On a Second-Order Difference Equation with Two Periodic Coefficients

Abstract

This work focuses on a class of nonlinear second-order difference equations with non-autonomous periodic coefficients. We obtain explicit solution formulas that allow for a constructive characterization of their behavior. In particular, we derive necessary and sufficient conditions for the existence of periodic solutions, providing a complete description of when they arise. Several numerical examples are also included to illustrate the theoretical results.

1. Introduction

Difference equations and their systems are an essential and fundamental tool for studying many discrete models in various fields such as applied mathematics, economics, and engineering. They provide an appropriate framework for describing discrete phenomena characterized by memory effects, delayed responses, or interactions between successive states. This explains the large number of studies dealing with these types of equations and systems. As examples, one may consult the following references [1,2,3,4,5,6,7,8].

Real discrete phenomena subject to periodic or seasonal variations can be modeled using difference equations and systems with periodic coefficients. For instance, in population dynamics, such models incorporate periodically varying parameters to reflect environmental seasonality, while in economics and engineering they capture cyclic behavior driven by internal or external periodic influences. A lot of studies have been devoted to such types of difference equations and systems; see, for example, references [9,10,11,12].

Solving nonlinear difference equations, especially those with non-autonomous coefficients, is challenging due to the lack of a general method; thus, researchers often rely on suitable variable transformations to reduce them to solvable forms and then reconstruct the original solutions. Once explicit formulas are derived, they facilitate a deeper analysis of key properties such as periodicity, asymptotic behavior, and oscillatory dynamics. Motivated by these observations, the present work is devoted to the study of a class of nonlinear second-order difference equations with non-autonomous periodic coefficients.

In realizing the present work, we are mainly motivated by a sequence of contributions that appeared progressively in the literature and that we briefly review below in chronological order. One of the earliest related contributions is due to Elabbasy et al. [13], in which the authors provide, among other things, the explicit formulas of the following nonlinear second-order rational difference equation:

$$x_{n+1}={\displaystyle \frac{x_n{x}_{n-1}}{x_n-1}}+1,n\in {\mathbb{N}}_0.$$

(1)

Haddad et al., in [14] and as a generalization of Equation (1), studied the following second-order rational system:

$$x_{n+1}={\displaystyle \frac{a{x}_n{y}_{n-1}}{y_n-\gamma }}+c,y_{n+1}={\displaystyle \frac{\alpha y_n{x}_{n-1}}{x_n-c}}+\gamma ,n\in {\mathbb{N}}_0,$$

(2)

where the parameters and the initial values are positive real numbers. Subsequently, Yazlik et al. [15] extended system (2) to the non-autonomous higher-order system

$$x_{n+1}={\displaystyle \frac{a_n{x}_{n-k+1}{y}_{n-k}}{y_n-{\gamma }_n}}+c_{n+1},y_{n+1}={\displaystyle \frac{{\alpha }_n{y}_{n-k+1}{x}_{n-k}}{x_n-c_n}}+{\gamma }_{n+1},n\in {\mathbb{N}}_0,$$

where $k\in \mathbb{N}$ and the sequences $\left(a_n\right)n\in {\mathbb{N}}_0$, $\left({\alpha }_n\right)n\in {\mathbb{N}}_0$, $\left(c_n\right)n\in {\mathbb{N}}_0$, and $\left({\gamma }_n\right)n\in {\mathbb{N}}_0$ are periodic sequences of real numbers. Inspired by the above-mentioned works, Touafek and AL-Juaid [16] recently investigated the solvability of the following autonomous system of difference equations:

$$x_{n+1}={\displaystyle \frac{a{x}_n{y}_{n-1}}{y_n-\beta y_{n-1}-\gamma }}+b{x}_n+c,y_{n+1}={\displaystyle \frac{\alpha y_n{x}_{n-1}}{x_n-b{x}_{n-1}-c}}+\beta y_n+\gamma ,n\in {\mathbb{N}}_0,$$

(3)

where a, $\alpha$ and the initial values $x_{-1}$, $y_{-1}$, $x_0$ and $y_0$ are non-zero real numbers, while b, c, $\beta$ and $\gamma$ are real numbers, and as a consequence of their obtained results, they present the formulas of the solutions of the following autonomous difference equation:

$$x_{n+1}={\displaystyle \frac{\alpha x_n{x}_{n-1}}{x_n-\beta b{x}_{n-1}-\gamma }}+\beta x_n+\gamma ,n\in {\mathbb{N}}_0.$$

(4)

A natural extension of Equation (4) is to ask whether explicit formulas can be derived for the following difference equation with non-autonomous periodic coefficients:

$$x_{n+1}={\displaystyle \frac{{\alpha }_n{x}_n{x}_{n-1}}{x_n-{\beta }_n{x}_{n-1}-{\gamma }_n}}+{\beta }_{n+1}{x}_n+{\gamma }_{n+1},$$

