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Theory and Modern Applications

On the solutions of a max-type system of difference equations with period-two parameters

Abstract

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

$$\textstyle\begin{cases}x_{n} = \max \{A_{n},\frac{y_{n-1}}{x_{n-2}} \},\\ y_{n} = \max \{B_{n} ,\frac{x_{n-1}}{y_{n-2}} \}, \end{cases}\displaystyle n\in \{0,1,2,\ldots\}, $$

where \(A_{n},B_{n}\in(0, +\infty)\) are periodic sequences with period 2 and the initial values \(x_{-1},y_{-1},x_{-2},y_{-2}\in (0,+\infty)\). We show that every solution of the above system is eventually periodic.

1 Introduction

Our purpose in this paper is to study eventual periodicity of the following max-type system of difference equations:

$$\begin{aligned} \textstyle\begin{cases}x_{n} = \max \{A_{n},\frac{y_{n-1}}{x_{n-2}} \},\\ y_{n} = \max \{B_{n} ,\frac{x_{n-1}}{y_{n-2}} \}, \end{cases}\displaystyle n\in {\mathbf{N}}_{0}\equiv\{0,1,\ldots\}, \end{aligned}$$
(1.1)

where \(A_{n},B_{n}\in\mathbf{R}_{+}\equiv(0,+\infty)\) are periodic sequences with period 2 and the initial values \(x_{-2},y_{-2}, x_{-1}, y_{-1}\in \mathbf{R}_{+}\).

In [1], Fotiades and Papaschinopoulos studied the following max-type system of difference equations:

$$\begin{aligned} \textstyle\begin{cases} x_{n} = \max \{A,\frac{y_{n-1}}{x_{n-2}} \},\\ y_{n} = \max \{B ,\frac{x_{n-1}}{y_{n-2}} \}, \end{cases}\displaystyle n\in {\mathbf{N}}_{0} \end{aligned}$$
(1.2)

with \(A,B\in\mathbf{R}_{+}\) and showed that every positive solution of (1.2) is eventually periodic.

In [2], we studied the eventually periodic solutions of the following max-type system of difference equations:

$$\begin{aligned} \textstyle\begin{cases} x_{n} = \max \{A,\frac{y_{n-k}}{x_{n-1}} \},\\ y_{n} = \max \{B,\frac{x_{n-k}}{y_{n-1}} \}, \end{cases}\displaystyle n\in{\mathbf{N}}_{0}, \end{aligned}$$
(1.3)

where \(A,B\in{\mathbf{R}}_{+}\), \(k\in{\mathbf{N}}\equiv\{1,2,\ldots\}\) and the initial values \(x_{-k},y_{-k},x_{-k+1},y_{-k+1}, \ldots, x_{-1}, y_{-1}\in{\mathbf{R}}_{+}\).

Recently, there has been a great interest in studying max-type systems of difference equations. In 2012, Stević in [3] obtained in an elegant way the general solution to the following max-type system of difference equations:

$$\begin{aligned} \textstyle\begin{cases} x_{n+1}=\max \{\frac{A}{x_{n}},\frac{y_{n}}{x_{n}} \} ,\\ y_{n+1}=\max \{\frac{A}{y_{n}},\frac{x_{n}}{y_{n}} \}, \end{cases}\displaystyle n\in\mathbf{N}_{0} \end{aligned}$$
(1.4)

for the case \(x_{0},y_{0}\geq A>0\) and \(y_{0}/x_{0}\geq\max\{A,1/A\}\). The solvability of various systems of difference equations has reattracted some recent interest, see, e.g., [46] and the references therein.

In 2016, we in [7] studied the following max-type system of difference equations:

$$\begin{aligned} \textstyle\begin{cases} x_{n} = \max \{\frac{1}{x_{n-m}}, \min \{1,\frac{A}{y_{n-r}} \} \},\\ y_{n} = \max \{\frac{1}{y_{n-m}}, \min \{1,\frac{B}{x_{n-t}} \} \}, \end{cases}\displaystyle n\in{\mathbf{N}}_{0}, \end{aligned}$$
(1.5)

where \(A,B\in{\mathbf{R}}_{+}\), \(m,r,t\in{\mathbf{N}}\) and the initial values \(x_{-d},y_{-d},x_{-d+1},y_{-d+1}, \ldots, x_{-1}, y_{-1}\in{\mathbf{\mathbf{R}}}_{+}\) with \(d=\max\{m,r,t\}\) and showed that every positive solution of (1.5) is eventually periodic with period 2m.

