Convergence Time and Phase Transition in a Non-monotonic Family of Probabilistic Cellular Automata

Abstract

In dynamical systems, some of the most important questions are related to phase transitions and convergence time. We consider a one-dimensional probabilistic cellular automaton where their components assume two possible states, zero and one, and interact with their two nearest neighbors at each time step. Under the local interaction, if the component is in the same state as its two neighbors, it does not change its state. In the other cases, a component in state zero turns into a one with probability $\alpha ,$ and a component in state one turns into a zero with probability $1-\beta$. For certain values of $\alpha$ and $\beta$, we show that the process will always converge weakly to $\delta _{0},$ the measure concentrated on the configuration where all the components are zeros. Moreover, the mean time of this convergence is finite, and we describe an upper bound in this case, which is a linear function of the initial distribution. We also demonstrate an application of our results to the percolation PCA. Finally, we use mean-field approximation and Monte Carlo simulations to show coexistence of three distinct behaviours for some values of parameters $\alpha$ and $\beta$.

Appendix

1.1 Transitions of $(i_t,j_t)$ and Process X

Now, we will study the random variables $i_t$ and $j_t$ (see Fig. 2), starting by its probabilities transition. Let $a,~b,~c,~d\in \mathbb N$. We denote by

$$\begin{aligned} \mathbb P((a,b)|(c,d))=\mathbb P((i_t=a,j_t=b)|(i_{t-1}=c,j_{t-1}=d)). \end{aligned}$$

$$\begin{aligned} \begin{array}{lll} \mathbb P((a,b)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 1&{}\hbox { if }b=a+1;\\ \mathbb P(L_2\cap R_2)&{}\hbox { in other cases. } \end{array} \right. \\ \mathbb P((a+1,b+1)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{} \hbox { if }b=a+1;\\ \mathbb P(R_1\cap E_1)&{}\hbox { if }b=a+2;\\ \mathbb P(L_1\cap R_3)&{} \hbox { otherwise. } \end{array} \right. \\ \mathbb P((a-1,b-1)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{}\hbox { if }b=a+1;\\ \mathbb P(L_1\cap E_2)&{}\hbox { if }b=a+2;\\ \mathbb P(L_3\cap R_1)&{} \hbox { otherwise. } \end{array} \right. \\ \mathbb P((a,b-1)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{}\hbox { if }b=a+1;\\ \mathbb P(L_2\cap R_1)&{}\hbox { if }b>a+2. \end{array} \right. \\ \mathbb P((a+1,b)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{} \hbox { if }b=a+1;\\ \mathbb P(L_1\cap R_2)&{} \hbox { if }b>a+2. \end{array} \right. \\ \mathbb P((a+1,b-1)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{}\hbox { if }b\le a+2;\\ \mathbb P(L_1\cap R_1)&{}\hbox { otherwise. } \end{array} \right. \\ \mathbb P((a,b+1)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{} \hbox { if }b=a+1;\\ \mathbb P(L_2\cap R_3)&{} \hbox { otherwise. } \end{array} \right. \\ \mathbb P((a-1,b)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{} \hbox { if }b=a+1;\\ \mathbb P(L_3\cap R_2)&{} \hbox { otherwise. } \end{array} \right. \\ \mathbb P((a-1,b+1)|(a,b))&{}=&{} \left\{ \begin{array}{ll} 0&{} \hbox { if }b=a+1;\\ \mathbb P(L_3\cap R_3)&{} \hbox { otherwise. } \end{array} \right. \end{array} \end{aligned}$$

On $\mathbb P((a,b-1)|(a,b))$ and $\mathbb P((a+1,b)|(a,b)),$ we do not indicate what happened when $b=a+2,$ because in this case just on the position $a+1$ we get the component on state 1 and in all the other components the state is 0. Moreover, in this case $\{(a, b-1)| (a, b)\}= \{(a+1, b)|(a, b)\},$ hence we reach the configuration “all zeros” and $\mathbb P((a+1,b)|(a,b))=\mathbb P(R_1\cap L_1).$

For $\mathbb P((a,b-1)|(a,b)),~\mathbb P((a+1,b)|(a,b))$ (resp. $\mathbb P((a+1,b+1)|(a,b)),~\mathbb P((a-1,b-1)|(a,b))$ ) if $b> a+2$ (resp. $b=a+2$) then $j_t-i_t-1$ describes the length of the island of ones which with non-null probability can: increase by one, increase by two, decrease by one, decrease by two, or stay the same.

