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Chaotic behavior of the p-adic Potts–Bethe mapping II

Published online by Cambridge University Press:  30 September 2021

OTABEK KHAKIMOV
Affiliation:
Department of Algebra and Analysis, Institute of Mathematics named after V.I. Romanovski, 4, University str., 100125, Tashkent, Uzbekistan AKFA University, 1st Deadlock 10, Kukcha Darvoza, 100095 Tashkent, Uzbekistan (e-mail: [email protected], [email protected])
FARRUKH MUKHAMEDOV*
Affiliation:
Department of Mathematical Sciences, College of Science, United Arab Emirates University, P.O.Box, 15551, Al Ain, Abu Dhabi, UAE (e-mail: [email protected])
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Abstract

The renormalization group method has been developed to investigate p-adic q-state Potts models on the Cayley tree of order k. This method is closely related to the examination of dynamical behavior of the p-adic Potts–Bethe mapping which depends on the parameters q, k. In Mukhamedov and Khakimov [Chaotic behavior of the p-adic Potts–Behte mapping. Discrete Contin. Dyn. Syst. 38 (2018), 231–245], we have considered the case when q is not divisible by p and, under some conditions, it was established that the mapping is conjugate to the full shift on $\kappa _p$ symbols (here $\kappa _p$ is the greatest common factor of k and $p-1$ ). The present paper is a continuation of the forementioned paper, but here we investigate the case when q is divisible by p and k is arbitrary. We are able to fully describe the dynamical behavior of the p-adic Potts–Bethe mapping by means of a Markov partition. Moreover, the existence of a Julia set is established, over which the mapping exhibits a chaotic behavior. We point out that a similar result is not known in the case of real numbers (with rigorous proofs).

Type
Original Article
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Copyright
© The Author(s), 2021. Published by Cambridge University Press

1 Introduction

The present paper is a continuation of [Reference Mukhamedov and Khakimov35], where we have started to investigate the chaotic behavior of the Potts–Bethe mapping over the p-adic field (here p is some prime number). Note that the mapping is governed by

(1.1) $$ \begin{align} f_{\theta,q,k}(x)=\bigg(\frac{\theta x+q-1}{x+\theta+q-2}\bigg)^k, \end{align} $$

where $k,q\in \mathbb N$ and $|\theta -1|_p<1$ . In [Reference Mukhamedov and Khakimov35], we have considered the case when q is not divisible by p, that is, $|q|_p=1$ . In that setting, under some conditions, we were able to prove that $f_{\theta ,q,k}$ is conjugate to the full shift on $\kappa _p$ symbols (here $\kappa _p$ is the greatest common factor (GCF) of k and $p-1$ ). In the current paper, we are going to study the same Potts–Bethe mapping when q is divisible by p, that is $|q|_p<1$ . It is known that the thermodynamic behavior of the central site of the Potts model with nearest-neighbor interactions on a Cayley tree is reduced to the recursive system which is given by (1.1). The existence of at least two non-trivial p-adic Gibbs measures indicates that the phase transition may exist. This is closely connected to the chaotic behavior of the associated dynamical system [Reference Fan and Liao12, Reference Khakimov16, Reference Khakimov17, Reference Monroe23, Reference Mukhamedov26, Reference Mukhamedov27]. Therefore, it is important to investigate the chaotic properties of (1.1).

We stress that the Potts–Ising mapping is a particular case of the Potts–Bethe mapping, which can be obtained from (1.1) by putting $q=2$ . Recently, in [Reference Mukhamedov, Akin and Dogan30, Reference Mukhamedov and Khakimov34] under some condition, a Julia set of the Potts–Ising mapping was described, and it was shown that restricted to its Julia set, the Potts–Ising mapping is conjugate to a full shift. Therefore, it is natural to consider the Potts–Bethe mapping for $q\geq 3$ with $|q|_p<1$ and $k\geq 2$ . In [Reference Rozikov and Khakimov43], all fixed points of $f_{\theta ,q,k}$ were found when $k=2$ and $|q|_p<1$ . Then, using these fixed points, the dynamics of (1.1) whenever $k=2$ and $|q|_p<1$ was investigated in [Reference Fan, Fan, Liao and Wang11, Reference Mukhamedov and Khakimov31, Reference Mukhamedov and Khakimov32]. Recently in [Reference Ahmad, Liao and Saburov1, Reference Saburov and Ahmad44], the Potts–Bethe mapping was studied for the case $k=3$ and $|q|_p<1$ . In the present paper, we are going to consider a more general case, that is, arbitrary $k\geq 2$ and $|q|_p<1$ . To formulate our main result, let us recall some necessary notions.

It is easy to notice that the function (1.1) is defined on $\mathbb Q_p\setminus \{x^{(\infty )}\}$ , where $x^{(\infty )}=2-q-\theta $ . For the sake of convenience, we write $\text {Dom}(f_{\theta ,q,k}):=\mathbb Q_p\setminus \{x^{(\infty )}\}$ . Let us denote

$$ \begin{align*}\mathcal P_{x^{(\infty)}}=\bigcup_{n=1}^\infty f_{\theta,q,k}^{-n}(x^{(\infty)}). \end{align*} $$

One can see that the set $\mathcal P_{x^{(\infty )}}$ is at most countable, and could be empty for some $k,q$ and $\theta $ (see §3). If it is not empty, then for any $x_0\in \mathcal P_{x^{(\infty )}}$ , there exists an $n\geq 1$ such that after n-times, we will ‘lose’ that point.

For a given mapping f on $\mathbb Q_p$ , we denote by $\text {Fix}(f)$ the set of all fixed points of f, that is,

$$ \begin{align*}\text{Fix}(f)=\{x\in\mathbb Q_p: f(x)=x\}. \end{align*} $$

Let f be an analytic function and $x^{(0)}\in \text {Fix}(f)$ . We define

$$ \begin{align*}\lambda=\frac{d}{d x}f(x^{(0)}). \end{align*} $$

The fixed point $x^{(0)}$ is called attractive if $0<|\lambda |_p<1$ , indifferent if $|\lambda |_p=1$ , and repelling if $|\lambda |_p>1$ .

For an attractive fixed point $x^{(0)}$ of f, its basin of attraction is defined by

$$ \begin{align*}A(x^{(0)})=\{x\in\mathbb Q_p: \lim_{n\to\infty}f^n(x)=x^{(0)}\}, \end{align*} $$

where $f^n=\underbrace {f\circ f\circ \cdots \circ f}_n$ .

The main result of the present paper is given in the following theorem.

Theorem 1.1. Let $p\geq 3$ , $k\geq 2$ , $|q|_p<1$ , $|\theta -1|_p<1$ , and $x_0^*=1$ . Then the dynamical structure of the system $(\mathbb Q_p, f_{\theta ,q,k})$ is described as follows.

  1. (A) If $|k|_p\leq |q+\theta -1|_p$ , then $\mathrm{Fix}(f_{\theta ,q,k})=\{x_0^*\}$ and

    $$ \begin{align*}A(x_0^*)=\mathrm{Dom}(f_{\theta,q,k}). \end{align*} $$
  2. (B) Assume that $|k|_p>|q+\theta -1|_p$ and $|\theta -1|_p<|q|_p^2$ . Then there exists a non-empty set $J_{f_{\theta ,q,k}}\subset \mathrm{Dom}(f_{\theta ,q,k})\setminus \mathcal P_{x^{(\infty )}}$ which is invariant with respect to $f_{\theta ,q,k}$ and

    $$ \begin{align*}A(x_0^*)=\mathrm{Dom}(f_{\theta,q,k})\setminus(\mathcal P_{x^{(\infty)}}\cup J_{f_{\theta,q,k}}). \end{align*} $$
    Moreover, if $\kappa _p$ is the GCF of k and $p-1$ , then the following hold:
    1. (B1) if $\kappa _p=1$ , then there exists $x_*\in \mathrm{Fix}(f_{\theta ,q,k})$ such that $x_*\neq x_0^*$ and $J_{f_{\theta ,q,k}}=\{x_*\}$ ;

    2. (B2) if $\kappa _p\geq 2$ , then $(J_{f_{\theta ,q,k}},f_{\theta ,q,k},|\cdot |_p)$ is topologically conjugate to the full shift dynamics on $\kappa _p$ symbols.

Remark 1.2. It is worth pointing out that, in the present paper, the condition $|\theta -1|_p<|q|_p^2$ is assumed to get essential estimations and calculations to prove the main result. The results of a recent paper [Reference Ahmad, Liao and Saburov1] show that such a condition could be loosened to $|\theta -1|_p<|q|_p$ , but only for the case $k=3$ where explicit expressions of the fixed points of the function $f_{\theta ,q,k}$ have essentially been used to get more exact estimations. However, in this paper, we are able to prove the chaoticity of the Potts–Bethe mapping for arbitrary values of k (under the condition $|\theta -1|_p<|q|_p^2$ ) and moreover, we are not even using the existence of the fixed points. Once we have proved that the Potts–Bethe mapping is conjugate to a full shift, then one concludes the existence of the fixed points. Roughly speaking, we are constructing (explicitly) a Markov partition of the mapping (1.1) which allows us to prove the main result of the current paper. However, the results of [Reference Ahmad, Liao and Saburov1] indicate that the chaoticity of the function (1.1) could be obtained even in the case of $|q|^2\leq |\theta -1|_p<|q|_p$ , but this will be a topic for another work. Here, it is better to emphasize that the results are valid when $p\geq 3$ . The case $p=2$ is considered pathological in the p-adic analysis (see for example [Reference Fan, Liao, Wang and Zhou10]). Indeed, in [Reference Ahmad, Liao and Saburov1], it was established that when $p=2$ and $k=3$ , the function (1.1) does not have chaotic behavior. For general values of k, owing to huge calculations and numerous technical issues, this case could be investigated elsewhere.

Remark 1.3. In [Reference Mukhamedov and Rozikov41, Reference Mukhamedov and Rozikov42], the authors established that the function (1.1) may have at least one fixed point and, moreover, they found a necessary condition (that is q is divisible by p) for the existence of more than one fixed point. Therefore, the following conjecture was formulated: Let $k\in {\mathbb N}$ , $q\in p{\mathbb N}$ , and $|\theta -1|_p<1$ , then the function (1.1) has at least two fixed points. The formulated Theorem 1.1(A) shows that the mentioned conjecture is not always true.

