📖 Definition: Chaos

• Long term aperiodic behaviour
• Deterministic system
• Sensitive dependence on initial conditions (SIC)

🗒️ Note: SIC means that neighbouring orbits separate exponentially fast on average

⚽ Goal: lets try to measure the sensitivity on initial conditions of a map

• Consider nearby initial condition $x_0$ and $x_0+\delta _0$

  $$ \left . \begin{aligned} x_n&=f^{(n)}(x_0) \\ x_n+\delta_n &=f^{(n)}(x_0+\delta _0)

  \;\; \end{aligned} \right \} \quad \Rightarrow \quad \delta_n=f^{(n)}(x_0 +\delta _0)-f^{(n)}(x_0) $$

  where $\delta_n$ described for each $n$ the distance between two orbitals

• Ansatz: $|\delta _n|\simeq |\delta_0|e^{n\lambda}$ where $\lambda$ is the Lyapunov exponent

  $$ \begin{aligned} \lambda &=\lim_{\delta_0\to 0}\frac{1}{n} \ln \left | \frac{\delta_n}{\delta_0} \right | \\ &=\lim_{\delta_0 \to 0}\frac 1n \ln \left | \frac{f^{(n)}(x_0+\delta_0)-f^{(n)}(x_0)}{\delta_0} \right | \\ &=\frac 1n \ln|(f^{(n)})'(x_0)|

  \end{aligned} $$

  Looking at the inside term and using the chain rule we get

  $$ (f^{(n)})'(x_0)=\prod^{n-1}_{i=0} f'(x_i) $$

  • Expand for details as to why this works

  Hence we can write

  $$ \lambda \simeq \frac{1}{n}\ln \left | \prod^{n-1}_{i=0} f'(x_i)\right | $$

• The limit of this expression as $n\to \infin$ is the Lyapunov exponent for the orbit starting at $x_0$

  $$ \begin{aligned} \lambda &= \lim_{n\to \infin} \frac{1}{n} \ln \left | \prod _{i=0}^{n-1}f'(x_i) \right | \\

  &=\lim_{n\to \infin} \frac 1n \sum^{n-1}_{i=0} \ln |f'(x_i) | \end{aligned} $$

  🗒️ Note: $\lambda$ depends on $x_0$, it takes the same value for all $x_0$ in the attraction region of a given attractor

  • $\lambda<0$ : stable fixed points and cycles
  • $\lambda >0$ : Chaotic attractors

💃 Example: Suppose $f$ has a stable $p$-cycle containing the point $x_0$, Show that the Lyapunov exponent $\lambda <0.$ If the cycle is super stable show that $\lambda \to -\infin$

1. From the question we know

   • $x_0$ is an element of the $p$-cycle of $f$ so $x_0$ is a fixed point of $f^{(p)}$
   • The cycle is stable so $|(f^{(p)})'(x_0)|<1\Rightarrow \ln|(f^{(p)})'(x_0)|<0$

2. $p$ term keeps appearing in the infinite sum so we can re-write $\lambda$ as

   $$ \lambda = \lim_{n\to \infin} \left [ \frac 1n \sum^{n-1}{i=0} \ln \left | f'(x_i) \right | \right ]= \frac 1p \sum^{p-1}{i=0} \ln |f'(x_i )| $$

3. Now we use our second condition to to rewrite it as

   $$ \lambda = \frac 1p\sum^{p-1}_{i=0}\ln |f'(x_i)|=\frac 1p \ln |(f^{(p)})'(x_0)|<0 $$

   This is the first part of the question completed

4. For the second part we care about the cycle being super stable ie $|(f^{(p)})'(x_0)|=0$ thus

   $$ \lambda = \frac{1}{p}\ln(0)\to -\infin $$

   Second part of question completed

Numerical computation of the Lyapunov exponent

🖥️ Algorithm

1. Choose $x_0$
2. Calculate $N\gg 1$ iterates ($N\sim 10^4$)
3. Neglect the transients before the the iterations reach the attractor ($n_T\sim300$)
4. Calculate your exponent $\lambda = \frac 1n \sum^{N}_{n=n_T}\ln |f'(x_n)|$
5. Update $x_0$ and repeat

🗒️ Note: while this looks and is very computational intense modern computers can easy handle it