On the solutions of a max-type system of difference equations of higher order

Abstract

In this paper, we study the following max-type system of difference equations of higher order:

$$ \textstyle\begin{cases} x_{n} = \max \{A ,\frac{y_{n-t}}{x_{n-s}} \}, \\ y_{n} = \max \{B ,\frac{x_{n-t}}{y_{n-s}} \},\end{cases}\displaystyle \quad n\in \{0,1,2,\ldots \}, $$

where $A,B\in (0, +\infty )$, $t,s\in \{1,2,\ldots \}$ with $\gcd (s,t)=1$, the initial values $x_{-d},y_{-d},x_{-d+1},y_{-d+1}, \ldots , x_{-1}, y_{-1}\in (0,+ \infty )$ and $d=\max \{t,s\}$.

1 Introduction

Concrete nonlinear difference equations and systems have attracted some recent attention (see, e.g., [1–39]). One of the classes of such equations/systems are max-type difference equations/systems. For some results of solutions of many max-type difference equations and systems, such as eventual periodicity, the boundedness character and attractivity, see, e.g. [1–5, 7–9, 11–16, 18–25, 28–30, 32–36, 38, 39] and the references therein. Our purpose in this paper is to study the eventual periodicity of the following max-type system of difference equation of higher order:

$$ \textstyle\begin{cases} x_{n} = \max \{A ,\frac{y_{n-t}}{x_{n-s}} \}, \\ y_{n} = \max \{B ,\frac{x_{n-t}}{y_{n-s}} \},\end{cases}\displaystyle \quad n\in { \mathbf{{N}}}_{0}\equiv \{0,1,\ldots \}, $$

(1.1)

where $A,B\in {\mathbf{{R}}}_{+}\equiv (0,+\infty )$, $t,s\in {\mathbf{{N}\equiv }\{\mathbf{1},\mathbf{2},\ldots \}}$ with $\gcd (s,t)=1$, the initial values $x_{-d},y_{-d}, x_{-d+1},y_{-d+1}, \ldots , x_{-1}, y_{-1}\in \mathbf{{R}}_{+}$ and $d=\max \{t,s\}$.

When $t=1$ and $s=2$, (1.1) reduces to the max-type system of difference equations

$$ \textstyle\begin{cases} x_{n} = \max \{A ,\frac{y_{n-1}}{x_{n-2}} \}, \\ y_{n} = \max \{B ,\frac{x_{n-1}}{y_{n-2}} \},\end{cases}\displaystyle \quad n\in { \mathbf{{N}}}_{0}. $$

(1.2)

Fotiades and Papaschinopoulos in [5] showed that every positive solution of (1.2) is eventually periodic.

In 2012, Stević [23] obtained in an elegant way the general solution to the following max-type system of difference equations:

$$ \textstyle\begin{cases} x_{n+1}=\max \{\frac{A}{x_{n}},\frac{y_{n}}{x_{n}} \}, \\ y_{n+1}=\max \{\frac{A}{y_{n}},\frac{x_{n}}{y_{n}} \},\end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0}, $$

(1.3)

for the case $x_{0},y_{0}\geq A>0$ and $y_{0}/x_{0}\geq \max \{A,1/A\}$.

In [35], Sun and Xi studied the following max-type system of difference equations:

$$ \textstyle\begin{cases} x_{n} = \max \{\frac{1}{x_{n-m}} , \min \{1,\frac{A}{y_{n-r}} \} \}, \\ y_{n} = \max \{\frac{1}{y_{n-m}} , \min \{1,\frac{B}{x_{n-t}} \} \},\end{cases}\displaystyle \quad n\in { \mathbf{{N}}}_{0}, $$

(1.4)

where $A,B\in { \mathbf{{R}}}_{+}$, $m,r,t\in {\mathbf{{N}}}$ and the initial values $x_{-d},y_{-d},x_{-d+1},y_{-d+1}, \ldots , x_{-1}, y_{-1}\in {\mathbf{{R}}}_{+}$ with $d=\max \{m,r,t\}$ and showed that every positive solution of (1.4) is eventually periodic with period 2m.

When $m=r=t=1$ and $A=B$, (1.4) reduces to the max-type system of difference equations

$$ \textstyle\begin{cases} x_{n}=\max \{\frac{1}{x_{n-1}},\min \{1,\frac{A}{y_{n-1}} \} \}, \\ y_{n}=\max \{\frac{1}{y_{n-1}},\min \{1,\frac{A}{x_{n-1}} \} \}, \end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0}. $$

(1.5)

Yazlik et al. [39] in 2015 obtained in an elegant way the general solution of (1.5).

