Is there an analytic solution for this partial differential equation? The Fokker-Planck equation for a probability distribution $P(\theta,t)$:
\begin{align}
\frac{\partial P(\theta,t)}{\partial t}=-\frac{\partial}{\partial\theta}\Big[[\sin(k\theta)+f]P(\theta,t)-D\frac{\partial P(\theta,t)}{\partial\theta}\Big].
\end{align}
where $f$, $k$, $D$ are constants, and the initial distribution is a delta function.
 A: Not a complete solution, but some ideas that may be helpful. Since the eigen-expansion of the solution $P(\theta,t)$ must be of the form $\displaystyle\sum_{i=1}^{n} c_{i} e^{-\lambda_{i}t}\psi_{i}(\theta)$, where $\left(\lambda_{i},\psi_{i}\right)$ are the $i$-th eigenvalue-eigenfunction pair, hence we can substutute the ansatz $e^{-\lambda t}\psi(\theta)$ for $P(\theta,t)$ in the Fokker-Planck PDE, and multiply both sides by $e^{\lambda t}$ to get the following second order homogeneous linear ODE
$$ D\psi^{\prime\prime} - \left(f + \sin(k\theta)\right)\psi^{\prime} + \left(\lambda - k\cos(k\theta)\right)\psi = 0 $$
where $^{\prime}$ denotes derivative w.r.t. $\theta$. From here, if we can find all pairs $(\lambda,\psi)$ those solve the above ODE, then the transient PDE solution can be constructed by linear combination of such $e^{-\lambda t}\psi(\theta)$. Notice that for $\lambda = 0$, the above ODE coincides with the one you'd get if you had set $\displaystyle\frac{\partial P}{\partial t} = 0$ in the original equation, which means the eigenfunction corresponding to $\lambda = 0$, is the stationary density. The constants $c_{i}$ would follow from the normalization condition $\displaystyle\int_{-\pi}^{\pi}P(\theta,t) d\theta = 1$ for all $t$. So the question now is whether one could solve the above ODE in closed form.
A: Starting from Abhishek Halder's derivation of an ODE, we can indeed get a closed form, involving the solutions of the double-confluent Heun equation.
$\phi \left( \theta \right) = 
C_1 {\it HeunD} \left({\frac{2}{kd}},{\frac { \left( 2\,k-4\,\lambda \right) d+{f}^{2}+1}{k^{2}d^{2}}},{\frac {4\,if}{{k}^{2}{d}^{2}}},{\frac { \left( 2k+4\lambda \right) d-{f}^{2}-1}{{k}^{2}{d}^{2}}},{\frac {i}{\tan \left(k\theta/2 \right) }}\right) 
{{\rm e}^{{\frac {1}{kd} \left( -i\sin\left( k\theta/2 \right) \cos \left( k\theta/2 \right) +f
\arctan \left( {\frac {\sin \left( k\theta/2 \right) }{\cos \left( 
k\theta/2 \right) }} \right) - \left( \cos \left( k\theta/2
 \right)  \right) ^{2} \right) }}} +
C_2\,{\it HeunD} \left({-\frac {2}{kd}},{\frac { \left( 2k-4\lambda \right) d+{f}^{2}+1}{{
k}^{2}{d}^{2}}},{\frac {4\,if}{{k}^{2}{d}^{2}}},{\frac { \left( 2k+4
\lambda \right) d-{f}^{2}-1}{{k}^{2}{d}^{2}}},{\frac {i}{\tan
 \left( k\theta/2 \right) }} \right) {{\rm e}^{{\frac {1}{kd}
 \left( i\sin \left( k\theta/2 \right) \cos \left( k\theta/2
 \right) +f\arctan \left( {\frac {\sin \left( k\theta/2 \right) }{
\cos \left( k\theta/2 \right) }} \right) - \left( \cos \left(
k\theta/2 \right)  \right) ^{2} \right) }}}
$
A change of variables, $\theta \mapsto t/k$, can simplify the results a fair bit.somewhat.
A: Closed form fundamental solutions in many case are obtained in Igor Tanski's paper

