On the General Solution of $x_{n+1}={\displaystyle \frac{a{x}_{n+1-2k}}{b+c{x}_{n+1-k}{x}_{n+1-2k}}}$

Abstract

This paper investigates the global dynamics of a broad class of nonlinear rational difference equations given by $x_{n+1}={\displaystyle \frac{a{x}_{n+1-2k}}{b+c{x}_{n+1-k}{x}_{n+1-2k}}},n=0,1,\dots ,$ which generalizes several known models in the literature. We establish the existence of exactly three equilibrium points and show that the trivial equilibrium is globally asymptotically stable when the parameter ratio $\alpha =(b/a)$ lies in $(-1,1)$. The nontrivial equilibria are shown to be always unstable. An explicit general solution is derived, enabling a detailed analysis of solution behavior in terms of initial conditions and parameters. Furthermore, we identify and classify minimal period $2k$ and $4k$ solutions, providing necessary and sufficient conditions for the occurrence of constant and periodic behaviors. These analytical results are supported by numerical simulations, confirming the theoretical predictions. The findings generalize and refine existing results by offering a unified framework for analyzing a wide class of rational difference equations.

1. Introduction

Difference equations have garnered significant attention in recent years due to their widespread application in modeling real-world problems across various scientific fields. In biology, for instance, they are employed to describe population dynamics, blood cell production, and the growth of annual plants. In economics, they are used to model commodity pricing and national income [1,2,3,4]. The class of rational difference equations studied in this paper also has potential applications in these areas as well as in engineering (see, for example, ref. [5] and the references therein), where global stability and periodicity results ensure predictable and reliable system behavior regardless of initial conditions. This motivates a detailed investigation of their dynamic behavior, including equilibrium points, stability, and periodic solutions.

Numerous researchers have investigated rational difference equations of various forms.

Cınar [6] studied the positive solutions of the rational difference equation

$$x_{n+1}={\displaystyle \frac{x_{n-1}}{1+x_n{x}_{n-1}}},n=0,1,\dots$$

In [7], he analyzed the solutions of the equation

$$x_{n+1}={\displaystyle \frac{x_{n-1}}{-1+a{x}_n{x}_{n-1}}},a>0,$$

and in [8], he examined the positive solutions of the equation

$$x_{n+1}={\displaystyle \frac{x_{n-1}}{1+a{x}_n{x}_{n-1}}},a>0.$$

Aloqeili [9] studied the dynamics, including stability and semi-cyclic behavior, of the solutions of the equation

$$x_{n+1}={\displaystyle \frac{x_{n-1}}{a-x_n{x}_{n-1}}},a>0.$$

Andruch-Sobi and Migda [10] investigated the asymptotic properties of the rational difference equation

$$x_{n+1}={\displaystyle \frac{a{x}_{n-1}}{b+c{x}_n{x}_{n-1}}},a,c>0,b<0,$$

with nonnegative initial values. The case with $b>0$ was later addressed in [11].

Elsayed [12] studied the solutions of the equation

$$x_{n+1}={\displaystyle \frac{x_{n-3}}{\pm 1\pm x_{n-1}{x}_{n-3}}}.$$

Abo-Zeid [13] analyzed the global behavior of the rational difference equation

$$x_{n+1}={\displaystyle \frac{a{x}_{n-3}}{b+c{x}_{n-1}{x}_{n-3}}}.$$

Ghazel et al. [14] examined the dynamics and behavior of the equation

$$x_{n+1}={\displaystyle \frac{C{x}_{n-5}}{A+B{x}_{n-2}{x}_{n-5}}},$$

while Karatas et al. [15] studied the positive solutions of

$$x_{n+1}={\displaystyle \frac{x_{n-5}}{1+x_{n-2}{x}_{n-5}}}.$$

Karatas [16] derived the general solution of the rational difference equation

$$x_{n+1}={\displaystyle \frac{a{x}_{n-(2k+3)}}{-a\mp x_{n-(k+1)}{x}_{n-(2k+3)}}}.$$

Karatas et al. [17] investigated the global stability of the equation

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

and provided solutions to special cases using the iteration method.

Al-Hdaibat [18] studied the general solution and dynamics of the equation

$$x_{n+1}={\displaystyle \frac{a{x}_{n-1}}{b+c{x}_n{x}_{n-1}}},n=0,1,\dots$$

with real parameters a, b, and c satisfying $bc>0$, and arbitrary real initial values $x_{-1}$ and $x_0$. The study also includes a bifurcation analysis of the equation.

In this paper, we study a class of rational difference equations of the form

$$x_{n+1}={\displaystyle \frac{a{x}_{n+1-2k}}{b+c{x}_{n+1-k}{x}_{n+1-2k}}},n=0,1,\dots .$$

(1)

This class include all equations of the form:

$$x_{n+1}={\displaystyle \frac{a{x}_{n-1}}{b+c{x}_n{x}_{n-1}}},x_{n+1}={\displaystyle \frac{a{x}_{n-3}}{b+c{x}_{n-1}{x}_{n-3}}},x_{n+1}={\displaystyle \frac{a{x}_{n-5}}{b+c{x}_{n-2}{x}_{n-5}}},x_{n+1}={\displaystyle \frac{a{x}_{n-7}}{b+c{x}_{n-3}{x}_{n-7}}},\dots .$$

This paper is motivated by the works in [12,14,16,17,18]. We extend their results by investigating a more general form of the class of equations given in Equation (1), where a, b, and c are real parameters satisfying $bc>0$, and the initial values are real.

