The stable equilibrium of a system of piecewise linear difference equations

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1 Tikjha et al. Advances in Difference Equations (2017) 2017:67 DOI /s R E S E A R C H Open Access The stable equilibrium of a system of piecewise linear difference equations Wirot Tikjha 1,2*, Evelina Lapierre 3 Thanin Sitthiwirattham 4 * Correspondence: wirottik@psru.ac.th 1 Faculty of Science Technology, Pibulsongkram Rajabhat University, Phitsanulok, Thail 2 Centre of Excellence in Mathematics, PERDO, CHE, Phitsanulok, Thail Full list of author information is available at the end of the article Abstract In this article we consider the global behavior of the system of first order piecewise linear difference equations: x n+1 = x n y n + b y n+1 = x n y n d where the parameters b d are any positive real numbers. We show that for any initial condition in R 2 the solution to the system is eventually the equilibrium, (2b + d, b). Moreover, the solutions of the system will reach the equilibrium within six iterations. MSC: 39A10; 65Q10 Keywords: difference equations; piecewise linear; equilibrium; stability 1 Introduction In applications, difference equations usually describe the evolution of a certain phenomenon over the course of time. In mathematics, a difference equation produces a sequence of numbers where each term of the sequence is defined as a function of the preceding terms. For the convenience of the reader we supply the following definitions. See [1, 2]. A system of difference equations of the first order is a system of the form x n+1 = f (x n, y n ), n =0,1,..., (1) y n+1 = g(x n, y n ), where f g are continuous functions which map R 2 into R. A solution of the system of difference equations (1)isasequence{(x n, y n )} n=0 which satisfies the system for all n 0. If we prescribe an initial condition (x 0, y 0 ) R 2 then x 1 = f (x 0, y 0 ), y 1 = g(x 0, y 0 ), x 2 = f (x 1, y 1 ), y 2 = g(x 1, y 1 ),. The Author(s) This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, reproduction in any medium, provided you give appropriate credit to the original author(s) the source, provide a link to the Creative Commons license, indicate if changes were made.

2 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 2 of 10 so the solution {(x n, y n )} n=0 of the system of difference equations (1) exists for all n 0 is uniquely determined by the initial condition (x 0, y 0 ). A solution of the system of difference equations (1) which is constant for all n 0is called an equilibrium solution.if (x n, y n )=( x, ȳ) for all n 0 isanequilibriumsolutionofthesystemofdifferenceequations (1), then ( x, ȳ) iscalledan equilibrium point,orsimplyanequilibrium of the system of difference equations (1). Known methods to determine the local asymptotic stability global stability are not easily applied to piecewise systems. This is why two of the most famous enigmatic systems of difference equations are piecewise: the Lozi Map x n+1 = a x n + y n +1, n =0,1,..., y n+1 = bx n, where the initial condition (x 0, y 0 ) R 2 the parameters a, b R, the Gingerbreadman map x n+1 = x n y n +1, n =0,1,..., y n+1 = x n, where the initial condition (x 0, y 0 ) R 2.See[3 5] for more information regarding the Lozi map Gingerbreadman map. In the last 30 years there has been progress in determining the local behavior of such systems but only limited progress in determining the global behavior. See [1, 6]. Ladas Grove developed the following family of 81 piecewise linear systems: x n+1 = x n + ay n + b, n =0,1,..., (2) y n+1 = x n + c y n + d, where the initial condition (x 0, y 0 ) R 2 the parameters a, b, c, d { 1,0,1}, in the hope of creating prototypes that will help us underst the global behavior of more complicated systems such as the Lozi map the Gingerbreadman map. See ([7 9]). In 2013, Lapierre found in [8] that the solutions of the following system of piecewise linear difference equations: x n+1 = x n y n +1, n =0,1,..., (3) y n+1 = x n y n 1, are eventually the unique equilibrium for every initial condition (x 0, y 0 ) R 2.Inthispaper we extend the results by examining a generalization of System (3), that is, x n+1 = x n y n + b, n =0,1,..., (4) y n+1 = x n y n d,