We stress that, in the p-adic setting, owing to the lack of a convex structure of the set of p-adic Gibbs measures, it was quite difficult to constitute a phase transition with some features of the set of p-adic Gibbs measures. However, Theorem 1.1(B2) yields that the set of p-adic Gibbs measures is huge which is a priori not clear (see [Reference Mukhamedov24, Reference Mukhamedov and Rozikov42]). Moreover, the method of the present work allows one to find lots of periodic p-adic Gibbs measures for the p-adic Potts model. Furthermore, Theorem 1.1(B) together with the results of [Reference Mukhamedov and Akin29, Reference Mukhamedov and Khakimov33] will open new perspectives in investigations of generalized p-adic self-similar sets.

On one hand, our results shed some light on the question of the investigation of dynamics of rational functions in the p-adic analysis, because a global dynamical structure of rational maps on $\mathbb {Q}_p$ remains unclear. Some particular rational functions have been considered in [Reference Benedetto4, Reference Benedetto5, Reference Diao and Silva7, Reference Ganikhodjaev, Mukhamedov and Rozikov8, Reference Fan, Liao, Wang and Zhou10, Reference Khakimov13Reference Khakimov15, Reference Khamraev and Mukhamedov18, Reference Khrennikov and Nilsson21, Reference Mukhamedov, Saburov and Khakimov39]. On the other hand, the obtained results may have potential applications in the cryptography to build pseudo-random codes (see [Reference Anashin2, Reference Anashin, Khrennikov and Yurova3, Reference Mukhamedov, Omirov and Saburov37, Reference Woodcock and Smart45]). We point out that some p-adic chaotic dynamical systems have been studied in [Reference Gyorgyi, Kondor, Sasvari and Tel9, Reference Woodcock and Smart45].

2 Preliminaries

2.1 p-adic numbers

Let $\mathbb {Q}$ be the field of rational numbers. For a fixed prime number p, every rational number $x\ne 0$ can be represented in the form $x = p^r{n\over m}$ , where $r, n\in \mathbb {Z}$ , m is a positive integer, and n and m are relatively prime with p: $(p, n) = 1$ , $(p, m) = 1$ . The p-adic norm of x is given by

$$ \begin{align*}|x|_p=\left\{\begin{array}{@{}ll} p^{-r} \quad& \text{for } x\ne 0,\\ 0 \quad& \text{for } x = 0. \end{array}\right. \end{align*} $$

This norm is non-Archimedean and satisfies the so-called strong triangle inequality

$$ \begin{align*}|x+y|_p\leq \max\{|x|_p,|y|_p\}.\end{align*} $$

The completion of $\mathbb {Q}$ with respect to the p-adic norm defines the p-adic field $\mathbb {Q}_p$ . Any p-adic number $x\ne 0$ can be uniquely represented in the canonical form

(2.1) $$ \begin{align} x = p^{\mathrm{ord}_p(x)}(x_0+x_1p+x_2p^2+\cdots), \end{align} $$

where $\mathrm{ord}_p(x)\in \mathbb Z$ and the integers $x_j$ satisfy: $0\leq x_j \leq p - 1$ , $x_0\neq 0$ . In this case, $|x|_p = p^{-\mathrm{ord}_p{x}}$ .

Recall that $\mathbb Q_p$ is not an ordered field. So, we may compare two p-adic numbers only with respect to their p-adic norms.

In what follows, to simplify our calculations, we are going to introduce new symbols ‘O’ and ‘o’ (roughly speaking, these symbols replace the notation ‘ $\operatorname {\mathrm{mod}}p^k$ ’ without noticing the power of k). Namely, for a given p-adic number x, by $O[x]$ , we mean a p-adic number with the norm $p^{-\mathrm{ord}_p(x)}$ , that is, $|x|_p=|O(x)|_p$ . By $o[x]$ , we mean a p-adic number with a norm strictly less than $p^{-\mathrm{ord}_p(x)}$ , that is, $|o(x)|_p<|x|_p$ . For instance, if $x=1-p+p^2$ , we can write $x-1+p=o[p]$ , $x-1=o[1]$ , or $x=O[1]$ . The symbols $O[\cdot ]$ and $o[\cdot ]$ will make our work easier when we need to calculate the p-adic norm of p-adic numbers. It is easy to see that $y=O[x]$ if and only if $x=O[y]$ .

We give some basic properties of $O[\cdot ]$ and $o[\cdot ]$ , which will be used later on.

Lemma 2.1. Let $x,y\in \mathbb Q_p$ . Then the following statements hold.

  1. (1) $O[x]O[y]=O[xy]$ .

  2. (2) $xO[y]=O[xy]$ , $O[y]x=O[xy]$ .

  3. (3) $O[x]o[y]=o[xy]$ .

  4. (4) $o[x]o[y]=o[xy]$ .

  5. (5) $xo[y]=o[xy]$ , $o[y]x=o[xy]$ .

  6. (6) If $y\neq 0$ , then ${O[x]}/{O[y]}=O[{x}/{y}]$ .

  7. (7) If $y\neq 0$ , then ${o[x]}/{O[y]}=o[{x}/{y}]$ .

For each $a\in {\mathbb Q}_p$ , $r>0$ , we denote

$$ \begin{align*}B_r(a)=\{x\in {\mathbb Q}_p : |x-a|_p< r\}.\end{align*} $$

We recall that $\mathbb {Z}_p=\{x\in \mathbb {Q}_p: |x|_p\leq 1\}$ and $\mathbb Z_p^*=\{x\in \mathbb Q_p: |x|_p=1\}$ are the set of all p-adic integers and p-adic units, respectively.

The following result is well known as Hensel’s lemma.

Lemma 2.2. [Reference Borevich and Shafarevich6, Reference Koblitz22]

Let $F(x)$ be a polynomial whose coefficients are p-adic integers. Let $x^*$ be a p-adic integer such that for some $i\geq 0$ ,

$$ \begin{align*}F(x^*)\equiv0\ (\operatorname{\mathrm{mod}}p^{2i+1}),\ \ \ F'(x^*)\equiv0\ (\operatorname{\mathrm{mod }}p^{i}),\ \ \ F'(x^*)\not\equiv0\ (\operatorname{\mathrm{mod }}p^{i+1}). \end{align*} $$

Then $F(x)$ has a p-adic integer root $x_*$ such that $x_*\equiv x^* (\operatorname {\mathrm{mod }}p^{i+1})$ .

The p-adic exponential is defined by

$$ \begin{align*}\exp_p(x) =\sum^\infty_{n=0}{x^n\over n!},\end{align*} $$

which converges for every $x\in B_{p^{-1/(p-1)}}(0)$ . Denote

$$ \begin{align*}\mathcal E_p=\{x\in\mathbb Q_p: |x-1|_p<p^{-1/(p-1)}\}. \end{align*} $$

This set is the range of the p-adic exponential function. The following fact is well known.

Lemma 2.3. [Reference Mukhamedov, Saburov and Khakimov40]

The set $\mathcal E_p$ has the following properties.

  1. (a) $\mathcal E_p$ is a group under multiplication.

  2. (b) If $a,b\in \mathcal E_p$ , then the following are true:

    $$ \begin{align*}|a-b|_p<\left\{\begin{array}{@{}ll} \tfrac{1}{2}, & p=2,\\ 1, & p\neq2, \end{array}\right.\ \ \ \ \ |a+b|_p=\left\{\begin{array}{@{}ll} \tfrac{1}{2}, & p=2,\\ 1, & p\neq2. \end{array}\right. \end{align*} $$
  3. (c) If $a\in \mathcal E_p$ , then there is an element $h\in B_{p^{-1/(p-1)}}(0)$ such that $a=\exp _p(h)$ .

Lemma 2.4. Let $k\geq 2$ and $\alpha ,\beta \in \mathcal E_p$ . Then there exists a unique $\gamma \in 1+p\mathbb Z_p$ such that

(2.2) $$ \begin{align} \sum_{j=0}^{k-1}\alpha^{k-j-1}\beta^j=k\gamma. \end{align} $$

Moreover, if $p\neq 2$ , then $\gamma \in \mathcal E_p$ .

Remark 2.5. We notice that Lemma 2.4 has been proved in [Reference Mukhamedov and Khakimov35] for $p\neq 2$ . The proof of the case $p=2$ is similar to that one. We notice that this lemma plays a crucial role in our further investigations. Especially, we will often use the fact $\gamma \in \mathcal E_p$ .

Corollary 2.6. Let $k\in \mathbb N$ . Then

$$ \begin{align*}\alpha^k-\beta^k=k(\alpha-\beta)+o[k(\alpha-\beta)]\quad\text{for all }\alpha,\beta\in\mathcal E_p. \end{align*} $$

Proof. Let $\alpha ,\beta \in \mathcal E_p$ . By Lemma 2.4,

$$ \begin{align*}\sum_{j=0}^{k-1}\alpha^{k-1-j}\beta^j=k+k(\gamma-1), \end{align*} $$

where $\gamma -1=o[1]$ .

Hence, Lemma 2.1 implies

$$ \begin{align*} \alpha^k-\beta^k&=(\alpha-\beta)\sum_{j=0}^{k-1}\alpha^{k-j-1}\beta^j\\ &=k(\alpha-\beta)+k(\alpha-\beta)(\gamma-1)\\ &=k(\alpha-\beta)+O[k(\alpha-\beta)]o[1]\\ &=k(\alpha-\beta)+o[k(\alpha-\beta)], \end{align*} $$

which is the required relation.

Remark 2.7. In our further investigations, we mainly use Corollary 2.6 in the following form. Namely, for $k\in \mathbb N$ ,

(2.3) $$ \begin{align} \alpha^k-1=k(\alpha-1)+o[k(\alpha-1)]\quad\text{for all }\alpha\in\mathcal E_p. \end{align} $$

We notice that a monomial equation $x^k=a$ over $\mathbb Q_p$ has been studied in [Reference Mukhamedov and Khakimov36, Reference Mukhamedov and Saburov38]. In our further investigations, we only need the following special case of that equation.

Theorem 2.8. [Reference Mukhamedov and Khakimov36]

Let $p\geq 3$ and $a\in \mathcal E_p$ . Then the following statements hold:

  1. (i) if $|k|_p\leq |a-1|_p$ , then the polynomial $x^k-a$ has no root;

  2. (ii) if $|k|_p>|a-1|_p$ , then for every $\xi \in \{y\in \mathbb F_p: y^k\equiv a(\operatorname {\mathrm{mod }}p)\}$ , the polynomial $x^k-a$ has a unique root in $B_1(\xi )$ .

Here $\mathbb F_p$ stands for the ring of integers modulo p.

