Open Access

The stable equilibrium of a system of piecewise linear difference equations

Advances in Difference Equations20172017:67

DOI: 10.1186/s13662-017-1117-2

Received: 1 October 2016

Accepted: 10 February 2017

Published: 28 February 2017

Abstract

In this article we consider the global behavior of the system of first order piecewise linear difference equations: \(x_{n+1} = \vert x_{n}\vert - y _{n} +b\) and \(y_{n+1} = x_{n} - \vert y_{n}\vert - d\) where the parameters b and 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.

Keywords

difference equations piecewise linear equilibrium stability

MSC

39A10 65Q10

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
$$ \textstyle\begin{cases} x_{n+1} = f(x_{n}, y_{n}), \\ y_{n+1} = g(x_{n}, y_{n}), \end{cases}\displaystyle n=0,1,\ldots, $$
(1)
where f and g are continuous functions which map \(\mathbf{R}^{2}\) into R.
A solution of the system of difference equations (1) is a sequence \(\{(x_{n},y_{n})\}_{n = 0}^{\infty}\) which satisfies the system for all \(n \geq0\). If we prescribe an initial condition
$$(x_{0}, y_{0}) \in\mathbf{R}^{2} $$
then
$$\begin{aligned}& \textstyle\begin{cases} x_{1} = f(x_{0}, y_{0}), \\ y_{1} = g(x_{0}, y_{0}), \end{cases}\displaystyle \\& \textstyle\begin{cases} x_{2} = f(x_{1}, y_{1}), \\ y_{2} = g(x_{1}, y_{1}), \end{cases}\displaystyle \\& \vdots \end{aligned}$$
and so the solution \(\{(x_{n},y_{n})\}_{n = 0}^{\infty}\) of the system of difference equations (1) exists for all \(n\geq0\) and 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 \(n\geq0\) is called an equilibrium solution. If
$$(x_{n}, y_{n}) = (\bar{x}, \bar{y})\quad \text{for all } n \geq0 $$
is an equilibrium solution of the system of difference equations (1), then \((\bar{x},\bar{y})\) is called an equilibrium point, or simply an equilibrium of the system of difference equations (1).
Known methods to determine the local asymptotic stability and global stability are not easily applied to piecewise systems. This is why two of the most famous and enigmatic systems of difference equations are piecewise: the Lozi Map
$$\begin{aligned}& \textstyle\begin{cases} x_{n+1} = -a\vert x_{n} \vert + y_{n} + 1, \\ y_{n+1} = bx_{n} , \end{cases}\displaystyle n=0,1,\ldots, \end{aligned}$$
where the initial condition \((x_{0},y_{0}) \in\mathbf{R}^{2}\) and the parameters \(a,b \in\mathbf{R}\), and the Gingerbreadman map
$$\begin{aligned}& \textstyle\begin{cases} x_{n+1} = \vert x_{n} \vert - y_{n} + 1, \\ y_{n+1} = x_{n}, \end{cases}\displaystyle n=0,1,\ldots, \end{aligned}$$
where the initial condition \((x_{0},y_{0}) \in\mathbf{R}^{2}\). See [35] for more information regarding the Lozi map and 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 and Grove developed the following family of 81 piecewise linear systems:
$$ \textstyle\begin{cases} x_{n+1} = \vert x_{n} \vert + ay_{n} + b, \\ y_{n+1} = x_{n} + c\vert y_{n}\vert + d, \end{cases}\displaystyle n = 0, 1, \ldots, $$
(2)
where the initial condition \((x_{0}, y_{0}) \in R^{2}\) and the parameters a, b, c, and \(d \in\{-1,0,1\}\), in the hope of creating prototypes that will help us understand the global behavior of more complicated systems such as the Lozi map and the Gingerbreadman map. See ([79]).
In 2013, Lapierre found in [8] that the solutions of the following system of piecewise linear difference equations:
$$ \textstyle\begin{cases} x_{n+1} = \vert x_{n} \vert - y_{n} +1, \\ y_{n+1} = x_{n} - \vert y_{n}\vert - 1, \end{cases}\displaystyle n = 0, 1, \ldots, $$
(3)
are eventually the unique equilibrium for every initial condition \((x_{0}, y_{0}) \in\mathbf{R}^{2}\). In this paper we extend the results by examining a generalization of System (3), that is,
$$ \textstyle\begin{cases} x_{n+1} = \vert x_{n} \vert - y_{n} + b, \\ y_{n+1} = x_{n} - \vert y_{n}\vert - d, \end{cases}\displaystyle n = 0, 1, \ldots, $$
(4)
where the initial condition \((x_{0}, y_{0}) \in\mathbf{R}^{2}\) and the parameters b and d are any positive real numbers.

