Theory and Modern Applications

On the solutions of some nonlinear systems of difference equations

Abstract

In this paper, we deal with the existence of solutions and the periodicity character of the following systems of rational difference equations with order three:

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(±1±{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(±1±{y}_{n}{x}_{n-2}\right)},$

where the initial conditions ${x}_{-2}$, ${x}_{-1}$, ${x}_{0}$, ${y}_{-2}$, ${y}_{-1}$ and ${y}_{0}$ are nonzero real numbers.

MSC:39A10.

1 Introduction

Recently, rational difference equations have attracted the attention of many researchers for various reasons. On the one hand, they provide examples of nonlinear equations which are, in some cases, treatable but their dynamics present some new features with respect to the linear case. On the other hand, rational equations frequently appear in some biological models. Hence, their study is of interest also due to their applications. A good example of both facts is Ricatti difference equations because the richness of the dynamics of Ricatti equations is very well known (see, e.g., [1, 2]), and a particular case of these equations provides the classical Beverton-Holt model on the dynamics of exploited fish populations . Obviously, higher-order rational difference equations and systems of rational equations have also been widely studied but still have many aspects to be investigated. The reader can find in the following books , and works cited therein, many results, applications, and open problems on higher-order equations and rational systems.

A preliminary study of planar rational systems can be found in the paper  by Camouzis et al. In the work, they give some results and provide some open questions for systems of equations of the following type:

${x}_{n+1}=\frac{{\alpha }_{1}+{\beta }_{1}{x}_{n}+{\gamma }_{1}{y}_{n}}{{A}_{1}+{B}_{1}{x}_{n}+{C}_{1}{y}_{n}},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{\alpha }_{2}+{\beta }_{2}{x}_{n}+{\gamma }_{2}{y}_{n}}{{A}_{2}+{B}_{2}{x}_{n}+{C}_{2}{y}_{n}},\phantom{\rule{1em}{0ex}}n=0,1,2,\dots ,$

where the parameters are taken to be nonnegative. As shown in the cited paper, some of these systems can be reduced to some Ricatti equations or to some previously studied second-order rational equations. Furthermore, for some choices of the parameters, one obtains a system which is equivalent to the case with some other parameters. Camouzis et al. arrived at a list of 325 non-equivalent systems that should be focused on. They list such systems as pairs k, l, where k and l make reference to the number of the corresponding equation in their Tables 3 and 4. There are many papers that are related to the difference equations systems. For example, the periodicity of the positive solutions of the rational difference equations systems

${x}_{n+1}=\frac{m}{{y}_{n}},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{p{y}_{n}}{{x}_{n-1}{y}_{n-1}}$

has been obtained by Cinar in .

In  Clark and Kulenovic investigated the global asymptotic stability of the system

${x}_{n+1}=\frac{{x}_{n}}{a+c{y}_{n}},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}}{b+d{x}_{n}}.$

The behavior of the positive solutions of the following system:

${x}_{n+1}=\frac{{x}_{n-1}}{1+{x}_{n-1}{y}_{n}},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n-1}}{1+{y}_{n-1}{x}_{n}}$

has been studied by Kurbanli et al. .

Touafek et al.  investigated the periodic nature and got the form of the solutions of the following systems of rational difference equations:

${x}_{n+1}=\frac{{x}_{n-3}}{±1±{x}_{n-3}{y}_{n-1}},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n-3}}{±1±{y}_{n-3}{x}_{n-1}}.$

In  Yalçınkaya investigated the sufficient conditions for the global asymptotic stability of the following system of difference equations:

${x}_{n+1}=\frac{{x}_{n}+{y}_{n-1}}{{x}_{n}{y}_{n-1}-1},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}+{x}_{n-1}}{{y}_{n}{x}_{n-1}-1}.$

Similarly, difference equations and nonlinear systems of the rational difference equations were investigated, see .

In this paper, we investigate the periodic nature and the form of the solutions of some nonlinear difference equations systems of order three

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(±1±{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(±1±{y}_{n}{x}_{n-2}\right)},$

where the initial conditions ${x}_{-2}$, ${x}_{-1}$, ${x}_{0}$, ${y}_{-2}$, ${y}_{-1}$ and ${y}_{0}$ are nonzero real numbers.

Definition (Periodicity)

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

2 On the solution of the system ${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1+{x}_{n}{y}_{n-2}\right)}$, ${y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(1+{y}_{n}{x}_{n-2}\right)}$

In this section, we investigate the solutions of the two difference equations system

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1+{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(1+{y}_{n}{x}_{n-2}\right)},$
(1)

where $n\in {\mathbb{N}}_{0}$ and the initial conditions are arbitrary nonzero real numbers.