Now, we define process X : 

$$\begin{aligned} \begin{array}{ll} \mathbb P(X_t=c|X_{t-1}=c)&{}=\mathbb P((a,b)|(a,b))+\mathbb P((a-1,b-1)|(a,b))\\ &{}~~ +\mathbb P((a+1,b+1)|(a,b));\\ \mathbb P(X_t=c-1|X_{t-1}=c)&{}= \left\{ \begin{array}{ll} \mathbb P((a+1,b)|(a,b)) +\mathbb P((a,b-1)|(a,b)) &{}\hbox { if }c>1;\\ \mathbb P((a,b-1)|(a,b))&{}\hbox { if }c=1. \end{array} \right. \\ \mathbb P(X_t=c-2|X_{t-1}=c)&{}=\mathbb P((a+1,b-1)|(a,b));\\ \mathbb P(X_t=c+1|X_{t-1}=c)&{}=\mathbb P((a-1,b)|(a,b))+ \mathbb P((a,b+1)|(a,b));\\ \mathbb P(X_t=c+2|X_{t-1}=c)&{}=\mathbb P((a-1,b+1)|(a,b)), \end{array} \end{aligned}$$

where $c=b-a-1$. Thus, we concluded the task to define the process X associated to G.

1.2 Proof of Lemmas 7 and 8

Proof of lemma 7

Item (a) results directly from the difference equation (10). Now, let us prove item (b); before we do that, several observations need to be established:

• (1) If $\alpha \in (0,1)$ and $\beta \le \phi (\alpha ),$ then $\gamma _1\ge 1$. Moreover, $\gamma _1= 1$ when $\beta = \phi (\alpha ).$

• (2) If $\alpha \in (0,1/3],$ then $\gamma _2< -1.$

• (3) If $\alpha \in (1/3,1)$ and $\beta \ge (\alpha -1)/(3\alpha -1),$ then $\gamma _2\le -1$. Moreover, $\gamma _2=-1$ when $\beta = (\alpha -1)/(3\alpha -1).$

As $(\alpha -1)/(3\alpha -1)$ is negative for $\alpha \in (1/3,1)$ and $\beta$ is non-negative, we can conclude that if $\alpha \in (1/3,1),$ then $\gamma _2< -1.$ Thus, (2) and (3) can be concatenated in the following way:

(2’) if $\alpha \in (0,1)$ then $\gamma _2< -1.$

Let us consider the case $\beta =\phi (\alpha )$. So, $\gamma _1=1,$ which implies that $h_i=1+C_2((\gamma _2)^{i}-1).$ But in that case, $\gamma _2<-1$. Thus, we get $|h_i|\rightarrow \infty$. However, $h_i\in [0,1]$ for all $i\in \mathbb N$ which will be satisfied just for $C_2=0.$

From (1) and (2’), if $\alpha \in (0,1)$ and $\beta < \phi (\alpha ),$ then

$$\begin{aligned} \lim _{i\rightarrow \infty }C_j((\gamma _j)^{2i}-1)= \left\{ \begin{array}{l} ~~\infty \quad \,\,\hbox { if } \quad C_j>0;\\ -\infty \quad \hbox { if } \quad C_j<0. \end{array} \right. \end{aligned}$$

(19)

$C_1$ and $C_2$ can not have the same sign. Because if $C_1$ and $C_2$ are both positive(respectively both are negative), then using the limit (19) $C_1((\gamma _1)^{2i}-1)$ and $C_2((\gamma _2)^{2i}-1)$ goes to infinity (respectively $-\infty$) when $i\rightarrow \infty .$ So $h_{2i}\rightarrow \infty$ (respectively $-\infty$) with $i\rightarrow \infty$. But $h_i\in [0,1]$ for all $i\in \mathbb N$, what will be satisfied only for $C_1=C_2=0.$

$C_1$ and $C_2$ can not have different signals. Because if $C_1>0$ and $C_2<0$(respectively $C_1<0$ and $C_2>0$), then using the limit (19) $C_1((\gamma _1)^{2i-1}-1)$ and $C_2((\gamma _2)^{2i-1}-1)$ goes to infinity (respectively $-\infty$) when $i\rightarrow \infty .$ So $h_{2i-1}\rightarrow \infty$ with $i\rightarrow \infty$ but $h_i\in [0,1]$ for all $i\in \mathbb N$, what will be satisfied just for $C_1=C_2=0.$

