\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 234, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/234\hfil Cyclic system of difference equations]
{Long-term behavior of a cyclic max-type system of difference equations}

\author[T. Stevi\'c, B. Iri\v{c}anin \hfil EJDE-2015/234\hfilneg]
{Tatjana Stevi\'c, Bratislav Iri\v{c}anin}

\address{Tatjana Stevi\'c \newline
Faculty of Electrical Engineering, Belgrade University, 
Bulevar Kralja Aleksandra 73, 11000 Beograd, Serbia}
\email{tanjas019@gmail.com}

\address{Bratislav Iri\v{c}anin \newline
Faculty of Electrical Engineering,
Belgrade University, Bulevar Kralja Aleksandra 73,
11000 Beograd, Serbia}
\email{iricanin@etf.rs}

\thanks{Submitted June 6 2015. Published September 11, 2015.}
\subjclass[2010]{39A10, 39A20}
\keywords{Max-type system of difference equations; cyclic system;
\hfill\break\indent positive solutions; boundedness character; global attractivity}

\begin{abstract}
 We study the long-term behavior of positive solutions of the cyclic system of
 difference equations
 $$
 x^{(i)}_{n+1}=\max\Big\{\alpha,\frac{(x^{(i+1)}_n)^p}{(x^{(i+2)}_{n-1})^q}\Big\},
 \quad i=1,\ldots,k,\; n\in\mathbb{N}_0,
 $$
 where $k\in\mathbb{N}$, $\min\{\alpha, p, q\}>0$ and where we
 regard that $x^{(i_1)}_n=x^{(i_2)}_n$ when $i_1\equiv i_2$
 (mod $k$). We determine the set of parameters $\alpha$, $p$
 and $q$ in $(0,\infty)^3$ for which all such solutions are
 bounded. In the other cases we show that the system has unbounded
 solutions. For the case $p=q$ we give some sufficient conditions
 which guaranty the convergence of all positive solutions. The main
 results in this paper generalize and complement some recent ones.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


\section{Introduction}

Unlike the linear difference equations and systems, there is no
unified theory for nonlinear ones. The lack of the theory,
among other reasons, motivated numerous experts to study various
concrete nonlinear equations and systems, which seem or look like
good prototypes. Since the beginning of 1990's there have been published
a lot of papers on such equations and systems (see, e.g.,
\cite{ad}-\cite{bs-jdea09}, \cite{dkk}, \cite{fp},
\cite{gl}-\cite{pss}, \cite{ps-cana}-\cite{yl}).
Many of these papers study only positive solutions of the equations and systems.
This is not so unexpected since many of the equations and systems
appear in some applications (see, e.g., \cite{bc, gr, rr1, ijpam3,37264}
and the references cited therein).

After studying numerous special cases of rational equations of the
form
\begin{equation}
x_{n+1}=\alpha+\frac{x_{n-k}}{x_{n-l}},\quad
n\in\mathbb{N}_0, \label{c1}
\end{equation}
where $\alpha>0$,
$k,l\in\mathbb{N}_0$, $k\ne l$ (see, e.g., \cite{ad, bs1, bs2,
ijcm1, bs0, dkk, ddnsgut, tjm2, jdea1010, 53890, cma1, sx}), the
topic began developing in a natural direction, that is, into the
study of special cases of the non-rational difference equation
\begin{equation} \label{s5}
x_{n+1}=\alpha+\frac{x^p_{n-k}}{x_{n-l}^q},\quad
n\in\mathbb{N}_0,
\end{equation}
where $k,l\in\mathbb{N}_0$, $k\ne l$, and $\min\{\alpha, p, q\}>0$.
The critical moment for investigating positive solutions of equation
\eqref{s5} seems the publication of paper \cite{jamc1}, where the
case $p=q>0$, $k=l+1=1$ was studied, which was the initial motivation for
further investigations. Since that time equation \eqref{s5} and
its extensions have been extensively studied by several authors
(see, e.g., \cite{jdea129, i-aaa2010, i-amc2010, amc213, pss,
53890, 40963, na2009, csf3})


The first problem studied was the boudedness character of positive
solutions of equation \eqref{s5}. One of the first steps in the
study was \cite[Theorem 2]{tjm2} where it was given an elegant
proof for the boundedness of positive solutions of the following
equation with a variable coefficient
\begin{equation}
x_{n+1}=\alpha_n+\frac{x_n}{x_{n-k}}, \quad n\in\mathbb{N}_0,\label{eqgen}
\end{equation}
where $k\in\mathbb{N}$ and $(\alpha_n)_{n\in\mathbb{N}_0}$
is a sequence of real numbers such that
$0<M_1\le \alpha_n\le M_2<+\infty$, $n\in\mathbb{N}_0$.
Another important source related to nonlinear difference equations
with variable coefficients is note \cite{dcdis1}, because it gives
some conditions which guarantee the monotonicity of the
subsequences $(x_{2n})$ and $(x_{2n+1})$ for all solutions of a
related difference equation, which along with the boundedness of
the solutions produce the eventual periodicity.

