3 ( 1 22 − 1 ) = − 1 4 and 22K ( −22K + 1 3 · 22K ) + 22K − 1 3 = −22K + 1 3 + 22K − 1 3 = 0 we have − 1 4 < yN+3K ≤ 0 and so it follows from Statement 1 of Lemma 5 that {(xn, yn)} ∞ n=N+3K+5 is the period-3 cycle P1 3 . (2) Suppose yn = − 1 3 . Note that (0, −1 3 ) ∈ P1 3 and so {(xn, yn)} ∞ n=N is the period-3 cycle P 1 3 . (3) Suppose −4 3 < yN ≤ − 1 3 . We shall first consider the case where −4 3 < yN ≤ −1. So suppose −4 3 < yN ≤ −1. For each integer n ≥ 0, let bn = −22n+2 + 1 3 · 22n . Observe that −1 = b0 > b1 > b2 > . . . > − 4 3 and lim n→∞ bn = − 4 3 . Thus there exists a unique integer K ≥ 1 such that yN ∈ (bK, bK−1]. We first consider the case K = 1; that is, yN ∈ ( −5 4 , −1 ] . Note that if yN = −1 then (xN, yN) = (0, −1) and {(xn, yn)} ∞ n=N is the period-3 cycle P 1 3 . So assume yN ∈ ( −5 4 , −1 ) . By Statements 1 and 2 of Lemma 4, we have xN+3 = 2yN + 2 < 0 and yN+3 = −2yN − 3 ≤ 0. Then xN+4 = |xN+3| − yN+3 − 1 = 0 yN+4 = xN+3 + |yN+3| = 4yN + 5 > 0 CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 121 and so it follows by Lemma 3 that {(xn, yn)} ∞ n=N+6 is the period-3 cycle P 1 3 . Hence without loss of generality, we may assume K ≥ 2. For each integer m ≥ 1, let Q(m) be the following statement: xN+3m+1 = 0 yN+3m+1 = 2 2myN + 22m+2 − 1 3 ≤ 0. Claim: Q(m) is true for 1 ≤ m ≤ K − 1. The proof of the Claim will be by induction on m. We shall first show that Q(1) is true. Recall that xN = 0 and yN ∈ (bK, bK−1] ⊂ ( −4 3 , −5 4 ] , and so by Statements 1 and 2 of Lemma 4 we have xN+3 = 2yN + 2 < 0 yN+3 = −2yN − 3 < 0 xN+3(1)+1 = |xN+3| − yN+3 − 1 = 0 yN+3(1)+1 = xN+3 + |yN+3| = 4yN + 5 ≤ 0 = 22(1)yN + 22(1)+2 − 1 3 ≤ 0 and so Q(1) is true. Thus if K = 2, then we have shown that for 1 ≤ m ≤ K − 1, Q(m) is true. It remains to consider the case K ≥ 3. So assume that K ≥ 3. Let m be an inte- ger such that 1 ≤ m ≤ K−2, and suppose Q(m) is true. We shall show that Q(m+1) is true. Since Q(m) is true we know xN+3m+1 = 0 yN+3m+1 = 2 2myN + 22m+2 − 1 3 ≤ 0 and so xN+3m+2 = |xN+3m+1| − yN+3m+1 − 1 = 0 − yN+3m+1 − 1. Recall that yN ∈ (bK, bK−1] = ( −22K+2 + 1 3 · 22K , −22K + 1 3 · 22K−2 ] . 122 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) In particular, xN+3m+2 = −2 2myN − ( 22m+2 − 1 3 ) − 1 < −22m ( −22K+2 + 1 3 · 22K ) − ( 22m+2 − 1 3 ) − 1 = 22K+2m+2 3 · 22K − 22m 3 · 22K − 22m+2 3 + 1 3 − 1 = − 1 3 ( 22m−2K+2 + 2 ) < 0 and yN+3m+2 = xN+3m+1 + |yN+3m+1| = 0 − yN+3m+1 ≥ 0. Hence xN+3m+3 = |xN+3m+2| − yN+3m+2 − 1 = yN+3m+1 + 1 − (−yN+3m+1) − 1 = 2yN+3m+1 ≤ 0 and yN+3m+3 = xN+3m+2 + |yN+3m+2| = −yN+3m+1 − 1 + (−yN+3m+1) = −2yN+3m+1 − 1. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 123 In particular, yN+3m+3 = −2 [ 22myN + 22m+2 − 1 3 ] − 1 < −2 [ 22m ( −22K+2 + 1 3 · 22K ) + 22m+2 − 1 3 ] − 1 = 22K+2m+3 3 · 22K − 22m+1 3 · 22K − 22m+3 3 + 2 3 − 1 = − 1 3 ( 22m−2K+1 + 1 ) < 0. Finally, xN+3(m+1)+1 = xN+3m+4 = |xN+3m+3| − yN+3m+1 − 1 = −2yN+3m+1 − (−2yN+3m+1 − 1) − 1 = 0 and yN+3(m+1)+1 = yN+3m+4 = xN+3m+3 + |yN+3m+3| = 2yN+3m+1 + 2yN+3m+1 + 1 = 4yN+3m+1 + 1 = 22(m+1)yN + 22(m+1)+2 − 1 3 . In particular, yN+3(m+1)+1 ≤ 2 2m+2 ( −22K + 1 3 · 22K−2 ) + 22m+4 − 1 3 124 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) = − 22K+2m+2 3 · 22K−2 + 22m+2 3 · 22K−2 + 22m+4 3 − 1 3 = 1 3 ( 22m−2K+4 − 1 ) ≤ 0 and so Q(m + 1) is true. Thus the proof of the Claim is complete. That is, Q(m) is true for 1 ≤ m ≤ K − 1. Specifically, Q(K − 1) is true, and so xN+3(K−1)+1 = 0 yN+3(K−1)+1 = 2 2(K−1)yN + 22(K−1)+2 − 1 3 ≤ 0. Note that 0 ≥ yN+3(K−1)+1 > 2 2(K−1) ( −22K+2 + 1 3 · 22K ) + 22K − 1 3 = − 24K 3 · 22K + 22K−2 3 · 22K + 22K 3 − 1 3 = 1 3 ( 1 4 − 1 ) = − 1 4 and so it follows by Statement 1 of Lemma 5 that {(xn, yn)} ∞ n=N+3K+3 is the period-3 cycle P1 3 . Suppose −1 < yN < − 1 2 . By Statement 3 of Lemma 5 we have {(xn, yn)} ∞ n=N+3 is the period-3 cycle P1 3 . To complete the proof of Statement 3 we shall now suppose that −1 2 ≤ yN < − 1 3 . For each integer n ≥ 1, let αn = −22n−1 − 1 3 · 22n−1 . Observe that − 1 2 = α1 < α2 < α3 < . . . < − 1 3 and lim n→∞ αn = − 1 3 . CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 125 Thus there exists a unique integer K ≥ 1 such that yN ∈ [αK, αK+1). We first consider the case K = 1; that is, yN ∈ [ −1 2 , −3 8 ) . By Statement 2 of Lemma 5 we have xN+5 = 8yN + 2 ≤ 0, yN+5 = −8yN − 3 > 0, and so it follows by Lemma 2 that {(xn, yn)} ∞ n=N+6 is the period-3 cycle P1 3 . Without loss of generality we may assume K ≥ 2. For each integer m ≥ 1, let R(m) be the following statement: xN+3m+2 = 2 2m+1yN + 22m+1 − 2 3 < 0 yN+3m+2 = −2 2m+1yN − ( 22m+1 + 1 3 ) ≤ 0. Claim: R(m) is true for 1 ≤ m ≤ K − 1. The proof of the Claim will be by induction on m. We shall first show that R(1) is true. Recall that xN = 0 and yN ∈ [αK, αK+1) ⊂ [ −3 8 , −1 3 ) , and so it follows from Statement 2 of Lemma 5 that xN+3(1)+2 = 8yN + 2 = 2 2(1)+1yN + 22(1)+1 − 2 3 < 0 yN+3(1)+2 = −8yN − 3 = −2 2(1)+1yN − ( 22(1)+1 + 1 3 ) ≤ 0 and so R(1) is true. Thus if K = 2, then we have shown that for 1 ≤ m ≤ K − 1, R(m) is true. It remains to consider the case K ≥ 3. So assume that K ≥ 3. Let m be an inte- ger such that 1 ≤ m ≤ K−2, and suppose R(m) is true. We shall show that R(m+1) is true. Since R(m) is true we know xN+3m+2 = 2 2m+1yN + 22m+1 − 2 3 < 0 yN+3m+2 = −2 2m+1yN − ( 22m+1 + 1 3 ) ≤ 0. Then xN+3m+3 = |xN+3m+2| − yN+3m+2 − 1 = −22m+1yN − 22m+1 − 2 3 − ( −22m+1yN − 22m+1 + 1 3 ) − 1 = 0 126 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) yN+3m+3 = xN+3m+2 + |yN+3m+2| = 22m+1yN + 22m+1 − 2 3 + 22m+1yN + 22m+1 + 1 3 = 22m+2yN + 22m+2 − 1 3 . Recall that yN ∈ [αK, αK+1) = [ −22K−1 − 1 3 · 22K−1 , −22(K+1)−1 − 1 3 · 22(K+1)−1 ) . In particular, yN+3m+3 < 2 2m+2 ( −22(K+1)−1 − 1 3 · 22(K+1)−1 ) + 22m+2 − 1 3 = − 22K+2m+3 3 · 22K+1 − 22m+2 3 · 22K+1 + 22m+2 3 − 1 3 = − 1 3 ( 1 + 22m−2K+1 ) < 0. Then xN+3m+4 = |xN+3m+3| − yN+3m+3 − 1 = 0 − yN+3m+3 − 1 = −yN+3m+3 − 1. In particular, xN+3m+4 = −2 2m+2yN − 22m+2 − 1 3 − 1 ≤ −22m+2 ( −22K−1 − 1 3 · 22K−1 ) − ( 22m+2 − 1 3 ) − 1 = 22m+2K+1 3 · 22K−1 + 22m+2 3 · 22K−1 − 22m+2 3 + 1 3 − 1 = − 2 3 ( 1 − 22m−2K+2 ) < 0. Hence yN+3m+4 = xN+3m+3 + |yN+3m+3| = 0 + (−yN+3m+3) = −yN+3m+3 > 0. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 127 Finally, xN+3(m+1)+2 = xN+3m+5 = |xN+3m+4| − yN+3m+4 − 1 = yN+3m+3 + 1 − (−yN+3m+3) − 1 = 2yN+3m+3 < 0 = 22(m+1)+1yN + 22(m+1)+1 − 2 3 < 0 and yN+3(m+1)+2 = yN+3m+5 = xN+3m+4 + |yN+3m+4| = −yN+3m+3 − 1 + (−yN+3m+3) = −2yN+3m+3 − 1 = −22(m+1)+1yN − ( 22(m+1)+1 + 1 3 ) . In particular, yN+3(m+1)+2 ≤ −2 2m+3 ( −22K−1 − 1 3 · 22K−1 ) − ( 22m+3 + 1 3 ) = 22m+2K+2 3 · 22K−1 + 22m+3 3 · 22K−1 − 22m+3 3 − 1 3 = 1 3 ( 22m−2K+4 − 1 ) ≤ 0 and so R(m + 1) is true. Thus the proof of the Claim is complete. That is, R(m) is true for 1 ≤ m ≤ K − 1. Specifically, R(K − 1) is true, and so xN+3(K−1)+2 = 2 2(K−1)+1yN + 22(K−1)+1 − 2 3 < 0 yN+3(K−1)+2 = −2 2(K−1)+1yN − ( 22(K−1)+1 + 1 3 ) ≤ 0. 128 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Then xN+3K = xN+3(K−1)+3 = |xN+3(K−1)+2| − yN+3(K−1)+2 − 1 = 0 and yN+3K = yN+3(K−1)+3 = xN+3(K−1)+2 + |yN+3(K−1)+2| = 22KyN + 22K − 1 3 . Note that 22K ( −22K−1 − 1 3 · 22K−1 ) + 22K − 1 3 ≤ yN+3K < 2 2K ( −22K+1 − 1 3 · 22K+1 ) + 22K − 1 3 . So as 22K ( −22K−1 − 1 3 · 22K−1 ) + 22K − 1 3 = −24K−1 3 · 22K−1 + 22K 3 · 22K−1 + 22K 3 − 1 3 = − 2 3 − 1 3 = −1 and 22K ( −22K+1 − 1 3 · 22K+1 ) + 22K − 1 3 = −24K+1 3 · 22K+1 + 22K 3 − 1 3 = − 1 6 − 1 3 = − 1 2 we have −1 ≤ yN+3K < − 1 2 and so it follows by