678 lines
22 KiB
TeX
678 lines
22 KiB
TeX
% Copyright 2017-2024 Ian Jauslin
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%
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% Licensed under the Apache License, Version 2.0 (the "License");
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% you may not use this file except in compliance with the License.
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% You may obtain a copy of the License at
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%
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% http://www.apache.org/licenses/LICENSE-2.0
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%
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% Unless required by applicable law or agreed to in writing, software
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% distributed under the License is distributed on an "AS IS" BASIS,
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% WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
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% See the License for the specific language governing permissions and
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% limitations under the License.
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\documentclass{ian}
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\usepackage{largearray}
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\usepackage{dsfont}
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\begin{document}
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\hbox{}
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\hfil{\bf\LARGE
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{\tt nstrophy}
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}
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\vfill
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\tableofcontents
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\vfill
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\eject
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\setcounter{page}1
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\pagestyle{plain}
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\section{Description of the computation}
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\subsection{Irreversible equation}
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\indent Consider the incompressible Navier-Stokes equation in 2 dimensions
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\begin{equation}
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\partial_tU=\nu\Delta U+G-(U\cdot\nabla)U,\quad
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\nabla\cdot U=0
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\label{ins}
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\end{equation}
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in which $G$ is the forcing term.
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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
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\begin{equation}
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\hat U_k(t):=\frac1{L^2}\int_{\mathbb T^2}dx\ e^{i\frac{2\pi}L kx}U(t,x)
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\end{equation}
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for $k\in\mathbb Z^2$, and rewrite~\-(\ref{ins}) as
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\begin{equation}
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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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-i\frac{2\pi}L\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
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(q\cdot\hat U_p)\hat U_q
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,\quad
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k\cdot\hat U_k=0
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\label{ins_k}
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\end{equation}
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We then reduce the equation to a scalar one, by writing
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\begin{equation}
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\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)
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\label{udef}
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\end{equation}
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in terms of which, multiplying both sides of the equation by $\frac L{i2\pi}\frac{k^\perp}{|k|}$,
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\begin{equation}
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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
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+\hat g_k
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+\frac{4\pi^2}{L^2|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
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\frac{(q\cdot p^\perp)(k^\perp\cdot q^\perp)}{|q||p|}\hat u_p\hat u_q
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\label{ins_k}
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\end{equation}
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with
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\begin{equation}
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\hat g_k:=\frac{Lk^\perp}{2i\pi|k|}\cdot\hat G_k
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.
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\label{gdef}
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\end{equation}
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Furthermore
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\begin{equation}
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(q\cdot p^\perp)(k^\perp\cdot q^\perp)
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=
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(q\cdot p^\perp)(q^2+p\cdot q)
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\end{equation}
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and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore,
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\begin{equation}
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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|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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\end{equation}
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with
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\begin{equation}
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T(\hat u,k):=
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\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
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\frac{(q\cdot p^\perp)|q|}{|p|}\hat u_p\hat u_q
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.
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\label{T}
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\end{equation}
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We truncate the Fourier modes and assume that $\hat u_k=0$ if $|k_1|>K_1$ or $|k_2|>K_2$. Let
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\begin{equation}
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\mathcal K:=\{(k_1,k_2),\ |k_1|\leqslant K_1,\ |k_2|\leqslant K_2\}
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.
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\end{equation}
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\bigskip
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{\bf Remark}:
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Since $U$ is real, $\hat U_{-k}=\hat U_k^*$, and so
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\begin{equation}
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\hat u_{-k}=\hat u_k^*
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.
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\label{realu}
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\end{equation}
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Similarly,
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\begin{equation}
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\hat g_{-k}=\hat g_k^*
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.
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\label{realg}
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\end{equation}
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Thus,
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\begin{equation}
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T(\hat u,-k)
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=
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T(\hat u,k)^*
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.
