The interaction between language and working memory: a systematic review of fMRI studies in the past two decades

Zoha Deldar; Carlos Gevers-Montoro; Ali Khatibi; Ladan Ghazi-Saidi; Zoha Deldar; Carlos Gevers-Montoro; Ali Khatibi; Ladan Ghazi-Saidi

doi:10.3934/Neuroscience.2021001

AIMS Neuroscience

2021, Volume 8, Issue 1: 1-32. doi: 10.3934/Neuroscience.2021001

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Review Topical Sections

The interaction between language and working memory: a systematic review of fMRI studies in the past two decades

1.
Department of Anatomy, Université du Québec à Trois-Rivières, Trois-Rivières, QC, Canada
2.
Madrid College of Chiropractic, Real Centro Universitario María Cristina, San Lorenzo de El Escorial, Madrid, Spain
3.
Centre of Precision Rehabilitation for Spinal Pain, University of Birmingham, Birmingham, UK
4.
Centre for Human Brain Health, University of Birmingham, Birmingham, UK
5.
Language and Cognition Laboratory, Department of Communication Disorders, College of Education, University of Nebraska at Kearney, USA
^† Zoha Deldar and Carlos Gevers-Montoro contributed equally to this work

Received: 03 September 2020 Accepted: 10 November 2020 Published: 16 November 2020

Language processing involves other cognitive domains, including Working Memory (WM). Much detail about the neural correlates of language and WM interaction remains unclear. This review summarizes the evidence for the interaction between WM and language obtained via functional Magnetic Resonance Imaging (fMRI) in the past two decades. The search was limited to PubMed, Google Scholar, Science direct and Neurosynth for working memory, language, fMRI, neuroimaging, cognition, attention, network, connectome keywords. The exclusion criteria consisted of studies including children, older adults, bilingual or multilingual population, clinical cases, music, sign language, speech, motor processing, review papers, meta-analyses, electroencephalography/event-related potential, and positron emission tomography. A total of 20 articles were included and discussed in four categories: language comprehension, language production, syntax, and networks. Studies on neural correlates of WM and language interaction are rare. Language tasks that involve WM activate common neural systems. Activated areas can be associated with cognitive concepts proposed by Baddeley and Hitch (1974), including the phonological loop of WM (mainly Broca and Wernicke's areas), other prefrontal cortex and right hemispheric regions linked to the visuospatial sketchpad. There is a clear, dynamic interaction between language and WM, reflected in the involvement of subcortical structures, particularly the basal ganglia (caudate), and of widespread right hemispheric regions. WM involvement is levered by cognitive demand in response to task complexity. High WM capacity readers draw upon buffer memory systems in midline cortical areas to decrease the WM demands for efficiency. Different dynamic networks are involved in WM and language interaction in response to the task in hand for an ultimate brain function efficiency, modulated by language modality and attention.

Keywords:

Citation: Zoha Deldar, Carlos Gevers-Montoro, Ali Khatibi, Ladan Ghazi-Saidi. The interaction between language and working memory: a systematic review of fMRI studies in the past two decades[J]. AIMS Neuroscience, 2021, 8(1): 1-32. doi: 10.3934/Neuroscience.2021001

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Abstract

1. Introduction and preliminaries

In this paper, we study the behavior of the solutions of the following three-dimensional system of difference equations of second order

$\begin{equation} x_{n+1} = f(y_{n}, y_{n-1}), \, y_{n+1} = g(z_{n}, z_{n-1}), \, z_{n+1} = h(x_{n}, x_{n-1}) \end{equation}$

(1)

where $n\in \mathbb{N}_{0}$ , the initial values $x_{-i}$ , $y_{-i}$ and $z_{-i}, i = 0, 1$ , are positive real numbers, the functions $f, \, g, \, h:\, \left(0, +\infty\right)^{2}\rightarrow\left(0, +\infty\right)$ are continuous and homogeneous of degree zero. We establish results on local and global stability of the unique positive equilibrium point. To do this we prove some general convergence theorems, that can be applied to generalize a lot of existing systems and to study new ones. Some results on existence of periodic and oscillatory solutions are also proved.

Now, we explain our motivation for doing this work. Clearly if we take $z_{-i} = x_{-i}, \, i = 0, 1$ , and $h\equiv g$ , then the system (1), will be

$\begin{equation} x_{n+1} = f(y_{n}, y_{n-1}), \, y_{n+1} = g(x_{n}, x_{n-1}). \end{equation}$

(2)

Noting also that if we choose $z_{-i} = y_{-i} = x_{-i}, \, i = 0, 1$ , and $h\equiv g\equiv f$ , then System (1) will be

$\begin{equation} x_{n+1} = f(x_{n}, x_{n-1}). \end{equation}$

(3)

In [23], the behavior of the solutions of System (2) has been investigated. System (2) is a generalization of Equation (3), studied in [17]. The present System (1) is the three-dimensional generalization of System (2).

In the literature there are many studies on difference equations defined by homogeneous functions, see for instance [1,2,5,7,11,16]. Noting that also a lot of studies are devoted to various models of difference equations and systems, not necessary defined by homogeneous functions, see for example [4,9,10,12,14,18,19,20,21,22,24,25,26,27,28,29].

Before we state our results, we recall the following definitions and results. For more details we refer to the following references [3,6,8,13].

Let $F: \left(0, +\infty\right)^{6}\rightarrow \left(0, +\infty\right)^{6}$ be a continuous function and consider the system of difference equations

$\begin{equation} Y_{n+1} = F(Y_{n}), \, n\in \mathbb{N}_{0}, \end{equation}$

(4)

where the initial value $Y_0\in \left(0, +\infty\right)^{6}$ . A point $\overline{Y } \in \left(0, +\infty\right)^{6}$ is an equilibrium point of (4), if it is a solution of $\overline{Y} = F(\overline{Y})$ .

Definition 1.1. Let $\overline{Y}$ be an equilibrium point of System (4), and let $\|.\|$ any convenient vector norm.

$1.$ We say that the equilibrium point $\overline{Y}$ is stable (or locally stable) if for every $\epsilon > 0$ there exists $\delta > 0$ such that for every initial condition $Y_0$ : $\|Y_0-\overline{Y}\|<\delta$ implies $\|Y_n-\overline{Y}\| <\epsilon$ . Otherwise, the equilibrium point $\overline{Y}$ is unstable.

$2.$ We say that the equilibrium point $\overline{Y}$ is asymptotically stable (or locally asymptotically stable) if it is stable and there exists $\gamma>0$ such that $\|Y_0-\overline{Y}\|<\gamma$ implies

$\begin{equation*} \lim\limits_{n\rightarrow \infty }Y_n = \overline{Y}. \end{equation*}$

$3.$ We say that the equilibrium point $\overline{Y}$ is a global attractor if for every $Y_0$ ,

$\begin{equation*} \lim\limits_{n\rightarrow \infty }Y_n = \overline{Y}. \end{equation*}$

$4.$ We say that the equilibrium point $\overline{Y}$ is globally (asymptotically) stable if it is stable and a global attractor.

Assume that $F$ is $C^{1}$ on $\left(0, +\infty\right)^{6}$ . To System (4), we associate a linear system, about the equilibrium point $\overline{Y}$ , given by

$\begin{equation*} Z_{n+1} = F_J(\overline{Y})Z_n, \quad n \in \mathbb{N}_{0}, \, Z_n = Y_n- \overline{Y}, \end{equation*}$

where $F_J$ is the Jacobian matrix of the function $F$ evaluated at the equilibrium point $\overline{Y}$ .

To study the stability of the equilibrium point $\overline{Y}$ , we need the following theorem.

Theorem 1.2. Let $\overline{Y}$ be an equilibrium point of System $(4)$ . Then, the following statements are true:

(i) If all the eigenvalues of the Jacobian matrix $F_J$ lie in the open unit disk $|\lambda| < 1$ , then the equilibrium $\overline{Y}$ is asymptotically stable.

(ii) If at least one eigenvalue of $F_J$ has absolute value greater than one, then the equilibrium $\overline{Y}$ is unstable.

Definition 1.3. A solution $\left(x_{n}, y_{n}, z_{n}\right)_{n\geq-1}$ of System (1) is said to be periodic of period $p\in \mathbb{N}$ if

$\begin{equation} x_{n+p} = x_n, \, y_{n+p} = y_n, \, z_{n+p} = z_n, \, n\geq-1. \end{equation}$

(5)

The solution $\left(x_{n}, y_{n}, z_{n}\right)_{n\geq-1}$ is said to be periodic with prime period $p\in \mathbb{N}$ , if it is periodic with period $p$ and $p$ is the least positive integer for which (1.3) holds.

Definition 1.4. Let $(x_n, y_n, z_n)_{n\geq-1}$ be a solution of System (1). We say that the sequence $(x_n)_{n\geq-1}$ (resp. $(y_n)_{n\geq-1}$ , $(z_n)_{n\geq-1}$ ) oscillates about $\overline{x}$ (resp. $\overline{y}$ , $\overline{z}$ ) with a semi-cycle of length one if: $(x_n-\overline{x})(x_{n+1}-\overline{x})<0, n\geq-1$ (resp. $(y_n-\overline{y })(y_{n+1}-\overline{y})<0, n\geq-1$ , $(z_n-\overline{z})(z_{n+1}-\overline{z} )<0, n\geq-1$ ).

Remark 1. For every term $x_{n_0}$ of the sequence $(x_n)_{n\geq-1}$ , the notation " $+$ " means $x_{n_0}-\overline{x}>0$ and the notation " $-$ " means $x_{n_0}- \overline{x}<0$ . The same notations will be used for the terms of the sequences $(y_n)_{n\geq-1}$ and $(z_n)_{n\geq-1}$ .

Definition 1.5. A function $\Phi:\, \left(0, +\infty\right)^{2}\rightarrow\left(0, +\infty \right)$ is said to be homogeneous of degree $m\in \mathbb{R}$ if we have

$\begin{equation*} \Phi(\lambda u, \lambda v) = \lambda^{m} \Phi(u, v) \end{equation*}$

for all $(u, v)\in \left(0, +\infty\right)^{2}$ and for all $\lambda>0$ .

Theorem 1.6. Let $\Phi:\, \left(0, +\infty\right)^{2}\rightarrow\left(0, + \infty\right)$ be a $C^{1}$ function on $\left(0, +\infty\right)^{2}$ .

1. Then, $\Phi$ is homogeneous of degree $m$ if and only if

$\begin{equation*} u\frac{\partial \Phi}{\partial u}(u, v)+v\frac{\partial \Phi}{\partial v} (u, v) = m \Phi(u, v), \, (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

(This statement is usually called Euler's Theorem).

2. If $\Phi$ is homogeneous of degree $m$ on $\left(0, +\infty\right)^{2}$ , then $\frac{\partial \Phi}{\partial u}$ and $\frac{\partial \Phi}{\partial v}$ are homogenous of degree $m-1$ on $\left(0, +\infty\right)^{2}$ .

2. Stability of the equilibrium points

A point $(\overline{x}, \overline{y}, \overline{z})\in \left(0, +\infty\right)^{3}$ is an equilibrium point of System (1) if it is a solution of the following system

$\begin{equation*} \overline{x} = f(\overline{y}, \overline{y}), \, \overline{y} = g(\overline{z}, \overline{z}), \, \overline{z} = g(\overline{x}, \overline{x}). \end{equation*}$

Using the fact that $f$ , $g$ and $h$ are homogeneous of degree zero, we get that

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)) \end{equation*}$

is the unique equilibrium point of System (1).

Let $F: \left(0, +\infty\right)^{6}\rightarrow \left(0, +\infty\right)^{6}$ be the function defined by

$\begin{equation*} F(W) = \left(f_{1}(W), f_{2}(W), g_{1}(W), g_{2}(W), h_{1}(W), h_{2}(W)\right), \, W = (u, v, w, t, r, s) \end{equation*}$

with

$\begin{equation*} f_{1}(W) = f(w, t), \, f_{2}(W) = u, \, g_{1}(W) = g(r, s), \, g_{2}(W) = w, \, h_{1}(W) = h(u, v), \, g_{2}(W) = r. \end{equation*}$

Then, System (1) can be written as follows

$\begin{equation*} W_{n+1} = F(W_{n}), \, W_{n} = \left(x_{n}, x_{n-1}, y_{n}, y_{n-1}, z_{n}, z_{n-1} \right)^{t}, \, n\in \mathbb{N}_{0}. \end{equation*}$

So, $(\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1))$ is an equilibrium point of system (1) if and only if

$\begin{equation*} \overline{W} = (\overline{x}, \overline{x}, \overline{y}, \overline{y}, \overline{z }, \overline{z}) = (f(1, 1), f(1, 1), g(1, 1), g(1, 1), h(1, 1), h(1, 1)) \end{equation*}$

is an equilibrium point of $W_{n+1} = F(W_{n})$ .

