22 Apr 2020 - Harikrishnan N B

Materials

The materials of this lecture is from the 9-th chapter of the following book - Devaney, Robert L., et al. “A first course in chaotic dynamical systems: theory and experiment.” Computers in Physics 7.4 (1993): 416-417.

Chaotic Dynamical Systems

A dynamical system $F$ is chaotic if :
  1. Periodic points for $F$ are dense.
  2. $F$ is transitive.
  3. $F$ depends sensitively on initial conditions.

Now we consider a dynamical system and prove the above three properties to show that the dynamical system is chaotic. Before proving the above three properties, we shall define a dynamical system and some important theorems.

Consider the sequence space $\Sigma = { (s_0, s_1, \ldots) : s_i = 0$ or $1}$.

The sequence $s_0, s_1, \ldots $ is a infinite binary sequence. For eg: $00101101\dots \in \Sigma$.

Properties of $\Sigma$

Distance Metric on $\Sigma$

In order to prove the three properties of chaos we need to define a distance metric. Let $s = (s_0, s_1, \ldots) \in \Sigma, t = (t_0, t_1, \ldots) \in \Sigma, w = (w_0,w_1,\ldots) \in \Sigma$. \(d(s,t) : = \sum_{i = 0}^{\infty}\frac{|s_i - t_i|}{2^{i}}.\) The distance metric satisfies the following three properties:

  1. $d(s,t) \geq 0, d(s,t) = 0$ iff $s=t$.
  2. Commutative: $d(s,t) = d(t,s)$.
  3. Triangular Inequality : $d(s,w) \leq d(s,t) + d(t,w)$.

Minimum and maximum value of $d(s,t)$

The minimum distance of $s$ and $t$ is $d(s,t) = 0$. The maximum distance happens when $s_i \neq t_i \forall i$, Using the distance metric, we have the following: \(d_{max}(s,t) = 1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \ldots.\\ d_{max}(s,t) = \frac{1}{1-\frac{1}{2}} = 2~\text{units}.\)

Proximity Theorem

Let $s,t \in \Sigma$, where $s \neq t$ but $s_i = t_i$ for i = $0,1,\ldots, n$ then the $d(s,t) \leq \frac{1}{2^n}$.

Proof

The distance metric is as follows: \(d(s,t) : = \sum_{i = 0}^{\infty}\frac{|s_i - t_i|}{2^{i}}.\)

In this case, from $i = 0,1,2, \ldots, n$, $s_i = t_i$, Therefore, the first $n+1$ terms will be zero. The remaining terms can be different. So the maximum distance happens when $s_i \neq t_i$, for $i = n+1, n+2, \ldots$. The distance for this case is as follows:

\[d(s,t) = 0 + \frac{1}{2^{n+1}} + \frac{1}{2^{n+2}}+ \ldots,\\ d(s,t) = \frac{1}{2^n}\Biggl(\frac{1}{2} + \frac{1}{2^2}+\ldots \Biggr),\\ d(s,t) = \frac{1}{2^n}\Biggl(\frac{\frac{1}{2}}{1 - \frac{1}{2}}\Biggr),\\ d(s,t) = \frac{1}{2^n}.\]

In general, according to proximity theorem $d(s,t) \leq \frac{1}{2^n}$.

Now we will introduce a map (shift map - $\sigma$ ) on $\Sigma$.

Shift Map ($\sigma$)

$\sigma : \Sigma \rightarrow \Sigma$. The map does a left shift operation as follows: \(\sigma(s_0s_1s_2s_3s_4\ldots) = s_1s_2s_3s_4\ldots \overset{\sigma}\rightarrow s_2s_3s_4\ldots \overset{\sigma}\rightarrow s_3s_4\ldots,\) Now we shall prove the dynamical system ($\Sigma, \sigma$) satisfies the three conditions of chaos and hence it is a chaotic dynamical system.

A set Y is a dense subset of X

Definition : Suppose $X$ is a set and $Y$ is a subset of $X$. We say that $Y$ is dense in $X$ if, for any point $x \in X$, there is a point $y$ in the subset $Y$ arbitrarily close to $x$.

Theorem : Prove that $Y$ which is a set of periodic sequence is dense in $\Sigma$.

Proof:

We consider $X \in \Sigma$ and $Y \subset X$ as the set of all periodic points. For an $s \in \Sigma$ and $\epsilon >0$, find a $t \in Y$, where $Y \subset X$ such that $d(s,t) \leq \epsilon$.

There are two cases in this proof. In case I, the given $s$ is periodic. In case II, the given $s$ is non periodic. We will analyze the two cases.

Case II : Find a $t \in Y$ such that $t$ is periodic. Now choose an $n$ such that $\frac{1}{2^n} < \epsilon$, $n>0$ and $n$ is an integer. Choose $t$ in such a way that the first $n+1$ terms of $t$ is same as the first $n+1$ terms of $s$, (i.e., $t = s_0,s_1,\dots,s_n,s_0,s_1 \ldots$).

