A Generalised Difference Equation and Its Dynamics and Solutions

Abstract

Rational difference equations have a wide range of applications in various fields of science. To illustrate, the equation $x_{n+1}={\displaystyle \frac{a+b{x}_n}{c+d{x}_n}},n=0,1,...$, known as the Riccati difference equation, has been applied in the field of optics. In this study, the global asymptotic stability of the difference equation $x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{B+C{x}_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}},n=0,1,...$, is proved. The solutions of this difference equation are obtained by applying the standard iteration method, and the periodicity of these solutions is determined. Furthermore, this difference equation represents a generalisation of the results obtained in previous studies.

1. Introduction

Our aim in this paper is to analyse the global behaviour of the non-negative equilibrium points of the difference equation

$$x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{B+C{x}_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}},n=0,1,...$$

(1)

where $A,B$, and C are non-negative real numbers, the initial conditions are non-negative, and k and j are non-negative integers. Also, we determined the solutions of some special cases of Equation (1).

The aim of our study of the difference equation, Equation (1), in this paper is to obtain a generalisation of many previous studies. With Equation (1), the general form of the difference equations in the references [1,2,3,4,5,6,7,8,9,10,11,12] has been reached. In addition, it will be possible to construct new difference equations with this equation. It is seen below how previous studies have reached difference equations.

In Equation (1), the following is obtained: when $k=0$ and $j=0$, the equation $x_{n+1}={\displaystyle \frac{a{x}_{n-1}}{b+c{x}_n{x}_{n-1}}}$ studied in [1]; when $k=3,$ $j=2,$ $A=1,B=\pm 1$, and $C=\pm 1$, the equation $x_{n+1}={\displaystyle \frac{x_{n-11}}{\pm 1\pm x_{n-5}{x}_{n-11}}}$ studied in [2]; when $k=2,$ $j=0,$ $A=1,B=\pm 1$, and $C=-1$, the equation $x_{n+1}={\displaystyle \frac{x_{n-5}}{\pm 1-x_{n-2}{x}_{n-5}}}$ studied in [3]; when $k=2,$ $j=2,$ $A=11,B=\pm 1$, and $C=\pm 1$, the equation $x_{n+1}={\displaystyle \frac{x_{n-9}}{\pm 1\pm x_{n-4}{x}_{n-9}}}$ studied in [4]; when $k=1,$ $j=0,$ $A=1,B=\pm 1$, and $C=\pm 1$, the equation $x_{n+1}={\displaystyle \frac{x_{n-3}}{\pm 1-x_{n-1}{x}_{n-3}}}$ studied in [5]; when $k=0,$ $j=0,$ $A=1,B=a$, and $C=-1$, the equation $x_{n+1}={\displaystyle \frac{x_{n-1}}{a-x_{n-1}{x}_n}}$ studied in [6]; when $k=2,$ $j=0,$ $A=C,B=A$, and $C=B$, the equation $x_{n+1}={\displaystyle \frac{C{x}_{n-5}}{A+B{x}_{n-2}{x}_{n-5}}}$ studied in [7]; when $k=1$ and $j=0$, the equation $x_{n+1}={\displaystyle \frac{a{x}_{n-3}}{b+c{x}_{n-1}{x}_{n-3}}}$ studied in [8]; when $j=0,A=a,B=-a$, and $C=1$, the equation $x_{n+1}={\displaystyle \frac{a{x}_{n-(2k+1)}}{-a+x_{n-k}{x}_{n-(2k+1)}}}$ studied in [9]; when $k=2,$ $j=0,$ $A=1,B=1$, and $C=1$, the equation $x_{n+1}={\displaystyle \frac{x_{n-5}}{1+x_{n-2}{x}_{n-5}}}$ studied in [10]; when $j=1,$ $A=a,B=-a$, and $C=\mp 1$, the equation $x_{n+1}={\displaystyle \frac{a{x}_{n-(2k+3)}}{-a\mp x_{n-(k+1)}{x}_{n-(2k+3)}}}$ studied in [11]; when $k=0,$ $j=0,$ $A=a,B=-1$, and $C=b$, the equation $x_{n+1}={\displaystyle \frac{a{x}_{n-1}}{-1+b{x}_n{x}_{n-1}}}$ studied in [12]. In addition, with Equation (1), new difference equations similar to the difference equations in the above-mentioned studies can be obtained. Namely, when $j=4$ in Equation (1), the difference equation $x_{n+1}={\displaystyle \frac{A{x}_{n-(2k+7)}}{B+C{x}_{n-(k+3)}{x}_{n-(2k+7)}}}$ can be obtained.

