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., [4–6] 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., [12–30] 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 2.1–2.4 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 \(T_{x},T_{y}\in\{2,4,6,12\}\).