By proximity theorem, we have $d(s,t) \leq \frac{1}{2^n} < \epsilon$ (since the first $n+1$ elements of $s$ and $t$ are the same). Therefore from case I and case II, $Y$ is a dense subset of $X$.

Transitivity ($\Sigma, \sigma$)

Definition : A dynamical system is transitive if for each two points $x, y \in \Sigma$ and for $\epsilon > 0$, there exist a $z \in \Sigma$ such that on finite number of iterations, $z$ reaches the $\epsilon$ neighbourhood of $x$ and $y$.

Proof: To prove this we provide a single orbit $s$ which is transitive (“UNIVERSAL ORBIT”). The point $s$ contains binary strings of all length ($s = 0$ $1$ $00$ $01$ $10$ $11$ $000$ $001$ $010$ $011$ $100\ldots$).

Let $x = x_0, x_1,\ldots x_n, \ldots$ and $y = y_0,y_1,\ldots, y_n, \ldots$. Now choose an $n$, such that $\frac{1}{2^n} < \epsilon$, $n>0$ and $n$ is an integer. Since $s$ contains binary strings of all length, so on finite number of iteration s reaches the first $n+1$ sequence of $x$ and $y$. Let $s$ takes $k$ and $r$ iterations to reach $x_0, x_1,\ldots, x_n$ and $y_0, y_1, \ldots y_n$ respectively.

From proximity theorem, we have $d(\sigma^{k}(s),x) \leq \frac{1}{2^{n}} < \epsilon$ and $d(\sigma^{r}(s),y) \leq \frac{1}{2^{n}} < \epsilon$.

There exist $k, r$ such the “UNIVERSAL ORBIT” reaches the $\epsilon$ neighbourhood of $x$ and $y$. Hence ($\Sigma, \sigma$) is transitive.

Example for Transitivity

Let us consider a period two point $x = 010101\ldots = \overline{01}$. We consider a positive integer $n>0$ such that $\frac{1}{2^{n}} < \epsilon$.

Let $s$ contains binary sequence of all lengths. After $k$ - iterations ($k > 0$), we have binary sequences of length $n+1$ (specifically we reach the sequence $01010101\ldots$ Here

$\underbrace{010101 \ldots01}\ldots$

the first $n+1$ elements of $\sigma^{k}(s)$ are the same as the first $n+1$ elements of $x$. Now according to the proximity theorem, we have

$d(\sigma^{k}(s),x) \leq \frac{1}{2^{n}} < \epsilon$. Similarly we can consider a non periodic sequence $y = 01001000100001\ldots$. Let after $r$ iterations over $s$ we reach a sequence $\sigma^{r}(s) = 01001000100001\ldots$, where the first $n+1$ elements of $\sigma^{r}(s)$ and $y$ are the same. From proximity theorem, we have the following:

$d(\sigma^{r}(s),y) \leq \frac{1}{2^{n}} < \epsilon$.

There exist $k, r$ such the “UNIVERSAL ORBIT” reaches the $\epsilon$ neighbourhood of $x$ and $y$. Hence ($\Sigma, \sigma$) is transitive.

Sensitivity

Definition: A dynamical system $F$ is said to be sensitively dependent on initial conditions, if there exist a $\beta > 0$ such that for any $x \in F$ and $\epsilon > 0$, there exist $y \in F$ within the $\epsilon$ of $x$ and there exist $k \in \mathbf{Z^{+}}$ such that $d(\sigma^{k}(x),\sigma^{k}(y)) \geq \beta$.

Proof: Given $s \in \Sigma$, $\beta = 1$ and $\epsilon > 0$.

Choose an $n$ such that $\frac{1}{2^n} \in \epsilon$. Choose $t$ such that $t_i = s_i$ for $i = 0,1,\ldots,n$. From proximity Theorem, the distance $d(t,s) \leq \frac{1}{2^n}$.

Since $s \neq t$, there must be a $k > n$ such that $s_k \neq t_k $. Now compute the distance between $\sigma^{k}(s)$ and $\sigma^{k}(t)$. The distance is as follows:

\[d(\sigma^{k}(s), \sigma^{k}(t)) = \frac{|s_k - t_k|}{2^0} + ~~ \text{remaining terms}, \\ d(\sigma^{k}(s), \sigma^{k}(t)) = 1 + ~~ \text{remaining terms}.\]

This implies $d(\sigma^{k}(s), \sigma^{k}(t)) > 1 (\beta)$. The orbit of $t$ diverges from the orbit of $s$ after $k$ iteration. Hence the dynamical system ($\Sigma, \sigma$) is sensitive to initial conditions and satisfies the three conditions of chaos (Note: In the sensitivity theorem, the orbit of t diverges from orbit of s after k iterations, but it can come back being close again. It diverges for the first time after k iterations). Therefore ($\Sigma, \sigma$) is a chaotic dynamical system.