# General solution to a partial differential equation

Given the partial differential equation:

$$\tau\partial_t\varPhi(x,t)=-\partial_x[(-x+A)\varPhi(x,t)]+D\partial_{xx}\varPhi(x,t)$$ where $\tau$ , $A$ and $D$ are constant parameters.

with the boundary conditions:

$$\varPhi(x,0)=\lim\limits_{n\to0^+}\frac{1}{2\sqrt{\pi n}}e^\frac{(x-x_0)^2}{4n}$$

and boundary conditions of the kind:

$$\varPhi(a,t)=0$$ and $$\varPhi(-\infty,t)=0$$

what is the distribution of the $\varPhi(x,t)$?

$\tau\partial_t\varPhi(x,t)=-\partial_x[(-x+A)\varPhi(x,t)]+D\partial_{xx}\varPhi(x,t)$

$\tau\partial_t\varPhi(x,t)=D\partial_{xx}\varPhi(x,t)+(x-A)\partial_x\varPhi(x,t)+\varPhi(x,t)$

Let $\begin{cases}x_1=x-A\\t_1=t\end{cases}$ ,

Then $\partial_x\varPhi=\partial_{x_1}\varPhi\partial_xx_1+\partial_{t_1}\varPhi\partial_xt_1=\partial_{x_1}\varPhi$

$\partial_{xx}\varPhi=\partial_x(\partial_{x_1}\varPhi)=\partial_{x_1}(\partial_{x_1}\varPhi)\partial_xx_1+\partial_{t_1}(\partial_{x_1}\varPhi)\partial_xt_1=\partial_{x_1x_1}\varPhi$

$\partial_t\varPhi=\partial_{x_1}\varPhi\partial_tx_1+\partial_{t_1}\varPhi\partial_tt_1=\partial_{t_1}\varPhi$

$\therefore\tau\partial_{t_1}\varPhi(x_1,t_1)=D\partial_{x_1x_1}\varPhi(x_1,t_1)+x_1\partial_{x_1}\varPhi(x_1,t_1)+\varPhi(x_1,t_1)$

With reference to Change variables into Fokker-Planck PDE,

Let $\begin{cases}x_2=x_1e^\frac{t_1}{\tau}\\t_2=t_1\end{cases}$ ,

Then $\partial_{x_1}\varPhi=\partial_{x_2}\varPhi\partial_{x_1}x_2+\partial_{t_2}\varPhi\partial_{x_1}t_2=e^\frac{t_1}{\tau}~\partial_{x_2}\varPhi=e^\frac{t_2}{\tau}~\partial_{x_2}\varPhi$

$\partial_{x_1x_1}\varPhi=\partial_{x_1}(e^\frac{t_2}{\tau}~\partial_{x_2}\varPhi)=\partial_{x_2}(e^\frac{t_2}{\tau}~\partial_{x_2}\varPhi)\partial_{x_1}x_2+\partial_{t_2}(e^\frac{t_2}{\tau}~\partial_{x_2}\varPhi)\partial_{x_1}t_2=e^\frac{2t_2}{\tau}~\partial_{x_2x_2}\varPhi$

$\partial_{t_1}\varPhi=\partial_{x_2}\varPhi\partial_{t_1}x_2+\partial_{t_2}\varPhi\partial_{t_1}t_2=\dfrac{x_1e^\frac{t_1}{\tau}}{\tau}\partial_{x_2}\varPhi+\partial_{t_2}\varPhi$

$\therefore\tau\biggl(\dfrac{x_1e^\frac{t_1}{\tau}}{\tau}\partial_{x_2}\varPhi(x_2,t_2)+\partial_{t_2}\varPhi(x_2,t_2)\biggr)=De^\frac{2t_2}{\tau}~\partial_{x_2x_2}\varPhi(x_2,t_2)+x_1e^\frac{t_1}{\tau}~\partial_{x_2}\varPhi(x_2,t_2)+\varPhi(x_2,t_2)$

$x_1e^\frac{t_1}{\tau}~\partial_{x_2}\varPhi(x_2,t_2)+\tau\partial_{t_2}\varPhi(x_2,t_2)=De^\frac{2t_2}{\tau}~\partial_{x_2x_2}\varPhi(x_2,t_2)+x_1e^\frac{t_1}{\tau}~\partial_{x_2}\varPhi(x_2,t_2)+\varPhi(x_2,t_2)$

$\tau\partial_{t_2}\varPhi(x_2,t_2)-\varPhi(x_2,t_2)=De^\frac{2t_2}{\tau}~\partial_{x_2x_2}\varPhi(x_2,t_2)$

Which is separable.

Can you take it from here?

• many many thank you for the help. I obtained this: $$\varPhi(x,t)=[a\sin \sqrt D\lambda (x-A)e^{t/\tau}+b\cos\sqrt D\lambda (x-A)e^{t/\tau}].[c\exp ({t\over\tau}-{\lambda ^2\over2}e^{2t\over\tau})]$$ is this right? and how to apply conditions on this now? Thanks once again for the help. – Malik Nov 16 '13 at 2:48
• waiting for your further help. Thanks – Malik Nov 17 '13 at 11:30