On a two dimensional dynamical system generated by the floor function

TL;DR

This paper studies a two-dimensional dynamical system created by the floor function, analyzing how its long-term behavior varies with different parameters and initial conditions.

Contribution

It provides a comprehensive description of all limit points of the system depending on the parameter and initial point, advancing understanding of floor function-based dynamics.

Findings

Characterization of limit points for various parameter values

Dependence of system behavior on initial conditions

Identification of stable and unstable regimes

Abstract

In this paper we investigate the two dimensional dynamical system generated by the floor function with a parameter $\lambda\in \R$. We describe all limit points of the dynamical system depending on $\lambda$ and on the initial point.

Equations

$$A:z=(x,y)\in {\mathbb{R}}^2\to z^{\prime }=(x^{\prime },y^{\prime })=A(z)\in {\mathbb{R}}^2$$

$$A(z)=\left\{\begin{array}[]{ll}x^{\prime}=\lfloor\lambda y\rfloor&\hbox{}\\ y^{\prime}=\lfloor\lambda x\rfloor&\hbox{},\end{array}\right.z=(x,y)\in\mathbb{R}^{2}$$

$$z=(x,y),A(z)=(\lfloor \lambda y\rfloor ,\lfloor \lambda x\rfloor ),A^2(z)=(\lfloor \lambda \lfloor \lambda x\rfloor \rfloor ,\lfloor \lambda \lfloor \lambda y\rfloor \rfloor ),\dots$$

$$\underset{n\to \infty }{\lim }{A}^n(z),$$

$$\mathrm{Fix}(A)=\{(x,\lfloor \lambda x\rfloor )|x\in \{0,-1,-2,...,-m\}\};$$

$$\mathrm{Fix}(A)=\{(x,\lfloor \lambda x\rfloor )|x\in \{0,1,2,...,m-1\}\}.$$

$$\underset{n\to \infty }{\lim }{A}^n(z)=(0,0)$$

$$\omega(z)=\left\{\begin{array}[]{ll}\{z,A(z)\},\ \ \ \ \ \ \ \mbox{if}\ \ z\in\mathbb{Z}\times\mathbb{Z};\\ \{A(z),A^{2}(z)\},\ \ \mbox{if}\ \ z\in\mathbb{R}^{2}\setminus\mathbb{Z}\times\mathbb{Z}.\end{array}\right.$$

$$\omega(z)=\left\{\begin{array}[]{ll}\{(0,0)\},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ if\ \ z\in U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] \{(\infty,\infty),(-\infty,-\infty)\},\ \ if\ \ z\in\{\mathbb{R}_{++}^{2}\cup\mathbb{R}_{--}^{2}\}\setminus U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] \{(\infty,-\infty),(-\infty,\infty)\},\ \ if\ \ z\in\{\mathbb{R}_{+-}^{2}\cup\mathbb{R}_{-+}^{2}\}\setminus U_{\frac{1}{|\lambda|}}^{-}(0).\end{array}\right.$$

$$z=(x_0,y_0),A(z)=(\lfloor \lambda y_0\rfloor ,\lfloor \lambda x_0\rfloor ),A^2(z)=(\lfloor \lambda \lfloor \lambda x_0\rfloor \rfloor ,\lfloor \lambda \lfloor \lambda y_0\rfloor \rfloor ),....$$

$$u_{n}=\left\{\begin{array}[]{ll}x_{n},\ \ \hbox{if}\ \ n-\hbox{even}\\ y_{n},\ \ \hbox{if}\ \ n-\hbox{odd}\end{array}\right.$$

$$v_{n}=\left\{\begin{array}[]{ll}x_{n},\ \ \hbox{if}\ \ n-\hbox{odd}\\ y_{n},\ \ \hbox{if}\ \ n-\hbox{even}.\end{array}\right.$$

$$\{u_n\}:u_0=x_0,u_1=y_1=\lfloor \lambda x_0\rfloor ,u_2=x_2=\lfloor \lambda \lfloor \lambda x_0\rfloor \rfloor ,u_3=y_3=\lfloor \lambda \lfloor \lambda \lfloor \lambda x_0\rfloor \rfloor \rfloor ,...$$

