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\documentclass{ian}

\usepackage{largearray}
\usepackage{dsfont}

\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 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)
  =:\mathfrak F_k(\hat u)
  \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}

\subsubsection{Runge-Kutta methods}.
To solve the equation numerically, we will use Runge-Kutta methods, which compute an approximate value $\hat u_k^{(n)}$ for $\hat u_k(t_n)$.
{\tt nstrophy} supports the 4th order Runge-Kutta ({\tt RK4}) and 2nd order Runge-Kutta ({\tt RK2}) algorithms.
In addition, several variable step methods are implemented:
\begin{itemize}
  \item the Runge-Kutta-Dormand-Prince method ({\tt RKDP54}), which is of 5th order, and adjusts the step by comparing to a 4th order method;
  \item the Runge-Kutta-Fehlberg method ({\tt RKF45}), which is of 4th order, and adjusts the step by comparing to a 5th order method;
  \item the Runge-Kutta-Bogacki-Shampine method ({\tt RKBS32}), which is of 3d order, and adjusts the step by comparing to a 2nd order method.
\end{itemize}
In these adaptive step methods, two steps are computed at different orders: $\hat u_k^{(n)}$ and $\hat U_k^{(n)}$, the step size is adjusted at every step in such a way that the error is small enough:
\begin{equation}
  D(\hat u^{(n)},\hat U^{(n)})
  <\epsilon_{\mathrm{target}}
\end{equation}
for some given {\it cost function} $D$, and $\epsilon_{\mathrm{target}}$, set using the {\tt adaptive\_tolerance} parameter.
The choice of the cost function matters, and will be discussed below.
If the error is larger than the target, then the step size is decreased.
How this is done depends on the order of algorithm.
If the order is $q$ (here we mean the smaller of the two orders, so 4 for {\tt RKDP54} and {\tt RKF45} and 2 for {\tt RKBS32}), then we expect (as long as the cost function is such that $D(\hat u,\hat u+\varphi)\sim\|\varphi\|$ in some norm)
\begin{equation}
  D(\hat u^{(n)},\hat U^{(n)})=\delta_n^qC_n
\end{equation}
for some number $C_n$.
We wish to set $\delta_{n+1}$ so that
\begin{equation}
  \delta_{n+1}^qC_n=\epsilon_{\mathrm{target}}
\end{equation}
so
\begin{equation}
  \delta_{n+1}
  =\left(\frac{\epsilon_{\mathrm{target}}}{C_n}\right)^{\frac1q}
  =\delta_n\left(\frac{\epsilon_{\mathrm{target}}}{D(\hat u^{(n)},\hat U^{(n)})}\right)^{\frac1q}
  .
  \label{adaptive_delta}
\end{equation}
(Actually, to be safe and ensure that $\delta$ decreases sufficiently, we multiply this by a safety factor that can be set using the {\tt adaptive\_factor} parameter.)
If the error is smaller than the target, we increase $\delta$ using\-~(\ref{adaptive_delta}) (without the safety factor).
To be safe, we also set a maximal value for $\delta$ via the {\tt max\_delta} parameter.
\bigskip

\indent
The choice of the cost function $D$ matters.
It can be made by specifying the parameter {\tt adaptive\_cost}.
\begin{itemize}
  \item
  For computations where the main focus is the enstrophy\-~(\ref{enstrophy}), one may want to set the cost function to the relative difference of the enstrophies:
  \begin{equation}
    D(\hat u,\hat U):=\frac{|\mathcal En(\hat u)-\mathcal En(\hat U)|}{\mathcal En(\hat u)}
    .
  \end{equation}
  This cost function is selected by choosing {\tt adaptive\_cost=enstrophy}.

  \item
  For computations where the main focus is the value of $\alpha$\-~(\ref{alpha}), one may want to set the cost function to the relative difference of $\alpha$:
  \begin{equation}
    D(\hat u,\hat U):=\frac{|\alpha(\hat u)-\alpha(\hat U)|}{|\alpha(\hat u)|}
    .
  \end{equation}
  This cost function is selected by choosing {\tt adaptive\_cost=alpha}.

  \item Alternatively, one my take $D$ to be the normalized $L_1$ norm:
  \begin{equation}
    D(\hat u,\hat U):=
    \frac1{\mathcal N}\sum_k|\hat u_k-\hat U_k|
    ,\quad
    \mathcal N:=\sum_k|\hat u_k|
    .
  \end{equation}
  This function is selected by choosing {\tt adaptive\_cost=L1}.

  \item Empirically, we have found that $|\hat u-\hat U|$ behaves like $k^{-3}$ for {\tt RKDP54} and {\tt RKF45}, and like $k^{-\frac32}$ for {\tt RKBS32}, so a cost function of the form
  \begin{equation}
    D(\hat u,\hat U):=\frac1{\mathcal N}\sum_k|\hat u_k-\hat U_k|k^{-3}
    ,\quad
    \mathcal N:=\sum_k|\hat u_k|k^{-3}
  \end{equation}
  or
  \begin{equation}
    D(\hat u,\hat U):=\frac1{\mathcal N}\sum_k|\hat u_k-\hat U_k|k^{-\frac32}
    ,\quad
    \mathcal N:=\sum_k|\hat u_k|k^{-\frac32}
  \end{equation}
  are sensible choices.
  These cost functions are selected by choosing {\tt adaptive\_cost=k3} and {\tt adaptive\_cost=k32} respectively.
\end{itemize}

\subsubsection{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}
In order to keep the computation as quick as possible, we only compute and store the values for $k_1\geqslant 0$.

