Computing Lyapunov Exponents

$$J_n=\mathcal{D}{\vec{f}}^n({\vec{v}}_0)$$

is typically difficult to write down for large $n$, so we are forced to calculate the ellipsoid $J_{n}U$ on ${\mathbb{R}}^m$ on the computer.

However, $J_{n}U$ has axes of length $r_k^n$ where $(r_k^n{)}^2$ are the eigenvalues of $J_n{J}_n^{⊺}$. These are difficult to calculate.

Large $n\to ellipsoid$ is quite long in some directions, and quite thin on others. i.e., $r_1^n>>r_m^n$ and $J_n{J}_n^{⊺}$ is ill-conditioned.

Fixing This with Chain Rule

$$J_{n}U=\mathcal{D}{\vec{f}}^n({\vec{v}}_0)U=\mathcal{D}\vec{f}({\vec{v}}_{n-1})\cdot \mathcal{D}\vec{f}({\vec{v}}_{n-2})\dots \mathcal{D}\vec{f}({\vec{v}}_0)$$

Superscript now indexes iterate, subscripts index direction of decreasing expansion.

Take a unit ball in ${\mathbb{R}}^m$ defined by an orthonormal basis:

$$\{{\hat{w}}_1^0,{\hat{w}}_2^0,\dots ,{\hat{w}}_m^0\}$$

around ${\vec{v}}_0$ and compute:

$$\begin{array}{rl}{\vec{z}}_1^0& =\mathcal{D}\vec{f}({\vec{v}}_0){\hat{w}}_1^0\\ {\vec{z}}_2^0& =\mathcal{D}\vec{f}({\vec{v}}_0){\hat{w}}_2^0\\ & ⋮\\ {\vec{z}}_m^0& =\mathcal{D}\vec{f}({\vec{v}}_0){\hat{w}}_m^0\end{array}$$

${\vec{z}}_i$ lie on surface of the ellipsoid $\mathcal{D}\vec{f}({\vec{v}}_0)U$, not necessarily Orthogonal.

To make Orthogonal vectors on an ellipsoid with the same volume as $\mathcal{D}\vec{f}({\vec{v}}_0)U$, we use Gram-Schmidt Process.

$$\begin{array}{rl}{\vec{y}}_1^1& ={\vec{z}}_1^0\\ {\vec{y}}_2^1& ={\vec{z}}_2^0-{\vec{z}}_2^0\cdot \left(\frac{{\vec{y}}_1^1}{||{\vec{y}}_1^1||}\right)\cdot \left(\frac{{\vec{y}}_1^1}{||{\vec{y}}_1^1||}\right)\end{array}$$

This leaves us wtih ${\vec{y}}_j^1$ as the component of ${\vec{z}}_j^0$ perpendicular to ${\vec{y}}_1^1,{\vec{y}}_2^1,\dots {\vec{y}}_{j-1}^1$ and the resulting set of vectors

$${\hat{w}}_1^1=\frac{{\vec{y}}_1^1}{||{\vec{y}}_1^1||}{\hat{w}}_2^1=\frac{{\vec{y}}_2^1}{||{\vec{y}}_2^1||}\dots {\hat{w}}_m^1=\frac{{\vec{y}}_m^1}{||{\vec{y}}_m^1||}$$

$\hat{w}$ are orthonormal vectors in the expanding and contracting directions around $v_0$ and $||{\vec{y}}_i^1||$ measure the one step growth rate in direction $i$.

Next, we apply $\mathcal{D}\vec{f}({\vec{v}}_1)$ to the ${\hat{w}}_i^1$ basis:

$$\begin{array}{rl}{\vec{z}}_1^1& =\mathcal{D}\vec{f}({\vec{v}}_1){\hat{w}}_1^1\\ {\vec{z}}_2^1& =\mathcal{D}\vec{f}({\vec{v}}_1){\hat{w}}_2^1\end{array}$$

Repeat: in the limit of this process:

$$r_i^{n}Z||{\hat{y}}_i^n||\cdot ||{\hat{y}}_i^{n-1}||\dots ||{\vec{y}}_i^1||$$

and the $i^{th}$ largest Lyapunov Exponent after $n$ steps is:

$$\frac{1}{N}\sum _{j=1}^N\ln ||{\vec{y}}_i^j||$$