Assume that the functions $f$ , $g$ and $h$ are $C^{1}$ on $\left(0, +\infty\right)^{2}$ . To System (1), we associate about the equilibrium point $\overline{W}$ the following linear system

$\begin{equation*} X_{n+1} = J_{F}X_{n}, \, n\in \mathbb{N}_{0} \end{equation*}$

where $J_{F}$ is the Jacobian matrix associated to the function $F$ evaluated at

$\begin{equation*} \overline{W} = (f(1, 1), f(1, 1), g(1, 1), g(1, 1), h(1, 1), h(1, 1)). \end{equation*}$

We have

$\begin{equation*} J_{F} = \begin{pmatrix} 0 & 0 & \frac{\partial f}{\partial w}(\overline{y}, \overline{y}) & \frac{ \partial f}{\partial t}(\overline{y}, \overline{y}) & 0 & 0 & \\ 1 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & \frac{\partial g}{\partial r}(\overline{z}, \overline{z}) & \frac{\partial g}{\partial s}(\overline{z}, \overline{z}) & \\ 0 & 0 & 1 & 0 & 0 & 0 & \\ \frac{\partial h}{\partial u}(\overline{x}, \overline{x}) & \frac{\partial h}{ \partial v}(\overline{x}, \overline{x}) & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 1 & 0 & \end{pmatrix} \end{equation*}$

As $f$ , $g$ and $h$ are homogeneous of degree $0$ , then using Part 1. of Theorem 1.6, we get

$\begin{equation*} \overline{y}\frac{\partial f}{\partial w}(\overline{y}, \overline{y}) + \overline{y} \frac{\partial f}{\partial t}(\overline{y}, \overline{y}) = 0 \end{equation*}$

which implies

$\begin{equation*} \frac{\partial f}{\partial t}(\overline{y}, \overline{y}) = -\frac{\partial f}{ \partial w}(\overline{y}, \overline{y}). \end{equation*}$

Similarly we get

$\begin{equation*} \frac{\partial g}{\partial s}(\overline{z}, \overline{z}) = -\frac{\partial g}{ \partial r}(\overline{z}, \overline{z}), \, \frac{\partial h}{\partial v}( \overline{x}, \overline{x}) = -\frac{\partial h}{\partial u}(\overline{x}, \overline{x}). \end{equation*}$

It follows that $J_{F}$ takes the form:

$\begin{equation*} J_{F} = \begin{pmatrix} 0 & 0 & \frac{\partial f}{\partial w}(\overline{y}, \overline{y}) & -\frac{ \partial f}{\partial w}(\overline{y}, \overline{y}) & 0 & 0 & \\ 1 & 0 & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & \frac{\partial g}{\partial r}(\overline{z}, \overline{z}) & - \frac{\partial g}{\partial r}(\overline{z}, \overline{z}) & \\ 0 & 0 & 1 & 0 & 0 & 0 & \\ \frac{\partial h}{\partial u}(\overline{x}, \overline{x}) & -\frac{\partial h }{\partial u}(\overline{x}, \overline{x}) & 0 & 0 & 0 & 0 & \\ 0 & 0 & 0 & 0 & 1 & 0 & \end{pmatrix} \end{equation*}$

The characteristic polynomial of the matrix $J_F$ is given by

$\begin{eqnarray*} P(\lambda)& = &\lambda^{6}-\frac{\partial h}{\partial u}(\overline{x}, \overline{x})\frac{\partial g}{\partial r}(\overline{z}, \overline{z})\frac{ \partial f}{\partial w}(\overline{y}, \overline{y})\lambda^{3}+3\frac{ \partial h}{\partial u}(\overline{x}, \overline{x})\frac{\partial g}{\partial r}(\overline{z}, \overline{z})\frac{\partial f}{\partial w}(\overline{y}, \overline{y})\lambda^{2} \\ &&-3\frac{\partial h}{\partial u}(\overline{x}, \overline{x})\frac{\partial g }{\partial r}(\overline{z}, \overline{z})\frac{\partial f}{\partial w}( \overline{y}, \overline{y})\lambda+\frac{\partial h}{\partial u}(\overline{x}, \overline{x})\frac{\partial g}{\partial r}(\overline{z}, \overline{z})\frac{ \partial f}{\partial w}(\overline{y}, \overline{y}). \end{eqnarray*}$

Now assume that

$\begin{equation*} \left|\frac{\partial h}{\partial u}(\overline{x}, \overline{x})\frac{\partial g}{\partial r}(\overline{z}, \overline{z})\frac{\partial f}{\partial w}( \overline{y}, \overline{y})\right| < \frac{1}{8} \end{equation*}$

and consider the two functions

$\begin{equation*} \Phi(\lambda) = \lambda^{6}, \end{equation*}$

$\begin{eqnarray*} \Psi(\lambda)& = &-\frac{\partial h}{\partial u}(\overline{x}, \overline{x}) \frac{\partial g}{\partial r}(\overline{z}, \overline{z})\frac{\partial f}{ \partial w}(\overline{y}, \overline{y})\lambda^{3}+3\frac{\partial h}{ \partial u}(\overline{x}, \overline{x})\frac{\partial g}{\partial r}( \overline{z}, \overline{z})\frac{\partial f}{\partial w}(\overline{y}, \overline{y})\lambda^{2} \\ &&-3\frac{\partial h}{\partial u}(\overline{x}, \overline{x})\frac{\partial g }{\partial r}(\overline{z}, \overline{z})\frac{\partial f}{\partial w}( \overline{y}, \overline{y})\lambda+\frac{\partial h}{\partial u}(\overline{x}, \overline{x})\frac{\partial g}{\partial r}(\overline{z}, \overline{z})\frac{ \partial f}{\partial w}(\overline{y}, \overline{y}). \end{eqnarray*}$

We have

$\begin{equation*} \left|\Psi(\lambda)\right|\leq 8\left|\frac{\partial h}{\partial u}( \overline{x}, \overline{x})\frac{\partial g}{\partial r}(\overline{z}, \overline{z})\frac{\partial f}{\partial w}(\overline{y}, \overline{y} )\right| < 1 = \left|\Phi(\lambda)\right|, \, \forall\, \lambda \in \mathbb{C} :\, \left|\lambda\right| = 1. \end{equation*}$

So, by Rouché's Theorem it follows that all roots of $P(\lambda)$ lie inside the unit disk.

Hence, by Theorem 1.2, we deduce from the above consideration that the equilibrium point $(\overline{x}, \overline{y}, \overline{z} ) = (f(1, 1), g(1, 1), h(1, 1))$ is locally asymptotically stable.

Using Part 2. of Theorem 1.6 and the fact that the function $f$ , $g$ , $h$ are homogeneous of degree zero, we get that $\frac{\partial f}{\partial w }$ , $\frac{\partial g}{\partial r}$ and $\frac{\partial h}{\partial u}$ are homogeneous of degree $-1$ . So, it follows that

$\begin{equation*} \frac{\partial f}{\partial w}(\overline{y}, \overline{y}) = \frac{\frac{ \partial f}{\partial w}(1, 1)}{\overline{y}}, \, \frac{\partial g}{\partial r}( \overline{z}, \overline{z}) = \frac{\frac{\partial g}{\partial r}(1, 1)}{ \overline{z}}, \, \frac{\partial h}{\partial u}(\overline{x}, \overline{x}) = \frac{\frac{\partial h}{\partial u}(1, 1)}{\overline{x}}. \end{equation*}$

In summary, we have proved the following result.

Theorem 2.1. Assume that $f(u, v)$ , $g(u, v)$ and $h(u, v)$ are $C^{1}$ on $\left(0, +\infty\right)^{2}$ . The equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)) \end{equation*}$

of System $(1)$ is locally asymptotically stable if

$\begin{equation*} \left|\frac{\partial f}{\partial u}(1, 1)\frac{\partial g}{\partial u}(1, 1) \frac{\partial h}{\partial u}(1, 1)\right| < \frac{f(1, 1)g(1, 1)h(1, 1)}{8}. \end{equation*}$

Now, we will prove some general convergence results. The obtained results allow us to deal with the global attractivity of the equilibrium point $( \overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1))$ and so the global stability.

Theorem 2.2. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v), \, g(u, v), \, h(u, v)$ are increasing in $u$ for all $v$ and decreasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(m_{2}, M_{2}), \, M_{1} = f(M_{2}, m_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$\begin{equation*} m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma \end{equation*}$

and for each $i = 0, 1, ...,$

$\begin{equation*} m_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}), \, M_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \end{equation*}$

$\begin{equation*} m_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \, M_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \end{equation*}$

$\begin{equation*} m_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}), \, M_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}). \end{equation*}$

We have

$\begin{equation*} a\leq f(\alpha, \beta)\leq f(\beta, \alpha)\leq b, \end{equation*}$

$\begin{equation*} \alpha\leq g(\lambda, \gamma)\leq g(\gamma, \lambda)\leq \beta, \end{equation*}$

$\begin{equation*} \lambda\leq h(a, b)\leq h(b, a)\leq \gamma, \end{equation*}$

and so,

$\begin{equation*} m_{1}^{0} = a\leq f(m_2^{0}, M_{2}^{0})\leq f(M_2^{0}, m_{2}^{0})\leq b = M_1^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0} = \alpha\leq g(m_3^{0}, M_{3}^{0})\leq g(M_3^{0}, m_{3}^{0})\leq \beta = M_2^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0} = \lambda\leq h(m_1^{0}, M_{1}^{0})\leq g(M_1^{0}, m_{1}^{0})\leq \gamma = M_3^{0}. \end{equation*}$

Hence,

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

Now, we have

$\begin{equation*} m_{1}^{1} = f(m_2^{0}, M_{2}^{0})\leq f(m_2^{1}, M_{2}^{1}) = m_1^{2}\leq f(M_2^{1}, m_{2}^{1}) = M_1^{2}\leq f(M_2^{0}, m_{2}^{0}) = M_1^{1}, \end{equation*}$

$\begin{equation*} m_{2}^{1} = g(m_3^{0}, M_{3}^{0})\leq g(m_3^{1}, M_{3}^{1}) = m_2^{2}\leq g(M_3^{1}, m_{3}^{1}) = M_2^{2}\leq g(M_3^{0}, m_{3}^{0}) = M_2^{1}, \end{equation*}$

$\begin{equation*} m_{3}^{1} = h(m_1^{0}, M_{1}^{0})\leq h(m_1^{1}, M_{1}^{1}) = m_3^{2}\leq h(M_1^{1}, m_{1}^{1}) = M_3^{2}\leq h(M_1^{0}, m_{1}^{0}) = M_3^{1}, \end{equation*}$

and it follows that

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

By induction, we get for $i = 0, 1, ...,$ that

$\begin{equation*} a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b, \end{equation*}$

$\begin{equation*} \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta, \end{equation*}$

and

$\begin{equation*} \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma. \end{equation*}$

It follows that the sequences $(m_1^{i})_{i\in\mathbb{N}_{0}}$ , $(m_2^{i})_{i\in\mathbb{N}_{0}}$ , $(m_3^{i})_{i\in\mathbb{N}_{0}}$ (resp. $(M_1^{i})_{i\in\mathbb{N}_{0}}$ , $(M_2^{i})_{i\in\mathbb{N}_{0}}$ , $(M_3^{i})_{i\in\mathbb{N}_{0}}$ ) are increasing (resp. decreasing) and also bounded, so convergent. Let

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$\begin{equation*} a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma. \end{equation*}$

By taking limits in the following equalities

$\begin{equation*} m_1^{i+1} = f(m_2^{i}, M_2^{i}), \, M_1^{i+1} = f(M_2^{i}, m_2^{i}), \end{equation*}$

$\begin{equation*} m_2^{i+1} = g(m_3^{i}, M_3^{i}), \, M_2^{i+1} = g(M_3^{i}, m_3^{i}), \end{equation*}$

$\begin{equation*} m_3^{i+1} = h(m_1^{i}, M_1^{i}), \, M_3^{i+1} = h(M_1^{i}, m_1^{i}), \end{equation*}$

and using the continuity of $f$ , $g$ and $h$ we obtain

$\begin{eqnarray*} m_1& = &f(m_2, M_2), \, M_1 = f(M_2, m_2), \\ m_2& = &g(m_3, M_3), \, M_2 = g(M_3, m_3), \\ m_3& = &h(m_1, M_1), \, M_3 = h(M_1, m_1) \end{eqnarray*}$

so it follows from $H_3$ that

$\begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*}$

From $H_1$ , for $n = 1, 2, \cdots$ , we get

$\begin{equation*} m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, . \end{equation*}$

For $n = 2, 3, ...,$ we have

$\begin{equation*} m_1^{1} = f(m_2^{0}, M_{2}^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{0}, m_{2}^{0}) = M_{1}^{1}, \end{equation*}$

$\begin{equation*} m_2^{1} = g(m_3^{0}, M_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{0}, m_{3}^{0}) = M_{2}^{1}, \end{equation*}$

and

$\begin{equation*} m_3^{1} = h(m_1^{0}, M_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{0}, m_{1}^{0}) = M_{3}^{1}, \end{equation*}$

that is

$\begin{equation*} m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots. \end{equation*}$

Now, for $n = 4, 5, ...,$ we have

$\begin{equation*} m_1^{2} = f(m_2^{1}, M_{2}^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{1}, m_{2}^{1}) = M_{1}^{2}, \end{equation*}$

and

$\begin{equation*} m_2^{2} = g(m_3^{1}, M_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{1}, m_{3}^{1}) = M_{2}^{2}, \end{equation*}$

$\begin{equation*} m_3^{2} = h(m_1^{1}, M_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{1}, m_{1}^{1}) = M_{3}^{2}, \end{equation*}$

that is

$\begin{equation*} m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots. \end{equation*}$

Similarly, for $n = 6, 7, ...,$ we have

$\begin{equation*} m_1^{3} = f(m_2^{2}, M_{2}^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{2}, m_{2}^{2}) = M_{1}^{3}, \end{equation*}$

$\begin{equation*} m_2^{3} = g(m_3^{2}, M_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{2}, m_{3}^{2}) = M_{2}^{3}, \end{equation*}$

and

$\begin{equation*} m_3^{3} = h(m_1^{2}, M_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{2}, m_{1}^{2}) = M_{3}^{3}, \end{equation*}$

that is

$\begin{equation*} m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots. \end{equation*}$

It follows by induction that for $i = 0, 1, ...$ we get

$\begin{equation*} m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1. \end{equation*}$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\begin{equation*} \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3. \end{equation*}$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$\begin{equation*} M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1). \end{equation*}$