2 Main results

Set
$$\begin{aligned}& \mbox{Condition (1)} =\bigl\{ (x,y): \vert x\vert - x + \vert y\vert - y + 2b - d \geq \bigl\vert x - \vert y\vert - d\bigr\vert - \bigl\vert \vert x\vert - y + b\bigr\vert \bigr\} , \\& \mbox{Condition (2)} =\bigl\{ (x,y): x+ \vert x\vert \geq y + \vert y \vert - b + 2d\bigr\} , \\& \mbox{Condition (3)} =\bigl\{ (x,y): x \geq \vert y\vert + d \bigr\} , \\& \mbox{Condition (4)} =\bigl\{ (x,y): x \geq0, y \geq0 \text{ and } x \geq \vert y\vert + d \bigr\} , \\& \mbox{Condition (5)} =\bigl\{ (x,y): x \geq0, y \geq0 \text{ and } x = y + b + d \bigr\} . \end{aligned}$$

The proof of the theorem below uses the result from the four lemmas that follow. They show that if \((x_{0}, y_{0})\in R^{2}\) then \((x_{1}, y _{1})\) is an element of Condition (1).

Theorem 1

Let \(\{(x_{n}, y_{n})\}_{n=0}^{\infty}\) be the solution of System (4).
$$\textstyle\begin{cases} x_{n+1} = \vert x_{n} \vert - y_{n} + b, \\ y_{n+1} = x_{n} - \vert y_{n}\vert - d, \end{cases}\displaystyle n = 0, 1, \ldots, $$
with \((x_{0}, y_{0}) \in\mathbf{R}^{2}\) and \(b,d \in(0, \infty)\). Then \(\{(x_{n}, y_{n})\}_{n=6}^{\infty}\) is the equilibrium \((2b + d, d)\).

Proof

Suppose \((x_{0}, y_{0})\in R^{2}\). First we will show that \((x_{2}, y _{2})\) is an element of Condition (2), that is,
$$x_{2} + \vert x_{2}\vert \geq y_{2} + \vert y_{2}\vert -b+2d. $$
By Lemmas 1 through 4 we know that \((x_{1}, y_{1})\) is an element of Condition (1), so we have
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d \geq \bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert -\bigl\vert \vert x_{1} \vert -y_{1}+b\bigr\vert . $$
Then
$$\vert x_{1}\vert - y_{1} + b + \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert \geq x_{1} - \vert y_{1}\vert - d + \bigl\vert x_{1}-\vert y_{1}\vert -d\bigr\vert - b +2d. $$
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
$$x_{3} \geq \vert y_{3}\vert + d. $$
Since \((x_{2}, y_{2})\) is an element of Condition (2), we have
$$x_{2} - \vert y_{2}\vert - d \geq- \vert x_{2}\vert +y_{2}-b+d $$
and we have
$$x_{2} - \vert y_{2}\vert - d \leq \vert x_{2} \vert - y_{2}+b-d. $$
Then
$$\bigl\vert x_{2} - \vert y_{2}\vert -d\bigr\vert \leq \vert x_{2}\vert - y_{2} + b - d. $$
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
$$x_{4} \geq0, \qquad y_{4} \geq0\quad \text{and} \quad x_{4} \geq \vert y_{4}\vert + d. $$
Since \((x_{3}, y_{3})\) is an element of Condition (3) and \(x_{4}=\vert x_{3}\vert -y_{3}+b\) and \(y_{4}= x_{3} - \vert y_{3}\vert -d\), we see that \(x_{4} \geq0\) and \(y_{4} \geq0\). Also since \((x_{3}, y_{3})\) is an element of Condition (3), we have
$$\vert x_{3}\vert + x_{3} \geq y_{3} + \vert y_{3}\vert - b + 2d, $$
and so
$$x_{3} - \vert y_{3}\vert - d \geq-\vert x_{3}\vert + y_{3} - b + d. $$
Note that
$$x_{3} - \vert y_{3}\vert - d \leq \vert x_{3}\vert - y_{3} + b - d. $$
Then
$$\bigl\vert x_{3} - \vert y_{3}\vert - d\bigr\vert \leq \vert x_{3}\vert - y_{3} + b - d. $$
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
$$x_{5} \geq0, \qquad y_{5} \geq0\quad \text{and}\quad x_{5} = y_{5} + b + d. $$
Since \((x_{4}, y_{4})\) is an element of Condition (4) and \(x_{5}=\vert x_{4}\vert -y_{4}+b\) and \(y_{5}= x_{4} - \vert y_{4}\vert -d\), we see that \(x_{5} \geq0 \) and \(y_{5} \geq0\). Consider
$$x_{5} - y_{5} = \vert x_{4} \vert -y_{4}+b - x_{4} + \vert y_{4}\vert +d = b + d \quad \text{and so}\quad x_{5} = y_{5} + b + d. $$