When \(m=r=t=1\) and \(A=B\), (1.5) reduces to the max-type system of difference equations

$$\begin{aligned} \textstyle\begin{cases} x_{n}=\max \{\frac{1}{x_{n-1}},\min \{1,\frac {A}{y_{n-1}} \} \},\\ y_{n}=\max \{\frac{1}{y_{n-1}},\min \{1,\frac{A}{x_{n-1}} \} \}, \end{cases}\displaystyle n\in\mathbf{N}_{0}. \end{aligned}$$
(1.6)

In 2015, the authors of [8] obtained the general solution of system (1.6).

Motivated by papers [9, 10], in 2014, Stević et al. in [11] investigated the following max-type system of difference equations:

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

For some results of some properties of many max-type difference equations and systems, such as eventual periodicity, the boundedness character, and attractivity, see, e.g., [1230] and the related references therein.

2 Main results and proofs

In this section, we study the eventual periodicity of positive solutions of system (1.1). Write \(x_{2n}=p_{n},x_{2n+1}=q_{n},y_{2n}=s_{n},y_{2n+1}=t_{n}\) for any \(n\in {\mathbf{N}}_{0}\). Then system (1.1) reduces to the system

$$\begin{aligned} \textstyle\begin{cases}p_{n} = \max \{A_{0},\frac{t_{n-1}}{p_{n-1}} \},\\ t_{n} = \max \{B_{1},\frac{p_{n}}{t_{n-1}} \},\\ q_{n} = \max \{A_{1},\frac{s_{n}}{q_{n-1}} \},\\ s_{n} = \max \{B_{0} ,\frac{q_{n-1}}{s_{n-1}} \}, \end{cases}\displaystyle n\in {\mathbf{N}}_{0}, \end{aligned}$$
(2.1)

where \(A_{0},A_{1},B_{0},B_{1}\in {\mathbf{R}}_{+}\) and the initial values \(s_{-1},t_{-1}, p_{-1},q_{-1}\in{\mathbf{R}}_{+}\).

The following lemma will be used in the proofs of our main results.

Lemma 2.1

Let \(\{x_{n}\}_{n\geq-1}\) be a solution of the following equation:

$$\begin{aligned} x_{n}=\max\biggl\{ A,\frac{B}{x_{n-1}}\biggr\} ,\quad n\in{ \mathbf{N}}_{0} \end{aligned}$$
(2.2)

with \(A,B\in {\mathbf{R}}_{+}\) and the initial value \(x_{-1}\in{\mathbf{R}}_{+}\). Then \(x_{n}\) is eventually periodic with period 2.

Proof

By (2.2) we see \(x_{n}x_{n-1}\geq B\) and \(x_{n}\geq A\) for \(n\in{\mathbf{N}}_{0}\) and for any \(n\geq2\),

$$\begin{aligned} A&\leq x_{n} = \max \biggl\{ A,\frac{B}{x_{n-1}} \biggr\} \\ &=\max \biggl\{ A,\frac{Bx_{n-2}}{x_{n-1}x_{n-2}} \biggr\} \\ &\leq \max\{A,x_{n-2}\}= x_{n-2}. \end{aligned}$$
(2.3)

Then, for every \(i\in\{0,1\}\), \(x_{2n+i}\) is eventually nonincreasing.

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

$$\begin{aligned} A&< x_{2k_{n+1}+i}=\frac{B}{x_{2k_{n+1}+i-1}} \\ &< x_{2 k_{n}+i}=\frac{B}{x_{2 k_{n}+i-1}}, \end{aligned}$$

which implies \(x_{2k_{n+1}+i-1}>x_{2k_{n}+i-1}\) for any \(n\in{\mathbf{N}}\). This is a contradiction. Thus \(x_{n}\) is eventually periodic with period 2. The proof is complete. □

From (2.1) we see that it suffices to consider the eventual periodicity of positive solutions of the following system:

$$\begin{aligned} \textstyle\begin{cases}u_{n} = \max \{A,\frac{v_{n-1}}{u_{n-1}} \},\\ v_{ n} = \max \{B ,\frac{u_{n}}{v_{n-1}} \}, \end{cases}\displaystyle n\in {\mathbf{N}}_{0}, \end{aligned}$$
(2.4)

where \(A,B\in{\mathbf{R}}_{+}\) and the initial values \(u_{-1}, v_{-1}\in{\mathbf{R}}_{+}\). Let \(\{(u_{n},v_{n})\}_{n\geq-1}\) be a solution of (2.4). From (2.4) it immediately follows that, for any \(n\in{\mathbf{\mathbf{N}}}_{0}\),

$$\begin{aligned} u_{n}\geq A \end{aligned}$$
(2.5)

and

$$\begin{aligned} v_{n}\geq B \end{aligned}$$
(2.6)

and

$$\begin{aligned} \textstyle\begin{cases} u_{n} = \max \{A,\frac{B}{u_{n-1}}, \frac {1}{v_{n-2}} \},\\ v_{n} = \max \{B,\frac{A }{v_{n-1}}, \frac{1}{u_{n-1}} \}. \end{cases}\displaystyle n\in {\mathbf{N}}, \end{aligned}$$
(2.7)

