From d69f386547d6f1170a9d6f49a14e3e914cf6f8e3 Mon Sep 17 00:00:00 2001 From: Ian Jauslin Date: Fri, 31 Mar 2023 15:40:15 -0400 Subject: Doc: reversible equation --- docs/nstrophy_doc/bibliography/conf.BBlog | 1 + docs/nstrophy_doc/nstrophy_doc.tex | 177 ++++++++++++++++++++---------- 2 files changed, 117 insertions(+), 61 deletions(-) diff --git a/docs/nstrophy_doc/bibliography/conf.BBlog b/docs/nstrophy_doc/bibliography/conf.BBlog index d1c22bc..abe5ad4 100644 --- a/docs/nstrophy_doc/bibliography/conf.BBlog +++ b/docs/nstrophy_doc/bibliography/conf.BBlog @@ -3,3 +3,4 @@ out_file: bibliography.tex filter:auth: s/([A-Z])[^, ]* /\1. /g; s/ ([^ ,]*),/_\1,_/g; s/ ([^ ,]*)$/_\1/g; s/ //g; s/_/ /g; filter:journal: s/([a-zA-Z]) ([0-9]+)/\1~\\-\2/g; aux_cmd: \\citation{ +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% diff --git a/docs/nstrophy_doc/nstrophy_doc.tex b/docs/nstrophy_doc/nstrophy_doc.tex index 35c42ec..ae571ef 100644 --- a/docs/nstrophy_doc/nstrophy_doc.tex +++ b/docs/nstrophy_doc/nstrophy_doc.tex @@ -22,38 +22,45 @@ \subsection{Irreversible equation} \indent Consider the incompressible Navier-Stokes equation in 2 dimensions \begin{equation} - \partial_tu=\nu\Delta u+g-(u\cdot\nabla)u,\quad - \nabla\cdot u=0 + \partial_tU=\nu\Delta U+G-(U\cdot\nabla)U,\quad + \nabla\cdot U=0 \label{ins} \end{equation} -in which $g$ is the forcing term and $w$ is the pressure. -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 +in which $G$ is the forcing term. +We take periodic boundary conditions, so, at every given time, $U(t,\cdot)$ is a function on the torus $\mathbb T^2:=\mathbb R^2/(L\mathbb Z)^2$. We represent $U(t,\cdot)$ using its Fourier series \begin{equation} - \hat u_k(t):=\frac1{L^2}\int_{\mathbb T^2}dx\ e^{i\frac{2\pi}L kx}u(t,x) + \hat U_k(t):=\frac1{L^2}\int_{\mathbb T^2}dx\ e^{i\frac{2\pi}L kx}U(t,x) \end{equation} for $k\in\mathbb Z^2$, and rewrite~\-(\ref{ins}) as \begin{equation} - \partial_t\hat u_k= - -\frac{4\pi^2}{L^2}\nu k^2\hat u_k+\hat g_k + \partial_t\hat U_k= + -\frac{4\pi^2}{L^2}\nu k^2\hat U_k+\hat G_k -i\frac{2\pi}L\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}} - (q\cdot\hat u_p)\hat u_q + (q\cdot\hat U_p)\hat U_q ,\quad - k\cdot\hat u_k=0 + k\cdot\hat U_k=0 \label{ins_k} \end{equation} We then reduce the equation to a scalar one, by writing \begin{equation} - \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) + \hat U_k=\frac{i2\pi k^\perp}{L|k|}\hat u_k\equiv\frac{i2\pi}{L|k|}(-k_y\hat u_k,k_x\hat u_k) + \label{udef} \end{equation} in terms of which, multiplying both sides of the equation by $\frac L{i2\pi}\frac{k^\perp}{|k|}$, \begin{equation} - \partial_t\hat \varphi_k= - -\frac{4\pi^2}{L^2}\nu k^2\hat \varphi_k+\frac{Lk^\perp}{2i\pi|k|}\cdot\hat g_k + \partial_t\hat