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The stable equilibrium of a system of piecewise linear difference equations

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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.

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.

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.

 □

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.

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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.

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Correspondence to Wirot Tikjha.

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The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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MSC

  • 39A10
  • 65Q10

Keywords

  • difference equations
  • piecewise linear
  • equilibrium
  • stability