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\section { Description of the computation}
\subsection { Irreversible equation}
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\indent Consider the incompressible Navier-Stokes equation in 2 dimensions
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\begin { equation}
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\partial _ tu=\nu \Delta u+g-(u\cdot \nabla )u,\quad
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\nabla \cdot u=0
\label { ins}
\end { equation}
in which $ g $ is the forcing term and $ w $ is the pressure.
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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
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\begin { equation}
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\hat u_ k(t):=\frac 1{ L^ 2} \int _ { \mathbb T^ 2} dx\ e^ { i\frac { 2\pi } L kx} u(t,x)
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\end { equation}
for $ k \in \mathbb Z ^ 2 $ , and rewrite~\- (\ref { ins} ) as
\begin { equation}
\partial _ t\hat u_ k=
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-\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} }
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(q\cdot \hat u_ p)\hat u_ q
,\quad
k\cdot \hat u_ k=0
\label { ins_ k}
\end { equation}
We then reduce the equation to a scalar one, by writing
\begin { equation}
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\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)
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\end { equation}
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in terms of which, multiplying both sides of the equation by $ \frac L { i 2 \pi } \frac { k ^ \perp } { |k| } $ ,
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\begin { equation}
\partial _ t\hat \varphi _ k=
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-\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} }
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\frac { (q\cdot p^ \perp )(k^ \perp \cdot q^ \perp )} { |q||p|} \hat \varphi _ p\hat \varphi _ q
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.
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\label { ins_ k}
\end { equation}
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Furthermore
\begin { equation}
(q\cdot p^ \perp )(k^ \perp \cdot q^ \perp )
=
(q\cdot p^ \perp )(q^ 2+p\cdot q)
\end { equation}
and $ q \cdot p ^ \perp $ is antisymmetric under exchange of $ q $ and $ p $ . Therefore,
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\begin { equation}
\partial _ t\hat \varphi _ k=
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-\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} }
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\frac { (q\cdot p^ \perp )|q|} { |p|} \hat \varphi _ p\hat \varphi _ q
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.
\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
\begin { equation}
\mathcal K:=\{ (k_ 1,k_ 2),\ |k_ 1|\leqslant K_ 1,\ |k_ 2|\leqslant K_ 2\}
.
\end { equation}
\bigskip
\point { \bf FFT} . We compute the last term in~\- (\ref { ins_ k} )
\begin { equation}
T(\hat \varphi ,k):=
\sum _ { \displaystyle \mathop { \scriptstyle p,q\in \mathbb Z^ 2} _ { p+q=k} }
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\frac { (q\cdot p^ \perp )|q|} { |p|} \hat \varphi _ q\hat \varphi _ p
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\end { equation}
using a fast Fourier transform, defined as
\begin { equation}
\mathcal F(f)(n):=\sum _ { m\in \mathcal N} e^ { -\frac { 2i\pi } { N_ 1} m_ 1n_ 1-\frac { 2i\pi } { N_ 2} m_ 2n_ 2} f(m_ 1,m_ 2)
\end { equation}
where
\begin { equation}
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\mathcal N:=\{ (n_ 1,n_ 2),\ 0\leqslant n_ 1< N_ 1,\ 0\leqslant n_ 2< N_ 2\}
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\end { equation}
for some fixed $ N _ 1 ,N _ 2 $ . The transform is inverted by
\begin { equation}
\frac 1{ N_ 1N_ 2} \mathcal F^ *(\mathcal F(f))(n)=f(n)
\end { equation}
in which $ \mathcal F ^ * $ is defined like $ \mathcal F $ but with the opposite phase.
\bigskip
\indent The condition $ p + q = k $ can be rewritten as
\begin { equation}
T(\hat \varphi ,k)
=
\sum _ { p,q\in \mathcal K}
\frac 1{ 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)}
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(q\cdot p^ \perp )\frac { |q|} { |p|} \hat \varphi _ q\hat \varphi _ p
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\end { equation}
provided
\begin { equation}
N_ i>4K_ i.
\end { equation}
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Indeed, $ \sum _ { n _ i = 0 } ^ { N _ i } e ^ { - \frac { 2 i \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 \in \mathcal K $ , then $ |p _ i + q _ i| \leqslant 2 K _ i $ .
Therefore,
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\begin { equation}
T(\hat \varphi ,k)
=
\textstyle
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\frac 1{ N_ 1N_ 2}
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\mathcal F^ *\left (
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\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_ y\hat \varphi _ p} { |p|} \right )(n)
\mathcal F\left (q_ x|q|\hat \varphi _ q\right )(n)
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\right )(k)
\end { equation}
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\bigskip
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\point { \bf Energy} .
We define the energy as
\begin { equation}
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E(t)=\frac 12\int \frac { dx} { L^ 2} \ u^ 2(t,x)=\frac 12\sum _ { k\in \mathbb Z^ 2} |\hat u_ k|^ 2
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.
\end { equation}
We have
\begin { equation}
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\partial _ t E=\int \frac { dx} { L^ 2} \ u\partial tu
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=
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\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
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.
\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=
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\int \frac { dx} { L^ 2} \ \left (-\nu |\nabla u|^ 2+ug\right )
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=
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\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 )
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.
\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}
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\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
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\leqslant
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-\frac { 8\pi ^ 2} { L^ 2} \nu E+\frac { 4\pi ^ 2} { L^ 2} \nu \hat u_ 0^ 2+
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\| \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}
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\partial _ t E\leqslant -\frac { 8\pi ^ 2} { L^ 2} \nu E+\| \hat g\| _ 2\sqrt { 2E}
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.
\end { equation}
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Now, if $ \frac { 8 \pi ^ 2 } { L ^ 2 } \nu \sqrt E< \sqrt 2 \| \hat g \| _ 2 $ , then
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\begin { equation}
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\frac { \partial _ t E} { -\frac { 8\pi ^ 2} { L^ 2} \nu E+\| \hat g\| _ 2\sqrt { 2E} } \leqslant 1
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\end { equation}
and so
\begin { equation}
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\frac { \log (1-\frac { 8\pi ^ 2\nu } { L^ 2\sqrt 2\| \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\sqrt 2\| \hat g\| _ 2} \sqrt { E(0)} )} { -\frac { 4\pi ^ 2} { L^ 2} \nu }
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\end { equation}
and
\begin { equation}
E(t)
\leqslant
\left (
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\frac { L^ 2\sqrt 2\| \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)}
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\right )^ 2
.
\end { equation}
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If $ \frac { 8 \pi ^ 2 } { L ^ 2 } \nu \sqrt E> \sqrt 2 \| \hat g \| _ 2 $ ,
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\begin { equation}
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\frac { \partial _ t E} { -\frac { 8\pi ^ 2} { L^ 2} \nu E+\| \hat g\| _ 2\sqrt { 2E} } \geqslant 1
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\end { equation}
and so
\begin { equation}
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\frac { \log (\frac { 8\pi ^ 2\nu } { L^ 2\sqrt 2\| \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\sqrt 2\| \hat g\| _ 2} \sqrt { E(0)} )-1} { -\frac { 4\pi ^ 2} { L^ 2} \nu }
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\end { equation}
and
\begin { equation}
E(t)
\leqslant
\left (
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\frac { L^ 2\sqrt 2\| \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)}
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\right )^ 2
.
\end { equation}
\bigskip
\point { \bf Enstrophy} .
The enstrophy is defined as
\begin { equation}
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\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
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.
\end { equation}
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\vfill
\eject
\begin { thebibliography} { WWW99}
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\IfFileExists { bibliography/bibliography.tex} { \input bibliography/bibliography.tex} { }
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\end { document}