Theorem 1 Assume that $\left\{{x}_{n},{y}_{n}\right\}$ are solutions of system (1). Then for $n=0,1,2,\dots$ , we see that all solutions of system (1) are given by the following formulas:

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i\right)cd\right)}{\left(1+\left(4i+2\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+1\right)cd\right)}{\left(1+\left(4i+3\right)af\right)},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+2\right)cd\right)}{\left(1+\left(4i+4\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}}{e{c}^{n}{d}^{n}\left(1+af\right)}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+3\right)cd\right)}{\left(1+\left(4i+5\right)af\right)}\hfill \end{array}$

and

$\begin{array}{c}{y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i\right)af\right)}{\left(1+\left(4i+2\right)cd\right)},\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+1\right)af\right)}{\left(1+\left(4i+3\right)cd\right)},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+2\right)af\right)}{\left(1+\left(4i+4\right)cd\right)},\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}}{b{a}^{n}{f}^{n}\left(1+cd\right)}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+3\right)af\right)}{\left(1+\left(4i+5\right)cd\right)},\hfill \end{array}$

where ${x}_{-2}=c$, ${x}_{-1}=b$, ${x}_{0}=a$, ${y}_{-2}=f$, ${y}_{-1}=e$ and ${y}_{0}=d$.

Proof For $n=0$ the result holds. Now suppose that $n>0$ and that our assumption holds for $n-1$, that is,

$\begin{array}{c}{x}_{4n-6}=\frac{{a}^{n-1}{f}^{n-1}}{{c}^{n-2}{d}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i\right)cd\right)}{\left(1+\left(4i+2\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-5}=\frac{b{a}^{n-1}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)cd\right)}{\left(1+\left(4i+3\right)af\right)},\hfill \\ {x}_{4n-4}=\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)cd\right)}{\left(1+\left(4i+4\right)af\right)},\hfill \\ {x}_{4n-3}=\frac{{a}^{n}{f}^{n}}{e{c}^{n-1}{d}^{n-1}\left(1+af\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+3\right)cd\right)}{\left(1+\left(4i+5\right)af\right)},\hfill \\ {y}_{4n-6}=\frac{{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-2}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i\right)af\right)}{\left(1+\left(4i+2\right)cd\right)},\phantom{\rule{2em}{0ex}}{y}_{4n-5}=\frac{e{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)af\right)}{\left(1+\left(4i+3\right)cd\right)},\hfill \\ {y}_{4n-4}=\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)af\right)}{\left(1+\left(4i+4\right)cd\right)},\hfill \\ {y}_{4n-3}=\frac{{c}^{n}{d}^{n}}{b{a}^{n-1}{f}^{n-1}\left(1+cd\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+3\right)af\right)}{\left(1+\left(4i+5\right)cd\right)}.\hfill \end{array}$

