solution to the Fokker-Planck equation In my textbook for Evolutionary Dynamics I came across the following Fokker-Planck equation:
$\partial_t\psi(p,t)=-m\partial_p\psi(p,t)+\frac{\nu}{2}\partial_p^2\psi(p,t)$.
I know that the solution is a gaussian but I can't really see how you get to this solution. In the text they say "for constant $m\neq 0$, a solution can be found by substituting 
$z=p-mt$ for $p$. 
Unfortunately, I don't really know how I can then derive a solution. I have tried an ansatz like 
$\psi(p,t)=P(p)T(t)$ 
as I usually do for such PDE's but then I somewhat do not use the given substitution. Can anyone help me with this?
Thanks a lot!
 A: I'll use dots for the $t$ derivative and primes for the $p$ derivative. Let's take your ansatz and plug it into the PDE. We get
$$\dot{T}P=-mTP'+\frac{\nu}{2}\,TP''.$$
We do the usual and divide through by $TP$ to get
$$\frac{\dot{T}}{T}=-\frac{mP'}{P}+\frac{\nu P''}{2P}=\frac{\nu P''-2mP'}{2P}.$$
Now we do the separation step and set each side equal to a constant. Because we have second-order in $P,$ the sign of the constant will matter. We'll just say it's $k,$ and let it be signed. We end up with the two ODE's
\begin{align*}
\dot{T}&=kT \\
\nu P''-2mP'-2kP&=0.
\end{align*}
The solution to the first is, of course, $T=T_0e^{kt}.$ If we do a $P=e^{rp}$ ansatz we'll get the usual characteristic polynomial $\nu r^2-2mr-2k=0,$ with discriminant
$4m^2+8\nu k. $ So the relative size of $m^2$ and $2\nu k$ will determine the type of solutions you get. You'll either get pure sinusoidal, if $m^2+2\nu k<0,$ or an exponential combination $A\,e^{rp}+B\,e^{-rp},$ if $m^2+2\nu k>0,$ or the unusual case $A\,e^{rp}+B\,p\,e^{rp},$ if $m^2+2\nu k=0.$ None of these is a Gaussian. Your final solution will be:
$$
\psi=T_0e^{kt}\times\begin{cases}A\sin(rp)+B\cos(rp),\;&m^2+2\nu k<0 \\A\,e^{rp}+B\,p\,e^{rp},&m^2+2\nu k=0 \\ A\,e^{rp}+B\,e^{-rp},&m^2+2\nu k>0 \end{cases}.
$$
One important caveat: these solutions assume that neither $m$ nor $\nu$ depend on $p$. They also assume that the solution you're seeking is separable into a $\psi=T(t)\,P(p).$ It might be that there are desirable solutions that don't look like that. Another note: in the $m^2+2\nu k>0$ case you could also take $$P=A\sinh(rp)+B\cosh(rp),$$ if you want to take advantage of hyperbolic trig identities. 
