Theory and Modern Applications

# On the behavior of solutions of the system of rational difference equations ${x}_{n+1}=\frac{{x}_{n-1}}{{y}_{n}{x}_{n-1}-1},\phantom{\rule{1em}{0ex}}{y}_{n+1}=\frac{{y}_{n-1}}{{x}_{n}{y}_{n-1}-1},\phantom{\rule{1em}{0ex}}{z}_{n+1}=\frac{1}{{y}_{n}{z}_{n}}$

## Abstract

In this article, we investigate the solutions of the system of difference equations ${x}_{n+1}=\frac{{x}_{n-1}}{{y}_{n}{x}_{n-1}-1}$, ${y}_{n+1}=\frac{{y}_{n-1}}{{x}_{n}{y}_{n-1}-1}$, ${z}_{n+1}=\frac{1}{{y}_{n}{z}_{n}}$ where x 0, x -1, y 0, y -1, z 0, z -1 real numbers such that y 0 x -1 ≠ 1, x 0 y -1 ≠ 1 and y 0 z 0 ≠ 0.

## 1. Introduction

In , Kurbanli et al. studied the behavior of positive solutions of the system of rational difference equations

${x}_{n+1}=\frac{{x}_{n-1}}{{y}_{n}{x}_{n-1}+1},\phantom{\rule{2.77695pt}{0ex}}{y}_{n+1}=\frac{{y}_{n-1}}{{x}_{n}{y}_{n-1}+1}.$

In , Cinar studied the solutions of the systems of difference equations

${x}_{n+1}=\frac{1}{{y}_{n}},\phantom{\rule{2.77695pt}{0ex}}{y}_{n+1}=\frac{{y}_{n}}{{x}_{n-1}{y}_{n-1}}.$

In , Kurbanli, studied the behavior of solutions of the system of rational difference equations

${x}_{n+1}=\frac{{x}_{n-1}}{{y}_{n}{x}_{n-1}-1},\phantom{\rule{2.77695pt}{0ex}}{y}_{n+1}=\frac{{y}_{n-1}}{{x}_{n}{y}_{n-1}-1},\phantom{\rule{2.77695pt}{0ex}}{z}_{n+1}=\frac{{z}_{n-1}}{{y}_{n}{z}_{n-1}-1}.$

In , Papaschinnopoulos and Schinas proved the boundedness, persistence, the oscillatory behavior, and the asymptotic behavior of the positive solutions of the system of difference equations

${x}_{n+1}=\sum _{i=0}^{k}{A}_{i}∕{y}_{n-i}^{{p}_{i}},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{y}_{n+1}=\sum _{i=0}^{k}{B}_{i}∕{x}_{n-i}^{{q}_{i}}$

In , Clark and Kulenović investigate the global stability properties and asymptotic behavior of solutions of the system of difference equations

${x}_{n+1}=\frac{{x}_{n}}{a+c{y}_{n}},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{y}_{n+1}=\frac{{y}_{n}}{b+d{x}_{n}}.$

In , Camouzis and Papaschinnopoulos studied the global asymptotic behavior of positive solutions of the system of rational difference equations

${x}_{n+1}=1+\frac{{x}_{n}}{{y}_{n-m}},\phantom{\rule{2.77695pt}{0ex}}{y}_{n+1}=1+\frac{{y}_{n}}{{x}_{n-m}}.$

In , Kulenović and Nurkanović studied the global asymptotic behavior of solutions of the system of difference equations

${x}_{n+1}=\frac{a+{x}_{n}}{b+{y}_{n}},\phantom{\rule{2.77695pt}{0ex}}{y}_{n+1}=\frac{c+{y}_{n}}{d+{z}_{n}},\phantom{\rule{2.77695pt}{0ex}}{z}_{n+1}=\frac{e+{z}_{n}}{f+{x}_{n}}.$

In , Özban studied the positive solutions of the system of rational difference equations

${x}_{n+1}=\frac{1}{{y}_{n-k}},\phantom{\rule{2.77695pt}{0ex}}{y}_{n+1}=\frac{{y}_{n}}{{x}_{n-m}{y}_{n-m-k}}.$

In , Zhang et al. investigated the behavior of the positive solutions of the system of the difference equations

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

In , Yalcinkaya studied the global asymptotic stability of the system of difference equations