Theorem 2.3. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v), \, g(u, v)$ are increasing in $u$ for all $v$ and decreasing in $v$ for all $u$ and $h(u, v)$ is decreasing in $u$ for all $v$ and increasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(m_{2}, M_{2}), \, M_{1} = f(M_{2}, m_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$\begin{equation*} m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma \end{equation*}$

and for each $i = 0, 1, ...,$

$\begin{equation*} m_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}), \, M_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \end{equation*}$

$\begin{equation*} m_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \, M_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \end{equation*}$

$\begin{equation*} m_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}), \, M_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}). \end{equation*}$

We have

$\begin{equation*} a\leq f(\alpha, \beta)\leq f(\beta, \alpha)\leq b, \end{equation*}$

$\begin{equation*} \alpha\leq g(\lambda, \gamma)\leq g(\gamma, \lambda)\leq \beta, \end{equation*}$

$\begin{equation*} \lambda\leq h(b, a)\leq h(a, b)\leq \gamma, \end{equation*}$

and so,

$\begin{equation*} m_{1}^{0} = a\leq f(m_2^{0}, M_{2}^{0})\leq f(M_2^{0}, m_{2}^{0})\leq b = M_1^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0} = \alpha\leq g(m_3^{0}, M_{3}^{0})\leq g(M_3^{0}, m_{3}^{0})\leq \beta = M_2^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0} = \lambda\leq h(M_1^{0}, m_{1}^{0})\leq h(m_1^{0}, M_{1}^{0})\leq \gamma = M_3^{0}. \end{equation*}$

Hence,

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

Now, we have

$\begin{equation*} m_{1}^{1} = f(m_2^{0}, M_{2}^{0})\leq f(m_2^{1}, M_{2}^{1}) = m_1^{2}\leq f(M_2^{1}, m_{2}^{1}) = M_1^{2}\leq f(M_2^{0}, m_{2}^{0}) = M_1^{1}, \end{equation*}$

$\begin{equation*} m_{2}^{1} = g(m_3^{0}, M_{3}^{0})\leq g(m_3^{1}, M_{3}^{1}) = m_2^{2}\leq g(M_3^{1}, m_{3}^{1}) = M_2^{2}\leq g(M_3^{0}, m_{3}^{0}) = M_2^{1}, \end{equation*}$

$\begin{equation*} m_{3}^{1} = h(M_1^{0}, m_{1}^{0})\leq h(M_1^{1}, m_{1}^{1}) = m_3^{2}\leq h(m_1^{1}, M_{1}^{1}) = M_3^{2}\leq h(m_1^{0}, M_{1}^{0}) = M_3^{1}, \end{equation*}$

and it follows that

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

By induction, we get for $i = 0, 1, ...,$ that

$\begin{equation*} a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b, \end{equation*}$

$\begin{equation*} \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta, \end{equation*}$

and

$\begin{equation*} \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma. \end{equation*}$

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$\begin{equation*} a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma. \end{equation*}$

By taking limits in the following equalities

$\begin{equation*} m_1^{i+1} = f(m_2^{i}, M_2^{i}), \, M_1^{i+1} = f(M_2^{i}, m_2^{i}), \end{equation*}$

$\begin{equation*} m_2^{i+1} = g(m_3^{i}, M_3^{i}), \, M_2^{i+1} = g(M_3^{i}, m_3^{i}), \end{equation*}$

$\begin{equation*} m_3^{i+1} = h(M_1^{i}, m_1^{i}), \, M_3^{i+1} = h(m_1^{i}, M_1^{i}), \end{equation*}$

and using the continuity of $f$ , $g$ and $h$ we obtain

$\begin{eqnarray*} m_1& = &f(m_2, M_2), \, m_2 = f(M_2, m_2), \, \\ m_2& = &g(m_3, M_3), \, M_2 = g(M_3, m_3), \, \\ m_3& = &h(M_1, m_1), \, M_3 = h(m_1, M_1) \end{eqnarray*}$

so it follows from $H_3$ that

$\begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*}$

From $H_1$ , for $n = 1, 2, \cdots$ , we get

$\begin{equation*} m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, . \end{equation*}$

For $n = 2, 3, ...,$ we have

$\begin{equation*} m_1^{1} = f(m_2^{0}, M_{2}^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{0}, m_{2}^{0}) = M_{1}^{1}, \end{equation*}$

$\begin{equation*} m_2^{1} = g(m_3^{0}, M_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{0}, m_{3}^{0}) = M_{2}^{1}, \end{equation*}$

and

$\begin{equation*} m_3^{1} = h(M_1^{0}, m_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{0}, M_{1}^{0}) = M_{3}^{1}, \end{equation*}$

that is

$\begin{equation*} m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots. \end{equation*}$

Now, for $n = 4, 5, ...,$ we have

$\begin{equation*} m_1^{2} = f(m_2^{1}, M_{2}^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{1}, m_{2}^{1}) = M_{1}^{2}, \end{equation*}$

and

$\begin{equation*} m_2^{2} = g(m_3^{1}, M_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{1}, m_{3}^{1}) = M_{2}^{2}, \end{equation*}$

$\begin{equation*} m_3^{2} = h(M_1^{1}, m_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{1}, M_{1}^{1}) = M_{3}^{2}, \end{equation*}$

that is

$\begin{equation*} m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots. \end{equation*}$

Similarly, for $n = 6, 7, ...,$ we have

$\begin{equation*} m_1^{3} = f(m_2^{2}, M_{2}^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{2}, m_{2}^{2}) = M_{1}^{3}, \end{equation*}$

$\begin{equation*} m_2^{3} = g(m_3^{2}, M_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{2}, m_{3}^{2}) = M_{2}^{3}, \end{equation*}$

and

$\begin{equation*} m_3^{3} = h(M_1^{2}, m_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{2}, M_{1}^{2}) = M_{3}^{3}, \end{equation*}$

that is

$\begin{equation*} m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots. \end{equation*}$

It follows by induction that for $i = 0, 1, ...$ we get

$\begin{equation*} m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1. \end{equation*}$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\begin{equation*} \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3. \end{equation*}$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$\begin{equation*} M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1). \end{equation*}$

Theorem 2.4. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v)$ is increasing in $u$ for all $v$ and decreasing in $v$ for all $u$ and $g(u, v)$ , $h(u, v)$ are decreasing in $u$ for all $v$ and increasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(m_{2}, M_{2}), \, M_{1} = f(M_{2}, m_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$\begin{equation*} m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma \end{equation*}$

and for each $i = 0, 1, ...,$

$\begin{equation*} m_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}), \, M_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \end{equation*}$

$\begin{equation*} m_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \, M_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \end{equation*}$

$\begin{equation*} m_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}), \, M_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}). \end{equation*}$

We have

$\begin{equation*} a\leq f(\alpha, \beta)\leq f(\beta, \alpha)\leq b, \end{equation*}$

$\begin{equation*} \alpha\leq g(\gamma, \lambda)\leq g(\lambda, \gamma)\leq \beta, \end{equation*}$

$\begin{equation*} \lambda\leq h(b, a)\leq h(a, b)\leq \gamma, \end{equation*}$

and so,

$\begin{equation*} m_{1}^{0} = a\leq f(m_2^{0}, M_{2}^{0})\leq f(M_2^{0}, m_{2}^{0})\leq b = M_1^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0} = \alpha\leq g(M_3^{0}, m_{3}^{0})\leq g(m_3^{0}, M_{3}^{0})\leq \beta = M_2^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0} = \lambda\leq h(M_1^{0}, m_{1}^{0})\leq h(m_1^{0}, M_{1}^{0})\leq \gamma = M_3^{0}. \end{equation*}$

Hence,

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

Now, we have

$\begin{equation*} m_{1}^{1} = f(m_2^{0}, M_{2}^{0})\leq f(m_2^{1}, M_{2}^{1}) = m_1^{2}\leq f(M_2^{1}, m_{2}^{1}) = M_1^{2}\leq f(M_2^{0}, m_{2}^{0}) = M_1^{1}, \end{equation*}$

$\begin{equation*} m_{2}^{1} = g(M_3^{0}, m_{3}^{0})\leq g(M_3^{1}, m_{3}^{1}) = m_2^{2}\leq g(m_3^{1}, M_{3}^{1}) = M_2^{2}\leq g(m_3^{0}, M_{3}^{0}) = M_2^{1}, \end{equation*}$

$\begin{equation*} m_{3}^{1} = h(M_1^{0}, m_{1}^{0})\leq h(M_1^{1}, m_{1}^{1}) = m_3^{2}\leq h(m_1^{1}, M_{1}^{1}) = M_3^{2}\leq h(m_1^{0}, M_{1}^{0}) = M_3^{1}, \end{equation*}$

and it follows that

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

By induction, we get for $i = 0, 1, ...,$ that

$\begin{equation*} a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b, \end{equation*}$

$\begin{equation*} \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta, \end{equation*}$

and

$\begin{equation*} \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma. \end{equation*}$

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$\begin{equation*} a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma. \end{equation*}$

By taking limits in the following equalities

$\begin{equation*} m_1^{i+1} = f(m_2^{i}, M_2^{i}), \, M_1^{i+1} = f(M_2^{i}, m_2^{i}), \end{equation*}$

$\begin{equation*} m_2^{i+1} = g(M_3^{i}, m_3^{i}), \, M_2^{i+1} = g(m_3^{i}, M_3^{i}), \end{equation*}$

$\begin{equation*} m_3^{i+1} = h(M_1^{i}, m_1^{i}), \, M_3^{i+1} = h(m_1^{i}, M_1^{i}), \end{equation*}$

and using the continuity of $f$ , $g$ and $h$ we obtain

$\begin{eqnarray*} m_1& = &f(m_2, M_2), \, M_1 = f(M_2, m_2), \, \\ m_2& = &g(M_3, m_3), \, M_2 = g(m_3, M_3), \, \\ m_3& = &h(M_1, m_1), \, M_3 = h(m_1, M_1) \end{eqnarray*}$

so it follows from $H_3$ that

$\begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*}$

From $H_1$ , for $n = 1, 2, \cdots$ , we get

$\begin{equation*} m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, . \end{equation*}$

For $n = 2, 3, ...,$ we have

$\begin{equation*} m_1^{1} = f(m_2^{0}, M_{2}^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{0}, m_{2}^{0}) = M_{1}^{1}, \end{equation*}$

$\begin{equation*} m_2^{1} = g(M_3^{0}, m_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{0}, M_{3}^{0}) = M_{3}^{1}, \end{equation*}$

and

$\begin{equation*} m_3^{1} = h(M_1^{0}, m_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{0}, M_{1}^{0}) = M_{3}^{1}, \end{equation*}$

that is

$\begin{equation*} m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots. \end{equation*}$

Now, for $n = 4, 5, ...,$ we have

$\begin{equation*} m_1^{2} = f(m_2^{1}, M_{2}^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{1}, m_{2}^{1}) = M_{1}^{2}, \end{equation*}$

and

$\begin{equation*} m_2^{2} = g(M_3^{1}, m_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{1}, M_{3}^{1}) = M_{2}^{2}, \end{equation*}$

$\begin{equation*} m_3^{2} = h(M_1^{1}, m_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{1}, M_{1}^{1}) = M_{3}^{2}, \end{equation*}$

that is

$\begin{equation*} m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots. \end{equation*}$

Similarly, for $n = 6, 7, ...,$ we have

$\begin{equation*} m_1^{3} = f(m_2^{2}, M_{2}^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{2}, m_{2}^{2}) = M_{1}^{3}, \end{equation*}$

$\begin{equation*} m_2^{3} = g(M_3^{2}, m_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{2}, M_{3}^{2}) = M_{2}^{3}, \end{equation*}$

and

$\begin{equation*} m_3^{3} = h(M_1^{2}, m_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{2}, M_{1}^{2}) = M_{3}^{3}, \end{equation*}$

that is

$\begin{equation*} m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots. \end{equation*}$

It follows by induction that for $i = 0, 1, ...$ we get

$\begin{equation*} m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1. \end{equation*}$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\begin{equation*} \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3. \end{equation*}$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$\begin{equation*} M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1). \end{equation*}$

Theorem 2.5. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v)$ , $h(u, v)$ are increasing in $u$ for all $v$ and decreasing in $v$ for all $u$ and $g(u, v)$ is decreasing in $u$ for all $v$ and increasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(m_{2}, M_{2}), \, M_{1} = f(M_{2}, m_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$\begin{equation*} m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma \end{equation*}$

and for each $i = 0, 1, ...,$

$\begin{equation*} m_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}), \, M_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \end{equation*}$

$\begin{equation*} m_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \, M_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \end{equation*}$

$\begin{equation*} m_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}), \, M_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}). \end{equation*}$

We have

$\begin{equation*} a\leq f(\alpha, \beta)\leq f(\beta, \alpha)\leq b, \end{equation*}$

$\begin{equation*} \alpha\leq g(\gamma, \lambda)\leq g(\lambda, \gamma)\leq \beta, \end{equation*}$

$\begin{equation*} \lambda\leq h(a, b)\leq h(b, a)\leq \gamma, \end{equation*}$

and so,

$\begin{equation*} m_{1}^{0} = a\leq f(m_2^{0}, M_{2}^{0})\leq f(M_2^{0}, m_{2}^{0})\leq b = M_1^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0} = \alpha\leq g(M_3^{0}, m_{3}^{0})\leq g(m_3^{0}, M_{3}^{0})\leq \beta = M_2^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0} = \lambda\leq h(m_1^{0}, M_{1}^{0})\leq h(M_1^{0}, m_{1}^{0})\leq \gamma = M_3^{0}. \end{equation*}$