(5)

where $n\in {\mathbb{N}}_0$, $\left({\alpha }_n\right)$, $\left({\beta }_n\right)$, $\left({\gamma }_n\right)$ are periodic sequences of real numbers of period two, and the initial values $x_{-1}$, $x_0$, are non-zero real numbers. In this work, we provide a positive answer to this question. Specifically, we derive explicit formulas for the solutions of the equation. These formulas allow us to establish necessary and sufficient conditions for the existence of periodic solutions, leading to a complete characterization of when such solutions occur. In addition, we present results concerning the limiting behavior of solutions. Our theoretical findings are illustrated by several numerical examples. More related works can be found in the following references: [17,18,19].

While the present work is mainly devoted to the solvability of Equation (5) and the existence of periodic solutions, the study of local and global stability remains essential for understanding the robustness and long-term behavior of such solutions. Moreover, bifurcation analysis is crucial for describing qualitative changes in the dynamics caused by variations in the periodic parameters, including transitions between stability and instability and the emergence of more complex behaviors. For some contributions on these important notions, we refer to [20,21,22,23,24,25].

In this work, for periodic solutions, we adopt the following definition.

Definition 1.

A solution ${\left(x_n\right)}_{n=-1}^{+\infty }$ of Equation (5) is said to be eventually p-periodic, $p\in \mathbb{N}$, if there exists $n_0\ge -1$ such that

$$x_{n+p}=x_n,n\ge n_0.$$

If $n_0=-1$, the solution is said p-periodic.

Remark 1.

If p in Definition 1 is the smallest $k\in \mathbb{N}$ for which

$$x_{n+k}=x_n,n\ge n_0,\mathit{respectively}(x_{n+k}=x_n,n\ge -1),$$

the solution is eventually prime p-periodic, respectively, prime p-periodic.

2. Explicit Representation of the Solutions of Equation (5)

In this section, we show that Equation (5) is solvable, and hence we provide explicit representation of its solutions. Following the approach in [14,16], we first transform (5) into a solvable first-order non-autonomous rational difference equation. Then, by considering the parity of $n\in {\mathbb{N}}_0$, we derive two solvable first-order non-autonomous linear difference equations whose solutions coincide exactly with those of (5).

Let ${\alpha }_n=\left\{\begin{array}{cc}{\alpha }_0,\hfill & \mathrm{if}n\mathrm{even}\hfill \\ {\alpha }_1\hfill & \mathrm{if}n\mathrm{odd}\hfill \end{array}\right.$, ${\beta }_n=\left\{\begin{array}{cc}{\beta }_0,\hfill & \mathrm{if}n\mathrm{even}\hfill \\ {\beta }_1\hfill & \mathrm{if}n\mathrm{odd}\hfill \end{array}\right.$, ${\gamma }_n=\left\{\begin{array}{cc}{\gamma }_0,\hfill & \mathrm{if}n\mathrm{even}\hfill \\ {\gamma }_1\hfill & \mathrm{if}n\mathrm{odd}.\hfill \end{array}\right.$

From (5), we can write

$${\displaystyle \frac{x_{n+1}-{\beta }_{n+1}{x}_n-{\gamma }_{n+1}}{x_n}}={\displaystyle \frac{{\alpha }_n}{\frac{x_n-{\beta }_n{x}_{n-1}-{\gamma }_n}{x_{n-1}}}},n\in {\mathbb{N}}_0.$$

(6)

To be well-defined, in identity (6), we must assume that

$$x_n-{\beta }_n{x}_{n-1}-{\gamma }_n\ne 0,x_{n-1}\ne 0,n\in {\mathbb{N}}_0.$$

(7)

Through the rest of our work, for each solution of (5), it is assumed that (7) holds.

Our first result is devoted to the explicit formulas of the solutions of Equation (5).

Theorem 1.

Let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5), where the following statements are true.

1. 

For all $n\in {\mathbb{N}}_0$, we have

$$x_{2n}=x_0{\displaystyle \prod _{k=0}^{n-1}}{A}_k+{\displaystyle \sum _{j=0}^{n-1}}\left({\displaystyle \prod _{k=j+1}^{n-1}}{A}_k\right)\left(\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^{j+1}{v}_0\right){\gamma }_1+{\gamma }_0\right),$$

$$x_{2n-1}=x_{-1}{\displaystyle \prod _{k=0}^{n-1}}{B}_k+{\displaystyle \sum _{j=0}^{n-1}}\left({\displaystyle \prod _{k=j+1}^{n-1}}{B}_k\right)\left(\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^j{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right),$$

where

$$A_k=\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^{k+1}{v}_0\right)\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^k{\displaystyle \frac{{\alpha }_0}{v_0}}\right),$$

$$B_k=\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^k{v}_0\right)\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^k{\displaystyle \frac{{\alpha }_0}{v_0}}\right).$$

2. 