Now we find from Eq. (1) that

$\begin{array}{r}{x}_{4n-2}=\frac{{x}_{4n-3}{y}_{4n-5}}{{y}_{4n-4}\left(1+{x}_{4n-3}{y}_{4n-5}\right)}\\ \phantom{{x}_{4n-2}}=\left(\frac{{a}^{n}{f}^{n}}{e{c}^{n-1}{d}^{n-1}\left(1+af\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+3\right)cd\right)}{\left(1+\left(4i+5\right)af\right)}\right)\left(\frac{e{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)af\right)}{\left(1+\left(4i+3\right)cd\right)}\right)\\ \phantom{{x}_{4n-2}=}/\left(\left(\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)af\right)}{\left(1+\left(4i+4\right)cd\right)}\right)\\ \phantom{{x}_{4n-2}=}×\left(1+\left(\frac{{a}^{n}{f}^{n}}{e{c}^{n-1}{d}^{n-1}\left(1+af\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+3\right)cd\right)}{\left(1+\left(4i+5\right)af\right)}\right)\\ \phantom{{x}_{4n-2}=}×\left(\frac{e{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)af\right)}{\left(1+\left(4i+3\right)cd\right)}\right)\right)\right)\\ \phantom{{x}_{4n-2}}=\frac{\left(\frac{af}{\left(1+af\right)}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)af\right)}{\left(1+\left(4i+5\right)af\right)}\right)}{\left(\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)af\right)}{\left(1+\left(4i+4\right)cd\right)}\right)\left(1+\frac{af}{\left(1+af\right)}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)af\right)}{\left(1+\left(4i+5\right)af\right)}\right)}\\ \phantom{{x}_{4n-2}}=\frac{\left(\frac{af}{\left(1+\left(4n-3\right)af\right)}\right)}{\left(\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)af\right)}{\left(1+\left(4i+4\right)cd\right)}\right)\left(1+\frac{af}{\left(1+\left(4n-3\right)af\right)}\right)}\\ \phantom{{x}_{4n-2}}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}\left(1+\left(4n-2\right)af\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+4\right)cd\right)}{\left(1+\left(4i+2\right)af\right)}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i\right)cd\right)}{\left(1+\left(4i+2\right)af\right)},\\ {y}_{4n-2}=\frac{{y}_{4n-3}{x}_{4n-5}}{{x}_{4n-4}\left(1+{y}_{4n-3}{x}_{4n-5}\right)}\\ \phantom{{y}_{4n-2}}=\left(\frac{{c}^{n}{d}^{n}}{b{a}^{n-1}{f}^{n-1}\left(1+cd\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+3\right)af\right)}{\left(1+\left(4i+5\right)cd\right)}\right)\left(\frac{b{a}^{n-1}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)cd\right)}{\left(1+\left(4i+3\right)af\right)}\right)\\ \phantom{{y}_{4n-2}=}/\left(\left(\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)cd\right)}{\left(1+\left(4i+4\right)af\right)}\right)\\ \phantom{{y}_{4n-2}=}×\left(1+\left(\frac{{c}^{n}{d}^{n}}{b{a}^{n-1}{f}^{n-1}\left(1+cd\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+3\right)af\right)}{\left(1+\left(4i+5\right)cd\right)}\right)\\ \phantom{{y}_{4n-2}=}×\left(\frac{b{a}^{n-1}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)cd\right)}{\left(1+\left(4i+3\right)af\right)}\right)\right)\right)\\ \phantom{{y}_{4n-2}}=\frac{\frac{cd}{\left(1+cd\right)}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)cd\right)}{\left(1+\left(4i+5\right)cd\right)}}{\left(\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)cd\right)}{\left(1+\left(4i+4\right)af\right)}\right)\left(1+\frac{cd}{\left(1+cd\right)}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+1\right)cd\right)}{\left(1+\left(4i+5\right)cd\right)}\right)}\\ \phantom{{y}_{4n-2}}=\frac{\left(\frac{cd}{\left(1+\left(4n-3\right)cd\right)}\right)}{\left(\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}{\prod }_{i=0}^{n-2}\frac{\left(1+\left(4i+2\right)cd\right)}{\left(1+\left(4i+4\right)af\right)}\right)\left(1+\frac{cd}{\left(1+\left(4n-3\right)cd\right)}\right)}\\ \phantom{{y}_{4n-2}}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}\left(1+\left(4n-2\right)cd\right)}\prod _{i=0}^{n-2}\frac{\left(1+\left(4i+4\right)af\right)}{\left(1+\left(4i+2\right)cd\right)}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i\right)af\right)}{\left(1+\left(4i+2\right)cd\right)}.\end{array}$

Also, we can prove the other relations. The proof is complete. □

The following theorem can be proved similarly.

Theorem 2 Assume that $\left\{{x}_{n},{y}_{n}\right\}$ are solutions of the following system:

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1+{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(1-{y}_{n}{x}_{n-2}\right)}.$

Then for $n=0,1,2,\dots$ ,

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}\prod _{i=0}^{n-1}\frac{\left(1-\left(4i\right)cd\right)}{\left(1+\left(4i+2\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{\left(1-\left(4i+1\right)cd\right)}{\left(1+\left(4i+3\right)af\right)},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{\left(1-\left(4i+2\right)cd\right)}{\left(1+\left(4i+4\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}}{e{c}^{n}{d}^{n}\left(1+af\right)}\prod _{i=0}^{n-1}\frac{\left(1-\left(4i+3\right)cd\right)}{\left(1+\left(4i+5\right)af\right)},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i\right)af\right)}{\left(1-\left(4i+2\right)cd\right)},\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+1\right)af\right)}{\left(1-\left(4i+3\right)cd\right)},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+2\right)af\right)}{\left(1-\left(4i+4\right)cd\right)},\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}}{b{a}^{n}{f}^{n}\left(1-cd\right)}\prod _{i=0}^{n-1}\frac{\left(1+\left(4i+3\right)af\right)}{\left(1-\left(4i+5\right)cd\right)}.\hfill \end{array}$

Example 1 For confirming the results of this section, we consider numerical example for the difference system (1) with the initial conditions ${x}_{-2}=3$, ${x}_{-1}=5$, ${x}_{0}=-1$, ${y}_{-2}=-0.4$, ${y}_{-1}=0.16$ and ${y}_{0}=7$.See Figure 1.

