From d37d6104e01897491412e2949db327e905d6b53a Mon Sep 17 00:00:00 2001
From: Ian Jauslin <ian@jauslin.org>
Date: Wed, 25 May 2022 10:56:48 -0400
Subject: Add size of box in docs

---
 docs/nstrophy_doc/nstrophy_doc.tex | 72 +++++++++++++++++++-------------------
 1 file changed, 36 insertions(+), 36 deletions(-)

(limited to 'docs/nstrophy_doc/nstrophy_doc.tex')

diff --git a/docs/nstrophy_doc/nstrophy_doc.tex b/docs/nstrophy_doc/nstrophy_doc.tex
index abc8f48..fd609a9 100644
--- a/docs/nstrophy_doc/nstrophy_doc.tex
+++ b/docs/nstrophy_doc/nstrophy_doc.tex
@@ -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}
 
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