Hence,

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

Now, we have

$\begin{equation*} m_{1}^{1} = f(m_2^{0}, M_{2}^{0})\leq f(m_2^{1}, M_{2}^{1}) = m_1^{2}\leq f(M_2^{1}, m_{2}^{1}) = M_1^{2}\leq f(M_2^{0}, m_{2}^{0}) = M_1^{1}, \end{equation*}$

$\begin{equation*} m_{2}^{1} = g(M_3^{0}, m_{3}^{0})\leq g(M_3^{1}, m_{3}^{1}) = m_2^{2}\leq g(m_3^{1}, M_{3}^{1}) = M_2^{2}\leq g(m_3^{0}, M_{3}^{0}) = M_2^{1}, \end{equation*}$

$\begin{equation*} m_{3}^{1} = h(m_1^{0}, M_{1}^{0})\leq h(m_1^{1}, M_{1}^{1}) = m_3^{2}\leq h(M_1^{1}, m_{1}^{1}) = M_3^{2}\leq h(M_1^{0}, m_{1}^{0}) = M_3^{1}, \end{equation*}$

and it follows that

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

By induction, we get for $i = 0, 1, ...,$ that

$\begin{equation*} a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b, \end{equation*}$

$\begin{equation*} \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta, \end{equation*}$

and

$\begin{equation*} \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma. \end{equation*}$

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$\begin{equation*} a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma. \end{equation*}$

By taking limits in the following equalities

$\begin{equation*} m_1^{i+1} = f(m_2^{i}, M_2^{i}), \, M_1^{i+1} = f(M_2^{i}, m_2^{i}), \end{equation*}$

$\begin{equation*} m_2^{i+1} = g(M_3^{i}, m_3^{i}), \, M_2^{i+1} = g(m_3^{i}, M_3^{i}), \end{equation*}$

$\begin{equation*} m_3^{i+1} = h(m_1^{i}, M_1^{i}), \, M_3^{i+1} = h(M_1^{i}, m_1^{i}), \end{equation*}$

and using the continuity of $f$ , $g$ and $h$ we obtain

so it follows from $H_3$ that

$\begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*}$

From $H_1$ , for $n = 1, 2, \cdots$ , we get

$\begin{equation*} m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, . \end{equation*}$

For $n = 2, 3, ...,$ we have

$\begin{equation*} m_1^{1} = f(m_2^{0}, M_{2}^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{0}, m_{2}^{0}) = M_{1}^{1}, \end{equation*}$

$\begin{equation*} m_2^{1} = g(M_3^{0}, m_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{0}, M_{3}^{0}) = M_{3}^{1}, \end{equation*}$

and

$\begin{equation*} m_3^{1} = h(m_1^{0}, M_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{0}, m_{1}^{0}) = M_{3}^{1}, \end{equation*}$

that is

$\begin{equation*} m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots. \end{equation*}$

Now, for $n = 4, 5, ...,$ we have

$\begin{equation*} m_1^{2} = f(m_2^{1}, M_{2}^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{1}, m_{2}^{1}) = M_{1}^{2}, \end{equation*}$

and

$\begin{equation*} m_2^{2} = g(M_3^{1}, m_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{1}, M_{3}^{1}) = M_{2}^{2}, \end{equation*}$

$\begin{equation*} m_3^{2} = h(m_1^{1}, M_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{1}, m_{1}^{1}) = M_{3}^{2}, \end{equation*}$

that is

$\begin{equation*} m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots. \end{equation*}$

Similarly, for $n = 6, 7, ...,$ we have

$\begin{equation*} m_1^{3} = f(m_2^{2}, M_{2}^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(M_2^{2}, m_{2}^{2}) = M_{1}^{3}, \end{equation*}$

$\begin{equation*} m_2^{3} = g(M_3^{2}, m_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_3^{2}, M_{3}^{2}) = M_{2}^{3}, \end{equation*}$

and

$\begin{equation*} m_3^{3} = h(m_1^{2}, M_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{2}, m_{1}^{2}) = M_{3}^{3}, \end{equation*}$

that is

$\begin{equation*} m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots. \end{equation*}$

It follows by induction that for $i = 0, 1, ...$ we get

$\begin{equation*} m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1. \end{equation*}$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\begin{equation*} \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3. \end{equation*}$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$\begin{equation*} M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1). \end{equation*}$

Theorem 2.6. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v), \, g(u, v), \, h(u, v)$ are decreasing in $u$ for all $v$ and increasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$\begin{equation*} m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma \end{equation*}$

and for each $i = 0, 1, ...,$

$\begin{equation*} m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}), \end{equation*}$

$\begin{equation*} m_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \, M_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \end{equation*}$

$\begin{equation*} m_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}), \, M_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}). \end{equation*}$

We have

$\begin{equation*} a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b, \end{equation*}$

$\begin{equation*} \alpha\leq g(\gamma, \lambda)\leq g(\lambda, \gamma)\leq \beta, \end{equation*}$

$\begin{equation*} \lambda\leq h(b, a)\leq h(a, b)\leq \gamma \end{equation*}$

and so,

$\begin{equation*} m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0} = \alpha\leq g(M_{3}^{0}, m_3^{0})\leq g(m_{3}^{0}, M_3^{0})\leq \beta = M_2^{0}, \end{equation*}$

$\begin{equation*} m_{3}^{0} = \lambda\leq h(M_{1}^{0}, m_1^{0})\leq h(m_{1}^{0}, M_1^{0})\leq \gamma = M_3^{0}. \end{equation*}$

Hence,

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

and

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

Now, we have

$\begin{equation*} m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1}, \end{equation*}$

$\begin{equation*} m_{2}^{1} = g(M_{3}^{0}, m_3^{0})\leq g(M_{3}^{1}, m_3^{1}) = m_2^{2}\leq g(m_{3}^{1}, M_3^{1}) = M_2^{2}\leq g(m_{3}^{0}, M_3^{0}) = M_2^{1}, \end{equation*}$

$\begin{equation*} m_{3}^{1} = h(M_{1}^{0}, m_1^{0})\leq h(M_{1}^{1}, m_1^{1}) = m_3^{2}\leq h(m_{1}^{1}, M_1^{1}) = M_3^{2}\leq h(m_{1}^{0}, M_1^{0}) = M_3^{1} \end{equation*}$

and it follows that

$\begin{equation*} m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0}, \end{equation*}$

$\begin{equation*} m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0}, \end{equation*}$

$\begin{equation*} m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}. \end{equation*}$

By induction, we get for $i = 0, 1, ...,$ that

$\begin{equation*} a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b, \end{equation*}$

$\begin{equation*} \alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta, \end{equation*}$

and

$\begin{equation*} \lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma. \end{equation*}$

It follows that the sequences $(m_1^{i})_{i\in\mathbb{N}_{0}}$ , $(m_2^{i})_{i\in\mathbb{N}_{0}}$ $(m_3^{i})_{i\in\mathbb{N}_{0}}$ (resp. $(M_1^{i})_{i\in\mathbb{N}_{0}}$ , $(M_2^{i})_{i\in\mathbb{N}_{0}}$ , $(M_3^{i})_{i\in\mathbb{N}_{0}}$ ) are increasing (resp. decreasing) and also bounded, so convergent. Let

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$\begin{equation*} a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma. \end{equation*}$

By taking limits in the following equalities

$\begin{equation*} m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}), \end{equation*}$

$\begin{equation*} m_2^{i+1} = g(M_3^{i}, m_3^{i}), \, M_2^{i+1} = g(m_3^{i}, M_3^{i}), \end{equation*}$

$\begin{equation*} m_3^{i+1} = h(M_1^{i}, m_1^{i}), \, M_3^{i+1} = h(m_1^{i}, M_1^{i}) \end{equation*}$

and using the continuity of $f$ , $g$ , and $h$ we obtain

so it follows from $H_3$ that

$\begin{equation*} m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3. \end{equation*}$

From $H_1$ , for $n = 1, 2, \cdots$ , we get

$\begin{equation*} m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, . \end{equation*}$

For $n = 2, 3, ...,$ we have

$\begin{equation*} m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1}, \end{equation*}$

$\begin{equation*} m_2^{1} = g(M_{3}^{0}, m_3^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{0}, M_3^{0}) = M_{2}^{1}, \end{equation*}$

$\begin{equation*} m_3^{1} = h(M_{1}^{0}, m_1^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_{1}^{0}, M_1^{0}) = M_{3}^{1}, \end{equation*}$

that is

$\begin{equation*} m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots. \end{equation*}$

Now, for $n = 4, 5, ...,$ we have

$\begin{equation*} m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2}, \end{equation*}$

$\begin{equation*} m_2^{2} = g(M_{3}^{1}, m_1^{3})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{1}, M_1^{3}) = M_{2}^{2}, \end{equation*}$

$\begin{equation*} m_3^{2} = h(M_{1}^{1}, m_1^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_{1}^{1}, M_1^{1}) = M_{3}^{2}, \end{equation*}$

that is

$\begin{equation*} m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots. \end{equation*}$

Similarly, for $n = 6, 7, ...,$ we have

$\begin{equation*} m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3}, \end{equation*}$

$\begin{equation*} m_2^{3} = g(M_{3}^{2}, m_3^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{2}, M_3^{2}) = M_{2}^{3}, \end{equation*}$

$\begin{equation*} m_3^{3} = h(M_{1}^{2}, m_1^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_{1}^{2}, M_1^{2}) = M_{3}^{3}, \end{equation*}$

that is

$\begin{equation*} m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots. \end{equation*}$

It follows by induction that for $i = 0, 1, ...$ we get

$\begin{equation*} m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1. \end{equation*}$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\begin{equation*} \lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3. \end{equation*}$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$\begin{equation*} M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1). \end{equation*}$

Theorem 2.7. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v), \, g(u, v)$ are decreasing in $u$ for all $v$ and increasing in $v$ for all $u$ , however $h(u, v)$ is increasing in $u$ for all $v$ and decreasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(M_{3}, m_{3}), \, M_{2} = g(m_{3}, M_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma$

and for each $i = 0, 1, ...,$

$m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}),$

$m_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}), \, M_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}),$

$m_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}), \, M_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}).$

We have

$a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b,$

$\alpha\leq g(\gamma, \lambda)\leq g(\lambda, \gamma)\leq \beta,$

$\lambda\leq h(a, b)\leq h(b, a)\leq \gamma,$

and so,

$m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0},$

$m_{2}^{0} = \alpha\leq g(M_{3}^{0}, m_3^{0})\leq g(m_{3}^{0}, M_3^{0})\leq \beta = M_2^{0},$

and

$m_{3}^{0} = \lambda\leq h(m_1^{0}, M_{1}^{0})\leq h(M_1^{0}, m_{1}^{0})\leq \gamma = M_3^{0}.$

Hence,

$m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0},$

$m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0},$

and

$m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}.$

Now, we have

$m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1},$

$m_{2}^{1} = g(M_{3}^{0}, m_3^{0})\leq g(M_{3}^{1}, m_3^{1}) = m_2^{2}\leq g(m_{3}^{1}, M_3^{1}) = M_2^{2}\leq g(m_{3}^{0}, M_3^{0}) = M_2^{1},$

$m_{3}^{1} = h(m_1^{0}, M_{1}^{0})\leq h(m_1^{1}, M_{1}^{1}) = m_3^{2}\leq h(M_1^{1}, m_{1}^{1}) = M_3^{2}\leq h(M_1^{0}, m_{1}^{0}) = M_3^{1},$

and it follows that

$m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0},$

$m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0},$

and

$m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}.$

By induction, we get for $i = 0, 1, ...,$ that

$a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b,$

$\alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta,$

and

$\lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma.$

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma.$

By taking limits in the following equalities

$m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}),$

$m_2^{i+1} = g(M_3^{i}, m_3^{i}), \, M_2^{i+1} = g(m_3^{i}, M_3^{i}),$

$m_3^{i+1} = h(m_1^{i}, M_1^{i}), \, M_3^{i+1} = h(M_1^{i}, m_1^{i}),$

and using the continuity of $f$ , $g$ , and $h$ we obtain

so it follows from $H_3$ that

$m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3.$

From $H_1$ , for $, n = 1, 2, \ldots$ , we get

$m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}.$

For $n = 2, 3, ...,$ we have

$m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1},$

$m_2^{1} = g(M_{3}^{0}, m_3^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{0}, M_3^{0}) = M_{2}^{1},$

and

$m_3^{1} = h(m_1^{0}, M_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{0}, m_{1}^{0}) = M_{3}^{1},$

that is

$m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots.$

Now, for $n = 4, 5, ...,$ we have

$m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2},$

$m_2^{2} = g(M_{3}^{1}, m_1^{3})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{1}, M_1^{3}) = M_{2}^{2},$

and

$m_3^{2} = h(m_1^{1}, M_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{1}, m_{1}^{1}) = M_{3}^{2},$

that is

$m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots.$

Similarly, for $n = 6, 7, ...,$ we have

$m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3},$

$m_2^{3} = g(M_{3}^{2}, m_3^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(m_{3}^{2}, M_3^{2}) = M_{2}^{3},$

and

$m_3^{3} = h(m_1^{2}, M_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{2}, m_{1}^{2}) = M_{3}^{3},$

that is

$m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots.$

It follows by induction that for $i = 0, 1, ...$ we get

$m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1.$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3.$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1).$

Theorem 2.8. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v)$ is decreasing in $u$ for all $v$ and increasing in $v$ for all $u$ , however $g(u, v), \, h(u, v)$ are increasing in $u$ for all $v$ and decreasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(m_{1}, M_{1}), \, M_{3} = h(M_{1}, m_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma$

and for each $i = 0, 1, ...,$

$m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}),$

$m_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \, M_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}),$

$m_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}), \, M_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}).$