Example 2 We assumethat the initial conditions for the difference system (1) are ${x}_{-2}=-0.2$, ${x}_{-1}=0.15$, ${x}_{0}=-0.51$, ${y}_{-2}=-0.23$, ${y}_{-1}=0.16$ and ${y}_{0}=-0.7$. See Figure 2.

3 On the solution of the system ${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1+{x}_{n}{y}_{n-2}\right)}$, ${y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1+{y}_{n}{x}_{n-2}\right)}$

In this section, we obtain the form of the solutions of the two difference equations system

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1+{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1+{y}_{n}{x}_{n-2}\right)},$
(2)

where $n\in {\mathbb{N}}_{0}$ and the initial conditions are arbitrary nonzero real numbers with ${x}_{-2}{y}_{0}\ne 1$.

Theorem 3 Let ${\left\{{x}_{n},{y}_{n}\right\}}_{n=-2}^{+\mathrm{\infty }}$ be solutions of system (2). Then ${\left\{{x}_{n}\right\}}_{n=-2}^{+\mathrm{\infty }}$ and ${\left\{{y}_{n}\right\}}_{n=-2}^{+\mathrm{\infty }}$ are given by the formula for $n=0,1,2,\dots$ ,

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1+\left(4i+2\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1+cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1+\left(4i+3\right)af\right)},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1+\left(4i+4\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}}{e{c}^{n}{d}^{n}\left(1+af\right)}\frac{{\left(-1+cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1+\left(4i+5\right)af\right)},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}\prod _{i=0}^{n-1}\left(1+\left(4i\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1+\left(4i+1\right)af\right)}{{\left(-1+cd\right)}^{n}},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\left(1+\left(4i+2\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}}{b{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1+\left(4i+3\right)af\right)}{{\left(-1+cd\right)}^{n+1}}.\hfill \end{array}$

Proof For $n=0$ the result holds. Now suppose that $n>0$ and that our assumption holds for $n-1$, that is,

$\begin{array}{c}{x}_{4n-5}=\frac{b{a}^{n-1}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\frac{{\left(-1+cd\right)}^{n-1}}{{\prod }_{i=0}^{n-2}\left(1+\left(4i+3\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-4}=\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\prod _{i=0}^{n-2}\frac{1}{\left(1+\left(4i+4\right)af\right)},\hfill \\ {x}_{4n-3}=\frac{{a}^{n}{f}^{n}}{e{c}^{n-1}{d}^{n-1}\left(1+af\right)}\frac{{\left(-1+cd\right)}^{n-1}}{{\prod }_{i=0}^{n-2}\left(1+\left(4i+5\right)af\right)},\hfill \\ {y}_{4n-5}=\frac{e{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-1}}\frac{{\prod }_{i=0}^{n-2}\left(1+\left(4i+1\right)af\right)}{{\left(-1+cd\right)}^{n-1}},\hfill \\ {y}_{4n-4}=\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}\prod _{i=0}^{n-2}\left(1+\left(4i+2\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n-3}=\frac{{c}^{n}{d}^{n}}{b{a}^{n-1}{f}^{n-1}}\frac{{\prod }_{i=0}^{n-2}\left(1+\left(4i+3\right)af\right)}{{\left(-1+cd\right)}^{n}}.\hfill \end{array}$