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\label{realT}
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\end{equation}
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In order to keep the computation as quick as possible, we only compute and store the values for $k_1\geqslant 0$.
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(In fact, if we do not enforce the reality conditions, the computation has been found to be unstable.)
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\subsection{Reversible equation}
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\indent The reversible equation is similar to\-~(\ref{ins}) but instead of fixing the viscosity, we fix the enstrophy\-~\cite{Ga22}.
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It is defined directly in Fourier space:
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\begin{equation}
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\partial_t\hat U_k=
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-\frac{4\pi^2}{L^2}\alpha(\hat U) k^2\hat U_k+\hat G_k
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-i\frac{2\pi}L\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
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(q\cdot\hat U_p)\hat U_q
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,\quad
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k\cdot\hat U_k=0
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\end{equation}
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where $\alpha$ is chosen such that the enstrophy is constant.
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In terms of $\hat u$\-~(\ref{udef}), (\ref{gdef}), (\ref{T}):
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\begin{equation}
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\partial_t\hat u_k=
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-\frac{4\pi^2}{L^2}\alpha(\hat u) k^2\hat u_k
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+\hat g_k
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+\frac{4\pi^2}{L^2|k|}T(\hat u,k)
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.
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\label{rns_k}
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\end{equation}
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To compute $\alpha$, we use the constancy of the enstrophy:
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\begin{equation}
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\sum_{k\in\mathbb Z^2}k^2\hat U_k\cdot\partial_t\hat U_k
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=0
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\end{equation}
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which, in terms of $\hat u$ is
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\begin{equation}
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\sum_{k\in\mathbb Z^2}k^2\hat u_k^*\partial_t\hat u_k
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=0
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\end{equation}
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that is
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\begin{equation}
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\frac{4\pi^2}{L^2}\alpha(\hat u)\sum_{k\in\mathbb Z^2}k^4|\hat u_k|^2
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=
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\sum_{k\in\mathbb Z^2}k^2\hat u_k^*\hat g_k
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+\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}|k|\hat u_k^*T(\hat u,k)
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\end{equation}
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and so
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\begin{equation}
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\alpha(\hat u)
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=\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}
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.
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\label{alpha}
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\end{equation}
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Note that, by\-~(\ref{realu})-(\ref{realT}),
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\begin{equation}
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\alpha(\hat u)\in\mathbb R
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.
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\end{equation}
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\subsection{Runge-Kutta methods}.
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To solve these equations numerically, we will use Runge-Kutta methods, which compute an approximate value $\hat u_k^{(n)}$ for $\hat u_k(t_n)$.
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These algorithms approximate the solution to an equation of the form $\dot u=f(t;u)$ with
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\begin{equation}
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\hat u_k^{(n+1)}=\hat u_k^{(n)}
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+\delta_n\sum_{i=1}^sb_ik_i(t_n;\hat u^{(n)})
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,\quad
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k_i(t_n;\hat u^{(n)}):=f\left(t_n+c_i\delta_n;\ \hat u+\delta_n\sum_{j=1}^{i-1}a_{i,j}k_j(t_n,\hat u^{(n)})\right)
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.
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\end{equation}
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The $c_i$ and $a_{i,j}$ are chosen in one of various ways, depending on the desired accuracy.
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\bigskip
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\indent
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{\tt nstrophy} supports the 4th order Runge-Kutta ({\tt RK4}) and 2nd order Runge-Kutta ({\tt RK2}) algorithms.
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In addition, several variable step methods are implemented:
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\begin{itemize}
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\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;
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\item the Runge-Kutta-Fehlberg method ({\tt RKF45}), which is of 4th order, and adjusts the step by comparing to a 5th order method;
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\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.
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\end{itemize}
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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:
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\begin{equation}
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D(\hat u^{(n)},\hat U^{(n)})
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<\epsilon_{\mathrm{target}}
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\end{equation}
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for some given {\it cost function} $D$, and $\epsilon_{\mathrm{target}}$, set using the {\tt adaptive\_tolerance} parameter.
