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\begin{document}

Consider the SDE
\begin{equation*}
  dU(t) = 2 \begin{pmatrix} -\sin(U_2(t)) \\ \phantom{+}\cos(U_1(t)) \end{pmatrix} dt
       + \sigma \,dB(t)
       \qquad \mbox{for $t\in[0,2\pi]$.}
\end{equation*}

\textbf{1.}
In \cite[section 9]{HaiStuaVo07} we conjecture, that the distribution
of this SDE coincides with the stationary distribution of the SPDE
\begin{equation*}
  \d_t u = \frac{1}{\sigma^2}\d_x^2 u
	     + \frac{1}{\sigma^2}\bigl(Df^*-Df\bigr)\d_x u
	     - \nabla \Phi(u) + \sqrt{2}\, \partial_t w
\end{equation*}
where $f(u_1, u_2) = \bigl( -2\sin(u_2), 2\cos(u_1) \bigr)$, $\Phi =
\frac12 \bigl(|f|^2/\sigma^2+\mathrm{div} f\bigr)$ and $Df$ is the
Jacobian matrix of~$f$.  In our case we have
\begin{equation*}
  Df(u) =
  2 \begin{pmatrix}
      0 & -\cos(u_2) \\
      -\sin(u_1) & 0
    \end{pmatrix},
\end{equation*}
$\mathrm{div}f = 0$ and
\begin{equation*}
  \nabla\Phi(u)
  = \frac{4}{2\sigma^2} \nabla \bigl(\sin(u_2)^2 + \cos(u_1)^2\bigr)
  = \frac{4}{\sigma^2} \begin{pmatrix}
    - \sin(u_1)\cos(u_1) \\
    \phantom{+} \cos(u_2)\sin(u_2)
  \end{pmatrix}.
\end{equation*}
Substituting these expressions we get
\begin{equation}\label{E:SPDE}
  \d_t u = \frac{1}{\sigma^2} \,\d_x^2 u
	     + \frac{2}{\sigma^2} \begin{pmatrix}
	       0 & \cos(u_2)-\sin(u_1) \\
	       \sin(u_1)-\cos(u_2) & 0
	     \end{pmatrix} \d_x u
	     + \frac{4}{\sigma^2} \begin{pmatrix}
	       \phantom{+} \sin(u_1)\cos(u_1) \\
	       - \cos(u_2)\sin(u_2)
	     \end{pmatrix}
	     + \sqrt{2}\, \partial_t w.
\end{equation}

\textbf{2.}
In our experiment we consider the SPDE \eqref{E:SPDE}, but with periodic
boundary conditions in order to not leave the framework of~\cite{HaiVo10}.
In \cite[conjecture 3.2]{HaiVo10}
we conjecture the additional drift term, when using the right hand derivative,
to be
\begin{equation*}
  - \frac{\tilde\sigma^2}{4\tilde\nu} \sum_{j=1}^2 \d_{j} G_{ij}(u)
  = \begin{pmatrix} \phantom{+}\sin(u_2) \\ -\cos(u_1) \end{pmatrix}
\end{equation*}
where $\tilde\sigma = \sqrt{2}$ and $\tilde\nu = 1/\sigma^2$
are $\sigma$ and $\nu$ from~\cite{HaiVo10}.

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