$$\{v_n\}:v_0=y_0,v_1=x_1=\lfloor \lambda y_0\rfloor ,v_2=y_2=\lfloor \lambda \lfloor \lambda y_0\rfloor \rfloor ,v_3=x_3=\lfloor \lambda \lfloor \lambda \lfloor \lambda y_0\rfloor \rfloor \rfloor ,...$$

$$\underset{n\to \infty }{\lim }{A}^n(z)=(0,0).$$

$$\omega(z)=\left\{\begin{array}[]{ll}\{z,A(z)\},\ \ \ \ \ \ \ \mbox{if}\ \ z\in\mathbb{Z}\times\mathbb{Z};\\[8.53581pt] \{A(z),A^{2}(z)\},\ \ \mbox{if}\ \ z\in\mathbb{R}^{2}\setminus\mathbb{Z}\times\mathbb{Z}.\end{array}\right.$$

$$u_1<|u_2|\le u_3<|u_4|\le u_5<\dots$$

$$|u_1|\le u_2<|u_3|\le u_4<|u_5|\le \dots .$$

$$\lim_{n\rightarrow\infty}A^{2n}(z)=\left\{\begin{array}[]{ll}(\infty,\infty),\ \ \ \ \ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{+,+}\setminus U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] (-\infty,-\infty),\ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{-,-}\setminus U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] (\infty,-\infty),\ \ \ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{+,-}\setminus U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] (-\infty,\infty),\ \ \ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{-,+}\setminus U^{-}_{\frac{1}{|\lambda|}}(0).\end{array}\right.$$

$$\lim_{n\rightarrow\infty}A^{2n+1}(z)=\left\{\begin{array}[]{ll}(-\infty,-\infty),\ \ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{+,+}\setminus U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] (\infty,\infty),\ \ \ \ \ \ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{-,-}\setminus U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] (-\infty,\infty),\ \ \ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{+,-}\setminus U^{-}_{\frac{1}{|\lambda|}}(0);\\[8.53581pt] (\infty,-\infty),\ \ \ \ \hbox{if}\ \ z\in\mathbb{R}^{2}_{-,+}\setminus U^{-}_{\frac{1}{|\lambda|}}(0).\end{array}\right.$$

$$\underset{n\to \infty }{\lim }{A}^n(z)=(0,0).$$

$$\omega (z)=\{(k,0),(0,k)\};$$

$$\omega (z)=\{(-m,0),(0,-m)\}.$$

$$\underset{n\to \infty }{\lim }{A}^n(z)=(-m,-m).$$

$$\omega (z)=\{(k,p),(p,k)\};$$

$$\omega (z)=\{(k,-m),(-m,k)\};$$

$$0\le \underset{n\to \infty }{\lim }{u}_n\le \underset{n\to \infty }{\lim }{\lambda }^n{u}_0=0,$$

$$0\le \underset{n\to \infty }{\lim }{v}_n\le \underset{n\to \infty }{\lim }{\lambda }^n{v}_0=0,$$

$$\underset{n\to \infty }{\lim }{u}_n=\underset{n\to \infty }{\lim }{v}_n=-m\Rightarrow \underset{n\to \infty }{\lim }{A}^n(z)=(-m,-m),$$

$$\underset{n\to \infty }{\lim }{u}_n=k,\underset{n\to \infty }{\lim }{v}_n=p\Rightarrow \omega (z)=\{(k,p),(p,k)\},$$

Full text

1. Introduction

Let $X\subset\mathbb{R}^{2}$ and $A$ be an operator from $X$ to itself. The main problem of the theory of dynamical systems is to study iterations of $A$ : if $A^{n}$ denotes the $n$-fold composition (iteration) of $A$ with itself, then for a given point $x$ one investigates the sequence $x,Ax,A^{2}x,A^{3}x$, and so on. This sequence is called two-dimensional discrete time dynamical system or the forward orbit of $x$, or just the orbit of $x$ for short ([1], [3]).

Definition 1**.**

A point $z\in\mathbb{R}^{2}$ is called a fixed point of $A$ operator if $A(z)=z$. The set of all fixed points is denoted by ${\rm Fix}(A)$.

Definition 2**.**

The point $z\in\mathbb{R}^{2}$ is a periodic point of period $n$ if $A^{n}(z)=z$ and $A^{n-1}(z)\neq z$. We denote the set of periodic points of period $n$ by ${\rm Per}_{n}(A)$. The set of all iterates of a periodic point form a periodic orbit.

