Change variables into Fokker-Planck PDE I have a question regarding a PDE (Fokker-Planck) and change of variables. I have a problem deciding what route to take after I use the chain route. I have an expression
$$\frac{\partial u}{\partial t} = \frac{\partial}{\partial x}\left(\frac{\partial}{\partial x}+x\right)u$$
and would like the variables substitution
$$x = Xe^{-t} \text{ and } u=ve^t.$$
This give
$$\frac{\partial v}{\partial t} = e^{2t} \frac{\partial^2 v}{\partial X^2}.$$
I thank you for your help.  
 A: We have $$\partial_t u=\partial_x(\partial_x u+xu)=\partial_{xx}u+u+x\partial_xu.$$
We also also $v=e^{-t}u$, hence 
\begin{align}
\partial_t v&=-e^{-t}u+e^{-t}\partial_t u\\
&=-e^{—t}u+e^{—t}\partial_{xx}u+e^{-t}u+e^{-t}x\partial_x u\\
&=e^{—t}\partial_{xx}u+e^{-t}x\partial_x u.
\end{align}
Now compute $\partial_X V$ with the chain rule. We have $v(X,t)=u(x,t),$
and 
$$\partial_X V=\partial_x u\frac{\partial x}{\partial X}+\partial_t u\frac{\partial t}{\partial X},$$
then substitute with the values of $\frac{\partial x}{\partial X}$ and $\frac{\partial t}{\partial X}$, then do the same for the second derivative. 
A: We should take high precautions about changing variables of PDEs. Don't replay the mistakes like Why solving $\dfrac{\partial u}{\partial x}=\dfrac{\partial^2u}{\partial y^2}$ like this is wrong? !
Note that
$\dfrac{\partial u}{\partial t}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial x}+x\right)u$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial}{\partial x}\left(\dfrac{\partial u}{\partial x}+xu\right)$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial^2u}{\partial x^2}+x\dfrac{\partial u}{\partial x}+u$
Let $\begin{cases}x=Xe^{-t}\\T=t\end{cases}$ ,
Then $\begin{cases}X=xe^t\\T=t\end{cases}$
$\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial X}\dfrac{\partial X}{\partial x}+\dfrac{\partial u}{\partial T}\dfrac{\partial T}{\partial x}=e^t\dfrac{\partial u}{\partial X}$
$\dfrac{\partial^2u}{\partial x^2}=\dfrac{\partial u}{\partial x}\left(e^t\dfrac{\partial u}{\partial X}\right)=\dfrac{\partial u}{\partial X}\left(e^t\dfrac{\partial u}{\partial X}\right)\dfrac{\partial X}{\partial x}+\dfrac{\partial u}{\partial T}\left(e^t\dfrac{\partial u}{\partial X}\right)\dfrac{\partial T}{\partial x}=e^t\dfrac{\partial^2u}{\partial X^2}e^t=e^{2t}\dfrac{\partial^2u}{\partial X^2}$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial X}\dfrac{\partial X}{\partial t}+\dfrac{\partial u}{\partial T}\dfrac{\partial T}{\partial t}=xe^t\dfrac{\partial u}{\partial X}+\dfrac{\partial u}{\partial T}$
$\therefore xe^t\dfrac{\partial u}{\partial X}+\dfrac{\partial u}{\partial T}=e^{2t}\dfrac{\partial^2u}{\partial X^2}+xe^t\dfrac{\partial u}{\partial X}+u$
$\dfrac{\partial u}{\partial t}=e^{2t}\dfrac{\partial^2u}{\partial X^2}+u$
Let $u=ve^t$ ,
Then $\dfrac{\partial u}{\partial X}=\dfrac{\partial(ve^t)}{\partial X}=e^t\dfrac{\partial v}{\partial X}$
$\dfrac{\partial^2u}{\partial X^2}=\dfrac{\partial}{\partial X}\left(e^t\dfrac{\partial v}{\partial X}\right)=e^t\dfrac{\partial^2v}{\partial X^2}$
$\dfrac{\partial u}{\partial t}=\dfrac{\partial(ve^t)}{\partial t}=e^t\dfrac{\partial v}{\partial t}+ve^t$
$\therefore e^t\dfrac{\partial v}{\partial t}+ve^t=e^{3t}\dfrac{\partial^2v}{\partial X^2}+ve^t$
$e^t\dfrac{\partial v}{\partial t}=e^{3t}\dfrac{\partial^2v}{\partial X^2}$
$\dfrac{\partial v}{\partial t}=e^{2t}\dfrac{\partial^2v}{\partial X^2}$
Of course the following PDE is then solved by using separation of variables:
Let $v=f(X)g(t)$ ,
Then $f(X)g'(t)=e^{2t}f''(X)g(t)$
$\dfrac{g'(t)}{e^{2t}g(t)}=\dfrac{f''(X)}{f(X)}=-s^2$
$\begin{cases}\dfrac{g'(t)}{g(t)}=-e^{2t}s^2\\f''(X)+s^2f(X)=0\end{cases}$
$\begin{cases}g(t)=c_3(s)e^{-\frac{e^{2t}s^2}{2}}\\f(X)=\begin{cases}c_1(s)\sin Xs+c_2(s)\cos Xs&\text{when}~s\neq0\\c_1X+c_2&\text{when}~s=0\end{cases}\end{cases}$
$\therefore v=c_1X+c_2+\int_sc_3(s)e^{-\frac{e^{2t}s^2}{2}}\sin Xs~ds+\int_sc_4(s)e^{-\frac{e^{2t}s^2}{2}}\cos Xs~ds$ or $v=c_1X+c_2+\sum_sc_3(s)e^{-\frac{e^{2t}s^2}{2}}\sin Xs+\sum_sc_4(s)e^{-\frac{e^{2t}s^2}{2}}\cos Xs$
$ue^{-t}=c_1xe^t+c_2+\int_sc_3(s)e^{-\frac{e^{2t}s^2}{2}}\sin xe^ts~ds+\int_sc_4(s)e^{-\frac{e^{2t}s^2}{2}}\cos xe^ts~ds$ or $ue^{-t}=c_1xe^t+c_2+\sum_sc_3(s)e^{-\frac{e^{2t}s^2}{2}}\sin xe^ts+\sum_sc_4(s)e^{-\frac{e^{2t}s^2}{2}}\cos xe^ts$
$u=c_1xe^{2t}+c_2e^t+\int_sc_3(s)e^{t-\frac{e^{2t}s^2}{2}}\sin xe^ts~ds+\int_sc_4(s)e^{t-\frac{e^{2t}s^2}{2}}\cos xe^ts~ds$ or $u=c_1xe^{2t}+c_2e^t+\sum_sc_3(s)e^{t-\frac{e^{2t}s^2}{2}}\sin xe^ts+\sum_sc_4(s)e^{t-\frac{e^{2t}s^2}{2}}\cos xe^ts$
