Prime period-6 solutions of a certain piecewise linear system of difference equations

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1 Prime period-6 solutions of a certain piecewise linear system of difference equations WIROT TIKJHA 1, YONGWIMON LENBURY 2,3, EVELINA GIUSTI LAPIERRE 4 1 Faculty of Science and Technology, Pibulsongkram Rajabhat University, THAILAND 2 Department of Mathematics, Faculty of Science, Mahidol University, THAILAND 3 Center of Ecellence in Mathematics, CHE, 328 Si Ayutthaya Road, Bangkok, THAILAND 4 Department of Mathematics, Johnson and Wales University, 8 Abbott Park Place, Providence, RI 02903, USA *Corresponding author: wirottik@gmailcom, yongwimonlen@mahidolacth Abstract: In this paper we consider a system of piecewise linear difference equations where the initial condition is in the closed first quadrant, the open second quadrant and the closed third quadrant We show that there eists eactly two prime period-6 solutions, and that every solution of the system is eventually one of the two prime period-6 solutions Key Words: periodic solution; systems of piecewise linear difference equations 1 Introduction During the last three years we have been interested in the global behavior of systems of piecewise linear difference equations This paper is part of a general project which involves the following system n+1 = n + ay n + b, n = 0, 1, N ) y n+1 = n + c y n + d where the parameters a, b, c, d { 1, 0, 1} and initial conditions 0, y 0 ) R 2 There are 81 special cases The system s number N is given by N = 2a + 1) + 9b + 1) + 3c + 1) + d + 1) + 1 This family of piecewise linear difference equations are the prototypes for more elaborate piecewise difference equations that, in many cases, ehibit complicated behavior Interest in the area began in 1984 when Davaney published his famous paper introducing the gingerbread man map: n+1 = n y n + 1, n = 0, 1, y n+1 = n with parameters a, b R and initial conditions 0, y 0 ) R 2 See [1, 2] The gingerbread man map was Devaneys response to the 198 generalized Lozi equation: n+1 = a n + y n + 1, n = 0, 1, y n+1 = b n with parameters a, b R and initial conditions 0, y 0 ) R 2 See [3, 4] The Lozi equation had been used to eamine an attractor that was observed by Lorenz in the Henon map, a non-linear system of difference equations n+1 = a 2 n + y n + 1, n = 0, 1, y n+1 = b n with parameters a, b R and initial conditions 0, y 0 ) R 2 that modeled weather patterns See [] More recently, piecewise linear difference equations have applications in neural networks and cellular neural networks, which are capable of high-speed parallel signal processing See [6] For other systems of this form, see [, 8, 9, 10, 11] 2 Problem Formulation Consider the system of piecewise linear difference equations n+1 = n y n 1 y n+1 = n y n 1, n = 0, 1, 1) where the initial condition 0, y 0 ) R 2 {, y) : > 0, y < 0} The unique equilibrium point of System1) is, ȳ) = 1, 1) We show that every solution of System1) is eventually one of the following period-6 cycles: ISBN:

