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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 \(d=\max \{t,s\}\).

Introduction

Concrete nonlinear difference equations and systems have attracted some recent attention (see, e.g., [139]). 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. [15, 79, 1116, 1825, 2830, 3236, 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.

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. (1)

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

  2. (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. (1)

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

    1. (i)

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

    2. (ii)

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

    3. (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. (2)

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

    1. (i)

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

    2. (ii)

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

    3. (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. (1)

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

  2. (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. (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. (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. (1)

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

  2. (2)

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

  3. (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\(\lim_{n\longrightarrow \infty }x_{n}=\lim_{n\longrightarrow \infty }y_{n}=1\).

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The authors would like to thank the referees for their valuable comments and suggestions.

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The research was supported by NNSF of China (11761011, 71862003) and NSF of Guangxi (2018GXNSFAA294010) and SF of Guangxi University of Finance and Economics (2019QNB10).

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Su, G., Han, C., Sun, T. et al. On the solutions of a max-type system of difference equations of higher order. Adv Differ Equ 2020, 213 (2020). https://doi.org/10.1186/s13662-020-02673-2

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MSC

  • 39A10
  • 39A11

Keywords

  • Max-type system of difference equations
  • Solution
  • Eventual periodicity