On the other hand, Papaschinopoulos and Schinas initiated the
study of symmetric systems of difference equations in the second
half of 1990's, and since that time there have been published a
lot of papers in the topic (see, e.g., \cite{jdea128, amc218-sde,
fp, p3, prs, ps1, ps2, ps3, ps4, ps-jdea2011, ps-cana, ijpam1,
amc218-sys1, amc-thos, amc-maxpsde, amc-218solsys, amc-mtsde,
amc219-ssmtde, ejqtde-fib, ejqtde1, amc219-nsosde, ejqtde-maxsde,
amc235-psmtdce}).

In 2006 appeared paper \cite{dcdis4} which could be the first
paper which suggested investigation of cyclic systems of
difference equations. There are just a few papers on the topic.
Recently, in \cite{jdea205} was studied the boundedness character
of the following cyclic system of difference equations, which is a
natural extension of the equation in \cite{40963} and the system
in \cite{ejqtde-maxsde}
\begin{equation}
x^{(i)}_{n+1}=\alpha+\frac{(x^{(i+1)}_n)^p}{(x^{(i+2)}_{n-1})^q},\quad
i=1,\ldots,k,\; n\in\mathbb{N}_0,\label{mf}
\end{equation}
where
$k\in\mathbb{N}$, $\min\{\alpha, p, q\}>0$, and where we regard
that
\begin{equation}
x^{(i_1)}_n=x^{(i_2)}_n,\quad\text{when}\quad i_1\equiv i_2\;
({\rm mod}\; k).\label{s2}
\end{equation}
Note that $i+2>k$, for $i\in\{k-1,k\}$, but we will also have some
situations in the paper where the superscript is bigger than $k+2$.

Another topic of recent interest is the investigation of, so
called, max-type difference equations and systems, which appeared
for the first time in the mid of 1990's. For some results in the
area up to 2004, see monograph \cite{gl}. A systematic study of
non-rational max-type difference equations started in the mid of
2000's (see, e.g., \cite{40963, aml6, amc216-max, na4, um83-2,
jmaa376, amc217-max, amc-maxpsde, amc-mtsde, amc219-ssmtde,
amc235-psmtdce, yl}).

The corresponding max-type system of difference equations to
\eqref{mf} is
\begin{equation}
x^{(i)}_{n+1}=\max\Big\{\alpha,\frac{(x^{(i+1)}_n)^p}{(x^{(i+2)}_{n-1})^q}\Big\},
\quad i=1,\ldots,k,\; n\in\mathbb{N}_0,\label{main}
\end{equation}
where $k\in\mathbb{N}$, $\min\{\alpha, p, q\}>0$, and where we
also accept the convention in \eqref{s2}.

Motivated by above mentioned line of investigations and especially
by papers \cite{amc219-ssmtde,amc219-nsosde,ejqtde-maxsde}
here we study the long-term behavior of
positive solutions of system \eqref{main}. Our results generalize
and complement some results in these papers.

For a solution $(\vec x_n)_{n\ge -1}=(x^{(1)}_n,\ldots,x^{(k)}_n)_{n\ge -1}$
of system \eqref{main} is said that is {\it bounded} if there is $L\ge 0$
such that
\begin{equation}
\sup_{n\ge -1}\|\vec x_n\|_2=\sup_{n\ge -1}
\Big(\sum_{i=1}^k(x^{(i)}_n)^2\Big)^{1/2}\le
L.\label{c6}
\end{equation}
Of course, in the definition, instead
of the Euclidean norm in ${\mathbb R}^k$ we could use any
equivalent one (for example, maximum norm). If we say that a
solution $(\vec x_n)_{n\ge -1}$ of system \eqref{main} is
positive, it will mean that $x^{(i)}_n>0$ for every $1\le i\le k$
and $n\ge -1$.