Assume that the sequence ${\left({\alpha }_n\right)}_{n=-1}^{+\infty }$ is constant, that is ${\alpha }_1={\alpha }_0$.

• If $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\ne 1$, then for all $n\in \mathbb{N}$, we have

  $$x_{2n-1}=x_{-1}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right){\displaystyle \frac{{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)}^n-1}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)-1}},$$

  (8)

  $$x_{2n}=x_0{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right){\displaystyle \frac{{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)}^n-1}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)-1}},$$

  (9)
• If $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=1$, then for all $n\in {\mathbb{N}}_0$, we have

  $$x_{2n-1}=x_{-1}+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right)n,$$

  $$x_{2n}=x_0+\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right)n,$$

  where

  $$v_0={\displaystyle \frac{x_0-{\beta }_0{x}_{-1}-{\gamma }_0}{x_{-1}}}.$$

Proof. 

Let

$$v_n={\displaystyle \frac{x_n-{\beta }_n{x}_{n-1}-{\gamma }_n}{x_{n-1}}},n=0,1,\dots$$

(10)

It follows from (6) that

$$v_{n+1}={\displaystyle \frac{{\alpha }_n}{v_n}},n\in {\mathbb{N}}_0.$$

(11)

It is not hard to see that

$$v_{2n}={\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^n{v}_0,v_{2n+1}={\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n{\displaystyle \frac{{\alpha }_0}{v_0}},n=0,1,\dots .$$

(12)

Using (10), we obtain

$$x_n=({\beta }_n+v_n)x_{n-1}+{\gamma }_n,n=0,1,\dots .$$

(13)

Depending on the parity of n, we get from (13)

$$x_{2n}=({\beta }_{2n}+v_{2n})x_{2n-1}+{\gamma }_{2n},x_{2n+1}=({\beta }_{2n+1}+v_{2n+1})x_{2n}+{\gamma }_{2n+1},n=0,1,\dots .$$

(14)

Using the formulas of ${\alpha }_n$, ${\beta }_n$, ${\gamma }_n$ and $v_n$, we get for all $n=0,1,\dots ,$

$$x_{2n}=\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^n{v}_0\right)x_{2n-1}+{\gamma }_0,$$

(15)

$$x_{2n+1}=\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n{\displaystyle \frac{{\alpha }_0}{v_0}}\right)x_{2n}+{\gamma }_1.$$

(16)

From (15) and (16), we obtain for all $n=0,1,\dots ,$

$$x_{2n+2}=\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^{n+1}{v}_0\right)\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n{\displaystyle \frac{{\alpha }_0}{v_0}}\right)x_{2n}+\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^{n+1}{v}_0\right){\gamma }_1+{\gamma }_0.$$

(17)

$$x_{2n+1}=\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^n{v}_0\right)\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n{\displaystyle \frac{{\alpha }_0}{v_0}}\right)x_{2n-1}+\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1.$$

(18)

Equation (17) is a non-autonomous first-order linear difference equation, in $x_{2n}$, and Equation (18) is a non-autonomous first-order linear difference equation, in $x_{2n-1}$; the formulas of their solutions are given for all $n=0,1,\dots ,$ by

$$x_{2n}=x_0{\displaystyle \prod _{k=0}^{n-1}}{A}_k+{\displaystyle \sum _{j=0}^{n-1}}\left({\displaystyle \prod _{k=j+1}^{n-1}}{A}_k\right)\left(\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^{j+1}{v}_0\right){\gamma }_1+{\gamma }_0\right),$$

(19)

$$x_{2n-1}=x_{-1}{\displaystyle \prod _{k=0}^{n-1}}{B}_k+{\displaystyle \sum _{j=0}^{n-1}}\left({\displaystyle \prod _{k=j+1}^{n-1}}{B}_k\right)\left(\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^j{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right),$$

(20)

where

$$A_k=\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^{k+1}{v}_0\right)\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^k{\displaystyle \frac{{\alpha }_0}{v_0}}\right),$$

$$B_k=\left({\beta }_0+{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^k{v}_0\right)\left({\beta }_1+{\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^k{\displaystyle \frac{{\alpha }_0}{v_0}}\right).$$

Now, in the case when $\left({\alpha }_n\right)$ is a constant sequence, that is ${\alpha }_1={\alpha }_0$, we obtain

$$A_k=B_k=\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right).$$

So, for all $n=0,1,\dots ,$ we get

$$x_{2n}=x_0{\displaystyle \prod _{k=0}^{n-1}}\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)+{\displaystyle \sum _{j=0}^{n-1}}\left({\displaystyle \prod _{k=j+1}^{n-1}}\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right),$$

$$x_{2n-1}=x_{-1}{\displaystyle \prod _{k=0}^{n-1}}\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)+{\displaystyle \sum _{j=0}^{n-1}}\left({\displaystyle \prod _{k=j+1}^{n-1}}\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right).$$