The structure of this paper is organized as follows. In Section 2, we show that Equation (1) has exactly three equilibrium points. The trivial equilibrium point ${\tilde{x}}_1$ is globally asymptotically stable, while the equilibrium points ${\tilde{x}}_2$ and ${\tilde{x}}_3$ are never linearly stable. In Section 3, we derive the analytical solution of Equation (1). We prove that if $(b/a)\in (-1,1)$, then every solution of Equation (1) converges to zero, even when the initial conditions are negative. Furthermore, we discuss the dynamic behavior of the solutions, including the existence of periodic solutions. In Section 4, we present numerical simulations to support and validate the theoretical results obtained.

2. Preliminaries

Here we present some known results that will be useful in the study of Equation (1), see, for example, [1,19]. Let $I\subseteq \mathbb{R}$ and let $f:I^{2k-1}⟶I$ be a continuously differentiable function. Then for any initial conditions $x_{-2k+1},x_{-2k+2},\dots ,x_0\in I$, the difference equation

$$x_{n+1}=f(x_n,x_{n-1},\dots ,x_{n+1-2k}),n=0,1,\dots .$$

(2)

has a unique solution ${\left\{x_n\right\}}_{n=-2k+1}^{\infty }$.

Definition 1.

A point $\tilde{x}$ is called an equilibrium point of Equation (2) if $f(\tilde{x},\tilde{x},\dots ,\tilde{x})=\tilde{x}$.

Definition 2.

A solution ${\left\{x_n\right\}}_{n=-2k+1}^{\infty }$ is said to be periodic with period t if

$$x_{n+t}=x_n\mathit{for}\mathit{all}n\ge -2k+1.$$

(3)

A solution ${\left\{x_n\right\}}_{n=-2k+1}^{\infty }$ is called periodic with prime period t if t is the smallest positive integer for which Equation (3) holds.

Definition 3.

Let $\tilde{x}$ be an equilibrium point of Equation (2).

1. 

$\tilde{x}$ is called stable if for every $\epsilon >0$, there exists $\delta >0$ such that for all $x_{-2k+1},x_{-k+1},\dots ,x_0\in I$ and ${\displaystyle \sum _{j=1}^{2k}}\left|x_{-2k+j}-\tilde{x}\right|<\delta$, we have $\left|x_n-\tilde{x}\right|<\epsilon$ for all $n\ge -2k+1$.

2. 

$\tilde{x}$ is called locally asymptotically stable if $\tilde{x}$ is stable and there exists $\gamma >0$, such that for all $x_{-2k+1},x_{-2k+2},\dots ,x_0\in I$ and ${\displaystyle \sum _{j=1}^{2k}}\left|x_{-2k+j}-\tilde{x}\right|<\gamma$, we have $\underset{n\to \infty }{lim}{x}_n=\tilde{x}$.

3. 

$\tilde{x}$ is called a global attractor if for all $x_{-2k+1},x_{-2k+2},\dots ,x_0\in I$, we have $\underset{n\to \infty }{lim}{x}_n=\tilde{x}$.

4. 

$\tilde{x}$ is called globally asymptotically stable if $\tilde{x}$ is stable and is a global attractor.

5. 

$\tilde{x}$ is called unstable if $\tilde{x}$ is not stable.

Let

$$y_{n+1}=\sum _{j=0}^{2k-1}{\rho }_j{y}_{n-j},n=0,1,\dots$$

(4)

where

$${\rho }_j={\displaystyle \frac{\partial f(\tilde{x},\tilde{x},\dots ,\tilde{x})}{\partial x_{n-j}}},$$

the function f is given in Equation (2), and $\tilde{x}$ is an equilibrium of Equation (2). Equation (4) is the linearized equation of Equation (2) about $\tilde{x}$. The characteristic equation of Equation (4) is

$${\lambda }^{2k}-\sum _{j=0}^{2k-1}{\rho }_j{\lambda }^{2k-1-j}=0.$$

(5)

Theorem 1.

Assume that f is a continuously differentiable function and let $\tilde{x}$ be an equilibrium point of Equation (2). Then, the following statements are true:

1. 

$\tilde{x}$ is locally asymptotically stable if all roots of Equation (5) (i.e., the eigenvalues) have absolute value less than 1.

2. 

$\tilde{x}$ is unstable if at least one root of Equation (5) has an absolute value greater than 1.

The change of variables $x_n\mapsto \sqrt{b/c}{x}_n$ (with $bc>0$) reduces Equation (1) to the rational difference equation

$$x_{n+1}={\displaystyle \frac{\alpha x_{n+1-2k}}{1+x_{n+1-k}{x}_{n+1-2k}}},n=0,1,2,\dots ,$$

(6)

where $\alpha =a/b$.

Theorem 2.

Equation (6) has exactly three equilibrium points, which are given by

$${\tilde{x}}_1=0\mathit{when}\alpha \in \mathbb{R},{\tilde{x}}_2=\sqrt{\alpha -1}\mathit{when}\alpha >1,{\tilde{x}}_3=\sqrt{\alpha -1}\mathit{when}\alpha >1.$$

Proof. 

For the equilibrium points of Equation (6), we can write $\tilde{x}=\left(\alpha \tilde{x}\right)/(1+\tilde{x}\tilde{x})$. Then we have $\tilde{x}\left[{\tilde{x}}^2-(\alpha -1)\right]=0$. Therefore, the equilibrium points of Equation (6) are ${\tilde{x}}_1=0$ when $\alpha \in \mathbb{R}$ and ${\tilde{x}}_{2,3}=\pm \sqrt{\alpha -1}$ when $\alpha >1$. □

The linearized equation associated with Equation (6) about the equilibrium point $\tilde{x}$ is given by the linear difference equation:

$$y_{n+1}={\rho }_{2k-1}{y}_{n+1-2k}+{\rho }_{k-1}{y}_{n+1-k}={\displaystyle \frac{\alpha }{{\left(1+{\tilde{x}}^2\right)}^2}}{y}_{n+1-2k}-{\displaystyle \frac{\alpha {\tilde{x}}^2}{{\left(1+{\tilde{x}}^2\right)}^2}}{y}_{n+1-k}.$$

(7)

The characteristic equation corresponding to Equation (7) is

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

(8)

The following corollary is a direct consequence of Theorem 1.