Lemma 2.2

If there exist \(k,p\in{\mathbf{N}}\) such that \(u_{p+k}=u_{k}\) and \(v_{p+k}=v_{k}\), then \(u_{n+p}=u_{n}\) and \(v_{n+p}=v_{n}\) for any \(n\geq k\).

Proof

It is easy to see that

$$u_{k+p+1} = \max \biggl\{ A,\frac{v_{k+p}}{u_{k+p}} \biggr\} =\max \biggl\{ A , \frac{v_{k}}{u_{k}} \biggr\} =u_{k+1} $$

and

$$v_{k+p+1} = \max \biggl\{ B ,\frac{u_{k+p+1}}{v_{k+p}} \biggr\} =\max \biggl\{ B , \frac{u_{k+1}}{v_{k}} \biggr\} =v_{k+1}. $$

Assume that, for some \(N\in{\mathbf{N}}\), we have \(u_{k+p+N}=u_{k+N}\) and \(v_{k+p+N}=v_{k+N}\). Then

$$u_{k+p+N+1} = \max \biggl\{ A,\frac{v_{k+p+N}}{u_{k+p+N}} \biggr\} =\max \biggl\{ A, \frac{v_{k+N}}{u_{k+N}} \biggr\} =u_{k+N+1} $$

and

$$v_{k+p+N+1} = \max \biggl\{ B ,\frac{u_{k+p+N+1}}{v_{k+p+N}} \biggr\} =\max \biggl\{ B , \frac{u_{k+N+1}}{v_{k+N}} \biggr\} =v_{k+N+1}. $$

By mathematical induction, we see that \(u_{n+p}=u_{n}\) and \(v_{n+p}=v_{n}\) for any \(n\geq k\). The proof is complete. □

Proposition 2.1

If \(A> B\geq1\), then \(u_{n}=A\) eventually and \(v_{n}\) is eventually periodic with period 2. If \(B\geq A\geq1\), then \(v_{n}=B\) eventually and \(u_{n}\) is eventually periodic with period 2.

Proof

If \(A> B\geq1\), then by (2.5)–(2.7) we see that, for \(n\geq 2\),

$$\begin{aligned} A&\leq u_{n} = \max \biggl\{ A,\frac{B}{u_{n-1}}, \frac {1}{v_{n-2}} \biggr\} \\ &\leq\max \biggl\{ A,\frac{B }{A }, \frac{1}{B } \biggr\} =A . \end{aligned}$$

Thus, for \(n\geq2\), we have \(u_{n}=A\) and

$$v_{n} = \max \biggl\{ B,\frac{A }{v_{n-1}} \biggr\} . $$

By Lemma 2.1 we see that \(v_{n}\) is eventually periodic with period 2.

If \(B\geq A\geq1\), then by (2.5)–(2.7) we see that, for \(n\geq 2\),

$$\begin{aligned} B&\leq v_{n} = \max \biggl\{ B,\frac{A }{v_{n-1}}, \frac {1}{u_{n-1}} \biggr\} \\ &\leq \max \biggl\{ B,\frac{A}{B}, \frac{1}{A} \biggr\} =B. \end{aligned}$$

Thus, for \(n\geq2\), we have \(v_{n}=B\) and

$$u_{n} = \max \biggl\{ A,\frac{B}{u_{n-1}} \biggr\} . $$

By Lemma 2.1 we see that \(u_{n}\) is eventually periodic with period 2. The proof is complete. □

Proposition 2.2

If \(B\geq1>A\geq1/B\), then \(v_{n}=B\) eventually and \(u_{n}\) is eventually periodic with period 2. If \(1/A>B\geq1>A\), then \(v_{n}=B\) eventually and \(u_{n}\) is eventually periodic with period 2 or \(u_{n},v_{n}\) are eventually periodic with period 3.