Therefore \((x_{5},y_{5})\) is an element of Condition (5), as required.

Finally, it is easy to show by direct computations that \((x_{6}, y_{6}) = (2b + d, b)\). This completes the proof of the theorem. □

The following four lemmas will show that if \((x_{0}, y_{0})\in R^{2}\) then \((x_{1}, y_{1})\) is an element of Condition (1). Set
$$\begin{aligned}& \mathcal{Q}_{1} = \bigl\{ (x, y)\vert x \geq0\mbox{ and }y \geq0 \bigr\} , \\& \mathcal{Q}_{2} = \bigl\{ (x, y)\vert x \leq0\mbox{ and }y \geq0 \bigr\} , \\& \mathcal{Q}_{3} = \bigl\{ (x, y)\vert x \leq0\mbox{ and }y \leq0 \bigr\} , \\& \mathcal{Q}_{4} = \bigl\{ (x, y)\vert x \geq0\mbox{ and }y \leq0 \bigr\} , \end{aligned}$$
and recall that
$$\mbox{Condition (1)} =\bigl\{ (x,y): \vert x\vert - x + \vert y\vert - y + 2b - d \geq\bigl\vert x - \vert y\vert - d\bigr\vert - \bigl\vert \vert x\vert - y + b\bigr\vert \bigr\} . $$

Lemma 1

Let \(\{ (x_{n},y_{n})\}_{n=0}^{\infty}\) be a solution of System (4) with \((x_{0}, y_{0})\) in \(\mathcal{Q}_{1}\). Then \((x_{1}, y_{1})\) is an element of Condition (1).

Proof

Suppose \((x_{0}, y_{0}) \in\mathcal{Q}_{1}\) then \(x_{0} \geq0\) and \(y_{0} \geq0\). Thus
$$\textstyle\begin{cases} x_{1} = \vert x_{0}\vert - y_{0} +b = x_{0} - y_{0} + b,\\ y_{1} = x_{0} - \vert y_{0}\vert - d = x_{0} - y_{0} - d. \end{cases} $$

Case 1

Suppose further \(x_{0} \geq y_{0} +d \). We have \(x_{1} = x_{0} - y_{0} + b > 0\) and \(y_{1} = x_{0} - y_{0} - d \geq0\). Note that
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = 2b-d $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = -b - d. $$

Hence \((x_{1},y_{1})\) is an element of Condition (1) and Case 1 is complete.