Corollary 1.

If $\alpha \in (-1,1)$, then the equilibrium point ${\tilde{x}}_1$ of Equation (6) is locally asymptotically stable.

Note that for the equilibrium points ${\tilde{x}}_{2,3}$, the characteristic Equation (8) can be rewritten as a quadratic in $\eta ={\lambda }^k$:

$${\eta }^2+{\displaystyle \frac{\alpha {\tilde{x}}^2}{{(1+{\tilde{x}}^2)}^2}}\eta -{\displaystyle \frac{\alpha }{{(1+{\tilde{x}}^2)}^2}}=0.$$

Solving this equation gives two roots ${\eta }_1$ and ${\eta }_2$, and the eigenvalues $\lambda$ are obtained as the $k^{\mathrm{th}}$ roots of these values. For all $\alpha >1$, one of these roots satisfies $|{\lambda }_1|=1$, while the other satisfies $|{\lambda }_2|<1$. Since at least one eigenvalue always lies on the unit circle, the equilibrium point is never locally asymptotically stable.

3. Main Results

In the following, we derive an analytical expression for the general solution ${\left\{x_n\right\}}_{n=1}^{\infty }$ of Equation (6) with arbitrary initial conditions ${\left\{x_{-2k+i}\right\}}_{i=1}^{2k}$.

Let $k\in \mathbb{N}$ be fixed. For each $p=k,k-1,\dots ,1$, define

$$y_{m-1}^{k,p}=x_{(m-1)k-p+1},m=0,1,2,\dots .$$

(9)

Assume that $n=(m+1)k-p$. Then, from Equation (6), we obtain

$$x_{(m+1)k-p+1}={\displaystyle \frac{\alpha x_{(m+1)k-p+1-2k}}{1+x_{(m+1)k-p+1-k}·x_{(m+1)k-p+1-2k}}}={\displaystyle \frac{\alpha x_{(m-1)k-p+1}}{1+x_{mk-p+1}·x_{(m-1)k-p+1}}}.$$

Therefore,

$$y_{m+1}^{k,p}={\displaystyle \frac{\alpha y_{m-1}^{k,p}}{1+y_m^{k,p}{y}_{m-1}^{k,p}}},m=0,1,2,\dots ,$$

(10)

with arbitrary initial conditions $y_{-1}^{k,p}$ and $y_0^{k,p}$. Hence, the $(2k+1)$-th-order Equation (6) can be decomposed into k independent second-order subequations given by Equation (10). The explicit general solution of Equation (10) is obtained in the following theorem.

Theorem 3.

For each $p=k,k-1,\dots ,1$, let $y_{-1}^{k,p},y_0^{k,p}\in \mathbb{R}$, and let ${\left\{y_m^{k,p}\right\}}_{m=1}^{\infty }$ be the solution of Equation (10). Then, for $m=1,2,\dots$, all solutions of Equation (10) are given by

$$y_m^{k,p}=\left\{\begin{array}{cc}{\displaystyle \frac{{\alpha }^{m/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\frac{m}{2}-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\frac{m}{2}-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}\hfill & \mathit{if}m\mathit{is}\mathit{even},\hfill \\ {\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{\frac{m+1}{2}-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\frac{m+1}{2}-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}\hfill & \mathit{if}m\mathit{is}\mathit{odd}.\hfill \end{array}\right.$$

(11)

Proof. 

Assume that m is even. Therefore, $m-1$ and $m+1$ are odd. Substituting Equation (11) into the L.H.S of Equation (10) gives

$$y_{m+1}^{k,p}={\displaystyle \frac{{\alpha }^{m/2+1}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{m/2}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{m/2}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}.$$

(12)

On the other hand, substituting Equation (11) into the R.H.S of Equation (10) gives

$$\begin{array}{cc}{\displaystyle \frac{\alpha y_{m-1}^{k,p}}{1+y_m^{k,p}{y}_{-1}^{k,p}}}\hfill & ={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{m/2+1}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m{\displaystyle \frac{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}{\displaystyle \frac{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{m/2+1}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m{\displaystyle \frac{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}}}={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{m/2+1}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m{\displaystyle \frac{1}{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l\right)}}}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{m/2+1}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}}{{\displaystyle \frac{1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m}{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l\right)}}}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\alpha }^{m/2+1}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{(m/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}{\displaystyle \frac{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l\right)}{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^m}{\alpha }^l\right)}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\alpha }^{m/2+1}{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{m/2}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{m/2}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}=y_{m+1}^{k,p}.\hfill \end{array}$$

Similarly, if we assume that m is odd, then $m-1$ and $m+1$ are even. If we substitute Equation (11) into the L.H.S of Equation (10), we obtain

$$y_{m+1}^{k,p}={\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\left(\right(m+1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m+1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}.$$

(13)

From the R.H.S of Equation (10), it follows that

$$\begin{array}{cc}{\displaystyle \frac{\alpha y_{m-1}^{k,p}}{1+y_m^{k,p}{y}_{-1}^{k,p}}}\hfill & ={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m{\displaystyle \frac{{\displaystyle \prod _{j=0}^{\left(\right(m+1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m+1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}{\displaystyle \frac{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m{\displaystyle \frac{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m+1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m{\displaystyle \frac{1}{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l\right)}}}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}}{{\displaystyle \frac{1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l+y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^m}{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l\right)}}}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m-1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}{\displaystyle \frac{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{m-1}}{\alpha }^l\right)}{\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^m}{\alpha }^l\right)}}\hfill \\ \hfill \\ & ={\displaystyle \frac{{\alpha }^{(m+1)/2}{y}_0^{k,p}{\displaystyle \prod _{j=0}^{\left(\right(m+1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{\left(\right(m+1)/2)-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}=y_{m+1}^{k,p}.\hfill \end{array}$$

