$p$-Adic Dynamical Systems of the Function $a x^{-2}$

Abstract

In this paper we study $p$-adic dynamical systems generated by the function $f(x)={a\over x^2}$ in the set of complex $p$-adic numbers. We find an explicit formula for the $n$-fold composition of $f$ for any $n\geq 1$. Using this formula we give fixed points, periodic points, basin of attraction and Siegel disk of each fixed (periodic) point depending on parameters $p$ and $a$.

1. Introduction

Nowadays the theory of $p$-adic numbers is one of very actively developing area in mathematics. It has numerous applications in many branches of mathematics, biology, physics and other sciences (see for example [4, 7, 12] and the references therein).

In this paper we continue our study of $p$-adic dynamical systems generated by rational functions (see [1-10]) and references therein for motivations and history of $p$-adic dynamical systems).

Let us recall the main definition of the paper:

$p$-Adic numbers. Denote by $(n,m)$ the greatest common divisor of the positive integers $n$ and $m$.

Let $\mathbb{Q}$ be the field of rational numbers.

For each fixed prime number $p$, every rational number $x\neq 0$ can be represented in the form $x=p^r\frac{n}{m}$, where $r,n\in\mathbb{Z}$, $m$ is a positive integer, $(p,n)=1$, $(p,m)=1$.

The $p$-adic norm of $x$ is given by

$$|x|_p=\left\{ \begin{array}{ll} p^{-r}, & \ \textrm{~for~} x\neq 0, \\ 0, &\ \textrm{~for~} x=0.\\ \end{array} \right.$$

It has the following properties:

1) $|x|_p\geq 0$ and $|x|_p=0$ if and only if $x=0$,

2) $|xy|_p=|x|_p|y|_p$,

3) the strong triangle inequality

$$|x+y|_p\leq\max\{|x|_p,|y|_p\},$$

3.1) if $|x|_p\neq |y|_p$ then $|x+y|_p=\max\{|x|_p,|y|_p\}$,

3.2) if $|x|_p=|y|_p$ then for $p=2$ we have $|x+y|_p\leq {1\over 2}|x|_p$ (see [12]).

The completion of $\mathbb{Q}$ with respect to $p$-adic norm defines the $p$-adic field which is denoted by $\mathbb{Q}_p$ (see [5]).

The algebraic completion of $\mathbb{Q}_p$ is denoted by ${\mathbb C}_p$ and it is called the set of complex $p$-adic numbers.

For any $a\in{\mathbb C}_p$ and $r>0$ denote

$$U_r(a)=\{x\in{\mathbb C}_p : |x-a|_p<r\},\ \ V_r(a)=\{x\in{\mathbb C}_p : |x-a|_p\leq r\},$$

$$S_r(a)=\{x\in{\mathbb C}_p : |x-a|_p= r\}.$$

Dynamical systems in ${\mathbb C}_p$. To define a dynamical system we consider a function $f: x\in U\to f(x)\in U$, (in this paper $U=U_r(a)$ or ${\mathbb C}_p$) (see for example [6]).

For $x\in U$ denote by $f^n(x)$ the $n$-fold composition of $f$ with itself (i.e. $n$ times iteration of $f$ to $x$):

$$f^n(x)=\underbrace{f(f(f\dots (f}_{n \,{\rm times}}(x)))\dots).$$

For arbitrary given $x_0\in U$ and $f:U\to U$ the discrete-time dynamical system (also called the trajectory) of $x_0$ is the sequence of points

$$ x_0, x_1=f(x_0), x_2=f^2(x_0), x_3=f^3(x_0), \dots$$

(1.1)

The main problem: Given a function $f$ and initial point $x_0$ what ultimately happens with the sequence (1.1). Does the limit $\lim_{n\to\infty} x_n$ exist? If not what is the set of limit points of the sequence?

A point $x\in U$ is called a fixed point for $f$ if $f(x)=x$. The point $x$ is a periodic point of period $m$ if $f^m(x) = x$. The least positive $m$ for which $f^m(x) = x$ is called the prime period of $x$.

A fixed point $x_0$ is called an attractor if there exists a neighborhood $U(x_0)$ of $x_0$ such that for all points $x\in U(x_0)$ it holds $\lim\limits_{n\to\infty}f^n(x)=x_0$. If $x_0$ is an attractor then its basin of attraction is

$$\mathcal A(x_0)=\{x\in {\mathbb C}_p :\ f^n(x)\to x_0, \ n\to\infty\}.$$

A fixed point $x_0$ is called repeller if there exists a neighborhood $U(x_0)$ of $x_0$ such that $|f(x)-x_0|_p>|x-x_0|_p$ for $x\in U(x_0)$, $x\neq x_0$.