Statement 3 of Lemma 5 and the fact (0, −1) ∈ P1 3 that the solution {(xn, yn)} ∞ n=N+3K+3 is the period-3 cycle P 1 3 . (4) Suppose yN = − 4 3 . By direct computations we have (xN+3, yN+3) = (− 2 3 , −1 3 ) ∈ P2 3 , and so {(xn, yn)} ∞ n=N+3 is the period-3 cycle P 2 3 . CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 129 (5) Suppose yN < − 4 3 . First consider the case −3 2 ≤ yN < − 4 3 . For each integer n ≥ 0, let βn = −22n+3 − 1 3 · 22n+1 . Observe that − 3 2 = β0 < β1 < β2 < . . . < − 4 3 and lim n→∞ βn = − 4 3 . Thus there exists a unique integer K ≥ 1 such that yN ∈ [βK−1, βK). We first consider the case K = 1; that is, yN ∈ [ −3 2 , −11 8 ) . By Statements 1 and 2 of Lemma 4 we have xN+3 = 2yN + 2 < 0 yN+3 = −2yN − 3 ≤ 0 and so xN+4 = |xN+3| − yN+3 − 1 = 0 yN+4 = xN+3 + |yN+3| = 4yN + 5 < 0. In particular, −1 ≤ yN+4 < − 1 2 . It follows by Statement 3 of Lemma 5 that the solution {(xn, yn)} ∞ n=N+7 is the period-3 cycle P 1 3 . Thus without loss of generality, we may assume that K ≥ 2. For each integer m ≥ 1, let S(m) be the following statement: xN+3m+3 = 2 2m+1yN + 22m+3 − 2 3 < 0 yN+3m+3 = −2 2m+1yN − ( 22m+3 − 2 3 ) − 1 ≤ 0. Claim: S(m) is true for 1 ≤ m ≤ K − 1. The proof of the Claim will be by induction on m. We shall first show that S(1) is true. Recall that xN = 0 and yN ∈ [βK−1, βK) ⊂ [ −11 8 , −4 3 ) , and so by Statements 1 and 2 of Lemma 4 we have xN+3 = 2yN + 2 < 0 130 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) yN+3 = −2yN − 3 < 0 xN+4 = |xN+3| − yN+3 − 1 = 0 yN+4 = xN+3 + |yN+3| = 4yN + 5 < 0 xN+5 = |xN+4| − yN+4 − 1 = −4yN − 6 < 0 yN+5 = xN+4 + |yN+4| = −4yN − 5 > 0. Finally, xN+3(1)+3 = xN+6 = |xN+5| − yN+5 − 1 = 8yN + 10 < 0 yN+3(1)+3 = yN+6 = xN+5 + |yN+5| = −8yN − 11 ≤ 0. It follows that S(1) is true. Thus if K = 2, then we have shown that for 1 ≤ m ≤ K − 1, S(m) is true. It remains to consider the case K ≥ 3. So assume that K ≥ 3. Let m be an in- teger such that 1 ≤ m ≤ K−2, and suppose S(m) is true. We shall show that S(m+1) is true. Since S(m) is true, we know xN+3m+3 = 2 2m+1yN + 22m+3 − 2 3 < 0 yN+3m+3 = −2 2m+1yN − ( 22m+3 − 2 3 ) − 1 ≤ 0. Note that yN+3m+3 = −xN+3m+3 − 1, and so −1 ≤ xN+3m+3 < 0. Thus xN+3m+4 = |xN+3m+3| − yN+3m+3 − 1 = −xN+3m+3 − (−xN+3m+3 − 1) − 1 = 0 CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 131 and yN+3m+4 = xN+3m+3 + |yN+3m+3| = xN+3m+3 + xN+3m+3 + 1 = 2xN+3m+3 + 1. Recall that yN ∈ [βK−1, βK) = [ −22(K−1)+3 − 1 3 · 22(K−1)+1 , −22K+3 − 1 3 · 22K+1 ) . In particular, yN+3m+4 = 2 [ 22m+1yN + 22m+3 − 2 3 ] + 1 < 2 [ 22m+1 ( −22K+3 − 1 3 · 22K+1 ) + 22m+3 − 2 3 ] + 1 = − 22K+2m+5 3 · 22K+1 − 22m+2 3 · 22K+1 + 22m+4 3 − 1 3 = − 1 3 ( 22m−2K+1 + 1 ) < 0. Also note that −1 < xN+3m+3 < − 1 2 . Thus xN+3m+5 = |xN+3m+4| − yN+3m+4 − 1 = 0 − (2xN+3m+3 + 1) − 1 = −2xN+3m+3 − 2 < 0 132 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) and yN+3m+5 = xN+3m+4 + |yN+3m+4| = 0 + (−2xN+3m+3 − 1) = −2xN+3m+3 − 1 > 0. Finally, xN+3(m+1)+3 = xN+3m+6 = |xN+3m+5| − yN+3m+5 − 1 = 2xN+3m+3 + 2 − (−2xN+3m+3 − 1) − 1 = 4xN+3m+3 + 2 < 0 = 4 [ 22m+1yN + ( 22m+3 − 2 3 )] + 2 < 0 = 22(m+1)+1yN + ( 22(m+1)+3 − 2 3 ) + 2 < 0 and yN+3(m+1)+3 = yN+3m+6 = xN+3m+5 + |yN+3m+5| = −2xN+3m+3 − 2 + (−2xN+3m+3 − 1) = −4xN+3m+3 − 3 = −4 [ 22m+1yN + ( 22m+3 − 2 3 )] − 3 = −22(m+1)+1yN − ( 22(m+1)+3 − 2 3 ) − 1. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 133 In particular, yN+3m+6 ≤ −4 [ 22m+1 ( −22(K−1)+3 − 1 3 · 22(K−1)+1 ) + 22m+3 − 2 3 ] − 3 = 22K+2m+4 3 · 22K−1 + 22m+3 3 · 22K−1 − 22m+5 3 − 1 3 = 1 3 ( 22m−2K+4 − 1 ) < 0 and so S(m + 1) is true. Thus the proof of the Claim is complete. That is, S(m) is true for 1 ≤ m ≤ K − 1. Specifically, S(K − 1) is true, and so xN+3(K−1)+3 = xN+3K = 2 2K−1yN + 22K+1 − 2 3 < 0 yN+3(K−1)+3 = yN+3K = −2 2K−1yN − ( 22K+1 − 2 3 ) − 1 < 0. Note that yN+3K = −xN+3K − 1. Thus xN+3K+1 = |xN+3K| − yN+3K − 1 = −xN+3K − (−xN+3K − 1) − 1 = 0 and yN+3K+1 = xN+3K + |yN+3K| = xN+3K + xN+3K + 1 = 2xN+3K + 1 = 2 ( 22K−1yN + 22K+1 − 2 3 ) + 1. 