Please let me first describe the general background.

The state of the system at time $t$ will be described by a scalar or phase $u=u^t$. Both $t$ and $u$ are discrete. $u$ take values from the fixed set $\mathfrak{S}=\left\{0,1,2,\ldots,e,e+1,e+2,\ldots,e+r\right\}$ of nonnegative integers and $t\geq 0$. Elements of the subset $\left\{1,2,\ldots,e\right\}$ will be identified as the "excited" states, elements of the subset $\left\{e+1,e+2,\ldots,e+r\right\}$ as the "refractory" states, and the singleton $\left\{0\right\}$ as the "rest" or "equilibrium" state. Assume throughout that $r\geq e$.

The dynamics of the model are specified by the rule $$ u^{t+1}=\mathfrak{E}(u^t), $$ where $$ \mathfrak{E}(k)=k+1,~1\leq k\leq e+r-1 $$ and $$ \mathfrak{E}(e+r)=\mathfrak{E}(0)=0. $$ Now we want to put the neighbour cells into consideration. In one-dimension this is done as follows: $$ u_k^{t+1}=\mathfrak{E}(u_k^t)+D(u_k^t; u_{k+1}^t, u_{k-1}^t),~~~(*) $$ where $$ D(u;v_1,v_2)=\begin{cases}1, & \text{if }u=0\text{ and }1\leq v_i\leq e\text{ for some }i=1,2\\0, & \text{otherwise}\end{cases}. $$ That is, if a cell is in rest and at least one of its two neighbors is excited, then the cell becomes excited.

--- Now consider the following special case.

Let $r+2\leqslant L\leqslant e+r$ and consider the initial conditions $$ u_k^0=k,~~0\leqslant k\leqslant L-1~~~(**) $$ with periodic boundary conditions, that is, we take the neighbours of cell $0$ to be cell $1$ and cell $L-1$ and the neighbours of cell $L-1$ to be cell $0$ and cell $L-2$. Apply the rule $(*)$ on this.

(The solution $u_k^t$, corresponding to the initial data $(**)$ has been dubbed a "caterpillar-wave" because the widths of the excited and refractory regions alternately expand and contract.)

Now the question is, which topological entropy this example has.

I looked at an example, namely $L=5, e=r=3$. What I get for the evolution is $$ 01234\\ 12345\\ 23456\\ 34560\\ 45601\\ 56012\\ 60123\\ 01234\\ $$ and so on, i.e. period $7$.

My idea therefore is to use that topological entropy is an invariant of conjugation.

Consider $$ Y\subset\left\{0,1,2,\ldots,e+r\right\}^{\mathbb{Z}}, $$ consisting of the sequences where

  • k has k+1 to its right for $0\leq k\leq e+r-1$
  • e+r has 0 to its right.

On $Y$ consider the left-shift $\sigma$.

As far as I see there is a homeomorphism $\varphi$ from $Y$ to the configurations $u_k^0, u_k^1,\ldots$ on $[0,\ldots L-1]$.

So that the topological entropy that I am searching for can be computed as the topological entropy of $(Y,\sigma)$.

But this is known to be $\ln\lambda$, where $\lambda$ here is the biggest (in absolute value) eigenvalue of the $(e+r+1)\times (e+r+1)$-dimensional adjacency-matrix. And if I am not wrong, it is $\lambda=1$.

So my question is, if I am right? Do the "caterpillar waves" indeed have topological entropy $$ \ln 1=0? $$

Am I right or do I think much too easy? I am very unsure!

With greetings

  • $\begingroup$ Which definition of topological entropy you use? The one that I know (from Wikipedia, for example) is not very useful in the context of finite-state spaces (because it is always zero). $\endgroup$ – demitau May 8 '15 at 9:09
  • $\begingroup$ I use the one using open covers (or, equivalently, the Bowen definition). $\endgroup$ – math12 May 8 '15 at 9:10
  • $\begingroup$ Sorry I still do not get it. The only definition of this kind I know is when you take the minimal number of open sets you have to use to cover certain set (depending on $n$), divide it by $n$ and take a limit for large $n$. Here you have a finite number of sets always because you topology is finite. Then the limit is always zero. $\endgroup$ – demitau May 8 '15 at 9:47
  • $\begingroup$ Maybe I missunderstood the task... maybe the task is to compute the top. entropy of another system... I only understood "Compute the top. entropy of the caterpillar waves"... maybe something different was meant... $\endgroup$ – math12 May 8 '15 at 9:54
  • 1
    $\begingroup$ Maybe you want to consider the dynamics not on the configurations themselves but on admissible sequences of configurations. This would make more sense (and it would reflect what people do for Markov shifts). $\endgroup$ – demitau May 8 '15 at 10:29

This is not a complete answer, I am writing here because SE does not like long lines of comments.

If you consider the dynamics on the space of admissible sequences of configurations i.e. on the space of all possible (one-sided) trajectories of your $\mathfrak{C}+D$ map on the space of configurations, then it is indeed corresponds to the shift on a certain subspace $A$ of $\{0,\ldots, e+r \}^\mathbb{N}$.

When you consider only special initial conditions you consider another subspace $B\subset A$. Then indeed you have to compute the adjacency matrix and (if the shift is irreducible) get information about its largest e.v.

So finally after all I think your idea is right but you need to consider one-sided sequences instead of two-sided. If you did your computation of the matrix and its largest e.v. (which I did not check) then this gives your the right answer.

  • $\begingroup$ But how does the space of admissible sequences look like here and what are the initial conditions for $k > L-1$? $\endgroup$ – math12 May 8 '15 at 11:16
  • $\begingroup$ The way you wrote it your caterpillary waves exist on "discrete circles" of length $L$, so the last question I do not understand. $\endgroup$ – demitau May 8 '15 at 11:45
  • $\begingroup$ When I understand you correcly, I have to distinguish between $A\subset\left\{0,\ldots,e+r\right\}^{\mathbb{N}}$ (which is the whole space) and $B\subset A$, where $B$ is the special case I am looking here. But how does B look like? Are this infinite sequences, too? They have, if $B\subset A$. But what are the position $k>L-1$? $\endgroup$ – math12 May 8 '15 at 11:49
  • $\begingroup$ Yes. How it looks like depends on what do authors of the task mean. For example you can consider a finite $r+2 < L_0 \leq e+r$ and consider all the caterpillar initial conditions for all $r+2 \leq L \leq L_0$. Each of these conditions generate a periodic trajectory of length $L+2$ on its own "discrete circle" of length $L$. So you have $L_0 - r - 2$ independent periodic motions, but it means that your space if finite again (and consequently the entropy is $0$ automatically). $\endgroup$ – demitau May 8 '15 at 11:54
  • $\begingroup$ Just a simple question I should have asked in the very beginning. Do you consider a finite lattice or the whole double infinite line of sites such that each of them can take values from $\{ 0, \ldots, e+r\}$? In the first case for caterpillary initial condition you do not have to define anything for $k>L-1$. $\endgroup$ – demitau May 8 '15 at 11:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.