Let $x_0$ be a fixed point of a function $f(x)$. Put $\lambda=f'(x_0)$. The point $x_0$ is attractive if $0<|\lambda|_p < 1$, indifferent if $|\lambda|_p = 1$, and repelling if $|\lambda|_p > 1$.

The ball $U_r(x_0)$ (contained in $V$) is said to be a Siegel disk if each sphere $S_{ \rho }(x_0)$, $\rho <r$ is an invariant sphere of $f(x)$, i.e. if $x\in S_{ \rho }(x_0)$ then all iterated points $f^n(x)\in S_{ \rho }(x_0)$ for all $n=1,2\dots$. The union of all Siegel disks with the center at $x_0$ is called a maximum Siegel disk and is denoted by $SI(x_0)$.

In Section 2 we consider the function $f(x)={a\over x^2}$ and study the dynamical systems generated by this function in ${\mathbb C}_p$. We give fixed points, periodic points, basin of attraction and Siegel disk of each fixed (and periodic) point.

2. The function $a/x^2$

Consider the dynamical system associated with the function $f:{\mathbb C}_p\to{\mathbb C}_p$ defined by

$$ f(x)=\frac{a}{x^2}, \ \ a\neq 0, \ \ a\in {\mathbb C}_p,$$

(2.1)

where $x\ne 0$.

Denote by $\theta_{j,n}$, $j = 1,\dots, n$, the $n$th root of unity in ${\mathbb C}_p$, while $\theta_{1,n} = 1$.

This function has three fixed points $x_k$, $k=1,2,3$, which are solutions to $x^3=a$ in ${\mathbb C}_p$.

For these fixed points we have

$$ x^3_k=a \ \ \Rightarrow \ \ x_k=\theta_{k,3}a^{1\over 3} \ \ \Rightarrow \ \ |x^3_k|_p=|a|_p \ \ \Rightarrow \ \ |x_k|_p=\alpha\equiv(|a|_p)^{1/3}.$$

(2.2)

Thus $x_k\in S_\alpha(0)$, $k=1,2,3$.

We have

$$f'(x)={-2a\over x^3}= {-2\over x}\cdot f(x).$$

Therefore at a fixed point we get

$$f'(x_k)={-2\over x_k}\cdot f(x_k)=-2.$$

$$|f'(x_k)|_p=\left\{\begin{array}{ll} 1/2, \ \ \mbox{if} \ \ p=2 \\ 1, \ \ \mbox{if} \ \ p\geq 3 \end{array}\right.$$

Hence the fixed point $x_k$ is an attractive for $p=2$ and an indifferent for $p\geq 3$. Therefore the fixed point is never repeller.

We can explicitly calculate $f^n$.

Lemma 2.1.

For any $x\in {\mathbb C}_p\setminus \{0\}$ we have

$$f^n(x)=a^{{1\over 3}(1-(-2)^n)}\cdot x^{(-2)^n}, \ \ n\geq 1.$$

Proof.

We use induction over $n$. For $n=1,2$ the formula is clear. Assume it is true for $n$ and show it for $n+1$:

$$\begin{aligned} \, \nonumber f^{n+1}(x)&=f^n(f(x))=a^{{1\over 3}(1-(-2)^n)}\cdot (f(x))^{(-2)^n} \\ &= a^{{1\over 3}(1-(-2)^n)}\cdot ({a\over x^2})^{(-2)^n}=a^{{1\over 3}(1-(-2)^{n+1})}\cdot x^{(-2)^{n+1}}. \nonumber \end{aligned}$$

This completes the proof. $\square$

Recall $\alpha=(|a|_p)^{1/3}$. For $r>0$, take $x\in S_r(0)$, i.e., $|x|_p=r$. Then we have

$$ |f^n(x)|_p=\left|a^{{1\over 3}(1-(-2)^n)}\cdot x^{(-2)^n}\right|_p =\alpha^{1-(-2)^n}\cdot r^{(-2)^n}, \ \ n\geq 1.$$

(2.3)

2.1. Dynamics on ${\mathbb C}_p\setminus S_\alpha(0)$

Lemma 2.2.

