On the behavior of solutions of the system of rational difference equations $x_{n+1}=\frac{x_{n-1}}{y_n{x}_{n-1}-1},y_{n+1}=\frac{y_{n-1}}{x_n{y}_{n-1}-1},z_{n+1}=\frac{1}{y_n{z}_n}$

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 [1], Kurbanli et al. studied the behavior of positive solutions of the system of rational difference equations

$$x_{n+1}=\frac{x_{n-1}}{y_n{x}_{n-1}+1},y_{n+1}=\frac{y_{n-1}}{x_n{y}_{n-1}+1}.$$

In [2], Cinar studied the solutions of the systems of difference equations

$$x_{n+1}=\frac{1}{y_n},y_{n+1}=\frac{y_n}{x_{n-1}{y}_{n-1}}.$$

In [3], Kurbanli, studied the behavior of solutions of the system of rational difference equations

$$x_{n+1}=\frac{x_{n-1}}{y_n{x}_{n-1}-1},y_{n+1}=\frac{y_{n-1}}{x_n{y}_{n-1}-1},z_{n+1}=\frac{z_{n-1}}{y_n{z}_{n-1}-1}.$$

In [4], Papaschinnopoulos and Schinas proved the boundedness, persistence, the oscillatory behavior, and the asymptotic behavior of the positive solutions of the system of difference equations

$$x_{n+1}=\sum _{i=0}^k{A}_i∕y_{n-i}^{p_i},y_{n+1}=\sum _{i=0}^k{B}_i∕x_{n-i}^{q_i}$$

In [5], Clark and Kulenović investigate the global stability properties and asymptotic behavior of solutions of the system of difference equations

$$x_{n+1}=\frac{x_n}{a+c{y}_n},y_{n+1}=\frac{y_n}{b+d{x}_n}.$$

In [6], Camouzis and Papaschinnopoulos studied the global asymptotic behavior of positive solutions of the system of rational difference equations

$$x_{n+1}=1+\frac{x_n}{y_{n-m}},y_{n+1}=1+\frac{y_n}{x_{n-m}}.$$

In [7], Kulenović and Nurkanović studied the global asymptotic behavior of solutions of the system of difference equations

$$x_{n+1}=\frac{a+x_n}{b+y_n},y_{n+1}=\frac{c+y_n}{d+z_n},z_{n+1}=\frac{e+z_n}{f+x_n}.$$

In [8], Özban studied the positive solutions of the system of rational difference equations

$$x_{n+1}=\frac{1}{y_{n-k}},y_{n+1}=\frac{y_n}{x_{n-m}{y}_{n-m-k}}.$$

In [9], Zhang et al. investigated the behavior of the positive solutions of the system of the difference equations

$$x_n=A+\frac{1}{y_{n-p}},y_n=A+\frac{y_{n-1}}{x_{n-r}{y}_{n-s}}.$$

In [10], Yalcinkaya studied the global asymptotic stability of the system of difference equations

$$z_{n+1}=\frac{t_n{z}_{n-1}+a}{t_n+z_{n-1}},t_{n+1}=\frac{z_n{t}_{n-1}+a}{z_n+t_{n-1}}$$

In [11], Irićanin and Stević studied the positive solutions of the system of difference equations

$$\begin{array}{c}{x}_{n+1}^{\left(1\right)}=\frac{1+x_n^{\left(2\right)}}{x_{n-1}^{\left(3\right)}},x_{n+1}^{\left(2\right)}=\frac{1+x_n^{\left(3\right)}}{x_{n-1}^{\left(4\right)}},\dots ,x_{n+1}^{\left(k\right)}=\frac{1+x_n^{\left(1\right)}}{x_{n-1}^{\left(2\right)}},\\ x_{n+1}^{\left(1\right)}=\frac{1+x_n^{\left(2\right)}+x_{n-1}^{\left(3\right)}}{x_{n-2}^{\left(4\right)}},x_{n+1}^{\left(2\right)}=\frac{1+x_n^{\left(3\right)}+x_{n-1}^{\left(4\right)}}{x_{n-2}^{\left(5\right)}},\dots ,x_{n+1}^{\left(k\right)}=\frac{1+x_n^{\left(1\right)}+x_{n-1}^{\left(2\right)}}{x_{n-2}^{\left(3\right)}}\end{array}$$

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. [12–34].

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

$$x_{n+1}=\frac{x_{n-1}}{y_n{x}_{n-1}-1},y_{n+1}=\frac{y_{n-1}}{x_n{y}_{n-1}-1},z_{n+1}=\frac{1}{y_n{z}_n}$$

(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

$$x_n=\left\{\frac{d}{{\left(ad-1\right)}^n}\right\},n---oddc{\left(cb-1\right)}^n,n---even$$

(1.2)

$$y_n=\left\{\frac{b}{{\left(cb-1\right)}^n}\right\},n---odda{\left(ad-1\right)}^n,n---even$$

(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

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

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

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

□

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

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

and

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

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

Hence, we obtain

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

and

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

□

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

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

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

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

and

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

□

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

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

and

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

Similarly, we have

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

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. □