2 0 = 3, y 0 = 3 1 =, y 1 = 1 2 =, y 2 = 3 P6 1 =, 3 = 1, y 3 = 1 4 = 1, y 4 = 1 = 1, y = 3 0 =, y 0 = 3 1 = 1, y 1 = 13 2 =, y 2 = 1 P6 2 = 3 = 21, y 3 = 19 4 = 3, y 4 = 3 = 1, y = 11 3 Result Set Q 1 = {, y) : 0, y 0} Q 2 = {, y) : < 0, y > 0} Q 3 = {, y) : 0, y 0} Theorem 1 Let { n, y n )} n=0 be a solution of System1) with 0, y 0 ) R 2 {, y) : > 0, y < 0} Then, there eists a non-negative integer N 0 such that the solution { n, y n )} n=n of System1) is either the prime period-6 cycle P6 1 or the prime period-6 cycle P6 2 The proof of Theorem 1 is a direct consequence of the following lemmas Lemma 2 Assume that there is a positive integer N such that N = y N 0 Then, { n, y n )} n=n+1 is the prime period-6 cycle P 1 6 Proof: Suppose that N, y N ) satisfies the hypothesis, N+1 = N N 1= 1, y N+1 = N N 1= 1 Hence, N+1, y N+1 ) = 1, 1) P 1 6 Lemma 3 Assume that there is a positive integer N such that N = y N 2 0 and y N 0 Then, { n, y n )} n=n+1 is the prime period-6 cycle P 1 6 Proof: Suppose that N, y N ) satisfies the hypothesis, N+1 = y N + 2 y N 1= 1, y N+1 = y N 2 + y N 1= 3 Hence, N+1, y N+1 ) = 1, 3) P6 1 Lemma 4 Let S := {, 3) R} Then, every solution { n, y n )} n=0 of System 1) with an initial condition in S is eventually a prime period-6 cycle P6 1 or P 6 2) Proof: Case 1: Suppose 0, y 0 ) S and 0 0 Then, 1 = > 0 and y 1 = 4 y 1 = 4 0, 3, y 3 ) = 1, 1) P6 1 Suppose that y 1 = 4 < 0, 2 = and y 2 = 2 3 Case 11: ) Suppose y 2 = < 4, 3 = and y 3 = = y 3 = ), 2 by Lemma 2, { n, y n )} n=3 is eventually prime period-6 P6 1) Suppose ) that 3 = y 3 = < 0 2 < 0 < 4 We will prove, by mathematical induction, that for 0 ) 2, 4 the solution is eventually prime period-6 For each n 0, let Pn) be the following statement: for 0 a n, b n ), 6n+4 = 2 4n+2 δ n, y 6n+4 = 2 4n δ n 2 < 0 0 a n, c n ], 6n+4 0 So, the solution is eventually prime period-6 0 c n, b n ), 6n+4 > 0 So, 6n+ = 2 4n+3 2δ n + 1 > 0, y 6n+ = 3, 6n+6 = 2 4n+3 2δ n + 3 > 0, y 6n+6 = 2 4n+3 2δ n 3 < 0, 6n+ =, y 6n+ = 2 4n+4 4δ n 1 0 [b n+1, b n ), y 6n+ 0 So, the solution is eventually prime period-6 P6 1) 0 c n, b n+1 ), y 6n+ < 0 So 6n+8 = 2 4n δ n + > 0, y 6n+8 = 2 4n+4 4δ n + 3 > 0, ISBN:

3 6n+9 = 2 4n δ n + 1, y 6n+9 = 2 4n δ n c n, a n+1 ], 6n+9 = y 6n+9 0 So, { n, y n )} n=0 is eventually prime period-6 P 1 6 ) 0 a n+1, b n+1 ), 6n+9 = y 6n+9 = 2 4n δ n + 1 < 0, where a n = 4n n+1, b n = 4n n, c n = 4n n+2, δ n = 4n+2 1 We shall show that P0) ) is true For 0 a 0, b 0 ) = 2, 4 and 3 = y 3 = < 0 Thus, we have 60)+4 = 4 = 4 1 = 2 40)+2 δ 0, y 60)+4 = y 4 = = 2 40) δ 0 2 < 0 0 a 0, c 0 ] = 2, 1 ], 4 = By Lemma 3, { n, y n )} n=4 is eventually prime period-6 P6 1) ) 1 0 c 0, b 0 ) = 4, 4, 4 = 4 1 > 0 Thus, we have 60)+ = 8 29 = 2 40)+3 2δ > 0, y 60)+ = 3, 60)+6 = 8 2 = 2 40)+3 2δ > 0, y 60)+6 = 8 33 = 2 40)+3 2δ 0 3 < 0, 60)+ =, y 60)+ = = 2 40)+4 4δ 0 1 [ ) 61 0 [b 1, b 0 ) = 16, 4, y 60)+ = , So 60)+8 = y 60)+8 = > 0 By Lemma 2, { n, y n )} n=9 is prime period c 0, b 1 ) = 4, 61 ), 16 60)+ = < 0 Thus, we have 60)+8 = = 2 40) δ 0 + > 0, y 60)+8 = 16 = 2 40)+4 4δ > 0, 60)+9 = = 2 40) δ 0 + 1, y 60)+9 = = 2 40) δ c 0, a 1 ] = 