Nstrophy/docs/nstrophy_doc.tex

532 lines
17 KiB
TeX
Raw Normal View History

2024-10-15 15:47:13 +00:00
% Copyright 2017-2024 Ian Jauslin
2023-05-10 23:33:29 +00:00
%
% Licensed under the Apache License, Version 2.0 (the "License");
% you may not use this file except in compliance with the License.
% You may obtain a copy of the License at
%
% http://www.apache.org/licenses/LICENSE-2.0
%
% Unless required by applicable law or agreed to in writing, software
% distributed under the License is distributed on an "AS IS" BASIS,
% WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
% See the License for the specific language governing permissions and
% limitations under the License.
2018-01-11 22:48:14 +00:00
\documentclass{ian}
\usepackage{largearray}
2024-02-19 16:41:35 +00:00
\usepackage{dsfont}
2018-01-11 22:48:14 +00:00
\begin{document}
\hbox{}
\hfil{\bf\LARGE
{\tt nstrophy}
}
\vfill
\tableofcontents
\vfill
\eject
\setcounter{page}1
\pagestyle{plain}
\section{Description of the computation}
\subsection{Irreversible equation}
2022-05-25 14:56:48 +00:00
\indent Consider the incompressible Navier-Stokes equation in 2 dimensions
2018-01-11 22:48:14 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
\partial_tU=\nu\Delta U+G-(U\cdot\nabla)U,\quad
\nabla\cdot U=0
2018-01-11 22:48:14 +00:00
\label{ins}
\end{equation}
2023-03-31 19:40:15 +00:00
in which $G$ is the forcing term.
We take periodic boundary conditions, so, at every given time, $U(t,\cdot)$ is a function on the torus $\mathbb T^2:=\mathbb R^2/(L\mathbb Z)^2$. We represent $U(t,\cdot)$ using its Fourier series
2018-01-11 22:48:14 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
\hat U_k(t):=\frac1{L^2}\int_{\mathbb T^2}dx\ e^{i\frac{2\pi}L kx}U(t,x)
2018-01-11 22:48:14 +00:00
\end{equation}
for $k\in\mathbb Z^2$, and rewrite~\-(\ref{ins}) as
\begin{equation}
2023-03-31 19:40:15 +00:00
\partial_t\hat U_k=
-\frac{4\pi^2}{L^2}\nu k^2\hat U_k+\hat G_k
2022-05-25 14:56:48 +00:00
-i\frac{2\pi}L\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
2023-03-31 19:40:15 +00:00
(q\cdot\hat U_p)\hat U_q
2018-01-11 22:48:14 +00:00
,\quad
2023-03-31 19:40:15 +00:00
k\cdot\hat U_k=0
2018-01-11 22:48:14 +00:00
\label{ins_k}
\end{equation}
We then reduce the equation to a scalar one, by writing
\begin{equation}
2023-03-31 19:40:15 +00:00
\hat U_k=\frac{i2\pi k^\perp}{L|k|}\hat u_k\equiv\frac{i2\pi}{L|k|}(-k_y\hat u_k,k_x\hat u_k)
\label{udef}
2018-01-11 22:48:14 +00:00
\end{equation}
2022-05-25 14:56:48 +00:00
in terms of which, multiplying both sides of the equation by $\frac L{i2\pi}\frac{k^\perp}{|k|}$,
2018-01-11 22:48:14 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
\partial_t\hat u_k=
-\frac{4\pi^2}{L^2}\nu k^2\hat u_k
+\hat g_k
2022-05-25 14:56:48 +00:00
+\frac{4\pi^2}{L^2|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
2023-03-31 19:40:15 +00:00
\frac{(q\cdot p^\perp)(k^\perp\cdot q^\perp)}{|q||p|}\hat u_p\hat u_q
2018-01-11 22:48:14 +00:00
\label{ins_k}
\end{equation}
2023-03-31 19:40:15 +00:00
with
\begin{equation}
\hat g_k:=\frac{Lk^\perp}{2i\pi|k|}\cdot\hat G_k
.