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The choice of the cost function matters, and will be discussed below.
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If the error is larger than the target, then the step size is decreased.
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How this is done depends on the order of algorithm.
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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 (as long as the cost function is such that $D(\hat u,\hat u+\varphi)\sim\|\varphi\|$ in some norm)
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\begin{equation}
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D(\hat u^{(n)},\hat U^{(n)})=\delta_n^qC_n
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\end{equation}
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for some number $C_n$.
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We wish to set $\delta_{n+1}$ so that
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\begin{equation}
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\delta_{n+1}^qC_n=\epsilon_{\mathrm{target}}
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\end{equation}
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so
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\begin{equation}
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\delta_{n+1}
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=\left(\frac{\epsilon_{\mathrm{target}}}{C_n}\right)^{\frac1q}
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=\delta_n\left(\frac{\epsilon_{\mathrm{target}}}{D(\hat u^{(n)},\hat U^{(n)})}\right)^{\frac1q}
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.
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\label{adaptive_delta}
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\end{equation}
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(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.)
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If the error is smaller than the target, we increase $\delta$ using\-~(\ref{adaptive_delta}) (without the safety factor).
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To be safe, we also set a maximal value for $\delta$ via the {\tt max\_delta} parameter.
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\bigskip
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\indent
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The choice of the cost function $D$ matters.
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It can be made by specifying the parameter {\tt adaptive\_cost}.
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\begin{itemize}
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\item
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For computations where the main focus is the enstrophy\-~(\ref{enstrophy}), one may want to set the cost function to the relative difference of the enstrophies:
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\begin{equation}
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D(\hat u,\hat U):=\frac{|\mathcal En(\hat u)-\mathcal En(\hat U)|}{\mathcal En(\hat u)}
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.
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\end{equation}
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This cost function is selected by choosing {\tt adaptive\_cost=enstrophy}.
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\item
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For computations where the main focus is the value of $\alpha$\-~(\ref{alpha}), one may want to set the cost function to the relative difference of $\alpha$:
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\begin{equation}
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D(\hat u,\hat U):=\frac{|\alpha(\hat u)-\alpha(\hat U)|}{|\alpha(\hat u)|}
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.
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\end{equation}
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This cost function is selected by choosing {\tt adaptive\_cost=alpha}.
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\item Alternatively, one my take $D$ to be the normalized $L_1$ norm:
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\begin{equation}
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D(\hat u,\hat U):=
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\frac1{\mathcal N}\sum_k|\hat u_k-\hat U_k|
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,\quad
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\mathcal N:=\sum_k|\hat u_k|
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.
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\end{equation}
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This function is selected by choosing {\tt adaptive\_cost=L1}.
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\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 cost function of the form
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\begin{equation}
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D(\hat u,\hat U):=\frac1{\mathcal N}\sum_k|\hat u_k-\hat U_k|k^{-3}
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,\quad
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\mathcal N:=\sum_k|\hat u_k|k^{-3}
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\end{equation}
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or
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\begin{equation}
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D(\hat u,\hat U):=\frac1{\mathcal N}\sum_k|\hat u_k-\hat U_k|k^{-\frac32}
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,\quad
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\mathcal N:=\sum_k|\hat u_k|k^{-\frac32}
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\end{equation}
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are sensible choices.
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These cost functions are selected by choosing {\tt adaptive\_cost=k3} and {\tt adaptive\_cost=k32} respectively.
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\end{itemize}
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\subsection{Computation of $T$: FFT}. We compute T using a fast Fourier transform, defined as
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\begin{equation}
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\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)
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\end{equation}
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where
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\begin{equation}
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\mathcal N:=\{(n_1,n_2),\ 0\leqslant n_1< N_1,\ 0\leqslant n_2< N_2\}
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\end{equation}
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for some fixed $N_1,N_2$. The transform is inverted by
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\begin{equation}
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\frac1{N_1N_2}\mathcal F^*(\mathcal F(f))(n)=f(n)
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\end{equation}
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in which $\mathcal F^*$ is defined like $\mathcal F$ but with the opposite phase.