We have

$a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b,$

$\alpha\leq g(\lambda, \gamma)\leq g(\gamma, \lambda)\leq \beta,$

$\lambda\leq h(a, b)\leq h(b, a)\leq \gamma,$

and so,

$m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0},$

$m_{2}^{0} = \alpha\leq g(m_3^{0}, M_{3}^{0})\leq g(M_3^{0}, m_{3}^{0})\leq \beta = M_2^{0},$

and

$m_{3}^{0} = \lambda\leq h(m_1^{0}, M_{1}^{0})\leq g(M_1^{0}, m_{1}^{0})\leq \gamma = M_3^{0}.$

Hence,

$m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0},$

$m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0},$

and

$m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}.$

Now, we have

$m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1},$

$m_{2}^{1} = g(m_3^{0}, M_{3}^{0})\leq g(m_3^{1}, M_{3}^{1}) = m_2^{2}\leq g(M_3^{1}, m_{3}^{1}) = M_2^{2}\leq g(M_3^{0}, m_{3}^{0}) = M_2^{1},$

$m_{3}^{1} = h(m_1^{0}, M_{1}^{0})\leq h(m_1^{1}, M_{1}^{1}) = m_3^{2}\leq h(M_1^{1}, m_{1}^{1}) = M_3^{2}\leq h(M_1^{0}, m_{1}^{0}) = M_3^{1},$

and it follows that

$m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0},$

$m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0},$

and

$m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}.$

By induction, we get for $i = 0, 1, ...,$ that

$a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b,$

$\alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta,$

and

$\lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma.$

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma.$

By taking limits in the following equalities

$m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}),$

$m_2^{i+1} = g(m_3^{i}, M_3^{i}), \, M_2^{i+1} = g(M_3^{i}, m_3^{i}),$

$m_3^{i+1} = h(m_1^{i}, M_1^{i}), \, M_3^{i+1} = h(M_1^{i}, m_1^{i}),$

and using the continuity of $f$ , $g$ and $h$ we obtain

so it follows from $H_3$ that

$m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3.$

From $H_1$ , for $n = 1, 2, \ldots$ , we get

$m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, .$

For $n = 2, 3, ...,$ we have

$m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1},$

$m_2^{1} = g(m_3^{0}, M_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{0}, m_{3}^{0}) = M_{3}^{1},$

and

$m_3^{1} = h(m_1^{0}, M_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{0}, m_{1}^{0}) = M_{3}^{1},$

that is

$m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots.$

Now, for $n = 4, 5, ...,$ we have

$m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2},$

$m_2^{2} = g(m_3^{1}, M_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{1}, m_{3}^{1}) = M_{2}^{2},$

and

$m_3^{2} = h(m_1^{1}, M_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{1}, m_{1}^{1}) = M_{3}^{2},$

that is

$m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots.$

Similarly, for $n = 6, 7, ...,$ we have

$m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3},$

$m_2^{3} = g(m_3^{2}, M_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{2}, m_{3}^{2}) = M_{2}^{3},$

and

$m_3^{3} = h(m_1^{2}, M_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(M_1^{2}, m_{1}^{2}) = M_{3}^{3},$

that is

$m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots.$

It follows by induction that for $i = 0, 1, ...$ we get

$m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1.$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3.$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1).$

Theorem 2.9. Consider System $(1)$ . Assume that the following statements are true:

1. $H_1$ : There exist $a, \, b, \, \alpha, \, \beta, \lambda, \, \gamma \in\left(0, +\infty\right)$ such that

$\begin{equation*} a\leq f(u, v)\leq b, \, \alpha\leq g(u, v)\leq\beta, \, \lambda \leq h(u, v)\leq \gamma, \, \forall (u, v)\in \left(0, +\infty\right)^{2}. \end{equation*}$

2. $H_2$ : $f(u, v), \, h(u, v)$ are decreasing in $u$ for all $v$ and increasing in $v$ for all $u$ , however $g(u, v)$ is increasing in $u$ for all $v$ and decreasing in $v$ for all $u$ .

3. $H_{3}$ : If $(m_{1}, M_{1}, m_{2}, M_{2}, m_{3}, M_{3})\in \left[ a, b\right] ^{2}\times \left[ \alpha , \beta \right] ^{2}\times \left[ \lambda , \gamma \right] ^{2}$ is a solution of the system

$\begin{eqnarray*} m_{1} & = &f(M_{2}, m_{2}), \, M_{1} = f(m_{2}, M_{2}), \, \\ m_{2} & = &g(m_{3}, M_{3}), \, M_{2} = g(M_{3}, m_{3}), \, \\ m_{3} & = &h(M_{1}, m_{1}), \, M_{3} = h(m_{1}, M_{1}) \end{eqnarray*}$

then

$\begin{equation*} m_{1} = M_{1}, \, m_{2} = M_{2}, \, m_{3} = M_{3}. \end{equation*}$

Then every solution of System $(1)$ converges to the unique equilibrium point

$\begin{equation*} (\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1)). \end{equation*}$

Proof. Let

$m_{1}^{0}: = a, \, M_{1}^{0}: = b, \, m_{2}^{0}: = \alpha, \, M_{2}^{0}: = \beta, \, m_{3}^{0}: = \lambda, \, M_{3}^{0}: = \gamma$

and for each $i = 0, 1, ...,$

$m_{1}^{i+1}: = f(M_2^{i}, m_{2}^{i}), \, M_{1}^{i+1}: = f(m_2^{i}, M_{2}^{i}),$

$m_{2}^{i+1}: = g(m_3^{i}, M_{3}^{i}), \, M_{2}^{i+1}: = g(M_3^{i}, m_{3}^{i}),$

$m_{3}^{i+1}: = h(M_1^{i}, m_{1}^{i}), \, M_{3}^{i+1}: = h(m_1^{i}, M_{1}^{i}).$

We have

$a\leq f(\beta, \alpha)\leq f(\alpha, \beta)\leq b,$

$\alpha\leq g(\lambda, \gamma)\leq g(\gamma, \lambda)\leq \beta,$

$\lambda\leq h(b, a)\leq h(a, b)\leq \gamma,$

and so,

$m_{1}^{0} = a\leq f(M_{2}^{0}, m_2^{0})\leq f(m_{2}^{0}, M_2^{0})\leq b = M_1^{0},$

$m_{2}^{0} = \alpha\leq g(m_3^{0}, M_{3}^{0})\leq g(M_3^{0}, m_{3}^{0})\leq \beta = M_2^{0},$

and

$m_{3}^{0} = \lambda\leq h(M_1^{0}, m_{1}^{0})\leq h(m_1^{0}, M_{1}^{0})\leq \gamma = M_3^{0}.$

Hence,

$m_{1}^{0}\leq m_{1}^{1}\leq M_{1}^{1}\leq M_{1}^{0},$

$m_{2}^{0}\leq m_{2}^{1}\leq M_{2}^{1}\leq M_{2}^{0},$

and

$m_{3}^{0}\leq m_{3}^{1}\leq M_{3}^{1}\leq M_{3}^{0}.$

Now, we have

$m_{1}^{1} = f(M_{2}^{0}, m_2^{0})\leq f(M_{2}^{1}, m_2^{1}) = m_1^{2}\leq f(m_{2}^{1}, M_2^{1}) = M_1^{2}\leq f(m_{2}^{0}, M_2^{0}) = M_1^{1},$

$m_{2}^{1} = g(m_3^{0}, M_{3}^{0})\leq g(m_3^{1}, M_{3}^{1}) = m_2^{2}\leq g(M_3^{1}, m_{3}^{1}) = M_2^{2}\leq g(M_3^{0}, m_{3}^{0}) = M_2^{1},$

$m_{3}^{1} = h(M_1^{0}, m_{1}^{0})\leq h(M_1^{1}, m_{1}^{1}) = m_3^{2}\leq h(m_1^{1}, M_{1}^{1}) = M_3^{2}\leq h(m_1^{0}, M_{1}^{0}) = M_3^{1},$

and it follows that

$m_{1}^{0}\leq m_{1}^{1}\leq m_{1}^{2} \leq M_{1}^{2} \leq M_{1}^{1}\leq M_{1}^{0},$

$m_{2}^{0}\leq m_{2}^{1}\leq m_{2}^{2} \leq M_{2}^{2} \leq M_{2}^{1}\leq M_{2}^{0},$

and

$m_{3}^{0}\leq m_{3}^{1}\leq m_{3}^{2} \leq M_{3}^{2} \leq M_{3}^{1}\leq M_{3}^{0}.$

By induction, we get for $i = 0, 1, ...,$ that

$a = m_{1}^{0}\leq m_{1}^{1}\leq...\leq m_1^{i-1} \leq m_1^{i}\leq M_1^{i}\leq M_1^{i-1}\leq...\leq M_{1}^{1}\leq M_{1}^{0} = b,$

$\alpha = m_{2}^{0}\leq m_{2}^{1}\leq...\leq m_2^{i-1} \leq m_2^{i}\leq M_2^{i}\leq M_2^{i-1}\leq...\leq M_{2}^{1}\leq M_{2}^{0} = \beta,$

and

$\lambda = m_{3}^{0}\leq m_{3}^{1}\leq...\leq m_3^{i-1} \leq m_3^{i}\leq M_3^{i}\leq M_3^{i-1}\leq...\leq M_{3}^{1}\leq M_{3}^{0} = \gamma.$

$\begin{equation*} m_1 = \lim\limits_{i\rightarrow+\infty}m_1^{i}, \, M_1 = \lim\limits_{i\rightarrow+\infty}M_1^{i}, \end{equation*}$

$\begin{equation*} m_2 = \lim\limits_{i\rightarrow+ \infty}m_2^{i}, \, M_2 = \lim\limits_{i\rightarrow+ \infty}M_2^{i}. \end{equation*}$

$\begin{equation*} m_3 = \lim\limits_{i\rightarrow+\infty}m_3^{i}, \, M_3 = \lim\limits_{i\rightarrow+\infty}M_3^{i}. \end{equation*}$

Then

$a\leq m_1\leq M_1\leq b, \, \alpha\leq m_2\leq M_2\leq \beta, \, \lambda\leq m_3\leq M_3\leq \gamma.$

By taking limits in the following equalities

$m_1^{i+1} = f(M_2^{i}, m_2^{i}), \, M_1^{i+1} = f(m_2^{i}, M_2^{i}),$

$m_2^{i+1} = g(m_3^{i}, M_3^{i}), \, M_2^{i+1} = g(M_3^{i}, m_3^{i}),$

$m_3^{i+1} = h(M_1^{i}, m_1^{i}), \, M_3^{i+1} = h(m_1^{i}, M_1^{i}),$

and using the continuity of $f$ , $g$ and $h$ we obtain

so it follows from $H_3$ that

$m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3.$

From $H_1$ , for $n = 1, 2, \ldots$ , we get

$m_1^{0} = a\leq x_n\leq b = M_1^{0}, \, m_2^{0} = \alpha\leq y_n\leq \beta = M_2^{0}, \, m_3^{0} = \lambda\leq z_n\leq \gamma = M_3^{0}\, .$

For $n = 2, 3, ...,$ we have

$m_1^{1} = f(M_{2}^{0}, m_2^{0})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{0}, M_2^{0}) = M_{1}^{1},$

$m_2^{1} = g(m_3^{0}, M_{3}^{0})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{0}, m_{3}^{0}) = M_{3}^{1},$

and

$m_3^{1} = h(M_1^{0}, m_{1}^{0})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{0}, M_{1}^{0}) = M_{3}^{1},$

that is

$m_1^{1}\leq x_{n} \leq M_{1}^{1}, \, m_2^{1}\leq y_{n} \leq M_{2}^{1}, \, m_3^{1}\leq z_{n} \leq M_{3}^{1}, \, n = 3, 4, \cdots.$

Now, for $n = 4, 5, ...,$ we have

$m_1^{2} = f(M_{2}^{1}, m_2^{1})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{1}, M_2^{1}) = M_{1}^{2},$

$m_2^{2} = g(m_3^{1}, M_{3}^{1})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{1}, m_{3}^{1}) = M_{2}^{2},$

and

$m_3^{2} = h(M_1^{1}, m_{1}^{1})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{1}, M_{1}^{1}) = M_{3}^{2},$

that is

$m_1^{2}\leq x_{n} \leq M_{1}^{2}, \, m_2^{2}\leq y_{n} \leq M_{2}^{2}, \, m_3^{2}\leq z_{n} \leq M_{3}^{2}, \, n = 5, 6, \cdots.$

Similarly, for $n = 6, 7, ...,$ we have

$m_1^{3} = f(M_{2}^{2}, m_2^{2})\leq x_{n+1} = f(y_n, y_{n-1})\leq f(m_{2}^{2}, M_2^{2}) = M_{1}^{3},$

$m_2^{3} = g(m_3^{2}, M_{3}^{2})\leq y_{n+1} = g(z_n, z_{n-1})\leq g(M_3^{2}, m_{3}^{2}) = M_{2}^{3},$

and

$m_3^{3} = h(M_1^{2}, m_{1}^{2})\leq z_{n+1} = h(x_n, x_{n-1})\leq h(m_1^{2}, M_{1}^{2}) = M_{3}^{3},$

that is

$m_1^{3}\leq x_{n} \leq M_{1}^{3}, \, m_2^{3}\leq y_{n} \leq M_{2}^{3}, \, m_3^{3}\leq z_{n} \leq M_{3}^{3}, \, n = 7, 8, \cdots.$

It follows by induction that for $i = 0, 1, ...$ we get

$m_1^{i}\leq x_{n}\leq M_1^{i}, \, m_2^{i}\leq y_{n}\leq M_2^{i}, \, m_3^{i}\leq z_{n}\leq M_3^{i}, \, n\geq2i+1.$

Using the fact that $i\rightarrow+\infty$ implies $n\rightarrow+\infty$ and $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ , we obtain that

$\lim\limits_{n\rightarrow +\infty} x_n = M_1, \, \lim\limits_{n\rightarrow +\infty} y_n = M_2, \, \, \lim\limits_{n\rightarrow +\infty} z_n = M_3.$

From (1) and using the fact that $f$ , $g$ and $h$ are continuous and homogeneous of degree zero, we get

$M_1 = f(M_2, M_2) = f(1, 1), \, M_2 = g(M_3, M_3) = g(1, 1), \, M_3 = h(M_1, M_1) = h(1, 1).$

The following theorem is devoted to global stability of the equilibrium point $(\overline{x}, \overline{y}, \overline{z}) = (f(1, 1), g(1, 1), h(1, 1))$ .