${z}_{n+1}=\frac{{t}_{n}{z}_{n-1}+a}{{t}_{n}+{z}_{n-1}},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{t}_{n+1}=\frac{{z}_{n}{t}_{n-1}+a}{{z}_{n}+{t}_{n-1}}$

In , Irićanin and Stević studied the positive solutions of the system of difference equations

$\begin{array}{c}{x}_{n+1}^{\left(1\right)}=\frac{1+{x}_{n}^{\left(2\right)}}{{x}_{n-1}^{\left(3\right)}},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{n+1}^{\left(2\right)}=\frac{1+{x}_{n}^{\left(3\right)}}{{x}_{n-1}^{\left(4\right)}},\phantom{\rule{2.77695pt}{0ex}}\dots ,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{n+1}^{\left(k\right)}=\frac{1+{x}_{n}^{\left(1\right)}}{{x}_{n-1}^{\left(2\right)}},\\ {x}_{n+1}^{\left(1\right)}=\frac{1+{x}_{n}^{\left(2\right)}+{x}_{n-1}^{\left(3\right)}}{{x}_{n-2}^{\left(4\right)}},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{n+1}^{\left(2\right)}=\frac{1+{x}_{n}^{\left(3\right)}+{x}_{n-1}^{\left(4\right)}}{{x}_{n-2}^{\left(5\right)}},\phantom{\rule{2.77695pt}{0ex}}\dots ,\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{n+1}^{\left(k\right)}=\frac{1+{x}_{n}^{\left(1\right)}+{x}_{n-1}^{\left(2\right)}}{{x}_{n-2}^{\left(3\right)}}\end{array}$

Although difference equations are very simple in form, it is extremely difficult to understand throughly the global behavior of their solutions, for example, see Refs. .

In this article, we investigate the behavior of the solutions of the difference equation system

${x}_{n+1}=\frac{{x}_{n-1}}{{y}_{n}{x}_{n-1}-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{y}_{n+1}=\frac{{y}_{n-1}}{{x}_{n}{y}_{n-1}-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}{z}_{n+1}=\frac{1}{{y}_{n}{z}_{n}}$
(1.1)

where x 0, x -1, y 0, y -1, z 0, z -1 real numbers such that y 0 x -1 ≠ 1, x 0 y -1 ≠ 1 and y 0 z 0 ≠ 0.

## 2. Main results

Theorem 1. Let y 0 = a, y -1 = b, x 0 = c, x -1 = d, z 0 = e, z -1 = f be real numbers such that y 0 x -1 ≠ 1, x 0 y -1 ≠ 1 and y 0 z 0 ≠ 0. Let {x n , y n , z n } be a solution of the system (1.1). Then all solutions of (1.1) are

${x}_{n}=\left\{\frac{d}{{\left(ad-1\right)}^{n}}\right\},\phantom{\rule{1em}{0ex}}n---odd\phantom{\rule{0.3em}{0ex}}c\phantom{\rule{0.3em}{0ex}}{\left(cb-1\right)}^{n},\phantom{\rule{1em}{0ex}}n---even$
(1.2)
${y}_{n}=\left\{\frac{b}{{\left(cb-1\right)}^{n}}\right\},\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}n---odd\phantom{\rule{0.3em}{0ex}}a\phantom{\rule{0.3em}{0ex}}{\left(ad-1\right)}^{n},\phantom{\rule{1em}{0ex}}n---even$
(1.3)
${z}_{n}=\left\{\begin{array}{cc}\hfill \frac{{b}^{n}-1}{{a}^{n}e{\left[\left(ad-1\right)\left(cd-1\right)\right]}^{{\sum }_{i=1}^{k}i}},\hfill & \hfill n---odd\hfill \\ \hfill \frac{ane{\left(ad-1\right)}^{{\sum }_{i=1}^{k}\left(i-1\right)}{\left(cb-1\right)}^{{\sum }_{i=1}^{k}i}}{{b}^{n}},\hfill & \hfill n---even\hfill \end{array}\right\$
(1.4)