Or equivalently, for all $n=0,1,\dots ,$

$$x_{2n}=x_0{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right){\displaystyle \sum _{j=0}^{n-1}}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^{n-j-1},$$

(21)

$$x_{2n-1}=x_{-1}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right){\displaystyle \sum _{j=0}^{n-1}}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^{n-j-1}.$$

(22)

Again, from (21) and (22), we get for all $n\in \mathbb{N}$

$$x_{2n-1}=x_{-1}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right){\displaystyle \frac{{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)}^n-1}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)-1}},$$

$$x_{2n}=x_0{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right){\displaystyle \frac{{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)}^n-1}{\left(\left({\beta }_0+v_0\right)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)\right)-1}},$$

if $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\ne 1$, and for all $n\in {\mathbb{N}}_0$

$$x_{2n-1}=x_{-1}+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right)n,$$

$$x_{2n}=x_0+\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right)n,$$

if $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=1$. □

Remark 2.

If $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\ne 0,$ the formulas in (8) and (9) are also valid for $n=0$.

3. The Behavior of the Solutions of Equation (5) When $\alpha$₁ = $\alpha$₀

From Theorem 1, the behavior of the solutions of Equation (5) depends on the quantities ${\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}$ and ${\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}$. When ${\alpha }_1\ne {\alpha }_0$, one of the terms ${\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n$, ${\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^n$ grows exponentially while the other decays exponentially as $n\to \infty$. More precisely, if ${\alpha }_0>{\alpha }_1$, then

$${\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n\to +\infty ,{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^n\to 0,$$

whereas if ${\alpha }_0<{\alpha }_1$, the reverse occurs:

$${\left({\displaystyle \frac{{\alpha }_0}{{\alpha }_1}}\right)}^n\to 0,{\left({\displaystyle \frac{{\alpha }_1}{{\alpha }_0}}\right)}^n\to +\infty .$$

Therefore, the asymptotic behavior of the solutions becomes difficult to characterize. Motivated by this observation, we focus in this section on the case where the sequence $\left({\alpha }_n\right)$ is constant, namely ${\alpha }_1={\alpha }_0$.

3.1. The Case $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=0$

The following result, which is a direct consequence of Theorem 1, describes explicitly the formulas of the solutions of Equation (5) when $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=0$.

Lemma 1.

Assume that $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=0$, and let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5). Then

$$x_{2n-1}=\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1,x_{2n}=\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0,n\in \mathbb{N};$$

(23)

that is, the solution is eventually 2-periodic.

Theorem 2.

Assume that $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=0$, and let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5). The following statements are true.

1. 

The solution ${\left(x_n\right)}_{n=-1}^{+\infty }$ is eventually prime 1-periodic if and only if

$$({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1=({\beta }_0+v_0){\gamma }_1+{\gamma }_0.$$

2. 

The solution ${\left(x_n\right)}_{n=-1}^{+\infty }$ is eventually prime 2-periodic if and only if

$$({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1\ne ({\beta }_0+v_0){\gamma }_1+{\gamma }_0.$$

Proof. 

1.

First, assume that

$$({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1=({\beta }_0+v_0){\gamma }_1+{\gamma }_0,$$

then from Lemma 1, the solution takes the form

$${\left(x_n\right)}_{n=-1}^{+\infty }=(x_{-1},x_0,\overbrace{({\beta }_0+v_0){\gamma }_1+{\gamma }_0},\overbrace{({\beta }_0+v_0){\gamma }_1+{\gamma }_0},\dots ),$$

which means that the solution is eventually prime 1-periodic. Second, assume that ${\left(x_n\right)}_{n=-1}^{+\infty }$ is eventually prime 1-periodic, then

$$\exists n_0\ge -1:x_{n+1}=x_n,n\ge n_0.$$

(24)

We have $n_0+2\ge 1$, so from the formulas of the solutions given in Lemma 1, we get

$$x_{2(n_0+2)-1}=\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1,x_{2(n_0+2)}=\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0.$$

(25)

It follows from (24) that

$$x_{2(n_0+2)}=x_{2(n_0+2)-1},$$

from which, and using (25), we obtain

$$({\beta }_0+v_0){\gamma }_1+{\gamma }_0=({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1.$$

2.