\section{Boundedness character of system \eqref{main}}

Prior to stating and proving our theorems we want to mention an
obvious estimate which will be frequently used from now on.
Namely, note that for any positive solution of system \eqref{main}
the following estimate holds
\begin{equation}
x^{(i)}_n\ge \alpha,\label{a7}
\end{equation}
for every $1\le i\le k$ and $n\in\mathbb{N}$.


\begin{theorem} \label{thm1}
If $2\sqrt q\leq p<1+q$ and $q\in(0,1)$, then all positive solutions of \eqref{main}
are bounded.
\end{theorem}

\begin{proof}
 It is easy to see that the conditions $2\sqrt q\leq
p<1+q$ and $q\in (0,1)$, imply that
$\lambda^2-p\lambda+q=(\lambda-\lambda_1)(\lambda-\lambda_2)$ for
$\lambda_1$ and $\lambda_2$ such that $0<\lambda_2\le\lambda_1<1$.
Hence
$$
x^{(i)}_{n+1}=\max\Big\{\alpha,\frac{(x^{(i+1)}_n)^{\lambda_1+\lambda_2}}
{(x^{(i+2)}_{n-1})^{\lambda_1\lambda_2}}\Big\}.
$$
From this and \eqref{a7}, for every positive solution of system
\eqref{main} we have 
\begin{equation}
\begin{aligned}
\frac{x^{(i)}_{n+1}}{(x^{(i+1)}_n)^{\lambda_1}}
&=\max\Big\{ \frac \alpha{(x^{(i+1)}_n)^{\lambda_1}}, 
\Big(\frac{x^{(i+1)}_{n}}{(x^{(i+2)}_{n-1})^{\lambda_1}} 
\Big)^{\lambda_2}\Big\} \\
&\leq \max\Big\{ \alpha^{1-\lambda_1}, 
\Big(\frac{x^{(i+1)}_{n}}{(x^{(i+2)}_{n-1})^{\lambda_1}} 
\Big)^{\lambda_2}\Big\}
\end{aligned}\label{mod}
\end{equation}
for every $1\le i\le k$ and $n\in\mathbb{N}$.
Let
$$
y^{(i)}_n=\frac{x^{(i)}_{n+1}}{(x^{(i+1)}_n)^{\lambda_1}},\quad 
i=1,\ldots,k,\quad n\in\mathbb{N}_0,
$$
and \begin{equation}
z_n=k\alpha^{1-\lambda_1}+kz_{n-1}^{\lambda_2},\quad 
n\in\mathbb{N},\label{ips}
\end{equation}
where 
\begin{equation} 
z_0=\sum_{i=1}^k y^{(i)}_0.\label{s3}
\end{equation}
From \eqref{mod} it follows that
\begin{equation}
\begin{aligned}
\sum_{i=1}^ky^{(i)}_n
&\le \sum_{i=1}^k\max\Big\{\alpha^{1-\lambda_1},(y^{(i)}_{n-1})^{\lambda_2}\Big\}
\le k\alpha^{1-\lambda_1} +\sum_{i=1}^k (y^{(i)}_{n-1})^{\lambda_2} \\
&\le k\alpha^{1-\lambda_1} +k\Big(\sum_{i=1}^k
y^{(i)}_{n-1}\Big)^{\lambda_2},\quad\text{for }
 n\in\mathbb{N}.
\end{aligned}\label{s6}
\end{equation}

Since the function
$f(x)=k\alpha^{1-\lambda_1}+kx^{\lambda_2}$
is nondecreasing on the interval $(0,\infty)$ and by using
\eqref{s3} in \eqref{s6} with $n=1$, we obtain
$$
\sum_{i=1}^ky^{(i)}_1\le z_1.
$$ 
From this, by using \eqref{s6}, a simple
inductive argument yields 
\begin{equation}
\sum_{i=1}^ky^{(i)}_n\leq z_n,\quad n\in \mathbb{N}_0.\label{c10}
\end{equation}

On the other hand, since $\lambda_2\in (0,1)$, function $f$ is
also concave on $(0,\infty)$. Hence, there is a unique solution
$x^*$ of the equation $f(x)=x$, and 
\begin{equation}
(f(x)-x)(x-x^*)<0,\quad x\in (0,\infty)\setminus\{x^*\}.
\end{equation}
 Moreover, from
$f(1)=k\alpha^{1-\lambda_1}+k>1$, it follows that $x^*>1$. So, if
$z_0\in (0,x^*]$, then $(z_n)_{n\in\mathbb{N}_0}$ is
nondecreasing and $z_n\le x^*$, $n\in\mathbb{N}_0$, while if
$z_0\geq x^*$, it is nonincreasing and $z_n\ge x^*$, $n\in\mathbb{N}_0$.
Hence, $(z_n)_{n\in\mathbb{N}_0}$ is bounded. This and
\eqref{c10} imply the existence of $M_1\ge x^*>1$ such that
$$
\sum_{i=1}^ky^{(i)}_n\le M_1,\quad n\in\mathbb{N}_0,
$$ 
and consequently 
$$
x^{(i)}_{n+1}\leq M_1(x^{(i+1)}_n)^{\lambda_1},
$$ 
for every $1\le i\le k$ and $n\in\mathbb{N}_0$.
Hence 
$$
\sum_{i=1}^kx^{(i)}_n\leq
kM_1\Big(\sum_{i=1}^kx^{(i)}_{n-1}\Big)^{\lambda_1},\quad
n\in\mathbb{N},
$$ 
from which we have
\[
\sum_{i=1}^kx^{(i)}_n
\leq (kM_1)^{\frac{1-{\lambda_1}^n}{1-{\lambda_1}}}
\Big(\sum_{i=1}^kx^{(i)}_0\Big)^{{\lambda_1}^n}
\leq(kM_1)^{\frac 1{1-{\lambda_1}}}\max\Big\{1,
\sum_{i=1}^kx^{(i)}_0\Big\},
\]
from which the
boundedness of any positive solution follows.
\end{proof}