Proof. For n = 0, 1, 2, 3, we have

$\begin{array}{c}{x}_{1}=\frac{{x}_{-1}}{{y}_{0}{x}_{-1}-1}=\frac{d}{ad-1},\\ {y}_{1}=\frac{{y}_{-1}}{{x}_{0}{y}_{-1}-1}=\frac{b}{cb-1},\\ {z}_{1}=\frac{1}{{y}_{0}{z}_{0}}=\frac{1}{ae},\\ {x}_{2}=\frac{{x}_{0}}{{y}_{1}{x}_{0}-1}=\frac{c}{\frac{b}{cb-1}c-1}=c\left(cb-1\right),\\ {y}_{2}=\frac{{y}_{0}}{{x}_{1}{y}_{0}-1}=\frac{a}{\frac{d}{ad-1}a-1}=a\left(ad-1\right)\\ {z}_{2}=\frac{1}{{y}_{1}{z}_{1}}=\frac{1}{\frac{b}{cb-1}\frac{1}{ae}}=\frac{\left(cb-1\right)ae}{b},\\ {x}_{3}=\frac{{x}_{1}}{{y}_{2}{x}_{1}-1}=\frac{\frac{d}{ad-1}}{a\left(ad-1\right)\frac{d}{ad-1}-1}=\frac{d}{{\left(ad-1\right)}^{2}},\\ {y}_{3}=\frac{{y}_{1}}{{x}_{2}{y}_{1}-1}=\frac{\frac{b}{cb-1}}{c\left(cb-1\right)\frac{b}{cb-1}-1}=\frac{b}{{\left(cb-1\right)}^{2}},\\ {z}_{3}=\frac{1}{{y}_{2}{z}_{2}}=\frac{1}{a\left(ad-1\right)\frac{\left(cb-1\right)ae}{b}}=\frac{b}{{a}^{2}e\left(ad-1\right)\left(cb-1\right)}\end{array}$

for n = k, assume that

$\begin{array}{c}{x}_{2k-1}=\frac{{x}_{2k-3}}{{y}_{2k-2}{x}_{2k-3}-1}=\frac{d}{{\left(ad-1\right)}^{k}},\\ {x}_{2k}=\frac{{x}_{2k-2}}{{y}_{2k-1}{x}_{2k-2}-1}=c{\left(cb-1\right)}^{k},\\ {y}_{2k-1}=\frac{{y}_{2k-3}}{{x}_{2k-2}{y}_{2k-3}-1}=\frac{b}{{\left(cb-1\right)}^{k}},\\ {y}_{2k}=\frac{{y}_{2k-2}}{{x}_{2k-1}{y}_{2k-2}-1}=a{\left(ad-1\right)}^{k}\end{array}$

and

$\begin{array}{c}{z}_{2k-1}=\frac{{b}^{k-1}}{{a}^{k}e{\left[\left(ad-1\right)\left(cb-1\right)\right]}^{\sum _{i=1}^{k}i}},\\ {z}_{2k}=\frac{{a}^{k}e{\left(ad-1\right)}^{\sum _{i=1}^{k}\left(i-1\right)}{\left(cb-1\right)}^{\sum _{i=1}^{k}i}}{{b}^{k}}\end{array}$

are true. Then, for n = k + 1 we will show that (1.2), (1.3), and (1.4) are true. From (1.1), we have

$\begin{array}{c}{x}_{2k+1}=\frac{{x}_{2k-1}}{{y}_{2k}{x}_{2k-1}-1}=\frac{\frac{d}{{\left(ad-1\right)}^{k}}}{a{\left(ad-1\right)}^{k}\frac{d}{{\left(ad-1\right)}^{k}}-1}=\frac{d}{{\left(ad-1\right)}^{k+1}},\\ {y}_{2k+1}=\frac{{y}_{2k-1}}{{x}_{2k}{y}_{2k-1}-1}=\frac{\frac{b}{{\left(cb-1\right)}^{k}}}{c{\left(cb-1\right)}^{k}\frac{b}{{\left(cb-1\right)}^{k}}-1}=\frac{b}{{\left(cb-1\right)}^{k+1}}.\end{array}$

Also, similarly from (1.1), we have

$\begin{array}{c}{z}_{2k+1}=\frac{1}{{y}_{2k}{z}_{2k}}=\frac{1}{a{\left(ad-1\right)}^{k}\frac{{a}^{k}e{\left(ad-1\right)}^{\sum _{i=1}^{k}\left(i-1\right)}{\left(cb-1\right)}^{\sum _{i=1}^{k}i}}{{b}^{k}}}\\ \phantom{\rule{1em}{0ex}}=\frac{{b}^{k}}{{a}^{k+1}e{\left(ad-1\right)}^{\sum _{i=1}^{k}i}{\left(cb-1\right)}^{\sum _{i=1}^{k}i}}.\end{array}$

