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