Remark 2.9. Thanks to Theorem 2.8, for every $a\in \mathcal E_p$ with $|a-1|_p<|k|_p$ , the equation $x^k=a$ has a single root belonging to $\mathcal E_p$ , which is called the principal kth root and denoted by $\sqrt [k]{a}$ . In what follows, the symbol $\sqrt [k]{a}$ (for $a\in \mathcal E_p$ ) always means the principal kth root of a. Therefore, for $|a-1|_p<|k|_p$ , all solutions of the monomial equation $x^k=a$ have the following form: $x_i=\xi _i\sqrt [k]{a}$ , where $\xi _i^k=1$ and $\sqrt [k]{a}$ is a principal kth root of a.

2.2 p-adic subshift

Let $f:X\to \mathbb Q_p$ be a map from a compact open set X of $\mathbb Q_p$ into $\mathbb Q_p$ . We assume that (i) $f^{-1}(X)\subset X$ ; (ii) $X=\cup _{j\in I}B_{r}(a_j)$ can be written as a finite disjoint union of balls of centers $a_j$ and of the same radius r such that for each $j\in I$ , there is an integer $\tau _j\in \mathbb Z$ such that

(2.4) $$ \begin{align} |f(x)-f(y)|_p=p^{\tau_j}|x-y|_p,\quad x,y\in B_r(a_j). \end{align} $$

For such a map f, define its Julia set by

(2.5) $$ \begin{align} J_f=\bigcap_{n=0}^\infty f^{-n}(X). \end{align} $$

It is clear that $f^{-1}(J_f)=J_f$ and then $f(J_f)\subset J_f$ . The triple $(X,J_f,f)$ is called a p-adic weak repeller if all $\tau _j$ in (2.4) are non-negative, but at least one is positive. We call it a p-adic repeller if all $\tau _j$ in (2.4) are positive. For any $i\in I$ , we let

$$ \begin{align*}I_i:=\{j\in I: B_r(a_j)\cap f(B_r(a_i))\neq\varnothing\}=\{j\in I: B_r(a_j)\subset f(B_r(a_i))\} \end{align*} $$

(the second equality holds because of the expansiveness of the ultrametric property). Then define a matrix $A=(a_{ij})_{I\times I}$ , called incidence matrix, as follows:

$$ \begin{align*}a_{ij}=\left\{\begin{array}{@{}ll} 1\quad& \text{if } j\in I_i,\\ 0\quad& \text{if } j\not\in I_i. \end{array} \right. \end{align*} $$

If A is irreducible, we say that $(X,J_f,f)$ is transitive. Here the irreducibility of A means that for any pair $(i,j)\in I\times I$ , there is a positive integer m such that $a_{ij}^{(m)}>0$ , where $a_{ij}^{(m)}$ is the entry of the matrix $A^m$ .

Given I and the irreducible incidence matrix A as above, we denote

$$ \begin{align*}\Sigma_A=\{(x_k)_{k\geq 0}: \ x_k\in I,\ A_{x_k,x_{k+1}}=1, \ k\geq 0\}, \end{align*} $$

which is the corresponding subshift space, and let $\sigma $ be the shift transformation on $\Sigma _A$ . We equip $\Sigma _A$ with a metric $d_f$ depending on the dynamics, which is defined as follows. First, for $i,j\in I,\ i\neq j$ , let $\kappa (i,j)$ be the integer such that $|a_i-a_j|_p=p^{-\kappa (i,j)}$ . It is clear that $\kappa (i,j)<\log _p(r)$ . By the ultrametric inequality,

$$ \begin{align*}|x-y|_p=|a_i-a_j|_p\ \ \ i\neq j\quad \text{for all } x\in B_r(a_i), \text{ for all } y\in B_r(a_j). \end{align*} $$

For $x=(x_0,x_1,\ldots ,x_n,\ldots )\in \Sigma _A$ and $y=(y_0,y_1,\ldots ,y_n,\ldots )\in \Sigma _A$ , define

$$ \begin{align*}d_f(x,y)=\left\{\begin{array}{@{}ll} p^{-\tau_{x_0}-\tau_{x_1}-\cdots-\tau_{x_{n-1}}-\kappa(x_{n},y_{n})}\quad& \text{if }n\neq 0,\\ p^{-\kappa(x_0,y_0)}\quad& \text{if }n=0 \end{array}\right. \end{align*} $$

where $n=n(x,y)=\min \{i\geq 0: x_i\neq y_i\}$ . It is clear that $d_f$ defines the same topology as the classical metric which is defined by $d(x,y)=p^{-n(x,y)}$ .

Theorem 2.10. [Reference Gyorgyi, Kondor, Sasvari and Tel9]

Let $(X,J_f,f)$ be a transitive p-adic weak repeller with incidence matrix A. Then the dynamics $(J_f,f,|\cdot |_p)$ is isometrically conjugate to the shift dynamics $(\Sigma _A,\sigma ,d_f)$ .

3 Proof of Theorem 1.1: part (A)

In what follows, we always assume that $p\geq 3$ and $|q|_p<1$ . To prove Theorem 1.1(A), we need the following auxiliary lemma.

Lemma 3.1. Let $p\geq 3$ and $k\in \mathbb N$ . If $a\in \mathcal E_p$ and $|a-1|_p\geq |k|_p$ , then $|x^k-a|_p\geq |a-1|_p$ for any $x\in \mathbb Q_p$ .

Proof. Take an arbitrary $a\in \mathcal E_p$ such that $|a-1|_p\geq |k|_p$ . We distinguish three cases.

Case $x\not \in \mathbb Z_p^*$ . Then we immediately get $|x^k-1|_p\geq 1$ . From $|a-1|_p<1$ , using the strong triangle inequality, one has $|x^k-a|_p\geq 1$ . This yields that $|x^k-a|_p>|a-1|_p$ .

Case $x\in \mathcal E_p$ . Then noting $|x-1|_p<1$ , owing to Corollary 2.6, we obtain $|x^k-1|_p<|k|_p$ . The last one together with $|a-1|_p\geq |k|_p$ implies $|x^k-a|_p=|a-1|_p$ .

Case $x\in \mathbb Z_p^*\setminus \mathcal E_p$ . In this case, x has the following canonical form:

$$ \begin{align*}x=x_0+x_1\cdot p+x_2\cdot p^2+\cdots, \end{align*} $$

where $2\leq x_0\leq p-1$ and $0\leq x_i\leq p-1$ , $i\geq 1$ . Then $({x}/{x_0})\in \mathcal E_p$ . According to Remark 2.7,

$$ \begin{align*}\bigg(\frac{x}{x_0}\bigg)^k-1=O[k(x-x_0)]=o[k]. \end{align*} $$

Consequently, $|x^k-x_0^k|_p<|k|_p$ , which yields $|x^k-1|_p=|x_0^k-1|_p$ . Now we need to check two subcases, $|x_0^k-1|_p=1$ and $|x_0^k-1|_p<1$ , separately.

Suppose that $|x_0^k-1|_p=1$ . Then, owing to $|a-1|_p<1$ , one has $|x_0^k-a|_p=1$ . Hence, $|x^k-a|_p>|a-1|_p$ .

Let us assume that $|x_0^k-1|_p<1$ . For convenience, let us write $k=mp^s$ , where $s\geq 1$ and $(m,p)=1$ . Then noting $x_0^{p}\equiv x_0(\operatorname {\mathrm{mod }}p)$ , from $x^{mp^s}\equiv 1\ (\operatorname {\mathrm{mod }}p)$ , we obtain $|x_0^m-1|_p<1$ . Thanks to Remark 2.7, one finds

$$ \begin{align*}x_0^{mp^s}-1=p^{s}(x_0^m-1)+o[p^s(x_0^m-1)], \end{align*} $$

which yields $|x_0^k-1|_p<|k|_p$ . Hence, from $|a-1|_p\geq |k|_p$ , it follows that $|x_0^k-a|_p=|a-1|_p$ . Consequently, $|x^k-a|_p=|a-1|_p$ . This completes the proof.

Remark 3.2. We notice that the set $\mathcal P_{x^{(\infty )}}$ is empty if $|k|_p\leq |q+\theta -1|_p$ . Indeed, from $x^{(\infty )}\in \mathcal E_p$ , where $x^{(\infty )}=2-q-\theta $ and $|x^{(\infty )}-1|_p\geq |k|_p$ , owing to Lemma 3.1, we infer that

$$ \begin{align*}|f_{\theta,q,k}^n(x)-x^{(\infty)}|_p\geq|x^{(\infty)}-1|_p\quad \text{for all } n\in\mathbb N, \text{ for all } x\in \mathrm{Dom}(f_{\theta,q,k}). \end{align*} $$

Hence, noting $x^{(\infty )}\neq 1$ , we can conclude that $\mathcal P_{x^{(\infty )}}=\varnothing $ .

Let us define

(3.1) $$ \begin{align} g_{\theta,q}(x)=\frac{\theta x+q-1}{x+\theta+q-2}. \end{align} $$

In our further investigations, we use the following simple property of the function $g_{\theta ,q}$ :

(3.2) $$ \begin{align} g_{\theta,q}(x)-1=\frac{(\theta-1)(x-1)}{x+\theta+q-2}. \end{align} $$

We notice that $f_{\theta ,q,k}(x)=(g_{\theta ,q}(x))^k$ for any $x\in \mathrm{Dom}(f_{\theta ,q,k})$ . It is clear that the function $f_{\theta ,q,k}$ has a fixed point $x_0^*=1$ .

Proof of Theorem 1.1: (A)

Let $|k|_p\leq |q+\theta -1|_p$ and denote

$$ \begin{align*}\begin{array}{@{}ll} K_1=\{x\in\mathbb Q_p: |x-1|_p<|q+\theta-1|_p\},\\[2mm] K_2=\{x\in\mathbb Q_p: |x-1|_p=|x-2+q+\theta|_p\}. \end{array} \end{align*} $$

First, we show that $f_{\theta ,q,k}(x)\in K_1\cup K_2$ for any $x\notin K_1\cup K_2$ . Then we prove that $f_{\theta ,q,k}(x)\in K_1$ for any $x\in K_2$ . Finally, we show that $f_{\theta ,q,k}^n(x)\to 1$ for any $x\in K_1$ .

Indeed, let $x\notin K_1\cup K_2$ . From $|q+\theta -1|_p<1$ , owing to Lemma 3.1,

$$ \begin{align*}|f_{\theta,q,k}(x)-2+q+\theta|_p\geq|q+\theta-1|_p, \end{align*} $$

which is equivalent to either $|f_{\theta ,q,k}(x)-1|_p<|q+\theta -1|_p$ or $|f_{\theta ,q,k}(x)-1|_p=|f_{\theta ,q,k}(x)-2+q+\theta |_p$ . This yields that $f_{\theta ,q,k}(x)\in K_1\cup K_2$ .