Before we state the next result we introduce a notation in order
to save some space in writing some long formulas. Namely, if
$a_j$, $j=1,\ldots,l$, are nonnegative numbers and $r$ is a
positive real number then the notation
$$
\max\{a_1,a_2,\ldots,a_l\}^r$$ has the same meaning as
$(\max\{a_1,a_2,\ldots,a_l\})^r$, that is, as
$\max\{a_1^r,a_2^r,\ldots,a_l^r\}.$\bigskip


\begin{theorem} \label{thm2} 
 If $p^2<4q$, then all positive solutions of system \eqref{main} are bounded.
\end{theorem}

\begin{proof} 
Since  \eqref{main} is cyclic it is sufficient to
prove the boundedness of $(x^{(1)}_n)_{n\ge -1}$. Let $p_0=0$ and
\begin{equation}
 p_{k+1}=\frac q{p-p_k},\quad k\in\mathbb{N}_0.\label{defpk}
\end{equation}
We have
\begin{equation} \label{p2r}
\begin{aligned}
x^{(1)}_{n+1}
&=\max\Big\{\alpha,\frac{(x^{(2)}_n)^p}{(x^{(3)}_{n-1})^q}\Big\}
=\max\Big\{\alpha,\Big(\frac{x^{(2)}_n}{(x^{(3)}_{n-1})^{\frac qp}}\Big)^p\Big\}\\
&=\max\Big\{\alpha,\Big(\frac{x^{(2)}_n}{(x^{(3)}_{n-1})^{p_1}}\Big)^p\Big\}\\
&=\max\Big\{\alpha,\max\Big\{\frac \alpha{(x^{(3)}_{n-1})^{p_1}},
 \frac{(x^{(3)}_{n-1})^{p-p_1}}{(x^{(4)}_{n-2})^q}\Big\}^p\Big\}\\
&=\max\Big\{\alpha,\max\Big\{\frac \alpha{(x^{(3)}_{n-1})^{p_1}},
 \Big(\frac{x^{(3)}_{n-1}}{(x^{(4)}_{n-2})^{\frac{q}{p-p_1}}}
 \Big)^{p-p_1}\Big\}^p\Big\}\\
&=\max\Big\{\alpha,\max\Big\{\frac \alpha{(x^{(3)}_{n-1})^{p_1}},
 \Big(\frac{x^{(3)}_{n-1}}{(x^{(4)}_{n-2})^{p_2}}\Big)^{p-p_1}\Big\}^p\Big\}\\
&=\max\Big\{\alpha,\max\Big\{\frac\alpha{(x^{(3)}_{n-1})^{p_1}},
 \max\Big\{\frac\alpha{(x^{(4)}_{n-2})^{p_2}},
 \frac{(x^{(4)}_{n-2})^{p-p_2}}{(x^{(5)}_{n-3})^q}\Big\}^{p-p_1}\Big\}^p\Big\}.
\end{aligned}
\end{equation}
Now assume that for some $m$ such that $1\le m\le n+1$ we have
proved the following
\begin{equation}
x^{(1)}_{n+1}
=\max\Big\{\alpha,\ldots,\max\Big\{\frac{\alpha}{(x^{(m)}_{n-m+2})^{p_{m-2}}},
\frac{(x^{(m)}_{n-m+2})^{p-p_{m-2}}}{(x^{(m+1)}_{n-m+1})^q}\Big\}^{p-p_{m-3}}
\ldots\Big\}.\label{s7}
\end{equation}
Then by using \eqref{main} in \eqref{s7} we have
\begin{align*}
&x^{(1)}_{n+1}\\
&=\max\Big\{\alpha,\ldots,\max\Big\{\frac{\alpha}{(x^{(m)}_{n-m+2})^{p_{m-2}}},
\Big(\frac{x^{(m)}_{n-m+2}}{(x^{(m+1)}_{n-m+1})^\frac{q}{p-p_{m-2}}}
 \Big)^{p-p_{m-2}}\Big\}^{p-p_{m-3}}\ldots\Big\}\\
&=\max\Big\{\alpha,\ldots,\max\Big\{\frac{\alpha}{(x^{(m)}_{n-m+2})^{p_{m-2}}},
\Big(\frac{x^{(m)}_{n-m+2}}{(x^{(m+1)}_{n-m+1})^{p_{m-1}}}\Big)^{p-p_{m-2}}
 \Big\}^{p-p_{m-3}}\ldots\Big\}\\
&=\max\Big\{\alpha,\ldots,
\max\Big\{\frac{\alpha}{(x^{(m+1)}_{n-m+1})^{p_{m-1}}},
 \frac{(x^{(m+1)}_{n-m+1})^{p-p_{m-1}}}{(x^{(m+2)}_{n-m})^q}
 \Big\}^{p-p_{m-2}}\ldots\Big\}.
\end{align*}
From this, \eqref{p2r} and the method of induction we see that
\eqref{s7} holds for every $1\le m\le n+2$. We have to say that if
$p_m=p$ for some $m\in\mathbb{N}$, then the above (iterating)
procedure is stopped.

