Document Lyapunov exponents

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Ian Jauslin 2024-02-19 11:41:35 -05:00
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\documentclass{ian} \documentclass{ian}
\usepackage{largearray} \usepackage{largearray}
\usepackage{dsfont}
\begin{document} \begin{document}
@ -86,6 +87,7 @@ and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore,
\partial_t\hat u_k= \partial_t\hat u_k=
-\frac{4\pi^2}{L^2}\nu k^2\hat u_k+\hat g_k -\frac{4\pi^2}{L^2}\nu k^2\hat u_k+\hat g_k
+\frac{4\pi^2}{L^2|k|}T(\hat u,k) +\frac{4\pi^2}{L^2|k|}T(\hat u,k)
=:\mathfrak F_k(\hat u)
\label{ins_k} \label{ins_k}
\end{equation} \end{equation}
with with
@ -375,6 +377,74 @@ The enstrophy is defined as
. .
\end{equation} \end{equation}
\subsubsection{Lyapunov exponents}
\indent
To compute the Lyapunov exponents, we must first compute the Jacobian of $\hat u^{(n)}\mapsto\hat u^{(n+1)}$.
This map is always of Runge-Kutta type, that is,
\begin{equation}
\hat u^{(n+1)}=\hat u^{(n)}+\delta\sum_{i=1}^q w_i\mathfrak F(\hat u^{(n)})
\end{equation}
(see\-~(\ref{ins_k})), so
\begin{equation}
D\hat u^{(n+1)}=\mathds 1+\delta\sum_{i=1}^q w_iD\mathfrak F(\hat u^{(n)})
.
\end{equation}
We then compute
\begin{equation}
(D\mathfrak F(\hat u))_{k,\ell}
=
-\frac{4\pi^2}{L^2}\nu k^2\delta_{k,\ell}
+\frac{4\pi^2}{L^2|k|}\partial_{\hat u_\ell}T(\hat u,k)
\end{equation}
and, by\-~(\ref{T}),
\begin{equation}
\partial_{\hat u_\ell}T(\hat u,k)
=
\sum_{\displaystyle\mathop{\scriptstyle q\in\mathbb Z^2}_{\ell+q=k}}
\left(
\frac{(q\cdot \ell^\perp)|q|}{|\ell|}
+
\frac{(\ell\cdot q^\perp)|\ell|}{|q|}
\right)\hat u_q
=
\sum_{\displaystyle\mathop{\scriptstyle q\in\mathbb Z^2}_{\ell+q=k}}
(q\cdot \ell^\perp)\left(
\frac{|q|}{|\ell|}
-
\frac{|\ell|}{|q|}
\right)\hat u_q
.
\end{equation}
\bigskip
\indent
The Lyapunov exponents are then defined as
\begin{equation}
\frac1n\log\mathrm{spec}(\pi_i)
,\quad
\pi_i:=\prod_{i=1}^nD\hat u^{(i)}
.
\end{equation}
However, the product of $D\hat u$ may become quite large or quite small (if the exponents are not all 1).
To avoid this, we periodically rescale the product.
We set $\mathfrak L_r\in\mathbb N_*$ (set by adjusting the {\tt lyanpunov\_reset} parameter), and, when $n$ is a multiple of $\mathfrak L_r$, we rescale the eigenvalues of $\pi_i$ to 1.
To do so, we perform a $QR$ decomposition:
\begin{equation}
\pi_{\alpha\mathfrak L_r}
=Q^{(\alpha)}R^{(\alpha)}
\end{equation}
where $Q^{(\alpha)}$ is diagonal and $R^{(\alpha)}$ is an orthogonal matrix, and we divide by $Q^{(\alpha)}$ (thus only keeping $R^{(\alpha)}$).
Thus, we replace
\begin{equation}
\pi_{\alpha\mathfrak L_r+i}\mapsto R^{(\alpha)}\prod_{j=1}^iD\hat u^{(j)}
.
\end{equation}
The Lyapunov exponents at time $\alpha\mathfrak L_r$ are then
\begin{equation}
\frac1{\alpha\mathfrak L_r}\sum_{\beta=1}^\alpha\log\mathrm{spec}(Q^{(\beta)})
.
\end{equation}
\subsubsection{Numerical instability}. \subsubsection{Numerical instability}.
In order to prevent the algorithm from blowing up, it is necessary to impose the reality of $u(x)$ by hand, otherwise, truncation errors build up, and lead to divergences. In order to prevent the algorithm from blowing up, it is necessary to impose the reality of $u(x)$ by hand, otherwise, truncation errors build up, and lead to divergences.
It is sufficient to ensure that the convolution term $T(\hat u,k)$ satisfies $T(\hat u,-k)=T(\hat u,k)^*$. It is sufficient to ensure that the convolution term $T(\hat u,k)$ satisfies $T(\hat u,-k)=T(\hat u,k)^*$.