Split real and imag part in jacobian

docs/nstrophy_doc.tex

@@ -382,62 +382,81 @@ The enstrophy is defined as
 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)})
+  \hat u(t_{n+1})=\mathfrak F_{t_n}(\hat u(t_n))
   .
 \end{equation}
-We then compute
+Let $D\mathfrak F_{t_{n}}$ be the Jacobian of this map, in which we split the real and imaginary parts: if
 \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)
+  \hat u_k(t_n)=:\rho_k+i\iota_k
+  ,\quad
+  \mathfrak F_{t_n}(\hat u(t_n))_k=:\phi_k+i\psi_k
 \end{equation}
-and, by\-~(\ref{T}),
+then
 \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
-  =
-  (k\cdot \ell^\perp)\left(
-    \frac{|k-\ell|}{|\ell|}
-    -
-    \frac{|\ell|}{|k-\ell|}
-  \right)\hat u_{k-\ell}
+  (D\mathfrak F_{t_n})_{k,p}:=\left(\begin{array}{cc}
+    \partial_{\rho_p}\phi_k&\partial_{\iota_p}\phi_k\\
+    \partial_{\rho_p}\psi_k&\partial_{\iota_p}\psi_k
+  \end{array}\right)
+\end{equation}
+We compute this Jacobian numerically using a finite difference, by computing
+\begin{equation}
+  (D\mathfrak F_{t_n})_{k,p}:=\frac1\epsilon
+  \left(\begin{array}{cc}
+    \phi_k(\hat u+\epsilon\delta_p)-\phi_k(\hat u)&\phi_k(\hat u+i\epsilon\delta_p)-\phi_k(\hat u)\\
+    \psi_k(\hat u+\epsilon\delta_p)-\psi_k(\hat u)&\psi_k(\hat u+i\epsilon\delta_p)-\psi_k(\hat u)
+  \end{array}\right)
   .
 \end{equation}
+The parameter $\epsilon$ can be set using the parameter {\tt D\_epsilon}.
+%, 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
+%  =
+%  (k\cdot \ell^\perp)\left(
+%    \frac{|k-\ell|}{|\ell|}
+%    -
+%    \frac{|\ell|}{|k-\ell|}
+%  \right)\hat u_{k-\ell}
+%  .
+%\end{equation}
 \bigskip
 
 \indent
 The Lyapunov exponents are then defined as
 \begin{equation}
-  \frac1n\log\mathrm{spec}(\pi_i)
+  \frac1{t_n}\log\mathrm{spec}(\pi_{t_n})
   ,\quad
-  \pi_i:=\prod_{i=1}^nD\hat u^{(i)}
+  \pi_{t_n}:=\prod_{i=1}^nD\hat u(t_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.
+We set $\mathfrak L_r>0$ (set by adjusting the {\tt lyanpunov\_reset} parameter), and, when $t_n$ crosses 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)}
-  .
+  =R^{(\alpha)}Q^{(\alpha)}
 \end{equation}
+where $Q^{(\alpha)}$ is orthogonal and $R^{(\alpha)}$ is a diagonal matrix, and we divide by $R^{(\alpha)}$ (thus only keeping $Q^{(\alpha)}$).
 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)})