In 2012, Stević [24] studied the following max-type system of difference equations:

$$ \textstyle\begin{cases} y^{(1)}_{n}=\max_{1\leq i\leq m_{1}} \{f_{1i}(y^{(1)}_{n-k_{i,1}^{(1)}},y^{(2)}_{n-k_{i,2}^{(1)}}, \ldots ,y^{(l)}_{n-k_{i,l}^{(1)}},n),y_{n-s}^{(1)} \}, \\ y^{(2)}_{n}=\max_{1\leq i\leq m_{2}} \{f_{2i}(y^{(1)}_{n-k_{i,1}^{(2)}},y^{(2)}_{n-k_{i,2}^{(2)}}, \ldots ,y^{(l)}_{n-k_{i,l}^{(2)}},n),y_{n-s}^{(2)} \}, \\ \ldots , \\ y^{(l)}_{n}=\max_{1\leq i\leq m_{l}} \{f_{li}(y^{(1)}_{n-k_{i,1}^{(l)}},y^{(2)}_{n-k_{i,2}^{(l)}}, \ldots ,y^{(l)}_{n-k_{i,l}^{(l)}},n),y_{n-s}^{(l)} \},\end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0}, $$

(1.6)

where $s,l,m_{j},k^{(j)}_{i,t}\in {\mathbf{{N}}}$ ($j,t\in \{1,2,\ldots ,l \}$) and $f_{ji}:\mathbf{{R}}_{+}^{l}\times {\mathbf{{ N}}}_{0}\longrightarrow \mathbf{{R}}_{+}$ ($j\in \{1,\ldots ,l\}$ and $i\in \{1,\ldots ,m_{j}\}$), and showed that every positive solution of (1.6) is eventually periodic with (not necessarily prime) period s if $f_{ji}$ satisfy some conditions.

Moreover, Stević et al. [29] in 2014 investigated the following max-type system of difference equations:

$$ \textstyle\begin{cases} y^{(1)}_{n}=\max_{1\leq i_{1}\leq m_{1}} \{f_{1i_{1}}(y^{(1)}_{n-k_{i_{1},1}^{(1)}},y^{(2)}_{n-k_{i_{1},2}^{(1)}}, \ldots ,y^{(l)}_{n-k_{i_{1},l}^{(1)}},n),y_{n-t_{1}s}^{(\sigma (1))} \}, \\ x^{(2)}_{n}=\max_{1\leq i_{2}\leq m_{2}} \{f_{2i_{2}}(y^{(1)}_{n-k_{i_{2},1}^{(2)}},y^{(2)}_{n-k_{i_{2},2}^{(2)}}, \ldots ,y^{(l)}_{n-k_{i_{2},l}^{(2)}},n),y_{n-t_{2}s}^{(\sigma (2))} \}, \\ \ldots, \\ y^{(l)}_{n}=\max_{1\leq i_{l}\leq m_{l}} \{f_{li_{l}}(y^{(1)}_{n-k_{i_{l},1}^{(l)}},y^{(2)}_{n-k_{i_{l},2}^{(l)}}, \ldots ,y^{(l)}_{n-k_{i_{l},l}^{(l)}},n),y_{n-t_{l}s}^{(\sigma (l))} \},\end{cases}\displaystyle \quad n\in \mathbf{{N}}_{0}, $$

(1.7)

where $s,l,m_{j},t_{j},k^{(j)}_{i_{j},h}\in { \mathbf{{N}}}$ ($j,h\in \{1,2,\ldots ,l\}$), $(\sigma (1),\ldots ,\sigma (l))$ is a permutation of $(1,\ldots ,l)$ and $f_{ji_{j}}:\mathbf{{R}}_{+}^{l}\times {\mathbf{{ N}}}_{0}\longrightarrow \mathbf{{R}}_{+}$ ($j\in \{1,\ldots ,l\}$ and $i_{j}\in \{1,\ldots ,m_{j}\}$). They showed that every positive solution of (1.7) is eventually periodic with period sT for some $T\in \mathbf{{N}}$ if $f_{ji_{j}}$ satisfy some conditions.

2 Main results and proofs

In this section, we study the eventual periodicity of positive solutions of system (1.1). Let $\{(x_{n},y_{n})\}_{n\geq -d}$ be a solution of (1.1) with the initial values $x_{-d},y_{-d},x_{-d+1},y_{-d+1},\ldots , x_{-1}, y_{-1}\in { \mathbf{{R}}}_{+}$.

Lemma 2.1

If$x_{n}=A$eventually, then$y_{n}$is a periodic sequence with period 2seventually. If$y_{n}=B$eventually, then$x_{n}$is a periodic sequence with period 2seventually.

Proof

Assume that $x_{n}=A$ eventually. By (1.1) we see

$$ y_{n} = \max \biggl\{ B ,\frac{A}{y_{n-s}} \biggr\} \quad \mbox{eventually}, $$

(2.1)

which implies $y_{n}y_{n-s}\geq A$ eventually and

$$ \begin{aligned}[b] B&\leq y_{n} = \max \biggl\{ B , \frac{A}{y_{n-s}} \biggr\} \\ &= \max \biggl\{ B ,\frac{Ay_{n-2s}}{y_{n-s}y_{n-2s}} \biggr\} \\ &\leq \max \{B,y_{n-2s}\}\leq y_{n-2s} \quad \mbox{eventually}.\end{aligned} $$

(2.2)

Then, for any $0\leq i\leq 2s-1$, $y_{2ns+i}$ is eventually nonincreasing.