Theorem 2.10. Under the hypotheses of Theorem 2.1 and one of Theorems 2.2– 2.9, the equilibrium point $(\overline{x}, \overline{y}, \overline{z} ) = (f(1, 1), g(1, 1), h(1, 1))$ is globally stable.

Now as an application of the previous results, we give an example.

Example 1. Consider the following system of difference equations

$\begin{equation} x_{n+1} = f(y_n, y_{n-1}), \, y_{n+1} = g(z_n, z_{n-1}), \, z_{n+1} = h(x_n, x_{n-1}), \, n \in\mathbb{N}_{0}, \end{equation}$

(6)

where $x_{-1}$ , $x_{0}$ , $y_{-1}$ , $y_{0}$ , $z_{-1}$ , $z_{0}$ $\in\left(0, +\infty\right)$ and

$\begin{equation*} f(u, v) = \frac{a_1u+b_1v}{\alpha_1u+\beta_1v}, \, g(u, v) = \frac{a_2u+b_2v}{ \alpha_2u+\beta_2v}, \, h(u, v) = \frac{a_3u+b_3v}{\alpha_3u+\beta_3v}. \end{equation*}$

Assume that $r_i: = a_i\beta_i-\alpha_ib_i, \, i = 1, 2, 3$ are positive. For all $(u, v)\in\left(0, +\infty\right)$ , we have

$\begin{equation*} \frac{\partial f}{\partial u}(u, v) = \frac{r_1v}{(\alpha_1u+\beta_1v)^{2}}, \, \frac{\partial f}{\partial v}(u, v) = -\frac{r_1u}{(\alpha_1u+\beta_1v)^{2}} , \, a: = \frac{b_1}{\beta_1}\leq f(u, v)\leq b: = \frac{a_1}{\alpha_1} \end{equation*}$

$\begin{equation*} \frac{\partial g}{\partial u}(u, v) = \frac{r_2v}{(\alpha_2u+\beta_2v)^{2}}, \, \frac{\partial g}{\partial v}(u, v) = -\frac{r_2u}{(\alpha_2u+\beta_2v)^{2}} , \, \alpha: = \frac{b_2}{\beta_2}\leq g(u, v)\leq \beta: = \frac{a_2}{\alpha_2} \end{equation*}$

$\begin{equation*} \frac{\partial h}{\partial u}(u, v) = \frac{r_3v}{(\alpha_3u+\beta_3v)^{2}}, \, \frac{\partial h}{\partial v}(u, v) = -\frac{r_3u}{(\alpha_3u+\beta_3v)^{2}} , \, \lambda: = \frac{b_3}{\beta_3}\leq h(u, v)\leq \gamma: = \frac{a_3}{\alpha_3} \end{equation*}$

It follows from Theorem 2.1 that the unique equilibrium point

$\begin{equation*} \left(f(1, 1), g(1, 1), h(1, 1)\right) = \left(\frac{a_1+b_1}{\alpha_1+\beta_1}, \frac{a_2+b_2}{\alpha_2+\beta_2}, \frac{a_3+b_3}{\alpha_3+\beta_3}\right) \end{equation*}$

of System (6) will be locally stable if

$\begin{equation*} \frac{\partial f}{\partial u}(1, 1)\frac{\partial g}{\partial u}(1, 1)\frac{ \partial h}{\partial u}(1, 1) < \frac{f(1, 1)g(1, 1)h(1, 1)}{8} \end{equation*}$

which is equivalent to

$\begin{equation*} \frac{r_1r_2r_3}{(\alpha_1+\beta_1)^{2}(\alpha_2+\beta_2)^{2}(\alpha_3+ \beta_3)^{2}} < \frac{(a_1+b_1)(a_2+b_2)(a_3+b_3)} {8(\alpha_1+\beta_1)( \alpha_2+\beta_2)(\alpha_3+\beta_3)}. \end{equation*}$

Also, we have conditions $H_1$ and $H_2$ of Theorem 2.2 are satisfied. So, to prove the global stability of the equilibrium point $(f(1, 1), g(1, 1), h(1, 1))$ it suffices to check condition $H_3$ of Theorem 2.2.

For this purpose, let $(m_1, M_1, m_2, M_2, m_3, M_3)\in[a, b]\times[\alpha, \beta] \times[\gamma, \lambda]$ such that

$\begin{equation} m_1 = f(m_2, M_2) = \frac{a_1m_2+b_1M_2}{\alpha_1m_2+\beta_1M_2}, \, M_1 = f(M_2, m_2) = \frac{a_1M_2+b_1m_2}{\alpha_1M_2+\beta_1m_2}, \end{equation}$

(7)

$\begin{equation} m_2 = g(m_3, M_3) = \frac{a_2m_3+b_2M_3}{\alpha_2m_3+\beta_2M_3}, \, M_2 = g(M_3, m_3) = \frac{a_2M_3+b_2m_3}{\alpha_2M_3+\beta_2m_3}, \end{equation}$

(8)

$\begin{equation} m_3 = h(m_1, M_1) = \frac{a_3m_1+b_3M_1}{\alpha_3m_1+\beta_3M_1}, \, M_3 = h(M_1, m_1) = \frac{a_3M_1+b_3m_1}{\alpha_3M_1+\beta_3m_1}. \end{equation}$

(9)

From (7)-(9), we get

$\begin{equation} m_1-M_1 = \frac{r_1(m_2-M_2)(m_2+M_2)}{(\alpha_1m_2+\beta_1M_2)(\alpha_1M_2+ \beta_1m_2)}, \end{equation}$

(10)

$\begin{equation} m_2-M_2 = \frac{r_2(m_3-M_3)(m_3+M_3)}{(\alpha_2m_3+\beta_2M_3)(\alpha_2M_3+ \beta_2m_3)}, \end{equation}$

(11)

$\begin{equation} m_3-M_3 = \frac{r_3(m_1-M_1)(m_1+M_1)}{(\alpha_3m_1+\beta_3M_1)(\alpha_3M_1+ \beta_3m_1)}. \end{equation}$

(12)

Now, from (10)-(12), we get

$\begin{equation} (m_1-M_1)(m_2-M_2)(m_3-M_3) = (m_1-M_1)(m_2-M_2)(m_3-M_3) \Theta(m_1, M_1, m_2, M_2, m_3, M_3), \end{equation}$

(13)

where $\Theta(m_1, M_1, m_2, M_2, m_3, M_3)$ is equal to

$\begin{equation*} \frac{r_1r_2r_3(m_1+M_1)(m_2+M_2)(m_3+M_3)} {(\alpha_3m_1+\beta_3M_1)(\alpha_3M_1+\beta_3m_1)(\alpha_1m_2+\beta_1M_2)(\alpha_1M_2+\beta_1m_2)(\alpha_2m_3+\beta_2M_3)(\alpha_2M_3+\beta_2m_3)}. \end{equation*}$

It is not hard to see that

$\begin{equation*} 8\prod\limits_{i = 1}^{3}\frac{\alpha_i^{2}b_ir_i}{a_i^{2}\beta_i(\alpha_i+ \beta_i)^{2}}\leq\Theta(m_1, M_1, m_2, M_2, m_3, M_3) \leq8\prod\limits_{i = 1}^{3}\frac{ a_i\beta_i^{2}r_i}{\alpha_ib_i^{2}(\alpha_i+\beta_i)^{2}}. \end{equation*}$

Using the fact

$\begin{equation*} r_i\leq a_i\beta_i, \, (\alpha_i+\beta_i)^2\geq 2\alpha_i\beta_i, \, \alpha_ib_i < a_i\beta_i, \, i = 1, 2, 3, \end{equation*}$

we get

$\begin{equation*} 8\prod\limits_{i = 1}^{3}\frac{\alpha_i^{2}b_ir_i}{a_i^{2}\beta_i(\alpha_i+ \beta_i)^{2}}\leq\prod\limits_{i = 1}^{3}\frac{\alpha_ib_i}{a_i\beta_i} < 1. \end{equation*}$

If we choose the parameters $a_i, \, b_i, \, \alpha_i, \, \beta_i, \, i = 1, 2, 3$ such that

$\begin{equation*} \prod\limits_{i = 1}^{3}\frac{a_i\beta_i^{2}r_i}{\alpha_ib_i^{2}(\alpha_i+\beta_i)^{2} } < \frac{1}{8}, \end{equation*}$

then we get that

$\begin{equation*} \Theta(m_1, M_1, m_2, M_2, m_3, M_3)\neq 1 \end{equation*}$

and so it follows from (13) that

$\begin{equation*} (m_1-M_1)(m_2-M_2)(m_3-M_3) = 0. \end{equation*}$

Using this and (10)-(12), we obtain $m_1 = M_1, \, m_2 = M_2, \, m_3 = M_3$ . That is the condition $H_3$ is satisfied.

In summary we have the following result.

Theorem 2.11. Assume that that parameters $a_i, \, b_i, \, \alpha_i, \, \beta_i, \, i = 1, 2, 3$ are such that

● $C_1$ : $a_i\beta_i>\alpha_ib_i, \, i = 1, 2, 3$ .

● $C_2$ :

$\begin{equation*} \prod\limits_{i = 1}^{3}\frac{a_i\beta_i-\alpha_ib_i}{(\alpha_i+\beta_i)^{2}} < \frac{1 }{8}\prod\limits_{i = 1}^{3}\frac{a_i+b_i}{\alpha_i+\beta_i}. \end{equation*}$

● $C_3$ :

$\begin{equation*} \prod\limits_{i = 1}^{3}\frac{a_i\beta_i^{2}r_i}{\alpha_ib_i^{2}(\alpha_i+\beta_i)^{2} } < \frac{1}{8}. \end{equation*}$

Then the equilibrium point $\left(f(1, 1), g(1, 1), h(1, 1)\right) = \left(\frac{ a_1+b_1}{\alpha_1+\beta_1}, \frac{a_2+b_2}{\alpha_2+\beta_2}, \frac{a_3+b_3}{ \alpha_3+\beta_3}\right)$ is globally stable.

We visualize the solutions of System (6) in Figures 1-3. In Figure 1 and Figure 2, we give the solution and corresponding global attractor of System (6) for $x_{-1} = 2.13, x_{0} = 3.1, y_{-1} = 4.03, y_{0} = 2.21, z_{-1} = 2.76, z_{0} = 3.12$ and $a_1 = 3, \alpha_1 = 3.2, b_{1} = 4, \beta_{1} = 5, a_2 = 1.2, \alpha_2 = 2, b_{2} = 3.2, \beta_{2} = 6, a_3 = 2.4, \alpha_3 = 2.3, b_{3} = 1.3, \beta_{3} = 1.3$ , respectively. Note that the conditions $C_1, C_2, C_3$ is satisfied for these values.

Figure 1. .

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Figure 2. .

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Figure 3. .

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However Figure 3 shows the unstable solution corresponding to the values $x_{-1} = 2.13, x_{0} = 3.1, y_{-1} = 4.03, y_{0} = 2.21, z_{-1} = 2.76, z_{0} = 3.12$ and $a_1 = 0.3, \alpha_1 = 4, b_{1} = 3, \beta_{1} = 1.1, a_2 = 1.2, \alpha_2 = 4, b_{2} = 3.7, \beta_{2} = 1, a_3 = 2.7, \alpha_3 = 0.2, b_{3} = 1.3, \beta_{3} = 3$ , which do not satisfy the conditions $C_1, C_2, C_3$ .

3. Existence of periodic solutions

Here, we are interested in existence of periodic solutions for System (1). In the following result we will established a necessary and sufficient condition for which there exist prime period two solutions of System (1).