If $p^2\leq q$, which is equivalent to $p\le p_1$, by using
\eqref{a7} and \eqref{p2r} for $n\ge 3$, we obtain
$$
x^{(1)}_{n+1}=\max\Big\{\alpha,
\max\Big\{\frac \alpha{(x^{(3)}_{n-1})^{\frac qp}},
 \frac{(x^{(3)}_{n-1})^{p-\frac qp}}{(x^{(4)}_{n-2})^q}\Big\}^p\Big\}
\leq \max\Big\{\alpha,\frac 1{\alpha^{q-p}},
\frac 1{\alpha^{q-p^2+pq}}\Big\},
$$ 
which proves the boundedness of $x_n^{(1)}$ in this case.

Since $0=p_0<p_1=q/p$ and the function $g(x)=q/(p-x)$ is
increasing for $x<p$, we have that $p_k$ is increasing as far as
$p_k<p$. Assume that $p_k<p$, $k\in\mathbb{N}_0$. Then as a
monotone and bounded sequence it would have a finite limit $p^*$
such that
\begin{equation}
(p^*)^2-pp^*+q=0.\label{a1}
\end{equation}
 On the other hand, since $p^2<4q$ then equation \eqref{a1} does not have
real zeros. This implies that there is a $k_0\in\mathbb{N}$ such
that $p_{k_0-1}<p$ and $p_{k_0}\geq p$.

Using \eqref{s7} for $m=k_0+2$ and \eqref{a7}, we obtain
\begin{align*}
x^{(1)}_{n+1}
&=\max\Big\{\alpha,\ldots,\max\Big\{\frac{\alpha}{(x^{(k_0+2)}_{n-k_0})^{p_{k_0}}},
\frac{(x^{(k_0+2)}_{n-k_0})^{p-p_{k_0}}}{(x^{(k_0+3)}_{n-k_0-1})^q}
 \Big\}^{p-p_{k_0-1}}\ldots\Big\}\\
&\leq \max\Big\{\alpha,\ldots,\max\Big\{\frac1{\alpha^{p_{k_0}-1}},
\frac1{\alpha^{q-p+p_{k_0}}}\Big\}^{p-p_{k_0-1}}\ldots\Big\},
\end{align*}
for $n\geq k_0+2$, as desired.
\end{proof}

\begin{theorem} \label{thm3} 
If $\alpha>0$, $p=q+1$ and $q\in(0,1)$, then all positive solutions of 
\eqref{main} are bounded.
\end{theorem}

\begin{proof} 
We may suppose $\alpha=1$, since the change of variables
$$
x^{(i)}_n=\alpha\hat x^{(i)}_n,\quad i=1,\ldots,k,\quad n\ge -1,
$$
transforms system \eqref{main} into the same with $\alpha=1$.

