Lyapunov Dimension

much easier to calculate than box counting, comes for free with Lyapunov spectrum.

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Let $\vec{f}$ be a map in ${\mathbb{R}}^m$ with $m\ge 2$. Consider an orbit with Lyapunov Exponent $h_1\ge h_2\ge \cdots \ge h_m$ and let $p$ be the largest integer such that:

$$\sum _{i=1}^p{h}_i\ge 0$$

The Lyapunov Dimension $D_L$ of the orbit is defined to be:

• $0$ if $p$ doesnt exist ( all directions contracting ). this would happen in the basin of a sink
• $p+\frac{1}{|h_{p+1}|}{\sum }_{i=1}^p{h}_i$ if $<m$
• $m$ if $p=m$

For the skinny Bakers Map, $h_1=\ln 2$, $h_2=-\ln 3$, so $h_1>0>h_2$ and $h_1+h_2<0\Rightarrow p=1$

$$D_L=1+\frac{h_1}{|h_2|}=1+\underset{\text{shrinking x}}{\underbrace{\frac{\ln 2}{\ln 3}}}$$

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The area after $K$ iterates is $d^2{e}^{K(h_1+h_2)}$. Assuming $h_1>0>h_2$ and $h_1+h_2<0$, area goes to $0$ as $K\to \infty$. But the dimensions is at least $1$ becuase $h_1>0$.

Area:

d^2e^{K(h_{1}+h_{2})}=\underbrace{ (de^{Kh_{2}})^2 }_{ \text{area of single box} } \underbrace{ \frac{de^{Kh_{1}}}{de^{Kh_{2}}} }_{ \begin{array} f\text{number of boxes of} \\ \text{side length } de^{Kh_{2}} \\ \text{required to cover set} \end{array} }

BCD:

\begin{align*} \lim_{ G \to 0 } \frac{\ln(N)}{\ln\left( \frac{1}{\epsilon} \right)} &= \lim_{ \epsilon \to 0 } \frac{-\ln N}{\ln \epsilon} \\ &= \frac{-\ln ( N(d)e^{K(h_{1}-h_{2})})}{\ln (de^{Kh_{2}})} \\ &= \frac{-\ln(N(d)) - K(h_{1}-h_{2})}{\ln d + Kh_{2}} \quad \text{in lim as }\epsilon\to 0 \end{align*} \quad \quad \begin{array} . N(d) = \text{number of boxes} \\ \text{of size d} \\ \text{I started with} \end{array}

expression reduced to:

$$\frac{K(h_2-h_1)}{K{h}_2}\text{because }N,d\text{ is negligible as }K\to \infty$$ $$\frac{h_2-h_1}{h_2}=1-\frac{h_1}{h_2}$$

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Fixed Point Theorem 2

Theorem

Let $\vec{f}$ be a continuous map on ${\mathbb{R}}^2$ and $S$ be a rectangular region such that as the boundary of $S$ is traversed, the net rotation of the vectors:

$$\vec{v}(\vec{x})=\vec{f}(\vec{x})-\vec{x}$$

is nonzero. Then $\vec{f}$ has a Fixed Point in $S$

To do so, there must come a point ${\vec{x}}^{\ast }$ as we shrink $S$ such that $\vec{v}(\vec{x})$