Doc: reversible equation
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@ -3,3 +3,4 @@ out_file: bibliography.tex
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filter:auth: s/([A-Z])[^, ]* /\1. /g; s/ ([^ ,]*),/_\1,_/g; s/ ([^ ,]*)$/_\1/g; s/ //g; s/_/ /g;
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filter:journal: s/([a-zA-Z]) ([0-9]+)/\1~\\-\2/g;
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aux_cmd: \\citation{
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extra: Gallavotti2022:Ga22:Ga22:\bibitem[%token%]{%citeref%}G.\-~Gallavotti - {\it Navier-Stokes and equivalence conjectures}, preprint, 2022\par\penalty10000%n%arxiv:{\tt\color{blue}\href{https://arxiv.org/abs/2211.02961}{2211.02961}}%n%%n%
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@ -22,38 +22,45 @@
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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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\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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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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\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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\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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(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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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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\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 \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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\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\varphi_p\hat\varphi_q
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.
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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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@ -62,14 +69,14 @@ Furthermore
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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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\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|}\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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\frac{(q\cdot p^\perp)|q|}{|p|}\hat u_p\hat u_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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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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@ -78,9 +85,10 @@ We truncate the Fourier modes and assume that $\hat\varphi_k=0$ if $|k_1|>K_1$ o
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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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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\varphi_q\hat\varphi_p
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\frac{(q\cdot p^\perp)|q|}{|p|}\hat u_q\hat u_p
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\label{T}
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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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@ -99,12 +107,12 @@ in which $\mathcal F^*$ is defined like $\mathcal F$ but with the opposite phase
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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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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\varphi_q\hat\varphi_p
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(q\cdot p^\perp)\frac{|q|}{|p|}\hat u_q\hat u_p
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\end{equation}
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provided
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\begin{equation}
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@ -113,16 +121,16 @@ provided
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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\varphi,k)
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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\varphi_p}{|p|}\right)(n)
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\mathcal F\left(q_y|q|\hat\varphi_q\right)(n)
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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\varphi_p}{|p|}\right)(n)
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\mathcal F\left(q_x|q|\hat\varphi_q\right)(n)
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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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\bigskip
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@ -130,32 +138,32 @@ Therefore,
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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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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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\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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\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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\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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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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-\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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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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@ -165,64 +173,64 @@ 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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\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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\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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\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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\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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-\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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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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\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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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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\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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\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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\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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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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\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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\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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\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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@ -232,17 +240,64 @@ and
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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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\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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It is sufficient to ensure that the convolution term $T(\hat u,k)$ satisfies $T(\hat u,-k)=T(\hat u,k)^*$.
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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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\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
|
||||
=
|
||||
\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_kT(\hat u,k)
|
||||
\end{equation}
|
||||
and so
|
||||
\begin{equation}
|
||||
\alpha(\hat u)
|
||||
=\frac{\frac{L^2}{4\pi^2}\sum_k k^2\hat u_k\hat g_k+\sum_k|k|\hat u_kT(\hat u,k)}{\sum_kk^4\hat u_k^2}
|
||||
.
|
||||
\end{equation}
|
||||
|
||||
|
||||
|
||||
\vfill
|
||||
\eject
|
||||
|
||||
|
||||
Loading…
Reference in New Issue
Block a user