Assume that sequences $(a_n)_{n\in\mathbb{N}_0}$ and
$(b_n)_{n\in\mathbb{N}_0}$ are defined by
\begin{equation} \label{a2}
\begin{gathered}
 a_0=q,\quad b_0=q+1,\\
a_{2n+1}=(q+1)b_{2n}-a_{2n},\quad b_{2n+1}=qb_{2n},\quad
n\in\mathbb{N}_0,\\
b_{2n+2}=(q+1)a_{2n+1}-b_{2n+1},\quad a_{2n+2}=qa_{2n+1},\quad
n\in\mathbb{N}_0.
\end{gathered}
\end{equation}
Using \eqref{main} and \eqref{a2} we have
\begin{align}
x^{(1)}_{n+1}
&= \max\Big\{1,\frac{(x^{(2)}_n)^{q+1}}{(x^{(3)}_{n-1})^q}\Big\}
 =\max\Big\{1,\frac{(x^{(2)}_n)^{b_0}}{(x^{(3)}_{n-1})^{a_0}}\Big\} \nonumber\\
& = \max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}
 \frac{(x^{(3)}_{n-1})^{(q+1)b_0-a_0}}{(x^{(4)}_{n-2})^{qb_0}}\Big\} \nonumber\\
&=\max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}
 \frac{(x^{(3)}_{n-1})^{a_1}}{(x^{(4)}_{n-2})^{b_1}}\Big\} \label{s10} \\
&=\max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},
\frac{(x^{(4)}_{n-2})^{(q+1)a_1-b_1}}{(x^{(5)}_{n-3})^{qa_1}}\Big\} \nonumber\\
&=\max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},
\frac{(x^{(4)}_{n-2})^{b_2}}{(x^{(5)}_{n-3})^{a_2}}\Big\} \label{s11}
\end{align}
Now assume that for some $m$,  $4\le 2m\le n$ we have
proved the following two equalities
\begin{align}
&x^{(1)}_{n+1} \nonumber\\
&=\max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},
 \ldots,\frac1{(x^{(2m-1)}_{n-2m+3})^{a_{2m-4}}},
\frac{(x^{(2m-1)}_{n-2m+3})^{a_{2m-3}}}{(x^{(2m)}_{n-2m+2})^{b_{2m-3}}}\Big\}
 \label{s8}\\
&=\max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},
 \ldots,\frac1{(x^{(2m)}_{n-2m+2})^{b_{2m-3}}},
\frac{(x^{(2m)}_{n-2m+2})^{b_{2m-2}}}{(x^{(2m+1)}_{n-2m+1})^{a_{2m-2}}}\Big\}
\label{s9}.
\end{align}

By using \eqref{main} in \eqref{s9} we obtain
\begin{align*}
&x^{(1)}_{n+1}\\
&= \max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},
 \ldots,\frac1{(x^{(2m)}_{n-2m+2})^{b_{2m-3}}},
\frac{(x^{(2m)}_{n-2m+2})^{b_{2m-2}}}{(x^{(2m+1)}_{n-2m+1})^{a_{2m-2}}}\Big\}\\
&= \max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},\ldots,
\frac1{(x^{(2m+1)}_{n-2m+1})^{a_{2m-2}}},\frac{(x^{(2m+1)}_{n-2m+1})^{(q+1)
 b_{2m-2}-a_{2m-2}}}{(x^{(2m+2)}_{n-2m})^{qb_{2m-2}}}\Big\}\\
&= \max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},\ldots,
\frac1{(x^{(2m+1)}_{n-2m+1})^{a_{2m-2}}},\frac{(x^{(2m+1)}_{n-2m+1})^{a_{2m-1}}}
{(x^{(2m+2)}_{n-2m})^{b_{2m-1}}}\Big\}\\
&= \max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},\ldots,
\frac1{(x^{(2m+2)}_{n-2m})^{b_{2m-1}}},\frac{(x^{(2m+2)}_{n-2m})^{(q+1)a_{2m-1}
 -b_{2m-1}}}{(x^{(2m+3)}_{n-2m-1})^{qa_{2m-1}}}\Big\}\\
&= \max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}\frac1{(x^{(4)}_{n-2})^{b_1}},\ldots,
\frac1{(x^{(2m+2)}_{n-2m})^{b_{2m-1}}},\frac{(x^{(2m+2)}_{n-2m})^{b_{2m}}}
 {(x^{(2m+3)}_{n-2m-1})^{a_{2m}}}.\Big\}
\end{align*}
From this, \eqref{s10},  \eqref{s11} and by the method of induction
we see that \eqref{s8} and \eqref{s9} hold for $4\le 2m\le n+2$.

From the relations in \eqref{a2} it is easy to see that
\begin{gather*}
b_{2n}=\frac{a_{2n+1}+a_{2n}}{q+1},\quad n\in\mathbb{N}_0, \\
a_{2n+3}-(q^2+1)a_{2n+1}+q^2a_{2n-1}=0,\quad n\in\mathbb{N},
\end{gather*}
from which we obtain
\begin{equation}
a_{2n+1}=\frac{1-q^{2n+3}}{1-q},\quad n\in\mathbb{N}_0.\label{a5}
\end{equation} 
Letting $n\to+\infty$ in \eqref{a5}
and \eqref{a2} we also obtain
\begin{gather}
\lim_{n\to+\infty}a_{2n}=\lim_{n\to+\infty}b_{2n+1}=\frac
q{1-q},\label{s17}\\
\lim_{n\to+\infty}b_{2n}=\lim_{n\to+\infty}a_{2n+1}=\frac
1{1-q},\label{a10}
\end{gather} 
(see \cite{ejqtde-maxsde} for a detailed explanation).