Now assume that $x\in K_2$ . Then by (3.2),

$$ \begin{align*}g_{\theta,q}(x)-1=(\theta-1)O[1] =O[\theta-1] =o[1] \end{align*} $$

which means $g_{\theta ,q}(x)\in \mathcal E_p$ . Then thanks to Remark 2.7,

$$ \begin{align*}|f_{\theta,q,k}(x)-1|_p<|k|_p. \end{align*} $$

The last one together with $|k|_p\leq |q+\theta -1|_p$ implies $|f_{\theta ,q,k}(x)-1|_p<|q+\theta -1|_p$ and hence $f_{\theta ,q,k}(x)\in K_1$ .

Finally, we suppose that $x\in K_1$ . It then follows from (3.2) that

$$ \begin{align*}g_{\theta,q}(x)-1=\frac{(\theta-1)(x-1)}{O[q+\theta-1]} =(\theta-1)o[1] =o[\theta-1] =o[1]. \end{align*} $$

This again means $g_{\theta ,q}(x)\in \mathcal E_p$ . By Remark 2.7,

$$ \begin{align*}f_{\theta,q,k}(x)-1=O\bigg[\frac{k(\theta-1)(x-1)}{q+\theta-1}\bigg]. \end{align*} $$

Noting $|q+\theta -1|_p>|k(\theta -1)|_p$ , from the last one,

$$ \begin{align*}|f_{\theta,q,k}(x)-1|_p<|x-1|_p. \end{align*} $$

Hence,

$$ \begin{align*}|f_{\theta,q,k}^n(x)-1|_p\leq\frac{1}{p^n}|x-1|_p, \end{align*} $$

which yields $f_{\theta ,q,k}^n(x)\to 1$ as $n\to \infty $ . This completes the proof.

4 Proof of Theorem 1.1: the first part of (B)

In this section, we are going to study the dynamics of $f_{\theta ,q,k}$ when $|\theta -1|_p<|q^2|_p$ and $|q|_p<|k|_p$ . In what follows, the following auxiliary fact is needed.

Proposition 4.1. Let $p\geq 3$ and $|\theta -1|_p<|q|_p<|k|_p$ . If $x\in \mathrm{Dom}(f_{\theta ,q,k})$ with $|x-2+q+\theta |_p>|\theta -1|_p$ , then $f_{\theta ,q,k}^n(x)\to 1$ as $n\to \infty $ .

Proof. First, we notice that $|x-2+q+\theta |_p>|\theta -1|_p$ implies $|x-1+q|_p>|\theta -1|_p$ . Owing to $|\theta -1|_p<|q|_p$ , we are going to consider two cases: (i) $|x-1+q|_p\geq |q|_p$ and (ii) $|\theta -1|_p<|x-1+q|_p<|q|_p$ .

Case (i). Let $|x-1+q|_p\geq |q|_p$ . This means that either $x\in B_{|q|_p}(1)$ or $|x-1+q|_p=|x-1|_p$ . First, we show that the condition $|x-1+q|_p=|x-1|_p$ yields $f_{\theta ,q,k}(x)\in B_{|q|_p}(1)$ . Furthermore, we establish that $f_{\theta ,q,k}^n(x)\to 1$ for any $x\in B_{|q|_p}(1)$ .

Let us pick $x\in \mathbb Q_p$ such that $|x-1+q|_p=|x-1|_p$ . Then $|q|_p\leq |x-1|_p$ . Keeping in mind $\theta -1=o[q]$ , one finds $\theta -1=o[x-1+q]$ and

$$ \begin{align*}x+\theta+q-2=O[x-1+q]=O[x-1].\end{align*} $$

So, by (3.2),

$$ \begin{align*}g_{\theta,q}(x)-1=\frac{(\theta-1)(x-1)}{O[x-1]}=o[q]O[1]=o[q]. \end{align*} $$

Because $|k|_p\leq 1$ , owing to Remark 2.7, we obtain $|f_{\theta ,q,k}(x)-1|_p<|q|_p$ , which implies $f_{\theta ,q,k}(x)\in B_{|q|_p}(1)$ .

Now let us suppose that $x\in B_{|q|_p}(1)$ . Then by (3.2),

$$ \begin{align*}g_{\theta,q}(x)-1=\frac{o[q](x-1)}{q+o[q]} =\frac{o[q](x-1)}{O[q]}=o[1](x-1)=o[x-1]. \end{align*} $$

Hence, again thanks to Remark 2.7, one has $|f_{\theta ,q,k}(x)-1|_p<|x-1|_p$ , which yields

$$ \begin{align*}|f_{\theta,q,k}^n(x)-1|_p\leq\frac{1}{p^n}|x-1|_p\quad\text{for all } n\in\mathbb N. \end{align*} $$

So, $f_{\theta ,q,k}^n(x)\to 1$ as $n\to \infty $ .

Case (ii). Let $|\theta -1|_p<|x-1+q|_p<|q|_p$ . Then

$$ \begin{align*}g_{\theta,q}(x)-1=\frac{o[x-1+q]O[q]}{O[x-1+q]} =o[1]O[q]=o[q]. \end{align*} $$

Again, Remark 2.7 yields $|f_{\theta ,q,k}(x)-1|_p<|q|_p$ . Hence, by (i), we have $f_{\theta ,q,k}^n(x)\to 1$ as $n\to \infty $ . This completes the proof.

Corollary 4.2. Let $p\geq 3$ and $|\theta -1|_p<|q|_p<|k|_p$ . If $|x-1+q|_p\geq |q|_p$ , then $f_{\theta ,q,k}^n(x)\to 1$ as $n\to \infty $ .

Proof. Let $|x-1+q|_p\geq |q|_p$ . By $|\theta -1|_p<|q|_p$ and the strong triangle inequality, one finds $|x-2+q+\theta |_p>|\theta -1|_p$ . Hence, the last one owing to Proposition 4.1 yields $f_{\theta ,q,k}^n(x)\to 1$ .

Lemma 4.3. Let $p\geq 3$ and $|\theta -1|_p<|q|_p<|k|_p$ . If $|x-2+q+\theta |_p<|q(\theta -1)|_p$ , then $f_{\theta ,q,k}^n(x)\to 1$ as $n\to \infty $ .

Proof. Take arbitrary $x\in \mathrm{Dom}(f_{\theta ,q,k})$ such that $|x-2+q+\theta |_p<|q(\theta -1)|_p$ . Then

$$ \begin{align*}\frac{\theta x+q-1}{x+q+\theta-2}=\theta-\frac{(q+\theta-1)(\theta-1)}{x+q+\theta-2}= \theta+\frac{O[q(\theta-1)]}{o[q(\theta-1)]}=\frac{O[q(\theta-1)]}{o[q(\theta-1)]}, \end{align*} $$

which yields $|f_{\theta ,q,k}(x)|_p>1$ . Hence, $|f_{\theta ,q,k}(x)-2+q+\theta |_p>|\theta -1|_p$ . Then by Proposition 4.1, we obtain the desired assertion.

Our aim is to construct a set $X\subset \mathrm{Dom}(f_{\theta ,q,k})$ for which a triple $(X,J_{f_{\theta ,q,k}},f_{\theta ,q,k})$ is a transitive p-adic repeller. Thanks to Proposition 4.1 and Lemma 4.3, the required set X should be a subset of the following set:

$$ \begin{align*}Y=\big\{x\in U:\ |q(\theta-1)|_p\leq|f_{\theta,q,k}(x)-2+q+\theta|_p\leq|\theta-1|_p\big\}, \end{align*} $$

where

$$ \begin{align*}U:=\bigcup_{\eta\in\mathbb Q_p:\atop{|q|_p\leq |\eta|_p\leq 1}} B_{|q(\theta-1)|_p}(2-q-\theta+\eta(\theta-1)). \end{align*} $$

One can see that for $|q|_p\leq |\eta |_p\leq 1$ , we have $x_\eta \in Y$ if and only if

(4.1) $$ \begin{align} \left\{\begin{array}{@{}ll} x_\eta=2-q-\theta+\eta(\theta-1)+o[q(\theta-1)],\\[2mm] |q(\theta-1)|_p\leq\left|f_{\theta,q,k}(x_\eta)-2+q+\theta\right|{}_p\leq|\theta-1|_p. \end{array}\right. \end{align} $$

Remark 4.4. We notice that if for $|q|_p\leq |\eta |_p\leq 1$ one of the assumptions of (4.1) does not hold, then $f^n_{\theta ,q,k}(x_\eta )\to 1$ as $n\to \infty $ .

Now we are going to find a necessary condition for $\eta \in \mathbb Q_p$ which yields (4.1).

Proposition 4.5. Let $p\geq 3$ , $|k|_p>|q|_p$ and $|\theta -1|_p<|q^2|_p$ . Assume that for $\eta \in \mathbb Q_p$ with $|q|_p\leq |\eta |_p\leq 1$ , (4.1) holds. Then the following statements are true:

  1. (1η) If $|\eta |_p=|q|_p$ , then $(({\eta -q})/{\eta })^k=1+o[1]$ ;

  2. (2η) If $|\eta |_p>|q|_p$ , then $\eta =k-({((k-1)q)}/{2})+(({(k-1)(k-2)q^2})/{6k})+o[q]$ .

Proof. Assume that (4.1) holds. Then

(4.2) $$ \begin{align} f_{\theta,q,k}(x_\eta)=2-q+\theta+O[\theta-1]=1-q+o[q]=1+o[1]. \end{align} $$

( $1_\eta $ ) Let $|\eta |_p=|q|_p$ . By (4.2), one finds

(4.3) $$ \begin{align} \bigg(1-\frac{q}{\eta}+\frac{(\eta-1)(\theta-1)}{\eta}\bigg)^k=1+o[1]. \end{align} $$

Noting $|\theta -1|_p<|q|_p$ , we obtain $({(\eta -1)(\theta -1)})/{\eta }=o[1]$ . Plugging the last one into (4.3),

(4.4) $$ \begin{align} \bigg(\frac{\eta-q}{\eta}+o[1]\bigg)^k=1+o[1]. \end{align} $$

Finally, keeping in mind $|k|_p\leq 1$ , from (4.4),

$$ \begin{align*}\bigg(\frac{\eta-q}{\eta}\bigg)^k=1+o[1]. \end{align*} $$

( $2_\eta $ ) Let $|\eta |_p>|q|_p$ . First, let us assume that $|k|_p\leq |\eta -k|_p$ . Then, using the strong triangle inequality, we can easily check

(4.5) $$ \begin{align} \frac{k}{\eta}\neq1+o[1]. \end{align} $$

From $|\theta -1|_p<|q^2|_p$ and $|q|_p<|\eta |_p$ ,

$$ \begin{align*}g_{\theta,q}(x_\eta)=1-\frac{q}{\eta}+o[q]. \end{align*} $$

Keeping in mind $f_{\theta ,q,k}(x_\eta )=(g_{\theta ,q}(x_\eta ))^k$ , by (2.3),

(4.6) $$ \begin{align} f_{\theta,q,k}(x_\eta)=1-\frac{kq}{\eta}+o\bigg[\frac{kq}{\eta}\bigg]. \end{align} $$

Plugging (4.5) into (4.6) yields

$$ \begin{align*}f_{\theta,q,k}(x_\eta)-1\neq-q+o[q], \end{align*} $$

but it contracts to $f_{\theta ,q,k}(x_\eta )-1+q=o[q]$ . This means that $|\eta |_p>|q|_p$ and (4.1) hold only for $|\eta -k|_p<|k|_p$ .

