Add size of box in docs

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Ian Jauslin 2022-05-25 10:56:48 -04:00
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@ -20,22 +20,22 @@
\section{Description of the computation}
\subsection{Irreversible equation}
\indent Consider the {\it irreversible} Navier-Stokes equation in 2 dimensions
\indent Consider the incompressible Navier-Stokes equation in 2 dimensions
\begin{equation}
\partial_tu=\nu\Delta u+g-\nabla w-(u\cdot\nabla)u,\quad
\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 unit torus $\mathbb T^2:=\mathbb R^2/\mathbb Z^2$. We represent $u(t,\cdot)$ using its Fourier series
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):=\int_{\mathbb T^2}dx\ e^{2i\pi 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=
-4\pi^2\nu k^2\hat u_k+\hat g_k-2i\pi k\hat w_k
-2i\pi\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=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
,\quad
k\cdot\hat u_k=0
@ -43,13 +43,13 @@ for $k\in\mathbb Z^2$, and rewrite~\-(\ref{ins}) as
\end{equation}
We then reduce the equation to a scalar one, by writing
\begin{equation}
\hat u_k=\frac{2i\pi k^\perp}{|k|}\hat\varphi_k\equiv\frac{2i\pi}{|k|}(-k_y\hat\varphi_k,k_x\hat\varphi_k)
\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)
\end{equation}
in terms of which, multiplying both sides of the equation by $\frac{k^\perp}{|k|}$,
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=
-4\pi^2\nu k^2\hat \varphi_k+\frac{k^\perp}{2i\pi|k|}\cdot\hat g_k
+\frac{4\pi^2}{|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
-\frac{4\pi^2}{L^2}\nu k^2\hat \varphi_k+\frac{Lk^\perp}{2i\pi|k|}\cdot\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
.
\label{ins_k}
@ -63,8 +63,8 @@ Furthermore
and $q\cdot p^\perp$ is antisymmetric under exchange of $q$ and $p$. Therefore,
\begin{equation}
\partial_t\hat \varphi_k=
-4\pi^2\nu k^2\hat \varphi_k+\frac{k^\perp}{2i\pi|k|}\cdot\hat g_k
+\frac{4\pi^2}{|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
-\frac{4\pi^2}{L^2}\nu k^2\hat \varphi_k+\frac{Lk^\perp}{2i\pi|k|}\cdot\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
.
\label{ins_k}
@ -130,16 +130,16 @@ Therefore,
\point{\bf Energy}.
We define the energy as
\begin{equation}
E(t)=\frac12\int dx\ 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 dx\ u\partial tu
\partial_t E=\int\frac{dx}{L^2}\ u\partial tu
=
\nu\int dx\ u\Delta u
+\int dx\ ug
-\int dx\ 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,
@ -165,9 +165,9 @@ and so $I=0$.
Thus,
\begin{equation}
\partial_t E=
\int dx\ \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(-4\pi^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,
@ -178,52 +178,52 @@ Furthermore,
\end{equation}
so
\begin{equation}
\partial_t E\leqslant -8\pi^2\nu E+4\pi^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
-8\pi^2\nu E+4\pi^2\nu\hat u_0^2+
-\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),
\begin{equation}
\partial_t E\leqslant -8\pi^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 $8\pi^2\nu\sqrt E<\sqrt2\|\hat g\|_2$, then
Now, if $\frac{8\pi^2}{L^2}\nu\sqrt E<\sqrt2\|\hat g\|_2$, then
\begin{equation}
\frac{\partial_t E}{-8\pi^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}{\sqrt2\|\hat g\|_2}\sqrt{E(t)})}{-4\pi^2\nu}\leqslant t+
\frac{\log(1-\frac{8\pi^2\nu}{\sqrt2\|\hat g\|_2}\sqrt{E(0)})}{-4\pi^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{\sqrt2\|\hat g\|_2}{8\pi^2\nu}(1-e^{-4\pi^2\nu t})
+e^{-4\pi^2\nu t}\sqrt{E(0)}
\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 $8\pi^2\nu\sqrt E>\sqrt2\|\hat g\|_2$,
If $\frac{8\pi^2}{L^2}\nu\sqrt E>\sqrt2\|\hat g\|_2$,
\begin{equation}
\frac{\partial_t E}{-8\pi^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}{\sqrt2\|\hat g\|_2}\sqrt{E(t)}-1)}{-4\pi^2\nu}\geqslant t+
\frac{\log(\frac{8\pi^2\nu}{\sqrt2\|\hat g\|_2}\sqrt{E(0)})-1}{-4\pi^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{\sqrt2\|\hat g\|_2}{8\pi^2\nu}(1-e^{-4\pi^2\nu t})
+e^{-4\pi^2\nu t}\sqrt{E(0)}
\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}
@ -232,8 +232,8 @@ and
\point{\bf Enstrophy}.
The enstrophy is defined as
\begin{equation}
\mathcal En(t)=\int dx\ |\nabla u|^2
=4\pi^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}