Now note that from \eqref{s8} and \eqref{s9} we have that
\begin{gather}
 x^{(1)}_{2n+1} =\max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}
 \frac1{(x^{(4)}_{n-2})^{b_1}},\ldots,\frac1{(x^{(2n+2)}_0)^{b_{2n-1}}},
\frac{(x^{(2n+2)}_0)^{b_{2n}}}{(x^{(2n+3)}_{-1})^{a_{2n}}}\Big\}\label{s12}
\\
x^{(1)}_{2n}=\max\Big\{1,\frac1{(x^{(3)}_{n-1})^{a_0}}
\frac1{(x^{(4)}_{n-2})^{b_1}},\ldots,
\frac1{(x^{(2n+1)}_0)^{a_{2n-2}}},
\frac{(x^{(2n+1)}_0)^{a_{2n-1}}}{(x^{(2n+2)}_{-1})^{b_{2n-1}}}\Big\},\label{s14}
\end{gather}
for $n\in\mathbb{N}$.

Using that
 $$
\min_{i\in\mathbb{N}}x^{(i)}_n=\min_{1\le i\le
k}x^{(i)}_n\ge 1,\quad\text{for } n\in\mathbb{N},
$$
\eqref{s17} and \eqref{a10}, in \eqref{s12} and \eqref{s14}, the
boundedness of $(x^{(1)}_n)_{n\ge -1}$ follows, which along with
the cyclicity of system \eqref{main} implies the boundedness of
$(x^{(i)}_n)_{n\ge -1}$ for every $1\le i\le k$, as
claimed.
\end{proof}


\begin{remark} \label{rmk1}\rm
Note that from the proofs of Theorems \ref{thm2} and \ref{thm3}
is seen that the value  of superscript $i\in\mathbb{N}$ in $x_n^{(i)}$ 
does not influence at  the many points.
 This is why we introduced the convention in \eqref{s2}. 
In fact, an important fact in the proof of Theorem \ref{thm2} is estimate \eqref{a7}, 
while the inequality $\min_{1\le i\le k}x^{(i)}_n\ge 1$, $n\in\mathbb{N}$,
is the corresponding important fact in the proof of Theorem \ref{thm3}.
\end{remark}

\begin{theorem} \label{thm4}
 If $\alpha>0$, $p^2\geq 4q>4$, or $p>1+q$, $q\leq 1$, or  
$p=q+1=2$, then system \eqref{main} has positive
unbounded solutions.
\end{theorem}

\begin{proof} 
Since from \eqref{main} we have 
\begin{equation}
x^{(i)}_{n+1}\ge \frac{(x^{(i+1)}_n)^p}{(x^{(i+2)}_{n-1})^q},\quad
i=1,\ldots,k,\; n\in\mathbb{N}_0,\label{ewa}
\end{equation} 
the same arguments as in the proof of \cite[Theorem 2]{jdea205} can
be used. So, we will only sketch the proof for the completeness.

Let $y_n=\ln \prod_{i=1}^k x^{(i)}_n$, $n\ge -1$, then by using
\eqref{ewa}, we obtain 
\begin{equation} 
y_{n+1}-py_n+qy_{n-1}\geq
0,\quad n\in\mathbb{N}_0.\label{ey}
\end{equation}
If $p^2\geq 4q>4$, then the polynomial $\lambda^2-p\lambda+q$ has
two zeroes $\lambda_1$ and $\lambda_2$, such that
$\lambda_1>1$ and $\lambda_2>0$, while if $p>1+q$, $q\leq 1$, then
$\lambda_1>1>\lambda_2>0$.

From \eqref{ey} and some simple iterations we obtain
\begin{equation}
\frac{\prod_{i=1}^k x^{(i)}_{n+1}}{(\prod_{i=1}^k
x^{(i)}_n)^{\lambda_1}}\geq \Big(\frac{\prod_{i=1}^k
x^{(i)}_0}{(\prod_{i=1}^k
x^{(i)}_{-1})^{\lambda_1}}\Big)^{\lambda_2^{n+1}},\quad
n\in\mathbb{N}_0.\label{xiter}
\end{equation}
If $x^{(i)}_{-1}, x^{(i)}_0$, $i=1,\ldots,k$, are chosen, for
example, such that 
\begin{equation}
\prod_{i=1}^k x^{(i)}_0>1\quad\text{and}\quad \prod_{i=1}^k x^{(i)}_0\ge
\Big(\prod_{i=1}^k
x^{(i)}_{-1}\Big)^{\lambda_1},\label{s15}
\end{equation} 
then by employing \eqref{xiter} and \eqref{s15}, it is obtained
\begin{equation} 
\prod_{i=1}^k x^{(i)}_n\ge \Big(\prod_{i=1}^k
x^{(i)}_0\Big)^{\lambda_1^n}\to+\infty\quad\text{as }
n\to\infty,\label{xiter1}
\end{equation} 
which along with the inequality between the arithmetic and geometric
 means gives $\|\vec x_n\|_2\to+\infty$ as $n\to\infty$, showing the existence
of unbounded solutions in these two cases. 
If $p = q + 1 = 2$, then we can get unbounded solutions by choosing 
the initial values satisfying the strict inequalities in \eqref{s15}.
\end{proof}


