📂Dynamics

Definition of Lyapunov Spectrum

Definition

Given a space $X = \mathbb{R}^{n}$ and a function $f : X \to X$, suppose that the following vector field is given by a differential equation. $$ \dot{x} = f(x) $$

Simple Definition

The Lyapunov numbers and Lyapunov exponents for a multi-dimensional map with respect to the time-$1$ map of the flow $F_{T} (v)$ are each defined as the Lyapunov number and Lyapunov exponent of $F_{T} (v)$¹.

Difficult Definition

Variational Equation: With respect to the Jacobian matrix $J$ of $f$, the following is called the variational equation. $$ \dot{Y} = J Y $$ Here, the initial condition of the matrix function $Y = Y(t) \in \mathbb{R}^{n \times n}$ is set as the identity matrix $Y(0) = I$. … Geometrically, $Y$ can be thought of as showing how the tangent vector itself acts while undergoing a small perturbation in the original system’s $x(0)$, transforming into $x(t)$.

$$ \lambda_{k} := \lim_{t \to \infty} \log \left[ \left( {\frac{ \left\| Y(t, v) v \right\| }{ \left\| v \right\| }} \right)^{1/t} \right] $$ The $\left\{ \lambda_{1} , \cdots , \lambda_{n} \right\}$ defined as above is called the Lyapunov spectrum, $$ \Lambda_{v} := \lim_{t \to \infty} \left[ Y(t)^{\ast} Y(t) \right]^{1/2t} $$ or alternatively, the Lyapunov spectrum is defined as the log of the eigenvalues $\mu_{1} , \cdots , \mu_{n}$ of the matrix $\Lambda_{v}$, as defined above².

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• $Y^{\ast}$ is the transpose conjugate matrix of the matrix $Y$.

Explanation

In fact, neither definition is particularly simple, and understanding and handling the concept of the Lyapunov spectrum in continuous systems is not an easy task.

Similar to the Lyapunov number in a one-dimensional map, the motivation for the Lyapunov spectrum stems from the desire to express the difference after some time between $\delta_{0}$ and $t$, resulting from the small difference $x_{0}$ and $x_{0} + \delta_{0}$, with respect to some $\lambda$. $$ \left| \delta_{t} \right| \approx \left| \delta_{0} \right| e^{t \lambda} $$ If we assume that an operator denoted $T_{N}$ acts as a map on $T_{N} : v_{N} \mapsto v_{N+1}$ at the point in time $t = N$, then the geometric mean of the rate at which $T_{N}$ expands or contracts the space is as follows. $$ \left( {\frac{ \left\| T_{1} v \right\| }{ \left\| v \right\| }} \cdot {\frac{ \left\| T_{2} v \right\| }{ \left\| T_{1} v \right\| }} \cdot \cdots \cdot \cdot {\frac{ \left\| T_{N} v \right\| }{ \left\| T_{N-1} v \right\| }} \right)^{1/N} = \left( {\frac{ \left\| T_{N} v \right\| }{ \left\| v \right\| }} \right)^{1/N} $$ In continuous systems, the $Y$ of the variational equation plays the role of $T_{n}$, and consequently, the $k$-th Lyapunov exponent $\lambda_{k}$ regarding the orthonormal set $\left\{ v_{1} , \cdots , v_{n} \right\}$ and $v_{k}$, is defined as follows³.

$$ \begin{align*} \lambda_{k} :=& \lim_{t \to \infty} \log \left[ \left( {\frac{ \left\| Y(t, v) v \right\| }{ \left\| v \right\| }} \right)^{1/t} \right] \\ =& \lim_{t \to \infty} {\frac{ 1 }{ t }} \log \left( \left\| Y(t, v) v \right\| - \left\| v \right\| \right) \\ =& \lim_{t \to \infty} {\frac{ 1 }{ t }} \log \left\| Y(t, v) v \right\| \end{align*} $$

Meanwhile, according to the singular value decomposition $Y = U \Sigma V^{\ast}$ of $Y(t,u)$, the $k$-th singular value $\sigma_{k} (t)$ can be expressed in relation to the $k$-th column vector $u_{k}, v_{k}$ of $U, V$ as follows⁴. $$ Y V = U \Sigma \implies Y v_{k} = \sigma_{k} (t) u_{k} $$

This provides a clue that the concept corresponding to the singular values, i.e., eigenvalues of $Y$ is related to the Lyapunov spectrum. In fact, by multiplying $Y^{\ast} = V \Sigma^{\ast} U^{\ast}$ on the left side of $Y$, $$ Y^{\ast} Y = V \Sigma^{2} V^{\ast} $$ one can readily surmise that the eigenvalue is $\sigma_{k}^{2} (t)$, $$ \Lambda_{v} := \lim_{t \to \infty} \left[ Y(t)^{\ast} Y(t) \right]^{1/2t} $$ and taking the logarithm of the eigenvalues $\mu_{k}$ of the matrix $\Lambda_{v}$ defined as such is the Lyapunov spectrum. When connected with the previously mentioned $Y v_{k} = \sigma_{k} (t) u_{k}$, $$ \begin{align*} \log \mu_{k} =& \log \lim_{t \to \infty} \left[ \sigma_{k}^{2} (t) \right]^{1/2t} \\ =& \lim_{t \to \infty} \log \left[ \sigma_{k} (t) \right]^{1/t} \\ =& \lim_{t \to \infty} {\frac{ 1 }{ t }} \log \sigma_{k} (t) \\ =& \lim_{t \to \infty} {\frac{ 1 }{ t }} \log \left\| \sigma_{k} (t) u \right\| \\ =& \lim_{t \to \infty} {\frac{ 1 }{ t }} \log \left\| Y(t, v) v \right\| \\ =& \lambda_{k} \end{align*} $$ one can intuitively accept that the two definitions are equivalent.