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\hfil{\bf\LARGE
{\tt nstrophy}
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\section{Description of the computation}
\subsection{Irreversible equation}
\indent Consider the incompressible Navier-Stokes equation in 2 dimensions
\begin{equation}
  \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.
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):=\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=
  -\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
  \label{ins_k}
\end{equation}
We then reduce the equation to a scalar one, by writing
\begin{equation}
  \hat U_k=\frac{i2\pi k^\perp}{L|k|}\hat u_k\equiv\frac{i2\pi}{L|k|}(-k_y\hat u_k,k_x\hat u_k)
  \label{udef}
\end{equation}
in terms of which, multiplying both sides of the equation by $\frac L{i2\pi}\frac{k^\perp}{|k|}$,
\begin{equation}
  \partial_t\hat u_k=
  -\frac{4\pi^2}{L^2}\nu k^2\hat u_k
  +\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 u_p\hat u_q
  \label{ins_k}
\end{equation}
with
\begin{equation}
  \hat g_k:=\frac{Lk^\perp}{2i\pi|k|}\cdot\hat G_k
  .
  \label{gdef}
\end{equation}
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,
\begin{equation}
  \partial_t\hat u_k=
  -\frac{4\pi^2}{L^2}\nu k^2\hat u_k+\hat g_k
  +\frac{4\pi^2}{L^2|k|}T(\hat u,k)
  \label{ins_k}
\end{equation}
with
\begin{equation}
  T(\hat u,k):=
  \sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
  \frac{(q\cdot p^\perp)|q|}{|p|}\hat u_p\hat u_q
  .
  \label{T}
\end{equation}
We truncate the Fourier modes and assume that $\hat u_k=0$ 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 Reality}.
Since $U$ is real, $\hat U_{-k}=\hat U_k^*$, and so
\begin{equation}
  \hat u_{-k}=\hat u_k^*
  .
  \label{realu}
\end{equation}
Similarly,
\begin{equation}
  \hat g_{-k}=\hat g_k^*
  .
  \label{realg}
\end{equation}
Thus,
\begin{equation}
  T(\hat u,-k)
  =
  T(\hat u,k)^*
  .
  \label{realT}
\end{equation}
\bigskip

\point{\bf FFT}. We compute T 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}
  \mathcal N:=\{(n_1,n_2),\ 0\leqslant n_1< N_1,\ 0\leqslant n_2< N_2\}
\end{equation}
for some fixed $N_1,N_2$. The transform is inverted by
\begin{equation}
  \frac1{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 u,k)
  =
  \sum_{p,q\in\mathcal K}
  \frac1{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)}
  (q\cdot p^\perp)\frac{|q|}{|p|}\hat u_q\hat u_p
\end{equation}
provided
\begin{equation}
  N_i>3K_i.
\end{equation}
Indeed, $\sum_{n_i=0}^{N_i}e^{-\frac{2i\pi}{N_i}n_im_i}$ vanishes unless $m_i=0\%N_i$ (in which $\%N_i$ means `modulo $N_i$'), and, if $p,q,k\in\mathcal K$, then $|p_i+q_i-k_i|\leqslant3K_i$, so, as long as $N_i>3K_i$, then $(p_i+q_i-k_i)=0\%N_i$ implies $p_i+q_i=k_i$.
Therefore,
\begin{equation}
  T(\hat u,k)
  =
  \textstyle
  \frac1{N_1N_2}
  \mathcal F^*\left(
    \mathcal F\left(\frac{p_x\hat u_p}{|p|}\right)(n)
    \mathcal F\left(q_y|q|\hat u_q\right)(n)
    -
    \mathcal F\left(\frac{p_y\hat u_p}{|p|}\right)(n)
    \mathcal F\left(q_x|q|\hat u_q\right)(n)
  \right)(k)
\end{equation}
\bigskip

\point{\bf Energy}.
We define the energy as
\begin{equation}
  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\frac{dx}{L^2}\ U\partial tU
  =
  \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,
\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\frac{dx}{L^2}\ \left(-\nu|\nabla U|^2+UG\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,
\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 -\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
  -\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 -\frac{8\pi^2}{L^2}\nu E+\|\hat G\|_2\sqrt{2E}
  .
\end{equation}
Now, if $\frac{8\pi^2}{L^2}\nu\sqrt E<\sqrt2\|\hat G\|_2$, then
\begin{equation}
  \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}{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{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 $\frac{8\pi^2}{L^2}\nu\sqrt E>\sqrt2\|\hat G\|_2$,
\begin{equation}
  \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}{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{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}
\bigskip

\point{\bf Enstrophy}.
The enstrophy is defined as
\begin{equation}
  \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}
\bigskip

\point{\bf Numerical instability}.
In order to prevent the algorithm from blowing up, it is necessary to impose the reality of $u(x)$ by hand, otherwise, truncation errors build up, and lead to divergences.
It is sufficient to ensure that the convolution term $T(\hat u,k)$ satisfies $T(\hat u,-k)=T(\hat u,k)^*$.
After imposing this condition, the algorithm no longer blows up, but it is still unstable (for instance, increasing $K_1$ or $K_2$ leads to very different results).

\subsection{Reversible equation}
\indent The reversible equation is similar to\-~(\ref{ins}) but instead of fixing the viscosity, we fix the enstrophy\-~\cite{Ga22}.
It is defined directly in Fourier space:
\begin{equation}
  \partial_t\hat U_k=
  -\frac{4\pi^2}{L^2}\alpha(\hat U) 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
\end{equation}
where $\alpha$ is chosen such that the enstrophy is constant.
In terms of $\hat u$\-~(\ref{udef}), (\ref{gdef}), (\ref{T}):
\begin{equation}
  \partial_t\hat u_k=
  -\frac{4\pi^2}{L^2}\alpha(\hat u) k^2\hat u_k
  +\hat g_k
  +\frac{4\pi^2}{L^2|k|}T(\hat u,k)
  .
  \label{rns_k}
\end{equation}
To compute $\alpha$, we use the constancy of the enstrophy:
\begin{equation}
  \sum_{k\in\mathbb Z^2}k^2\hat U_k\cdot\partial_t\hat U_k
  =0
\end{equation}
which, in terms of $\hat u$ is
\begin{equation}
  \sum_{k\in\mathbb Z^2}k^2\hat u_k^*\partial_t\hat u_k
  =0
\end{equation}
that is
\begin{equation}
  \frac{4\pi^2}{L^2}\alpha(\hat u)\sum_{k\in\mathbb Z^2}k^4|\hat u_k|^2
  =
  \sum_{k\in\mathbb Z^2}k^2\hat u_k^*\hat g_k
  +\frac{4\pi^2}{L^2}\sum_{k\in\mathbb Z^2}|k|\hat u_k^*T(\hat u,k)
\end{equation}
and so
\begin{equation}
  \alpha(\hat u)
  =\frac{\frac{L^2}{4\pi^2}\sum_k k^2\hat u_k^*\hat g_k+\sum_k|k|\hat u_k^*T(\hat u,k)}{\sum_kk^4|\hat u_k|^2}
  .
\end{equation}
Note that, by\-~(\ref{realu})-(\ref{realT}),
\begin{equation}
  \alpha(\hat u)\in\mathbb R
  .
\end{equation}



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