We claim that, for every $0\leq i\leq 2s-1$, $y_{2ns+i}$ is a constant sequence eventually. Assume on the contrary that, for some $0\leq i\leq 2s-1$, $y_{2ns+i}$ is not a constant sequence eventually. Then there exists a sequence of positive integers $k_{1}< k_{2}<\cdots$ such that, for any $n\in { \mathbf{{N}}}$, we have

$$ \begin{aligned}[b] B&< y_{2sk_{n+1}+i}=\frac{A}{y_{2sk_{n+1}+i-s}} \\ &< y_{2sk_{n}+i}=\frac{A}{y_{2sk_{n}+i-s}},\end{aligned} $$

(2.3)

which implies $y_{2sk_{n+1}+i-s}>y_{2sk_{n}+i-s}$ for any $n\in { \mathbf{{N}}}$. This is a contradiction. Thus $y_{n}$ is a periodic sequence with period 2s eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

Lemma 2.2

If$A\geq B\geq 1/A$, then$x_{2(n+1)t+i}\leq x_{2nt+i}$for any$n\geq t+s$and$i\in { {\mathbf{{N}}}}_{0}$. If$B\geq A\geq 1/B$, then$y_{2(n+1)t+i}\leq y_{2nt+i}$for any$n\geq t+s$and$i\in { {\mathbf{{N}}}}_{0}$.

Proof

Assume that $A\geq B\geq 1/A$. By (1.1) we see that $x_{n}\geq A$ and $y_{n}\geq B$ for any $n\in { \mathbf{{N}}}_{0}$, and

$$ x_{2(n+1)t+i} = \max \biggl\{ A,\frac{B}{x_{2(n+1)t+i-s}}, \frac{x_{2nt+i}}{x_{2(n+1)t+i-s}y_{2(n+1)t+i-t-s}} \biggr\} . $$

(2.4)

Since $B/x_{2(n+1)t+i-s}\leq B/A\leq 1\leq A$ and $x_{2(n+1)t+i-s}y_{2(n+1)t+i-t-s}\geq AB\geq 1$ for $2(n+1)t+i\geq t+s$, we obtain

$$ x_{2(n+1)t+i}\leq \max \{A, x_{2nt+i}\}\leq x_{2nt+i}. $$

(2.5)

The second case follows from the previously proved one by interchanging letters. The proof is complete. □

Theorem 2.1

Let$AB>1$. If$A\geq B$, then$x_{n}=A$eventually and$y_{n}$is a periodic sequence with period 2seventually. If$B> A$, then$y_{n}=B$eventually and$x_{n}$is a periodic sequence with period 2seventually.

Proof

Assume that $A\geq B$. For any $0\leq i\leq 2t-1$ and $n\in { \mathbf{{N}}}_{0}$, we have

$$ A\leq x_{2(n+1)t+i}= \max \biggl\{ A, \frac{x_{2nt+i}}{x_{2(n+1)t-s+i}y_{2(n+1)t-s-t+i}} \biggr\} . $$

(2.6)

By Lemma 2.2 we may let $\lim_{n\longrightarrow \infty }x_{2nt+i}=A_{i}$. Note that

$$ \lim_{n\longrightarrow \infty } \frac{x_{2nt+i}}{x_{2(n+1)t-s+i}y_{2(n+1)t-s-t+i}}\leq \lim _{n \longrightarrow \infty }\frac{x_{2nt+i}}{AB}=\frac{A_{i}}{AB}< A_{i} $$

(2.7)

and

$$ \lim_{n\longrightarrow \infty }x_{2(n+1)t+i}=A_{i}. $$

(2.8)

Thus we have $x_{n}=A$ eventually. By Lemma 2.1, we see that $y_{n}$ is a periodic sequence with period 2s eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

In the following, we assume $AB=1$. For any $i\in { \mathbf{{N}}}_{0}$, let

$$ \lim_{n\longrightarrow \infty } x_{2nt+i}=A_{i} \quad \mbox{if } A \geq B $$

(2.9)

and

$$ \lim_{n\longrightarrow \infty }y_{2nt+i}=B_{i}\quad \mbox{if } B \geq A. $$

(2.10)

Then $A_{i}\geq A$ and $B_{i}\geq B$.

Lemma 2.3

If$A\geq B=1/A$and$A_{i}>A$for some$i\in { {\mathbf{{N}}}}_{0}$, then, for any$k\in { {\mathbf{{N}}}}$, $x_{2nt+ks+i}$and$y_{2nt-t+ks+i}$are constant sequences eventually. If$B\geq A=1/B$and$B_{i}>B$for some$i\in { {\mathbf{{N}}}}_{0}$, then, for any$k\in { {\mathbf{{N}}}}$, $y_{2nt+ks+i}$and$x_{2nt-t+ks+i}$are constant sequences eventually.