In [1,6,12], solutions to second-order difference equations are presented. In this study, solutions of difference equations of order $2(k+j+1)$ are obtained. In [2,3,4,5,9,10,11], the investigation of solutions of lower-order difference equations of order $2(k+j+1)$ is conducted. The present study examines both the solutions of Equation (1) and the global behaviour of Equation (1). Furthermore, while [7,8] examine the dynamics of lower-order difference equations, this study focuses on the dynamics of a higher-order difference equation.

Studies on difference equations continue to increase day by day. The reason for this is that difference equations are used in modelling real-life problems in physics, biology, economics, statistics, etc. (see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26]).

2. Preliminaries

Let I be some interval of real numbers and let $f:I^{k+1}\to I$ be a continuously differentiable function. Then, for every set of initial conditions $x_{-k},x_{-(k-1)},...,x_0\in I,$ the difference equation

$$x_{n+1}=f\left(x_n,x_{n-1},...,x_{n-k}\right),n=0,1,...$$

(2)

has a unique solution ${\left\{x_n\right\}}_{n=-k}^{\infty }.$

Definition 1.

An equilibrium point for Equation (2) is a point $\overline{x}\in I$ such that $\overline{x}=f\left(\overline{x},\overline{x},...,\overline{x}\right).$

Definition 2.

A sequence ${\left\{x_n\right\}}_{n=-k}^{\infty }$ is said to be periodic with period p if $x_{n+p}=x_n$ for all $n\ge -k.$

Definition 3.

(i) 

The equilibrium point $\overline{x}$ of Equation (2) is locally stable if for given $\epsilon >0,$ there exists $\delta >0$ such that for all $x_{-k},x_{-(k-1)},...,x_0\in I$ with $\left|x_{-k}-\overline{x}\right|+\left|x_{-(k-1)}-\overline{x}\right|+...+\left|x_0-\overline{x}\right|<\delta ,$ we have $\left|x_n-\overline{x}\right|<\epsilon$ for all $n\ge -k.$

(ii) 

The equilibrium point $\overline{x}$ of Equation (2) is locally asymptotically stable if $\overline{x}$ is the locally stable solution of Equation (2) and there exists $\gamma >0,$ such that for all $x_{-k},x_{-(k-1)},...,x_0\in I$ with $\left|x_{-k}-\overline{x}\right|+\left|x_{-(k-1)}-\overline{x}\right|+...+\left|x_0-\overline{x}\right|<\gamma ,$ we have ${\displaystyle \underset{n\to \infty }{lim}}{x}_n=\overline{x}.$

(iii) 

The equilibrium point $\overline{x}$ of Equation (2) is a global attractor if for all

$x_{-k},x_{-(k-1)},...,x_0\in I,$ we have ${\displaystyle \underset{n\to \infty }{lim}}{x}_n=\overline{x}.$

(iv) 

The equilibrium point $\overline{x}$ of Equation (2) is globally asymptotically stable if $\overline{x}$ is locally stable, and $\overline{x}$ is also a global attractor of Equation (2).

(v) 

The equilibrium point $\overline{x}$ of Equation (2) is unstable if $\overline{x}$ is not locally stable.

The linearized equation associated with Equation (2) is

$$y_{n+1}=\sum _{i=0}^k{\displaystyle \frac{\partial f}{\partial x_{n-i}}}\left(\overline{x},\overline{x},...,\overline{x}\right)y_{n-i},n=0,1,...$$

(3)

The characteristic equation associated with Equation (3) is

$${\lambda }^{k+1}-\sum _{i=0}^k{\displaystyle \frac{\partial f}{\partial x_{n-i}}}\left(\overline{x},\overline{x},...,\overline{x}\right){\lambda }^{k-i}=0.$$

(4)

The following theorem which is commonly referred to as the Linear Stability Theorem in the academic literature is presented for consideration.

Theorem 1 ([13]).

Assume that f is a $C^1$ function and let $\overline{x}$ be an equilibrium point of Equation (2). Then, the following statements are true.