Also, we have

$\begin{array}{c}{x}_{2k+2}=\frac{{x}_{2k}}{{y}_{2k+1}{x}_{2k}-1}=\frac{c{\left(cb-1\right)}^{k}}{\frac{b}{{\left(cb-1\right)}^{k+1}}c{\left(cb-1\right)}^{k}-1}=\frac{c{\left(cb-1\right)}^{k}}{\frac{b}{\left(cb-1\right)}c-1}=c{\left(cb-1\right)}^{k+1},\\ {y}_{2k+2}=\frac{{y}_{2k}}{{x}_{2k+1}{y}_{2k}-1}=\frac{a{\left(ad-1\right)}^{k}}{\frac{d}{{\left(ad-1\right)}^{k+1}}a{\left(ad-1\right)}^{k}-1}=\frac{a{\left(ad-1\right)}^{k}}{\frac{d}{\left(ad-1\right)}a-1}=a{\left(ad-1\right)}^{k+1}\end{array}$

and

$\begin{array}{c}{z}_{2k+2}=\frac{1}{{y}_{2k+1}{z}_{2k+1}}=\frac{1}{\frac{b}{{\left(cb-1\right)}^{k+1}}\frac{{b}^{k}}{{a}^{k+1}e{\left(ad-1\right)}^{\sum _{i=1}^{k}i}{\left(cb-1\right)}^{\sum _{i=1}^{k}i}}}\\ \phantom{\rule{1em}{0ex}}=\frac{{a}^{k+1}e{\left(ad-1\right)}^{\sum _{i=1}^{k}i}{\left(cb-1\right)}^{\sum _{i=1}^{k+1}i}}{{b}^{k+1}}=\frac{{a}^{k+1}e{\left(ad-1\right)}^{\sum _{i=1}^{k+1}\left(i-1\right)}{\left(cb-1\right)}^{\sum _{i=1}^{k+1}i}}{{b}^{k+1}}.\end{array}$

Corollary 1. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0 and b ≠ 0. Also, if ad, cb (1, 2) and b > a then we have

$\underset{n\to \infty }{lim}{x}_{2n-1}=\underset{n\to \infty }{lim}{y}_{2n-1}=\underset{n\to \infty }{lim}{z}_{2n-1}=\infty$

and

$\underset{n\to \infty }{lim}{x}_{2n}=\underset{n\to \infty }{lim}{y}_{2n}=\underset{n\to \infty }{lim}{z}_{2n}=0.$

Proof. From ad, cb (1, 2) and b > a we have 0 < ad -1 < 1 and 0 < cb - 1 < 1.

Hence, we obtain

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n-1}=\underset{n\to \infty }{lim}\frac{d}{{\left(ad-1\right)}^{n}}=d\underset{n\to \infty }{lim}\frac{1}{{\left(ad-1\right)}^{n}}=d.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}\infty =\left\{\begin{array}{c}\hfill -\infty ,\phantom{\rule{2.77695pt}{0ex}}d<0\hfill \\ \hfill +\infty ,\phantom{\rule{2.77695pt}{0ex}}d>0\phantom{\rule{2.77695pt}{0ex}}\hfill \end{array}\right\,\\ \underset{n\to \infty }{lim}{y}_{2n-1}=\underset{n\to \infty }{lim}\frac{b}{{\left(cb-1\right)}^{n}}=b\underset{n\to \infty }{lim}\frac{1}{{\left(cb-1\right)}^{n}}=b.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}\infty =\left\{\begin{array}{c}\hfill -\infty ,\phantom{\rule{2.77695pt}{0ex}}b<0\hfill \\ \hfill +\infty ,\phantom{\rule{2.77695pt}{0ex}}b>0\hfill \end{array}\right\\end{array}$

and

$\underset{n\to \infty }{lim}{z}_{2n-1}=\underset{n\to \infty }{lim}\frac{{b}^{n-1}}{{a}^{n}e\phantom{\rule{2.77695pt}{0ex}}{\left[\left(ad-1\right)\left(cb-1\right)\right]}^{\sum _{i=1}^{k}i}}=\frac{1}{e}.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}\infty =\left\{\begin{array}{c}\hfill -\infty ,\phantom{\rule{2.77695pt}{0ex}}e<0\hfill \\ \hfill +\infty ,\phantom{\rule{2.77695pt}{0ex}}e>0\hfill \end{array}\right\$

Similarly, from ad, cb (1, 2) and b > a, we have 0 < ad - 1 < 1 and 0 < cb - 1 < 1.