Proof

Assume that $A\geq B$ and $A_{i}>A$ for some $i\in { \mathbf{{N}}}_{0}$. Since $A_{i}>A$, it follows from Lemma 2.2 and (1.1) that

$$ \begin{aligned}[b] x_{2nt+i}& =\max \biggl\{ A, \frac{x_{2(n-1)t+i}}{x_{2nt-s+i}y_{2nt-t-s+i}} \biggr\} \\ &=\frac{x_{2(n-1)t+i}}{x_{2nt-s+i}y_{2nt-t-s+i}} \quad \mbox{eventually}.\end{aligned} $$

(2.11)

From this we have

$$ \begin{aligned}[b] B&\leq \lim_{n\longrightarrow \infty }y_{2nt-t-s+i} \\ &=\lim_{n\longrightarrow \infty } \frac{x_{2(n-1)t+i}}{x_{2nt+i}x_{2nt-s+i}} \\ &=\frac{1}{A_{-s+i}}\leq \frac{1}{A}=B.\end{aligned} $$

(2.12)

This implies

$$ \lim_{n\longrightarrow \infty }x_{2nt-s+i}=A $$

(2.13)

and

$$ \lim_{n\longrightarrow \infty }y_{2nt-t-s+i}=B $$

(2.14)

and

$$ \lim_{n\longrightarrow \infty }y_{2nt-t+i}=\lim_{n\longrightarrow \infty }x_{2nt+i}x_{2nt-s+i}=A_{i}A. $$

(2.15)

Since

$$ x_{2nt+s+i} = \max \biggl\{ A, \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}} \biggr\} $$

(2.16)

and

$$ \lim_{n\longrightarrow \infty } \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}}= \frac{A_{s+i}}{A_{i}^{2}A}< A_{s+i}, $$

(2.17)

we see that $x_{2nt+s+i} =A$ eventually. Note that

$$ \lim_{n\longrightarrow \infty }\frac{x_{2(n-1)t+s+i}}{y_{2nt-t+i}}= \frac{A}{AA_{i}}< B, $$

(2.18)

from which it follows that

$$ \begin{aligned}[b] y_{2nt-t+s+i} &= \max \biggl\{ B, \frac{x_{2(n-1)t+s+i}}{y_{2nt-t+i}} \biggr\} \\ &=B \quad \mbox{eventually},\end{aligned} $$

(2.19)

and

$$ \begin{aligned}[b] y_{2nt-t+2s+i} &= \max \biggl\{ B, \frac{x_{2(n-1)t+2s+i}}{y_{2nt-t+s+i}} \biggr\} \\ &=\frac{x_{2(n-1)t+2s+i}}{B}\quad \mbox{eventually}, \end{aligned} $$

(2.20)

and

$$ \begin{aligned}[b] x_{2nt+2s+i}& = \max \biggl\{ A, \frac{y_{2nt-t+2s+i}}{x_{2nt+s+i}} \biggr\} \\ &=\max \{A,x_{2(n-1)t+2s+i}\} \\ &=x_{2(n-1)t+2s+i} \quad \mbox{eventually}.\end{aligned} $$

(2.21)

If $x_{2nt+2s+i}>A$ eventually, then, in a similar fashion, we obtain:

• (1) $x_{2nt+3s+i} =A$ eventually and $y_{2nt-t+3s+i} =B$ eventually.

• (2) $x_{2nt+4s+i}$ and $y_{2nt-t+4s+i}$ are constant sequences eventually.

If $x_{2nt+2s+i}=A$ eventually, then $y_{2nt-t+2s+i} =A/B$ eventually, and

$$ \begin{aligned}[b] y_{2nt-t+3s+i}& = \max \biggl\{ B, \frac{x_{2(n-1)t+3s+i}}{y_{2nt-t+2s+i}} \biggr\} \\ &=\max \biggl\{ B,\frac{x_{2(n-1)t+3s+i}B}{A} \biggr\} \\ &=\frac{x_{2(n-1)t+3s+i}B}{A}\quad \mbox{eventually},\end{aligned} $$

(2.22)

and

$$ \begin{aligned}[b] x_{2nt+3s+i}& = \max \biggl\{ A, \frac{y_{2nt-t+3s+i}}{x_{2nt+2s+i}} \biggr\} \\ &=\max \biggl\{ A,\frac{x_{2(n-1)t+3s+i}B}{A^{2}}\biggr\} \quad \mbox{eventually}.\end{aligned} $$

(2.23)

From this we see that if $A=B$, then

$$ x_{2nt+3s+i}=x_{2(n-1)t+3s+i} \quad \mbox{eventually}, $$

(2.24)

and if $A>B$, then

$$ x_{2nt+3s+i}=A\quad \mbox{ eventually}, $$

(2.25)

since

$$ \lim_{n\longrightarrow \infty }\frac{x_{2(n-1)t+3s+i}B}{A^{2}}= \frac{A_{3s+i}B}{A^{2}}< A_{3s+i}. $$

(2.26)

Using induction and arguments similar to the ones developed in the above given proof, we can show that, for any $k\in { \mathbf{{N}}}$, $x_{2nt+ks+i}$ and $y_{2nt-t+ks+i}$ are constant sequences eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

Lemma 2.4

If$A=1/B>B$and for some$i\in { {\mathbf{{N}}}}_{0}$, $x_{2nt+i}>A$eventually and$A_{i}=A$, then, for any$k\in { {\mathbf{{N}}}}$, $x_{2nt+ks+i}$and$y_{2nt-t+ks+i}$are constant sequences eventually. If$B=1/A>A$and for some$i\in { {\mathbf{{N}}}}_{0}$, $y_{2nt+i}>B$eventually and$B_{i}=B$, then, for any$k\in { {\mathbf{{N}}}}$, $y_{2nt+ks+i}$and$x_{2nt-t+ks+i}$are constant sequences eventually.