Proof

Assume that \(B\geq1>A\geq1/B\). By (2.5)–(2.7) we see that, for \(n\geq1\),

$$v_{n} = \max \biggl\{ B,\frac{A}{v_{n-1}}, \frac{1}{u_{n-1}} \biggr\} =B $$

since \(A/v_{n-1}\leq1\) and \(1/u_{n-1}\leq B\). By Lemma 2.1 we see that \(u_{n}\) is eventually periodic with period 2.

Assume that \(1/A>B\geq1>A\). Then by (2.5)–(2.7) we obtain

$$\begin{aligned} v_{n}& = \max \biggl\{ B, \frac{A }{v_{n-1}}, \frac{1}{u_{n-1}} \biggr\} \\ &=\max \biggl\{ B, \frac{1}{u_{n-1}} \biggr\} \quad (n\geq1) \end{aligned}$$
(2.8)

since \(A/v_{n-1}\leq1\).

If \(v_{n}=1/u_{n-1}\) eventually, then \(v_{n}u_{n-1}=1\) eventually and by (2.4) we have

$$\begin{aligned} u_{n}& = \max \biggl\{ A, \frac{v_{n-1}}{u_{n-1}} \biggr\} = \max \{A, v_{n}v_{n-1} \} \\ &=v_{n}v_{n-1}\quad \mbox{eventually} \end{aligned}$$
(2.9)

since \(v_{n}v_{n-1}\geq B^{2}> A\). Thus from (2.9) it follows that

$$\begin{aligned} u_{n+3}&= v_{n+3}v_{n+2}=\frac {v_{n+3}v_{n+2}v_{n+1}}{v_{n+1}} \\ &=\frac{v_{n+3}u_{n+2}v_{n+2}v_{n+1}}{u_{n+2}v_{n+1}u_{n}}{u_{n}} \\ &=\frac{1\times u_{n+2}}{u_{n+2}\times1}{u_{n}}=u_{n} \quad\mbox{eventually}, \end{aligned}$$

which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

If \(v_{n}=B\) eventually, then by (2.4) we have

$$u_{n} = \max \biggl\{ A, \frac{B}{u_{n-1}} \biggr\} \quad \mbox{eventually}. $$

By Lemma 2.1 we see that \(u_{n}\) is eventually periodic with period 2.

If there exists some \(k\in{\mathbf{ N}}\) such that

$$\begin{aligned} v_{k}=B\geq \frac{1}{u_{k-1}}\quad \mbox{and} \quad v_{k+1}= \frac{1}{u_{k}}>B, \end{aligned}$$
(2.10)

then by (2.4), (2.6), (2.8), and (2.10) it follows

$$\begin{aligned} u_{k+1}& = \max \biggl\{ A, \frac{v_{k}}{u_{k}} \biggr\} = \max \{A, v_{k+1}v_{k} \}=v_{k+1}v_{k}, \\ v_{k+2}& = \max \biggl\{ B, \frac{1}{u_{k+1}} \biggr\} =B, \\ u_{k+2}& = \max \biggl\{ A, \frac{v_{k+1}}{u_{k+1}} \biggr\} = \max \biggl\{ A, \frac{v_{k+1}}{v_{k+1}v_{k}} \biggr\} \\ &=\max \biggl\{ A, \frac{1}{v_{k}} \biggr\} =\frac{1}{B}, \\ v_{k+3}& = \max \biggl\{ B, \frac{1}{u_{k+2}} \biggr\} =B, \\ u_{k+3}& = \max \biggl\{ A, \frac{v_{k+2}}{u_{k+2}} \biggr\} =B^{2}, \\ v_{k+4}& = \max \biggl\{ B, \frac{1}{u_{k+3}} \biggr\} =B, \\ u_{k+4}& = \max \biggl\{ A,\frac{v_{k+3}}{u_{k+3}} \biggr\} = \frac{1}{B}, \\ v_{k+5}& = \max \biggl\{ B, \frac{1}{u_{k+4}} \biggr\} =B, \\ u_{k+5}& = \max \biggl\{ A,\frac{v_{k+4}}{u_{k+4}} \biggr\} = \frac{1}{B}. \end{aligned}$$

By Lemma 2.2 we see that \(v_{n}=B\ (n\geq k+2)\) and \(u_{k+2n}=1/B\ (n\geq1)\) and \(u_{k+2n+1}=B^{2}\ (n\geq1)\), which implies that \(v_{n}=B\) eventually and \(u_{n}\) is eventually periodic with period 2. The proof is complete. □

Proposition 2.3

If \(A\geq1>B\), then \(u_{n}=A\) eventually and \(v_{n}\) is eventually periodic with period 2.