\label{gdef}
\end{equation}
2018-01-12 19:20:59 +00:00
Furthermore
\begin{equation}
(q\cdot p^\perp)(k^\perp\cdot q^\perp)
=
(q\cdot p^\perp)(q^2+p\cdot q)
\end{equation}
and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore,
2018-01-11 22:48:14 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
\partial_t\hat u_k=
-\frac{4\pi^2}{L^2}\nu k^2\hat u_k+\hat g_k
2023-03-31 20:58:52 +00:00
+\frac{4\pi^2}{L^2|k|}T(\hat u,k)
2024-02-19 16:41:35 +00:00
=:\mathfrak F_k(\hat u)
2023-03-31 20:58:52 +00:00
\label{ins_k}
\end{equation}
with
\begin{equation}
T(\hat u,k):=
\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
2023-03-31 19:40:15 +00:00
\frac{(q\cdot p^\perp)|q|}{|p|}\hat u_p\hat u_q
2018-01-11 22:48:14 +00:00
.
2023-03-31 20:58:52 +00:00
\label{T}
2018-01-11 22:48:14 +00:00
\end{equation}
2023-03-31 19:40:15 +00:00
We truncate the Fourier modes and assume that $\hat u_k=0$ if $|k_1|>K_1$ or $|k_2|>K_2$. Let
2018-01-11 22:48:14 +00:00
\begin{equation}
\mathcal K:=\{(k_1,k_2),\ |k_1|\leqslant K_1,\ |k_2|\leqslant K_2\}
.
\end{equation}
2024-02-18 19:16:48 +00:00
\subsubsection{Runge-Kutta methods}.
2023-06-13 22:45:19 +00:00
To solve the equation numerically, we will use Runge-Kutta methods, which compute an approximate value $\hat u_k^{(n)}$ for $\hat u_k(t_n)$.
{\tt nstrophy} supports the 4th order Runge-Kutta ({\tt RK4}) and 2nd order Runge-Kutta ({\tt RK2}) algorithms.
In addition, several variable step methods are implemented:
\begin{itemize}
\item the Runge-Kutta-Dormand-Prince method ({\tt RKDP54}), which is of 5th order, and adjusts the step by comparing to a 4th order method;
\item the Runge-Kutta-Fehlberg method ({\tt RKF45}), which is of 4th order, and adjusts the step by comparing to a 5th order method;
\item the Runge-Kutta-Bogacki-Shampine method ({\tt RKBS32}), which is of 3d order, and adjusts the step by comparing to a 2nd order method.
\end{itemize}
In these adaptive step methods, two steps are computed at different orders: $\hat u_k^{(n)}$ and $\hat U_k^{(n)}$, the step size is adjusted at every step in such a way that the error is small enough:
\begin{equation}
\|\hat u^{(n)}-\hat U^{(n)}\|
<\epsilon_{\mathrm{target}}
\end{equation}
for some given $\epsilon_{\mathrm{target}}$, set using the {\tt adaptive\_tolerance} parameter.
The choice of the norm matters, and will be discussed below.
If the error is larger than the target, then the step size is decreased.
How this is done depends on the order of algorithm.
If the order is $q$ (here we mean the smaller of the two orders, so 4 for {\tt RKDP54} and {\tt RKF45} and 2 for {\tt RKBS32}), then we expect
\begin{equation}
\|\hat u^{(n)}-\hat U^{(n)}\|=\delta_n^qC_n
.
\end{equation}
We wish to set $\delta_{n+1}$ so that
\begin{equation}
\delta_{n+1}^qC_n=\epsilon_{\mathrm{target}}
\end{equation}
so
\begin{equation}
\delta_{n+1}
=\left(\frac{\epsilon_{\mathrm{target}}}{C_n}\right)^{\frac1q}
=\delta_n\left(\frac{\epsilon_{\mathrm{target}}}{\|\hat u^{(n)}-\hat U^{(n)}\|}\right)^{\frac1q}
.
\label{adaptive_delta}
\end{equation}
(Actually, to be safe and ensure that $\delta$ decreases sufficiently, we multiply this by a safety factor that can be set using the {\tt adaptive\_factor} parameter.)
If the error is smaller than the target, we increase $\delta$ using\-~(\ref{adaptive_delta}) (without the safety factor).
To be safe, we also set a maximal value for $\delta$ via the {\tt max\_delta} parameter.