(i) 

If all roots of Equation (4) lie in open disk $\left|\lambda \right|<1,$ then $\overline{x}$ is locally asymptotically stable.

(ii) 

If at least one root of Equation (4) has an absolute value greater than one, then $\overline{x}$ is unstable.

3. Dynamics of Equation (1)

In this section, we investigate the dynamics of Equation (1) under the assumptions that all parameters are non-negative real numbers, the initial conditions are non-negative real numbers, and k is a non-negative integer.

The change in the variables $x_n=\sqrt{{\displaystyle \frac{B}{C}}}{y}_n$ reduces Equation (1) to the difference equation

$$y_{n+1}={\displaystyle \frac{\gamma y_{n-\left[2\left(k+j\right)+1\right]}}{1+y_{n-(k+j)}{y}_{n-\left[2\left(k+j\right)+1\right]}}},n=0,1,...$$

(5)

where $\gamma ={\displaystyle \frac{A}{B}}.$ We can see that ${\overline{y}}_1=0$ is always an equilibrium point of Equation (5). When $\gamma >1,$ Equation (5) also possesses the unique positive equilibrium ${\overline{y}}_2=\sqrt{\gamma -1}.$

Theorem 2.

The following statements are true:

(i) 

If $\gamma <1$, then the equilibrium point ${\overline{y}}_1=0$ of Equation (5) is locally asymptotically stable;

(ii) 

If $\gamma >1,$ then the equilibrium points ${\overline{y}}_1=0$ and ${\overline{y}}_2=\sqrt{\gamma -1}$ are unstable.

Proof. 

The linearized equation associated with Equation (5) about $\overline{y}$ is

$$z_{n+1}+{\displaystyle \frac{\gamma {\overline{y}}^2}{{\left(1+{\overline{y}}^2\right)}^2}}{z}_{n-(k+j)}-{\displaystyle \frac{\gamma }{{\left(1+{\overline{y}}^2\right)}^2}}{z}_{n-\left[2\left(k+j\right)+1\right]}=0,n=0,1,...$$

The characteristic equation associated with this equation is

$${\lambda }^{2\left(k+j+1\right)}+{\displaystyle \frac{\gamma {\overline{y}}^2}{1+{\overline{y}}^2}}{\lambda }^{k+j+1}-{\displaystyle \frac{\gamma }{{\left(1+{\overline{y}}^2\right)}^2}}=0.$$

Then, the linearized equation of Equation (5) about the equilibrium point ${\overline{y}}_1=0$ is

$$z_{n+1}-\gamma z_{n-\left[2\left(k+j\right)+1\right]}=0,n=0,1,...$$

The characteristic equation of Equation (5) about the equilibrium point ${\overline{y}}_1=0$ is

$${\lambda }^{2\left(k+j+1\right)}-\gamma =0.$$

So, $\lambda =\sqrt[2\left(k+j+1\right)]{\gamma }.$ The below information thus follows from the statement of Theorem 1:

If $\gamma <1,$ then $\left|\lambda \right|<1$ for all roots and the equilibrium point ${\overline{y}}_1=0$ is locally asymptotically stable.

If $\gamma >1,$ it follows that the equilibrium point ${\overline{y}}_1=0$ is unstable.

The linearized equation of Equation (5) about the equilibrium point ${\overline{y}}_2=\sqrt{\gamma -1}$ becomes

$$z_{n+1}+\left(1-{\displaystyle \frac{1}{\gamma }}\right)z_{n-(k+j)}-{\displaystyle \frac{1}{\gamma }}{z}_{n-\left[2\left(k+j\right)+1\right]}=0,n=0,1,...$$

The characteristic equation of Equation (5) about the equilibrium point ${\overline{y}}_2=\sqrt{\gamma -1}$ is

$${\lambda }^{2\left(k+j+1\right)}+\left(1-{\displaystyle \frac{1}{\gamma }}\right){\lambda }^{k+j+1}-{\displaystyle \frac{1}{\gamma }}=0.$$

It is clear that this equation has a root in the interval $\left(-\infty ,-1\right)$. Then, the equilibrium point ${\overline{y}}_2=\sqrt{\gamma -1}$ is unstable. □

Theorem 3.

Suppose that $\gamma <1,$ then the equilibrium point ${\overline{y}}_1=0$ of Equation (5) is globally asymptotically stable.