Proof

If \(A\geq1>B\geq1/A\), then by (2.5)–(2.7) we see that, for \(n\geq2\),

$$\begin{aligned} u_{n} = \max \biggl\{ A,\frac{B}{u_{n-1}}, \frac {1}{v_{n-2}} \biggr\} =A \end{aligned}$$

since \(1/v_{n-2}\leq A \) and \(B < u_{n-1}\). Thus from (2.4) it follows

$$\begin{aligned} v_{n} = \max \biggl\{ B,\frac{A }{v_{n-1}} \biggr\} \quad\mbox{eventually}. \end{aligned}$$

By Lemma 2.1 we see that \(v_{n}\) is eventually periodic with period 2.

Now assume that \(1/B>A\geq1>B\). We claim that there exists a sequence of positive integers \(n_{1} < n_{2} < \cdots\) such that \(u_{n_{k}}=A\). Indeed, if \(u_{n}=v_{n-1}/u_{n-1}>A\) eventually, then

$$\begin{aligned} A^{2}&< u_{n}u_{n-1}=v_{n-1}=\max \biggl\{ B,\frac {u_{n-1}}{v_{n-2}} \biggr\} \\ &=\frac{u_{n-1}}{v_{n-2}} =\max \biggl\{ \frac{A}{v_{n-2}},\frac{1}{u_{n-2}} \biggr\} \\ &=\frac{1}{u_{n-2}} \quad\mbox{eventually}, \end{aligned}$$

which implies \(1 \leq A^{3}< u_{n}u_{n-1}u_{n-2}=1\), a contradiction.

If \(u_{n}=A\) eventually, then by Lemma 2.1 we see that \(v_{n}\) is eventually periodic with period 2.

If there exists some \(k\in{\mathbf{ N}}\) such that

$$\begin{aligned} u_{k}=A\geq \frac{v_{k-1}}{u_{k-1}} \quad\mbox{and}\quad u_{k+1}= \frac {v_{k}}{u_{k}}=\frac{v_{k}}{A}>A, \end{aligned}$$
(2.11)

then \(v_{k}=u_{k+1}u_{k}> A^{2}\) and by (2.4) and (2.11) it follows

$$\begin{aligned} v_{k+1}& = \max \biggl\{ B, \frac{u_{k+1}}{v_{k}} \biggr\} = \max \biggl\{ B, \frac {1}{u_{k}} \biggr\} =\frac{1}{A}, \\ u_{k+2}& = \max \biggl\{ A, \frac{v_{k+1}}{u_{k+1}} \biggr\} =\max \biggl\{ A, \frac{1}{v_{k}} \biggr\} =A, \\ v_{k+2}& = \max \biggl\{ B, \frac{u_{k+2}}{v_{k+1}} \biggr\} =A^{2}, \\ u_{k+3}& = \max \biggl\{ A, \frac{v_{k+2}}{u_{k+2}} \biggr\} =A, \\ v_{k+3}& = \max \biggl\{ B, \frac{u_{k+3}}{v_{k+2}} \biggr\} = \frac{1}{A}, \\ u_{k+4}& = \max \biggl\{ A, \frac{v_{k+3}}{u_{k+3}} \biggr\} =A, \\ v_{k+4}& = \max \biggl\{ B, \frac{u_{k+4}}{v_{k+3}} \biggr\} =A^{2}. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n}=A\ (n\geq k+2)\) and \(v_{k+2n-1}=1/A\ (n\geq1)\) and \(v_{k+2n}=A^{2}\ (n\geq1)\), which implies that \(u_{n}=A\) eventually and \(v_{n}\) is eventually periodic with period 2. The proof is complete. □

Proposition 2.4

If \(A< 1\) and \(B<1\), then \(u_{n}=A\) eventually and \(v_{n}\) is eventually periodic with period 2 or \(v_{n}=B\) eventually and \(u_{n}\) is eventually periodic with period 2 or \(u_{n},v_{n}\) are eventually periodic with period 3.

Proof

Note

$$u_{n} = \max \biggl\{ A,\frac{v_{n-1}}{u_{n-1}} \biggr\} . $$

There are three cases to consider.

Case 1. Assume that \(u_{n}=A\). By Lemma 2.1 we see that \(v_{n}\) is eventually periodic with period 2.

Case 2. Assume that

$$\begin{aligned} u_{n}=v_{n-1}/u_{n-1}>A \quad \mbox{eventually}. \end{aligned}$$
(2.12)

Then by (2.7) it follows

$$\begin{aligned} v_{n}=\max \biggl\{ B,\frac{A}{v_{n-1}},\frac{1}{u_{n-1}} \biggr\} = \max \biggl\{ B,\frac{1}{u_{n-1}} \biggr\} \quad\mbox{eventually}. \end{aligned}$$
(2.13)

If \(v_{n}=B\) eventually, then by Lemma 2.1 we see that \(u_{n}\) is eventually periodic with period 2.