\subsubsection{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}

\subsubsection{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
  .
  \label{enstrophy}
\end{equation}

\subsubsection{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}

\subsubsection{Lyapunov exponents}
\indent
To compute the Lyapunov exponents, we must first compute the Jacobian of $\hat u^{(n)}\mapsto\hat u^{(n+1)}$.
This map is always of Runge-Kutta type, that is,
\begin{equation}
  \hat u(t_{n+1})=\mathfrak F_{t_n}(\hat u(t_n))
  .
\end{equation}
Let $D\mathfrak F_{t_{n}}$ be the Jacobian of this map, in which we split the real and imaginary parts: if
\begin{equation}
  \hat u_k(t_n)=:\rho_k+i\iota_k
  ,\quad
  \mathfrak F_{t_n}(\hat u(t_n))_k=:\phi_k+i\psi_k
\end{equation}
then
\begin{equation}
  (D\mathfrak F_{t_n})_{k,p}:=\left(\begin{array}{cc}
    \partial_{\rho_p}\phi_k&\partial_{\iota_p}\phi_k\\
    \partial_{\rho_p}\psi_k&\partial_{\iota_p}\psi_k
  \end{array}\right)
\end{equation}
We compute this Jacobian numerically using a finite difference, by computing
\begin{equation}
  (D\mathfrak F_{t_n})_{k,p}:=\frac1\epsilon
  \left(\begin{array}{cc}
    \phi_k(\hat u+\epsilon\delta_p)-\phi_k(\hat u)&\phi_k(\hat u+i\epsilon\delta_p)-\phi_k(\hat u)\\
    \psi_k(\hat u+\epsilon\delta_p)-\psi_k(\hat u)&\psi_k(\hat u+i\epsilon\delta_p)-\psi_k(\hat u)
  \end{array}\right)
  .
\end{equation}
The parameter $\epsilon$ can be set using the parameter {\tt D\_epsilon}.
%, so
%\begin{equation}
%  D\hat u^{(n+1)}=\mathds 1+\delta\sum_{i=1}^q w_iD\mathfrak F(\hat u^{(n)})
%  .
%\end{equation}
%We then compute
%\begin{equation}
%  (D\mathfrak F(\hat u))_{k,\ell}
%  =
%  -\frac{4\pi^2}{L^2}\nu k^2\delta_{k,\ell}
%  +\frac{4\pi^2}{L^2|k|}\partial_{\hat u_\ell}T(\hat u,k)
%\end{equation}
%and, by\-~(\ref{T}),
%\begin{equation}
%  \partial_{\hat u_\ell}T(\hat u,k)
%  =
%  \sum_{\displaystyle\mathop{\scriptstyle q\in\mathbb Z^2}_{\ell+q=k}}
%  \left(
%    \frac{(q\cdot \ell^\perp)|q|}{|\ell|}
%    +
%    \frac{(\ell\cdot q^\perp)|\ell|}{|q|}
%  \right)\hat u_q
%  =
%  (k\cdot \ell^\perp)\left(
%    \frac{|k-\ell|}{|\ell|}
%    -
%    \frac{|\ell|}{|k-\ell|}
%  \right)\hat u_{k-\ell}
%  .
%\end{equation}
\bigskip

\indent
The Lyapunov exponents are then defined as
\begin{equation}
  \frac1{t_n}\log\mathrm{spec}(\pi_{t_n})
  ,\quad
  \pi_{t_n}:=\prod_{i=1}^nD\hat u(t_i)
  .
\end{equation}
However, the product of $D\hat u$ may become quite large or quite small (if the exponents are not all 1).
To avoid this, we periodically rescale the product.
We set $\mathfrak L_r>0$ (set by adjusting the {\tt lyanpunov\_reset} parameter), and, when $t_n$ crosses a multiple of $\mathfrak L_r$, we rescale the eigenvalues of $\pi_i$ to 1.
To do so, we perform a $QR$ decomposition:
\begin{equation}
  \pi_{\alpha\mathfrak L_r}
  =R^{(\alpha)}Q^{(\alpha)}
\end{equation}
where $Q^{(\alpha)}$ is orthogonal and $R^{(\alpha)}$ is a diagonal matrix, and we divide by $R^{(\alpha)}$ (thus only keeping $Q^{(\alpha)}$).
The Lyapunov exponents at time $\alpha\mathfrak L_r$ are then
\begin{equation}
  \frac1{\alpha\mathfrak L_r}\sum_{\beta=1}^\alpha\log\mathrm{spec}(Q^{(\beta)})
  .
\end{equation}

\subsubsection{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}
  .
  \label{alpha}
\end{equation}
Note that, by\-~(\ref{realu})-(\ref{realT}),
\begin{equation}
  \alpha(\hat u)\in\mathbb R
  .
\end{equation}



\vfill
\eject

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