Proof

Assume that $A=1/B>B$ and for some $i\in { \mathbf{{N}}}_{0}$, $x_{2nt+i}>A$ eventually and $A_{i}=A$. By (1.1) we have

$$ \begin{aligned}[b] x_{2nt+i}& = \max \biggl\{ A, \frac{y_{2nt-t+i}}{x_{2nt-s+i}} \biggr\} \\ &=\frac{y_{2nt-t+i}}{x_{2nt-s+i}} \quad \mbox{eventually}, \end{aligned} $$

(2.27)

and

$$ x_{2nt+s+i} = \max \biggl\{ A, \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}} \biggr\} $$

(2.28)

and

$$ \begin{aligned}[b] \lim_{n\longrightarrow \infty } \frac{x_{2(n-1)t+s+i}}{x_{2nt+i}y_{2nt-t+i}} &=\lim_{n \longrightarrow \infty } \frac{x_{2(n-1)t+s+i}}{x^{2}_{2nt+i}x_{2nt-s+i}} \\ &\leq \frac{A_{s+i}}{A^{3}}< A_{s+i}. \end{aligned} $$

(2.29)

Then we see that $x_{2nt+s+i} =A$ eventually. From this and $y_{2nt-t+i}\geq A^{2}$ eventually it follows that

$$ \begin{aligned}[b] y_{2nt-t+s+i}& = \max \biggl\{ B, \frac{x_{2(n-1)t+s+i}}{y_{2nt-t+i}} \biggr\} \\ &=\max \biggl\{ B,\frac{A}{y_{2nt-t+i}} \biggr\} \\ &=B \quad \mbox{eventually}, \end{aligned} $$

(2.30)

and

$$ \begin{aligned}[b] y_{2nt-t+2s+i} &= \max \biggl\{ B, \frac{x_{2(n-1)t+2s+i}}{y_{2nt-t+s+i}} \biggr\} \\ &=\frac{x_{2(n-1)t+2s+i}}{B} \quad \mbox{eventually},\end{aligned} $$

(2.31)

and

$$ \begin{aligned}[b] x_{2nt+2s+i}& = \max \biggl\{ A, \frac{y_{2nt-t+2s+i}}{x_{2nt+s+i}} \biggr\} \\ &=\max \{A,x_{2(n-1)t+2s+i}\} \\ &=x_{2(n-1)t+2s+i} \quad \mbox{eventually}.\end{aligned} $$

(2.32)

Thus $x_{2nt+2s+i}$ and $y_{2nt-t+2s+i}$ are constant sequences eventually. Using arguments similar to the ones developed in the proof of Lemma 2.3, we can show that, for any $k\in { \mathbf{{N}}}$, $x_{2nt+ks+i}$ and $y_{2nt-t+ks+i}$ are constant sequences eventually. The second case follows from the previously proved one by interchanging letters. The proof is complete. □

Theorem 2.2

• (1) Assume$A=1/B>B$. Then one of the following statements holds.

  • (i) $x_{n}=A$eventually and$y_{n}$is a periodic sequence with period 2seventually.

  • (ii) Ifsis odd, then$x_{n}$, $y_{n}$are periodic sequences with period 2teventually.

  • (iii) Ifsis even, then$x_{n}$is a periodic sequence with period 2teventually and$y_{n}$is a periodic sequence with period$2st$eventually.

• (2) Assume$B=1/A>A$. Then one of the following statements holds.

  • (i) $y_{n}=B$eventually and$x_{n}$is a periodic sequence with period 2seventually.

  • (ii) Ifsis odd, then$x_{n}$, $y_{n}$are periodic sequences with period 2teventually.

  • (iii) Ifsis even, then$y_{n}$is a periodic sequence with period 2teventually and$x_{n}$is a periodic sequence with period$2st$eventually.

Proof

Assume that $A=1/B>B$. If $x_{n}=A$ eventually, then by Lemma 2.1 we see that $y_{n}$ is a periodic sequence with period 2s eventually. Now we assume that $x_{n}\neq A$ eventually. Then we have $A_{i}>A$ (or $x_{2nt+i}>A$ eventually and $A_{i}=A$) for some $0\leq i\leq 2t-1$.

If s is odd, then $\gcd (2t,s)=1$. Thus, for every $j\in \{0,1,2,\ldots ,2t-1\}$, there exist some $1\leq i_{j}\leq 2t$ and integer $\lambda _{j}$ such that $i_{j}s=\lambda _{j} 2t+j$ since $\{rs:0\leq r\leq 2t-1\}=\{0,1,2,\ldots ,2t-1\}$ (mod 2t). By Lemma 2.3 and Lemma 2.4 we see that, for any $k\in { \mathbf{{N}}}$, $x_{2nt+ks+i}$ and $y_{2nt-t+ks+i}$ are constant sequences eventually. Thus, for any $0\leq r\leq 2t-1$, $x_{2nt+r}$ and $y_{2nt+r}$ are constant sequences eventually, which implies that $x_{n}$, $y_{n}$ are periodic sequences with period 2t eventually.