For $\alpha$ defined in ( 2.2 ) the following assertions hold:

• 1. The sphere $S_\alpha(0)$ is invariant with respect to $f$ , (i.e., $f(S_\alpha(0))\subset S_\alpha(0)$ );

• 2. $f(U_\alpha(0))\subset {\mathbb C}_p\setminus V_\alpha(0)$ ;

• 3. $f({\mathbb C}_p\setminus V_\alpha(0))\subset U_\alpha(0)$ .

Proof.

1. If $x\in S_\alpha(0)$, i.e., $|x|_p=\alpha$, then

$$|f(x)|_p=|\frac{a}{x^2}|_p={|a|_p\over \alpha^2}=\alpha.$$

2. If $x\in U_\alpha(0)$, i.e., $|x|_p<\alpha$, then

$$|f(x)|_p=|\frac{a}{x^2}|_p>{|a|_p\over \alpha^2}=\alpha.$$

Therefore, $f(x)\in {\mathbb C}_p\setminus V_\alpha(0)$. Proof of the part 3 is similar. $\square$

Lemma 2.3.

The function ( 2.1 ) does not have any periodic point in ${\mathbb C}_p\setminus S_\alpha(0)$ .

Proof.

We know that all three fixed points belong to $S_\alpha(0)$. Let $x\in {\mathbb C}_p\setminus S_\alpha(0)$ be a $m$-periodic ($m\geq 2$) point for (2.1), i.e., $x$ satisfies $f^m(x)=x$. Then it is necessary that $|f^m(x)|_p=|x|_p$. But for any $x\in {\mathbb C}_p\setminus S_\alpha(0)$ (i.e. $|x|_p=r\ne \alpha$), by (2.3) we get

$$ |f^m(x)|_p=\alpha^{1-(-2)^m}\cdot r^{(-2)^m}=\alpha \cdot \left(r\over \alpha\right)^{(-2)^m}\ne r, \ \ \forall r\ne \alpha.$$

(2.4)

Therefore, $f^m(x)=x$ is not satisfied for any $x\in {\mathbb C}_p\setminus S_\alpha(0)$. $\square$

For given $r>0$, denote

$$r_n= \alpha^{1-(-2)^n}\cdot r^{(-2)^n}.$$

Then by (2.3) one can see that the trajectory $f^n(x)$, $n\geq 1$ of $x\in S_r(0)$ has the following sequence of spheres on its route:

$$S_r(0) \rightarrow S_{r_1}(0)\rightarrow S_{r_2}(0)\rightarrow S_{r_3}(0)\rightarrow\dots$$

Now we calculate the limits of $r_n$.

Case of even $n$. From (2.3) it is easy to see that

$$\lim_{n\to \infty} |f^n(x)|_p=\lim_{n\to \infty} r_n=\left\{\begin{array}{lll} 0, \ \ \mbox{if} \ \ r<\alpha \\ \alpha, \ \ \mbox{if} \ \ r=\alpha \\ +\infty, \ \ \mbox{if} \ \ r>\alpha . \end{array} \right.$$

Case of odd $n$. In this case we have

$$\lim_{n\to \infty} |f^n(x)|_p=\lim_{n\to \infty} r_n=\left\{\begin{array}{lll} +\infty, \ \ \mbox{if} \ \ r<\alpha \\ \alpha, \ \ \mbox{if} \ \ r=\alpha \\ 0, \ \ \mbox{if} \ \ r>\alpha . \end{array} \right.$$

Summarizing above-mentioned results we obtain the following theorem:

Theorem 2.4.

Let $\alpha$ be defined by ( 2.2 ). Then

• 1. if $x\in U_\alpha(0)$ then

  $$\lim_{k\to \infty}f^{2k}(x)=0, \ \ \ \lim_{k\to \infty}|f^{2k-1}(x)|_p=+\infty.$$

• 2. if $x\in S_\alpha(0)$ then $f^n(x)\in S_\alpha(0), \, n\geq 1.$

• 3. if $x\in {\mathbb C}_p\setminus V_\alpha(0)$ then

  $$\lim_{k\to \infty}|f^{2k}(x)|_p=+\infty, \ \ \ \lim_{k\to \infty}f^{2k-1}(x)=0.$$

Remark 2.5.