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\bigskip
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\indent The condition $p+q=k$ can be rewritten as
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\begin{equation}
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T(\hat u,k)
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=
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\sum_{p,q\in\mathcal K}
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\frac1{N_1N_2}
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\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)}
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(q\cdot p^\perp)\frac{|q|}{|p|}\hat u_p\hat u_q
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\end{equation}
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provided
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\begin{equation}
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N_i>3K_i.
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\end{equation}
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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 $p_i+q_i=k_i$.
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Therefore,
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\begin{equation}
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T(\hat u,k)
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=
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\textstyle
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\frac1{N_1N_2}
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\mathcal F^*\left(
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\mathcal F\left(\frac{p_x\hat u_p}{|p|}\right)(n)
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\mathcal F\left(q_y|q|\hat u_q\right)(n)
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-
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\mathcal F\left(\frac{p_y\hat u_p}{|p|}\right)(n)
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\mathcal F\left(q_x|q|\hat u_q\right)(n)
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\right)(k)
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\end{equation}
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\subsection{Observables}
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\indent
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We define the following observables.
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\bigskip
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\subsubsection{Energy}.
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We define the energy as
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\begin{equation}
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E(t)=\frac12\int\frac{dx}{L^2}\ U^2(t,x)=\frac12\sum_{k\in\mathbb Z^2}|\hat U_k|^2
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.
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\end{equation}
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We have
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\begin{equation}
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\partial_t E=\int\frac{dx}{L^2}\ U\partial tU
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=
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\nu\int\frac{dx}{L^2}\ U\Delta U
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+\int\frac{dx}{L^2}\ UG
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-\int\frac{dx}{L^2}\ U(U\cdot\nabla)U
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.
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\end{equation}
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Since we have periodic boundary conditions,
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\begin{equation}
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\int dx\ U\Delta U=-\int dx\ |\nabla U|^2
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.
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\end{equation}
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Furthermore,
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\begin{equation}
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I:=\int dx\ U(U\cdot\nabla)U
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=\sum_{i,j=1,2}\int dx\ U_iU_j\partial_jU_i
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=
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-\sum_{i,j=1,2}\int dx\ (\partial_jU_i)U_jU_i
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-\sum_{i,j=1,2}\int dx\ U_i(\partial_jU_j)U_i
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\end{equation}
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and since $\nabla\cdot U=0$,
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\begin{equation}
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I
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=
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-I
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\end{equation}
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and so $I=0$.
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Thus,
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\begin{equation}
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\partial_t E=
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\int\frac{dx}{L^2}\ \left(-\nu|\nabla U|^2+UG\right)
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=
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\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)
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.
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\end{equation}
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Furthermore,
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\begin{equation}
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\sum_{k\in\mathbb Z^2}k^2|\hat U_k|^2\geqslant
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\sum_{k\in\mathbb Z^2}|\hat U_k|^2-|\hat U_0|^2
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=2E-|\hat U_0|^2
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\end{equation}
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so
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\begin{equation}
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\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
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\leqslant
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-\frac{8\pi^2}{L^2}\nu E+\frac{4\pi^2}{L^2}\nu\hat U_0^2+
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\|\hat G\|_2\sqrt{2E}
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.
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\end{equation}
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In particular, if $\hat U_0=0$ (which corresponds to keeping the center of mass fixed),
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\begin{equation}
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\partial_t E\leqslant -\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}
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.
