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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.

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What is the expanded form of $\dfrac{\partial}{\partial x}\left(\dfrac{\partial}{\partial x}+x\right)u$ ? – doraemonpaul Aug 2 '12 at 22:02
This question has been solved perfectly. Hope that the asker has been diving enough and accept the answer at an early date. – doraemonpaul Sep 10 '12 at 1:21

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.

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>I thank you for your response – REULAND Aug 3 '12 at 17:42
> I cannot compute dV/dX and d^2V/dX^2 – REULAND Aug 3 '12 at 17:45
> I can use :differentiation of X=xu/V - find by X=xe^t and u=v*e^t – REULAND Aug 3 '12 at 17:53
> May I have a complete proof of dv/dt=e^(2*t) d^2v/dX^2 - Thanking you in advance for your help. – REULAND Aug 3 '12 at 17:58
I've added a step, but not all the details. – Davide Giraudo Aug 3 '12 at 18:10

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 $f(X)g'(t)=e^{2t}f''(X)g(t)$



$\begin{cases}g(t)=c_3(s)e^{-\frac{e^{2t}s^2}{2}}\\f(X)=\begin{cases}c_1(s)\sin Xs+c_2(s)\cos Xs&\text{when}~s\neq0\\c_1X+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore v=c_1X+c_2+\int_sc_3(s)e^{-\frac{e^{2t}s^2}{2}}\sin Xs~ds+\int_sc_4(s)e^{-\frac{e^{2t}s^2}{2}}\cos Xs~ds$ 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 $u=c_1xe^{2t}+c_2e^t+\sum_sc_3(s)e^{t-\frac{e^{2t}s^2}{2}}\sin xe^ts+\sum_sc_4(s)e^{t-\frac{e^{2t}s^2}{2}}\cos xe^ts$

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> Many thanks once again ! – REULAND Aug 8 '12 at 12:54
> du/dX instead of du/dx - then du/dX= d(ve^t)/dX = e^t dv/dX and d^2u/dX^2= d/dX(e^t dv/dX)= e^t d^2v/dX^2- du/dt=d(ve^t)/dt=e^tdv/dt +ve^t - How do you have : e^t dv/dt +ve^t= e^3*t d^2/dX^2 + v*e^t - I do not understand the way to obtain: e^3*t d^2v/dX^2 ? I thank you for your help ( I am a beginner in Maths ) – REULAND Aug 16 '12 at 13:08
@REULAND: Just substitute the results of $\dfrac{\partial u}{\partial X}$ , $\dfrac{\partial^2u}{\partial X^2}$ and $\dfrac{\partial u}{\partial t}$ into the previous PDE $\dfrac{\partial u}{\partial t}=e^{2t}\dfrac{\partial^2u}{\partial X^2}+u$ . Reminder that the term $\dfrac{\partial^2u}{\partial X^2}$ also contain the coefficient $e^{2t}$ , so we get $e^{3t}\dfrac{\partial^2v}{\partial X^2}$ . Please use TEX for your comment. – doraemonpaul Aug 16 '12 at 17:55

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