Note that Theorem 2.4 is true for more general function: $f(x)={a\over x^q}$, where $q$ is a natural number, $q\geq 2$. In this case $\alpha=|a|_p^{1/(q+1)}$. The case $q=1$ is simple: in this case any point $x\in {\mathbb C}_p\setminus \{0\}$ is 2-periodic. That is $f(f(x))=x$. Indeed,

$$f(f(x))={a\over {a\over x}}=a\cdot {x\over a}=x.$$

2.2. Dynamics on $S_\alpha(0)$

By Theorem 2.4 it remains to study the dynamical system of $f:S_\alpha(0)\to S_\alpha(0)$. Recall that all fixed points $x_k$, $k=1,2,3$ are in $S_\alpha(0)$.

Lemma 2.6.

The distance between fixed points is

$$ |x_1-x_2|_p= |x_1-x_3|_p=|x_2-x_3|_p=\left\{\begin{array}{ll} \alpha, \ \ \mbox{if} \ \ p\ne 3 \\ {\alpha\over \sqrt{3}}, \ \ \mbox{if} \ \ p=3 . \end{array} \right.$$

(2.5)

Proof.

Since $x_i^3=a$, $i=1,2,3$, for $x_i\ne x_j$ we have

$$0=x_i^3-x_j^3=(x_i-x_j)(x_i^2+x_ix_j+x_j^2) \ \ \Rightarrow \ \ x_i^2+x_ix_j+x_j^2=0$$

$$\Leftrightarrow \ \ (x_i-x_j)^2=-3x_ix_j \ \ \Rightarrow \ \ |x_i-x_j|_p^2=|3x_ix_j|_p.$$

From the last equality, using $|x_i|_p=|x_j|_p=\alpha$, we get (2.5). $\square$

Take $x\in S_\alpha(0)$ such that $|x-x_1|_p=\rho$, i.e., $x=x_1+\gamma$, with $|\gamma|_p=\rho$. Note that $\rho\leq \alpha$. Then by Lemma 2.1 we have

$$ |f^n(x)-x_1|_p= |f^n(x)-f^n(x_1)|_p=\alpha^{1-(-2)^n}|x^{(-2)^n}-x_1^{(-2)^n}|_p.$$

(2.6)

Now we use the following formula

$$x^{2^n}-y^{2^n}=(x-y)\prod_{j=0}^{n-1}(x^{2^j}+y^{2^j}).$$

Then from (2.6) we get

$$ |f^n(x)-x_1|_p= \alpha^{1-(-2)^n}\cdot \left\{\begin{array}{ll} \rho\prod_{j=0}^{n-1}|(x_1+\gamma)^{2^j}+x_1^{2^j}|_p, \ \ \mbox{if} \ \ n \ \ \mbox{is even} \\ {\rho\over |xx_1|_p}\prod_{j=0}^{n-1}|(x_1+\gamma)^{-2^j}+x_1^{-2^j}|_p, \ \ \mbox{if} \ \ n \ \ \mbox{is odd}. \end{array}\right.$$

(2.7)

We have

$$ |(x_1+\gamma)^{2^j}+x_1^{2^j}|_p=\left|2x_1^{2^j}+\sum_{s=1}{2^j\choose s}x_1^{2^j-s}\gamma^s\right|_p=\left\{\begin{array}{ll} |2|_p\alpha^{2^j}, \ \ \mbox{if} \ \ \rho<\alpha \\ \leq |2|_p\alpha^{2^j}, \ \ \mbox{if} \ \ \rho=\alpha. \end{array} \right.$$

(2.8)

Here we used that

$$\left|{2^j\choose s}\right|_p\leq \left\{\begin{array}{ll} {1\over 2}, \ \ \mbox{if} \ \ p=2 \\ 1, \ \ \mbox{if} \ \ p\geq 3 . \end{array}\right.$$

Using (2.8) we get

$$ |(x_1+\gamma)^{-2^j}+x_1^{-2^j}|_p= {|(x_1+\gamma)^{2^j}+x_1^{2^j}|_p \over |(x_1+\gamma)^{2^j}x_1^{2^j}|_p}= \left\{\begin{array}{lll} |2|_p\alpha^{-2^j}, \ \ \mbox{if} \ \ \rho<\alpha \\ \leq |2|_p {1\over |(x_1+\gamma)^{2^j}|_p}, \ \ \mbox{if} \ \ \rho=\alpha. \end{array} \right.$$

(2.9)