Proof. 

Let ${\left\{y_n\right\}}_{n=-\left[2\left(k+j\right)+1\right]}^{\infty }$ be a solution of Equation (5). As a result of Theorem 2, we know that the equilibrium point ${\overline{y}}_1=0$ of Equation (5) is locally asymptotically stable. So, it is sufficient to show that

$$\underset{n\to \infty }{lim}{y}_n=0.$$

Since

$$y_{n+1}={\displaystyle \frac{\gamma y_{n-\left[2\left(k+j\right)+1\right]}}{1+y_{n-(k+j)}{y}_{n-\left[2\left(k+j\right)+1\right]}}}\le \gamma y_{n-\left[2\left(k+j\right)+1\right]},$$

we obtain

$$y_{n+1}\le \gamma y_{n-\left[2\left(k+j\right)+1\right]}.$$

Then, it can be written for $l=0,1,...$ that

$$y_{2l\left(k+j+1\right)+i}\le {\gamma }^{l+1}{y}_{-\left[2\left(k+j+1\right)-i\right]},\left(i=\overline{1,2\left(k+j+1\right)}\right).$$

If $\gamma <1,$ then $\underset{l\to \infty }{lim}{\gamma }^{l+1}=0$

and

$$\underset{n\to \infty }{lim}{y}_n=0.$$

The proof is complete. □

Corollary 1.

Assume that $\gamma =1.$ Then, every solution of Equation (5) is bounded.

Proof. 

Let ${\left\{y_n\right\}}_{n=-\left[2\left(k+j\right)+1\right]}^{\infty }$ be a solution of Equation (5). It follows from Equation (5) that

$$y_{n+1}={\displaystyle \frac{\gamma y_{n-\left[2\left(k+j\right)+1\right]}}{1+y_{n-(k+j)}{y}_{n-\left[2\left(k+j\right)+1\right]}}}\le \gamma y_{n-\left[2\left(k+j\right)+1\right]}.$$

Then, in light of the proof of Theorem 3, we have for $l=0,1,...$

$$y_{2l\left(k+j+1\right)+1}\le y_{-\left[2\left(k+j\right)+1\right]},$$

$$y_{2l\left(k+j+1\right)+2}\le y_{-2(k+j)},$$

$$...$$

$$y_{2l\left(k+j+1\right)+2\left[\left(k+j\right)+1\right]}\le y_0.$$

So, every solution of Equation (5) is bounded from above by

$$A=max\left\{y_{-\left[2\left(k+j\right)+1\right]},y_{-2(k+j)},...,y_0,\right\}.$$

 □

3.1. The Difference Equation $x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{-A\mp x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}}$

When we take $B=-A$, $C=1$, and the + sign between B and C as ∓ in Equation (1), we obtain the equation

$$x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{-A\mp x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}},$$

(6)

where k is a positive integer and the initial conditions are non-zero real numbers with $x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}\ne A.$

Theorem 4.

Let ${\left\{y_n\right\}}_{n=-\left[2\left(k+j\right)+1\right]}^{\infty }$ be a solution of Equation (6). Then, for $n=0,1,...$ all solutions of Equation (6) are of the form

$$x_{2(k+j+1)n+i}=\left\{\begin{array}{c}{\displaystyle \frac{A^{n+1}{x}_{-\left[2\left(k+j+1\right)-i\right]}}{{\left[-A\mp x_{-(k+j+1-i)}{x}_{-\left[2\left(k+j+1\right)-i\right]}\right]}^{n+1}}},\left(i=\overline{1,k+j+1}\right)\hfill \\ {\displaystyle \frac{1}{A^{n+1}}}{x}_{-\left[2\left(k+j+1\right)-i\right]}{\left[-A\mp x_{-\left[2\left(k+j+1\right)-i\right]}{x}_{-\left[3\left(k+j+1\right)-i\right]}\right]}^{n+1}.\hfill \\ \left(i=\overline{k+j+2,2\left(k+j+1\right)}\right)\hfill \end{array}\right.$$

Proof. 