Now, we obtain from Eq. (2) that

$\begin{array}{rcl}{x}_{4n-2}& =& \frac{{x}_{4n-3}{y}_{4n-5}}{{y}_{4n-4}\left(1+{x}_{4n-3}{y}_{4n-5}\right)}\\ =& \left(\frac{{a}^{n}{f}^{n}}{e{c}^{n-1}{d}^{n-1}\left(1+af\right)}\frac{{\left(-1+cd\right)}^{n-1}}{{\prod }_{i=0}^{n-2}\left(1+\left(4i+5\right)af\right)}\right)\left(\frac{e{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-1}}\frac{{\prod }_{i=0}^{n-2}\left(1+\left(4i+1\right)af\right)}{{\left(-1+cd\right)}^{n-1}}\right)\\ /\left(\left(\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}\prod _{i=0}^{n-2}\left(1+\left(4i+2\right)af\right)\right)\\ ×\left(1+\left(\frac{{a}^{n}{f}^{n}}{e{c}^{n-1}{d}^{n-1}\left(1+af\right)}\frac{{\left(-1+cd\right)}^{n-1}}{{\prod }_{i=0}^{n-2}\left(1+\left(4i+5\right)af\right)}\right)\\ ×\left(\frac{e{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-1}}\frac{{\prod }_{i=0}^{n-2}\left(1+\left(4i+1\right)af\right)}{{\left(-1+cd\right)}^{n-1}}\right)\right)\right)\\ =& \frac{\left(\frac{af}{\left(1+\left(4n-3\right)af\right)}\right)}{\left(\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}{\prod }_{i=0}^{n-2}\left(1+\left(4i+2\right)af\right)\right)\left(1+\frac{af}{\left(1+\left(4n-3\right)af\right)}\right)}\\ =& \frac{{a}^{n-1}{f}^{n-1}af}{\left({c}^{n-1}{d}^{n}{\prod }_{i=0}^{n-2}\left(1+\left(4i+2\right)af\right)\right)\left(1+\left(4n-3\right)af+af\right)}\\ =& \frac{{a}^{n}{f}^{n}}{\left({c}^{n-1}{d}^{n}{\prod }_{i=0}^{n-2}\left(1+\left(4i+2\right)af\right)\right)\left(1+\left(4n-2\right)af\right)}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}{\prod }_{i=0}^{n-1}\left(1+\left(4i+2\right)af\right)}.\end{array}$

Also, we can prove the other relations. This completes the proof. □

We consider the following systems, and the proof of the theorems is similar to the above theorem, and so it is left to the reader,

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1+{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1-{y}_{n}{x}_{n-2}\right)},$
(3)
${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1-{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1+{y}_{n}{x}_{n-2}\right)},$
(4)
${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(1-{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1-{y}_{n}{x}_{n-2}\right)}.$
(5)

The following theorems are devoted to the expressions of the form of the solutions of systems (3), (4), (5) with ${x}_{-2}=c$, ${x}_{-1}=b$, ${x}_{0}=a$, ${y}_{-2}=f$, ${y}_{-1}=e$ and ${y}_{0}=d$.

Theorem 4 Let ${\left\{{x}_{n},{y}_{n}\right\}}_{n=-2}^{+\mathrm{\infty }}$ be solutions of system (3) and ${x}_{-2}{y}_{0}\ne -1$. Then for $n=0,1,2,\dots$ ,

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1+\left(4i+2\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1-cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1+\left(4i+3\right)af\right)},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1+\left(4i+4\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}}{e{c}^{n}{d}^{n}\left(1+af\right)}\frac{{\left(-1-cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1+\left(4i+5\right)af\right)},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}\prod _{i=0}^{n-1}\left(1+\left(4i\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1+\left(4i+1\right)af\right)}{{\left(-1-cd\right)}^{n}},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\left(1+\left(4i+2\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}}{b{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1+\left(4i+3\right)af\right)}{{\left(-1-cd\right)}^{n+1}}.\hfill \end{array}$

Theorem 5 Assume that $\left\{{x}_{n},{y}_{n}\right\}$ are solutions of system (4) with ${x}_{-2}{y}_{0}\ne 1$. Then for $n=0,1,2,\dots$ ,

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1-\left(4i+2\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1+cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1-\left(4i+3\right)af\right)},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1-\left(4i+4\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}}{e{c}^{n}{d}^{n}\left(1+af\right)}\frac{{\left(-1+cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1-\left(4i+5\right)af\right)},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}\prod _{i=0}^{n-1}\left(1-\left(4i\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1-\left(4i+1\right)af\right)}{{\left(-1+cd\right)}^{n}},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\left(1-\left(4i+2\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}}{b{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1-\left(4i+3\right)af\right)}{{\left(-1+cd\right)}^{n+1}}.\hfill \end{array}$

Theorem 6 Suppose that $\left\{{x}_{n},{y}_{n}\right\}$ are solutions of system (5) where ${x}_{-2}{y}_{0}\ne -1$. Then for $n=0,1,2,\dots$ ,