If \(v_{n}=1/u_{n-1}>B\) eventually, then by (2.12) we have

$$\begin{aligned} u_{n+3} & = \frac{v_{n+2}}{u_{n+2}}=\frac{1}{u_{n+2}u_{n+1}} \\ &=\frac{u_{n}}{u_{n+2}u_{n+1}u_{n}} \\ &=\frac{u_{n}}{v_{n+1}u_{n}} \\ &=u_{n} \quad\mbox{eventually}, \end{aligned}$$

which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

In the following, we assume that there exists some \(k\in {\mathbf{ N}}\) such that, for every \(n\geq k\),

$$\begin{aligned} u_{n}= \frac{v_{n-1}}{u_{n-1}},\qquad v_{k}=B,\qquad v_{k+1}= \frac {1}{u_{k}}>B. \end{aligned}$$
(2.14)

Thus by (2.13) and (2.14) it follows

$$\begin{aligned} &u_{k+1} = \frac{B}{u_{k}}, \\ &u_{k+2} = \frac{v_{k+1}}{u_{k+1}}=\frac{1}{B}, \end{aligned}$$
(2.15)
$$\begin{aligned} &v_{k+2} = \max \biggl\{ B, \frac{1}{u_{k+1}} \biggr\} =\max \biggl\{ B, \frac{u_{k}}{B} \biggr\} . \end{aligned}$$
(2.16)

If \(v_{k+2}=B\geq u_{k}/B\), then by (2.13)–(2.16) we have

$$\begin{aligned} &u_{k+3} =\frac{v_{k+2}}{u_{k+2}}=B^{2}, \\ &v_{k+3} = \max \biggl\{ B, \frac{1}{u_{k+2}} \biggr\} =B, \\ &u_{k+4} = \frac{v_{k+3}}{u_{k+3}}=\frac{1}{B}, \\ &v_{k+4} = \max \biggl\{ B, \frac{1}{u_{k+3}} \biggr\} = \frac{1}{B^{2}}, \\ &u_{k+5} = \frac{v_{k+4}}{u_{k+4}}=\frac{1}{B}, \\ &v_{k+5} = \max \biggl\{ B, \frac{1}{u_{k+4}} \biggr\} =B. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+2)\), which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

If \(v_{k+2}= u_{k}/B>B\), then by (2.13)–(2.16) we have

$$\begin{aligned} &u_{k+3} = \frac{v_{k+2}}{u_{k+2}}=u_{k}, \\ &v_{k+3} = \max \biggl\{ B, \frac{1}{u_{k+2}} \biggr\} =B, \\ &u_{k+4} =\frac{v_{k+3}}{u_{k+3}}=\frac{B}{u_{k}}, \\ &v_{k+4} = \max \biggl\{ B, \frac{1}{u_{k+3}} \biggr\} = \frac{1}{u_{k}}, \\ &u_{k+5} = \frac{v_{k+4}}{u_{k+4}}=\frac{1}{B}, \\ &v_{k+5} = \max \biggl\{ B, \frac{1}{u_{k+4}} \biggr\} = \frac{u_{k}}{B}. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+2)\), which also implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

Case 3. Assume that there exists some \(k\in{\mathbf{ N}}\) such that

$$\begin{aligned} u_{k}=A\geq\frac{v_{k-1}}{u_{k-1}},\qquad u_{k+1}= \frac{v_{k}}{u_{k}}= \frac{v_{k}}{A}>A. \end{aligned}$$
(2.17)

Then \(v_{k}=u_{k+1}u_{k}>A^{2}\) and by (2.7) and (2.17) we have

$$\begin{aligned} v_{k+1} = \max \biggl\{ B, \frac{A}{v_{k}}, \frac{1}{u_{k}} \biggr\} =\max \biggl\{ B, \frac{1}{A} \biggr\} =\frac{1}{A} \end{aligned}$$
(2.18)

and

$$\begin{aligned} u_{k+2} = \max \biggl\{ A, \frac{v_{k+1}}{u_{k+1}} \biggr\} =\max \biggl\{ A, \frac{1}{v_{k}} \biggr\} . \end{aligned}$$
(2.19)