This completes the proof. □

By applying the change of variables (9), the original $(2k+1)$-th-order difference Equation (6) is decomposed into k independent second-order subequations given in Equation (10). This decomposition allows the general solution of each subequation to be determined explicitly through the application of Theorem 3. The general solution of the original Equation (6) is then obtained by combining the resulting k independent subsequences, corresponding to $p=k,k-1,\dots ,1$, thereby reconstructing the original Equation (6).

The following theorems summarize the dynamic and rich periodic behaviors exhibited by Equation (6).

Theorem 4.

Let $\alpha \in (-1,1)$. Then every solution $\left\{x_n\right\}$ of Equation (6) converges to zero.

Proof. 

Since Equation (6) can be decomposed into k independent subequations given by Equation (10), it is sufficient to show that the solution of Equation (10) converges to zero for $\alpha \in (-1,1)$ in order to guarantee that $x_n\to 0$ for $\alpha \in (-1,1)$.

Let $\left\{y_m^{k,p}\right\}$ be a solution of Equation (10) with the initial conditions $y_{-1}^{k,p}$ and $y_0^{k,p}$ being real numbers. If $y_{-1}^{k,p}=y_0^{k,p}=0$, then Equation (11) gives $y_m^{k,p}=0$ for all $m\ge 1$. If $y_0^{k,p}=0$ and $y_{-1}^{k,p}\ne 0$, then, from Equation (11), we can easily obtain that

$$y_m^{k,p}=\left\{\begin{array}{ccc}0\hfill & \mathrm{if}\hfill & m\mathrm{is}\mathrm{even},\hfill \\ y_{-1}^{k,p}{\alpha }^{(m+1)/2}\hfill & \mathrm{if}\hfill & m\mathrm{is}\mathrm{odd}.\hfill \end{array}\right.$$

Since $\alpha \in (-1,1)$, it is clear that $\underset{m\to \infty }{lim}{y}_m^{k,p}=0$. Similarly, if $y_{-1}^{k,p}=0$ and $y_0^{k,p}\ne 0$, then Equation (11) yields

$$y_m^{k,p}=\left\{\begin{array}{ccc}{y}_0^{k,p}{\alpha }^{m/2}\hfill & \mathrm{if}\hfill & m\mathrm{is}\mathrm{even},\hfill \\ 0\hfill & \mathrm{if}\hfill & m\mathrm{is}\mathrm{odd}.\hfill \end{array}\right.$$

Hence, $\underset{m\to \infty }{lim}{y}_m^{k,p}=0$ for all $\alpha \in (-1,1)$. If $y_{-1}^{k,p}\ne 0$ and $y_0^{k,p}\ne 0$, then let $y_{-1}^{k,p}{y}_0^{k,p}\ne 0$. It is enough to show that the subsequences $\left\{y_{2m}^{k,p}\right\}$ and $\left\{y_{2m-1}^{k,p}\right\}$ converge to 0 as $m\to \infty$. From Equation (11), we obtain

$$\begin{array}{cc}|y_{2m}^{k,p}|\hfill & =\left|y_0^{k,p}{\alpha }^m{\displaystyle \frac{{\displaystyle \prod _{j=0}^{m-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{m-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j+1}}{\alpha }^l\right)}}\right|\hfill \\ & =\left|y_0^{k,p}{\alpha }^m\right|exp\left({\displaystyle \sum _{j=0}^{m-1}}ln\left({\displaystyle \frac{1}{1+r\left(j\right)}}\right)\right),\hfill \end{array}$$

where

$$r\left(j\right)={\displaystyle \frac{y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^{2j+1}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l}}.$$

For $\alpha \in (-1,1)$, we can write

$$r\left(j\right)={\displaystyle \frac{y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^{2j+1}(\alpha -1)}{y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^{2j+1}-y_0^{k,p}{y}_{-1}^{k,p}+\alpha -1}},$$

and hence $\underset{j\to \infty }{lim}\left|{\displaystyle \frac{r(j+1)}{r\left(j\right)}}\right|=\left|{\alpha }^2\right|<1$. Thus, the series ${\displaystyle \sum _{j=0}^{\infty }}r\left(j\right)$ is convergent. Therefore,

$$|y_0^{k,p}{\alpha }^m|exp(-2m)exp\left(-\sum _{j=0}^{m-1}r\left(j\right)\right)\to 0\mathrm{as}m\to \infty ,$$

so $\left|y_{2m}^{k,p}\right|\to 0$ as $m\to \infty$, and hence $\left\{y_{2m}^{k,p}\right\}\to 0$.