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1-\left(4i+2\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1-cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1-\left(4i+3\right)af\right)},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}}\prod _{i=0}^{n-1}\frac{1}{\left(1-\left(4i+4\right)af\right)},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}}{e{c}^{n}{d}^{n}\left(1-af\right)}\frac{{\left(-1-cd\right)}^{n}}{{\prod }_{i=0}^{n-1}\left(1-\left(4i+5\right)af\right)},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}\prod _{i=0}^{n-1}\left(1-\left(4i\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1-\left(4i+1\right)af\right)}{{\left(-1-cd\right)}^{n}},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}}\prod _{i=0}^{n-1}\left(1-\left(4i+2\right)af\right),\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}}{b{a}^{n}{f}^{n}}\frac{{\prod }_{i=0}^{n-1}\left(1-\left(4i+3\right)af\right)}{{\left(-1-cd\right)}^{n+1}}.\hfill \end{array}$

Example 3 We consider an interesting numerical example for the difference system (2) with the initial conditions ${x}_{-2}=0.2$, ${x}_{-1}=0.15$, ${x}_{0}=-0.11$, ${y}_{-2}=0.23$, ${y}_{-1}=0.16$ and ${y}_{0}=0.17$. SeeFigure 3.

Example 4 See Figure 4, where we take system (3) with the initial conditions ${x}_{-2}=0.12$, ${x}_{-1}=0.15$, ${x}_{0}=0.11$, ${y}_{-2}=0.3$, ${y}_{-1}=-0.6$ and ${y}_{0}=0.17$.

4 On the solution of the system ${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(-1+{x}_{n}{y}_{n-2}\right)}$, ${y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1+{y}_{n}{x}_{n-2}\right)}$

In this section, we get the form of the solutions of the difference equations system

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(-1+{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1+{y}_{n}{x}_{n-2}\right)},$
(6)

where $n=0,1,2,\dots$ and the initial conditions ${x}_{-2}$, ${x}_{-1}$, ${x}_{0}$, ${y}_{-2}$, ${y}_{-1}$ and ${y}_{0}$ are arbitrary nonzero real numbers with ${x}_{0}{y}_{-2}$, ${x}_{-2}{y}_{0}\ne 1$.

Theorem 7 If $\left\{{x}_{n},{y}_{n}\right\}$ are solutions of difference equation system (6), then for $n=0,1,2,\dots$ ,

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1+cd\right)}^{n}}{{\left(-1+af\right)}^{n}},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}{\left(-1+cd\right)}^{n}}{e{c}^{n}{d}^{n}{\left(-1+af\right)}^{n+1}},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}},\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\left(-1+af\right)}^{n}}{{\left(-1+cd\right)}^{n}},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}},\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}{\left(-1+af\right)}^{n}}{b{a}^{n}{f}^{n}{\left(-1+cd\right)}^{n+1}}.\hfill \end{array}$

Proof For $n=0$ the result holds. Now, suppose that $n>1$ and that our assumption holds for $n-1$, that is,

$\begin{array}{c}{x}_{4n-6}=\frac{{a}^{n-1}{f}^{n-1}}{{c}^{n-2}{d}^{n-1}},\phantom{\rule{2em}{0ex}}{x}_{4n-5}=\frac{b{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n-1}}{{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n-1}},\hfill \\ {x}_{4n-4}=\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}},\phantom{\rule{2em}{0ex}}{x}_{4n-3}=\frac{{a}^{n}{f}^{n}{\left(-1+cd\right)}^{n-1}}{e{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n}},\hfill \\ {y}_{4n-6}=\frac{{c}^{n-1}{d}^{n-1}}{{a}^{n-1}{f}^{n-2}},\phantom{\rule{2em}{0ex}}{y}_{4n-5}=\frac{e{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n-1}}{{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n-1}},\hfill \\ {y}_{4n-4}=\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}},\phantom{\rule{2em}{0ex}}{y}_{4n-3}=\frac{{c}^{n}{d}^{n}{\left(-1+af\right)}^{n-1}}{b{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n}}.\hfill \end{array}$

