2018-01-11 22:48:14 +00:00
\documentclass { ian}
\usepackage { largearray}
\begin { document}
\hbox { }
\hfil { \bf \LARGE
{ \tt nstrophy}
}
\vfill
\tableofcontents
\vfill
\eject
\setcounter { page} 1
\pagestyle { plain}
\section { Description of the computation}
\subsection { Irreversible equation}
\indent Consider the { \it irreversible} Navier-Stokes equation in 2 dimensions
\begin { equation}
\partial _ tu=\nu \Delta u+g-\nabla w-(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
\begin { equation}
\hat u_ k(t):=\int _ { \mathbb T^ 2} dx\ e^ { 2i\pi 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} }
(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}
\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)
\end { equation}
in terms of which, multiplying both sides of the equation by $ \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
2018-01-12 19:20:59 +00:00
+\frac { 4\pi ^ 2} { |k|} \sum _ { \displaystyle \mathop { \scriptstyle p,q\in \mathbb Z^ 2} _ { p+q=k} }
2022-05-12 07:32:58 +00:00
\frac { (q\cdot p^ \perp )(k^ \perp \cdot q^ \perp )} { |q||p|} \hat \varphi _ p\hat \varphi _ q
2018-01-12 19:20:59 +00:00
.
2018-01-11 22:48:14 +00:00
\label { ins_ k}
\end { equation}
2018-01-12 19:20:59 +00:00
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,
2018-01-11 22:48:14 +00:00
\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} }
2018-01-12 19:20:59 +00:00
\frac { (q\cdot p^ \perp )|q|} { |p|} \hat \varphi _ p\hat \varphi _ q
2018-01-11 22:48:14 +00:00
.
\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} }
2018-01-12 19:20:59 +00:00
\frac { (q\cdot p^ \perp )|q|} { |p|} \hat \varphi _ q\hat \varphi _ p
2018-01-11 22:48:14 +00:00
\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}
2022-05-12 07:32:58 +00:00
\mathcal N:=\{ (n_ 1,n_ 2),\ 0\leqslant n_ 1< N_ 1,\ 0\leqslant n_ 2< N_ 2\}
2018-01-11 22:48:14 +00:00
\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)}
2018-01-12 19:20:59 +00:00
(q\cdot p^ \perp )\frac { |q|} { |p|} \hat \varphi _ q\hat \varphi _ p
2018-01-11 22:48:14 +00:00
\end { equation}
provided
\begin { equation}
N_ i>4K_ i.
\end { equation}
2022-05-12 07:32:58 +00:00
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,
2018-01-11 22:48:14 +00:00
\begin { equation}
T(\hat \varphi ,k)
=
\textstyle
2018-01-12 19:20:59 +00:00
\frac 1{ N_ 1N_ 2}
2018-01-11 22:48:14 +00:00
\mathcal F^ *\left (
2018-01-12 19:20:59 +00:00
\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)
2018-01-11 22:48:14 +00:00
\right )(k)
\end { equation}
2018-01-12 19:20:59 +00:00
\bigskip
2018-01-11 22:48:14 +00:00
\vfill
\eject
\begin { thebibliography} { WWW99}
\small
\IfFileExists { bibliography/bibliography.tex} { \input bibliography/bibliography.tex} { }
\end { thebibliography}
\end { document}