In case of even $n$, by (2.8) from (2.7), we get

$$|f^n(x)-x_1|_p= \alpha^{1-2^n}\cdot \rho\prod_{j=0}^{n-1}|(x_1+\gamma)^{2^j}+x_1^{2^j}|_p$$

$$ =\rho\cdot \alpha^{1-2^n}\cdot |2|_p^n \prod_{j=0}^{n-1}\alpha^{2^j}\cdot\left\{\begin{array}{ll} 1, \ \ \mbox{if} \ \ \rho<\alpha \\ \leq 1, \ \ \mbox{if} \ \ \rho=\alpha \end{array}\right. =\rho\cdot |2|_p^n\cdot\left\{\begin{array}{ll} 1, \ \ \mbox{if} \ \ \rho<\alpha \\ \leq 1, \ \ \mbox{if} \ \ \rho=\alpha . \end{array}\right.$$

(2.10)

Similarly, in case of odd $n$, by (2.9) from (2.7) we get

$$ |f^n(x)-x_1|_p= \alpha^{1+2^n}\cdot {\rho\over \alpha^2}\cdot |2|_p^n\prod_{j=0}^{n-1}\alpha^{-2^j} =\rho \cdot |2|_p^n\ \ \mbox{if} \ \ \rho<\alpha.$$

(2.11)

The same formulas are also true for $x_2$ and $x_3$.

For fixed $\alpha$ (defined in (2.2)) and $t\in S_\alpha(0)$ denote

$$\mathcal S_{\rho,t}=S_\alpha(0)\cap S_\rho(t)=\{x\in S_\alpha(0): |x-t|_p=\rho\}.$$

Thus we have proved the following lemma

Lemma 2.7.

Let $\rho<\alpha$ . Then for any $x\in \mathcal S_{\rho, x_i}$ ( $i=1,2,3$ ) we have

• • if $p=2$ then

  $$f^n(x)\in \mathcal S_{2^{-n}\rho, x_i}.$$

• • if $p\geq 3$ then

  $$f^n(x)\in \mathcal S_{\rho, x_i}, \ \ n\geq 1.$$

  In particular, the set $\mathcal S_{\rho, x_i}$ is invariant with respect to $f$ for any $\rho< \alpha$ .

Denote

$$\mathcal V_{\rho, t}=\bigcup_{0\leq r< \rho}\mathcal S_{r,t}=\{x\in S_\alpha(0): |x-t|_p<\rho\}.$$

Lemma 2.8.

If $x\in \mathcal S_{\rho, x_i}$ , for some $i=1,2,3$ , then:

• i. If $\rho$ is such that

  $$\rho<\left\{\begin{array}{ll} \alpha, \ \ \mbox{if} \ \ p\ne 3 \\ {\alpha\over \sqrt{3}}, \ \ \mbox{if} \ \ p=3. \end{array}\right.$$

  then

  $$x\in \left\{\begin{array}{ll} \mathcal S_{{\alpha\over \sqrt{3}}, x_j}, \ \ \mbox{for} \ \ p=3 \\ \mathcal S_{\alpha, x_j}, \ \ \mbox{for} \ \ p\ne 3, \end{array}\right. \ \ j\ne i.$$

• ii. If $p=3$ and $\rho\geq {\alpha\over \sqrt{3}}$ then

  $$x\in \left\{\begin{array}{ll} \mathcal V_{\rho, x_j}, \ \ \mbox{for} \ \ \rho= {\alpha\over \sqrt{3}} \\ \mathcal S_{\rho, x_j}, \ \ \mbox{for} \ \ \rho> {\alpha\over \sqrt{3}}, \end{array}\right. \ \ j\ne i.$$

Proof.

For $x\in \mathcal S_{\rho, x_i}$, using property of $p$-adic norm and formula (2.5) we get

$$|x-x_j|_p=|x-x_i+x_i-x_j|_p=\left\{\begin{array}{llll} \alpha, \ \ \mbox{if} \ \ p\ne 3 \\ {\alpha\over \sqrt{3}}, \ \ \mbox{if} \ \ p=3, \ \ \rho<{\alpha\over \sqrt{3}} \\ \leq\rho, \ \ \mbox{if} \ \ p=3, \ \ \rho={\alpha\over \sqrt{3}} \\ \rho, \ \ \mbox{if} \ \ p=3, \ \ \rho>{\alpha\over \sqrt{3}} \end{array} \right.$$

This completes the proof. $\square$

Denote

$$\mathcal U_\alpha=\{x\in S_\alpha(0): |x-x_1|_p=|x-x_2|_p=|x-x_3|_p=\alpha\}.$$

As a corollary of Lemma 2.8 we have

Lemma 2.9.