The case when $C_1=0$ and $C_2$ is different from zero (respectively $C_2=0$ and $C_2\not =0$) is trivial. $\square$

Lemma 9

Given real values $C_1,~C_2,~a$ and b where $b<-a\le -1$:

• (a) If $C_1$ and $C_2$ are negative, then $C_1 a^{2i}+C_2 b^{2i}\le C_1a^{2i}.$

• (b) If $C_1<0$ and $C_2>0,$ then $C_1 a^{2i-1}+C_2 b^{2i-1}\le C_1 a^{2i-1}.$

• (c) If $C_1$ and $C_2$ are positive, then

  $$\begin{aligned} C_1 a^{2i-1}+C_2 b^{2i-1}\rightarrow -\infty \hbox { and } \frac{2i-1}{C_1 a^{2i-1}+C_2 b^{2i-1}}\rightarrow 0\hbox { when }i\rightarrow \infty . \end{aligned}$$

• (d) If $C_1>0$ and $C_2<0,$ then

  $$\begin{aligned} C_1 a^{2i}+C_2 b^{2i}\rightarrow -\infty \hbox { and } \frac{2i}{C_1 a^{2i}+C_2 b^{2i}}\rightarrow 0\hbox { when }i\rightarrow \infty . \end{aligned}$$

Proof

Items (a) and (b) are trivial. The case $a=1$ is trivial; therefore, we shall consider when $a>1.$ Next, let us prove item (c). Note that if $1<-b/a,$ then $(-b/a)^{2i-1}\rightarrow \infty$ when $i\rightarrow \infty .$ Also, $a^{2i-1}\rightarrow \infty$ when $i\rightarrow \infty .$ So, for each negative M value, there is a natural value $i_M$ such that $i>i_M$ implies that

$$\begin{aligned} \frac{C_1}{C_2}<\frac{M}{C_2a^{2i-1}}+\left( \frac{-b}{a}\right) ^{2i-1}. \end{aligned}$$

(20)

Thus, ${C_1a^{2i-1}}<M-C_2b^{2i-1}.$ We proved the first part of item (c). Now, let us prove the second part. For this task, it is sufficient to prove

$$\begin{aligned} \frac{C_1 a^{2i-1}+C_2 b^{2i-1}}{2i-1}\rightarrow -\infty \hbox { when }i\rightarrow \infty . \end{aligned}$$

This is what we shall do. We get $(-b/a)^{2i-1}\rightarrow \infty$ and $a^{2i-1}\rightarrow \infty$ when $i\rightarrow \infty .$ So, for given M value, $M(2i-1)/a^{2i-1}\rightarrow 0$ when $i\rightarrow \infty .$ Therefore, for each negative M value, there is a natural value $i_M,$ such that $i>i_M$ implying that

$$\begin{aligned} C_1<\frac{M(2i-1)}{a^{2i-1}}+C_2\left( \frac{-b}{a}\right) ^{2i-1}. \end{aligned}$$

Thus, $C_1a^{2i-1}<M(2i-1)-C_2b^{2i-1}.$ Item (c) is proved. $\square$

Let us prove item (d). Note that $1<b^2/a^2,$ then $(b/a)^{2i}\rightarrow \infty$ when $i\rightarrow \infty .$ Also, $a^{2i}\rightarrow \infty$ when $i\rightarrow \infty .$ So, for each negative M value, there is a natural value $i_M$ such that $i>i_M,$ implying that

$$\begin{aligned} C_1<\frac{M}{a^{2i}}-C_2\left( \frac{b}{a}\right) ^{2i} \end{aligned}$$

Thus, ${C_1a^{2i}}<M-C_2b^{2i}.$ We have now finished the first part of item (d). Now, let us prove the second part, which essentially proves that

$$\begin{aligned} \frac{C_1 a^{2i}+C_2 b^{2i}}{2i}\rightarrow -\infty \hbox { when }i\rightarrow \infty . \end{aligned}$$

Once $b<-a\le -1,$ we get that for each negative M value there is natural value $i_M$ such that $i>i_M,$ implying that

$$\begin{aligned} C_1<\frac{M(2i)}{a^{2i}}-C_2\left( \frac{b}{a}\right) ^{2i}. \end{aligned}$$