Hence, we obtain

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n}=\underset{n\to \infty }{lim}c{\left(cd-1\right)}^{n}=c\underset{n\to \infty }{lim}{\left(cd-1\right)}^{n}=c.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}0=0,\\ \underset{n\to \infty }{lim}{y}_{2n}=\underset{n\to \infty }{lim}a\phantom{\rule{2.77695pt}{0ex}}{\left(af-1\right)}^{n}=a\underset{n\to \infty }{lim}\phantom{\rule{2.77695pt}{0ex}}{\left(af-1\right)}^{n}=a.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}0=0.\end{array}$

and

$\underset{n\to \infty }{lim}{z}_{2n}=\underset{n\to \infty }{lim}\frac{{a}^{n}e{\left(ad-1\right)}^{\sum _{i=1}^{k}\left(i-1\right)}{\left(cb-1\right)}^{\sum _{i=1}^{k}i}}{{b}^{n}}=0.e.\phantom{\rule{1em}{0ex}}0=0.$

Corollary 2. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0 and b ≠ 0. If a = b and cb = ad = 2 then we have

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n-1}=d,\\ \underset{n\to \infty }{lim}{y}_{2n-1}=b,\\ \underset{n\to \infty }{lim}{z}_{2n-1}=\frac{1}{ae}\end{array}$

and

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n}=c,\\ \underset{n\to \infty }{lim}{y}_{2n}=a,\\ \underset{n\to \infty }{lim}{z}_{2n}=e.\end{array}$

Proof. From a = b and cb = ad = 2 then we have, cb - 1 = ad - 1 = 1. Hence, we have

$\underset{n\to \infty }{lim}{\left(cb-1\right)}^{n}=1$

and

$\underset{n\to \infty }{lim}{\left(ad-1\right)}^{n}=1.$

Also, we have

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n-1}=\underset{n\to \infty }{lim}\frac{d}{{\left(ad-1\right)}^{n}}=d\underset{n\to \infty }{lim}\frac{1}{{\left(ad-1\right)}^{n}}=d.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}1=d,\\ \underset{n\to \infty }{lim}{y}_{2n-1}=\underset{n\to \infty }{lim}\frac{b}{{\left(cb-1\right)}^{n}}=b\underset{n\to \infty }{lim}\frac{1}{{\left(cb-1\right)}^{n}}=b.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}1=b\end{array}$

and

$\underset{n\to \infty }{lim}{z}_{2n-1}=\underset{n\to \infty }{lim}\frac{{b}^{n-1}}{{a}^{n}e{\left[\left(ad-1\right)\left(cb-1\right)\right]}^{{\sum }_{i=1}^{K}i}}=\underset{n\to \infty }{lim}\frac{1}{ae}\frac{{b}^{n-1}}{{a}^{n-1}{\left[\left(ad-1\right)\left(cb-1\right)\right]}^{{\sum }_{i=1}^{k}i}}=\frac{1}{ae}.$

Similarly, we have

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n}=\underset{n\to \infty }{lim}c{\left(cb-1\right)}^{n}=c\underset{n\to \infty }{lim}{\left(cb-1\right)}^{n}=c.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}1=c,\\ \underset{n\to \infty }{lim}{y}_{2n}=\underset{n\to \infty }{lim}a{\left(ad-1\right)}^{n}=a\underset{n\to \infty }{lim}{\left(ad-1\right)}^{n}=a.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}1=a.\end{array}$

and

$\underset{n\to \infty }{lim}{z}_{2n}=\underset{n\to \infty }{lim}\frac{{a}^{n}e{\left(ad-1\right)}^{\sum _{i=1}^{k}\left(i-1\right)}{\left(cb-1\right)}^{\sum _{i=1}^{k}i}}{{b}^{n}}=1.\phantom{\rule{2.77695pt}{0ex}}e=e.$

Corollary 3. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0 and b ≠ 0. Also, if 0 < a, b, c, d, e, f < 1 then we have