Theorem 3.1. Assume that $\alpha$ , $\beta$ and $\gamma$ are positive real numbers such that $(\alpha-1)(\beta-1)(\gamma-1)\neq0$ . Then, System $(1)$ has a prime period two solution in the form of

$\begin{equation*} ..., (\alpha p, \beta q, \gamma r), \, (p, q, r), (\alpha p, \beta q, \gamma r), \, (p, q, r), ... \end{equation*}$

if and only if

$\begin{equation*} f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1). \end{equation*}$

Proof. Let $\alpha$ , $\beta$ , $\gamma$ be positive real numbers such that $(\alpha-1)(\beta-1)(\gamma-1)\neq0$ and assume that

$\begin{equation*} ..., (\alpha p, \beta q, \gamma r), \, (p, q, r), (\alpha p, \beta q, \gamma r), \, (p, q, r), ... \end{equation*}$

is a solution for System (1). Then, we have

$\begin{equation} \alpha p = f(q, \beta q) = f(1, \beta) \end{equation}$

(14)

$\begin{equation} p = f(\beta q, q) = f(\beta, 1) \end{equation}$

(15)

$\begin{equation} \beta q = g(r, \gamma r) = g(1, \gamma) \end{equation}$

(16)

$\begin{equation} q = g(\gamma r, r) = g(\gamma, 1) \end{equation}$

(17)

$\begin{equation} \gamma r = h(p, \alpha p) = h(1, \alpha) \end{equation}$

(18)

$\begin{equation} r = h(\alpha p, p) = h(\alpha, 1). \end{equation}$

(19)

From (14)-(19), it follows that

$\begin{equation*} f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1). \end{equation*}$

Now, assume that

$\begin{equation*} f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1). \end{equation*}$

and let

$\begin{equation*} x_{0} = f(\beta, 1), \, x_{-1} = f(1, \beta), \, y_{0} = g(\gamma, 1), \, y_{-1} = g(1, \gamma), \, z_{0} = h(\alpha, 1), \, z_{-1} = h(1, \alpha). \end{equation*}$

We have

$\begin{equation*} x_{1} = f(y_0, y_{-1}) = f(g(\gamma, 1), g(1, \gamma)) = f(g(\gamma, 1), \beta g(\gamma, 1)) = f(1, \beta) = x_{-1}, \end{equation*}$

$\begin{equation*} y_{1} = g(z_0, z_{-1}) = g(h(\alpha, 1), h(1, \alpha)) = g(h(\alpha, 1), \gamma h(\alpha, 1)) = g(1, \gamma) = y_{-1}, \end{equation*}$

$\begin{equation*} z_{1} = h(x_0, x_{-1}) = h(f(\beta, 1), f(1, \beta)) = h(f(\beta, 1), \alpha f(\beta, 1)) = h(1, \alpha) = z_{-1}, \end{equation*}$

$\begin{equation*} x_{2} = f(y_1, y_{0}) = f(g(1, \gamma), g(\gamma, 1)) = f(\beta g(\gamma, 1), g(\gamma, 1)) = f(\beta, 1) = x_{0}, \end{equation*}$

$\begin{equation*} y_{2} = g(z_1, z_{0}) = g(h(1, \alpha), h(\alpha, 1)) = g(\gamma h(\alpha, 1), h(\alpha, 1)) = g(1, \gamma) = y_{0}, \end{equation*}$

$\begin{equation*} z_{1} = h(x_1, x_{0}) = h(f(1, \beta), f(\beta, 1)) = h(\alpha f(\beta, 1), f(\beta, 1)) = h(1, \alpha) = z_{0}. \end{equation*}$

By induction we get

$\begin{equation*} x_{2n-1} = x_{-1}, \, x_{2n} = x_0, \, y_{2n-1} = y_{-1}, \, y_{2n} = y_0, \, z_{2n-1} = z_{-1}, \, z_{2n} = z_0, \, n\in\mathbb{N}_{0}. \end{equation*}$

In the following, we apply our result in finding prime period two solutions of two special Systems.

Consider the three dimensional system of difference equations

$\begin{equation} \left\{ \begin{array}{c} x_{n+1} = a_{1}+b_{1}\frac{y_{n}}{y_{n-1}}+ c_{1}\frac{y_{n-1}}{y_{n}}, \\ y_{n+1} = a_{2}+b_{2}\frac{z_{n}}{z_{n-1}}+ c_{2}\frac{z_{n-1}}{z_{n}}, \\ z_{n+1} = a_{3}+b_{3}\frac{x_{n}}{x_{n-1}}+ c_{3}\frac{x_{n-1}}{x_{n}} \end{array} \right. \end{equation}$

(20)

where $n\in \mathbb{N}_{0}$ , the initial values $x_{-i}, \, y_{-i}, \, z_{-i}, \, i = 0, 1$ and the $a_{i}, b_i, \, c_i, \, i = 1, 2, 3$ are positive real numbers.

System (20) can be seen as a generalization of the system

$\begin{equation*} x_{n+1} = a_{1}+b_{1}\frac{y_{n}}{y_{n-1}}+ c_{1}\frac{y_{n-1}}{y_{n}} , \, y_{n+1} = a_{2}+b_{2}\frac{x_{n}}{x_{n-1}}+ c_{2}\frac{x_{n-1}}{x_{n}}, \end{equation*}$

studied in [23]. This last one is also a generalization of the equation

$\begin{equation*} x_{n+1} = a_{1}+b_{1}\frac{x_{n}}{x_{n-1}}+ c_{1}\frac{x_{n-1}}{x_{n}} \end{equation*}$

studied in [7] and [15].

Corollary 1. Assume that $(\alpha-1)(\beta-1)(\gamma-1)\neq0$ , then System $(20)$ has prime period two solution of the form

$\begin{equation*} ..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, ( f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ... \end{equation*}$

if and only if

$\begin{equation} \left\{ \begin{array}{c} (b_{1}\alpha-c_{1})\beta^2+a_{1}\beta(\alpha-1)+c_{1}\alpha-b_{1} = 0, \\ (b_{2} \beta-c_{2})\gamma^2+a_{2}\gamma(\beta-1)+c_{2}\beta-b_{2} = 0, \\ (b_{3}\gamma-c_{3})\alpha^2+a_{3}\alpha(\gamma-1)+c_{3}\gamma-b_{3} = 0. \end{array} \right . \end{equation}$

(21)

Proof. System (20) can be written as

$x_{n+1} = f(y_n, y_{n-1}), \, y_{n+1} = g(z_{n}, z_{n-1}), \, z_{n+1} = h(x_{n}, x_{n-1}),$

where

$f(u, v) = a_{1}+b_{1}\frac{u}{v}+ c_{1}\frac{v}{u}, \, g(u, v) = a_{2}+b_{2}\frac{u}{v}+ c_{2}\frac{v}{u}, \, h(u, v) = a_{3}+b_{3}\frac{u}{v}+ c_{3}\frac{v}{u}.$

So, from Theorem 3.1,

$\begin{equation*} ..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, ( f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ... \end{equation*}$

will be a period prime two solution of System (20) if and only if

$f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1).$

Clearly this condition is equivalent to (21).

Example 2. If we choose $\alpha = 2, \, \beta = 3, \, \gamma = \frac{1}{2}$ then, condition (21) will be

$\begin{equation*} 3a_{1}+17b_{1}-7c_{1} = 0, \, 4a_{2}-b_{2}+11c_{2} = 0, \, -2a_{3}+2b_{3}-7c_{3} = 0. \end{equation*}$

The last condition is satisfied for the choice

$\begin{equation*} a_1 = \frac{4}{3}, \, b_1 = 1, \, c_1 = 3, \, a_2 = \frac{1}{4}, \, b_2 = 2, \, c_2 = \frac{1}{ 11}, \, a_3 = \frac{1}{2}, \, b_3 = 4, \, c_3 = 1 \end{equation*}$

of the parameters. The corresponding prime period two solution will be

$\begin{equation*} x_{2n-1} = x_{-1} = \alpha f(\beta, 1) = \frac{32}{3}, \end{equation*}$

$\begin{equation*} y_{2n-1} = y_{-1} = \beta g(\gamma, 1) = \frac{189}{44}, \end{equation*}$

$\begin{equation*} z_{2n-1} = z_{-1} = \gamma h(\alpha, 1) = \frac{9}{2}, \end{equation*}$

and

$\begin{equation*} x_{2n} = x_{0} = f(\beta, 1) = \frac{16}{3}, \end{equation*}$

$\begin{equation*} y_{2n} = y_{0} = g(\gamma, 1) = \frac{63}{44}, \end{equation*}$

$\begin{equation*} z_{2n} = z_{0} = h(\alpha, 1) = 9, \end{equation*}$

that is

$\begin{equation*} \left\{\left(\frac{32}{3}, \frac{189}{44}, \frac{9}{2} \right), \, \left(\frac{ 16}{3}, \frac{63}{44}, 9 \right), \, \left(\frac{32}{3}, \frac{189}{44}, \frac{9 }{2} \right), \, \left(\frac{16}{3}, \frac{63}{44}, 9 \right), \cdots\right\}. \end{equation*}$

Now, consider the following system of difference equations

$\begin{equation} \left\{ \begin{array}{c} x_{n+1} = a_{1}+b_{1}\frac{y_{n}}{y_{n-1}}+ c_{1}\left(\frac{y_{n-1}}{y_{n}} \right)^{2}, \\ y_{n+1} = a_{2}+b_{2}\frac{z_{n-1}}{z_{n}}+ c_{2}\left(\frac{ z_{n-1}}{z_{n}}\right)^{2}, \\ z_{n+1} = a_{3}+b_{3}\frac{x_{n-1}}{x_{n}}+ 3_{2}\left(\frac{x_{n-1}}{x_{n}}\right)^{2}, \end{array} \, n\in \mathbb{N}_{0} \right. \end{equation}$

(22)

where the initial values $x_{-i}, \, y_{-i}, \, z_{-i}, \, i = 0, 1$ and $a_{i}, b_i, \, c_i, \, i = 1, 2, 3$ are positive real numbers.

Corollary 2. Assume that $(\alpha-1)(\beta-1)(\gamma-1)\neq0$ , then System $(22)$ has prime period two solution of the form

$\begin{equation*} ..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, (f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ... \end{equation*}$

if and only if

$\begin{equation} \left\{ \begin{array}{c} a_1\beta^2(\alpha-1)+b_1\beta(\alpha\beta^2-1)+c_1(\alpha-\beta^4) = 0, \\ a_2\gamma^2(\beta-1)+b_2\gamma(\beta\gamma^2-1)+c_2(\beta-\gamma^4) = 0, \\ a_3\alpha^2(\gamma-1)+b_3\alpha(\gamma\alpha^2-1)+c_3(\gamma-\alpha^4) = 0. \end{array} \right. \end{equation}$

(23)

Proof. System (22) can be written as

$x_{n+1} = f(y_n, y_{n-1}), \, y_{n+1} = g(z_{n}, z_{n-1}), \, z_{n+1} = h(x_{n}, x_{n-1}),$

where

$\begin{equation*} \left\{ \begin{array}{c} f(u, v) = a_{1}+b_{1}\frac{u}{v}+ c_{1}\left(\frac{v}{u}\right)^{2}, \\ g(u, v) = a_{2}+b_{2}\frac{v}{u}+ c_{2}\left(\frac{v}{u}\right)^{2}, \\ h(u, v) = a_{3}+b_{3}\frac{v}{u}+ c_{3}\left(\frac{v}{u}\right)^{2}. \end{array} \right. \end{equation*}$

So, from Theorem 3.1,

$..., ( \alpha f(\beta, 1), \beta g(\gamma, 1), \gamma h(\alpha, 1)), \, ( f(\beta, 1), g(\gamma, 1), h(\alpha, 1)), ...$

will be a period prime two solution of System (20) if and only if

$f(1, \beta) = \alpha f(\beta, 1), \, g(1, \gamma) = \beta g(\gamma, 1), \, h(1, \alpha) = \gamma h(\alpha, 1).$

Clearly this condition is equivalent to

$\begin{equation*} \left\{ \begin{array}{c} a_1\beta^2(\alpha-1)+b_1\beta(\alpha\beta^2-1)+c_1(\alpha-\beta^4) = 0, \\ a_2\gamma^2(\beta-1)+b_2\gamma(\beta\gamma^2-1)+c_2(\beta-\gamma^4) = 0, \\ a_3\alpha^2(\gamma-1)+b_3\alpha(\gamma\alpha^2-1)+c_3(\gamma-\alpha^4) = 0. \end{array} \right. \end{equation*}$

Example 3. For $\alpha = 3, \, \beta = 2, \, \gamma = \frac{1}{3}$ then, condition (23) will be

$\begin{equation*} 8a_1+22b_1-13c_1 = 0, \, 9a_2-21b_2+161c_2 = 0, \, 18a_3-18b_3+242c_3 = 0 . \end{equation*}$

The last condition is satisfied for the choice

$\begin{equation*} a_1 = 1, \, b_1 = 1, \, c_1 = \frac{30}{13}, \, a_2 = \frac{19}{9}, \, b_2 = 2, \, c_2 = \frac{1}{ 7}, \, a_3 = \frac{1}{6}, \, b_3 = \frac{7}{9}, \, c_3 = \frac{1}{22} \end{equation*}$

of the parameters. The corresponding prime period two solution will be

$\begin{equation*} x_{2n-1} = x_{-1} = 3 f(2, 1) = \frac{297}{26}, \end{equation*}$

$\begin{equation*} y_{2n-1} = y_{-1} = 2 g(\frac{1}{3}, 1) = \frac{512}{63}, \end{equation*}$

$\begin{equation*} z_{2n-1} = z_{-1} = \frac{1}{3} h(3, 1) = \frac{248}{297}, \end{equation*}$

and

$\begin{equation*} x_{2n} = x_{0} = f(2, 1) = \frac{93}{26}, \end{equation*}$

$\begin{equation*} y_{2n} = y_{0} = g(\frac{1}{3}, 1) = \frac{256}{63}, \end{equation*}$

$\begin{equation*} z_{2n} = z_{0} = h(3, 1) = \frac{744}{297}, \end{equation*}$

that is

$\begin{equation*} \left\{\left(\frac{297}{26}, \frac{512}{63}, \frac{248}{297} \right), \, \left( \frac{93}{26}, \frac{256}{63}, \frac{744}{297} \right), \, \left(\frac{297}{26}, \frac{512}{63}, \frac{248}{297} \right), \, \cdots\right\}. \end{equation*}$

4. Existence of oscillatory solutions

Here, we are interested in the oscillation of the solutions of System (1) about the equilibrium point $(\overline{x}, \overline{y}, \overline{z }) = (f(1, 1), g(1, 1),$ $h(1, 1))$ .

Theorem 4.1. Let $\left(x_n, y_n, z_n\right)_{n\geq-1}$ be a solution of System $(1)$ and assume that $f(x, y)$ , $g(x, y)$ $h(x, y)$ are decreasing in $x$ for all $y$ and are increasing in $y$ for all $x$ .

