255 lines
7.7 KiB
TeX
255 lines
7.7 KiB
TeX
\documentclass{ian}
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\usepackage{largearray}
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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 and $w$ is the pressure.
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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\varphi_k\equiv\frac{i2\pi}{L|k|}(-k_y\hat\varphi_k,k_x\hat\varphi_k)
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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 \varphi_k=
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-\frac{4\pi^2}{L^2}\nu k^2\hat \varphi_k+\frac{Lk^\perp}{2i\pi|k|}\cdot\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\varphi_p\hat\varphi_q
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.
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\label{ins_k}
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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 \varphi_k=
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-\frac{4\pi^2}{L^2}\nu k^2\hat \varphi_k+\frac{Lk^\perp}{2i\pi|k|}\cdot\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)|q|}{|p|}\hat\varphi_p\hat\varphi_q
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.
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\label{ins_k}
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\end{equation}
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We truncate the Fourier modes and assume that $\hat\varphi_k=0$ 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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\point{\bf FFT}. We compute the last term in~\-(\ref{ins_k})
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\begin{equation}
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T(\hat\varphi,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\varphi_q\hat\varphi_p
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\end{equation}
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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\varphi,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\varphi_q\hat\varphi_p
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\end{equation}
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provided
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\begin{equation}
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N_i>4K_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 $p,q\in\mathcal K$, then $|p_i+q_i|\leqslant2K_i$.
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Therefore,
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\begin{equation}
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T(\hat\varphi,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\varphi_p}{|p|}\right)(n)
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\mathcal F\left(q_y|q|\hat\varphi_q\right)(n)
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-
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\mathcal F\left(\frac{p_y\hat\varphi_p}{|p|}\right)(n)
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\mathcal F\left(q_x|q|\hat\varphi_q\right)(n)
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\right)(k)
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\end{equation}
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\bigskip
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\point{\bf 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(
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\frac{L^2\sqrt2\|\hat g\|_2}{8\pi^2\nu}(1-e^{-\frac{4\pi^2}{L^2}\nu t})
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+e^{-\frac{4\pi^2}{L^2}\nu t}\sqrt{E(0)}
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\right)^2
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.
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\end{equation}
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If $\frac{8\pi^2}{L^2}\nu\sqrt E>\sqrt2\|\hat g\|_2$,
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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}}\geqslant1
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\end{equation}
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and so
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\begin{equation}
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\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+
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\frac{\log(\frac{8\pi^2\nu}{L^2\sqrt2\|\hat g\|_2}\sqrt{E(0)})-1}{-\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(
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\frac{L^2\sqrt2\|\hat g\|_2}{8\pi^2\nu}(1-e^{-\frac{4\pi^2}{L^2}\nu t})
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+e^{-\frac{4\pi^2}{L^2}\nu t}\sqrt{E(0)}
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\right)^2
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.
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\end{equation}
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\bigskip
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\point{\bf Enstrophy}.
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The enstrophy is defined as
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\begin{equation}
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\mathcal En(t)=\int\frac{dx}{L^2}\ |\nabla u|^2
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=\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}k^2|\hat u_k|^2
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.
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\end{equation}
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\bigskip
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\point{\bf 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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It is sufficient to ensure that the convolution term $T(\hat\varphi,k)$ satifies $T(\hat\varphi,-k)=T(\hat\varphi,k)^*$.
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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).
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\vfill
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\eject
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\begin{thebibliography}{WWW99}
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\small
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\IfFileExists{bibliography/bibliography.tex}{\input bibliography/bibliography.tex}{}
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\end{thebibliography}
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\end{document}
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