4, 121 ], 32 60)+9 = y 60)+9 = So, by Lemma 2, { n, y n )} n=10 is eventually prime period-6 P 6 1) a 1, b 1 ) = 32, 61 ), 16 60)+9 = y 60)+9 = < 0 Hence P0) is true Net, we assume that PN) is true for some positive integer N 1 We shall show that PN +1) is true Since PN) is true, 6N+9 = y 6N+9 = 2 4N+ 0 +8δ N + 1 < 0, where 0 a N+1, b N+1 ) = 4N N+, 19 ) 24N N+4 Then 6N+1)+4 = 6N+10 = 2 4N+6 16δ N 3 = 2 4N+1)+2 δ N+1, y 6N+1)+4 = y 6N+10 = 2 4N δ N + 1 = 2 4N+1) δ N+1 2 = 2 4N+6 4N+6 ) < 0 Note that δ N+1 = 4N+6 1 = 4N = 16δ N a N+1, c N+1 ] = 4N N+, 19 ] 24N N+6, 6N+10 = 2 4N+6 4N+6 ) 1 0 By Lemma 3, { n, y n )} n=6n+10 is eventually prime period-6 P6 1) 0 c N+1, b N+1 ) = 4N N+6, 19 ) 24N N+4, 6N+10 = 2 4N+6 4N+6 ) 1 > 0 Thus, we have 6N+1)+ = 6N+11 = 2 4N+1)+3 2δ N > 0, y 6N+1)+ = y 6N+11 = 3, 6N+1)+6 = 6N+12 = 2 4N+1)+3 2δ N > 0, y 6N+1)+6 = y 6N+12 = 2 4N+1)+3 2δ N+1 3 = 2 4N+ 4N+ ) + 13 < 0, 6N+1)+ = 6N+13 =, y 6N+1)+ = y 6N+13 = 2 4N+1)+4 4δ N+1 1 [ y 0 [b N+2, b N+1 ) = 4N N+8, 19 ) 24N N+4, y 6N+13 = 2 4N+1)+4 4δ N+1 1 ISBN:

4 = 2 4N+8 4N+8 ) + 1 0, 6N+14 = 2 4N δ N+1 + = 2 4N+8 4N+8 ) > 0 y 6N+14 = 2 4N δ N+1 + > 0 By Lemma 2, { n, y n )} n=6n+14 is eventually prime period-6 0 c N+1, b N+2 ) = 4N N+6, 19 ) 24N N+8, y 6N+13 = 2 4N+1)+4 0 4δ N+1 1 = 2 4N+8 4N+8 ) + 1 < 0 Thus, we have 6N+1)+8 = 6N+14 = 2 4N+1) δ N+1 + > 0, y 6N+1)+8 = y 6N+14 = 2 4N+1)+4 0 4δ N = 2 4N+8 4N+8 ) 19 > 0, 6N+1)+9 = 6N+1 = 2 4N+1)+ 0 +8δ N+1 +1, y 6N+1)+9 = y 6N+1 = 2 4N+1) δ N c N+1, a N+2 ] = 4N N+6, 4N N+9 6N+1 = y 6N+1 = 2 4N+1) δ N = 2 4N+9 4N+9 ) By Lemma 2, { n, y n )} n=6n+1 is eventually prime period-6p6 1) 0 a N+2, b N+2 ) = 4N N+9, 19 ) 24N N+8, 6N+1 = y 6N+1 = 2 4N+1) δ N = 2 4N+9 4N+9 ) < 0 Hence, PN + 1) is true By mathematical induction, Pn) is true for all n 0 Note that lim a n = lim b n = lim c n = 19 n n n We also note that if 0, y 0 ) = 3, y 3 ) = 3 ), 3 P6 2 ], 19, 3 ), Case 12: Suppose y 2 = 2 3 < < 3 ), 4 = and y 4 = , ], 4 = y 4 = So we apply Lemma 2 and { n, y n )} n=4 is eventually prime period-6 P6 1) Suppose that 0 4, 3 ) We will prove, 2 for 0 4, 3 ), that the solution is eventually 2 prime period-6 by mathematical induction For each n 0, let Qn) be the following statement: for 0 d n, e n ), 6n+ = 2 4n+3 σ n, y 6n+ = 2 4n σ n 2 < 0 0 d n, f n ], 6n+ 0 So, { n, y n )} n=0 is eventually prime period-6 P 6 1) 0 f n, e n ), 6n+ > 0 So 6n+6 = 2 4n+4 2σ n + 1 > 0, y 6n+6 = 3, 6n+ = 2 4n+4 2σ n + 3 > 0, y 6n+ = 2 4n+4 2σ n 3 < 0, 6n+8 =, y 6n+8 = 2 4n+ 4σ n 1 0 [e n+1, e n ), y 6n+8 0 So, the solution is eventually prime period-6 P6 1) 0 f n, e n+1 ), y 6n+8 < 0 Thus, we have 6n+9 = 2 4n σ n + > 0, y 6n+9 = 2 4n+ 4σ n + 3 > 0, 6n+10 = 2 4n σ n + 1, y 6n+10 = 2 4n σ n f n, d n+1 ], 6n+10 = y 6n+10 0, and so the solution is eventually prime period-6 P6 1) 0 d n+1, e n+1 ), 6n+10 = y 6n+10 < 0, where d n = 24n n+2, e n = 24n n+1, f n = 24n n+3, σ n = 24n+3 1 The proof is similar to the previous case We can conclude that Qn) is true for all n 0 Note that lim d n = lim e n = lim f n = n n ) n We also note that, 3 P6 2 Case 2: Let 0, y 0 ) S and 0 < 0 Then, y 0 = 3 Thus, 2 = > 0 and y 2 = 3 We see that 2, y 2 ) S and 2 > 0 By Case 1, every solution { n, y n )} n=2 is eventually prime period-6 P6 1 or P 6 2) ISBN:

