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\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
  +\frac{4\pi^2}{|k|}\sum_{\displaystyle\mathop{\scriptstyle p,q\in\mathbb Z^2}_{p+q=k}}
  (q\cdot p^\perp)(k^\perp\cdot q^\perp)\hat\varphi_p\hat\varphi_q
  .
  \label{ins_k}
\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 \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}}
  \frac{(q\cdot p^\perp)|q|}{|p|}\hat\varphi_p\hat\varphi_q
  .
  \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}}
  \frac{(q\cdot p^\perp)|q|}{|p|}\hat\varphi_q\hat\varphi_p
\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}
  \mathcal N:=\{(n_1,n_2),\ 0\leqslant n_1\leqslant N_1,\ 0\leqslant n_2\leqslant 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\varphi,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\varphi_q\hat\varphi_p
\end{equation}
provided
\begin{equation}
  N_i>4K_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 $p,q\in\mathcal K$, then $|p_i+q_i|\leqslant2K_i$. Therefore,
\begin{equation}
  T(\hat\varphi,k)
  =
  \textstyle
  \frac1{N_1N_2}
  \mathcal F^*\left(
    \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)
  \right)(k)
\end{equation}
\bigskip


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