\section{Global attractivity}

For the case $p=q$, we give here some sufficient conditions which
guaranty the global attractivity of all positive solutions of
system \eqref{main}.

\begin{theorem} \label{thm5} 
If $p\in(0,1)$ and  $\alpha\in (0,1)$, then every positive solution 
of \eqref{main} converges to the $k$-dimensional vector $(1,\ldots,1)$.
\end{theorem}

\begin{proof} 
Using \eqref{a7} in the equality 
\begin{equation}
x^{(i)}_{n+1}
=\max\Big\{\alpha,\frac{\alpha^p}{(x^{(i+2)}_{n-1})^p},
\frac1{((x^{(i+2)}_{n-1})^{1-p}(x^{(i+3)}_{n-2})^p)^p}\Big\},\quad
i=1,\ldots,k,\; n\in\mathbb{N},\label{s16}
\end{equation} 
which is obtained by iterating the relations in \eqref{main}, we obtain
\begin{equation} 
\alpha\leq x^{(i)}_{n+1}\leq \max\Big\{\alpha,
1, \frac 1{\alpha^p}\Big\}=\frac 1{\alpha^p},\quad\text{for }
n\geq 3.\label{step1}
\end{equation}
We write the equations in \eqref{main} as follows 
\begin{equation}
\frac{x^{(i)}_{n+1}}{x^{(i+1)}_n}=\max\Big\{\frac
\alpha{x^{(i+1)}_n},
\frac1{(x^{(i+1)}_n)^{1-p}(x^{(i+2)}_{n-1})^p}\Big\},\quad
n\in\mathbb{N}_0,\label{main2}
\end{equation} for $i=1,\ldots,k$.
Using \eqref{a7} and \eqref{step1} in \eqref{main2}, and since
$p\in(0,1)$, we obtain 
\begin{equation} 
\alpha^p\leq \frac{x^{(i)}_{n+1}}{x^{(i+1)}_n}\leq \frac
1{\alpha},\quad\text{for } n\geq 5;\label{maing}
\end{equation}
for $i=1,\ldots,k$.

Using $p,\alpha\in(0,1)$ and \eqref{maing} in \eqref{main}, we obtain
\begin{equation}
 \alpha^{p^2}\leq x^{(i)}_{n+1}\leq \frac
1{\alpha^p},\quad\text{for } n\geq
6;\label{step2}\end{equation} for $i=1,\ldots,k$.


From  \eqref{a7}, \eqref{main2} and \eqref{step2} it follows that
\begin{equation} \alpha^p\leq \frac{x^{(i)}_{n+1}}{x^{(i+1)}_n}\leq \frac
1{\alpha^{p^2}},\quad\text{for } n\geq 8,
\label{step31}
\end{equation} 
for $i=1,\ldots,k$.
Using \eqref{step31} in \eqref{main} it follows that
$$
\alpha^{p^2}\leq x^{(i)}_{n+1}\leq \frac 1{\alpha^{p^3}},\quad\text{for } n\geq 9,
$$
for $i=1,\ldots,k$.

Assume that for some $m\in\mathbb{N}$
\begin{equation}
\alpha^{p^{2m}}\leq \min_{1\le i\le k}x^{(i)}_{n+1}\le \max_{1\le
i\le k}x^{(i)}_{n+1}\leq \frac
1{\alpha^{p^{2m+1}}},\label{k1}
\end{equation} 
for $n\geq 6m+3$, and
\begin{equation} 
\alpha^{p^{2m+2}}\leq \min_{1\le i\le
k}x^{(i)}_{n+1}\le \max_{1\le i\le k}x^{(i)}_{n+1}\leq \frac
1{\alpha^{p^{2m+1}}},\label{k2}
\end{equation} 
for $n\geq 6m+6$.
Then, an induction argument shows that \eqref{k1} and \eqref{k2}
hold for every $m\in\mathbb{N}_0$.

Letting $m\to\infty$ in \eqref{k1}, \eqref{k2} and using
$p\in(0,1)$, we obtain
$$
\lim_{n\to\infty}\|(x^{(1)}_n,\ldots,x^{(k)}_n)-(1,\ldots,1)\|_2=0,
$$ as desired.
\end{proof}

\subsection*{Acknowledgements} The work of the
second author was supported by the Serbian Ministry of Education
and Science, projects III 41025 and OI 171007.

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\end{document}