It is clear that for $n=0$, the equation is satisfied. Now, suppose that $n>0$ and our claim is true for $n-1$, then

$$x_{2(k+j+1)\left(n-1\right)+i}=\left\{\begin{array}{c}{\displaystyle \frac{A^n{x}_{-\left[2\left(k+j+1\right)-i\right]}}{{\left[-A\mp x_{-(k+j+1-i)}{x}_{-\left[2\left(k+j+1\right)-i\right]}\right]}^n}},\left(i=\overline{1,k+j+1}\right)\hfill \\ {\displaystyle \frac{1}{A^n}}{x}_{-\left[2\left(k+j+1\right)-i\right]}{\left[-A\mp x_{-\left[2\left(k+j+1\right)-i\right]}{x}_{-\left[3\left(k+j+1\right)-i\right]}\right]}^n.\hfill \\ \left(i=\overline{k+j+2,2\left(k+j+1\right)}\right)\hfill \end{array}\right.$$

It follows from Equation (6) and the above equalities that for $i=1,$

$$\begin{array}{c}{x}_{2(k+j+1)\left(n-1\right)+1}={\displaystyle \frac{A{x}_{2\left(k+j+1\right)n-\left[2\left(k+j\right)+1\right]}}{-A\mp x_{2(k+j+1)n-\left(k+j\right)}{x}_{2\left(k+j+1\right)n-\left[2\left(k+j\right)+1\right]}}}\hfill \\ ={\displaystyle \frac{A\frac{A^n{x}_{-\left[2\left(k+j\right)+1\right]}}{{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^n}}{-A\mp \frac{1}{A^n}{x}_{-(k+j)}{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^n\frac{A^n{x}_{-\left[2\left(k+j\right)+1\right]}}{{\left[-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right]}^n}}}\hfill \\ ={\displaystyle \frac{\frac{A^{n+1}{x}_{-\left[2\left(k+j\right)+1\right]}}{{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^n}}{-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}}}={\displaystyle \frac{A^{n+1}{x}_{-\left[2\left(k+j\right)+1\right]}}{{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^{n+1}}}.\hfill \end{array}$$

Hence, we have

$$x_{2(k+j+1)n+1}={\displaystyle \frac{A^{n+1}{x}_{-\left[2\left(k+j\right)+1\right]}}{{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^{n+1}}}.$$

Similarly, one can obtain $i=2,3,...,k+j+1.$

Now, we are going to show our claim for $i=k+j+2$. We obtain from Equation (6) that

$$x_{2(k+j+1)n+k+j+2}={\displaystyle \frac{A{x}_{2\left(k+j+1\right)n-\left(k+j\right)}}{-A\mp x_{2(k+j+1)n+1}{x}_{2\left(k+j+1\right)n-\left(k+j\right)}}}.$$

We have the below equalities

$$\begin{array}{c}{x}_{2(k+j+1)n+k+j+2}={\displaystyle \frac{A\frac{1}{A^n}{x}_{-\left(k+j\right)}{\left[-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right]}^n}{-A\mp \frac{A^{n+1}{x}_{-\left[2\left(k+j\right)+1\right]}}{{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^{n+1}}\frac{1}{A^n}{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^n}}\hfill \\ ={\displaystyle \frac{\frac{A{x}_{-\left(k+j\right)}{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^n}{A^n}}{-A\mp \frac{A{x}_{-\left(k+j\right)}{x}_{-\left[2\left(k+j\right)+1\right]}}{-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}}}}={\displaystyle \frac{A{x}_{-\left(k+j\right)}{\left(-A\mp x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}\right)}^{n+1}}{A^{n+2}}}.\hfill \end{array}$$

That is,

$$x_{2(k+j+1)n+k+j+2}={\displaystyle \frac{1}{A^{n+1}}}{x}_{-\left(k+j\right)}{\left[-A\mp x_{-\left(k+j\right)}{x}_{-\left[2\left(k+j\right)+1\right]}\right]}^{n+1}.$$

Similarly, one can obtain the other cases. Thus, the proof is complete. □

Corollary 2.

Assume that $x_0{x}_{-\left(k+j+1\right)}=x_{-1}{x}_{-\left(k+j+2\right)}=...=x_{-\left(k+j\right)}{x}_{-\left[2\left(k+j\right)+1\right]}=2A.$ Then, every solution of Equation (6) is periodic with period $2\left(k+j+1\right).$

Proof. 