So, suppose $|\eta -k|_p<|k|_p$ , which implies $|\eta |_p=|k|_p$ . Now we prove our assertion by contradiction. Suppose in contrary,

(4.7) $$ \begin{align} \bigg|\eta-k+\frac{(k-1)q}{2}-\frac{(k-1)(k-2)q^2}{6k}\bigg|_p\geq|q|_p. \end{align} $$

Noting $|q|_p<|k|_p\leq 1$ , we then can easily check the following:

$$ \begin{align*} &\bigg|\frac{(k-1)q}{2}\bigg|_p\leq|q|_p,\\ &\bigg|\frac{(k-1)(k-2)q^2}{6k}\bigg|_p\leq|q|_p. \end{align*} $$

These inequalities together with (4.7) yield

(4.8) $$ \begin{align} \bigg|\eta-k+\frac{(k-1)q}{2}-\frac{(k-1)(k-2)q^2}{6k}\bigg|_p=\max\{|\eta-k|_p, |q|_p\}. \end{align} $$

Owing to $|\theta -1|_p<|q^2|_p$ and $|\eta |_p=|k|_p$ , we have

(4.9) $$ \begin{align} f_{\theta,q,k}(x_\eta)&=\bigg(1-\frac{q}{\eta}+\frac{(\eta-1)(\theta-1)}{\eta}\bigg)^k\nonumber\\ &=\bigg(1-\frac{q}{\eta}+o\bigg[\frac{q^2}{k}\bigg]\bigg)^k\nonumber\\[-12pt] & \nonumber \\ &=\bigg(1-\frac{q}{\eta}\bigg)^k+o\bigg[\frac{q^2}{k}\bigg]\nonumber\\[2mm] &=\bigg(1-\frac{q}{k}\sum_{n=0}^\infty\bigg(\frac{k-\eta}{k}\bigg)^n\bigg)^k+o\bigg[\frac{q^2}{k}\bigg]\nonumber\\[2mm] &=1-q\sum_{n=0}^\infty\bigg(\frac{k-\eta}{k}\bigg)^n+\frac{(k-1)q^2}{2k}-\frac{(k-1)(k-2)q^3}{6k^2}+o\bigg[\frac{q^2}{k}\bigg]\nonumber\\[2mm] &=1-q+\frac{q}{k}\bigg(\eta-k+\frac{(k-1)q}{2}-\frac{(k-1)(k-2)q^2}{6k}\bigg)\nonumber\\ &\quad -q\sum_{n=2}^\infty\bigg(\frac{k-\eta}{k}\bigg)^n+o\bigg[\frac{q^2}{k}\bigg]. \end{align} $$

We calculate the norm of $q\sum _{n=2}^\infty (({k-\eta })/{k})^n.$ Keeping in mind $|\eta -k|_p<|k|_p$ , by the strong triangle inequality,

(4.10) $$ \begin{align} \bigg|q\sum_{n=2}^\infty\bigg(\frac{k-\eta}{k}\bigg)^n\bigg|_p=\bigg|\frac{q(k-\eta)^2}{k^2}\bigg|_p. \end{align} $$

So, we need to calculate the norm of $({q(k-\eta )^2})/{k^2}$ . One can see that

$$ \begin{align*}\bigg|\frac{(k-\eta)^2}{k}\bigg|_p<|\eta-k|_p\leq\max\{|\eta-k|_p, |q|_p\}. \end{align*} $$

The last inequality together with (4.10) yields

(4.11) $$ \begin{align} \bigg|q\sum_{n=2}^\infty\bigg(\frac{k-\eta}{k}\bigg)^n\bigg|_p <\bigg|\frac{q}{k}\bigg|_p\cdot\max\{|\eta-k|_p, |q|_p\}. \end{align} $$

Then by (4.8) and (4.11), using the strong triangle inequality, one finds

$$ \begin{align*} &\bigg|\frac{q}{k}\bigg(\eta-k+\frac{(k-1)q}{2}-\frac{(k-1)(k-2)q^2}{6k}\bigg) -q\sum_{n=2}^\infty\bigg(\frac{k-\eta}{k}\bigg)^n\bigg|_p\\&\qquad=\bigg|\frac{q}{k}\bigg|_p\cdot\max\{|\eta-k|_p,|q|_p\}. \end{align*} $$

From the last equality,

(4.12) $$ \begin{align} \bigg|\frac{q}{k}\bigg(\eta-k+\frac{(k-1)q}{2}-\frac{(k-1)(k-2)q^2}{6k}\bigg) -q\sum_{n=2}^\infty\bigg(\frac{k-\eta}{k}\bigg)^n\bigg|_p\geq\frac{|q^2|_p}{|k|_p}. \end{align} $$

Hence, plugging (4.12) into (4.9) and noting $|k|_p\leq 1$ , one finds

$$ \begin{align*}|f_{\theta,q,k}(x_\eta)-1+q|_p\geq|q^2|_p, \end{align*} $$

which together with $|\theta -1|_p<|q^2|_p$ implies $|f_{\theta ,q,k}(x_\eta )-2+q+\theta |_p>|\theta -1|_p$ , which contradicts (4.1). This means that if for $|\eta |_p>|q|_p$ , (4.1) holds, then

$$ \begin{align*}\eta=k-\frac{(k-1)q}{2}+\frac{(k-1)(k-2)q^2}{6k}+o[q].\\[-42pt] \end{align*} $$

Remark 4.6. One can see that if $|\eta |_p=|q|_p$ and $(({\eta -q})/{\eta })^k\in \mathcal E_p$ , then $(({\eta -q})/{\eta })\in \mathbb Z_p^*\setminus \mathcal E_p$ . This means that there exists a root of unity $\xi \neq 1$ such that $({\eta -q})/{\eta }=\xi +o[1]$ , which yields $\eta ={q}/({1-\xi })+o[q]$ . Without loss of generality for $\xi =1$ , we put $\eta =k-(({(k-1)q})/{2})+(({(k-1)(k-2)q^2})/{6k})+o[q]$ . Consequently, we have found a relation between all roots of unity and all $\eta \in \mathbb Q_p$ for which (4.1) holds.

Let us denote

$$ \begin{align*}\mathrm{Sol}_p(x^k-1)=\{\xi\in\mathbb Z_p^*: \xi^k=1\},\ \ \ \kappa_p=\mathrm{card}(\mathrm{Sol}_p(x^k-1)), \end{align*} $$

where $\mathrm {card}(A)$ is the cardinality of a set A.

We point out that $\kappa _p$ is the number of solutions of the equation $x^k=1$ in ${\mathbb Q}_p$ . From the results of [Reference Mukhamedov and Saburov38], we infer that $\kappa _p$ is the GCF of k and $p-1$ . Therefore, it is clear that $1\leq \kappa _p\leq k$ .

For a given $\xi _i\in \mathrm {Sol}_p(x^k-1),\ i\in \{1,\ldots ,\kappa _p\}$ , we define

(4.13) $$ \begin{align} {x}_{\xi_i}=\left\{ \begin{array}{@{}ll} 1-q+(k-1)\bigg(1-\dfrac{q}{2}+\dfrac{(k-2)q^2}{6k}\bigg)(\theta-1) & \text{if }\ \xi_i=1,\\[5mm] 2-q-\theta+\dfrac{q}{1-\xi_i}(\theta-1) & \text{if }\ \xi_i\neq1, \end{array}\right. \end{align} $$

and

(4.14) $$ \begin{align} X=\bigcup\limits_{i=1}^{\kappa_p}B_r({x}_{\xi_i}),\quad r=|q(\theta-1)|_p. \end{align} $$

By construction, the set X is a subset of $\mathcal E_p\setminus \{1\}$ .

Thanks to Remark 4.4, as a corollary of Proposition 4.5, we can formulate the following result which describes the basin of attraction of $x_0^*=1$ .

Proposition 4.7. Let $p\geq 3$ and $|k|_p>|q|_p$ . If $|\theta -1|_p<|q^2|_p$ , then

$$ \begin{align*}\lim\limits_{n\to\infty}f_{\theta,q,k}^n(x)=1 \quad\text{for all } x\in \mathrm{Dom}(f_{\theta,q,k})\setminus X. \end{align*} $$

The next result shows that the set X (given by (4.14)) consists of disjoint balls.

Lemma 4.8. Let $p\geq 3$ and $|\theta -1|_p<|q^2|_p<|k^2|_p$ . If $x_{\xi _i}$ is given by (4.13) and $r=|q(\theta -1)|_p$ , then $B_r({x}_{\xi _i})\cap B_r({x}_{\xi _j})=\varnothing $ if $i\neq j$ .

Proof. Let $x_{\xi _i}$ and $x_{\xi _j}$ be given by (4.13), where $i\neq j$ . We consider two cases.

Case $\xi _i=1$ and $\xi _j\neq 1$ . Then from (4.13),

$$ \begin{align*} x_{\xi_i}-x_{\xi_j}&=\bigg(k-\frac{(k-1)q}{2}+\frac{(k-1)(k-2)q^2}{6k}-\frac{q}{1-\xi_j}\bigg)(\theta-1)\\[2mm] &=(k+o[k])(\theta-1)\\[2mm] &=O[k(\theta-1)], \end{align*} $$

which implies that $|x_{\xi _i}-x_{\xi _j}|_p>|q(\theta -1)|_p$ . Hence, $B_r({x}_{\xi _i})\cap B_r({x}_{\xi _j})=\varnothing $ .