$\underset{n\to \infty }{lim}{x}_{2n}=\underset{n\to \infty }{lim}{y}_{2n}=\underset{n\to \infty }{lim}{z}_{2n}=0$

and

$\underset{n\to \infty }{lim}{x}_{2n-1}=\underset{n\to \infty }{lim}{y}_{2n-1}=\underset{n\to \infty }{lim}{z}_{2n-1}=\infty .$

Proof. From 0 < a, b, c, d, e, f < 1 we have -1 < ad - 1 < 0 and - 1 < cb - 1 < 0. Hence, we obtain

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n}=\underset{n\to \infty }{lim}c{\left(bc-1\right)}^{n}=c\underset{n\to \infty }{lim}{\left(bc-1\right)}^{n}=c.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}0=0,\\ \underset{n\to \infty }{lim}{y}_{2n}=\underset{n\to \infty }{lim}a{\left(ad-1\right)}^{n}=a\underset{n\to \infty }{lim}{\left(ad-1\right)}^{n}=a.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}0=0\end{array}$

and

$\underset{n\to \infty }{lim}{z}_{2n}=\underset{n\to \infty }{lim}\frac{{a}^{n}e{\left(ad-1\right)}^{\sum _{i=1}^{k}\left(i-1\right)}{\left(cb-1\right)}^{\sum _{i=1}^{k}i}}{{b}^{n}}=e.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}0=0.$

Similarly, we have

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n-1}=\underset{n\to \infty }{lim}\frac{d}{{\left(ad-1\right)}^{n}}=d\underset{n\to \infty }{lim}\frac{1}{{\left(ad-1\right)}^{n}}=d\underset{n\to \infty }{lim}\frac{1}{{\left(ad-1\right)}^{n}}=d.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}\infty =\left\{\begin{array}{c}\hfill -\infty ,\phantom{\rule{2.77695pt}{0ex}}n-odd\hfill \\ \hfill +\infty ,\phantom{\rule{2.77695pt}{0ex}}n\phantom{\rule{2.77695pt}{0ex}}-even\phantom{\rule{2.77695pt}{0ex}}\hfill \end{array}\right\,\\ \phantom{\rule{1em}{0ex}}\underset{n\to \infty }{lim}{y}_{2n-1}=\underset{n\to \infty }{lim}\frac{b}{{\left(bc-1\right)}^{n}}=b\underset{n\to \infty }{lim}\frac{1}{{\left(bc-1\right)}^{n}}=b.\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{1em}{0ex}}\infty =\left\{\begin{array}{c}\hfill -\infty ,\phantom{\rule{2.77695pt}{0ex}}n-odd\hfill \\ \hfill +\infty ,\phantom{\rule{2.77695pt}{0ex}}n-even\hfill \end{array}\right\.\end{array}$

and

$\underset{n\to \infty }{lim}{z}_{2n-1}=\underset{n\to \infty }{lim}\frac{{b}^{n-1}}{{a}^{n}e{\left[\left(ad-1\right)\left(cb-1\right)\right]}^{{\sum }_{i=1}^{k}i}}=+\infty .$

Corollary 4. Let {x n , y n , z n } be a solution of the system (1.1). Let a, b, c, d, e, f be real numbers such that ad ≠ 1, cb ≠ 1, ae ≠ 0, and b ≠ 0. Also, if 0 < a, b, c, d, e, f < 1 then we have

$\begin{array}{c}\underset{n\to \infty }{lim}{x}_{2n}{y}_{2n-1}=cb,\\ \underset{n\to \infty }{lim}{x}_{2n-1}{y}_{2n}=ad\end{array}$

and

$\underset{n\to \infty }{lim}{z}_{2n-1}{z}_{2n}=\infty .$

Proof. The proof is clear from Theorem 1. □

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Kurbanli, A.S. On the behavior of solutions of the system of rational difference equations ${x}_{n+1}=\frac{{x}_{n-1}}{{y}_{n}{x}_{n-1}-1},\phantom{\rule{1em}{0ex}}{y}_{n+1}=\frac{{y}_{n-1}}{{x}_{n}{y}_{n-1}-1},\phantom{\rule{1em}{0ex}}{z}_{n+1}=\frac{1}{{y}_{n}{z}_{n}}$ . Adv Differ Equ 2011, 40 (2011). https://doi.org/10.1186/1687-1847-2011-40

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### Keywords

• Differential Equation
• Real Number
• Partial Differential Equation
• Ordinary Differential Equation
• Functional Analysis 