u_k= + -\frac{4\pi^2}{L^2}\nu k^2\hat u_k + +\hat g_k +\frac{4\pi^2}{L^2|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}} - \frac{(q\cdot p^\perp)(k^\perp\cdot q^\perp)}{|q||p|}\hat\varphi_p\hat\varphi_q - . + \frac{(q\cdot p^\perp)(k^\perp\cdot q^\perp)}{|q||p|}\hat u_p\hat u_q \label{ins_k} \end{equation} +with +\begin{equation} + \hat g_k:=\frac{Lk^\perp}{2i\pi|k|}\cdot\hat G_k + . + \label{gdef} +\end{equation} Furthermore \begin{equation} (q\cdot p^\perp)(k^\perp\cdot q^\perp) @@ -62,14 +69,14 @@ Furthermore \end{equation} and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore, \begin{equation} - \partial_t\hat \varphi_k= - -\frac{4\pi^2}{L^2}\nu k^2\hat \varphi_k+\frac{Lk^\perp}{2i\pi|k|}\cdot\hat g_k + \partial_t\hat u_k= + -\frac{4\pi^2}{L^2}\nu k^2\hat u_k+\hat g_k +\frac{4\pi^2}{L^2|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}} - \frac{(q\cdot p^\perp)|q|}{|p|}\hat\varphi_p\hat\varphi_q + \frac{(q\cdot p^\perp)|q|}{|p|}\hat u_p\hat u_q . \label{ins_k} \end{equation} -We truncate the Fourier modes and assume that $\hat\varphi_k=0$ if $|k_1|>K_1$ or $|k_2|>K_2$. Let +We truncate the Fourier modes and assume that $\hat u_k=0$ if $|k_1|>K_1$ or $|k_2|>K_2$. Let \begin{equation} \mathcal K:=\{(k_1,k_2),\ |k_1|\leqslant K_1,\ |k_2|\leqslant K_2\} . @@ -78,9 +85,10 @@ We truncate the Fourier modes and assume that $\hat\varphi_k=0$ if $|k_1|>K_1$ o \point{\bf FFT}. We compute the last term in~\-(\ref{ins_k}) \begin{equation} - T(\hat\varphi,k):= + T(\hat u,k):= \sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}} - \frac{(q\cdot p^\perp)|q|}{|p|}\hat\varphi_q\hat\varphi_p + \frac{(q\cdot p^\perp)|q|}{|p|}\hat u_q\hat u_p + \label{T} \end{equation} using a fast Fourier transform, defined as \begin{equation} @@ -99,12 +107,12 @@ in which $\mathcal F^*$ is defined like $\mathcal F$ but with the opposite phase \indent The condition $p+q=k$ can be rewritten as \begin{equation} - T(\hat\varphi,k) + T(\hat u,k) = \sum_{p,q\in\mathcal K} \frac1{N_1N_2} \sum_{n\in\mathcal N}e^{-\frac{2i\pi}{N_1}n_1(p_1+q_1-k_1)-\frac{2i\pi}{N_2}n_2(p_2+q_2-k_2)} - (q\cdot p^\perp)\frac{|q|}{|p|}\hat\varphi_q\hat\varphi_p + (q\cdot p^\perp)\frac{|q|}{|p|}\hat u_q\hat u_p \end{equation} provided \begin{equation} @@ -113,16 +121,16 @@ provided 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 $p_i+q_i=k_i$. Therefore, \begin{equation} - T(\hat\varphi,k) + T(\hat u,k) = \textstyle \frac1{N_1N_2} \mathcal F^*\left( - \mathcal F\left(\frac{p_x\hat\varphi_p}{|p|}\right)(n) - \mathcal F\left(q_y|q|\hat\varphi_q\right)(n) + \mathcal F\left(\frac{p_x\hat u_p}{|p|}\right)(n) + \mathcal F\left(q_y|q|\hat u_q\right)(n) - - \mathcal F\left(\frac{p_y\hat\varphi_p}{|p|}\right)(n) - \mathcal