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

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).

1 Introduction

The present paper is a continuation of [35], 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 [35], 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 [12, 16, 17, 23, 26, 27]. 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 [30, 34] 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 [43], 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 [11, 31, 32]. Recently in [1, 44], 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 [1] 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 [1] 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 [10]). Indeed, in [1], 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 [41, 42], 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 [24, 42]). 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 [29, 33] 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 [4, 5, 7, 8, 10, 13–15, 18, 21, 39]. On the other hand, the obtained results may have potential applications in the cryptography to build pseudo-random codes (see [2, 3, 37, 45]). We point out that some p-adic chaotic dynamical systems have been studied in [9, 45].

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. [6, 22]

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. [40]

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 [35] 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 [36, 38]. In our further investigations, we only need the following special case of that equation.

Theorem 2.8. [36]

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. [9]

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

2. (2_η) 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 [38], 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 [36], 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 [36, 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 [38], 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.