Case 2

Suppose \(x_{0} < y_{0} +d \) but \(x_{0} +b \geq y _{0} \). We have \(x_{1} = x_{0} - y_{o} + b \geq0\) and \(y_{1} = x_{0} - y_{0} - d < 0\). Note that
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = -2x_{0}+2y_{0}+2b+d $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = \vert 2x_{0}-2y_{0}+b-2d\vert - 2b-d. $$

Case 2A

Suppose further \(2x_{0}-2y_{0}+b-2d \geq0\). Then
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = 2x_{0}-2y_{0}-2d - b -d. $$
Since \(y_{1} = x_{0} - y_{0} - d < 0\), we have \(2x_{0}-2y_{0}-2d - b < 0\). Also note that \(\vert y_{1}\vert - y_{1} + 2b > 0\), so
$$\begin{aligned} \vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d =&\vert y_{1}\vert - y_{1} + 2b - d \\ >& 2x_{0}-2y_{0}-2d - b -d \\ = &\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert . \end{aligned}$$
Case 2A is complete.

Case 2B

Suppose \(2x_{0}-2y_{0}+b-2d < 0\). Then
$$\begin{aligned} \bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = & -2x_{0}+2y_{0} - 3b +d \\ < &-2x_{0}+2y_{0}+2b+d \\ =&\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d. \end{aligned}$$
Hence \((x_{1},y_{1})\) is an element of Condition (1) and Case 2 is complete.

Case 3

Finally suppose \(x_{0} < y_{0} +d \) and \(x_{0} +b < y_{0} \). We have \(x_{1} = x_{0} - y_{o} + b < 0\) and \(y_{1} = x_{0} - y_{0} - d < 0\). Note that
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = -4x_{0}+4y_{0}+d $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = d - b. $$
Since \(x_{0} + b < y_{0}\), we have \(y_{0} > x_{0}\). Thus \(4y_{0} - 4x _{0} > 0\). We note that \(-b < 0\). Then
$$\begin{aligned} \vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d =& -4x_{0}+4y_{0}+d \\ >& d - b \\ =& \bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert . \end{aligned}$$
Hence \((x_{1},y_{1})\) is an element of Condition (1) and Case 3 is complete.

 □

Lemma 2

Let \(\{ (x_{n},y_{n})\}_{n=0}^{\infty}\) be a solution of System (4) with \((x_{0}, y_{0})\) in \(\mathcal{Q}_{2}\). Then \((x_{1}, y_{1})\) is an element of Condition (1).

Proof

Suppose \((x_{0}, y_{0}) \in\mathcal{Q}_{2}\) then \(x_{0} \leq0\) and \(y_{0} \geq0\). Thus
$$\textstyle\begin{cases} x_{1} = \vert x_{0}\vert - y_{0} +b = - x_{0} - y_{0} + b,\\ y_{1} = x_{0} - \vert y_{0}\vert - d = x_{0} - y_{0} - d < 0. \end{cases} $$

Case 1

Suppose further \(-x_{0} + b < y_{0}\). We have \(x_{1} = - x_{0} - y_{0} + b < 0\) and \(y_{1}= x_{0} - y_{0} - d < 0\). Note that
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = -2x_{1} - 2y_{1} + 2b-d = 4y _{0}+d $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = d - b. $$
Since \(y_{0} \geq0\) and \(-b < 0\), we see that \((x_{1}, y_{1})\) is an element of Condition (1) and so Case 1 is complete.

Case 2

Suppose \(-x_{0} + b \geq y_{0}\). We have \(x_{1} = - x_{0} - y_{0} + b \geq0\) and \(y_{1} = x_{0} - y_{0} - d < 0\). Note that
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = -2x_{0} + 2y_{0} + 2b+d > 0 $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = \vert {-}2y_{0} + b -2d\vert + 2x_{0} -2b -d. $$

Case 2A

Suppose further \(b \geq2y_{0}+d\). Then
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = 2x_{0}-2y_{0}- b -3d. $$
Since \(y_{1} = x_{0} - y_{0} - d < 0\), we have \(2x_{0}-2y_{0}- b -3d< 0\). Hence \((x_{1}, y_{1})\) is an element of Condition (1). Case 2A is complete.

Case 2B

Finally suppose \(b < 2y_{0}+d\). Then
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = 2x_{0}+2y_{0}-3b + d. $$
Since \(x_{1} = -x_{0} - y_{0} + b \geq0\), we have \(2x_{0}+2y_{0}-3b < 0\). Hence \((x_{1}, y_{1})\) is an element of Condition (1) and the proof to Lemma 2 is complete.