For a given operator $A:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ the $\omega$-limit set of $z\in\mathbb{R}^{2}$, denoted by $\omega(z,A)$ or $\omega(z)$, is the set of limit points of the forward orbit $\{A^{n}(z)\}_{n\in\mathbb{N}}$. Hence, $y\in\omega(z)$ if and only if there is a strictly increasing sequence of natural numbers $\{n_{k}\}_{k\in\mathbb{N}}$ such that $A^{n_{k}}(z)\rightarrow y$ as $k\rightarrow\infty$.

In this paper we will study the dynamical system generated by operator

[TABLE]

defined by

[TABLE]

where $\lambda\in\mathbb{R}$ is parameter and $\lfloor x\rfloor$ denotes the integer part of $x$.

In our case the dynamical system is

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The main problem is to investigate the following limit

[TABLE]

for any $z\in\mathbb{R}^{2}$.

2. The main results

2.1. Fixed points

The following lemma gives all fixed points of this operator.

Lemma 1**.**

For the set of fixed points the followings hold:

If $\lambda<0(\lambda\neq-1)$, then ${\rm Fix}(A)=\{(0,0)\}$;

• 2)

If $\lambda=-1$, then ${\rm Fix}(A)=\{(m,-m)\,|\,m\in\mathbb{Z}\}$;

• 3)

If $\frac{m-1}{m}<\lambda\leq\frac{m}{m+1}$ for some $m\in\mathbb{N}$, then

[TABLE]

• 4)

If $\lambda=1$, then ${\rm Fix}(A)=\{(m,m)\,|\,m\in\mathbb{Z}\}$;

• 5)

If $\frac{m+1}{m}\leq\lambda<\frac{m}{m-1}$ for some $m\in\mathbb{N}$, then

[TABLE]

Proof. For finding fixed points of the operator we need solve $A(z)=z$, i.e., that system \left\{\begin{array}[]{ll}x=\lfloor\lambda y\rfloor&\\ y=\lfloor\lambda x\rfloor.&\end{array}\right. By expressing $y$ we get the equation $\lfloor\lambda\lfloor\lambda x\rfloor\rfloor=x$. Roots of $\lfloor\lambda x\rfloor=x$ are also roots of $\lfloor\lambda\lfloor\lambda x\rfloor\rfloor=x$. That’s why fixed points given in parts 1-5 of Lemma took by solving $\lfloor\lambda x\rfloor=x$ (see e.g. [2]). But there may be some fixed points of $g(x)=\lfloor\lambda\lfloor\lambda x\rfloor\rfloor$ those are roots of $\lfloor\lambda\lfloor\lambda x\rfloor\rfloor=x$ and are not roots of $\lfloor\lambda x\rfloor=x$. We shall prove that fixed points of operator are just roots of $\lfloor\lambda x\rfloor=x$.

In [2] all limit points $\omega(x)$ of the floor function $f(x)=\lfloor\lambda x\rfloor$ found for $\forall\lambda,x\in R$. It was proved that the limit of $\{f^{n}(x)\}$ converges: to fixed points; to $\infty$, or -$\infty$(case $\lambda=-1$ exception). For the case $\lambda=-1$ it was proved that ${\rm Per}_{2}(f)=\mathbb{Z}$.

By ${\rm Per}_{2}(f)\subset{\rm Fix}(g)$ we have part 2 of Lemma. Because of $\{g^{n}(x)\}=\{f^{2n}(x)\}$, if $\{f^{n}(x)\}$ is convergent then $\{g^{n}(x)\}$ also is convergent, if $\{f^{n}(x)\}$ converges to $\infty$ then $\{g^{n}(x)\}$ also converges to $\infty$. Thus fixed points of $g(x)$ for $\lambda\neq-1$ consist of roots of $\lfloor\lambda x\rfloor=x$ only. Now we prove part 2, $x=\lfloor-y\rfloor=\lfloor-\lfloor-x\rfloor\rfloor=x$.

2.2. The limit points

Now we shall describe the set $\omega(z)$ for each given $z\in\mathbb{R}^{2}$.