Now, we conclude from Eq. (6) that

$\begin{array}{c}\begin{array}{rl}{x}_{4n-2}& =\frac{{x}_{4n-3}{y}_{4n-5}}{{y}_{4n-4}\left(-1+{x}_{4n-3}{y}_{4n-5}\right)}=\frac{\left(\frac{{a}^{n}{f}^{n}{\left(-1+cd\right)}^{n-1}}{e{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n}}\right)\left(\frac{e{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n-1}}{{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n-1}}\right)}{\left(\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}\right)\left(-1+\left(\frac{{a}^{n}{f}^{n}{\left(-1+cd\right)}^{n-1}}{e{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n}}\right)\left(\frac{e{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n-1}}{{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n-1}}\right)\right)}\\ =\frac{\left(\frac{af}{\left(-1+af\right)}\right)}{\left(\frac{{c}^{n-1}{d}^{n}}{{a}^{n-1}{f}^{n-1}}\right)\left(-1+\frac{af}{\left(-1+af\right)}\right)}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}},\end{array}\hfill \\ \begin{array}{rl}{y}_{4n-2}& =\frac{{y}_{4n-3}{x}_{4n-5}}{{x}_{4n-4}\left(-1+{y}_{4n-3}{x}_{4n-5}\right)}=\frac{\left(\frac{{c}^{n}{d}^{n}{\left(-1+af\right)}^{n-1}}{b{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n}}\right)\left(\frac{b{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n-1}}{{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n-1}}\right)}{\left(\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\right)\left(-1+\left(\frac{{c}^{n}{d}^{n}{\left(-1+af\right)}^{n-1}}{b{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n}}\right)\left(\frac{b{a}^{n-1}{f}^{n-1}{\left(-1+cd\right)}^{n-1}}{{c}^{n-1}{d}^{n-1}{\left(-1+af\right)}^{n-1}}\right)\right)}\\ =\frac{\left(\frac{cd}{\left(-1+cd\right)}\right)}{\left(\frac{{a}^{n}{f}^{n-1}}{{c}^{n-1}{d}^{n-1}}\right)\left(-1+\left(\frac{cd}{\left(-1+cd\right)}\right)\right)}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}.\end{array}\hfill \end{array}$

Also, we can prove the other relations. This completes the proof. □

We consider the following system, and the proof of the theorem is similar to the above mentioned theorem and so it is left to the reader,

${x}_{n+1}=\frac{{x}_{n}{y}_{n-2}}{{y}_{n-1}\left(-1+{x}_{n}{y}_{n-2}\right)},\phantom{\rule{2em}{0ex}}{y}_{n+1}=\frac{{y}_{n}{x}_{n-2}}{{x}_{n-1}\left(-1-{y}_{n}{x}_{n-2}\right)}.$
(7)

Theorem 8 Let ${\left\{{x}_{n},{y}_{n}\right\}}_{n=-2}^{+\mathrm{\infty }}$ be solutions of system (7) and ${x}_{0}{y}_{-2}\ne 1$, ${x}_{-2}{y}_{0}\ne -1$. Then for $n=0,1,2,\dots$ ,

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1-cd\right)}^{n}}{{\left(-1+af\right)}^{n}},\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}{\left(-1-cd\right)}^{n}}{e{c}^{n}{d}^{n}{\left(-1+af\right)}^{n+1}},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}},\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\left(-1+af\right)}^{n}}{{\left(-1-cd\right)}^{n}},\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}},\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}{\left(-1+af\right)}^{n}}{b{a}^{n}{f}^{n}{\left(-1-cd\right)}^{n+1}}.\hfill \end{array}$

Lemma 1 The solution of system (6) is unbounded except in the following case.

Theorem 9 System (6) has a periodic solution of period four iff $cd=af=2$ and it will take the following form: $\left\{{x}_{n}\right\}=\left\{c,b,a,\frac{af}{e},c,b,a,\dots \right\}$, $\left\{{y}_{n}\right\}=\left\{f,e,d,\frac{cd}{b},f,e,d,\dots \right\}$.

Proof First, suppose that a prime period four solution exists

$\left\{{x}_{n}\right\}=\left\{c,b,a,\frac{af}{e},c,b,a,\dots \right\},\phantom{\rule{2em}{0ex}}\left\{{y}_{n}\right\}=\left\{f,e,d,\frac{cd}{b},f,e,d,\dots \right\}$

of system (6). We see from the form of the solution of system (6) that

$\begin{array}{c}{x}_{4n-2}=c=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}},\phantom{\rule{2em}{0ex}}{x}_{4n-1}=b=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1+cd\right)}^{n}}{{\left(-1+af\right)}^{n}},\hfill \\ {x}_{4n}=a=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}},\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{af}{e}=\frac{{a}^{n+1}{f}^{n+1}{\left(-1+cd\right)}^{n}}{e{c}^{n}{d}^{n}{\left(-1+af\right)}^{n+1}},\hfill \\ {y}_{4n-2}=f=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}},\phantom{\rule{2em}{0ex}}{y}_{4n-1}=e=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\left(-1+af\right)}^{n}}{{\left(-1+cd\right)}^{n}},\hfill \\ {y}_{4n}=d=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}},\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{cd}{b}=\frac{{c}^{n+1}{d}^{n+1}{\left(-1+af\right)}^{n}}{b{a}^{n}{f}^{n}{\left(-1+cd\right)}^{n+1}}.\hfill \end{array}$