\bigskip
\indent
The choice of the norm $\|\cdot\|$ matters.
2023-06-14 03:56:35 +00:00
It can be made by specifying the parameter {\tt adaptive\_norm}.
2023-06-13 22:45:19 +00:00
\begin{itemize}
\item A naive choice is to take $\|\cdot\|$ to be the normalized $L_1$ norm:
\begin{equation}
\|f\|:=
\frac1{\mathcal N}\sum_k|f_k|
,\quad
\mathcal N:=\sum_k|\hat u_k^{(n)}-\hat u_k^{(n-1)}|
.
\end{equation}
2023-06-14 03:56:35 +00:00
This norm is selected by choosing {\tt adaptive\_norm=L1}.
2023-06-13 22:45:19 +00:00
\item Empirically, we have found that $|\hat u-\hat U|$ behaves like $k^{-3}$ for {\tt RKDP54} and {\tt RKF45}, and like $k^{-\frac32}$ for {\tt RKBS32}, so a norm of the form
\begin{equation}
\|f\|:=\frac1{\mathcal N}\sum_k|f_k|k^{-3}
,\quad
\mathcal N:=\sum_k|\hat u_k^{(n)}-\hat u_k^{(n-1)}|k^{-3}
\end{equation}
or
\begin{equation}
\|f\|:=\frac1{\mathcal N}\sum_k|f_k|k^{-\frac32}
,\quad
\mathcal N:=\sum_k|\hat u_k^{(n)}-\hat u_k^{(n-1)}|k^{-\frac32}
\end{equation}
are sensible choices.
2023-06-14 03:56:35 +00:00
These norms are selected by choosing {\tt adaptive\_norm=k3} and {\tt adaptive\_norm=k32} respectively.
2023-06-13 22:45:19 +00:00
\item
Another option is to define a norm based on the expression of the enstrophy\-~(\ref{enstrophy}):
\begin{equation}
\|f\|:=\frac1{\mathcal N}\sqrt{\sum_k k^2|f_k|^2}
2023-09-17 21:37:14 +00:00
,\quad
2023-09-19 15:51:55 +00:00
\mathcal N:=\frac{\sqrt{\sum_k k^2|\hat u_k^{(n)}|^2}+\sqrt{\sum_k k^2|\hat U_k^{(n)}|^2}}{\sum_k k^2|\hat u_k^{(n)}|^2}
2023-06-13 22:45:19 +00:00
.
\end{equation}
Doing so controls the error of the enstrophy through
\begin{equation}
2023-09-19 15:51:55 +00:00
\frac1{\mathcal N^2}|\mathcal En(\hat u)-\mathcal En(\hat U)|\equiv|\|\hat u\|^2-\|\hat U\|^2|\leqslant \|\hat u-\hat U\|(\|\hat u\|+\|\hat U\|)
\end{equation}
so
\begin{equation}
2023-09-19 15:51:55 +00:00
\frac1{\mathcal N^2}
|\mathcal En(\hat u)-\mathcal En(\hat U)|\leqslant
\|\hat u-\hat U\|\frac1{\mathcal N}\left(\sqrt{\sum_k k^2|\hat u_k|^2}+\sqrt{\sum_k k^2|\hat U_k|^2}\right)
\end{equation}
and thus
\begin{equation}
\frac{|\mathcal En(\hat u)-\mathcal En(\hat U)|}{\mathcal En(\hat u)}\leqslant
\|\hat u-\hat U\|
2023-06-13 22:45:19 +00:00
.
\end{equation}
2023-06-14 03:56:35 +00:00
This norm is selected by choosing {\tt adaptive\_norm=enstrophy}.
2023-06-13 22:45:19 +00:00
\end{itemize}
2024-02-18 19:16:48 +00:00
\subsubsection{Reality}.
2023-03-31 20:58:52 +00:00
Since $U$ is real, $\hat U_{-k}=\hat U_k^*$, and so
2018-01-11 22:48:14 +00:00
\begin{equation}
2023-03-31 20:58:52 +00:00
\hat u_{-k}=\hat u_k^*
.