In consideration of the solutions presented in Theorem 4 and the aforementioned assumption, the following can be written:

$$x_{2(k+j+1)n+1}=x_{-\left[2(k+j)+1\right]},x_{2(k+j+1)n+2}=x_{-2(k+j)},...,x_{2(k+j+1)n+2(k+j+1)}=x_0.$$

It is obvious that every solution of Equation (6) is periodic with period $2\left(k+j+1\right).$ The proof is complete. □

Corollary 3.

Let ${\left\{y_n\right\}}_{n=-\left[2\left(k+j\right)+1\right]}^{\infty }$ be a solution of the equation $x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{-A+x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}}.$ Assume that $x_{-\left[2\left(k+j\right)+1\right]},x_{-2\left(k+j\right)},...,x_0>0$ and $x_0{x}_{-\left(k+j+1\right)}>2A,$ $x_{-1}{x}_{-\left(k+j+2\right)}>2A,...,$ $x_{-(k+j)}{x}_{-\left[2\left(k+j\right)+1\right]}>2A.$ Then, ${\displaystyle \underset{n\to \infty }{lim}}{x}_{2(k+j+1)n+i}=0$ $(i=1,2,...,k+j+1)$ and ${\displaystyle \underset{n\to \infty }{lim}}{x}_{2(k+j+1)n+i}=0$ for $(i=\overline{k+j+2,2\left(k+j\right)+1}).$

Proof. 

This is evident from the proof presented in Theorem 4. □

3.2. The Difference Equation $x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{A\mp x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}}$

In this section, we obtain a form of the solutions of the equation

$$x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{A\mp x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}}$$

(7)

where k is a positive integer and the initial conditions are non-zero real numbers with $x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}\ne A.$

Theorem 5.

Let ${\left\{y_n\right\}}_{n=-\left[2\left(k+j\right)+1\right]}^{\infty }$ be a solution of Equation (7). Then, for $n=0,1,...$ all solutions of Equation (7) are of the form

$$x_{2(k+j+1)n+i}=\left\{\begin{array}{c}{\displaystyle \frac{A{x}_{-\left[2\left(k+j+1\right)-i\right]}{\displaystyle \prod _{p=1}^n}\left[A\mp 2p{x}_{-\left(k+j+1-i\right)}{x}_{-\left[2\left(k+j+1\right)-i\right]}\right]}{\left(A\mp x_{-\left(k+j+1-i\right)}{x}_{-\left[2\left(k+j+1\right)-i\right]}\right){\displaystyle \prod _{p=1}^n}\left[A\mp \left(2p+1\right)x_{-\left(k+j+1-i\right)}{x}_{-\left[2\left(k+j+1\right)-i\right]}\right]}},\left(i=\overline{1,k+j+1}\right)\hfill \\ {\displaystyle \frac{x_{-\left[2\left(k+j+1\right)-i\right]}\left[-A\mp x_{-\left[2\left(k+j+1\right)-i\right]}{x}_{-\left[3\left(k+j+1\right)-i\right]}\right]{\displaystyle \prod _{p=1}^n}\left[A\mp \left(2p+1\right)x_{-\left[2\left(k+j+1\right)-i\right]}{x}_{-\left[3\left(k+j+1\right)-i\right]}\right]}{\left[A\mp 2x_{-\left[2\left(k+j+1\right)-i\right]}{x}_{-\left[3\left(k+j+1\right)-i\right]}\right]{\displaystyle \prod _{p=1}^n}\left[A\mp \left(2p+2\right)x_{-\left[2\left(k+j+1\right)-i\right]}{x}_{-\left[3\left(k+j+1\right)-i\right]}\right]}}.\hfill \\ \left(i=\overline{k+j+2,2\left(k+j+1\right)}\right)\hfill \end{array}\right.$$

Proof. 

Similarly, someone can prove as the proof of Theorem 4. □

Corollary 4.

Assume that $x_{-(k+j)}=x_{-\left(k+j-1\right)}=...=x_0=0$ or $x_{-\left[2\left(k+j\right)+1\right]}=x_{-2(k+j)}=...=x_{-(k+j+1)}=0.$ Then, every solution of Equation (7) is periodic with period $2\left(k+j+1\right).$

Proof. 