134 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Note that 2 [ 22K−1 ( −22(K−1)+3 − 1 3 · 22(K−1)+1 ) + 22K+1 − 2 3 ] + 1 ≤ yN+3K+1 < 2 [ 22K−1 ( −22K+3 − 1 3 · 22K+1 ) + 22K+1 − 2 3 ] + 1. So as 2 [ 22K−1 ( −22(K−1)+3 − 1 3 · 22(K−1)+1 ) + 22K+1 − 2 3 ] + 1 = −24K+1 3 · 22K−1 − 22K 3 · 22K−1 + 22K+2 3 − 1 3 = − 1 3 (2 + 1) = −1 and 2 [ 22K−1 ( −22K+3 − 1 3 · 22K+1 ) + 22K+1 − 2 3 ] + 1 = −22K+3 3 · 2 − 1 6 + 22K+2 3 − 1 3 = − 1 6 − 1 3 = − 1 2 we have −1 ≤ yN+3K+1 < − 1 2 and hence it follows from case 3 of this Lemma and the fact that (0, −1) ∈ P1 3 that the solution {(xn, yn)} ∞ n=N+3K+5 is eventually the period-3 cycle P 1 3 . Finally, suppose yN < − 3 2 . Then by Statement 3 of Lemma 4 the solution {(xn, yn)} ∞ n=N+4 is the period-3 cycle P1 3 . Lemma 7. Suppose there exists a non-negative integer N ≥ 0 such that (xN, yN) ∈ Q1. Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 or the period-3 cycle P2 3 . CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 135 Proof. We have xN+1 = |xN| − yN − 1 = xN − yN − 1 yN+1 = xN + |yN| = xN + yN > 0. If xN+1 ≥ 0 then xN+2 = |xN+1| − yN+1 − 1 = −2yN − 2 < 0 yN+2 = xN+1 + |yN+1| = 2xN − 1 > 0 xN+3 = |xN+2| − yN+2 − 1 = −2xN + 2yN + 2 ≤ 0 yN+3 = xN+2 + |yN+2| = 2xN − 2yN − 3 xN+4 = |xN+3| − yN+3 − 1 = 0 and so (xN+4, yN+4) ∈ l2 ∪ l4. By Lemmas 3 and 6, the solution {(xn, yn)} ∞ n=N is eventually the period-3 cycle P1 3 or the period-3 cycle P2 3 . If xN+1 < 0 then xN+2 = |xN+1| − yN+1 − 1 = −2xN < 0 yN+2 = xN+1 + |yN+1| = 2xN − 1 xN+3 = |xN+2| − yN+2 − 1 = 0 and so (xN+3, yN+3) ∈ l2 ∪ l4. By Lemmas 3 and 6, the solution {(xn, yn)} ∞ n=N is eventually the period-3 cycle P1 3 or the period-3 cycle P2 3 . Lemma 8. Suppose there exists a non-negative integer N ≥ 0 such that (xN, yN) ∈ Q2. Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 or the period-3 cycle P2 3 . Proof. We have xN+1 = |xN| − yN − 1 = −xN − yN − 1 yN+1 = xN + |yN| = xN + yN. 136 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Case 1: Suppose yN+1 ≥ 0. Then by Lemma 2, the solution {(xn, yn)} ∞ n=N+2 is the period-3 cycle P1 3 . Case 2: Suppose yN+1 < 0 and xN+1 ≤ 0. Then xN+2 = |xN+1| − yN+1 − 1 = 0 and so (xN+2, yN+2) ∈ l2 ∪l4. By Lemmas 3 and 6, the solution {(xn, yn)} ∞ n=N is eventually the period-3 cycle P1 3 or the period-3 cycle P2 3 . Case 3: Suppose yN+1 < 0 and xN+1 > 0. Then xN+2 = |xN+1| − yN+1 − 1 = −2xN − 2yN − 2 > 0 yN+2 = xN+1 + |yN+1| = −2xN − 2yN − 1 > 0 xN+3 = |xN+2| − yN+2 − 1 = −2 yN+3 = xN+2 + |yN+2| = −4xN − 4yN − 3 > 0 xN+4 = |xN+3| − yN+3 − 1 = 4xN + 4yN + 4 < 0 yN+4 = xN+3 + |yN+3| = −4xN − 4yN − 5 xN+5 = |xN+4| − yN+4 − 1 = 0 and so (xN+5, yN+5) ∈ l2 ∪ l4. By Lemmas 3 and 6, the solution {(xn, yn)} ∞ n=N is eventually the period-3 cycle P1 3 or the period-3 cycle P2 3 . Lemma 9. Suppose there exists a non-negative integer N ≥ 0 such that (xN, yN) ∈ Q4. Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 or the period-3 cycle P2 3 . Proof. We have xN+1 = |xN| − yN − 1 = xN − yN − 1 yN+1 = xN + |yN| = xN − yN > 0 Case 1: Suppose xN+1 > 0. Then (xN+1, yN+1) ∈ Q1 and so by Lemma 7, the solution {(xn, yn)} ∞ n=N+2 is eventually the period-3 cycle P1 3 or the period-3 cycle P2 3 . Case 2: Suppose xN+1 = 0. Then (xN+1, yN+1) ∈ l2 and so by Lemma 3, the solution {(xn, yn)} ∞ n=N+4 is the period-3 cycle P1 3 . CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 137 Case 3: Suppose xN+1 < 0. Then (xN+1, yN+1) ∈ Q2 and so by Lemma 8, the solution {(xn, yn)} ∞ n=N+1 is eventually the period-3 cycle P1 3 or the period-3 cycle P2 3 . Lemma 10. Suppose there exists a non-negative integer N ≥ 0 such that (xN, yN) ∈ l1. Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 or P2 3 . Proof. We have xN+1 = |xN| − yN − 1 = xN − 1 yN+1 = xN + |yN| = xN Case 1: Suppose xN = 0. Then (xN+1, yN+1) = (−1, 0), and so (xN+2, yN+2) = (0, −1). Hence the solution {(xn, yn)} ∞ n=N+2 is the period-3 cycle P 1 3 . Case 2: Suppose 0 < xN ≤ 1. Then xN+1 ≤ 0 and yN+1 > 0. Thus (xN+1, yN+1) ∈ Q2 ∪ l2, and hence by Lemmas 3 and 8, the solution {(xn, yn)} ∞ n=N+1 is eventually the period-3 cycle P 1 3 or the period-3 cycle P2 3 . Case 3: Suppose xN > 1. Then xN+1 > 0 and yN+1 > 0. Thus (xN+1, yN+1) ∈ Q1 and by Lemma 7, the solution {(xn, yn)} ∞ n=N+1 is eventually the period-3 cycle P 1 3 or the period-3 cycle P2 3 . Lemma 11. Suppose there exists a non-negative integer N ≥ 0 such that (xN, yN) ∈ l3. Then {(xn, yn)} ∞ n=N is eventually the period-3 cycle P 1 3 or the period-3 cycle P2 3 . Proof. We have xN+1 = |xN| − yN − 1 = −xN − 1 yN+1 = xN + |yN| = xN < 0. Case 1: Suppose −1 < xN ≤ 0. Then xN+2 = |xN+1|−yN+1 −1 = 0, and so (xN+2, yN+2) ∈ l2 ∪l4. It follows by Lemmas 3 and 6, that the solution {(xn, yn)} ∞ n=N+2 is eventually the period-3 cycle P1 3 or the period-3 cycle P2 3 . Case 2: Suppose xN = −1. Then (xN+1, yN+1) = (0, −1) ∈ P 1 3 , and so the solution {(xn, yn)} ∞ n=N+1 is the period-3 cycle P1 3 . Case 3: Suppose xN < −1. Then (xN+1, yN+1) ∈ Q4 ∪ l1. It follows by Lemmas 9 and 10, the solution {(xn, yn)} ∞ n=N+2 is eventually the period-3 cycle P 1 3 or the period-3 cycle P2 3 . To complete the proof of Theorem 2.1 it remains to consider the case where the initial condition (x0, y0) ∈ Q3. 138 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Lemma 12. Suppose (x0, y0) ∈ Q3. Then {(xn, yn)} ∞ n=0 is the unique equilibrium solution (x̄, ȳ) = ( −2 5 , −1 5 ) , or else is eventually either the period-3 cycle P1 3 or the period-3 cycle P2 3 . Proof. If (x0, y0) = ( −2 5 , −1 5 ) , then the solution {(xn, yn)} ∞ n=0 is the equilibrium. So suppose (x0, y0) ∈ Q3 \ {( −2 5 , −1 5 )} . It suffices to show that there exists an integer N ≥ 0 such that {(xn, yn)} ∞ n=N is either the period-3 cycle P 1 3 or the period-3 cycle P2 3 . For the sake of contradiction, assume that it is false that there exists an integer N ≥ 0 such that {(xn, yn)} ∞ n=N is either the period-3 cycle P 1 3 or the period-3 cycle P2 3 . It follows from the previous lemmas that xn < 0 and yn < 0 for every integer n ≥ 0. Case 1: Suppose x0 ≤ −2 and y0 < 0. Then x1 = |x0| − y0 − 1 = −x0 − y0 − 1 > 0 which is a contradiction, and the proof is complete. Case 2: Suppose −2 < x0 < 0 and y0 ≤ −1. Then x1 = |x0| − y0 − 1 = −x0 − y0 − 1 > 0 which is a contradiction, and the proof is complete. Case 3: It remains to consider the case (x0, y0) ∈ (−2, 0) × (−1, 0). For each integer n ≥ 0, let an = −24n−2 − 1 5 · 24n−3 , bn = −24n + 1 5 · 24n−1 , cn = −24n−2 − 1 5 · 24n−2 , dn = −24n + 1 5 · 24n and Dn = 24n − 1 5 . Observe that −2 = a0 < a1 < a2 < . . . < − 2 5 and lim n→∞ an = − 2 5 0 = b0 > b1 > b2 > . . . > − 2 5 and lim n→∞ bn = − 2 5 −1 = c0 < c1 < c2 < . . . < − 1 5 and lim n→∞ cn = − 1 5 0 = d0 > d1 > d2 > . . . > − 1 5 and lim n→∞ dn = − 1 5 . CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 139 There exists a unique integer K ≥ 0 such that (x0, y0) ∈ [aK, bK] × [cK, dK] \ [aK+1, bK+1] × [cK+1, dK+1]. We first consider the case K = 0; that is, (x0, y0) ∈ [−2, 0] × [−1, 0] \ [− 1 2 , −3 8 ] × [−1 4 , − 3 16 ]. Note that by Lemmas 6 and 11, and by Case 1 and Case 2 of this lemma, we know that the solution {(xn, yn)} ∞ n=0 is eventually either the period-3 cycle P 1 3 or the period-3 cycle P2 3 when (x0, y0) is an element of the outer boundaries of [−2, 0] × [−1, 0]. Recall by assumption that xn < 0 and yn < 0 for every integer n ≥ 0. So suppose (x0, y0) ∈ (−2, 0) × (−1, 0) \ [ − 1 2 , − 3 8 ] × [ − 1 4 , − 3 16 ] . Then x1 = |x0| − y0 − 1 = −x0 − y0 − 1 y1 = x0 + |y0| = x0 − y0 x2 = |x1| − y1 − 1 = (x0 + y0 + 1) − (x0 − y0) − 1 = 2y0 y2 = x1 + |y1| = (−x0 − y0 − 1) + (−x0 + y0) = −2x0 − 1. If −2 < x0 < − 1 2 , then y2 > 0 which is a contradiction. Thus −1 2 ≤ x0 < 0. Then x3 = |x2| − y2 − 1 = (−2y0) − (−2x0 − 1) − 1 = 2x0 − 2y0 y3 = x2 + |y2| = (2y0) + (2x0 + 1) = 2x0 + 2y0 + 1 x4 = |x3| − y3 − 1 = (−2x0 + 2y0) − (2x0 + 2y0 + 1) − 1 = −4x0 − 2 y4 = x3 + |y3| = (2x0 − 2y0) + (−2x0 − 2y0 − 1) = −4y0 − 1. If −1 < y0 < − 1 4 , then y4 > 0 which is a contradiction. Hence −1 4 ≤ y0 < 0. Then x5 = |x4| − y4 − 1 = (4x0 + 2) − (−4y0 − 1) − 1 = 4x0 + 4y0 + 2 y5 = x4 + |y4| = (−4x0 − 2) + (4y0 + 1) = −4x0 + 4y0 − 1 x6 = |x5| − y5 − 1 = (−4x0 − 4y0 − 2) − (−4x0 + 4y0 − 1) − 1 = −8y0 − 2 y6 = x5 + |y5| = (4x0 + 4y0 + 2) + (4x0 − 4y0 + 1) = 8x0 + 3. 140 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) If −3 8 < x0 < 0, then y6 > 0 which is a contradiction. Hence −1 2 < x0 ≤ − 3 8 . Thus x7 = |x6| − y6 − 1 = (8y0 + 2) − (8x0 + 3) − 1 = −8x0 + 8y0 − 2 y7 = x6 + |y6| = (−8y0 − 2) + (−8x0 − 3) = −8x0 − 8y0 − 5 x8 = |x7| − y7 − 1 = (8x0 − 8y0 + 2) − (−8x0 − 8y0 − 5) − 1 = 16x0 + 6 y8 = x7 + |y7| = (−8x0 + 8y0 − 2) + (8x0 + 8y0 + 5) = 16y0 + 3 > 0, which is a contradiction. Thus the case K = 0 is complete. Next consider the case K ≥ 1. Recall that xn < 0 and yn < 0 for all n ≥ 0. For each integer m such that 0 ≤ m ≤ K − 1, let P(m) be the following proposition: x8m+1 = −2 4mx0 − 2 4my0 − 3Dm − 1 y8m+1 = 2 4mx0 − 2 4my0 + Dm x8m+2 = 2 4m+1y0 + 2Dm y8m+2 = −2 4m+1x0 − 4Dm − 1 x8m+3 = 2 4m+1x0 − 2 4m+1y0 + 2Dm y8m+3 = 2 4m+1x0 + 2 4m+1y0 + 6Dm + 1 x8m+4 = −2 4m+2x0 − 8Dm − 2 y8m+4 = −2 4m+2y0 − 4Dm − 1 x8m+5 = 2 4m+2x0 + 2 4m+2y0 + 12Dm + 2 y8m+5 = −2 4m+2x0 + 2 4m+2y0 − 4Dm − 1 x8m+6 = −2 4m+3y0 − 8Dm − 2 y8m+6 = 2 4m+3x0 + 16Dm + 3 CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 141 x8m+7 = −2 4m+3x0 + 2 4m+3y0 − 8Dm − 2 y8m+7 = −2 4m+3x0 − 2 4m+3y0 − 24Dm − 5 x8m+8 = 2 4m+4x0 + 32Dm + 6 y8m+8 = 2 4m+4x0 + 16Dm + 3. Claim: P(m) is true for 0 ≤ m ≤ K − 1. The proof of the Claim will be by induction on m. We shall first show that P(0) is true. x8(0)+1 = −x0 − y0 − 1 = −2 4(0)x0 − 2 4(0)y0 − 3D0 − 1 y8(0)+1 = x0 − y0 = 2 4(0)x0 − 2 4(0)y0 − D0 x8(0)+2 = 2y0 = 2 4(0)+1y0 + 2D0 y8(0)+2 = −2x0 − 1 = −2 4(0)+1x0 − 4D0 − 1 x8(0)+3 = 2x0 − 2y0 = 2 4(0)+1x0 − 2 4(0)+1y0 + 2D0 y8(0)+3 = 2x0 + 2y0 + 1 = 2 4(0)+1x0 + 2 4(0)+1y0 + 6D0 + 1 x8(0)+4 = −4x0 − 2 = −2 4(0)+2x0 − 8D0 − 2 y8(0)+4 = −4y0 − 1 = −2 4(0)+2x0 − 4D0 − 1 x8(0)+5 = 4x0 + 4y0 + 2 = 2 4(0)+1x0 − 2 4(0)+2y0 + 12D0 + 2 y8(0)+5 = −4x0 + 4y0 − 1 = −2 4(0)+2x0 + 2 4(0)+2y0 − 4D0 − 1 x8(0)+6 = −8x0 − 2 = −2 4(0)+3x0 − 8D0 − 2 y8(0)+6 = 8y0 + 3 = 2 4(0)+3x0 + 16D0 + 3 x8(0)+7 = −8x0 + 8y0 − 2 = −2 4(0)+3x0 + 2 4(0)+3y0 − 8D0 − 2 y8(0)+7 = −8x0 − 8y0 − 5 = −2 4(0)+3x0 − 2 4(0)+3y0 − 24D0 + 5 x8(0)+8 = 16x0 + 6 = 2 4(0)+4x0 + 32D0 + 6 y8(0)+8 = 16y0 + 3 = 2 4(0)+4x0 + 16D0 + 3 and so P(0) is true. Thus if K = 1, then we have shown that for 0 ≤ m ≤ K − 1, P(m) is true. It remains to consider the case K ≥ 2. So assume that K ≥ 2. Suppose that m is an integer such that 0 ≤ m ≤ K − 2, and that P(m) is true. We shall show that P(m + 1) is true. 142 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Since P(m) is true, we know x8m+8 = 2 4m+4x0 + 32Dm + 6 y8m+8 = 2 4m+4x0 + 16Dm + 3. Hence x8(m+1)+1 = x8m+9 = |x8m+8| − y8m+8 − 1 = −(24m+4x0 + 32Dm + 6) − (2 4m+4y0 + 16Dm + 3) − 1 = −24m+4x0 − 2 4m+4y0 − 48Dm − 10 = −24m+4x0 − 2 4m+4y0 − 48 ( 24m − 1 5 ) − 10 = −24(m+1)x0 − 2 4(m+1)y0 − 3 ( 24(m+1) − 1 5 ) − 1 = −24(m+1)x0 − 2 4(m+1)y0 − 3Dm+1 − 1 and y8(m+1)+1 = y8m+9 = x8m+8 + |y8m+8| = 24m+4x0 + 32Dm + 6 + (−2 4m+4y0 − 16Dm − 3) = 24m+4x0 − 2 4m+4y0 + 16Dm + 3 = 24(m+1)x0 − 2 4(m+1)y0 + 16 ( 24m − 1 5 ) + 3 = 24(m+1)x0 − 2 4(m+1)y0 + Dm+1. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 143 Thus x8(m+1)+2 = x8m+10 = |x8m+9| − y8m+9 − 1 = −(−24(m+1)x0 − 2 4(m+1)y0 − 3Dm+1 − 1) −(24(m+1)x0 − 2 4(m+1)y0 + Dm+1) − 1 = 24(m+1)+1y0 + 2Dm+1 and y8(m+1)+2 = y8m+10 = x8m+9 + |y8m+9| = −24(m+1)x0 − 2 4(m+1)y0 − 3Dm+1 − 1 + (−2 4(m+1)x0 + 2 4(m+1)y0 − Dm+1) = −24(m+1)+1x0 − 4Dm+1 − 1. Then x8(m+1)+3 = x8m+11 = |x8m+10| − y8m+10 − 1 = −24(m+1)+1y0 − 2Dm+1 + 2 4(m+1)+1x0 + 4Dm+1 + 1 − 1 = 24(m+1)+1x0 − 2 4(m+1)+1y0 + 2Dm+1 and y8(m+1)+3 = y8m+11 = x8m+10 + |y8m+10| = 24(m+1)+1y0 + 2Dm+1 + 2 4(m+1)+1x0 + 4Dm+1 + 1 = 24(m+1)+1x0 + 2 4(m+1)+1y0 + 6Dm+1 + 1. 144 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Hence x8(m+1)+4 = x8m+12 = |x8m+11| − y8m+11 − 1 = −24(m+1)+1x0 + 2 4(m+1)+1y0 − 2Dm+1 − 2 4(m+1)+1x0 −24(m+1)+1y0 − 6Dm+1 − 2 = −24(m+1)+2x0 − 8Dm+1 − 2 and y8(m+1)+4 = y8m+12 = x8m+11 + |y8m+11| = 24(m+1)+1x0 − 2 4(m+1)+1y0 + 2Dm+1 − 2 4(m+1)+1x0 −24(m+1)+1y0 − 6Dm+1 − 1 = −24(m+1)+2y0 − 4Dm+1 − 1. Thus x8(m+1)+5 = x8m+13 = |x8m+12| − y8m+12 − 1 = 24(m+1)+2x0 + 8Dm+1 + 2 + 2 4(m+1)+2y0 + 4Dm+1 + 1 − 1 = 24(m+1)+2x0 + 2 4(m+1)+2y0 + 12Dm+1 + 2 and y8(m+1)+5 = y8m+13 = x8m+12 + |y8m+12| = −24(m+1)+2x0 − 8Dm+1 − 2 + 2 4(m+1)+2y0 + 4Dm+1 + 1 = −24(m+1)+2x0 + 2 4(m+1)+2y0 − 4Dm+1 − 1. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 145 Hence x8(m+1)+6 = x8m+14 = |x8m+13| − y8m+13 − 1 = −24(m+1)+2x0 − 2 4(m+1)+2y0 − 12Dm+1 − 2 +24(m+1)+2x0 − 2 4(m+1)+2y0 + 4Dm+1 + 1 − 1 = −24(m+1)+3y0 − 8Dm+1 − 2 and y8(m+1)+6 = y8m+14 = x8m+13 + |y8m+13| = 24(m+1)+2x0 + 2 4(m+1)+2y0 + 12Dm+1 + 2 + 2 4(m+1)+2x0 −24(m+1)+2y0 + 4Dm+1 + 1 = 24(m+1)+3x0 + 16Dm+1 + 3. Then x8(m+1)+7 = x8m+15 = |x8m+14| − y8m+14 − 1 = 24(m+1)+3y0 + 8Dm+1 + 2 − 2 4(m+1)+3x0 − 16Dm+1 − 3 − 1 = −24(m+1)+3x0 + 2 4(m+1)+3y0 − 8Dm+1 − 2 and y8(m+1)+7 = y8m+15 = x8m+14 + |y8m+14| = −24(m+1)+3y0 − 8Dm+1 − 2 − 2 4(m+1)+3x0 − 16Dm+1 − 3 = −24(m+1)+3x0 − 2 4(m+1)+3y0 − 24Dm+1 − 5. 146 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Thus x8(m+1)+8 = x8m+16 = |x8m+15| − y8m+15 − 1 = 24(m+1)+3x0 − 2 4(m+1)+3y0 + 8Dm+1 + 2 +24(m+1)+3x0 + 2 4(m+1)+3y0 + 24Dm+1 + 5 − 1 = 24(m+1)+4x0 + 32Dm+1 + 6 and y8(m+1)+8 = y8m+16 = x8m+15 + |y8m+15| = −24(m+1)+3x0 + 2 4(m+1)+3y0 − 8Dm+1 − 2 +24(m+1)+3x0 + 2 4(m+1)+3y0 + 24Dm+1 + 5 = 24(m+1)+4y0 + 16Dm+1 + 3 and so P(m + 1) is true. Thus the proof of the Claim is complete. That is, P(m) is true for 0 ≤ m ≤ K − 1. In particular, P(K − 1) is true. Thus x8K = x8(K−1)+8 = 2 4(K−1)+4x0 + 32DK−1 + 6 and y8K = y8(K−1)+8 = 2 4(K−1)+4y0 + 16DK−1 + 3. CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 147 Hence x8K+1 = |x8K| − y8K − 1 = −24Kx0 − 32DK−1 − 6 − 2 4Ky0 − 16DK−1 − 3 − 1 = −24Kx0 − 2 4Ky0 − 48DK−1 − 10 = −24Kx0 − 2 4Ky0 − 48 ( 24(K−1) − 1 5 ) − 10 = −24Kx0 − 2 4Ky0 − 3 · 24K 5 + 3 5 − 1 = −24Kx0 − 2 4Ky0 − 3DK − 1 and y8K+1 = x8K + |y8K| = 24Kx0 + 32DK−1 + 6 − 2 4Ky0 − 16DK−1 − 3 = 24Kx0 − 2 4Ky0 + 16DK−1 + 3 = 24Kx0 − 2 4Ky0 + 16 ( 24(K−1) − 1 5 ) + 3 = 24Kx0 − 2 4Ky0 + 24K 5 − 24 5 + 3 = 24Kx0 − 2 4Ky0 + DK. Hence x8K+2 = |x8K+1| − y8K+1 − 1 = 24Kx0 + 2 4Ky0 + 3DK + 1 − 2 4Kx0 + 2 4Ky0 − DK − 1 = 24K+1y0 + 2DK 148 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) and y8K+2 = x8K+1 + |y8K+1| = −24Kx0 − 2 4Ky0 − 3DK − 1 − 2 4Kx0 + 2 4Ky0 − DK = −24K+1x0 − 4DK − 1. Recall that (x0, y0) ∈ [aK, bK] × [cK, dK] \ [aK+1, bK+1] × [cK+1, dK+1] = [ −24K−2 − 1 5 · 24K−3 , −24K + 1 5 · 24K−1 ] × [ −24K−2 − 1 5 · 24K−2 , −24K + 1 5 · 24K ] \ [ −24(K+1)−2 − 1 5 · 24(K+1)−3 , −24(K+1) + 1 5 · 24(K+1)−1 ] × [ −24(K+1)−2 − 1 5 · 24(K+1)−2 , −24(K+1) + 1 5 · 24(K+1) ] . Suppose (x0, y0) ∈ [aK, aK+1) × [cK, dK]. Hence y8K+2 > −2 4K+1 (aK+1) − 4DK − 1 = −24K+1 ( −24(K+1)−2 − 1 5 · 24(K+1)−3 ) − 4DK − 1 = 28K+3 5 · 24K+1 + 24(K+1) − 1 5 · 24(K+1)−3 − 24(K+2) 5 + 4 5 − 1 = 0 which is a contradiction. Next suppose (x0, y0) ∈ [aK+1, bK] × [cK, cK+1). Then x8K+3 = |x8K+2| − y8K+2 − 1 = −24K+1y0 − 2DK + 2 4K+1x0 + 4DK + 1 − 1 = 24K+1x0 − 2 4K+1y0 + 2DK CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 149 and y8K+3 = x8K+2 + |y8K+2| = 24K+1y0 + 2DK + 2 4K+1x0 + 4DK + 1 = 24K+1x0 + 2 4K+1y0 + 6DK + 1. Hence x8K+4 = |x8K+3| − y8K+3 − 1 = −24K+1x0 + 2 4K+1y0 − 2DK − 2 4K+1x0 − 2 4K+1y0 − 6DK − 1 − 1 = −24K+2x0 − 8DK − 2 and y8K+4 = x8K+3 + |y8K+3| = 24K+1x0 − 2 4K+1y0 + 2DK − 2 4K+1x0 − 2 4K+1y0 − 6DK − 1 = −24K+2y0 − 4DK − 1. Recall that (x0, y0) ∈ [aK+1, bK] × [cK, cK+1). Thus y8K+4 > −2 4K+2(cK+1) − 4DK − 1 = −24K+2 ( −24K+2 − 1 5 · 24K+2 ) − 4 ( 24K − 1 5 ) − 1 = 28K+4 5 · 24K+2 + 24K+2 5 · 24K+2 − 24K+2 5 + 4 5 − 1 = 0 which is a contradiction. Now suppose that (x0, y0) ∈ (bK+1, bK] × [cK+1, dK]. 150 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Hence x8K+5 = |x8K+4| − y8K+4 − 1 = 24K+2x0 + 8DK + 2 + 2 4K+2y0 + 4DK + 1 − 1 = 24K+2x0 + 2 4K+2y0 + 12DK + 2 and y8K+5 = x8K+4 + |y8K+4| = −24K+2x0 − 8DK − 2 + 2 4K+2y0 + 4DK + 1 = −24K+2x0 + 2 4K+2y0 − 4DK − 1. Then x8K+6 = |x8K+5| − y8K+5 − 1 = −24K+2x0 − 2 4K+2y0 − 12DK − 2 + 2 4K+2x0 − 2 4K+2y0 + 4DK + 1 − 1 = −24K+3y0 − 8DK − 2 and y8K+6 = x8K+5 + |y8K+5| = 24K+2x0 + 2 4K+2y0 + 12DK + 2 + 2 4K+2x0 − 2 4K+2y0 + 4DK + 1 = 24K+3x0 + 16DK + 3. Recall that (x0, y0) ∈ (bK+1, bK] × [cK+1, dK] . Thus y8K+6 > 2 4K+3 (bK+1) + 16 ( 24K − 1 5 ) + 3 = 24K+3 ( −24(K+1) + 1 5 · 24(K+1)−1 ) + 16 ( 24K − 1 5 ) + 3 CUBO 14, 2 (2012) On the global behavior of xn+1 = |xn| − yn − 1 and yn+1 = xn + |yn| 151 = −24K+4 5 + 1 5 + 24K+4 5 − 16 5 + 3 = 0 which is a contradiction. Finally, suppose (x0, y0) ∈ [aK+1, bK+1] × (dK+1, dK]. Thus x8K+7 = |x8K+6| − y8K+6 − 1 = 24K+3y0 + 8DK + 2 − 2 4K+3x0 − 16DK − 3 − 1 = −24K+3x0 + 2 4K+3y0 − 8DK − 2 and y8K+7 = x8K+6 + |y8K+6| = −24K+3y0 − 8DK − 2 − 2 4K+3x0 − 16DK − 3 = −24K+3x0 − 2 4K+3y0 − 24DK − 5. Hence x8K+8 = |x8K+7| − y8K+7 − 1 = 24K+3x0 − 2 4K+3y0 + 8DK + 2 + 2 4K+3x0 + 2 4K+3y0 + 24DK + 5 − 1 = 24K+3x0 + 32DK + 6 and y8K+8 = x8K+7 + |y8K+7| = −24K+3x0 + 2 4K+3y0 − 8DK − 2 + 2 4K+3x0 + 2 4K+3y0 + 24DK + 5 = 24K+4y0 + 16DK + 3. Recall that (x0, y0) ∈ [aK+1, bK+1] × (dK+1, dK]. 152 E.A. Grove, E. Lapierre and W. Tikjha CUBO 14, 2 (2012) Thus y8K+8 > 2 4K+4 (dK+1) + 16 ( 24K − 1 5 ) + 3 > 24K+4 ( −24(K+1) + 1 5 · 24(K+1) ) + 16 ( 24K − 1 5 ) + 3 = − 24K+4 5 + 1 5 + 24K+4 5 − 16 5 + 3 = 0 which is a contradiction. The proof is complete. Received: November 2011. Revised: November 2011. References [1] E. Camouzis, and G. Ladas, Dynamics of Third-Order Rational Difference Equations with Open Problems and Conjectures, Chapman & Hall/CRC, New York, 2008. [2] R.L. Devaney, A piecewise linear model of the the zones of instability of an area-preserving map, Physica 10D (1984), 387-393. [3] M.R.S. Kulenovic, and O. Merino, Discrete Dynamical Systems and Difference Equations with Mathematica, Chapman & Hall/CRC, New York, 2002. [4] H.O. Peitgen and D. Saupe, (eds.) The Science of Fractal Images, Springer-Verlog, New York, 1991. [5] W. Tikjha, Y. Lenbury, and E. G. Lapierre, On the Global Character of the System of Piecewise Linear Difference Equations xn+1 = |xn|−yn −1 and yn+1 = xn − |yn|, Advances in Difference Equations, Volume 2010 (2010), Article ID 573281. Introduction The Global Behavior Of The Solutions Of System(1.1)