 □

Lemma 3

Let \(\{ (x_{n},y_{n})\}_{n=0}^{\infty}\) be a solution of System (4) with \((x_{0}, y_{0})\) in \(\mathcal{Q}_{3}\). Then \((x_{1}, y_{1})\) is an element of Condition (1).

Proof

Suppose \((x_{0}, y_{0}) \in\mathcal{Q}_{3}\) then \(x_{0} \leq0\) and \(y_{0} \leq0\). Thus
$$\textstyle\begin{cases} x_{1} = \vert x_{0}\vert - y_{0} +b = -x_{0} - y_{0} + b > 0, \\ y_{1} = x_{0} - \vert y_{0}\vert - d = x_{0} + y_{0} - d < 0. \end{cases} $$
Then
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = -2x_{0} - 2y_{0}+2b+d >0 $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert =\vert b-2d\vert -(-2x_{0}-2y_{0}+2b+d). $$

Case 1

Suppose \(b-2d \geq0\). Then
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = 2x_{0} + 2y_{0} -b-3d < 0. $$
Hence \((x_{1}, y_{1})\) is an element of Condition (1) and Case 1 is complete.

Case 2

Suppose further \(b-2d < 0\). Then
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = -b + 2d + 2x_{0} + 2y_{0} - 2b - d = 2x_{0} + 2y_{0} - 3b +d. $$
Since \(-2x_{0} - 2y_{0} + 2b >0\), we have \(2x_{0} + 2y_{0} - 3b < 0\). Hence \((x_{1}, y_{1})\) is an element of Condition (1) and the proof to Lemma 3 is complete.

 □

Lemma 4

Let \(\{ (x_{n},y_{n})\}_{n=0}^{\infty}\) be a solution of System (4) with \((x_{0}, y_{0})\) in \(\mathcal{Q}_{4}\). Then \((x_{1}, y_{1})\) is an element of Condition (1).

Proof

Suppose \((x_{0}, y_{0}) \in\mathcal{Q}_{4}\) then \(x_{0} \geq0\) and \(y_{0} \leq0\). Thus
$$\textstyle\begin{cases} x_{1} = \vert x_{0}\vert - y_{0} +b = x_{0} - y_{0} + b > 0,\\ y_{1} = x_{0} - \vert y_{0}\vert - d = x_{0} + y_{0} - d. \end{cases} $$

Case 1

Suppose further \(y_{1} = x_{0} + y_{0} - d \geq0\). Then
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = 2b - d $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = -b -d. $$
Hence \((x_{1}, y_{1})\) is an element of Condition (1) and Case 1 is complete.

Case 2

Suppose \(y_{1} = x_{0} + y_{0} - d < 0\). Then
$$\vert x_{1}\vert - x_{1} + \vert y_{1} \vert -y_{1}+2b-d = -2y_{1} + 2b -d = -2x_{0} -2y _{0} +2b +d $$
and
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = \vert 2x_{0}+b-2d\vert +2y_{0}-2b-d. $$

Case 2A

Suppose further \(2x_{0}+b-2d \geq0\). Then
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = 2x_{0} + b - 2d + 2y_{0} - 2b - d = 2x_{0} + 2y_{0} -3d - b. $$
Since \(2x_{0}+b-2d \geq0\), \(b > -2x_{0}\). Thus \(-2x_{0} -2y_{0} +2b +d > 0\). Since \(y_{1} = x_{0} + y_{0} - d < 0\), we have \(2x_{0} + 2y_{0} -3d - b <0\). Hence \((x_{1}, y_{1})\) is an element of Condition (1) and Case 2A is complete.

Case 2B

Now suppose \(2x_{0}+b-2d < 0\). Then
$$\bigl\vert x_{1} - \vert y_{1}\vert -d\bigr\vert - \bigl\vert \vert x_{1}\vert -y_{1}+b\bigr\vert = -2x_{0} - b + 2d + 2y_{0} - 2b - d = -2x_{0} + 2y_{0} - 3b +d. $$
Since \(y_{0} \leq0\) and \(b > 0\), we see that \((x_{1}, y_{1})\) is an element of Condition (1) and the proof of Lemma 4 is complete.