From Lemma 1, the solution takes the form

$${\left(x_n\right)}_{n=-1}^{+\infty }=(x_{-1},x_0,\overbrace{({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1,({\beta }_0+v_0){\gamma }_1+{\gamma }_0},\overbrace{({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1,({\beta }_0+v_0){\gamma }_1+{\gamma }_0},\dots ).$$

That is, the solution repeats with the cycle of the two terms

$$({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1,({\beta }_0+v_0){\gamma }_1+{\gamma }_0.$$

So the solution will eventually be prime 2-periodic if and only if the terms of its cycle are different; that is, the solution is eventually prime 2-periodic if and only if

$$({\beta }_0+v_0){\gamma }_1+{\gamma }_0\ne ({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1.$$

□

3.2. The Case $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=-1$

The formulas of the solutions of Equation (5) when $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=-1$ are represented in the following result.

Lemma 2.

Let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5). Assume that $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=-1.$ Then, for all $n\in {\mathbb{N}}_0$ we have

$$x_{4n-1}=x_{-1},x_{4n}=x_0,x_{4n+1}=-x_{-1}+\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1,x_{4n+2}=-x_0+\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0,$$

and the solution is 4-periodic.

Proof. 

By replacing with $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=-1$ in Theorem 1, we obtain

$$x_{2n-1}=x_{-1}{(-1)}^n+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right)\left({\displaystyle \frac{{\left(-1\right)}^n-1}{-2}}\right),n\in {\mathbb{N}}_0,$$

(26)

$$x_{2n}=x_0{(-1)}^n+\left(({\beta }_0+v_0){\gamma }_1+{\gamma }_0\right)\left({\displaystyle \frac{{\left(-1\right)}^n-1}{-2}}\right),n\in {\mathbb{N}}_0.$$

(27)

Depending on the parity of n, it follows from (26) and (27) that

$$x_{4n-1}=x_{-1}{(-1)}^{2n}+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right)\left({\displaystyle \frac{{\left(-1\right)}^{2n}-1}{-2}}\right)=x_{-1},$$

$$x_{4n+1}=x_{-1}{(-1)}^{2n+1}+\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right)\left({\displaystyle \frac{{\left(-1\right)}^{2n+1}-1}{-2}}\right)=-x_{-1}+\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1,$$

$$x_{4n}=x_0{(-1)}^{2n}+\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right)\left({\displaystyle \frac{{\left(-1\right)}^{2n}-1}{-2}}\right)=x_0,$$

$$x_{4n+2}=x_0{(-1)}^{2n+1}+\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right)\left({\displaystyle \frac{{\left(-1\right)}^{2n+1}-1}{-2}}\right)=-x_0+\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0.$$

The proof is complete. □

Theorem 3.

Assume that $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=-1,$ and let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5). Then:

1. 

The solution ${\left(x_n\right)}_{n=-1}^{+\infty }$ is prime 1-periodic if and only if

$$x_0=x_{-1}={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1=\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0.$$

(28)

2. 

The solution ${\left(x_n\right)}_{n=-1}^{+\infty }$ is prime 2-periodic if and only if

$$x_{-1}={\displaystyle \frac{\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right){\gamma }_0+{\gamma }_1}{2}},x_0={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\ne \left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0.$$

(29)

Proof. 

1.

First assume that

$$x_0=x_{-1}={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1=\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0.$$

So, from Lemma 2, we obtain

$$x_{4n-1}=x_{4n}={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},$$

$$x_{4n+1}=-{\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}}+\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},$$

$$x_{4n+2}=-{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}}+\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}}.$$

That is,

$$x_n={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}},n\in {\mathbb{N}}_0,$$

which means that the solution is prime 1-periodic.

Second, assume that the solution is prime 1-periodic, then

$$x_{n+1}=x_n,n=-1,0,\dots ,$$

so it follows that

$$x_{4n+2}=x_{4n+1}=x_{4n}=x_{4n-1},n\in {\mathbb{N}}_0.$$

(30)

Now, using Lemma 2 and (30), we get

$$-x_0+\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0=-x_{-1}+\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1=x_0=x_{-1},$$

from which we get

$$x_0=x_{-1}={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1=({\beta }_0+v_0){\gamma }_1+{\gamma }_0.$$

2.

First assume that

$$x_{-1}={\displaystyle \frac{\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right){\gamma }_0+{\gamma }_1}{2}},x_0={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\ne \left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0.$$

So, from Lemma 2, we obtain

$$x_{4n+1}=x_{4n-1}={\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{2}},n\in {\mathbb{N}}_0,$$

$$x_{4n+2}=x_{4n}={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}},n\in {\mathbb{N}}_0.$$

Or equivalently

$${\left(x_n\right)}_{n=-1}^{+\infty }=\left(\overbrace{{\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{2}},{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}}},\overbrace{{\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{2}},{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}}},\dots \right).$$

That is, the solution repeats with the cycle of the two different terms

$${\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{2}},{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}},$$

and the solution will be prime 2-periodic.