If $p\ne 3$ then $S_\alpha(0)$ has the following partition

$$S_\alpha(0)=\mathcal U_\alpha\cup \bigcup_{i=1}^3\mathcal V_{\alpha, x_i}.$$

Lemma 2.10.

Let $\alpha$ be defined by ( 2.2 ). Then:

• 1. If $p=2$ then the set $\mathcal U_\alpha$ is invariant with respect to $f$ .

• 2. If $p\geq 3$ and $x\in \mathcal U_\alpha$ then one of the following assertions holds:

• 2.a) There exists $n_0$ and $\mu_{n_0}< \alpha$ such that

  $$\begin{array}{ll} f^n(x)\in \mathcal U_\alpha, \ \ \forall n\leq n_0, \\ f^n(x)\in \mathcal S_{\mu_{n_0}}(x_i), \ \ \forall n>n_0\ \ \mbox{for some} \ \ i=1,2,3. \end{array}$$

• 2.b) $f^n(x)\in \mathcal U_\alpha, \ \ \forall n\geq 1.$

Proof.

1. For any $x\in \mathcal U_\alpha$ we have

$$|f(x)-x_i|_p=\left|{a\over x^2}-{a\over x_i^2}\right|_p=|a|_p\left|{(x_i-x)(x_i+x)\over x^2x_i^2}\right|_p$$

$$ =\alpha^3\cdot {\alpha |x+x_i|_p\over \alpha^4}=|x+x_i|_p=|x-x_i+2x_i|_p=\left\{\begin{array}{ll} \alpha, \ \ \mbox{if} \ \ p=2 \\ \mu_{1,i}, \ \ \mbox{if} \ \ p\geq 3, \end{array}\right.$$

(2.12)

where $\mu_{1,i}\leq \alpha.$ The part 1 follows from this equality.

2. If in (2.12) there exists $i$ such that $\mu_{1,i}=|x+x_i|_p<\alpha$, then $f(x)\in \mathcal S_{\mu_{1,i}, x_i}$. The set $\mathcal S_{\mu_{1,i}, x_i}$ is invariant with respect to $f$. In case of all $\mu_{1,i}=\alpha$ we have $f(x)\in \mathcal U_\alpha$. Then we note that

$$|f^2(x)-x_i|_p=|f(x)-x_i+2x_i|_p=\left\{\begin{array}{ll} \alpha, \ \ \mbox{if} \ \ p=2 \\ \mu_{2,i}\leq \alpha, \ \ \mbox{if} \ \ p\geq 3 . \end{array}\right.$$

Thus we can repeat the above argument: if there exists $i$ such that $\mu_{2,i}<\alpha$, then $f^2(x)\in \mathcal S_{\mu_{2,i}, x_i}$ which is invariant with respect to $f$. If all $\mu_{2,i}=\alpha$ then $f^2(x)\in \mathcal U_\alpha$. Iterating this argument one proves the part 2. $\square$

Lemma 2.11.

For $k\in \{1,2,3\}$ , $j\in \{1,2,3\}\setminus \{k\}$ and fixed points $x_k$ , $x_j$ we have

• 1. $x_j\notin \mathcal V_{\rho, x_k},$ if and only if

  $$\rho\leq\left\{\begin{array}{ll} \alpha, \ \ \mbox{if} \ \ p\ne 3 \\ {\alpha\over \sqrt{3}}, \ \ \mbox{if} \ \ p=3. \end{array} \right.$$

• 2. if $p=2$ then

  $$\mathcal V_{\alpha, x_j}\cap \mathcal V_{\alpha, x_k}=\emptyset, \ \ \mbox{for all} \ \ j, k\in \{1,2,3\}, \, j\ne k,$$

Proof.

Follows from (2.5) and Lemma 2.7. $\square$

Summarizing above mentioned results we get

Theorem 2.12.

If $\alpha$ is defined by ( 2.2 ). Then for the dynamical system generated by $f: S_\alpha(0)\to S_\alpha(0)$ given in ( 2.1 ) the following assertions hold.