F\left(q_x|q|\hat\varphi_q\right)(n) + \mathcal F\left(\frac{p_y\hat u_p}{|p|}\right)(n) + \mathcal F\left(q_x|q|\hat u_q\right)(n) \right)(k) \end{equation} \bigskip @@ -130,32 +138,32 @@ Therefore, \point{\bf Energy}. We define the energy as \begin{equation} - E(t)=\frac12\int\frac{dx}{L^2}\ u^2(t,x)=\frac12\sum_{k\in\mathbb Z^2}|\hat u_k|^2 + E(t)=\frac12\int\frac{dx}{L^2}\ U^2(t,x)=\frac12\sum_{k\in\mathbb Z^2}|\hat U_k|^2 . \end{equation} We have \begin{equation} - \partial_t E=\int\frac{dx}{L^2}\ u\partial tu + \partial_t E=\int\frac{dx}{L^2}\ U\partial tU = - \nu\int\frac{dx}{L^2}\ u\Delta u - +\int\frac{dx}{L^2}\ ug - -\int\frac{dx}{L^2}\ u(u\cdot\nabla)u + \nu\int\frac{dx}{L^2}\ U\Delta U + +\int\frac{dx}{L^2}\ UG + -\int\frac{dx}{L^2}\ U(U\cdot\nabla)U . \end{equation} Since we have periodic boundary conditions, \begin{equation} - \int dx\ u\Delta u=-\int dx\ |\nabla u|^2 + \int dx\ U\Delta U=-\int dx\ |\nabla U|^2 . \end{equation} Furthermore, \begin{equation} - I:=\int dx\ u(u\cdot\nabla)u - =\sum_{i,j=1,2}\int dx\ u_iu_j\partial_ju_i + I:=\int dx\ U(U\cdot\nabla)U + =\sum_{i,j=1,2}\int dx\ U_iU_j\partial_jU_i = - -\sum_{i,j=1,2}\int dx\ (\partial_ju_i)u_ju_i - -\sum_{i,j=1,2}\int dx\ u_i(\partial_ju_j)u_i + -\sum_{i,j=1,2}\int dx\ (\partial_jU_i)U_jU_i + -\sum_{i,j=1,2}\int dx\ U_i(\partial_jU_j)U_i \end{equation} -and since $\nabla\cdot u=0$, +and since $\nabla\cdot U=0$, \begin{equation} I = @@ -165,64 +173,64 @@ and so $I=0$. Thus, \begin{equation} \partial_t E= - \int\frac{dx}{L^2}\ \left(-\nu|\nabla u|^2+ug\right) + \int\frac{dx}{L^2}\ \left(-\nu|\nabla U|^2+UG\right) = - \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) + \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) . \end{equation} Furthermore, \begin{equation} - \sum_{k\in\mathbb Z^2}k^2|\hat u_k|^2\geqslant - \sum_{k\in\mathbb Z^2}|\hat u_k|^2-|\hat u_0|^2 - =2E-|\hat u_0|^2 + \sum_{k\in\mathbb Z^2}k^2|\hat U_k|^2\geqslant + \sum_{k\in\mathbb Z^2}|\hat U_k|^2-|\hat U_0|^2 + =2E-|\hat U_0|^2 \end{equation} so \begin{equation} - \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 + \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 \leqslant - -\frac{8\pi^2}{L^2}\nu E+\frac{4\pi^2}{L^2}\nu\hat u_0^2+ - \|\hat g\|_2\sqrt{2E} + -\frac{8\pi^2}{L^2}\nu E+\frac{4\pi^2}{L^2}\nu\hat U_0^2+ + \|\hat G\|_2\sqrt{2E} . \end{equation} -In particular, if $\hat u_0=0$ (which corresponds to keeping the center of mass fixed), +In particular, if $\hat U_0=0$ (which corresponds to keeping the center of mass fixed), \begin{equation} - \partial_t E\leqslant -\frac{8\pi^2}{L^2}\nu E+\|\hat g\|_2\sqrt{2E} + \partial_t E\leqslant -\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E} . \end{equation} -Now, if $\frac{8\pi^2}{L^2}\nu\sqrt E<\sqrt2\|\hat g\|_2$, then +Now, if $\frac{8\pi^2}{L^2}\nu\sqrt E<\sqrt2\|\hat G\|_2$, then \begin{equation} - \frac{\partial_t E}{-\frac{8\pi^2}{L^2}\nu E+\|\hat g\|_2\sqrt{2E}}\leqslant1 + \frac{\partial_t E}{-\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}}\leqslant1 \end{equation} and so \begin{equation} - \frac{\log(1-\frac{8\pi^2\nu}{L^2\sqrt2\|\hat g\|_2}\sqrt{E(t)})}{-\frac{4\pi^2}{L^2}\nu}\leqslant t+ - \frac{\log(1-\frac{8\pi^2\nu}{L^2\sqrt2\|\hat g\|_2}\sqrt{E(0)})}{-\frac{4\pi^2}{L^2}\nu} + \frac{\log(1-\frac{8\pi^2\nu}{L^2\sqrt2\|\hat G\|_2}\sqrt{E(t)})}{-\frac{4\pi^2}{L^2}\nu}\leqslant t+ + \frac{\log(1-\frac{8\pi^2\nu}{L^2\sqrt2\|\hat G\|_2}\sqrt{E(0)})}{-\frac{4\pi^2}{L^2}\nu} \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}) + \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 $\frac{8\pi^2}{L^2}\nu\sqrt E>\sqrt2\|\hat g\|_2$, +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 + \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} + \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}) + \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 . @@ -232,17 +240,64 @@ and \point{\bf 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 + \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} \bigskip \point{\bf Numerical instability}. 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. -It is sufficient to ensure that the convolution term $T(\hat\varphi,k)$ satifies $T(\hat\varphi,-k)=T(\hat\varphi,k)^*$. +It is sufficient to ensure that the convolution term $T(\hat u,k)$ satisfies $T(\hat u,-k)=T(\hat u,k)^*$. 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). +\subsection{Reversible equation} +\indent The reversible equation is similar to\-~(\ref{ins}) but instead of fixing the viscosity, we fix the enstrophy\-~\cite{Ga22}. +It is defined directly in Fourier space: +\begin{equation} + \partial_t\hat U_k= + -\frac{4\pi^2}{L^2}\alpha(\hat U) k^2\hat U_k+\hat G_k + -i\frac{2\pi}L\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}} + (q\cdot\hat U_p)\hat U_q + ,\quad + k\cdot\hat U_k=0 +\end{equation} +where $\alpha$ is chosen such that the enstrophy is constant. +In terms of $\hat u$\-~(\ref{udef}), (\ref{gdef}), (\ref{T}): +\begin{equation} + \partial_t\hat u_k= + -\frac{4\pi^2}{L^2}\alpha(\hat u) k^2\hat u_k + +\hat g_k + +\frac{4\pi^2}{L^2|k|}T(\hat u,k) + . + \label{rns_k} +\end{equation} +To compute $\alpha$, we use the constancy of the enstrophy: +\begin{equation} + \sum_{k\in\mathbb Z^2}k^2\hat U_k\cdot\partial_t\hat U_k + =0 +\end{equation} +which, in terms of $\hat u$ is +\begin{equation} + \sum_{k\in\mathbb Z^2}k^2\hat u_k\partial_t\hat u_k + =0 +\end{equation} +that is +\begin{equation} + \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 -- cgit v1.2.3-54-g00ecf