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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – Jay Aug 15 '15 at 4:06
  • $\begingroup$ I meant the answer given (right now, above) my answer. $\endgroup$ – Jacques Carette Aug 15 '15 at 13:58
  • $\begingroup$ Hi, how to transform Abhishek Halder's ODE to that of double-confluent Heun equation you mentioned? I tried z=\sin(\theta), but it seems not work. $\endgroup$ – Jay Sep 6 '15 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. $\endgroup$ – Jacques Carette Sep 6 '15 at 16:03
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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.

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Closed form fundamental solutions in many case are obtained in Igor Tanski's paper

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  • $\begingroup$ the cases discussed in the paper does not include my question. Thank you! $\endgroup$ – Jay Apr 15 '15 at 4:37

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