In the following, we assume that s is even with $s=2s'$. Then $\gcd (t,s')=1$ and t is odd. Thus, for every $j\in \{0,1,2,\ldots ,t-1\}$, there exist some $1\leq i_{j}\leq t$ and integer $\lambda _{j}$ such that $i_{j}s'=\lambda _{j} t+j$ and $i_{j}s=\lambda _{j} 2t+2j$.

If $x_{2nt+i}\neq A$ eventually for some $i\in \{0,2,\ldots \}$ and $x_{2nt+l}\neq A$ eventually for some $l\in \{1,3,\ldots \}$, then by Lemma 2.3 and Lemma 2.4 we see that, for any $k\in { \mathbf{{N}}}$, $x_{2nt+ks+i}$, $y_{2nt-t+ks+i}$, $x_{2nt+ks+l}$ and $y_{2nt-t+ks+l}$ are constant sequences eventually. Thus, for any $0\leq r\leq 2t-1$, $x_{2nt+r}$ and $y_{2nt+r}$ are constant sequences eventually, which implies that $x_{n}$, $y_{n}$ are periodic sequences with period 2t eventually.

If $x_{2nt+i}\neq A$ eventually for some $i\in \{0,2,\ldots \}$ and $x_{2nt+l}=A$ eventually for any $l\in \{1,3,\ldots \}$, then by Lemma 2.3 and Lemma 2.4 we see that, for any $k\in { \mathbf{{N}}}$, $x_{2nt+ks+i}$ and $y_{2nt-t+ks+i}$ are constant sequences eventually. This implies that, for every $r\in \{0,1,2,\ldots ,2t-1\}$, $x_{2nt+r}$ is constant sequence eventually and for every $l\in \{1,3,\ldots \}$, $y_{2nst+l}$ is constant sequence eventually. By (1.1) we see that there exists $N\in { \mathbf{{N}}}$ such that, for any $n\geq N$ and $r\in \{0,2,\ldots \}$,

$$ y_{2nt+r} = \max \biggl\{ B ,\frac{A}{y_{2nt+r-s}} \biggr\} . $$

(2.33)

Then we have $y_{2nt+r}y_{2nt+r-s}\geq A$. Thus, for any $n\geq N$ and $l\in \{1,3,\ldots \}$ and $k\in { \mathbf{{N}}}$,

$$\begin{aligned} B&\leq y_{2nt+t+2ks+l} = \max \biggl\{ B , \frac{x_{2nt+2ks+l}}{y_{2nt+t+2ks+l-s}} \biggr\} \\ &= \max \biggl\{ B , \frac{Ay_{2nt+t+2ks+l-2s}}{y_{2nt+t+2ks+l-s}y_{2nt+t+2ks+l-2s}} \biggr\} \\ &\leq \max \{B,y_{2nt+t+2ks-2s+l}\} \\ &= y_{2nt+t+2ks-2s+l}\quad \mbox{eventually}. \end{aligned}$$

(2.34)

Then, for every $n\geq N$ and $l\in \{1,3,\ldots \}$, we have

$$ B\leq \cdots \leq y_{2nt+t+2ks+l}\leq y_{2nt+t+2ks-2s+l}\leq y_{2nt+t+2s+l} \leq y_{2nt+t+l}. $$

(2.35)

We claim that, for every $n\geq N$ and $l\in \{1,3,\ldots \}$, $\{y_{2nt+t+2ks+l}\}_{k\in { \mathbf{{N}}}}$ is a constant sequence eventually. Assume on the contrary that, for some $n\geq N$ and some $l\in \{1,3,\ldots \}$, $\{y_{2nt+t+2ks+l}\}_{k\in { \mathbf{{N}}}}$ is not a constant sequence eventually. Then there exists a sequence of positive integers $k_{1}< k_{2}<\cdots$ such that, for any $r\in { \mathbf{{N}}}$, we have

$$ \begin{aligned}[b] B&< y_{2nt+t+2k_{r}s+l} = \frac{A}{y_{2nt+t+2k_{r}s+l-s}} \\ &< y_{2nt+t+2k_{r-1}s+l}=\frac{A}{ y_{2nt+t+2k_{r-1}s+l-s}}, \end{aligned} $$

(2.36)

which implies $y_{2nt+t+2k_{r}s+l-s}>y_{2nt+t+2k_{r-1}s+l-s}$ for any $r\in { \mathbf{{N}}}$. This is a contradiction. Take $ps>N$. Then $y_{2nst+t+l}=y_{2pst+t+2(nt-pt)s+l}$ is a constant sequence eventually for any $l\in \{1,3,\ldots \}$. From the above we see that $y_{n}$ is a periodic sequence with period $2st$ eventually.

In a similar fashion, we can show that if $x_{2nt+i}=A$ eventually for any $i\in \{0,2,\ldots \}$ and $x_{2nt+l}\neq A$ eventually for some $l\in \{1,3,\ldots \}$, then also statement (1(iii)) holds.