 □

3 Discussion and conclusion

In this paper we showed that for any initial value \((x_{0}, y_{0}) \in R^{2}\) we have the following sequence:
$$\begin{aligned}& (x_{1}, y_{1}) \in\bigl\{ (x,y): \vert x\vert - x + \vert y\vert - y + 2b - d \geq\bigl\vert x - \vert y\vert - d\bigr\vert - \bigl\vert \vert x\vert - y + b\bigr\vert \bigr\} , \\& (x_{2}, y_{2}) \in\bigl\{ (x,y): x+ \vert x\vert \geq y + \vert y\vert - b + 2d\bigr\} , \\& (x_{3}, y_{3}) \in\bigl\{ (x,y): x \geq \vert y\vert + d \bigr\} , \\& (x_{4}, y_{4}) \in\bigl\{ (x,y): x \geq0, y \geq0 \text{ and } x \geq \vert y\vert + d \bigr\} , \\& (x_{5}, y_{5}) \in\bigl\{ (x,y): x \geq0, y \geq0 \text{ and } x = y + b + d\bigr\} , \\& (x_{6}, y_{6}) = (\bar{x}, \bar{y})=(2b + d, b). \end{aligned}$$
In addition, if we begin with an initial condition that is an element of Condition (N) for \(N \in\{1,2,3,4,5\}\), then it requires \(6-N\) iterations to reach the equilibrium point.

The generalized system of piecewise linear difference equations examined in this paper was created as a prototype to understand the global behavior of more complicated systems. We believe that this paper contributes broadly to the overall understanding of systems whose global behavior still remains unknown.

Declarations

Acknowledgements

This work was supported by the Thailand Research Fund [MRG5980053], National Research Council of Thailand and Pibulsongkram Rajabhat University. The first author is supported by the Centre of Excellence in Mathematics, CHE, Thailand.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
Faculty of Science and Technology, Pibulsongkram Rajabhat University
(2)
Centre of Excellence in Mathematics, PERDO, CHE
(3)
Department of Mathematics, Johnson and Wales University
(4)
King Mongkut’s University of Technology North Bangkok

References

  1. Grove, EA, Ladas, G: Periodicities in Nonlinear Difference Equations. Chapman Hall, New York (2005) MATHGoogle Scholar
  2. Kocic, VL, Ladas, G: Global Behavior of Nonlinear Difference Equations of Higher Order with Applications. Kluwer Academic, Boston (1993) View ArticleMATHGoogle Scholar
  3. Barnsley, MF, Devaney, RL, Mandelbrot, BB, Peitgen, HO, Saupe, D, Voss, RF: The Science of Fractal Images. Springer, New York (1991) MATHGoogle Scholar
  4. Devaney, RL: A piecewise linear model of the zones of instability of an area-preserving map. Physica 10D, 387-393 (1984) MATHGoogle Scholar
  5. Lozi, R: Un attracteur etrange du type attracteur de Henon. J. Phys. (Paris) 39, 9-10 (1978) View ArticleGoogle Scholar
  6. Botella-Soler, V, Castelo, JM, Oteo, JA, Ros, J: Bifurcations in the Lozi map. J. Phys. A, Math. Theor. 44, 1-17 (2011) View ArticleMATHGoogle Scholar
  7. Grove, EA, Lapierre, E, Tikjha, W: On the global behavior of \(x_{n+1} = \vert x_{n}\vert - y_{n} -1 \) and \(y_{n+1} = x_{n} + \vert y_{n}\vert \). CUBO 14, 125-166 (2012) View ArticleMATHGoogle Scholar
  8. Lapierre, EG: On the global behavior of some systems of difference equation. Doctoral dissertation, University of Rhode Island (2013)
  9. Tikjha, W, Lapierre, EG, Lenbury, Y: On the global character of the system of piecewise linear difference equations \(x_{n+1} = \vert x_{n}\vert - y_{n} - 1\) and \(y_{n+1} = x_{n} - \vert y_{n}\vert \). Adv. Differ. Equ. 2010 (2010)

Copyright

© The Author(s) 2017