Case $\xi _i\neq 1$ and $\xi _j\neq 1$ . In this case,

$$ \begin{align*}x_{\xi_i}-x_{\xi_j}=\frac{(\xi_i-\xi_j)q(\theta-1)}{(1-\xi_i)(1-\xi_j)} =\frac{O[1]q(\theta-1)}{O[1]} =O[q(\theta-1)], \end{align*} $$

which means $|x_{\xi _i}-x_{\xi _j}|_p=r$ . Hence, we infer that $B_r({x}_{\xi _i})\cap B_r({x}_{\xi _j})=\varnothing $ .

To prove the first part of (B) of Theorem 1.1, we define the following set:

(4.15) $$ \begin{align} J_{f_{\theta,q,k}}=\bigcap\limits_{n=1}^\infty f_{\theta,q,k}^{-n}(X). \end{align} $$

Remark 4.9. In [Reference Mukhamedov and Khakimov36], we have considered the following function over $\mathbb Q_p$ ( $p\geq 3$ ):

$$ \begin{align*}f_{b,c,d}(x)=\bigg(\frac{bx-c}{x-d}\bigg)^k,\quad b,c,d\in\mathcal E_p,\ c\neq bd. \end{align*} $$

It was proved that the mapping $f_{b,c,d}$ has exactly $\kappa _p+1$ fixed points belonging to $\mathcal E_p$ if $d=1-b+c$ and $|b-1|_p<|c-1|_p^2<|k|_p^2$ (see [Reference Mukhamedov and Khakimov36, Theorem 4.5]). One can see that if one takes $b=\theta $ , $c=1-q$ , and $d=2-q-\theta $ , then the function $f_{b,c,d}$ reduces to $f_{\theta ,q,k}$ . So, as a corollary of the mentioned result and noting that $\mathrm{Fix}(f_{\theta ,q,k})\cap (\mathbb Q_p\setminus X)=\{1\}$ , we conclude that if $|\theta -1|_p<|q|_p^2<|k|_p^2$ , then $f_{\theta ,q,k}$ has exactly $\kappa _p$ fixed points belonging to X. This yields $J_{f_{\theta ,q,k}}\neq \varnothing $ for $|\theta -1|_p<|q|_p^2<|k|_p^2$ . Moreover, we may check that for every $i\in \{1,2,\ldots ,\kappa _p\}$ , there exists a unique fixed point of $f_{\theta ,q,k}$ in $B_r(x_{\xi _i})$ (see Proposition 5.5).

Proof of Theorem 1.1: (B)

By Proposition 4.7, the set $\mathcal P_{x^{(\infty )}}$ can not belong to $\mathrm{Dom}(f_{\theta ,q,k})\setminus X$ . Then $\mathcal P_{x^{(\infty )}}\subset X$ . According to the construction of $J_{f_{\theta ,q,k}}$ (see (4.15)), we conclude that $J_{f_{\theta ,q,k}}\cap \mathcal P_{x^{(\infty )}}=\varnothing $ . However, owing to Remark 4.9, the set $J_{f_{\theta ,q,k}}$ is not empty and by the construction, it is invariant with respect to $f_{\theta ,q,k}$ . Then for any $x\not \in J_{f_{\theta ,q,k}}\cup \mathcal P_{x^{(\infty )}}$ , there exists a number $m\geq 1$ such that $f_{\theta ,q,k}^m(x)\not \in X$ . Hence, owing to Proposition 4.7, we infer that $f_{\theta ,q,k}^n(x)\to 1$ as $n\to \infty $ . The proof is complete.

5 Proof of Theorem 1.1: parts (B1) and (B2)

In the following, we need some auxiliary facts.

Lemma 5.1. Let $p\geq 3$ and $|k|_p>|q|_p$ . Then for any $a\in B_{|q^2|_p}(1-q)$ , the equation $x^k=a$ has a unique solution $x_*$ on $\mathcal E_p$ . Moreover, this solution satisfies

(5.1) $$ \begin{align} x_*-1+\frac{q}{k}+\frac{(k-1)q^2}{2k^2}-\frac{(k-1)(k-2)q^3}{6k^3}=o\bigg[\frac{q^2}{k^2}\bigg]. \end{align} $$

Proof. Let $|k|_p>|q|_p$ and $a\in B_{|q^2|_p}(1-q)$ . For convenience, we use the canonical form of a:

$$ \begin{align*}a=1+a_tp^t+a_{t+1}p^{t+1}+\cdots \end{align*} $$

We note that $|k|_p>p^{-t}$ . Let us put $x_t=1$ and define a sequence $\{x_{n+t-1}\}_{n\geq 1}$ as follows:

(5.2) $$ \begin{align} x_{n+t}=x_{n+t-1}+\frac{a-x_{n+t-1}^k}{k}. \end{align} $$

First, by induction, let us show that $x_{n+t-1}\in \mathcal E_p$ for any $n\geq 1$ . It is clear that $x_t\in \mathcal E_p$ and, therefore, we assume that $x_{n+t-1}\in \mathcal E_p$ for some $n\geq 1$ . Then, owing to Remark 2.7, we obtain

$$ \begin{align*}x_{n+t-1}^k-1=k(x_{n+t-1}-1)+o[k(x_{n+t-1}-1)], \end{align*} $$

which is equivalent to

$$ \begin{align*}|x_{n+t-1}^k-1|_p<|k|_p. \end{align*} $$

The last inequality together with $|a-1|_p<|k|_p$ implies that $|x_{n+t}-x_{n+t-1}|_p<1$ . Consequently, from $x_{n+t-1}\in \mathcal E_p$ , we find $x_{n+t}\in \mathcal E_p$ . Hence, $x_{n+t}\in \mathcal E_p$ for any $n\geq 1$ .

Owing to Corollary 2.6, by (5.2), we have

$$ \begin{align*} x_{n+t}^k-x_{n+t-1}^k&=k(x_{n+t}-x_{n+t-1})+o[k(x_{n+t}-x_{n+t-1})]\\ &=a-x_{n+t-1}^k+o[a-x_{n+t-1}], \end{align*} $$

which means

$$ \begin{align*}|x_{n+t}^k-a|_p<|x_{n+t-1}^k-a|_p. \end{align*} $$

Hence, there exists a number $n_0\geq 1$ such that

$$ \begin{align*}|x_{n_0+t}^k-a|_p\leq|(a-1)^2|_p.\end{align*} $$

Now, let us consider a polynomial $F(x)=x^k-a$ . It is easy to check that

$$ \begin{align*}|F'(x_{n_0+t-1})|_p=|k|_p, \quad \text{and} \quad |F(x_{n_0+t-1})|_p\leq|(a-1)^2|_p.\end{align*} $$

So by $|k^2|_p>|(a-1)^2|_p$ and Hensel’s lemma, we conclude that F has a root $x_*$ such that

$$ \begin{align*}|x_*-x_{n_0+t-1}|_p\leq|(a-1)^2|_p.\end{align*} $$

From $x_{n_0+t-1}\in \mathcal E_p$ , we infer that $x_*\in \mathcal E_p$ . The uniqueness of the solution on $\mathcal E_p$ immediately follows from Remark 2.7.

Suppose that $x_*\in \mathcal E_p$ is a solution of $x^k-a=0$ . Let us show that it can be represented by (5.1). It can be checked that $x_*$ has the following form:

(5.3) $$ \begin{align} x_*=1-\frac{q}{k}+\alpha_*, \end{align} $$

where $\alpha _*=o[{q}/{k}]$ . Indeed, because $x_*\in \mathcal E_p$ , there exists $y_*\in p\mathbb Z_p$ such that $x_*$ can be represented as follows: $x_*=1+y_*+o[y_*]$ . Then by Remark 2.7, we have $x_*^k=1+ky_*+o[ky_*]$ . By assumption, $a=1-q+o[q^2]$ . Hence, we obtain the following implications:

$$ \begin{align*}x_*^k-a=0&\ \ \ \Longrightarrow\ \ \ ky_*+q=o[q]\ \ \ \Longrightarrow\ \ \ y_*=-\frac{q}{k}+o\bigg[\frac{q}{k}\bigg] \\ &\ \ \ \Longrightarrow\ \ \ x_*=1-\frac{q}{k}+o\bigg[\frac{q}{k}\bigg]. \end{align*} $$

Furthermore, from (5.3), one finds

$$ \begin{align*} a=x_*^k&=1+k\bigg(-\frac{q}{k}+\alpha_*\bigg)+\frac{k(k-1)}{2}\bigg(-\frac{q}{k}+\alpha_*\bigg)^2\\[2mm] &\quad +\frac{k(k-1)(k-2)}{6} \bigg(-\frac{q}{k}+\alpha_*\bigg)^3+o\bigg[\frac{q^2}{k}\bigg]\\[2mm] &=1-q+k\alpha_*+\frac{(k-1)q^2}{2k}-\frac{(k-1)(k-2)q^3}{6k^2}+o\bigg[\frac{q^2}{k}\bigg]. \end{align*} $$

Plugging $a=1-q+o[q^2]$ into the last equality,

$$ \begin{align*}k\alpha_*+\frac{(k-1)q^2}{2k}-\frac{(k-1)(k-2)q^3}{6k^2}=o\bigg[\frac{q^2}{k}\bigg]. \end{align*} $$

Hence,

$$ \begin{align*}\alpha_*=-\frac{(k-1)q^2}{2k^2}+\frac{(k-1)(k-2)q^3}{6k^3}+o\bigg[\frac{q^2}{k^2}\bigg]. \end{align*} $$

Putting the last one into (5.3) yields (5.1), which completes the proof.

Remark 5.2. We point out that in [Reference Mukhamedov and Saburov38], the existence of solutions of the equation $x^k=a$ on $\mathbb Z^*_p$ has been obtained, but an advantage of Lemma 5.1 is that it provides the uniqueness of solution in ${\mathcal E}_p$ with an explicit expression which is essential in our investigation.

On the set X (see (4.14)), the mapping $f_{\theta ,q,k}$ has exactly $\kappa _p$ inverse branches:

$$ \begin{align*}h_{i}(x)=\frac{(q+\theta-2)\xi_i\sqrt[k]{x}-q+1}{\theta-\xi_i\sqrt[k]{x}}, \end{align*} $$

where $\xi _i^k=1$ , $i\in \{1,\ldots ,\kappa _p\}$ and $\sqrt [k]{x}$ , as before, is a principal root of $x\in X$ (see Remark 2.9).

Proposition 5.3. Let $p\geq 3$ and $|k|_p>|q|_p$ . If $|\theta -1|_p<|q^2|_p$ , then

(5.4) $$ \begin{align} h_{i}(X)\subset B_r(x_{\xi_i}). \end{align} $$

Proof. Let $x\in X$ . We consider two cases: $\xi _i=1$ and $\xi _i\neq 1$ .