Thus, $C_1a^{2i}<M(2i)-C_2b^{2i-1}.$

Proof of Lemma 8

Item (a) results directly from the difference equation (11). Here, we need to observe: (4) if $\alpha ,\beta \in (0,1),$ then $-1\ge -\gamma _1>\gamma _2 .$

If we consider $C_1$ and $C_2$ negative, then by Lemma 9 item (a), we get

$$\begin{aligned} k_{2i}^Y\le \frac{2i}{\eta }-C_1-C_2+C_1(\gamma _1)^{2i}. \end{aligned}$$

Moreover,

$$\begin{aligned} C_1(\gamma _1)^{2i}\rightarrow -\infty \hbox { and } \frac{2i}{\eta C_1(\gamma _1)^{2i}}\rightarrow 0, \hbox { when }i\rightarrow \infty , \end{aligned}$$

i.e., there is $i_0$ such that $i>i_0,$ implying that $k_{2i}^Y<0$ is impossible. Then, $C_1$ and $C_2$ can not be simultaneously negative.

If we consider $C_1<0$ and $C_2>0,$ then by Lemma 9 item (b), we get

$$\begin{aligned} k_{2i-1}^Y\le \frac{2i-1}{\eta }-C_1-C_2+C_1(\gamma _1)^{2i-1}. \end{aligned}$$

Moreover

$$\begin{aligned} C_1(\gamma _1)^{2i-1}\rightarrow -\infty \hbox { and } \frac{2i-1}{\eta C_1(\gamma _1)^{2i-1}}\rightarrow 0, \hbox { when }i\rightarrow \infty , \end{aligned}$$

i.e., there is $i_0$ such that $i>i_0,$ implying that $k_{2i-1}^Y<0$ is impossible. So, the case $C_1<0$ and $C_2>0$ is impossible.

If we consider $C_1=0$ and $C_2\not =0,$ then for $C_2>0,~$ $C_2(\gamma _2)^{2i-1}\rightarrow -\infty$ when $i\rightarrow \infty$ and for $C_2<0,~$ $C_2(\gamma _2)^{2i}\rightarrow -\infty$ when $i\rightarrow \infty$. In both cases, there is $i_0$ such that $i>i_0,$ implying that $k_{2i-1}^Y<0$ in the first case or $k_{2i}^Y<0$ in the second case. Both cases are impossible. Then, the case $C_1=0$ and $C_2\not =0$ is impossible.

If we consider $C_2=0$ and $C_1<0,$ then $C_1(\gamma _1)^{i}\rightarrow -\infty$ when $i\rightarrow \infty .$ So, there is $i_0$ such that $i>i_0$ implying that $k_i^Y<0$ is impossible. Then, the case $C_2=0$ and $C_1<0$ is impossible.

If we consider $C_1$ and $C_2$ positive, then by Lemma 9 item (c) we conclude that

$$\begin{aligned} C_1((\gamma _1)^i-1)+C_2((\gamma _2)^i-1)\rightarrow -\infty \hbox { and }\frac{i}{C_1((\gamma _1)^i-1)+C_2((\gamma _2)^i-1)}\rightarrow 0 \hbox { as }i\rightarrow \infty . \end{aligned}$$

Thus, there is $i_0$ such that $i>i_0,$ implying that $k_{2i-1}^Y<0$ is impossible. Then, the case of $C_1$ and $C_2$ being positive is impossible.

Analogously to the previous case, $C_1$ and $C_2$ positive, if we consider $C_1>0$ and $C_2<0,$ then by Lemma 9 item (d) we conclude that there is $i_0$ such that $i>i_0,$ implies that $k_{2i}^Y<0$ is impossible. Then, the case $C_1>0$ and $C_2<0$ is impossible.

Until this point, we proved that $C_2=0$ and $C_1\ge 0.$ Thus,

$$\begin{aligned} k_i^Y=\frac{i}{\eta }+C_1((\gamma _1)^i-1) \hbox { for }i\in \mathbb N. \end{aligned}$$

However, we get $(\gamma _1)^i>1\Rightarrow C_1((\gamma _1)^i-1))\ge 0\Rightarrow k_i^Y\ge i/\eta > 0$ when $\beta < \phi (\alpha )$. By theorem 1.3.5 in [22], the mean hitting time is the minimal non-negative solution to system (11). This happens in our case when $C_1=0.$ $\square$