3 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 3 of 10 where the initial condition (x 0, y 0 ) R 2 the parameters b d are any positive real numbers. 2 Main results Set Condition (1) = { (x, y): x x + y y +2b d } x y d x y + b, Condition (2) = { (x, y):x + x y + y b +2d }, Condition (3) = { (x, y):x y + d }, Condition (4) = { (x, y):x 0, y 0x y + d }, Condition (5) = { (x, y):x 0, y 0x = y + b + d }. The proof of the theorem below uses the result from the four lemmas that follow. They show that if (x 0, y 0 ) R 2 then (x 1, y 1 ) is an element of Condition (1). Theorem 1 Let {(x n, y n )} n=0 be the solution of System (4). x n+1 = x n y n + b, n =0,1,..., y n+1 = x n y n d, with (x 0, y 0 ) R 2 b, d (0, ). Then {(x n, y n )} n=6 is the equilibrium (2b + d, d). Proof Suppose (x 0, y 0 ) R 2.Firstwewillshowthat(x 2, y 2 ) is an element of Condition (2), that is, x 2 + x 2 y 2 + y 2 b +2d. By Lemmas 1 through 4 we know that (x 1, y 1 ) is an element of Condition (1), so we have x 1 x 1 + y 1 y 1 +2b d x 1 y 1 d x 1 y 1 + b. Then x 1 y 1 + b + x1 y 1 + b x1 y 1 d + x1 y 1 d b +2d. Therefore (x 2, y 2 ) is an element of Condition (2), as required. Next, we will show that (x 3, y 3 ) is an element of Condition (3), that is x 3 y 3 + d. Since (x 2, y 2 ) is an element of Condition (2), we have x 2 y 2 d x 2 + y 2 b + d

4 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 4 of 10 we have x 2 y 2 d x 2 y 2 + b d. Then x 2 y 2 d x 2 y 2 + b d. Therefore (x 3, y 3 ) is an element of Condition (3), as required. Next, we will show that (x 4, y 4 ) is an element of Condition (4), that is x 4 0, y 4 0 x 4 y 4 + d. Since (x 3, y 3 ) is an element of Condition (3) x 4 = x 3 y 3 + b y 4 = x 3 y 3 d, weseethatx 4 0y 4 0. Also since (x 3, y 3 ) is an element of Condition (3), we have x 3 + x 3 y 3 + y 3 b +2d, so x 3 y 3 d x 3 + y 3 b + d. Note that x 3 y 3 d x 3 y 3 + b d. Then x 3 y 3 d x 3 y 3 + b d. Therefore (x 4, y 4 ) is an element of Condition (4), as required. Next, we will show that (x 5, y 5 ) is an element of Condition (5), that is x 5 0, y 5 0 x 5 = y 5 + b + d. Since (x 4, y 4 ) is an element of Condition (4) x 5 = x 4 y 4 + b y 5 = x 4 y 4 d,we see that x 5 0y 5 0. Consider x 5 y 5 = x 4 y 4 + b x 4 + y 4 + d = b + d so x 5 = y 5 + b + d. Therefore (x 5, y 5 ) is an element of Condition (5), as required. Finally,itiseasytoshowbydirectcomputationsthat(x 6, y 6 )=(2b+d, b). This completes the proof of the theorem.

5 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 5 of 10 The following four lemmas will show that if (x 0, y 0 ) R 2 then (x 1, y 1 )isanelementof Condition (1). Set Q 1 = { (x, y) x 0y 0 }, Q 2 = { (x, y) x 0y 0 }, Q 3 = { (x, y) x 0y 0 }, Q 4 = { (x, y) x 0y 0 }, recall that Condition (1) = { (x, y): x x + y y +2b d x y d x y + b }. Lemma 1 Let {(x n, y n )} n=0 be a solution of System (4) with (x 0, y 0 ) in Q 1. Then (x 1, y 1 ) is an element of Condition (1). Proof Suppose (x 0, y 0 ) Q 1 then x 0 0y 0 0. Thus x 1 = x 0 y 0 + b = x 0 y 0 + b, y 1 = x 0 y 0 d = x 0 y 0 d. Case 1 Suppose further x 0 y 0 + d.wehavex 1 = x 0 y 0 + b >0y 1 = x 0 y 0 d 0. Note that x 1 x 1 + y 1 y 1 +2b d =2b d x1 y 1 d x1 y 1 + b = b d. Hence (x 1, y 1 ) is an element of Condition (1) Case 1 is complete. Case 2 Suppose x 0 < y 0 + d but x 0 + b y 0.Wehavex 1 = x 0 y o + b 0y 1 = x 0 y 0 d <0.Notethat x 1 x 1 + y 1 y 1 +2b d = 2x 0 +2y 0 +2b + d x1 y 1 d x1 y 1 + b = 2x0 2y 0 + b 2d 2b d. Case 2A Suppose further 2x 0 2y 0 + b 2d 0. Then x1 y 1 d x1 y 1 + b =2x0 2y 0 2d b d.