5 Lemma Every solution { n, y n )} n=0 of System 1) with an initial condition in Q 1 is eventually prime period-6 P6 1 or P 6 2) Proof: Let 0, y 0 ) Q 1 Then 0 0 and y 0 0 Hence, 1 = y 0 1, y 1 = y = y 1 = y , by Lemma 2, 2, y 2 ) P6 1 1 = y 1 = y < 0, 2 = y 0 + 1, y 2 = 2 2y 0 3 < 0 2 = y = y 2 2 0, by Lemma 3, 2, y 2 ) P6 1 2 = y > 0, 3 = y 0 + 3, y 3 = 3 Hence by Lemma 4, the solution is eventually prime period-6 P6 1 or P 6 2) Lemma 6 Every solution { n, y n )} n=0 of System 1) with an initial condition in Q 3 is eventually prime period-6 P6 1 or P 6 2) Proof: Let 0, y 0 ) Q 3 Then 0 0 and y 0 0 Hence, 1 = y 0 1, y 1 = 0 + y 0 1 < 0, y 2 = 3 By Lemma 4, the solution is eventually prime period- 6 P6 1 or P 6 2 ) Lemma Every solution { n, y n )} n=0 of System 1) with an initial condition in Q 2 is eventually prime period-6 P6 1 or P 6 2) Proof: Let 0, y 0 ) Q 2 Then 0 < 0 and y 0 > 0 Hence, 1 = y 0 1, y 1 = y 0 1 < 0 Case 1: Suppose 1 = y Then, 2 = 2y > 0, y 2 = 2y 0 3 < 0, y 3 = 3 By Lemma 4, the solution is eventually prime period- 6 P 1 6 or P 2 6 ) Case 2: Suppose 1 = y 0 1 > 0 and so 0 > 1 Then, 2 = 2 1 > 0, y 2 = 2y 0 3 < 0, 3 = y > 0, y 3 = 2 2y 0 y 3 = 2 2y 0 0, 4 = 4y 0 + > 0, y 4 = 4y 0 + > 0 By Lemma 2, the solution is eventually prime period- 6 P6 1) y 3 = 2 2y 0 < 0, 4 = 4y 0 + > 0, y 4 = 4 y 4 = 4 0 ) 4 0 < 1, = y > 0, y = y 0 1 > 0 By Lemma, the solution is eventually prime period- 6 P6 1 or P 6 2) y 4 = 4 > 0 0 < ), 4 = y 0 + 9, y = y = y 0, by Lemma, the solution is eventually prime period-6 P 1 6 or P 2 6 ) = y < 0, by Lemma 6, the solution is eventually prime period-6 P6 1 or P 6 2) 4 Conclusion and Discussion We have presented the prime period-6 character of the solutions of System1) We utilized mathematical induction, and direct computations to show that every solution of System1) with an initial condition in Q 1 Q 2 Q 3 is eventually either the prime period-6 solution P6 1, or the prime period-6 solution P6 2 The proofs involve careful consideration of the various caseswe conjecture that every solution of System1)with an initial condition in R 2 is eventually either the prime period-6 solution P6 1, the prime period-6 solution P6 2 or the unique equilibrium point, 1, -1), from the beginning Acknowledgements: The research was supported by the Thailand Research Fund and Pibulsongkram Rajabhat UniversityMRG80088) The second author is supported by the Centre of Ecellence in Mathematics, CHE, Thailand References: [1] HO Peitgen, D Saupe, Y Fisher, M McGuire, RFVoss, MF Barnsley, RL Devaney, BB Mandelbrot, The science of fractal images, Springer Verlag, New York 1991 [2] RL Devanney, A piecewise linear model of the the zones of instability of an area-preserving map, Physica D, Vol10, 1984, pp [3] R Lozi, Un attracteur étrange du type attracteur de Henon, J Phys Colloq,Vol39, 198, ppc- 9 c-10 ISBN:

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