Correctly deal with complex values in lyapunov exponents documentation

docs/nstrophy_doc.tex

@@ -446,18 +446,32 @@ Consider an equation of the form
   \dot u=f(t;u)
   .
 \end{equation}
+Now, the flow may not be complex-differentiable, so the tangent flow should be computed on the real and imaginary parts.
+Let
+\begin{equation}
+  u=\zeta+i\xi
+  ,\quad
+  f(t;u)=\theta(t;\zeta,\xi)+i\psi(t;\zeta,\xi)
+  .
+\end{equation}
 The tangent flow is given by
 \begin{equation}
-  \dot\delta=Df(t;u(t))\delta
+  \dot\delta=\left(\begin{array}{cc}
+    D_\zeta\theta&D_\xi\theta\\
+    D_\zeta\psi&D_\xi\psi
+  \end{array}\right)\delta
 \end{equation}
-where $Df$ is the Jacobian of $f$ with respect to $u$.
+where $D_\zeta\theta$ is the Jacobian of $\theta$ with respect to $\zeta$ and so forth...
 The flow of this equation is denoted by
 \begin{equation}
   \varphi_{t_0,t_1}(\delta_0)
 \end{equation}
 and defined by
 \begin{equation}
-  \frac d{dt}\varphi_{t_0,t}(\delta_0)=Df(t;\varphi_{t_0,t}(\delta_0))\varphi_{t_0,t}(\delta_0)
+  \frac d{dt}\varphi_{t_0,t}(\delta_0)=\left(\begin{array}{cc}
+    D_\zeta\theta(t;\zeta,\xi)&D_\xi\theta(t;\zeta,\xi)\\
+    D_\zeta\psi(t;\zeta,\xi)&D_\xi\psi(t;\zeta,\xi)
+  \end{array}\right)\varphi_{t_0,t}(\delta_0)
   ,\quad
   \varphi_{t_0,t_0}(\delta_0)=\delta_0
   .
@@ -510,26 +524,36 @@ The choice of the times $t_i$ can be done either by fixed-length intervals, spec
 \bigskip
 
 \indent
-To compute the Lyapunov exponents, we thus need the Jacobian of $f$.
+To compute the Lyapunov exponents, we thus need the Jacobians of $\theta$ and $\psi$.
+Note that, by the linearity of the tangent flow equation,
+\begin{equation}
+  ((D \theta(\hat u))\delta)_{k}
+  =
+  \mathcal Re(Df(\hat u)(\delta_{\mathrm r}+i\delta_{\mathrm i}))
+  ,\quad
+  ((D \psi(\hat u))\delta)_{k}
+  =
+  \mathcal Im(Df(\hat u)(\delta_{\mathrm r}+i\delta_{\mathrm i}))
+  .
+\end{equation}
 For the irreversible equation,
 \begin{equation}
   f(\hat u)=
   -\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)
 \end{equation}
-and so
+and
 \begin{equation}
-  ((D f(\hat u))\delta)_k
+  ((D f(\hat u))\delta)_{k}
   =
-  -\frac{4\pi^2}{L^2}\nu k^2\delta_k
+  -\frac{4\pi^2}{L^2}\nu k^2\delta_{k}
   +\frac{4\pi^2}{L^2|k|}DT(\hat u,k)\delta
 \end{equation}
-where
 \begin{equation}
   DT(\hat u,k)\delta
   =
   \sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
-  \left(\frac{(q\cdot p^\perp)|q|}{|p|}+\frac{(p\cdot q^\perp)|p|}{|q|}\right)\hat u_p\hat \delta_q
+  \left(\frac{(q\cdot p^\perp)|q|}{|p|}+\frac{(p\cdot q^\perp)|p|}{|q|}\right)\hat u_p(\delta_{q,\mathrm r}+i\delta_{q,\mathrm i})
   .
 \end{equation}
 %and, by\-~(\ref{T}),
@@ -575,26 +599,6 @@ where
 \end{equation}
 
 
-%\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(t_{n+1})=\mathfrak F_{t_n}(\hat u(t_n))
-%  .
-%\end{equation}
-%Let $D\mathfrak F_{t_{n}}$ be the Jacobian of this map, in which we split the real and imaginary parts: if
-%\begin{equation}
-%  \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}
-%then
-%\begin{equation}
-%  (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