2.2.1. The case $\lambda\leq 0$.

Theorem 1**.**

If $\lambda<0$, then the dynamical system generated by operator $A$ has the following properties:

• (1)

If $-1<\lambda<0$, then

[TABLE]

for all $z\in\mathbb{R}^{2}$.

• (2)

If $\lambda=-1$, then each pairs of integer numbers has period two and

[TABLE]

• (3)

If $\lambda<-1$, then $A(z)=0$ for $z\in U_{\frac{1}{|\lambda|}}^{-}(0)=\{(x,y)\,|\,\frac{1}{\lambda}<x\leq 0,\frac{1}{\lambda}<y\leq 0\}$ and

[TABLE]

where

$\mathbb{R}_{++}^{2}=\{(x,y)\,|\,x,y\in\mathbb{R},x>0,y>0\}$,  $\mathbb{R}_{--}^{2}=\{(x,y)\,\ |\,\ x,y\in\mathbb{R},x<0,y<0\}$,

$\mathbb{R}_{+-}^{2}=\{(x,y)\,\ |\,\ x,y\in\mathbb{R},x>0,y<0\}$,  $\mathbb{R}_{-+}^{2}=\{(x,y)\,\ |\,\ x,y\in\mathbb{R},x<0,y>0\}$.

Proof. (1) Let $-1<\lambda<0$. For all $z=(x_{0},y_{0})\in\mathbb{R}^{2}$ we have

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We can separate sequence (2.1) to two subsequences $\{u_{n}\}$ and $\{v_{n}\}$:

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[TABLE]

Let’s write several terms of those sequences,

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[TABLE]

In Theorem 2 of [2] was proved $\lim_{n\rightarrow\infty}u_{n}=0$ for any $\lambda\in(-1,0)$ and for all $x_{0}\in\mathbb{R}$. That’s why we have

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(2) If $\lambda=-1$, then $A^{2}(z)=z$ and $A(z)\neq z$ for all $z\in\mathbb{Z}\times\mathbb{Z}$. Thus each pairs of integer numbers has period two. If $z\in\mathbb{R}^{2}\setminus Z\times\mathbb{Z}$, then $A(z)\in\mathbb{Z}\times\mathbb{Z}$. So we have

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(3) Let $\lambda<-1$ and $z=(x_{0},y_{0})\in U^{-}_{\frac{1}{|\lambda|}}(0)$. Then we have $\lfloor\lambda x_{0}\rfloor=0$ and $\lfloor\lambda y_{0}\rfloor=0$, i.e., $A(z)=0$. If $x_{0}\leq\frac{1}{\lambda}$ then for sequence (2.2) we have

[TABLE]

for all $z\in\mathbb{R}^{2}\setminus U^{-}_{\frac{1}{|\lambda|}}(0)$ and, if $x_{0}>0$ then

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Since $\{|u_{n}|\}_{n\geq 1}\subset\mathbb{N}$ and $\lim_{n\rightarrow\infty}|u_{n}|=\infty$ the following hold

[TABLE]

[TABLE]

2.2.2. The case $0<\lambda<1$.

Note that for each $\lambda\in(0,1)$ there exists $m\in\mathbb{N}$ such that ${m-1\over m}<\lambda\leq{m\over m+1}$.

Theorem 2**.**

Let $\frac{m-1}{m}<\lambda\leq\frac{m}{m+1}$ for some $m\in\mathbb{N}$. Then the following hold:

• (1)

If $z\in\{(x_{0},\,y_{0})\,|\,x_{0}\geq 0,\,y_{0}\geq 0\}$, then

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• (2)

If $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},y_{0}\geq 0\}\bigcup\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq y_{0}<\frac{k+1}{\lambda},x_{0}\geq 0\}$, then

[TABLE]

where $k\in\{-1,-2,...,-m\}.$

• (3)

If $z\in\{(x_{0},y_{0})\,|\,x_{0}<\frac{-m}{\lambda},y_{0}\geq 0\}\bigcup\{(x_{0},y_{0})\,|\,y_{0}<\frac{-m}{\lambda},x_{0}\geq 0\}$, then

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• (4)

If $z\in\{(x_{0},y_{0})\,|\,x_{0}<\frac{-m+1}{\lambda},y_{0}<\frac{-m+1}{\lambda}\}$, then

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• (5)

If $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},\frac{p}{\lambda}\leq y_{0}<\frac{p+1}{\lambda}\}$, then

[TABLE]

where $k\in\{-1,-2,...,-m\},p\in\{-1,-2,...,-m\}$.