\label{realu}
2018-01-11 22:48:14 +00:00
\end{equation}
2023-03-31 20:58:52 +00:00
Similarly,
\begin{equation}
\hat g_{-k}=\hat g_k^*
.
\label{realg}
\end{equation}
Thus,
\begin{equation}
T(\hat u,-k)
=
T(\hat u,k)^*
.
\label{realT}
\end{equation}
2023-04-11 22:45:45 +00:00
In order to keep the computation as quick as possible, we only compute and store the values for $k_1\geqslant 0$.
2023-03-31 20:58:52 +00:00
2024-02-18 19:16:48 +00:00
\subsubsection{FFT}. We compute T using a fast Fourier transform, defined as
2018-01-11 22:48:14 +00:00
\begin{equation}
\mathcal F(f)(n):=\sum_{m\in\mathcal N}e^{-\frac{2i\pi}{N_1}m_1n_1-\frac{2i\pi}{N_2}m_2n_2}f(m_1,m_2)
\end{equation}
where
\begin{equation}
2022-05-12 07:32:58 +00:00
\mathcal N:=\{(n_1,n_2),\ 0\leqslant n_1< N_1,\ 0\leqslant n_2< N_2\}
2018-01-11 22:48:14 +00:00
\end{equation}
for some fixed $N_1,N_2$. The transform is inverted by
\begin{equation}
\frac1{N_1N_2}\mathcal F^*(\mathcal F(f))(n)=f(n)
\end{equation}
in which $\mathcal F^*$ is defined like $\mathcal F$ but with the opposite phase.
\bigskip
\indent The condition $p+q=k$ can be rewritten as
\begin{equation}
2023-03-31 19:40:15 +00:00
T(\hat u,k)
2018-01-11 22:48:14 +00:00
=
\sum_{p,q\in\mathcal K}
\frac1{N_1N_2}
\sum_{n\in\mathcal N}e^{-\frac{2i\pi}{N_1}n_1(p_1+q_1-k_1)-\frac{2i\pi}{N_2}n_2(p_2+q_2-k_2)}
2023-03-31 19:40:15 +00:00
(q\cdot p^\perp)\frac{|q|}{|p|}\hat u_q\hat u_p
2018-01-11 22:48:14 +00:00
\end{equation}
provided
\begin{equation}
N_i>3K_i.
2018-01-11 22:48:14 +00:00
\end{equation}
Indeed, $\sum_{n_i=0}^{N_i}e^{-\frac{2i\pi}{N_i}n_im_i}$ vanishes unless $m_i=0\%N_i$ (in which $\%N_i$ means `modulo $N_i$'), and, if $p,q,k\in\mathcal K$, then $|p_i+q_i-k_i|\leqslant3K_i$, so, as long as $N_i>3K_i$, then $(p_i+q_i-k_i)=0\%N_i$ implies $p_i+q_i=k_i$.
2022-05-12 07:32:58 +00:00
Therefore,
2018-01-11 22:48:14 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
T(\hat u,k)
2018-01-11 22:48:14 +00:00
=
\textstyle
2018-01-12 19:20:59 +00:00
\frac1{N_1N_2}
2018-01-11 22:48:14 +00:00
\mathcal F^*\left(
2023-03-31 19:40:15 +00:00
\mathcal F\left(\frac{p_x\hat u_p}{|p|}\right)(n)
\mathcal F\left(q_y|q|\hat u_q\right)(n)
2018-01-12 19:20:59 +00:00
-
2023-03-31 19:40:15 +00:00
\mathcal F\left(\frac{p_y\hat u_p}{|p|}\right)(n)
\mathcal F\left(q_x|q|\hat u_q\right)(n)
2018-01-11 22:48:14 +00:00
\right)(k)
\end{equation}
2024-02-18 19:16:48 +00:00
\subsubsection{Energy}.
2022-05-19 17:04:30 +00:00
We define the energy as
\begin{equation}
2023-03-31 19:40:15 +00:00
E(t)=\frac12\int\frac{dx}{L^2}\ U^2(t,x)=\frac12\sum_{k\in\mathbb Z^2}|\hat U_k|^2
2022-05-19 17:04:30 +00:00
.