If \(u_{k+2} =A\geq1/v_{k}\) and \(\sqrt{B}\geq A\geq B^{2}\), then by (2.4), (2.18), and (2.19) it follows

$$\begin{aligned} &v_{k+2} = \max \biggl\{ B, \frac{u_{k+2}}{v_{k+1}} \biggr\} =B, \\ &u_{k+3} = \max \biggl\{ A, \frac{v_{k+2}}{u_{k+2}} \biggr\} = \frac{B}{A}, \\ &v_{k+3} = \max \biggl\{ B, \frac{u_{k+3}}{v_{k+2}} \biggr\} = \frac{1}{A}, \\ &u_{k+4} = \max \biggl\{ A, \frac{v_{k+3}}{u_{k+3}} \biggr\} = \frac{1}{B}, \\ &v_{k+4} = \max \biggl\{ B, \frac{u_{k+4}}{v_{k+3}} \biggr\} = \frac{A}{B}, \\ &u_{k+5} = \max \biggl\{ A, \frac{v_{k+4}}{u_{k+4}} \biggr\} = A, \\ &v_{k+5} = \max \biggl\{ B, \frac{u_{k+5}}{v_{k+4}} \biggr\} =B. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+2)\), which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

If \(u_{k+2} =A\geq1/v_{k}\) and \(\sqrt{B}>B^{2}> A \), then by (2.4), (2.18), and (2.19) it follows

$$\begin{aligned} &v_{k+2} = B,\qquad u_{k+3} = \frac{B}{A}, \\ &v_{k+3} = \frac{1}{A},\qquad u_{k+4} =\frac{1}{B}, \\ &v_{k+4} = \max \biggl\{ B, \frac{u_{k+4}}{v_{k+3}} \biggr\} =B, \\ &u_{k+5} = \max \biggl\{ A, \frac{v_{k+4}}{u_{k+4}} \biggr\} =B^{2}, \\ &v_{k+5} = \max \biggl\{ B, \frac{u_{k+5}}{v_{k+4}} \biggr\} =B, \\ &u_{k+6} = \max \biggl\{ A, \frac{v_{k+5}}{u_{k+5}} \biggr\} = \frac{1}{B}, \\ &v_{k+6} = \max \biggl\{ B, \frac{u_{k+6}}{v_{k+5}} \biggr\} = \frac{1}{B^{2}}, \\ &u_{k+7} = \max \biggl\{ A, \frac{v_{k+6}}{u_{k+6}} \biggr\} = \frac{1}{B}, \\ &v_{k+7} = \max \biggl\{ B, \frac{u_{k+7}}{v_{k+6}} \biggr\} = B. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+4)\), which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

If \(u_{k+2} =A\geq1/v_{k}\) and \(\sqrt{B}< A \), then by (2.4), (2.18), and (2.19) it follows

$$\begin{aligned} &v_{k+2} = \max \biggl\{ B, \frac{u_{k+2}}{v_{k+1}} \biggr\} =A^{2}, \\ &u_{k+3} = \max \biggl\{ A, \frac{v_{k+2}}{u_{k+2}} \biggr\} = A, \\ &v_{k+3} = \max \biggl\{ B, \frac{u_{k+3}}{v_{k+2}} \biggr\} = \frac{1}{A}, \\ &u_{k+4} = \max \biggl\{ A, \frac{v_{k+3}}{u_{k+3}} \biggr\} = \frac{1}{A^{2}}, \\ &v_{k+4} = \max \biggl\{ B, \frac{u_{k+4}}{v_{k+3}} \biggr\} = \frac{1}{A}, \\ &u_{k+5} = \max \biggl\{ A, \frac{v_{k+4}}{u_{k+4}} \biggr\} =A, \\ &v_{k+5} = \max \biggl\{ B, \frac{u_{k+5}}{v_{k+4}} \biggr\} =A^{2}. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+2)\), which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

If \(u_{k+2} =1/v_{k}>A\geq Bv_{k}\), then by (2.4), (2.18), and (2.19) it follows

$$\begin{aligned} &v_{k+2} = \max \biggl\{ B, \frac{u_{k+2}}{v_{k+1}} \biggr\} = \frac{A}{v_{k}}, \\ &u_{k+3} = \max \biggl\{ A, \frac{v_{k+2}}{u_{k+2}} \biggr\} = A, \\ &v_{k+3} = \max \biggl\{ B, \frac{u_{k+3}}{v_{k+2}} \biggr\} =v_{k}, \\ &u_{k+4} = \max \biggl\{ A, \frac{v_{k+3}}{u_{k+3}} \biggr\} = \frac{v_{k}}{A}, \\ &v_{k+4} = \max \biggl\{ B, \frac{u_{k+4}}{v_{k+3}} \biggr\} = \frac{1}{A}, \\ &u_{k+5} = \max \biggl\{ A, \frac{v_{k+4}}{u_{k+4}} \biggr\} = \frac{1}{v_{k}}, \\ &v_{k+5} = \max \biggl\{ B, \frac{u_{k+5}}{v_{k+4}} \biggr\} = \frac{A}{v_{k}}. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+2)\), which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

