I am trying to solve the Smoluchowski equation for diffusing particles under the influence of a potential $W(x)$. $$\frac{\partial P(x,t)}{\partial t} = D\frac{\partial^2 P(x,t)}{\partial x^2} + \beta D \frac{dW(x)}{dx}\frac{\partial P(x,t)}{\partial x} + \beta D \frac{d^2 W(x)}{dx^2} P(x,t) $$ Where $P(x,t)$ is the probability of finding a particle at position $x$, at time $t$. The problem is 1-D over domain 0 to 2b.
The boundary conditions are given by $$ P(0,t) = 0 , P(2b,t)=0 $$ and initial condition is $$P(x,0) = \delta(x-b)$$
What I have done so far is solve this for a potential given by $W(x) = vx$. This makes the last term on the right of the PDE disappear, and the coefficients constant, reducing the PDE to:
$$\frac{\partial P(x,t)}{\partial t} = D\frac{\partial^2 P(x,t)}{\partial x^2} + \beta D v\frac{\partial P(x,t)}{\partial x} $$
Which I then was able to solve via Seperation of Variables $P(x,t) = X(x)T(t)$.
Giving me $$ P(x,t) = \frac{1}{b} e^{vb/2D}e^{-vx/2D}\sum_{n=1}^\infty \sin(\frac{n\pi}{2})\sin(\frac{n\pi x}{2})e^{-(n^2\pi^2D/4b^2+v^2/4D)t} $$
However, I would like to solve for $W(x) = -\frac{1}{2}(\rho_L-\rho_v)\tanh(\frac{x-x_0}{s})$.
Now the most right hand term does not disappear, and also the coefficients are now non-constant. Is this solve-able? If so, how?
Thank you.