6 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 6 of 10 Since y 1 = x 0 y 0 d <0,wehave2x 0 2y 0 2d b <0.Alsonotethat y 1 y 1 +2b >0, so x 1 x 1 + y 1 y 1 +2b d = y 1 y 1 +2b d >2x 0 2y 0 2d b d = x 1 y 1 d x 1 y 1 + b. Case 2A is complete. Case 2B Suppose 2x 0 2y 0 + b 2d <0.Then x 1 y 1 d x 1 y 1 + b = 2x 0 +2y 0 3b + d < 2x 0 +2y 0 +2b + d = x 1 x 1 + y 1 y 1 +2b d. Hence (x 1, y 1 ) is an element of Condition (1) Case 2 is complete. Case 3 Finally suppose x 0 < y 0 + d x 0 + b < y 0.Wehavex 1 = x 0 y o + b <0y 1 = x 0 y 0 d <0.Notethat x 1 x 1 + y 1 y 1 +2b d = 4x 0 +4y 0 + d x 1 y 1 d x 1 y 1 + b = d b. Since x 0 + b < y 0,wehavey 0 > x 0.Thus4y 0 4x 0 >0.Wenotethat b <0.Then x 1 x 1 + y 1 y 1 +2b d = 4x 0 +4y 0 + d > d b = x1 y 1 d x1 y 1 + b. Hence (x 1, y 1 ) is an element of Condition (1) Case 3 is complete. Lemma 2 Let {(x n, y n )} n=0 be a solution of System (4) with (x 0, y 0 ) in Q 2. Then (x 1, y 1 ) is an element of Condition (1). Proof Suppose (x 0, y 0 ) Q 2 then x 0 0y 0 0. Thus x 1 = x 0 y 0 + b = x 0 y 0 + b, y 1 = x 0 y 0 d = x 0 y 0 d <0.

7 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 7 of 10 Case 1 Suppose further x 0 + b < y 0.Wehavex 1 = x 0 y 0 + b <0y 1 = x 0 y 0 d <0. Note that x 1 x 1 + y 1 y 1 +2b d = 2x 1 2y 1 +2b d =4y 0 + d x1 y 1 d x1 y 1 + b = d b. Since y 0 0 b <0,weseethat(x 1, y 1 ) is an element of Condition (1) so Case 1 is complete. Case 2 Suppose x 0 + b y 0.Wehavex 1 = x 0 y 0 + b 0y 1 = x 0 y 0 d <0.Note that x 1 x 1 + y 1 y 1 +2b d = 2x 0 +2y 0 +2b + d >0 x 1 y 1 d x 1 y 1 + b = 2y 0 + b 2d +2x 0 2b d. Case 2A Suppose further b 2y 0 + d.then x1 y 1 d x1 y 1 + b =2x0 2y 0 b 3d. Since y 1 = x 0 y 0 d <0,wehave2x 0 2y 0 b 3d <0.Hence(x 1, y 1 )isanelementof Condition (1). Case 2Aiscomplete. Case 2B Finally suppose b <2y 0 + d.then x1 y 1 d x1 y 1 + b =2x0 +2y 0 3b + d. Since x 1 = x 0 y 0 + b 0, we have 2x 0 +2y 0 3b <0.Hence(x 1, y 1 )isanelementof Condition (1) the proof to Lemma 2 is complete. Lemma 3 Let {(x n, y n )} n=0 be a solution of System (4) with (x 0, y 0 ) in Q 3. Then (x 1, y 1 ) is an element of Condition (1). Proof Suppose (x 0, y 0 ) Q 3 then x 0 0y 0 0. Thus x 1 = x 0 y 0 + b = x 0 y 0 + b >0, y 1 = x 0 y 0 d = x 0 + y 0 d <0. Then x 1 x 1 + y 1 y 1 +2b d = 2x 0 2y 0 +2b + d >0