Case $\xi _i=1$ . In this case, we have

(5.5) $$ \begin{align} &h_{i}(x)-x_{\xi_i}\nonumber\\&=\frac{(q+\theta-2)\sqrt[k]{x}-q+1}{\theta-\sqrt[k]{x}}- \bigg(1-q+(k-1)\bigg(1-\frac{q}{2}+\frac{(k-2)q^2}{6k}\bigg)(\theta-1)\bigg)\nonumber\\[2mm] &=\frac{(\theta-1)(q+\theta-1+(k-(({(k-1)q})/{2})+(({(k-1)(k-2)q^2})/{6k}))(\sqrt[k]{x}-\theta))}{\theta-\sqrt[k]{x}}\nonumber\\[2mm] &=\frac{(\theta-1)(q+(k-(({(k-1)q})/{2})+(({(k-1)(k-2)q^2})/{6k}))(\sqrt[k]{x}-1)+o[q^2])}{1-\sqrt[k]{x}+o[q^2]}. \end{align} $$

However, owing to Lemma 5.1,

(5.6) $$ \begin{align} \sqrt[k]{x}-1=-\frac{q}{k}-\frac{(k-1)q^2}{2k^2}+\frac{(k-1)(k-2)q^3}{6k^3}+o\bigg[\frac{q^2}{k^2}\bigg]. \end{align} $$

Furthermore, keeping in mind $|q|_p<|k|_p$ , we can easily check the following:

(5.7) $$ \begin{align} q+\bigg(k-\frac{(k-1)q}{2}+\frac{(k-1)(k-2)q^2}{6k}\bigg)(\sqrt[k]{x}-1)=o\bigg[\frac{q^2}{k}\bigg]. \end{align} $$

Plugging (5.6), (5.7) into (5.5), one has

$$ \begin{align*}h_{i}(x)-x_{\xi_i}= \frac{(\theta-1)o[{q^2}/{k}]}{O[{q}/{k}]} =(\theta-1)o[q] =o[q(\theta-1)]. \end{align*} $$

This means $h_{i}(x)\in B_r(x_{\xi _i})$ . The arbitrariness of $x\in X$ yields (5.4).

Case $\xi _i\neq 1$ . Then,

$$ \begin{align*} h_{i}(x)-x_{\xi_i}&=\frac{(q+\theta-2)\xi_i\sqrt[k]{x}-q+1}{\theta-\xi_i\sqrt[k]{x}}-\bigg(2-q-\theta+\frac{q}{1-\xi_i}(\theta-1)\bigg)\\[2mm] &=\frac{(\theta-1)(q-({\theta q}/({1-\xi_i}))+({\xi_i\sqrt[k]{x}q}/({1-\xi_i}))+\theta-1)}{\theta-1+1-\xi_i\sqrt{x}}\\[2mm] &=\frac{(\theta-1)(q-({q}/({1-\xi_i}))+({\xi_iq}/({1-\xi_i}))+o[q])}{O[1]}\\[2mm] &=\frac{(\theta-1)o[q]}{O[1]}\\[2mm] &=o[q(\theta-1)]. \end{align*} $$

The last one implies $h_{i}(x)\in B_r(x_{\xi _i})$ . Again, owing to the arbitrariness of $x\in X$ , we obtain (5.4).

Corollary 5.4. Let $p\geq 3$ and $|k|_p>|q|_p$ . If $|\theta -1|_p<|q^2|_p$ and X is the set given by (4.14), then the following statements hold:

  1. (i) $f_{\theta ,q,k}^{-1}(X)\subset X$ ;

  2. (ii) $B_r({x}_{\xi _i})\subset f_{\theta ,q,k}(B_{r}(x_{\xi _j}))$ for any $i,j\in \{1,\ldots ,\kappa _p\}$ .

Proposition 5.5. Let $p\geq 3$ , $|k|_p>|q|_p$ . If $|\theta -1|_p<|q^2|_p$ and X is the set given by (4.14), then the following statements hold:

  1. (a) if $\xi _i=1$ , then

    (5.8) $$ \begin{align} |f_{\theta,q,k}(x)-f_{\theta,q,k}(y)|_p=\frac{|q(x-y)|_p}{|k(\theta-1)|_p}\quad \text{for any }x,y\in B_r(x_{\xi_i}); \end{align} $$
  2. (b) if $\xi _i\neq 1$ , then

    (5.9) $$ \begin{align} |f_{\theta,q,k}(x)-f_{\theta,q,k}(y)|_p=\frac{|k(x-y)|_p}{|q(\theta-1)|_p}\quad \text{for any }x,y\in B_r(x_{\xi_i}). \end{align} $$

Proof. (a) Recall that for $\xi _i=1$ ,

$$ \begin{align*}x_{\xi_i}=1-q+(k-1)\bigg(1-\frac{q}{2}+\frac{(k-2)q^2}{6k}\bigg)(\theta-1). \end{align*} $$

Thus for $x\in B_r(x_{\xi _i})$ , by (3.2), we have

$$ \begin{align*}g_{\theta,q}(x)-1=\frac{(\theta-1)(-q+o[q])}{k(\theta-1)+o[k(\theta-1)]} =O\bigg[\frac{q}{k}\bigg]. \end{align*} $$

This means $g_{\theta ,q}(x)\in \mathcal E_p$ . Then, owing to Corollary 2.6, for any $x,y\in B_r(x_{\xi _i})$ ,

(5.10) $$ \begin{align} |f_{\theta,q,k}(x)-f_{\theta,q,k}(y)|_p=|k(g_{\theta,q}(x)-g_{\theta,q}(y))|_p. \end{align} $$

However,

$$ \begin{align*} g_{\theta,q}(x)-g_{\theta,q}(y)&=\frac{(\theta-1)(q+\theta-1)(x-y)}{(x-2+q+\theta)(y-2+q+\theta)}\\[2mm] &=\frac{O[q(\theta-1)](x-y)}{(k(\theta-1)+o[k(\theta-1)])^2}\\[2mm] &=\frac{O[q](x-y)}{O[k^2(\theta-1)]}. \end{align*} $$

Plugging the last one into (5.10) implies (5.8).

(b) Recall that for $\xi _i\neq 1$ ,

$$ \begin{align*}x_{\xi_i}=2-q-\theta+\frac{q(\theta-1)}{1-\xi_i}. \end{align*} $$

Then for $x\in B_r(x_{\xi _i})$ ,

$$ \begin{align*}g_{\theta,q}(x)=1+\frac{(\theta-1)(x-1)}{x-2+q+\theta} =1+\frac{(\theta-1)(-q+o[q])}{({q(\theta-1)})/({1-\xi_i})+o[q(\theta-1)]} =\xi_i+o[1]. \end{align*} $$

So, $|g_{\theta ,q}(x)|_p=1$ . Moreover, $({g_{\theta ,q}(x)}/{g_{\theta ,q}(y)})\in \mathcal E_p$ for any $x,y\in B_r(x_{\xi _i})$ . Then, owing to Corollary 2.6,

(5.11) $$ \begin{align} |f_{\theta,q,k}(x)-f_{\theta,q,k}(y)|_p=|k(g_{\theta,q}(x)-g_{\theta,q}(y))|_p. \end{align} $$

However, from

$$ \begin{align*} g_{\theta,q}(x)-g_{\theta,q}(y)&=\frac{(\theta-1)(q+\theta-1)(x-y)}{(x-2+q+\theta)(y-2+q+\theta)}\\[2mm] &=\frac{O[q(\theta-1)](x-y)}{(({q(\theta-1)})/({1-\xi_i})+o[q(\theta-1)])^2}\\[2mm] &=\frac{(x-y)}{O[q(\theta-1)]} \end{align*} $$

and (5.11), we arrive at (5.9).

Proof of Theorem 1.1

(B1) Assume that $x^k=1$ has only one solution. Then the set X given by (4.14) consists of only one ball $B_r(x_1)$ , where

$$ \begin{align*}x_1=1-q+(k-1)\bigg(1-\frac{q}{2}\bigg)(\theta-1),\quad r=|q(\theta-1)|_p. \end{align*} $$

By the proof for the case (B),

$$ \begin{align*}A(x_0^*)=\mathrm{Dom}(f_{\theta,q,k})\setminus(J_{f_{\theta,q,k}}\cup\mathcal P_{x^{(\infty)}}), \end{align*} $$

where $x_0^*=1$ , and $J_{f_{\theta ,q,k}}$ is given by (4.15). By Proposition 5.5, for any $x,y\in B_r(x_1)$ , we have

$$ \begin{align*}|f_{\theta,q,k}(x)-f_{\theta,q,k}(y)|_p>p^2|x-y|_p, \end{align*} $$

which implies that $|J_{f_{\theta ,q,k}}|\leq 1$ . Thanks to Remark 4.9, we have $J_{f_{\theta ,q,k}}\neq \varnothing $ . Because $|J_{f_{\theta ,q,k}}|\leq 1$ , one has $J_{f_{\theta ,q,k}}=\{x_*\}$ , where $x_*\in \mathrm{Fix}(f_{\theta ,q,k})\cap (\mathcal E_p\setminus \{1\})$ .

(B2) Assume that $x^k=1$ has $\kappa _p$ ( $\kappa _p\geq 2$ ) solutions. Consider the set X defined by (4.14). Then according to Corollary 5.4(i) and Proposition 5.5, the triple $(X,J_{f_{\theta ,q,k}},f_{\theta ,q,k})$ is a p-adic repeller. Owing to Corollary 5.4(ii), the corresponding incidence matrix A has dimension $\kappa _p\times \kappa _p$ and can be written as follows:

$$ \begin{align*}A=\left(\begin{array}{@{}llll} 1 & 1 & \cdots & 1\\ 1 & 1 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & \cdots & 1 \end{array} \right). \end{align*} $$

This means that the triple $(X,J_{f_{\theta ,q,k}},f_{\theta ,q,k})$ is transitive. Hence, Theorem 2.10 implies that the dynamics $(J_{f_{\theta ,q,k}},f_{\theta ,q,k},|\cdot |_p)$ is topologically conjugate to the full shift dynamics on $\kappa _p$ symbols.

Acknowledgments

The present work is supported by the UAEU UPAR Grant No. G00003247 (Fund No. 31S391). We are grateful to an anonymous referee whose useful suggestions and comments allowed us to improve the presentation of this paper.