If \(u_{k+2} =1/v_{k}>A\) and \(A/B< v_{k}\leq1/B^{2}\), then by (2.4), (2.18), and (2.19) it follows

$$\begin{aligned} &v_{k+2} = \max \biggl\{ B, \frac{u_{k+2}}{v_{k+1}} \biggr\} =B, \\ &u_{k+3} = \max \biggl\{ A, \frac{v_{k+2}}{u_{k+2}} \biggr\} =Bv_{k}, \\ &v_{k+3} = \max \biggl\{ B, \frac{u_{k+3}}{v_{k+2}} \biggr\} =v_{k}, \\ &u_{k+4} = \max \biggl\{ A, \frac{v_{k+3}}{u_{k+3}} \biggr\} = \frac{1}{B}, \\ &v_{k+4} = \max \biggl\{ B, \frac{u_{k+4}}{v_{k+3}} \biggr\} =\max \biggl\{ B, \frac{1}{Bv_{k}} \biggr\} =\frac{1}{Bv_{k}}, \\ &u_{k+5} = \max \biggl\{ A, \frac{v_{k+4}}{u_{k+4}} \biggr\} = \frac{1}{v_{k}}, \\ &v_{k+5} = \max \biggl\{ B, \frac{u_{k+5}}{v_{k+4}} \biggr\} =B. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+2)\), which implies that \(u_{n},v_{n}\) are eventually periodic with period 3.

If \(u_{k+2} =1/v_{k}>A\) and \(v_{k}> 1/B^{2}\), then \(A< B^{2}\) and by (2.4), (2.18), and (2.19) it follows

$$\begin{aligned} &v_{k+2} = B,\qquad u_{k+3} = Bv_{k}, \\ &v_{k+3} = v_{k}, \qquad u_{k+4} = \frac{1}{B}, \\ &v_{k+4} = \max \biggl\{ B, \frac{u_{k+4}}{v_{k+3}} \biggr\} =\max \biggl\{ B, \frac{1}{Bv_{k}} \biggr\} =B, \\ &u_{k+5} = \max \biggl\{ A, \frac{v_{k+4}}{u_{k+4}} \biggr\} =B^{2}, \\ &v_{k+5} = \max \biggl\{ B, \frac{u_{k+5}}{v_{k+4}} \biggr\} =B, \\ &u_{k+6} = \max \biggl\{ A, \frac{v_{k+5}}{u_{k+5}} \biggr\} = \frac{1}{B}, \\ &v_{k+6} = \max \biggl\{ B, \frac{u_{k+6}}{v_{k+5}} \biggr\} = \frac{1}{B^{2}}, \\ &u_{k+7} = \max \biggl\{ A, \frac{v_{k+6}}{u_{k+6}} \biggr\} = \frac{1}{B}, \\ &v_{k+7} = \max \biggl\{ B, \frac{u_{k+7}}{v_{k+6}} \biggr\} = B. \end{aligned}$$

By Lemma 2.2 we see that \(u_{n+3}=u_{n}\) and \(v_{n+3}=v_{n}\ (n\geq k+4)\), which implies that \(u_{n},v_{n}\) are eventually periodic with period 3. The proof is complete. □

Combining (2.1) with (2.3), from Propositions 2.12.4 we obtain the following theorem.

Theorem 2.1

Let \(\{(x_{n},y_{n})\}_{n\geq-2}\) be a positive solution of (1.1). Then \(x_{n}\) and \(y_{n}\) are eventually periodic with periods \(T_{x}\) and \(T_{y}\), respectively, and \(T_{x},T_{y}\in\{2,4,6,12\}\).

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Acknowledgements

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

Funding

The research was supported by NNSF of China (11761011) and NSF of Guangxi (2016GXNSFBA380235, 2016GXNSFAA380286) and YMTBAPP of Guangxi Colleges (2017KY0598) and PIT of Guangxi University of Finance and Economics.

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Su, G., Sun, T. & Qin, B. On the solutions of a max-type system of difference equations with period-two parameters. Adv Differ Equ 2018, 358 (2018). https://doi.org/10.1186/s13662-018-1826-1

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