8 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 8 of 10 x1 y 1 d x1 y 1 + b = b 2d ( 2x0 2y 0 +2b + d). Case 1 Suppose b 2d 0. Then x 1 y 1 d x 1 y 1 + b =2x 0 +2y 0 b 3d <0. Hence (x 1, y 1 ) is an element of Condition (1) Case 1 is complete. Case 2 Suppose further b 2d <0.Then x1 y 1 d x1 y 1 + b = b +2d +2x0 +2y 0 2b d =2x 0 +2y 0 3b + d. Since 2x 0 2y 0 +2b >0,wehave2x 0 +2y 0 3b <0.Hence(x 1, y 1 ) is an element of Condition (1) the proof to Lemma 3 is complete. Lemma 4 Let {(x n, y n )} n=0 be a solution of System (4) with (x 0, y 0 ) in Q 4. Then (x 1, y 1 ) is an element of Condition (1). Proof Suppose (x 0, y 0 ) Q 4 then x 0 0y 0 0. Thus x 1 = x 0 y 0 + b = x 0 y 0 + b >0, y 1 = x 0 y 0 d = x 0 + y 0 d. Case 1 Suppose further y 1 = x 0 + y 0 d 0. Then x 1 x 1 + y 1 y 1 +2b d =2b d x 1 y 1 d x 1 y 1 + b = b d. Hence (x 1, y 1 ) is an element of Condition (1) Case 1 is complete. Case 2 Suppose y 1 = x 0 + y 0 d <0.Then x 1 x 1 + y 1 y 1 +2b d = 2y 1 +2b d = 2x 0 2y 0 +2b + d x 1 y 1 d x 1 y 1 + b = 2x 0 + b 2d +2y 0 2b d. Case 2A Suppose further 2x 0 + b 2d 0. Then x1 y 1 d x1 y 1 + b =2x0 + b 2d +2y 0 2b d =2x 0 +2y 0 3d b.

9 Tikjha et al. Advances in Difference Equations (2017) 2017:67 Page 9 of 10 Since 2x 0 + b 2d 0, b > 2x 0.Thus 2x 0 2y 0 +2b + d >0.Sincey 1 = x 0 + y 0 d <0, we have 2x 0 +2y 0 3d b <0.Hence(x 1, y 1 ) is an element of Condition (1) Case 2A is complete. Case 2B Now suppose 2x 0 + b 2d <0.Then x1 y 1 d x1 y 1 + b = 2x0 b +2d +2y 0 2b d = 2x 0 +2y 0 3b + d. Since y 0 0b >0,weseethat(x 1, y 1 ) is an element of Condition (1) the proof of Lemma 4 is complete. 3 Discussion conclusion In this paper we showed that for any initial value (x 0, y 0 ) R 2 we have the following sequence: (x 1, y 1 ) { (x, y): x x + y y +2b d } x y d x y + b, (x 2, y 2 ) { (x, y):x + x y + y b +2d }, (x 3, y 3 ) { (x, y):x y + d }, (x 4, y 4 ) { (x, y):x 0, y 0x y + d }, (x 5, y 5 ) { (x, y):x 0, y 0x = y + b + d }, (x 6, y 6 )=( x, ȳ)=(2b + d, b). In addition, if we begin with an initial condition that is an element of Condition (N) for N {1, 2, 3, 4, 5},thenitrequires6 N iterations to reach the equilibrium point. The generalized system of piecewise linear difference equations examined in this paper was created as a prototype to underst the global behavior of more complicated systems. We believe that this paper contributes broadly to the overall understing of systems whose global behavior still remains unknown. Competing interests The authors declare that they have no competing interests. Authors contributions All authors contributed equally to the writing of this paper. All authors read approved the final manuscript. Author details 1 Faculty of Science Technology, Pibulsongkram Rajabhat University, Phitsanulok, Thail. 2 Centre of Excellence in Mathematics, PERDO, CHE, Phitsanulok, Thail. 3 Department of Mathematics, Johnson Wales University, 8 Abbott Park Place, Providence, RI, USA. 4 King Mongkut s University of Technology North Bangkok, 1518 Pracharaj 1 Rd., Bangsue, Bangkok, 10800, Thail. Acknowledgements This work was supported by the Thail Research Fund [MRG ], National Research Council of Thail Pibulsongkram Rajabhat University. The first author is supported by the Centre of Excellence in Mathematics, CHE, Thail. Received: 1 October 2016 Accepted: 10 February 2017 References 1. Grove, EA, Ladas, G: Periodicities in Nonlinear Difference Equations. Chapman Hall, New York (2005) 2. Kocic, VL, Ladas, G: Global Behavior of Nonlinear Difference Equations of Higher Order with Applications. Kluwer Academic, Boston (1993)

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