\end{equation}
We have
\begin{equation}
2023-03-31 19:40:15 +00:00
\partial_t E=\int\frac{dx}{L^2}\ U\partial tU
2022-05-19 17:04:30 +00:00
=
2023-03-31 19:40:15 +00:00
\nu\int\frac{dx}{L^2}\ U\Delta U
+\int\frac{dx}{L^2}\ UG
-\int\frac{dx}{L^2}\ U(U\cdot\nabla)U
2022-05-19 17:04:30 +00:00
.
\end{equation}
Since we have periodic boundary conditions,
\begin{equation}
2023-03-31 19:40:15 +00:00
\int dx\ U\Delta U=-\int dx\ |\nabla U|^2
2022-05-19 17:04:30 +00:00
.
\end{equation}
Furthermore,
\begin{equation}
2023-03-31 19:40:15 +00:00
I:=\int dx\ U(U\cdot\nabla)U
=\sum_{i,j=1,2}\int dx\ U_iU_j\partial_jU_i
2022-05-19 17:04:30 +00:00
=
2023-03-31 19:40:15 +00:00
-\sum_{i,j=1,2}\int dx\ (\partial_jU_i)U_jU_i
-\sum_{i,j=1,2}\int dx\ U_i(\partial_jU_j)U_i
2022-05-19 17:04:30 +00:00
\end{equation}
2023-03-31 19:40:15 +00:00
and since $\nabla\cdot U=0$,
2022-05-19 17:04:30 +00:00
\begin{equation}
I
=
-I
\end{equation}
and so $I=0$.
Thus,
\begin{equation}
\partial_t E=
2023-03-31 19:40:15 +00:00
\int\frac{dx}{L^2}\ \left(-\nu|\nabla U|^2+UG\right)
2022-05-19 17:04:30 +00:00
=
2023-03-31 19:40:15 +00:00
\sum_{k\in\mathbb Z^2}\left(-\frac{4\pi^2}{L^2}\nu k^2|\hat U_k|^2+\hat U_{-k}\hat G_k\right)
2022-05-19 17:04:30 +00:00
.
\end{equation}
Furthermore,
\begin{equation}
2023-03-31 19:40:15 +00:00
\sum_{k\in\mathbb Z^2}k^2|\hat U_k|^2\geqslant
\sum_{k\in\mathbb Z^2}|\hat U_k|^2-|\hat U_0|^2
=2E-|\hat U_0|^2
2022-05-19 17:04:30 +00:00
\end{equation}
so
\begin{equation}
2023-03-31 19:40:15 +00:00
\partial_t E\leqslant -\frac{8\pi^2}{L^2}\nu E+\frac{4\pi^2}{L^2}\nu\hat U_0^2+\sum_{k\in\mathbb Z^2}\hat U_{-k}\hat G_k
2022-05-19 17:04:30 +00:00
\leqslant
2023-03-31 19:40:15 +00:00
-\frac{8\pi^2}{L^2}\nu E+\frac{4\pi^2}{L^2}\nu\hat U_0^2+
\|\hat G\|_2\sqrt{2E}
2022-05-19 17:04:30 +00:00
.
\end{equation}
2023-03-31 19:40:15 +00:00
In particular, if $\hat U_0=0$ (which corresponds to keeping the center of mass fixed),
2022-05-19 17:04:30 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
\partial_t E\leqslant -\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}
2022-05-19 17:04:30 +00:00
.
\end{equation}
2023-03-31 19:40:15 +00:00
Now, if $\frac{8\pi^2}{L^2}\nu\sqrt E<\sqrt2\|\hat G\|_2$, then
2022-05-19 17:04:30 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
\frac{\partial_t E}{-\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}}\leqslant1
2022-05-19 17:04:30 +00:00
\end{equation}
and so
\begin{equation}
2023-03-31 19:40:15 +00:00
\frac{\log(1-\frac{8\pi^2\nu}{L^2\sqrt2\|\hat G\|_2}\sqrt{E(t)})}{-\frac{4\pi^2}{L^2}\nu}\leqslant t+
\frac{\log(1-\frac{8\pi^2\nu}{L^2\sqrt2\|\hat G\|_2}\sqrt{E(0)})}{-\frac{4\pi^2}{L^2}\nu}
2022-05-19 17:04:30 +00:00
\end{equation}
and
\begin{equation}
E(t)
\leqslant
\left(
2023-03-31 19:40:15 +00:00
\frac{L^2\sqrt2\|\hat G\|_2}{8\pi^2\nu}(1-e^{-\frac{4\pi^2}{L^2}\nu t})
2022-05-25 14:56:48 +00:00
+e^{-\frac{4\pi^2}{L^2}\nu t}\sqrt{E(0)}
2022-05-19 17:04:30 +00:00
\right)^2
.
