Energy ineuqalities in doc

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Ian Jauslin 2022-05-19 19:04:30 +02:00
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@ -125,7 +125,117 @@ Therefore,
\mathcal F\left(q_x|q|\hat\varphi_q\right)(n) \mathcal F\left(q_x|q|\hat\varphi_q\right)(n)
\right)(k) \right)(k)
\end{equation} \end{equation}
\bigskip
\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
.
\end{equation}
We have
\begin{equation}
\partial_t E=\int dx\ u\partial tu
=
\nu\int dx\ u\Delta u
+\int dx\ ug
-\int dx\ 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
.
\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
=
-\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 $\nabla\cdot u=0$,
\begin{equation}
I
=
-I
\end{equation}
and so $I=0$.
Thus,
\begin{equation}
\partial_t E=
\int dx\ \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)
.
\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
\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
\leqslant
-8\pi^2\nu E+4\pi^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}
.
\end{equation}
Now, if $8\pi^2\nu\sqrt E<\sqrt2\|\hat g\|_2$, then
\begin{equation}
\frac{\partial_t E}{-8\pi^2\nu E+\|\hat g\|_2\sqrt{2E}}\leqslant1
\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}
\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)}
\right)^2
.
\end{equation}
If $8\pi^2\nu\sqrt E>\sqrt2\|\hat g\|_2$,
\begin{equation}
\frac{\partial_t E}{-8\pi^2\nu E+\|\hat g\|_2\sqrt{2E}}\geqslant1
\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}
\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)}
\right)^2
.
\end{equation}
\bigskip
\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
.
\end{equation}
\vfill \vfill
\eject \eject
@ -135,5 +245,4 @@ Therefore,
\IfFileExists{bibliography/bibliography.tex}{\input bibliography/bibliography.tex}{} \IfFileExists{bibliography/bibliography.tex}{\input bibliography/bibliography.tex}{}
\end{thebibliography} \end{thebibliography}
\end{document} \end{document}