1. If

$\begin{equation*} x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}, \end{equation*}$

then we get

$\begin{equation*} x_{2n} < \overline{x}, \, x_{2n-1} > \overline{x}, \, y_{2n} < \overline{y}, \, y_{2n-1} > \overline{y}, \, z_{2n} < \overline{z}, \, z_{2n-1} > \overline{z}, \, n\in\mathbb{N} _0. \end{equation*}$

That is for the sequences $(x_n)_{n\geq-1}$ , $(y_n)_{n\geq-1}$ and $(z_n)_{n\geq-1}$ we have semi-cycles of length one of the form

$\begin{equation*} +-+-+-\cdots. \end{equation*}$

2. If

$\begin{equation*} x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}, \end{equation*}$

then we get

$\begin{equation*} x_{2n} > \overline{x}, \, x_{2n-1} < \overline{x}, \, y_{2n} > \overline{y}, \, y_{2n-1} < \overline{y}, \, z_{2n} > \overline{z}, \, z_{2n-1} < \overline{z}, \, n\in\mathbb{N} _0. \end{equation*}$

That is for the sequences $(x_n)_{n\geq-1}$ , $(y_n)_{n\geq-1}$ and $(z_n)_{n\geq-1}$ we have semi-cycles of length one of the form

$\begin{equation*} -+-+-+\cdots. \end{equation*}$

Proof. 1. Assume that

$\begin{equation*} x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}. \end{equation*}$

We have

$\begin{equation*} x_{1} = f(y_0, y_{-1}) > f(\overline{y}, y_{-1}) > f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}$

$\begin{equation*} y_{1} = g(z_0, z_{-1}) > g(\overline{z}, z_{-1}) > g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}$

$\begin{equation*} z_{1} = h(x_0, x_{-1}) > h(\overline{x}, x_{-1}) > h(\overline{x}, \overline{x}) = h(1, 1) = \overline{z}, \end{equation*}$

$\begin{equation*} x_{2} = f(y_1, y_{0}) < f(\overline{y}, y_{0}) < f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}$

$\begin{equation*} y_{2} = g(z_1, z_{0}) < g(\overline{z}, z_{0}) < g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}$

$\begin{equation*} z_{2} = h(x_1, x_{0}) < h(\overline{x}, x_{0}) < h(\overline{x}, \overline{x}) = h(1, 1) = \overline{z}. \end{equation*}$

By induction, we get

2. Assume that

$\begin{equation*} x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}. \end{equation*}$

We have

$\begin{equation*} x_{1} = f(y_0, y_{-1}) < f(\overline{y}, y_{-1}) < f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}$

$\begin{equation*} y_{1} = g(z_0, z_{-1}) < g(\overline{z}, z_{-1}) < g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}$

$\begin{equation*} z_{1} = h(x_0, x_{-1}) < h(\overline{x}, x_{-1}) < h(\overline{x}, \overline{x}) = h(1, 1) = \overline{z}, \end{equation*}$

$\begin{equation*} x_{2} = f(y_1, y_{0}) > f(\overline{y}, y_{0}) > f(\overline{y}, \overline{y}) = f(1, 1) = \overline{x}, \end{equation*}$

$\begin{equation*} y_{2} = g(z_1, z_{0}) > g(\overline{z}, z_{0}) > g(\overline{z}, \overline{z}) = g(1, 1) = \overline{y}, \end{equation*}$

$\begin{equation*} x_{2} = h(x_1, x_{0}) > h(\overline{x}, x_{0}) > h(\overline{x}, \overline{x}) = h(1, 1) = \overline{y}. \end{equation*}$

By induction, we get

So, the proof is completed. Now, we will apply the results of this section on the following particular system.

Example 4. Consider the system of difference equations

$\begin{equation} x_{n+1} = a_{1}+b_{1}\left(\frac{y_{n-1}}{y_{n}}\right)^{p}, \, y_{n+1} = a_{2}+ b_{2}\left(\frac{z_{n-1}}{z_{n}}\right)^{q}, \, z_{n+1} = a_{3}+ b_{3}\left( \frac{x_{n-1}}{x_{n}}\right)^{r}, \, n\in \mathbb{N}_{0}, \end{equation}$

(24)

where $p, \, q, \, r\in \mathbb{N}$ , the initial values $x_{-i}, \, y_{-i}, \, z_{-i}, \, i = 0, 1$ and the parameters $a_i, \, b_i, \, i = 1, 2, 3$ are positive real numbers.

Let $f$ , $g$ and $h$ be the functions defined by

$\begin{equation*} f(u, v) = a_{1}+b_{1}\left(\frac{v}{u}\right)^{p}, \, g(u, v) = a_{2}+b_{2}\left( \frac{v}{u}\right)^{q}, \, h(u, v) = a_{3}+b_{3}\left(\frac{v}{u} \right)^{r}, \, u, v\in (0, +\infty). \end{equation*}$

It is not hard to see that

$\begin{equation*} \frac{\partial f}{\partial u}(u, v) < 0, \, \frac{\partial f}{\partial v} (u, v) > 0, \, \frac{\partial g}{\partial u}(u, v) < 0, \, \frac{\partial g}{\partial v }(u, v) > 0\, \frac{\partial h}{\partial u}(u, v) < 0, \, \frac{\partial h}{\partial v }(u, v) > 0. \end{equation*}$

System (24) has the unique equilibrium point $(\overline{x}, \overline{y }, \overline{z}) = (a_1+b_1, a_2+b_2, a_3+b_3)$ .

Corollary 3. Let $\left(x_n, y_n, z_n\right)_{n\geq-1}$ be a solution of System $(24)$ . The following statements holds true:

1. Let

$\begin{equation*} x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}. \end{equation*}$

Then the sequences $(x_n)_n$ (resp. $(y_n)_n$ , $(z_n)_n$ ) oscillates about $\overline{x}$ (resp. about $\overline{y}$ , $\overline{z}$ ) with semi-cycle of length one and every semi-cycle is in the form

$\begin{equation*} +-+-+-\cdots. \end{equation*}$

2. Let

$\begin{equation*} x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}. \end{equation*}$

$\begin{equation*} -+-+-+\cdots. \end{equation*}$

Proof. 1. Let

$x_{0} < \overline{x}, \, x_{-1} > \overline{x}, \, y_{0} < \overline{y}, \, y_{-1} > \overline{y}, \, z_{0} < \overline{z}, \, z_{-1} > \overline{z}.$

We have

$\frac{y_{-1}}{y_{0}} > \frac{\overline{y}}{\overline{y}} = 1,$

which implies that

$x_1 = a_1+b_1\left(\frac{y_{-1}}{y_{0}}\right)^{p} > a_1+b_1 = \overline{x}.$

Using the fact that

$\frac{z_{-1}}{z_{0}} > \frac{\overline{z}}{\overline{z}} = 1,$

we get

$y_1 = a_2+b_2\left(\frac{z_{-1}}{z_{0}}\right)^{q} > a_2+b_2 = \overline{y}.$

Also as,

$\frac{x_{-1}}{x_{0}} > \frac{\overline{x}}{\overline{x}} = 1,$

we get

$z_1 = a_3+b_3\left(\frac{x_{-1}}{x_{0}}\right)^{r} > a_3+b_3 = \overline{z}.$

Now, as

$\frac{y_{0}}{y_{1}} < \frac{\overline{y}}{\overline{y}} = 1,$

we obtain

$x_2 = a_1+b_1\left(\frac{y_{0}}{y_{1}}\right)^{p} < a_1+b_1 = \overline{x}.$

Similarly,

$\frac{z_{0}}{z_{1}} < \frac{\overline{z}}{\overline{z}} = 1\Rightarrow y_2 = a_2+b_2\left(\frac{z_{0}}{z_{1}}\right)^{q} < a_2+b_2 = \overline{y},$

and

$\frac{x_{0}}{x_{1}} < \frac{\overline{x}}{\overline{x}} = 1\Rightarrow z_2 = a_3+b_3\left(\frac{x_{0}}{x_{1}}\right)^{r} < a_3+b_3 = \overline{z},$

and by induction we get that

$x_{2n}-\overline{x} < 0, \, y_{2n}-\overline{y} < 0, \, z_{2n}-\overline{z} < 0,$

$x_{2n-1}-\overline{x} > 0, \, y_{2n-1}-\overline{y} > 0, \, z_{2n-1}-\overline{z} > 0,$

for $n\in \mathbb{N}_{0}$ . That is, the sequences $(x_n)_n$ (resp. $(y_n)_n$ , $(z_n)_n$ ) oscillates about $\overline{x}$ (resp. about $\overline{y}$ , $\overline{z}$ ) with semi-cycle of length one and every semi-cycle is in the form

$+-+-+-\cdots$

and this is the statement of Part 1. of Theorem 4.1.

$2.$ Let

$x_{0} > \overline{x}, \, x_{-1} < \overline{x}, \, y_{0} > \overline{y}, \, y_{-1} < \overline{y}, \, z_{0} > \overline{z}, \, z_{-1} < \overline{z}.$

We have

$\frac{y_{-1}}{y_{0}} < \frac{\overline{y}}{\overline{y}} = 1,$

which implies that

$x_1 = a_1+b_1\left(\frac{y_{-1}}{y_{0}}\right)^{p} < a_1+b_1 = \overline{x}.$

Using the fact that

$\frac{z_{-1}}{z_{0}} < \frac{\overline{z}}{\overline{z}} = 1,$

we get

$y_1 = a_2+b_2\left(\frac{z_{-1}}{z_{0}}\right)^{q} < a_2+b_2 = \overline{y}.$

Also as,

$\frac{x_{-1}}{x_{0}} < \frac{\overline{x}}{\overline{x}} = 1,$

we get

$z_1 = a_3+b_3\left(\frac{x_{-1}}{x_{0}}\right)^{r} < a_3+b_3 = \overline{z}.$

Now, as

$\frac{y_{0}}{y_{1}} > \frac{\overline{y}}{\overline{y}} = 1,$

we obtain

$x_2 = a_1+b_1\left(\frac{y_{0}}{y_{1}}\right)^{p} > a_1+b_1 = \overline{x}.$

Similarly,

$\frac{z_{0}}{z_{1}} > \frac{\overline{z}}{\overline{z}} = 1\Rightarrow y_2 = a_2+b_2\left(\frac{z_{0}}{z_{1}}\right)^{q} > a_2+b_2 = \overline{y},$

and

$\frac{x_{0}}{x_{1}} > \frac{\overline{x}}{\overline{x}} = 1\Rightarrow z_2 = a_3+b_3\left(\frac{x_{0}}{x_{1}}\right)^{r} > a_3+b_3 = \overline{z}.$

Thus, by induction we get that

$x_{2n}-\overline{x} > 0, \, y_{2n}-\overline{y} > 0, \, z_{2n}-\overline{z} > 0,$

$x_{2n-1}-\overline{x} < 0, \, y_{2n-1}-\overline{y} < 0, \, z_{2n-1}-\overline{z} < 0$

for $n\in \mathbb{N}_{0}$ . That is the sequences $(x_n)_n$ (resp. $(y_n)_n$ , $(z_n)_n$ ) oscillates about $\overline{x}$ (resp. about $\overline{y}$ , $\overline{z}$ ) with semi-cycle of length one and every semi-cycle is in the form

$-+-+-+\cdots$

and this is the statement of Part 2. of Theorem 4.1.

5. Conclusion

In this study, the global stability of the unique positive equilibrium point of a three-dimensional general system of difference equations defined by positive and homogeneous functions of degree zero was studied. For this, general convergence theorems were given considering all possible monotonicity cases in arguments of functions $f$ , $g$ and $h$ . In addition, the periodic nature and oscillation of the general system considered was also discussed and successful results were obtained. It is noteworthy that the results obtained on our general three-dimensional system have high applicability.

Acknowledgments

The authors thanks the two referees for their comments and suggestions. The work of N. Touafek and Y. Akrour was supported by DGRSDT (MESRS-DZ).

Abbreviation WM: Working memory; fMRI: functional Magnetic Resonance Imaging; IFG: Inferior Frontal Gyrus; BA: Brodmann area; STG: Superior Temporal Gyrus; SMA: Supplementary motor area; Pre-SMA: Pre-Supplementary motor area; AF: Arcuate Fasciculus; DLPFC: Dorsolateral prefrontal cortex; ACC: Anterior cingulate cortex; DMN: Default-Mode network; EEG/ERP: Electroencephalography/Event-Related potential; PET: Positron Emission Tomography; MTG: Middle Temporal Gyrus; PCC: Posterior cingulate cortex; OFC: Orbitofrontal cortex; MD: Multiple demand;

Conflicts of interest

The authors declare no competing interests and no relationship that may lead to any conflict of interest.

¹ The n-back task is administered broadly to study WM. The task consists of a list of visual or auditory stimuli that is presented to participants. They are instructed to indicate whether each stimulus is a correct match with the n^th stimulus presented before. The n-back task demands constantly storing, updating information and inhibiting distractors. The cognitive load can be altered in this task by changing the value of n, which correspondingly may affect accuracy and reaction times. In the n-back task, visual or auditory stimuli are stored in the phonological loop or visuospatial sketchpad loop while selecting the correct answer depends on the central executive function. The central executive function is responsible for updating information and inhibiting the processing of irrelevant information. Brain functions linked to different types of WM (such as visuospatial WM) can be examined by using n-back tasks.

² The WM span tasks are divided into two main categories: simple and complex span tasks. The simple span task requires encoding a list of items (e.g., words) and maintaining that information during short delay periods and then recalling them. On the other hand, the complex span task includes presenting a list of items and asking participants to memorize them while performing another cognitive task (e.g., mathematics problem-solving, reading) and then recalling them. Performing the distractive task while maintaining task-relevant information makes the maintaining phase more difficult by directing attention towards task-irrelevant information. The span tasks, particularly complex span tasks, need storing, updating information and inhibiting distractors.

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