Similarly, we obtain

$$\begin{array}{cc}|y_{2m-1}^{k,p}|\hfill & =\left|y_{-1}^{k,p}{\alpha }^m\right|{\displaystyle \frac{{\displaystyle {\displaystyle \prod _{j=0}^{m-1}}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l\right)}{{\displaystyle \prod _{j=0}^{m-1}}\left(1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j}}{\alpha }^l\right)}}\hfill \\ & =|y_0^{k,p}{\alpha }^m|exp\left({\displaystyle \sum _{j=0}^{m-1}}ln\left({\displaystyle \frac{1}{1+s\left(j\right)}}\right)\right),\hfill \end{array}$$

where

$$s\left(j\right)={\displaystyle \frac{y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^{2j}}{1+y_0^{k,p}{y}_{-1}^{k,p}{\displaystyle \sum _{l=0}^{2j-1}}{\alpha }^l}}.$$

For $\alpha \in (-1,1)$, we have

$$s\left(j\right)={\displaystyle \frac{y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^{2j}(\alpha -1)}{y_0^{k,p}{y}_{-1}^{k,p}{\alpha }^{2j}-y_0^{k,p}{y}_{-1}^{k,p}+\alpha -1}},$$

so that $\underset{j\to \infty }{lim}\left|{\displaystyle \frac{s(j+1)}{s\left(j\right)}}\right|=\left|{\alpha }^2\right|<1$. Hence, the series ${\displaystyle {\displaystyle \sum _{j=0}^{\infty }}}s\left(j\right)$ is convergent. Therefore,

$$|y_0^{k,p}{\alpha }^m|exp(-2m)exp\left(-{\displaystyle \sum _{j=0}^{m-1}}s\left(j\right)\right)\to 0\mathrm{as}m\to \infty ,$$

so $\left\{y_{2m-1}^{k,p}\right\}\to 0$ as $m\to \infty$.

This completes the proof. □

Theorem 5.

If $\alpha \in (-1,1)$, then the equilibrium point ${\tilde{x}}_1$ of Equation (6) is globally asymptotically stable.

Proof. 

It follows from Theorem 4 and Corollary 1. □

Theorem 6.

Let $\xi \in \mathbb{R}$ and $\alpha ={\xi }^2+1$. Suppose the initial conditions of Equation (6) satisfy

$$x_{-2k+1}=x_{-2k+2}=\cdots =x_0=\xi .$$

(14)

Then the solution $\left\{x_n\right\}$ of Equation (6) is equal to ξ.

Proof. 

Since the sequence $\left\{x_n\right\}$ can be decomposed into k independent subsequences $\left\{y_m^{k,p}\right\}$, it is sufficient to show that $y_m^{k,p}=\xi$ for all $m\ge 1$.

Using the change of variables (9), the initial conditions (14) imply

$$y_{-1}^{k,p}=y_0^{k,p}=\xi \mathrm{for}\mathrm{each}p=1,2,\dots ,k,$$

We show that for the corresponding subsequence $\left\{y_{2m}^{k,p}\right\}$ and $\left\{y_{2m-1}^{k,p}\right\}$, one has

$$y_{2m}^{k,p}=y_{2m-1}^{k,p}=\xi \mathrm{for}\mathrm{all}m\ge 1.$$

Using Equation (11), we have

$$\begin{array}{cc}\hfill y_{2m}^{k,p}& ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi {\displaystyle \prod _{j=0}^{m-1}}\left(1+{\xi }^2{\displaystyle \frac{{({\xi }^2+1)}^{2j+1}-1}{\left({\xi }^2+1\right)-1}}\right)}{{\displaystyle \prod _{j=0}^{m-1}}\left(1+{\xi }^2{\displaystyle \frac{{({\xi }^2+1)}^{2j+2}-1}{\left({\xi }^2+1\right)-1}}\right)}}\hfill \\ & ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi {\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j+1}}{{\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j+2}}}\hfill \\ & ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi {\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j+1}}{{\displaystyle \prod _{j=0}^{m-1}}\left({\xi }^2+1\right){\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j+1}}}\hfill \\ & ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi }{{\left({\xi }^2+1\right)}^m}}=\xi .\hfill \end{array}$$

Similarly,

$$\begin{array}{cc}\hfill y_{2m-1}^{k,p}& ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi {\displaystyle \prod _{j=0}^{m-1}}\left(1+{\xi }^2{\displaystyle \frac{{({\xi }^2+1)}^{2j}-1}{\left({\xi }^2+1\right)-1}}\right)}{{\displaystyle \prod _{j=0}^{m-1}}\left(1+{\xi }^2{\displaystyle \frac{{({\xi }^2+1)}^{2j+1}-1}{\left({\xi }^2+1\right)-1}}\right)}}\hfill \\ & ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi {\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j}}{{\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j+1}}}\hfill \\ & ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi {\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j}}{{\displaystyle \prod _{j=0}^{m-1}}\left({\xi }^2+1\right){\displaystyle \prod _{j=0}^{m-1}}{\left({\xi }^2+1\right)}^{2j}}}\hfill \\ & ={\displaystyle \frac{{\left({\xi }^2+1\right)}^m\xi }{{\left({\xi }^2+1\right)}^m}}=\xi .\hfill \end{array}$$

Hence, $y_m^{k,p}=\xi$ for all $m\ge 1$. Therefore, the corresponding solution $\left\{x_n\right\}$ of Equation (6) is equal to $\xi$, for all $n\ge 1$. □

Theorem 7.

Let $k\in \mathbb{N}$ be fixed, and suppose that for each $p=k,k-1,\dots ,1$, the initial conditions $y_{-1}^{k,p},y_0^{k,p}\in \mathbb{R}$ satisfy one of the following cases:

1. 

$y_{-1}^{k,p}\ne 0$, $y_0^{k,p}={\displaystyle \frac{\alpha -1}{y_{-1}^{k,p}}}$, with $\alpha \ne 1$.

2. 

$y_{-1}^{k,p}={\displaystyle \frac{\alpha -1}{y_0^{k,p}}}$, $y_0^{k,p}\ne 0$, with $\alpha \ne 1$.

3. 

$y_{-1}^{k,p}\ne 0$, $y_0^{k,p}=0$, and $\alpha =1$.

4. 

$y_{-1}^{k,p}=0$, $y_0^{k,p}\ne 0$, and $\alpha =0$.