• (6)

If $z\in\{(x_{0},y_{0})\,|\,x_{0}<\frac{-m+1}{\lambda},\frac{k}{\lambda}\leq y_{0}<\frac{k+1}{\lambda}\}\bigcup\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},y_{0}<\frac{-m+1}{\lambda}\}$, then

[TABLE]

where $k\in\{-1,-2,...,-m\}$.

Proof. (1) For subsequences (2.2) and (2.3) of (2.1) we have

[TABLE]

[TABLE]

for all $z=(x_{0},y_{0})\in\mathbb{R}^{2}_{++}$, i.e. $\lim_{n\rightarrow\infty}u_{n}=\lim_{n\rightarrow\infty}v_{n}=0$. Then $\lim_{n\rightarrow\infty}A^{n}(z)=(0,0)$.

(2) If $z\in\{(x_{0},y_{0})\ \ |\ \ \frac{k}{\lambda}\leq y_{0}<\frac{k+1}{\lambda},y_{0}\geq 0\}$ then $\lfloor\lambda x_{0}\rfloor=k$, where $k\in\{-1,-2,...,-m\}$. Since $k$ is a fixed point of $f(x)=\lfloor\lambda x\rfloor$, then $\lim_{n\rightarrow\infty}u_{n}=k$ and by proof of 1st part we have $\lim_{n\rightarrow\infty}v_{n}=0$ for $y_{0}\geq 0$.

In case $\forall z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq y_{0}<\frac{k+1}{\lambda},\,x_{0}\geq 0\}$ we can write $\lim_{n\rightarrow\infty}u_{n}=0$ and $\lim_{n\rightarrow\infty}v_{n}=k$ as above. That means, $\omega(z)=\{(k,0),(0,k)\}$.

(3) $u_{1}<-m$ and $u_{1}>u_{0}$ are true for all $z\in\{(x_{0},y_{0})\,|\,x_{0}<\frac{-m}{\lambda},\,y_{0}\geq 0\}$ . For terms of (2) we see that $u_{n+1}>u_{n}$, i.e. $\{u_{n}\}$ is an increasing sequence, which is bounded from above by $-m$. Since $-m$ is the unique fixed point of $f(x)=\lfloor\lambda x\rfloor$ in $(-\infty,-m]$, we have $\lim_{n\rightarrow\infty}u_{n}=-m$. For $\{v_{n}\}$ we have $\lim_{n\rightarrow\infty}v_{n}=0$.

In case $z\in\{(x_{0},y_{0})\,|\,y_{0}<\frac{-m}{\lambda},x_{0}\geq 0\}$ we can write $\lim_{n\rightarrow\infty}u_{n}=0$ and $\lim_{n\rightarrow\infty}v_{n}=-m$, then $\omega(z)=\{(-m,0),(0,-m)\}$.

(4) Like the proof of 3rd part we may write

[TABLE]

for all $z\in\{(x_{0},y_{0})\,|\,x_{0}<\frac{-m+1}{\lambda},y_{0}<\frac{-m+1}{\lambda}\}$.

(5) If $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},\frac{p}{\lambda}\leq x_{0}<\frac{p+1}{\lambda}\}$ then $\lfloor\lambda x_{0}\rfloor=k,\lfloor\lambda y_{0}\rfloor=p$. Since $k$ and $p$ are fixed points of $f(x)=\lfloor\lambda x\rfloor$ we have

[TABLE]

where $k\in\{-1,-2,...,-m\},p\in\{-1,-2,...,-m\}$.

(6) The proof is based on parts 3-4.

2.2.3. The case $\lambda\geq 1$.

In case $\lambda=1$, the form of operator $A$ is A(z)=\left\{\begin{array}[]{ll}x^{\prime}=\lfloor y\rfloor&\hbox{}\\ y^{\prime}=\lfloor x\rfloor&\hbox{}\end{array}\right. and

[TABLE]

for all $z=(x,y)\in\mathbb{R}^{2}$.