A Appendix

A.1 p-adic measure

Let $(X,{\mathcal B})$ be a measurable space, where ${\mathcal B}$ is an algebra of subsets X. A function $\mu :{\mathcal B}\to {\mathbb Q}_p$ is said to be a p-adic measure if for any $A_1,\ldots ,A_n\subset {\mathcal B}$ such that $A_i\cap A_j=\varnothing $ ( $i\neq j$ ), the equality holds

$$ \begin{align*}\mu\bigg(\bigcup_{j=1}^{n} A_j\bigg)=\sum_{j=1}^{n}\mu(A_j). \end{align*} $$

A p-adic measure is called a probability measure if $\mu (X)=1$ . A p-adic probability measure $\mu $ is called bounded if $\sup \{|\mu (A)|_p : A\in {\mathcal B}\}<\infty $ . For more detailed information about p-adic measures, we refer to [Reference Khrennikov19Reference Khrennikov and Nilsson21].

A.2 Cayley tree

Let $\Gamma ^k_+ = (V,L)$ be a semi-infinite Cayley tree of order $k\geq 1$ with the root $x^0$ (whose each vertex has exactly $k+1$ edges, except for the root $x^0$ , which has k edges). Here, V is the set of vertices and L is the set of edges. The vertices x and y are called nearest neighbors and they are denoted by $l=\langle x,y\rangle $ if there exists an edge connecting them. A collection of the pairs $\langle x,x_1\rangle ,\ldots ,\langle x_{d-1},y\rangle $ is called a path from the point x to the point y. The distance $d(x,y)$ , $x,y\in V$ , on the Cayley tree, is the length of the shortest path from x to y.

$$ \begin{align*}W_{n}=\{ x\in V\mid d(x,x^{0})=n\}, \quad V_n=\overset{n}{\underset{m=0}{\bigcup}}W_{m}, \quad L_{n}=\{ l=\langle x,y\rangle\in L\mid x,y\in V_{n}\}. \end{align*} $$

The set of direct successors of x is defined by

$$ \begin{align*}S(x)=\{ y\in W_{n+1}:d(x,y)=1\},\quad x\in W_{n}. \end{align*} $$

Observe that any vertex x has k direct successors.

A.3 p-adic quasi-Gibbs measure

In this section, we recall the definition of p-adic quasi-Gibbs measure (see [Reference Mukhamedov25]).

Let $q\geq 2$ and $\Phi =\{1,2,\ldots ,q\}$ . Here, $\Phi $ is called a state space and is assigned to the vertices of the tree $\Gamma ^k_+=(V,\Lambda )$ . A configuration $\sigma $ on V is then defined as a function $x\in V\to \sigma (x)\in \Phi $ ; in a similar manner, one defines configurations $\sigma _n$ and $\omega $ on $V_n$ and $W_n$ . The set of all configurations on V (respectively $V_n$ , $W_n$ ) coincides with $\Omega =\Phi ^{V}$ (respectively $\Omega _{V_n}=\Phi ^{V_n},\ \ \Omega _{W_n}=\Phi ^{W_n}$ ). One can see that $\Omega _{V_n}=\Omega _{V_{n-1}}\times \Omega _{W_n}$ . Using this, for given configurations $\sigma _{n-1}\in \Omega _{V_{n-1}}$ and $\omega \in \Omega _{W_{n}}$ , we define their concatenation by

$$ \begin{align*}(\sigma_{n-1}\vee\omega)(x)= \left\{ \begin{array}{@{}ll} \sigma_{n-1}(x) \quad& \text{if } x\in V_{n-1},\\ \omega(x) \quad& \text{if } x\in W_n.\\ \end{array} \right. \end{align*} $$

It is clear that $\sigma _{n-1}\vee \omega \in \Omega _{V_n}$ .

The Hamiltonian of the p-adic Potts model on $\Omega _{V_n}$ is

(A.1) $$ \begin{align} H_n(\sigma)=J\sum_{\langle x,y\rangle\in L_n} \delta_{\sigma(x)\sigma(y)}, \end{align} $$

where $J\in B(0, p^{-1/(p-1)})$ is a coupling constant and $\delta _{ij}$ is the Kroneker’s symbol.

A construction of a generalized p-adic quasi-Gibbs measure corresponding to the model is given below.

Assume that ${\mathbf {h}}: V\setminus \{x^{(0)}\}\to {\mathbb Q}_p^{\Phi }$ is a mapping, that is, ${\mathbf {h}}_x=(h_{1,x},h_{2,x},\ldots ,h_{q,x})$ , where $h_{i,x}\in {\mathbb Q}_p$ ( $i\in \Phi $ ) and $x\in V\setminus \{x^{(0)}\}$ . Given $n\in {\mathbb N}$ , we consider a p-adic probability measure $\mu ^{(n)}_{{\mathbf {h}},\rho }$ on $\Omega _{V_n}$ defined by

(A.2) $$ \begin{align} \mu^{(n)}_{{\mathbf{h}}}(\sigma)=\frac{1}{Z_{n}^{({\mathbf{h}})}}\exp_p\{H_n(\sigma)\}\prod_{x\in W_n}h_{\sigma(x),x}. \end{align} $$

Here, $\sigma \in \Omega _{V_n}$ , and $Z_{n}^{({\mathbf {h}})}$ is the corresponding normalizing factor

(A.3) $$ \begin{align} Z_{n}^{({\mathbf{h}})}=\sum_{\sigma\in\Omega_{V_n}}\exp_p\{H_n(\sigma)\}\prod_{x\in W_n}h_{\sigma(x),x}. \end{align} $$

We are interested in the construction of an infinite volume distribution with given finite-dimensional distributions. More exactly, we would like to find a p-adic probability measure $\mu _{\mathbf {h}}$ on $\Omega $ which is compatible with the given ones $\mu _{{\mathbf {h}}}^{(n)}$ , that is,

(A.4) $$ \begin{align} \mu_{\mathbf{h}}(\{\sigma\in\Omega: \sigma|_{V_n}\equiv\sigma_n\})=\mu^{(n)}_{{\mathbf{h}}}(\sigma_n) \quad \text{for all } \sigma_n\in\Omega_{V_n}, \ n\in{\mathbb N}. \end{align} $$

We say that the p-adic probability distributions (A.2) are compatible if for all $n\geq 1$ and $\sigma \in \Phi ^{V_{n-1}}$ :

(A.5) $$ \begin{align} \sum_{\omega\in\Omega_{W_n}}\mu^{(n)}_{{\mathbf{h}}}(\sigma_{n-1}\vee\omega)=\mu^{(n-1)}_{{\mathbf{h}}}(\sigma_{n-1}). \end{align} $$

This condition, according to the Kolmogorov extension theorem (see [Reference Khrennikov and Ludkovsky20]), implies the existence of a unique p-adic measure $\mu _{{\mathbf {h}}}$ defined on $\Omega $ with a required condition (A.4). Such a measure $\mu _{{\mathbf {h}}}$ is said to be a generalized p-adic Gibbs measure corresponding to the model [Reference Mukhamedov25, Reference Mukhamedov26]. If one has $h_x\in {\mathcal E}_p$ for all $x\in V\setminus \{x^{(0)}\}$ , then the corresponding measure $\mu _{\mathbf {h}}$ is called a p-adic Gibbs measure (see [Reference Mukhamedov and Rozikov41]).

By $Q{\mathcal G}(H)$ , we denote the set of all generalized p-adic Gibbs measures associated with functions ${\mathbf {h}}=\{{\mathbf {h}}_x,\ x\in V\}$ . If there are at least two distinct generalized p-adic Gibbs measures such that at least one of them is unbounded, then we say that a phase transition occurs.

The following statement describes conditions on $h_x$ guaranteeing compatibility of $\mu _{\textbf {h}}^{(n)}(\sigma )$ .

Theorem A.1. [Reference Mukhamedov25]

The measures $\mu ^{(n)}_{{\mathbf {h}}}$ , $ n=1,2,\ldots $ (see (A.2)) associated with the q-state Potts model (A.1) satisfy the compatibility condition (A.5) if and only if for any $n\in {\mathbb N}$ , the following equation holds:

(A.6) $$ \begin{align} \hat h_{x}=\prod_{y\in S(x)}{\mathbf{F}}(\hat {\mathbf{h}}_{y},\theta). \end{align} $$

Here and below, a vector $\hat {\mathbf {h}}=(\hat h_1,\ldots ,\hat h_{q-1})\in {\mathbb Q}_p^{q-1}$ is defined by a vector ${\mathbf {h}}=(h_1,h_2,\ldots ,h_{q})\in {\mathbb Q}_p^{q}$ as follows:

(A.7) $$ \begin{align} \hat h_i=\frac{h_i}{h_q}, \ \ \ i=1,2,\ldots,q-1 \end{align} $$

and the mapping ${\mathbf {F}}:{\mathbb Q}_p^{q-1}\times {\mathbb Q}_p\to {\mathbb Q}_p^{q-1}$ is defined by ${\mathbf {F}}({\mathbf {x}};\theta )=(F_1({\mathbf {x}};\theta ),\ldots , F_{q-1} ({\mathbf {x}};\theta ))$ with

(A.8) $$ \begin{align} F_i({\mathbf{x}};\theta)=\frac{(\theta-1)x_i+\displaystyle\sum\limits_{j=1}^{q-1}x_j+1} {\displaystyle\sum\limits_{j=1}^{q-1}x_j+\theta}, \quad {\mathbf{x}}=\{x_i\}\in{\mathbb Q}_p^{q-1}, \quad i=1,2,\ldots,q-1. \end{align} $$

Let us first observe that the set $(\underbrace {1,\ldots ,1,h}_m,1,\ldots ,1)$ ( $m=1,\ldots ,q-1$ ) is invariant for the equation (A.6). Therefore, in what follows, we restrict ourselves to one of such vectors, let us say $(h,1,\ldots ,1)$ .

In [Reference Mukhamedov and Khakimov32], to establish the phase transition, we considered translation-invariant (that is, ${\mathbf {h}}=\{{\mathbf {h}}_x\}_{x\in V\setminus \{x^0\}}$ such that ${\mathbf {h}}_x={\mathbf {h}}_y$ for all $x,y$ ) solutions of (A.6). Then the equation (A.6) is reduced to the following one:

(A.9) $$ \begin{align} h=f_{\theta,q,k}(h), \end{align} $$

where $f_{\theta ,q,k}$ is given by (1.1).

Hence, to establish the existence of the phase transition, when $k=2$ , we showed in [Reference Mukhamedov and Rozikov41] that (A.9) has three non-trivial solutions if q is divisible by p. Note that the full description of all solutions of the last equation has been carried out in [Reference Rozikov and Khakimov43] when $k=2$ . Certain periodic points of $f_{\theta ,q,k}$ have been carried out in [Reference Ahmad, Liao and Saburov1, Reference Mukhamedov and Akin28, Reference Mukhamedov and Khakimov31].

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