\end{equation}
2023-03-31 19:40:15 +00:00
If $\frac{8\pi^2}{L^2}\nu\sqrt E>\sqrt2\|\hat G\|_2$,
2022-05-19 17:04:30 +00:00
\begin{equation}
2023-03-31 19:40:15 +00:00
\frac{\partial_t E}{-\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}}\geqslant1
2022-05-19 17:04:30 +00:00
\end{equation}
and so
\begin{equation}
2023-03-31 19:40:15 +00:00
\frac{\log(\frac{8\pi^2\nu}{L^2\sqrt2\|\hat G\|_2}\sqrt{E(t)}-1)}{-\frac{4\pi^2}{L^2}\nu}\geqslant t+
\frac{\log(\frac{8\pi^2\nu}{L^2\sqrt2\|\hat G\|_2}\sqrt{E(0)})-1}{-\frac{4\pi^2}{L^2}\nu}
2022-05-19 17:04:30 +00:00
\end{equation}
and
\begin{equation}
E(t)
\leqslant
\left(
2023-03-31 19:40:15 +00:00
\frac{L^2\sqrt2\|\hat G\|_2}{8\pi^2\nu}(1-e^{-\frac{4\pi^2}{L^2}\nu t})
2022-05-25 14:56:48 +00:00
+e^{-\frac{4\pi^2}{L^2}\nu t}\sqrt{E(0)}
2022-05-19 17:04:30 +00:00
\right)^2
.
2023-06-13 22:45:19 +00:00
\label{enstrophy}
2022-05-19 17:04:30 +00:00
\end{equation}
2024-02-18 19:16:48 +00:00
\subsubsection{Enstrophy}.
2022-05-19 17:04:30 +00:00
The enstrophy is defined as
\begin{equation}
2023-03-31 19:40:15 +00:00
\mathcal En(t)=\int\frac{dx}{L^2}\ |\nabla U|^2
=\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}k^2|\hat U_k|^2
2022-05-19 17:04:30 +00:00
.
\end{equation}
2022-05-26 18:12:18 +00:00
2024-02-19 16:41:35 +00:00
\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}
2024-03-11 22:54:11 +00:00
\hat u(t_{n+1})=\mathfrak F_{t_n}(\hat u(t_n))
.
2024-02-19 16:41:35 +00:00
\end{equation}
2024-03-11 22:54:11 +00:00
Let $D\mathfrak F_{t_{n}}$ be the Jacobian of this map, in which we split the real and imaginary parts: if
2024-02-19 16:41:35 +00:00
\begin{equation}
2024-03-11 22:54:11 +00:00
\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
2024-02-19 16:41:35 +00:00
\end{equation}
2024-03-11 22:54:11 +00:00
then
2024-02-19 16:41:35 +00:00
\begin{equation}
2024-03-11 22:54:11 +00:00
(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)
2024-02-19 16:41:35 +00:00
\end{equation}
2024-03-11 22:54:11 +00:00
We compute this Jacobian numerically using a finite difference, by computing
2024-02-19 16:41:35 +00:00
\begin{equation}
2024-03-11 22:54:11 +00:00
(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)
2024-02-19 16:41:35 +00:00
.
\end{equation}
2024-03-11 22:54:11 +00:00
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}
2024-02-19 16:41:35 +00:00
\bigskip
\indent
The Lyapunov exponents are then defined as
\begin{equation}
2024-03-11 22:54:11 +00:00
\frac1{t_n}\log\mathrm{spec}(\pi_{t_n})
2024-02-19 16:41:35 +00:00
,\quad
2024-03-11 22:54:11 +00:00
\pi_{t_n}:=\prod_{i=1}^nD\hat u(t_i)
2024-02-19 16:41:35 +00:00
.