Second, assume that the solution is prime 2-periodic, then

$$x_{n+2}=x_n,n=-1,0,\dots ,$$

so it follows that

$$x_{4n+2}=x_{4n},x_{4n+1}=x_{4n-1},n\in {\mathbb{N}}_0.$$

(31)

Now, using Lemma 2 and (31), we get

$$-x_0+\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0=x_0,-x_{-1}+\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1=x_{-1},$$

from which we get

$$x_{-1}={\displaystyle \frac{\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right){\gamma }_0+{\gamma }_1}{2}},x_0={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},$$

and the solution takes the form

$$x_{2n-1}={\displaystyle \frac{\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right){\gamma }_0+{\gamma }_1}{2}},x_{2n}={\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},n\in {\mathbb{N}}_0.$$

As the solution is prime 2-periodic, it follows that

$${\displaystyle \frac{\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right){\gamma }_0+{\gamma }_1}{2}}\ne {\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{2}},$$

or equivalently

$$\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\ne \left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0.$$

□

As a consequence of Theorem 3, we get the following result.

Corollary 1.

Assume that $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=-1,$ and let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5). Then, the solution will be prime 4-periodic if and only if neither (28) nor (29) are satisfied.

3.3. The Case $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=1$

Now, we investigate the behavior of the solutions of Equation (5) in the case $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=1.$ The following result is a direct consequence of Theorem 1, so its proof will be omitted.

Theorem 4.

Assume that $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)=1$, and let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5). Then, the following statements are true.

1. 

If $\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1=\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0=0,$ we get

$$x_{2n-1}=x_{-1},x_{2n}=x_0,n\in {\mathbb{N}}_0,$$

and the solution will be prime 2-periodic provided that $x_{-1}\ne x_0.$

2. 

If $\left(\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right){\gamma }_0+{\gamma }_1\right)\left(\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0\right)\ne 0,$ we obtain

$$\underset{n\to +\infty }{lim}|x_{2n-1}|=\underset{n\to +\infty }{lim}\left|x_{2n}\right|=+\infty .$$

3.4. The Case $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\ne -1,0,1$

Here, we investigate the periodicity of the solutions of (5) when $\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\ne -1,0,1$.

Theorem 5.

Let ${\left(x_n\right)}_{n=-1}^{+\infty }$ be a solution of (5). The following statements are true.

1. 

Assume that

$$x_0+{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}}=x_{-1}+{\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}}=0,$$

(32)

then

$$x_{2n}=x_0={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}},x_{2n-1}=x_{-1}={\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}},n\in {\mathbb{N}}_0,$$

that is, the solution is 2-periodic and it will be prime 2-periodic provided that $x_{-1}\ne x_0$.

2. 

Assume that

$$|\left({\beta }_0+v_0\right)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)|<1,$$

(33)

then, if

$$x_0+{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}}\ne 0\ne x_{-1}+{\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}},$$

(34)

we obtain

$$\underset{n\to +\infty }{lim}{x}_{2n}={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}},\underset{n\to +\infty }{lim}{x}_{2n-1}={\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}}.$$

Proof. 

From Theorem 1, we have for all $n\in {\mathbb{N}}_0.$

$$x_{2n-1}=\left(x_{-1}+{\displaystyle \frac{\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right){\gamma }_0+{\gamma }_1}{({\beta }_0+v_0)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)-1}}\right){\left(({\beta }_0+v_0)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+{\displaystyle \frac{\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right){\gamma }_0+{\gamma }_1}{1-({\beta }_0+v_0)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)}},$$

(35)

$$x_{2n}=\left(x_0+{\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{({\beta }_0+v_0)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)-1}}\right){\left(({\beta }_0+v_0)\left({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}\right)\right)}^n+{\displaystyle \frac{\left({\beta }_0+v_0\right){\gamma }_1+{\gamma }_0}{1-({\beta }_0+v_0)\left({\beta }_1+\frac{{\alpha }_0}{v_0}\right)}}.$$

(36)

1.

Taking in (35) and (36),

$$x_0+{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}}=x_{-1}+{\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}}=0,$$

we obtain

$$x_{2n}={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}}=x_0,x_{2n-1}={\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}}=x_{-1},n\in {\mathbb{N}}_0,$$

that is, the solution is 2-periodic, and it will be prime provided that $x_{-1}\ne x_0$.

2.

Now, if

$$x_0+{\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}}\ne 0\ne x_{-1}+{\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})-1}},$$

using (33) and taking limits in (35) and (36), we get

$$\underset{n\to +\infty }{lim}{x}_{2n}={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}},\underset{n\to +\infty }{lim}{x}_{2n-1}={\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}}.$$

□

4. Numerical Examples

In this section, we present a series of numerical examples, together with their corresponding graphical representations, to illustrate the qualitative behavior of the solutions to Equation (5) when ${\alpha }_1={\alpha }_0$. The numerical results are in full agreement with the theoretical analysis, confirming that the periodicity and asymptotic behavior of the solutions depend on the values of the parameter $({\beta }_0+v_0)({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}})$.

