Energy ineuqalities in doc
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@ -125,7 +125,117 @@ Therefore,
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\mathcal F\left(q_x|q|\hat\varphi_q\right)(n)
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\right)(k)
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\end{equation}
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\bigskip
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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 dx\ 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 dx\ u\partial tu
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=
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\nu\int dx\ u\Delta u
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+\int dx\ ug
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-\int dx\ 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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.
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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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=
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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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\begin{equation}
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I
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=
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-I
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\end{equation}
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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 dx\ \left(-\nu|\nabla u|^2+ug\right)
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=
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\sum_{k\in\mathbb Z^2}\left(-4\pi^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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\end{equation}
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so
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\begin{equation}
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\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
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\leqslant
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-8\pi^2\nu E+4\pi^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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\begin{equation}
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\partial_t E\leqslant -8\pi^2\nu E+\|\hat g\|_2\sqrt{2E}
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.
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\end{equation}
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Now, if $8\pi^2\nu\sqrt E<\sqrt2\|\hat g\|_2$, then
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\begin{equation}
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\frac{\partial_t E}{-8\pi^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}{\sqrt2\|\hat g\|_2}\sqrt{E(t)})}{-4\pi^2\nu}\leqslant t+
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\frac{\log(1-\frac{8\pi^2\nu}{\sqrt2\|\hat g\|_2}\sqrt{E(0)})}{-4\pi^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{\sqrt2\|\hat g\|_2}{8\pi^2\nu}(1-e^{-4\pi^2\nu t})
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+e^{-4\pi^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 $8\pi^2\nu\sqrt E>\sqrt2\|\hat g\|_2$,
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\begin{equation}
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\frac{\partial_t E}{-8\pi^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}{\sqrt2\|\hat g\|_2}\sqrt{E(t)}-1)}{-4\pi^2\nu}\geqslant t+
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\frac{\log(\frac{8\pi^2\nu}{\sqrt2\|\hat g\|_2}\sqrt{E(0)})-1}{-4\pi^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{\sqrt2\|\hat g\|_2}{8\pi^2\nu}(1-e^{-4\pi^2\nu t})
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+e^{-4\pi^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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\bigskip
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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 dx\ |\nabla u|^2
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=4\pi^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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\vfill
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\eject
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@ -135,5 +245,4 @@ Therefore,
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\IfFileExists{bibliography/bibliography.tex}{\input bibliography/bibliography.tex}{}
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\end{thebibliography}
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\end{document}
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