\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.
2024-03-11 22:54:11 +00:00
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.
2024-02-19 16:41:35 +00:00
To do so, we perform a $QR$ decomposition:
\begin{equation}
\pi_{\alpha\mathfrak L_r}
2024-03-11 22:54:11 +00:00
=R^{(\alpha)}Q^{(\alpha)}
2024-02-19 16:41:35 +00:00
\end{equation}
2024-03-11 22:54:11 +00:00
where $Q^{(\alpha)}$ is orthogonal and $R^{(\alpha)}$ is a diagonal matrix, and we divide by $R^{(\alpha)}$ (thus only keeping $Q^{(\alpha)}$).
2024-02-19 16:41:35 +00:00
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}
2024-02-18 19:16:48 +00:00
\subsubsection{Numerical instability}.
2022-05-26 18:12:18 +00:00
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.
2023-03-31 19:40:15 +00:00
It is sufficient to ensure that the convolution term $T(\hat u,k)$ satisfies $T(\hat u,-k)=T(\hat u,k)^*$.
2022-05-26 18:12:18 +00:00
After imposing this condition, the algorithm no longer blows up, but it is still unstable (for instance, increasing $K_1$ or $K_2$ leads to very different results).
2018-01-11 22:48:14 +00:00
2023-03-31 19:40:15 +00:00
\subsection{Reversible equation}
\indent The reversible equation is similar to\-~(\ref{ins}) but instead of fixing the viscosity, we fix the enstrophy\-~\cite{Ga22}.
It is defined directly in Fourier space:
\begin{equation}
\partial_t\hat U_k=
-\frac{4\pi^2}{L^2}\alpha(\hat U) k^2\hat U_k+\hat G_k
-i\frac{2\pi}L\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
(q\cdot\hat U_p)\hat U_q
,\quad
k\cdot\hat U_k=0
\end{equation}
where $\alpha$ is chosen such that the enstrophy is constant.
In terms of $\hat u$\-~(\ref{udef}), (\ref{gdef}), (\ref{T}):
\begin{equation}
\partial_t\hat u_k=
-\frac{4\pi^2}{L^2}\alpha(\hat u) k^2\hat u_k
+\hat g_k
+\frac{4\pi^2}{L^2|k|}T(\hat u,k)
.
\label{rns_k}
\end{equation}
To compute $\alpha$, we use the constancy of the enstrophy:
\begin{equation}
\sum_{k\in\mathbb Z^2}k^2\hat U_k\cdot\partial_t\hat U_k
=0
\end{equation}
which, in terms of $\hat u$ is
\begin{equation}
2023-03-31 20:58:52 +00:00
\sum_{k\in\mathbb Z^2}k^2\hat u_k^*\partial_t\hat u_k
2023-03-31 19:40:15 +00:00
=0
\end{equation}
that is
\begin{equation}
2023-03-31 20:58:52 +00:00
\frac{4\pi^2}{L^2}\alpha(\hat u)\sum_{k\in\mathbb Z^2}k^4|\hat u_k|^2
2023-03-31 19:40:15 +00:00
=
2023-03-31 20:58:52 +00:00
\sum_{k\in\mathbb Z^2}k^2\hat u_k^*\hat g_k
+\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}|k|\hat u_k^*T(\hat u,k)
2023-03-31 19:40:15 +00:00
\end{equation}
and so
\begin{equation}
\alpha(\hat u)
2023-03-31 20:58:52 +00:00
=\frac{\frac{L^2}{4\pi^2}\sum_k k^2\hat u_k^*\hat g_k+\sum_k|k|\hat u_k^*T(\hat u,k)}{\sum_kk^4|\hat u_k|^2}
.
\end{equation}
Note that, by\-~(\ref{realu})-(\ref{realT}),
\begin{equation}
\alpha(\hat u)\in\mathbb R
2023-03-31 19:40:15 +00:00
.
\end{equation}
2018-01-11 22:48:14 +00:00
\vfill
\eject
\begin{thebibliography}{WWW99}
\small
\IfFileExists{bibliography/bibliography.tex}{\input bibliography/bibliography.tex}{}
\end{thebibliography}
\end{document}