Example 1.

Consider Equation (5), and choose the parameters and initial values as follows:

$${\alpha }_0=1,{\beta }_0=1,{\beta }_1=2,{\gamma }_0=5,{\gamma }_1=7,x_0=5,x_{-1}=-1-\sqrt{2}.$$

For this choice, we have $({\beta }_0+v_0)({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}})=0$, and the solution takes the form

$$x_{2n-1}=({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1=12,n=1,2,\dots ,$$

$$x_{2n}=({\beta }_0+v_0){\gamma }_1+{\gamma }_0=5,n=0,1,\dots ,$$

that is, the solution is eventually prime 2-periodic and this confirms the results of Lemma 1 and Theorem 2, case 2. The graph of the solution is presented in Figure 1.

Example 2.

Consider Equation (5), and choose the parameters and initial values as follows:

$${\alpha }_0=1,{\beta }_0=-{\displaystyle \frac{4}{3}},{\beta }_1=2,{\gamma }_0=-{\displaystyle \frac{4}{3}},{\gamma }_1=2,x_0=-1,x_{-1}=-1.$$

For this choice, we have $({\beta }_0+v_0)({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}})=-1$, and the solution takes the form

$$x_n={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{2}}=-1,n=0,1,\dots ,$$

that is, the solution is prime 1-periodic and this confirms the results of Theorem 3, case 1. The graph of the solution is presented in Figure 2.

Example 3.

Consider Equation (5), and choose the parameters and initial values as follows:

$${\alpha }_0=-5,{\beta }_0=2,{\beta }_1=-6,{\gamma }_0=-2,{\gamma }_1=4,x_0=1,x_{-1}=3.$$

For this choice, we have $({\beta }_0+v_0)({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}})=-1$, condition (29) is satisfied, and the solution takes the form

$$x_{2n-1}=3,x_{2n}=1,n=0,1,\dots ,$$

that is, the solution is prime 2-periodic and this confirms the results of Theorem 3, case 2. The graph of the solution is presented in Figure 3.

Example 4.

Consider Equation (5), and choose the parameters and initial values as follows:

$${\alpha }_0=-5,{\beta }_0=7,{\beta }_1=-4,{\gamma }_0=2,{\gamma }_1=-{\displaystyle \frac{17}{14}}+{\displaystyle \frac{2}{7}}\sqrt{29},x_0={\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{4}}\sqrt{29},x_{-1}=-{\displaystyle \frac{41}{14}}+{\displaystyle \frac{4}{7}}\sqrt{29}.$$

For this choice, we have $({\beta }_0+v_0)({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}})=-1$, conditions (28) and (29) are not satisfied, and the solution takes the form

$$x_{4n-1}=-{\displaystyle \frac{41}{14}}+{\displaystyle \frac{4}{7}}\sqrt{29},x_{4n}={\displaystyle \frac{3}{2}}+{\displaystyle \frac{1}{4}}\sqrt{29},x_{4n+1}=-{\displaystyle \frac{12}{7}}+{\displaystyle \frac{2}{7}}\sqrt{29},x_{4n+2}=1,n=0,1,\dots ,$$

that is, the solution is prime 4-periodic and this confirms the results of Corollary 1. The graph of the solution is presented in Figure 4.

Example 5.

Consider Equation (5), and choose the parameters and initial values as follows:

$${\alpha }_0=7,{\beta }_0=2,{\beta }_1=8,{\gamma }_0={\displaystyle \frac{1}{2}},{\gamma }_1=-{\displaystyle \frac{1}{2}},x_0={\displaystyle \frac{1}{4}},x_{-1}=-{\displaystyle \frac{1}{4}}.$$

For this choice, we have $({\beta }_0+v_0)({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}})=1$, $({\beta }_0+v_0){\gamma }_1+{\gamma }_0=0=({\beta }_1+{\displaystyle \frac{{\alpha }_0}{v_0}}){\gamma }_0+{\gamma }_1$, and the solution takes the form

$$x_{2n-1}=-{\displaystyle \frac{1}{4}},x_{2n}={\displaystyle \frac{1}{4}},n=0,1,\dots ,$$

that is, the solution is prime 2-periodic and this confirms the results of Theorem 4. The graph of the solution is presented in Figure 5.

Example 6.