• 1. If $p=2$ then $\mathcal A(x_j)= \mathcal V_{\alpha, x_j}$ , i.e.,

  $$\lim_{n\to \infty}f^{n}(x)=x_j, \ \ \mbox{for any} \ \ x\in \mathcal V_{\alpha, x_j}.$$
  $$f^n(x)\in \mathcal U_\alpha, \ \ n\geq 1, \ \ \mbox{for all} \ \ x\in \mathcal U_\alpha.$$

• 2. If $p\geq 3$ then

  $$SI(x_j)=\mathcal V_{\alpha, x_j}, \ \ j\in \{1,2,3\}.$$

  Moreover,

  $$SI(x_1)=SI(x_2)=SI(x_3), \ \ \mbox{if} \ \ p=3.$$
  $$SI(x_j)\cap SI(x_k)=\emptyset, \ \ \mbox{if} \ \ p>3.$$

• 3. If $p\geq 3$ and $x\in \mathcal U_\alpha$ then one of the following assertions holds

• 3.a) There exists $n_0$ and $\mu_{n_0}< \alpha$ such that

  $$\begin{array}{ll} f^n(x)\in \mathcal U_\alpha, \ \ \forall n\leq n_0, \\ f^n(x)\in \mathcal S_{\mu_{n_0}}(x_i), \ \ \forall n>n_0\ \ \mbox{for some} \ \ i=1,2,3. \end{array}$$

• 3.b) $f^n(x)\in \mathcal U_\alpha, \ \ \forall n\geq 1.$

This theorem does not give behavior of $f^n(x)\in \mathcal U_\alpha, \ \ n\geq 1$, i.e., in the case when the trajectory remains in $\mathcal U_\alpha$ (that is when $p=2$ and in the case part 3.b of Theorem 2.12). Since there is not any fixed point of $f$ in $\mathcal U_\alpha$, below we are interested to periodic points of $f$ in $\mathcal U_\alpha$: for a given natural $m\geq 2$ the $m$-periodic points of this set are solutions of the following system of equations

$$ \begin{array}{ll} f^m(x)=a^{{1\over 3}(1-(-2)^m)}\cdot x^{(-2)^m}=x, \\ |x-x_1|_p=|x-x_2|_p=|x-x_3|_p=\alpha. \end{array}$$

(2.13)

Remark 2.13.

Note that in case $m=2$, there is no any solution to the first equation of (2.13) (except fixed points). Therefore below we consider the case $m\geq 3$.

Denote

$$M_m=\left\{\begin{array}{ll} \left\{(j,p): \left|\theta_{k,3}-\theta_{j, 2^m-1}\right|_p=1, \ \ \forall k=1,2,3\right\} \ \ \mbox{if} \ \ m \ \ \mbox{is even} , \\ \left\{(j,p): \left|\theta_{k,3}-\theta_{j, 2^m+1}\right|_p=1, \ \ \forall k=1,2,3\right\} \ \ \mbox{if} \ \ m \ \ \mbox{is odd}. \end{array} \right.$$

Lemma 2.14.

The solutions of the system ( 2.13 ) in ${\mathbb C}_p$ are

$$ \hat x_j=a^{1\over 3}\cdot \left\{\begin{array}{ll} \theta_{j, 2^m-1}, \ \ \mbox{if} \ \ m \ \ \mbox{is even} , \\ 1/\theta_{j, 2^m+1}, \ \ \mbox{if} \ \ m \ \ \mbox{is odd} , \end{array} \right.$$

(2.14)

where $(j,p)\in M_m$ .

Proof.

From (2.13) we get

$$\left({x\over a^{1/3}}\right)^{(-2)^m-1}=1.$$

Which has solutions (2.14). The condition $(j,p)\in M_m$ is needed to satisfy the second equation of the system (2.13). $\square$

Remark 2.15.

We note that:

• • In the case $p=2$, by part 1 of Theorem 2.12, it follows that all $m$-periodic points (except fixed ones) mentioned in (2.14) belong to $\mathcal U_\alpha$.

• • In the case $m\geq 3$ and $p\geq 3$ it is not clear to see $M_m\ne \emptyset$. It is known that (see [2, Corollary 2.2.]) the equation $x^k = 1$ has $g = (k, p - 1)$ different roots in $\mathbb Q_p$. Using this fact and assuming that $a\in \mathbb Q_p$ and $a^{1\over 3}$ exists in $\mathbb Q_p$, one can see how many periodic solutions of (2.13) exist in $\mathbb Q_p$. For example, if $p=31$ then $t^3=1$ (with $t={x\over a^{1/3}}$) has $g=(3, 30)=3$, i.e., all possible solutions in $\mathbb Q_p$ and for $m=4$ the equation $t^{2^4-1}=1$ has $g=(15,30)=15$ distinct solutions in $\mathbb Q_p$. Three of 15 solutions coincide with solutions of $t^3=1$, therefore remains 12 distinct solutions to satisfy the second equation of (2.13). For these solutions one can check the condition $M_m\ne \emptyset$.