Then the corresponding solution $\left\{x_n\right\}$ of Equation (6) is nontrivial and periodic with period $2k$.

Proof. 

We prove case (1); the remaining cases follow by similar computations.

Assume that $y_{-1}^{k,p}\ne 0$ and $y_0^{k,p}={\displaystyle \frac{\alpha -1}{y_{-1}^{k,p}}}$ with $\alpha \ne 1$. We show that for the corresponding subsequence $\left\{y_{2m}^{k,p}\right\}$ and $\left\{y_{2m-1}^{k,p}\right\}$ one has

$$y_{2m}^{k,p}=y_0^{k,p}\mathrm{and}{y}_{2m-1}^{k,p}=y_{-1}^{k,p}\mathrm{for}\mathrm{all}m\ge 1.$$

Using Equation (11), we get

$$y_{2m}^{k,p}={\displaystyle \frac{{\alpha }^m{y}_0^{k,p}{\displaystyle \prod _{j=0}^{m-1}}\left(1+\left(\alpha -1\right){\displaystyle \frac{{\alpha }^{2j+1}-1}{\alpha -1}}\right)}{{\displaystyle \prod _{j=0}^{m-1}}\left(1+\left(\alpha -1\right){\displaystyle \frac{{\alpha }^{2j+2}-1}{\alpha -1}}\right)}}={\displaystyle \frac{{\alpha }^m{y}_0^{k,p}{\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j+1}}{{\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j+2}}}={\displaystyle \frac{{\alpha }^m{y}_0^{k,p}{\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j+1}}{{\displaystyle \prod _{j=0}^{m-1}}\alpha {\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j+1}}}=y_0^{k,p}.$$

Similarly,

$$y_{2m-1}^{k,p}={\displaystyle \frac{{\alpha }^m{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{m-1}}\left(1+\left(\alpha -1\right){\displaystyle \frac{{\alpha }^{2j}-1}{\alpha -1}}\right)}{{\displaystyle \prod _{j=0}^{m-1}}\left(1+\left(\alpha -1\right){\displaystyle \frac{{\alpha }^{2j+1}-1}{\alpha -1}}\right)}}={\displaystyle \frac{{\alpha }^m{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j}}{{\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j+1}}}={\displaystyle \frac{{\alpha }^m{y}_{-1}^{k,p}{\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j}}{{\displaystyle \prod _{j=0}^{m-1}}\alpha {\displaystyle \prod _{j=0}^{m-1}}{\alpha }^{2j}}}=y_{-1}^{k,p}.$$

Therefore each sequence ${\left\{y_m^{k,p}\right\}}_{m=1}^{\infty }$ is 2-periodic with pattern

$$\left\{y_{-1}^{k,p},y_0^{k,p},y_{-1}^{k,p},y_0^{k,p},\dots \right\}.$$

Reconstructing the original sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ from the k independent subsequences ${\left\{y_m^{k,p}\right\}}_{m=1}^{\infty }$ yields a nontrivial solution $\left\{x_n\right\}$ that is periodic with period $2k$. This completes the proof for case (1). □

Note that the period $2k$ is minimal if the initial conditions $y_{-1}^{k,p}$ and $y_0^{k,p}$ are not the same for all p. If all subsequences $\left\{y_m^{k,p}\right\}$ are identical, the resulting sequence $\left\{x_n\right\}$ may have a smaller period.

Theorem 8.

Let $k\in \mathbb{N}$ be fixed, and suppose that for each $p=k,k-1,\dots ,1$, the initial conditions $y_{-1}^{k,p},y_0^{k,p}\in \mathbb{R}$ satisfy one of the following cases:

1. 

$y_{-1}^{k,p}=0$, $y_0^{k,p}\ne 0$, and $\alpha =-1$.

2. 

$y_{-1}^{k,p}\ne 0$, $y_0^{k,p}=0$, and $\alpha =-1$.

Then the corresponding solution $\left\{x_n\right\}$ of Equation (6) is nontrivial and periodic with period $4k$.

Proof. 

We prove case (1); case (2) follows by similar computations.

Assume that $y_{-1}^{k,p}=0$, $y_0^{k,p}\ne 0$, and $\alpha =-1$. We show that for the corresponding subsequence $\left\{y_{2m}^{k,p}\right\}$ and $\left\{y_{2m-1}^{k,p}\right\}$, one has

$$y_{2m}^{k,p}={(-1)}^m{y}_0^{k,p},y_{2m-1}^{k,p}=0\mathrm{for}\mathrm{all}m\ge 1.$$

Using Equation (11),

$$y_{2m}^{k,p}={\displaystyle \frac{{(-1)}^m{y}_0^{k,p}{\displaystyle \prod _{j=0}^{m-1}}\left(1\right)}{{\displaystyle \prod _{j=0}^{m-1}}\left(1\right)}}={(-1)}^m{y}_0^{k,p}.$$

Similarly,

$$y_{2m-1}^{k,p}={\displaystyle \frac{{(-1)}^m\left(0\right){\displaystyle {\displaystyle \prod _{j=0}^{m-1}}}\left(1\right)}{{\displaystyle {\displaystyle \prod _{j=0}^{m-1}}}\left(1\right)}}=0.$$

Therefore, each sequence ${\left\{y_m^{k,p}\right\}}_{m=1}^{\infty }$ follows the pattern

$$\{0,-y_0^{k,p},0,y_0^{k,p},0,-y_0^{k,p},0,y_0^{k,p},0,\dots \},$$

which is periodic of period 4. Reconstructing the original sequence ${\left\{x_n\right\}}_{n=1}^{\infty }$ from the k independent subsequences ${\left\{y_m^{k,p}\right\}}_{m=1}^{\infty }$ yields a nontrivial solution $\left\{x_n\right\}$ that is periodic with period $4k$. This completes the proof for case (1). □

Note that the period $4k$ is minimal provided that the initial conditions $y_{-1}^{k,p}$ and $y_0^{k,p}$ differ across all p. If the subsequences $\left\{y_m^{k,p}\right\}$ are identical, the overall sequence $\left\{x_n\right\}$ may exhibit a smaller period.