Theorem 3**.**

Let $\frac{m+1}{m}\leq\lambda<\frac{m}{m-1}$ for some $m\in\mathbb{N}$. Then the following hold:

• (1)

If $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},y_{0}<0\}\bigcup\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq y_{0}<\frac{k+1}{\lambda},x_{0}<0\}$, then

[TABLE]

where $k\in\{0,1,2,...,m-1\}$.

• (2)

If $z\in\{(x_{0},y_{0})\,|\,x_{0}<0,y_{0}\geq\frac{m}{\lambda}\}\bigcup\{(x_{0},y_{0})\,|\,y_{0}<0,x_{0}\geq\frac{m}{\lambda}\}$, then

[TABLE]

• (3)

If $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},\frac{p}{\lambda}\leq x_{0}<\frac{p+1}{\lambda}\}$, then

[TABLE]

where $k\in\{0,1,2,...,m-1\},p\in\{0,1,2,...,m-1\}$.

• (4)

If $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},y_{0}\geq\frac{m}{\lambda}\}\bigcup\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq y_{0}<\frac{k+1}{\lambda},x_{0}\geq\frac{m}{\lambda}\}$, then

[TABLE]

where $k\in\{0,1,2,...,m-1\}$.

• (5)

If $z\in\{(x_{0},y_{0})\,|\,x_{0}<0,y_{0}<0\}$, then

[TABLE]

• (6)

If $z\in\{(x_{0},y_{0})\,|\,x_{0}\geq\frac{m}{\lambda},y_{0}\geq\frac{m}{\lambda}\}$, then

[TABLE]

Proof. (1) Since $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},y_{0}<0\}$ then $\lim_{n\rightarrow\infty}u_{n}=k$. $y_{0}>\lambda y_{0}\geq\lfloor\lambda y_{0}\rfloor=f(y_{0})$ because $y_{0}<0$ and $\lambda y_{0}<0$ for $\lambda>1$. Using this inequality we get $f^{n}(y_{0})>f^{n+1}(y_{0})(y_{n}>y_{n+1})$. Due to lack of fixed points of $f(x)=\lfloor\lambda x\rfloor$ in $(-\infty,0)$, we have $\lim_{n\rightarrow\infty}f^{n}(y_{0})=\lim_{n\rightarrow\infty}v_{n}=-\infty$.

In this case also $\forall z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq y_{0}<\frac{k+1}{\lambda},x_{0}<0\}$, we may write $\lim_{n\rightarrow\infty}v_{n}=k$ and $\lim_{n\rightarrow\infty}u_{n}=-\infty$ as above. Thus $\omega(z)=\{(k,-\infty),(-\infty,k)\}$.

(2) We showed that $\lim_{n\rightarrow\infty}u_{n}=-\infty$ is true for all $z\in\{(x_{0},y_{0})\,|\,x_{0}<0,y_{0}\geq\frac{m}{\lambda}\}$ in part 2. If $y_{0}\geq\frac{m}{\lambda}$, we have $v_{n}<v_{n+1}$ and $\{v_{n}\}$ is an increasing sequence, that’s why $\lim_{n\rightarrow\infty}v_{n}=\infty$. In case when $z\in\{(x_{0},y_{0})\,|\,y_{0}<0,x_{0}\geq\frac{m}{\lambda}\}$ we get $\lim_{n\rightarrow\infty}u_{n}=\infty$ and $\lim_{n\rightarrow\infty}v_{n}=-\infty$, i.e. $\omega(z)=\{(-\infty,\infty),(\infty,-\infty)\}$.

(3) If $z\in\{(x_{0},y_{0})\,|\,\frac{k}{\lambda}\leq x_{0}<\frac{k+1}{\lambda},\frac{p}{\lambda}\leq x_{0}<\frac{p+1}{\lambda}\}$ then $\lfloor\lambda x_{0}\rfloor=k$ and $\lfloor\lambda y_{0}\rfloor=p$. Since $k$ and $p$ are fixed points of $f(x)=\lfloor\lambda x\rfloor$ , we get $\lim_{n\rightarrow\infty}u_{n}=k,\lim_{n\rightarrow\infty}v_{n}=p\Rightarrow\omega(z)=\{(k,p),(p,k)\}$.

Proofs of parts 4-6 are directly come from above results.