Lemma 2.16.

If $x_*$ is a solution to ( 2.13 ) then

$$x_* \ is \ \left\{\begin{array}{ll} {\rm attracting}, \ \ \mbox{if} \ \ p=2 \\ {\rm indifferent}, \ \ \mbox{if} \ \ p\geq 3. \end{array} \right.$$

Proof.

We have

$$\left|(f^m)'(x_*)\right|_p=\left|(-2)^m\cdot a^{{1\over 3}(1-(-2)^m)}\cdot x_*^{(-2)^m-1}\right|_p$$

$$=\left|(-2)^m\cdot {f^m(x_*)\over x_*}\right|_p=\left\{\begin{array}{ll} 1/2^m, \ \ \mbox{if} \ \ p=2 \\ 1, \ \ \mbox{if} \ \ p\geq 3 . \end{array}\right.$$

This completes the proof. $\square$

Consider a $m$-periodic point $x_*$. It is clear that this point is a fixed point for the function $\varphi(x)\equiv f^m(x)$. The point $x_*$ generates $m$-cycle:

$$x_*, x^{(1)}=f(x_*), \dots, x^{(m-1)}=f^{m-1}(x_*).$$

Clearly, each element of this cycle is fixed point for function $\varphi$. We use the following

Theorem 2.17.

[2] Let $x_0$ be a fixed point of an analytic function $\varphi:U\to U$. The following assertions hold:

• 1. if $x_0$ is an attractive point of $\varphi$ and if $r>0$ satisfies the inequality

  $$Q=\max_{1\leq n<\infty}\bigg|\frac{1}{n!}\frac{d^n\varphi}{dx^n}(x_0)\bigg|_pr^{n-1}<1$$

  and $U_r(x_0)\subset U$ then $U_r(x_0)\subset \mathcal A(x_0)$;

• 2. if $x_0$ is an indifferent point of $\varphi$ then it is the center of a Siegel disk. If $r$ satisfies the inequality

  $$S=\max_{2\leq n<\infty}\bigg|\frac{1}{n!}\frac{d^n\varphi}{dx^n}(x_0)\bigg|_pr^{n-1}<|\varphi'(x_0)|_p$$

  and $U_r(x_0)\subset U$ then $U_r(x_0)\subset SI(x_0)$.

Lemma 2.16 suggests the following

Theorem 2.18.

• • If $p=2$ then for any $m = 2, 3,\dots$ , the $m$ -cycles are attractors and open balls with radius $\alpha$ are contained in the basins of attraction.

• • If $p\geq 3$ then for any $m = 2, 3,\dots$ , every $m$ -cycle is a center of a Siegel disk with radius $\alpha$ .

Proof.

Let $x_*$ be a $m$-periodic point. Recall that $|x_*|_p=\alpha$. We use Theorem 2.17, by Lemma 2.1 we get:

$$\begin{aligned} \, \nonumber Q=\max_{1\leq n<\infty}\bigg|\frac{1}{n!}\frac{d^n\varphi}{dx^n}(x_*)\bigg|_pr^{n-1} &= \max_{1\leq n<\infty}\bigg|\frac{1}{n!}a^{{1\over 3}(1-(-2)^m)}\cdot \prod_{s=0}^{n-1}\left((-2)^{m}-s\right)\cdot x_*^{(-2)^m-n}\bigg|_pr^{n-1} \nonumber \\ &= \max_{1\leq n<\infty}\bigg|\frac{1}{n!}\cdot \prod_{s=0}^{n-1}\left((-2)^{m}-s\right)\cdot {x_*\over x_*^n}\bigg|_pr^{n-1} \nonumber \\ &= \max_{1\leq n<\infty}\bigg|\frac{1}{n!}\cdot \prod_{s=0}^{n-1}\left((-2)^{m}-s\right)\bigg|_p\left({r\over \alpha}\right)^{n-1} \nonumber \\ &=\max_{1\leq n<\infty}\left({r\over \alpha}\right)^{n-1}\cdot\left\{ \begin{array}{ll}\left|{2^m\choose n}\right|_p, \ \ \mbox{if} \ \ m-{\rm even} \\ \left|{2^m+n\choose 2^m}\right|_p, \ \ \mbox{if} \ \ m-{\rm odd} \end{array}\right.<1. \end{aligned}$$

(2.15)

If $r <\alpha$, this condition is satisfied. The second part is similar. $\square$