4. Numerical Simulations

4.1. General Solution

Example 1.

For $k=3$, Equation (2) becomes the rational difference equation

$$x_{n+1}={\displaystyle \frac{\alpha x_{n-5}}{1+x_{n-2}{x}_{n-5}}},n=0,1,2,\dots$$

with initial values $x_{-5},x_{-4},x_{-3},x_{-2},x_{-1},x_0$. To solve this equation, we set the initial conditions as $x_{-5}=2.25$, $x_{-4}=1$, $x_{-3}=-3.25$, $x_{-2}=-1.5$, $x_{-1}=0.25$, and $x_0=-1$, corresponding to Equation (6), with the parameter $\alpha =2$. We compute the solution using a direct iteration of Equation (6) and compare the results with those obtained from the explicit formula presented in Theorem 3. The results are summarized in Table 1.

Table 1. The solution of Equation (6) for $k=6$, $\alpha =2$, and $x_{-5}=2.25$, $x_{-4}=1$, $x_{-3}=-3.25$, $x_{-2}=-1.5$, $x_{-1}=0.25$, and $x_0=-1$.

Computed ${\mathit{x}}_{\mathit{n}}$ Using Equation (6)	Computed ${\mathit{y}}_{\mathit{m}}^{3,\mathit{p}}$ Using Formula (11)	Relative Error
$x_1=-1.8947368420$	$y_1^{3,3}=-1.8947368420$	0
$x_2=1.6000000000$	$y_1^{3,2}=1.6000000000$	0
$x_3=-1.5294117650$	$y_1^{3,1}=-1.5294117650$	0
$x_4=-0.7808219178$	$y_2^{3,3}=-0.7808219178$	0
$x_5=0.3571428571$	$y_2^{3,2}=0.3571428571$	0
$x_6=-0.7906976743$	$y_2^{3,1}=-0.7906976744$	$1.26\times {10}^{-10}$
$x_7=-1.5283512650$	$y_3^{3,3}=-1.5283512650$	0
$x_8=2.0363636370$	$y_3^{3,2}=2.0363636360$	$4.91\times {10}^{-10}$
$x_9=-1.3845201240$	$y_3^{3,1}=-1.3845201240$	0
$x_{10}=-0.7119837136$	$y_4^{3,3}=-0.7119837135$	$1.40\times {10}^{-10}$
$x_{11}=0.4135338346$	$y_4^{3,2}=0.4135338346$	0
$x_{12}=-0.7549374782$	$y_4^{3,1}=-0.7549374781$	$1.32\times {10}^{-10}$
$x_{13}=-1.4638249750$	$y_5^{3,3}=-1.4638249750$	0
$x_{14}=2.2109090920$	$y_5^{3,2}=2.2109090910$	$4.52\times {10}^{-10}$
$x_{15}=-1.3539042000$	$y_5^{3,1}=-1.3539042000$	0
$x_{16}=-0.6972646171$	$y_6^{3,3}=-0.6972646172$	$1.43\times {10}^{-10}$
$x_{17}=0.4320502748$	$y_6^{3,2}=0.4320502749$	$2.31\times {10}^{-10}$
$x_{18}=-0.7466817827$	$y_6^{3,1}=-0.7466817822$	$6.70\times {10}^{-10}$
$x_{19}=-1.4488486890$	$y_7^{3,3}=-1.4488486890$	0
$x_{20}=2.2615405970$	$y_7^{3,2}=2.2615405970$	0
$x_{21}=-1.3465415790$	$y_7^{3,1}=-1.3465415800$	$7.43\times {10}^{-10}$
$x_{22}=-0.6937159388$	$y_8^{3,3}=-0.6937159386$	$2.88\times {10}^{-10}$
$x_{23}=0.4370547183$	$y_8^{3,2}=0.4370547183$	0
$x_{24}=-0.7446570351$	$y_8^{3,1}=-0.7446570349$	$2.68\times {10}^{-10}$
$x_{25}=-1.4451711420$	$y_9^{3,3}=-1.4451711420$	0
$x_{26}=2.2747146200$	$y_9^{3,2}=2.2747146190$	$4.40\times {10}^{-10}$
$x_{27}=-1.3447183690$	$y_9^{3,1}=-1.3447183700$	$7.43\times {10}^{-10}$
$x_{28}=-0.6928366410$	$y_{10}^{3,3}=-0.6928366406$	$5.77\times {10}^{-10}$
$x_{29}=0.4383314116$	$y_{10}^{3,2}=0.4383314120$	$9.13\times {10}^{-10}$
$x_{30}=-0.7441532455$	$y_{10}^{3,1}=-0.7441532452$	$4.03\times {10}^{-10}$

Example 2.

For $k=6$, Equation (2) becomes the rational difference equation

$$x_{n+1}={\displaystyle \frac{\alpha x_{n-11}}{1+x_{n-5}{x}_{n-11}}},n=0,1,2,\dots$$

with initial values $x_{-11},x_{-10},\dots ,x_0$. Let $x_{-11}=4.75$, $x_{-10}=1.5$, $x_{-9}=4.25$, $x_{-8}=1$, $x_{-7}=1.75$, $x_{-6}=2.5$, $x_{-5}=2.25$, $x_{-4}=3$, $x_{-3}=0.75$, $x_{-2}=-1.5$, $x_{-1}=2.25$, and $x_0=-1$ be the initial conditions of Equation (6) with $\alpha =-0.25$. Then Theorem 4 implies that the solution converges to zero. Table 2 sets forth the values of $\left\{x_n\right\}$ and $y_m^{6,p}$. See also Figure 1.

