The FokkerPlanck 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.
3 Answers
Starting from Abhishek Halder's derivation of an ODE, we can indeed get a closed form, involving the solutions of the doubleconfluent Heun equation.
$\phi \left( \theta \right) = C_1 {\it HeunD} \left({\frac{2}{kd}},{\frac { \left( 2\,k4\,\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( 2k4\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.

$\begingroup$ Hi, thanks for your reply. Could you give the reference for "Abhishek Halder's derivation of an ODE" so that I can learn the details? $\endgroup$– JayAug 15, 2015 at 4:06

$\begingroup$ I meant the answer given (right now, above) my answer. $\endgroup$ Aug 15, 2015 at 13:58

$\begingroup$ Hi, how to transform Abhishek Halder's ODE to that of doubleconfluent Heun equation you mentioned? I tried z=\sin(\theta), but it seems not work. $\endgroup$– JaySep 6, 2015 at 7:41

$\begingroup$ I don't know the details  the solution above was derived by Maple. I am sure that either Maple or Mathematica should be quite helpful in deriving the ODE transformations you need. $\endgroup$ Sep 6, 2015 at 16:03
Not a complete solution, but some ideas that may be helpful. Since the eigenexpansion 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 eigenvalueeigenfunction pair, hence we can substutute the ansatz $e^{\lambda t}\psi(\theta)$ for $P(\theta,t)$ in the FokkerPlanck 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.
Closed form fundamental solutions in many case are obtained in Igor Tanski's paper

$\begingroup$ the cases discussed in the paper does not include my question. Thank you! $\endgroup$– JayApr 15, 2015 at 4:37