The second case follows from the previously proved one by interchanging letters. The proof is complete. □

Now we assume that $A=B=1$. Then, for any $0\leq i\leq 2t-1$ and $n\in { \mathbf{{N}}}_{0}$, we have $1\leq x_{2(n+1)t+i}\leq x_{2nt+i}$ eventually and $1\leq y_{2(n+1)t+i}\leq y_{2nt+i}$ eventually.

Lemma 2.5

Let$A=B=1$and$s\geq t$. Then the following statements hold.

• (1) If$A_{i}=1$, then$B_{t+i}=1$. If$B_{i}=1$, then$A_{t+i}=1$.

• (2) If$x_{N}=1$for some$N\in { {\mathbf{{N}}}}$and$A_{2nt+N+ks}=1$for any$k,n\in { {\mathbf{{N}}}}$, then$x_{2nt+N+ks}=y_{2nt+t+ks+N}=1$for any$k,n\in { {\mathbf{{N}}}}$. If$y_{N}=1$for some$N\in { {\mathbf{{N}}}}$and$B_{2nt+N+ks}=1$for any$k,n\in { {\mathbf{{N}}}}$, then$y_{2nt+N+ks}=x_{2nt+t+ks+N}=1$for any$k,n\in { {\mathbf{{N}}}}$.

• (3) Ifsis even and$\gcd (s,t)=1$, then$1\in \{x_{n}:n\in \{0,2,\ldots \}\}\cup \{y_{t+n}:n\in \{0,2,\ldots \}\}$and$1\in \{x_{n}:n\in \{1,3,\ldots \}\}\cup \{y_{t+n}:n\in \{1,3,\ldots \}\}$.

• (4) $1\in \{x_{n}:n\in { {\mathbf{{N}}}}\}\cup \{y_{n}:n\in { {\mathbf{{N}}}}\}$.

Proof

(1) Assume that $A_{i}=1$. Assume on the contrary that $B_{t+i}>1$. It follows from (1.1) that

$$ y_{2nt+t+i} =\frac{x_{2nt+i}}{y_{2nt+t-s+i}}. $$

(2.37)

This implies

$$ 1\leq \lim_{n\longrightarrow \infty }y_{2nt+t-s+i}=\frac{1}{B_{t+i}}< 1. $$

(2.38)

This is a contradiction. The second case follows from the previously proved one by interchanging letters.

(2) If $x_{N}=1$ for some $N\in { \mathbf{{N}}}$, then $x_{2nt+N}=1$ for any $n\in { \mathbf{{N}}}$. It follows from (1.1) that

$$ \begin{aligned}[b] y_{2nt+t+N}& = \max \biggl\{ 1, \frac{x_{2nt+N}}{y_{2nt+t-s+N}} \biggr\} \\ &=\max \biggl\{ 1,\frac{1}{y_{2nt+t-s+N}}\biggr\} =1\end{aligned} $$

(2.39)

and

$$ \begin{aligned}[b] y_{2nt+t+s+N}& = \max \biggl\{ 1, \frac{x_{2nt+s+N}}{y_{2nt+t+N}} \biggr\} \\ &=x_{2nt+s+N}\end{aligned} $$

(2.40)

and

$$ \begin{aligned}[b] x_{2(n+1)t+N+s}& = \max \biggl\{ 1, \frac{y_{2nt+t+s+N}}{x_{2(n+1)t+N}} \biggr\} \\ &=\max \{1,y_{2nt+t+s+N}\} \\ &=y_{2nt+t+s+N} \\ &=x_{2nt+N+s}.\end{aligned} $$

(2.41)

Thus $x_{2nt+N+s}=y_{2nt+t+s+N}=1$ for any $n\in { \mathbf{{N}}}$. In a similar fashion, we can show that $x_{2nt+N+ks}=y_{2nt+t+ks+N}=1$ for any $k,n\in { \mathbf{{N}}}$. The second case follows from the previously proved one by interchanging letters.

(3) If s is even and $\gcd (s,t)=1$, then t is odd. Assume on the contrary that $1\notin \{x_{n}:n\in \{0,2,\ldots \}\}\cup \{y_{t+n}:n\in \{0,2, \ldots \}\}$. Then it follows from (1.1) that, for any $n\in { \mathbf{{N}}}$,

$$ \begin{aligned}[b] y_{2nt+t}& = \max \biggl\{ 1, \frac{x_{2nt}}{y_{2nt+t-s}} \biggr\} \\ &=\frac{x_{2nt}}{y_{2nt+t-s}}>1 \end{aligned} $$

(2.42)

and

$$ \begin{aligned}[b] x_{2nt+2t-s}& = \max \biggl\{ 1, \frac{y_{2nt+t-s}}{x_{2nt+2t-2s}} \biggr\} \\ &=\frac{y_{2nt+t-s}}{x_{2nt+2t-2s}}>1.\end{aligned} $$

(2.43)

Thus

$$ \begin{aligned}[b] x_{2nt}& > x_{2nt-2(s-t)} \\ &>x_{2nt-4(s-t)} \\ &\quad \cdots \\ &>x_{2t(n-t+s)}. \end{aligned} $$

(2.44)

This is a contradiction.