Firstly, let $x_{-(k+j)}=x_{-\left(k+j-1\right)}=...=x_0=0.$ In view of Theorem 5 and from our assumption, we obtain $x_{2(k+j+1)n+1}=x_{-\left[2(k+j)+1\right]},x_{2(k+j+1)n+2}=x_{-2(k+j)},...,x_{2(k+j+1)n+k+j+1}=x_{-\left(k+j+1\right)}$ and $x_{2(k+j+1)n+k+j+2}=x_{2(k+j+1)n+k+j+3}=...=x_{2(k+j+1)n+2(k+j+1)}=0.$

Secondly, $x_{-\left[2\left(k+j\right)+1\right]}=x_{-2(k+j)}=...=x_{-(k+j+1)}=0.$ Then, we obtain $x_{2(k+j+1)n+1}=x_{2(k+j+1)n+2}=...=x_{2(k+j+1)n+k+j+1}=0$ and $x_{2(k+j+1)n+k+j+2}=x_{-(k+j)},x_{2(k+j+1)n+k+j+3}=x_{-\left(k+j-1\right)},...,x_{2(k+j+1)n+2(k+j+1)}=x_0.$

Thus, every solution of Equation (7) is periodic with period $2\left(k+j+1\right).$ □

4. Numerical Results

In this section, we give a few numerical results for some special values of the parameters.

Example 1.

In Figure 1, we take Equation (5) with $\gamma =0.5,k=1,j=1,y_{-5}=4,y_{-4}=2,$ $y_{-3}=1.5,y_{-2}=1,y_{-1}=3$, and $y_0=5$.

Figure 1. The figure shows that the equilibrium ${\overline{y}}_1=0$ of Equation (5) is globally asymptotically stable.

Example 2.

In Figure 2, we take Equation (5) with $\gamma =1.5,k=1,j=1,y_{-5}=2,y_{-4}=3,$ $y_{-3}=2.5,y_{-2}=1,y_{-1}=4$, and $y_0=3$.

Figure 2. The figure shows that the equilibrium ${\overline{y}}_1=0$ of Equation (5) is unstable.

Example 3.

In Figure 3, we take Equation (5) with $\gamma =1,k=1,j=1,y_{-5}=1,y_{-4}=3,y_{-3}=2,$ $y_{-2}=1,y_{-1}=1.5$, and $y_0=4$.

Figure 3. The figure shows that the solution of Equation (5) is bounded.

Example 4.

In Figure 4, we take Equation (6) with $A=2,k=1,j=1,x_{-5}=1,x_{-4}=2,x_{-3}=4,$ $x_{-2}=4,x_{-1}=2$, and $x_0=1$.

Figure 4. The figure shows that the solution of Equation (6) is periodic with the period 6.

Example 5.

In Figure 5, we take Equation (7) with $A=2,k=1,j=1,x_{-5}=5,x_{-4}=3,x_{-3}=2$, and $x_{-2}=x_{-1}=x_0=0$.

Figure 5. The figure shows that the solution of Equation (7) is periodic with the period 6.

Example 6.

In Figure 6, we take Equation (7) with $A=2,k=1,j=1,x_{-5}=x_{-4}=x_{-3}=0$, $x_{-2}=3,x_{-1}=2$, and $x_0=1.5$.

Figure 6. The figure shows that the solution of Equation (7) is periodic with the period 6.

5. Conclusions

We defined the generalised difference equation $x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{B+C{x}_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}}$. By the change in the variables $x_n=\sqrt{{\displaystyle \frac{B}{C}}}{y}_n$, we obtained $y_{n+1}={\displaystyle \frac{\gamma y_{n-\left[2\left(k+j\right)+1\right]}}{1+y_{n-(k+j)}{y}_{n-\left[2\left(k+j\right)+1\right]}}}$ with $\gamma ={\displaystyle \frac{A}{B}}$. We showed that the equilibrium point ${\overline{y}}_1=0$ was globally asymptotically stable when $\gamma <1$, and when $\gamma >1$, the equilibrium points ${\overline{y}}_1=0$ and ${\overline{y}}_2=\sqrt{\gamma -1}$ were unstable. If $\gamma =1$, every solution was bounded. Additionally, we investigated the special case of Equation (1) as $x_{n+1}={\displaystyle \frac{A{x}_{n-\left[2\left(k+j\right)+1\right]}}{-A\mp x_{n-(k+j)}{x}_{n-\left[2\left(k+j\right)+1\right]}}}.$ We gave the explicit solutions of this equation. One may define more general forms of other difference equations and investigate their behaviour.