Then we get $cd=af=2$. Second, assume that $cd=af=2$. Then we see from the form of the solution of system (6) that

$\begin{array}{c}{x}_{4n-2}=\frac{{a}^{n}{f}^{n}}{{c}^{n-1}{d}^{n}}=c,\phantom{\rule{2em}{0ex}}{x}_{4n-1}=\frac{b{a}^{n}{f}^{n}}{{c}^{n}{d}^{n}}\frac{{\left(-1+cd\right)}^{n}}{{\left(-1+af\right)}^{n}}=b,\hfill \\ {x}_{4n}=\frac{{a}^{n+1}{f}^{n}}{{c}^{n}{d}^{n}}=a,\phantom{\rule{2em}{0ex}}{x}_{4n+1}=\frac{{a}^{n+1}{f}^{n+1}{\left(-1+cd\right)}^{n}}{e{c}^{n}{d}^{n}{\left(-1+af\right)}^{n+1}}=\frac{af}{e},\hfill \\ {y}_{4n-2}=\frac{{c}^{n}{d}^{n}}{{a}^{n}{f}^{n-1}}=f,\phantom{\rule{2em}{0ex}}{y}_{4n-1}=\frac{e{c}^{n}{d}^{n}}{{a}^{n}{f}^{n}}\frac{{\left(-1+af\right)}^{n}}{{\left(-1+cd\right)}^{n}}=e,\hfill \\ {y}_{4n}=\frac{{c}^{n}{d}^{n+1}}{{a}^{n}{f}^{n}}=c,\phantom{\rule{2em}{0ex}}{y}_{4n+1}=\frac{{c}^{n+1}{d}^{n+1}{\left(-1+af\right)}^{n}}{b{a}^{n}{f}^{n}{\left(-1+cd\right)}^{n+1}}=\frac{cd}{b}.\hfill \end{array}$

Thus, we have a periodic solution of period four and the proof is complete. □

Lemma 2 The solution of system (7) is unbounded except in the following case.

Theorem 10 System (7) has a periodic solution of period eight iff $cd=-2$, $af=2$ and it will take the following form: $\left\{{x}_{n}\right\}=\left\{c,b,a,\frac{af}{e},-c,-b,-a,-\frac{af}{e},c,b,a,\dots \right\}$, $\left\{{y}_{n}\right\}=\left\{f,e,d,\frac{cd}{b},-f,-e,-d,-\frac{cd}{b},f,e,d,\dots \right\}$.

Example 5 We consider a numerical example for the difference system (6) when we put the initial conditions ${x}_{-2}=0.12$, ${x}_{-1}=0.5$, ${x}_{0}=0.31$, ${y}_{-2}=0.23$, ${y}_{-1}=-0.6$ and ${y}_{0}=0.7$.See Figure 5.

Example 6 Figure 6 shows the behavior of the solution of the difference system (6)with the initial conditions ${x}_{-2}=4$, ${x}_{-1}=-7$, ${x}_{0}=-0.5$, ${y}_{-2}=-4$, ${y}_{-1}=6$ and ${y}_{0}=0.5$.

Example 7 We consider a numerical example for the difference system(7) when we put the initial conditions ${x}_{-2}=0.32$, ${x}_{-1}=0.25$, ${x}_{0}=-0.31$, ${y}_{-2}=0.23$, ${y}_{-1}=0.26$ and ${y}_{0}=0.17$. See Figure 7.

Example 8 Figure 8 shows the periodicity of the solution of the difference system (7) with the initial conditions ${x}_{-2}=9$, ${x}_{-1}=-7$, ${x}_{0}=0.4$, ${y}_{-2}=5$, ${y}_{-1}=8$ and ${y}_{0}=-2/9$.

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Acknowledgements

This work was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, under grant No. (130-028-D1433). The authors, therefore, acknowledge with thanks DSR technical and financial support.

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Correspondence to Elsayed M Elsayed.

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Elsayed, E.M., El-Metwally, H. On the solutions of some nonlinear systems of difference equations. Adv Differ Equ 2013, 161 (2013). https://doi.org/10.1186/1687-1847-2013-161

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• DOI: https://doi.org/10.1186/1687-1847-2013-161

Keywords

• difference equations
• periodic solution
• solution of difference equation
• system of difference equations 