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\end{equation}
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Now, if $\frac{8\pi^2}{L^2}\nu\sqrt E<\sqrt2\|\hat G\|_2$, then
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\begin{equation}
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\frac{\partial_t E}{-\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}}\leqslant1
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\end{equation}
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and so
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\begin{equation}
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\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+
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\frac{\log(1-\frac{8\pi^2\nu}{L^2\sqrt2\|\hat G\|_2}\sqrt{E(0)})}{-\frac{4\pi^2}{L^2}\nu}
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\end{equation}
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and
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\begin{equation}
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E(t)
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\leqslant
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\left(
|
|
\frac{L^2\sqrt2\|\hat G\|_2}{8\pi^2\nu}(1-e^{-\frac{4\pi^2}{L^2}\nu t})
|
|
+e^{-\frac{4\pi^2}{L^2}\nu t}\sqrt{E(0)}
|
|
\right)^2
|
|
.
|
|
\end{equation}
|
|
If $\frac{8\pi^2}{L^2}\nu\sqrt E>\sqrt2\|\hat G\|_2$,
|
|
\begin{equation}
|
|
\frac{\partial_t E}{-\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}}\geqslant1
|
|
\end{equation}
|
|
and so
|
|
\begin{equation}
|
|
\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}
|
|
\end{equation}
|
|
and
|
|
\begin{equation}
|
|
E(t)
|
|
\leqslant
|
|
\left(
|
|
\frac{L^2\sqrt2\|\hat G\|_2}{8\pi^2\nu}(1-e^{-\frac{4\pi^2}{L^2}\nu t})
|
|
+e^{-\frac{4\pi^2}{L^2}\nu t}\sqrt{E(0)}
|
|
\right)^2
|
|
.
|
|
\label{enstrophy}
|
|
\end{equation}
|
|
|
|
\subsubsection{Enstrophy}.
|
|
The enstrophy is defined as
|
|
\begin{equation}
|
|
\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
|
|
.
|
|
\end{equation}
|
|
|
|
\subsection{Lyapunov exponents}
|
|
\indent
|
|
The Lyapunov are defined from the {\it tangent flow} of the dynamics.
|
|
Consider an equation of the form
|
|
\begin{equation}
|
|
\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=\left(\begin{array}{cc}
|
|
D_\zeta\theta&D_\xi\theta\\
|
|
D_\zeta\psi&D_\xi\psi
|
|
\end{array}\right)\delta
|
|
\end{equation}
|
|
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)=\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
|
|
.
|
|
\end{equation}
|
|
The flow $\varphi_{t_0,t}(\delta_0)$ is linear in $\delta_0$, and so can be represented as a matrix.
|
|
The Lyapnuov exponents are defined from the $QR$ decomposition of $\varphi_{0,t}$: if
|
|
\begin{equation}
|
|
\varphi_{0,t}=Q_tR_t
|
|
\end{equation}
|
|
where $Q_t$ is orthogonal, and $R_t$ is triangular with positive diagonal entries $(r_t^{(j)})_j$, then the Lyapunov exponents are
|
|
\begin{equation}
|
|
\lim_{t\to\infty}\frac 1t\log|r_t^{(j)}|
|
|
.
|
|
\end{equation}
|
|
\bigskip
|
|
|
|
\indent
|
|
One problem to compute the Lyapunov exponents numerically is that the spectrum depends exponentially on time, and so has a tendency to blow up or shrink very rapidly, which leads to large truncation errors.
|
|
To avoid these, we compute the tangent flow {\it in parts}: we split time into intervals:
|
|
\begin{equation}
|
|
[0,t)=\bigcup_{i=0}^{N-1}[t_i,t_{i+1})
|
|
,\quad
|
|
\varphi_{0,t}=\prod_{i=N-1}^0\varphi_{t_{i},t_{i+1}}
|
|
.
|
|
\end{equation}
|
|
At the thresholds between these intervals, we perform a QR decomposition:
|
|
\begin{equation}
|
|
Q_0R_0:=\varphi_{0,t_0}
|
|
,\quad
|
|
Q_iR_i:=\varphi_{t_{i-1},t_i}Q_{i-1}
|
|
\end{equation}
|
|
where $Q_i$ is orthogonal and $R_i$ is upper triangular with $\geqslant 0$ diagonal entries (in addition, we assume these are not $0$).