Table 2. The solution of Equation (6) for $k=6$, $\alpha =-0.25$, and initials $x_{-11}=4.75$, $x_{-10}=1.5$, $x_{-9}=4.25$, $x_{-8}=1$, $x_{-7}=1.75$, $x_{-6}=2.5$, $x_{-5}=2.25$, $x_{-4}=3$, $x_{-3}=0.75$, $x_{-2}=-1.5$, $x_{-1}=2.25$, and $x_0=-1$.

Computed ${\mathit{x}}_{\mathit{n}}$ Using Equation (6)	Computed ${\mathit{y}}_{\mathit{m}}^{6,\mathit{p}}$ Using Formula (11)
$x_1=-0.10160427810$	$y_1^{6,6}=-0.10160427810$
$x_2=-0.06818181818$	$y_1^{6,5}=-0.06818181818$
$x_{10}=1.50000000000$	$y_2^{6,3}=1.50000000000$
$x_{50}=-0.00024922910$	$y_9^{6,5}=-0.00024922910$
$x_{150}=2.08533824700\times {10}^{-8}$	$y_{25}^{6,1}=2.085338246\times {10}^{-8}$
$x_{200}=-2.23119038100\times {10}^{-10}$	$y_{34}^{6,5}-2.231190381\times {10}^{-10}$

Figure 1. Solution of Equation (6) converges to zero for for $k=6$, $\alpha =-0.25$, and initials $x_{-11}=4.75$, $x_{-10}=1.5$, $x_{-9}=4.25$, $x_{-8}=1$, $x_{-7}=1.75$, $x_{-6}=2.5$, $x_{-5}=2.25$, $x_{-4}=3$, $x_{-3}=0.75$, $x_{-2}=-1.5$, $x_{-1}=2.25$, and $x_0=-1$.

4.2. Periodic Solutions

Example 3.

Let $k=4$. Then Equation (6) becomes

$$x_{n+1}={\displaystyle \frac{\alpha x_{n-7}}{1+x_{n-3}{x}_{n-7}}},n=0,1,\dots$$

We consider the parameters $\alpha =2.5$, $x_{-7}=-0.5$, $x_{-6}=1.5$, $x_{-5}=-0.25$, $x_{-4}=2.5$, $x_{-3}=-3$, $x_{-2}=1$, $x_{-1}=-6$, and $x_0=6$. According to Theorem 7, the solution of this equation is periodic with period 8; see Figure 2.

Figure 2. Periodic solution of period 8 for Equation (6) with $k=4$, $\alpha =2.5$, and initial values $x_{-7}=-0.5$, $x_{-6}=1.5$, $x_{-5}=-0.25$, $x_{-4}=2.5$, $x_{-3}=-3$, $x_{-2}=1$, $x_{-1}=-6$, and $x_0=6$.

Example  4.

Let $k=6$. Then Equation (6) becomes

$$x_{n+1}={\displaystyle \frac{\alpha x_{n-11}}{1+x_{n-5}{x}_{n-11}}},n=0,1,\dots$$

By Theorem 8, this equation admits a periodic solution of length 24 when the parameters are chosen as $\alpha =-1$, $x_{-11}=x_{-10}=x_{-9}=x_{-8}=x_{-7}=x_{-6}=0$, $x_{-5}=2$, $x_{-4}=-1$, $x_{-3}=0.75$, $x_{-2}=-2$, $x_{-1}=0.5$, and $x_0=-0.25$. The periodic solution is illustrated in Figure 3, and its repeating cycle is given by

Figure 3. Periodic solution of period 24 for Equation (6) with $k=6$, $\alpha =-1$, and initial values $x_{-11}=x_{-10}=x_{-9}=x_{-8}=x_{-7}=x_{-6}=0$, $x_{-5}=2$, $x_{-4}=-1$, $x_{-3}=0.75$, $x_{-2}=-2$, $x_{-1}=0.5$, and $x_0=-0.25$.

5. Conclusions

In this paper, we investigated the global dynamics of a broad class of rational difference equations of the form

$$x_{n+1}={\displaystyle \frac{a{x}_{n+1-2k}}{b+c{x}_{n+1-k}{x}_{n+1-2k}}},n=0,1,\dots ,$$

which generalizes several well-known cases studied in the literature. We showed that this class has exactly three equilibrium points: the trivial equilibrium ${\tilde{x}}_1$, which is globally asymptotically stable, and the nontrivial equilibria ${\tilde{x}}_2$ and ${\tilde{x}}_3$, which are never linearly stable. By deriving an explicit formula for the general solution, we provided a comprehensive analytical framework characterizing solution behavior in terms of system parameters and initial conditions. Under the condition $\alpha =(b/a)\in (-1,1)$, every solution converges to the zero equilibrium, establishing global asymptotic stability. We also identified and classified nontrivial periodic solutions with minimal periods $2k$ and $4k$, depending on the initial conditions and $\alpha$, supported by theorems providing necessary and sufficient conditions for constant and periodic behaviors. Numerical simulations confirmed the analytical predictions. The results extend and unify previous studies, offering sharper criteria for stability and periodicity. The presented study provides a unified framework for analyzing this class of nonlinear rational difference equations, offering predictive power, generality, and applicability in biology, economics, and engineering, while its main limitations lie in dependence on specific parameter conditions for global stability and restriction to a particular class of equations. Although a detailed bifurcation analysis is beyond the scope of this study, the equations considered can exhibit bifurcation phenomena under certain parameter variations, which will be explored in future research.