(4) Case (4) is treated similarly to case (3). The proof is complete. □

Theorem 2.3

Let$A=B=1$and$s\geq t$. Then one of the following statements holds.

• (1) $x_{n}=1$eventually and$y_{n}$is a periodic sequence with period 2seventually.

• (2) $y_{n}=1$eventually and$x_{n}$is a periodic sequence with period 2seventually.

• (3) $x_{n}$, $y_{n}$are periodic sequences with period 2teventually.

Proof

If $x_{n}=1$ (or $y_{n}=1$) eventually, then by Lemma 2.1 we see that $y_{n}$ (or $x_{n}$) is a periodic sequence with period 2s eventually. Now we assume that $x_{n}\neq1$ eventually. Then we have $A_{i}>1$ for some $0\leq i\leq 2t-1$ or $\lim_{n\longrightarrow \infty }x_{n}=1$.

If s is odd, then $\gcd (2t,s)=1$. Thus, for every $j\in \{0,1,2,\ldots ,2t-1\}$, there exist some $1\leq i_{j}\leq 2t$ and integer $\lambda _{j}$ such that $i_{j}s=\lambda _{j} 2t+j$. By Lemma 2.3 and Lemma 2.5 we see that, for any $k\in { \mathbf{{N}}}$, $x_{2nt+ks+i}$ and $y_{2nt-t+ks+i}$ are constant sequences eventually, or for some $N\in { \mathbf{{N}}}$, $x_{2nt+N+ks}=y_{2nt+t+ks+N}=1$ for any $k,n\in { \mathbf{{N}}}$, or for some $N\in { \mathbf{{N}}}$, $y_{2nt+N+ks}=x_{2nt+t+ks+N}=1$ for any $k,n\in { \mathbf{{N}}}$. Thus, for any $0\leq r\leq 2t-1$, $x_{2nt+r}$ and $y_{2nt+r}$ are constant sequences eventually, which implies that $x_{n}$, $y_{n}$ are periodic sequences with period 2t eventually.

In the following, we assume that s is even with $s=2s'$, then $\gcd (t,s')=1$ and t is odd. Thus, for every $j\in \{0,1,2,\ldots ,t-1\}$, there exist some $1\leq i_{j}\leq t$ and integer $\lambda _{j}$ such that $i_{j}s=\lambda _{j} 2t+2j$.

If $A_{i}>1$ for some $i\in \{0,2,\ldots \}$, then by Lemma 2.3 we see that, for any $k\in { \mathbf{{N}}}$, $x_{2nt+ks+i}$ and $y_{2nt-t+ks+i}$ are constant sequences eventually. If $A_{i}=1$ for any $i\in \{0,2,\ldots \}$, then by Lemma 2.5 we have $B_{t+i}=1$ for any $i\in \{0,2,\ldots \}$ and $x_{2nt+i+ks}=y_{2nt-t+ks+i}=1$ for any $k\in { \mathbf{{N}}}$ eventually. In a similar fashion, also we can show that, for any $i\in \{1,3,\ldots \}$, $x_{2nt+ks+i}$ and $y_{2nt-t+ks+i}$ are constant sequences eventually for any $k\in { \mathbf{{N}}}$, or $x_{2nt+i+ks}=y_{2nt-t+ks+i}=1$ for any $k\in { \mathbf{{N}}}$ eventually for any $i\in \{1,3,\ldots \}$ and $k\in { \mathbf{{N}}}$. Thus, for any $0\leq r\leq 2t-1$, $x_{2nt+r}$ and $y_{2nt+r}$ are constant sequences eventually. This implies that $x_{n}$, $y_{n}$ are periodic sequences with period 2t eventually.

Using the previously proved one by interchanging letters, also we can show that if $y_{n}\neq1$ eventually, then $x_{n}$, $y_{n}$ are periodic sequences with period 2t eventually. The proof is complete. □

In Example 3.1 of [37], we showed that the equation

$$ x_{n}=\frac{x_{n-t}}{x_{n-s}} (t>s) $$

(2.45)

has a positive solution $z_{n}$ ($n\geq -t$) with $1< z_{n+1}< z_{n}$ for any $n\geq -t$ and $\lim_{n\longrightarrow \infty }z_{n}=1$.

From Example 3.1 of [37], we obtain the following theorem.

Theorem 2.4

Let$A\leq 1$and$B\leq 1$and$s< t$. Assume$z_{n}$ ($n\geq -t$) is a positive solution of (2.45) with$1< z_{n+1}< z_{n}$for any$n\geq -t$and$\lim_{n\longrightarrow \infty }z_{n}=1$. Then equation (1.1) have a solution$(x_{n},y_{n})$with$1< x_{n+1}=y_{n+1}=z_{n+1}< x_{n}=y_{n}=z_{n}$for any$n\geq -t$and$\lim_{n\longrightarrow \infty }x_{n}=\lim_{n\longrightarrow \infty }y_{n}=1$.