|
|
Doing so, we find
|
|
\begin{equation}
|
|
\varphi_{0,t}=Q_{N-1}\prod_{i=N-1}^0R_i
|
|
\end{equation}
|
|
and since the product of triangular matrices is triangular and the diagonal elements multiply, we find that the Lyapunov exponents are given by
|
|
\begin{equation}
|
|
\lim_{t_{N-1}\to\infty}\frac1{t_{N-1}}\sum_{i=N-1}^0\log|r_i^{(j)}|
|
|
.
|
|
\end{equation}
|
|
To do the computation numerically, we drop the limit, and compute the logarithm of the product of the diagonal entries of $R_i$.
|
|
\bigskip
|
|
|
|
\indent
|
|
In practice, we approximate $\varphi_{t_{i-1},t_i}$ by running a Runge-Kutta algorithm for the tangent flow equation.
|
|
To obtain the full matrix, we consider every element of the canonical basis as an initial condition $\delta_0$.
|
|
We then iterate the Runge-Kutta algorithm until the time $t_0$ (chosen in one of two ways, see below), at which point we perform a QR decomposition, save the diagonal entries of $R$, replace the family of initial conditions with the columns of $Q$, and continue the flow from there.
|
|
The choice of the times $t_i$ can be done either by fixed-length intervals, specified with the option {\tt lyapunov\_reset}, or the QR decomposition can be triggered whenever $\|\delta\|_1$ exceeds a threshold, specified in {\tt lyapunov\_maxu} (after all, the intervals are used to prevent $\delta$ from becoming too large).
|
|
\bigskip
|
|
|
|
\indent
|
|
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
|
|
\begin{equation}
|
|
((D f(\hat u))\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}
|
|
\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(\delta_{q,\mathrm r}+i\delta_{q,\mathrm i})
|
|
.
|
|
\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}
|
|
For the reversible equation,
|
|
\begin{equation}
|
|
f(\hat u)=
|
|
-\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)
|
|
\end{equation}
|
|
so
|
|
\begin{equation}
|
|
((D f(\hat u))\delta)_k
|
|
=
|
|
-\frac{4\pi^2}{L^2}\alpha(\hat u) k^2\delta_k
|
|
-\frac{4\pi^2}{L^2}k^2\hat u_k D\alpha(\hat u)\delta
|
|
+\frac{4\pi^2}{L^2|k|}DT(\hat u,k)\delta
|
|
\end{equation}
|
|
where
|
|
\begin{equation}
|
|
D\alpha(\hat u)\delta
|
|
=
|
|
\frac{\frac{L^2}{4\pi^2}\sum_k k^2\hat \delta_k^*\hat g_k+\sum_k|k|\hat (\delta_k^*T(\hat u,k)+\hat u_k^*DT(\hat u,k)\delta)}{\sum_kk^4|\hat u_k|^2}
|
|
-\alpha(\hat u)\frac{2\sum_kk^4\delta_k^*\hat u_k}{\sum_kk^4|\hat u_k|^2}
|
|
.
|
|
\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}
|
|
% \frac1{t_n}\log\mathrm{spec}(\pi_{t_n})
|
|
% ,\quad
|
|
% \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 $\mathfrak L_r>0$ (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}
|
|
% =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)})
|
|
% .
|
|
%\end{equation}
|
|
|
|
|
|
|
|
\vfill
|
|
\eject
|
|
|
|
\begin{thebibliography}{WWW99}
|
|
\small
|
|
\IfFileExists{bibliography/bibliography.tex}{\input bibliography/bibliography.tex}{}
|
|
\end{thebibliography}
|
|
|
|
\end{document}
|