Consider Equation (5), and choose the parameters and initial values as follows:

$${\alpha }_0=1,{\beta }_0=6,{\beta }_1={\displaystyle \frac{4}{5}},{\gamma }_0=-3,{\gamma }_1=-7,x_0=8,x_{-1}=1.$$

For this choice, condition (32) is satisfied, and the solution takes the form

$$x_{2n-1}=1,x_{2n}=8,n=0,1,\dots ,$$

that is, the solution is prime 2-periodic and this confirms the results of Theorem 5, case 1. The graph of the solution is presented in Figure 6.

Example 7.

Consider Equation (5), and choose the parameters and initial values as follows:

$${\alpha }_0={\displaystyle \frac{1}{25}},{\beta }_0={\displaystyle \frac{1}{5}},{\beta }_1={\displaystyle \frac{8}{17}},{\gamma }_0=3,{\gamma }_1=-12,x_0=9,x_{-1}=5.$$

For this choice, conditions (33) and (34) are satisfied, and the solution satisfies

$$\underset{n\to +\infty }{lim}{x}_{2n-1}={\displaystyle \frac{({\beta }_1+\frac{{\alpha }_0}{v_0}){\gamma }_0+{\gamma }_1}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}}=-{\displaystyle \frac{22245}{823}}\simeq -27.02916160,$$

$$\underset{n\to +\infty }{lim}{x}_{2n}={\displaystyle \frac{({\beta }_0+v_0){\gamma }_1+{\gamma }_0}{1-({\beta }_0+v_0)({\beta }_1+\frac{{\alpha }_0}{v_0})}}=-{\displaystyle \frac{24225}{823}}\simeq -29.43499392,n=0,1,\dots ,$$

and this confirms the results of Theorem 5, case 2. The graphs of the solution are presented in Figure 7 and Figure 8.

5. Conclusions

In this work, explicit formulas for the solutions of Equation (5) are derived. This representation enables a detailed analysis of both the existence of periodic solutions and their asymptotic behavior. Several numerical examples are also presented to illustrate and confirm the validity of the results.

Open Problem: For interested readers and as a generalization of Equation (5), we propose to study the following second-order system of non-linear difference equations defined as follows:

$$x_{n+1}={\displaystyle \frac{a_n{x}_n{y}_{n-1}}{y_n-{\beta }_n{y}_{n-1}-{\gamma }_n}}+b_{n+1}{x}_n+c_{n+1},y_{n+1}={\displaystyle \frac{{\alpha }_n{y}_n{x}_{n-1}}{x_n-b_n{x}_{n-1}-c_n}}+{\beta }_{n+1}{y}_n+{\gamma }_{n+1},$$

(37)

where $n\in {\mathbb{N}}_0$, $\left(a_n\right)$, $\left({\alpha }_n\right)$, $\left(b_n\right)$, $\left({\beta }_n\right)$, $\left(c_n\right)$$\left({\gamma }_n\right)$ are periodic sequences of real numbers of period two, and the initial values $x_{-1}$, $y_{-1}$, $x_0$, $y_0$ are non-zero real numbers. System (37) can be written as follows:

$${\displaystyle \frac{x_{n+1}-b_{n+1}{x}_n-c_{n+1}}{x_n}}={\displaystyle \frac{a_n}{\frac{y_n-{\beta }_n{y}_{n-1}-{\gamma }_n}{y_{n-1}}}},{\displaystyle \frac{y_{n+1}-{\beta }_{n+1}{y}_n-{\gamma }_{n+1}}{y_n}}={\displaystyle \frac{{\alpha }_n}{\frac{x_n-b_n{x}_{n-1}-c_n}{x_{n-1}}}},n\in {\mathbb{N}}_0,$$

(38)

which leads to the following equivalent system:

$$u_{n+1}={\displaystyle \frac{a_n}{v_n}},v_{n+1}={\displaystyle \frac{{\alpha }_n}{u_n}},n\in {\mathbb{N}}_0,$$

(39)

where

$$u_n={\displaystyle \frac{x_n-b_n{x}_{n-1}-c_n}{x_{n-1}}},v_n={\displaystyle \frac{y_n-{\beta }_n{y}_{n-1}-{\gamma }_n}{y_{n-1}}}.$$

It is not hard to see that the solutions of system (39) are

$$u_{2n}={\left({\displaystyle \frac{a_1}{{\alpha }_0}}\right)}^n{u}_0,u_{2n+1}={\left({\displaystyle \frac{a_0}{{\alpha }_1}}\right)}^n{\displaystyle \frac{a_0}{v_0}},v_{2n}={\left({\displaystyle \frac{{\alpha }_1}{a_0}}\right)}^n{v}_0,v_{2n+1}={\left({\displaystyle \frac{{\alpha }_0}{a_1}}\right)}^n{\displaystyle \frac{{\alpha }_0}{u_0}},n\in {\mathbb{N}}_0.$$

Following the same approach applied for Equation (5), we can get explicit formulas of the solutions of system (37) and then use them to characterize their behavior.