How to solve the following linear PDE?

Question

I want to solve the following equation

how to solve it? I know that the evolution equation can be changed to a heat equation, but the boundary condition and initial condition changes to the new space which is a moving boundary.

Answer 1

Hint:

With reference the substitution concept in How to solve a PDE with in which the coefficient which depends on time but singular at a point?,

Let $\begin{cases}z_1=e^{\int^v\xi(v)~dv}z\\v_1=v\end{cases}$ ,

Then $\dfrac{\partial g}{\partial z}=\dfrac{\partial g}{\partial z_1}\dfrac{\partial z_1}{\partial z}+\dfrac{\partial g}{\partial v_1}\dfrac{\partial v_1}{\partial z}=e^{\int^v\xi(v)~dv}\dfrac{\partial g}{\partial z_1}$

$\dfrac{\partial^2g}{\partial z^2}=\dfrac{\partial}{\partial z}\left(e^{\int^v\xi(v)~dv}\dfrac{\partial g}{\partial z_1}\right)=\dfrac{\partial}{\partial z_1}\left(e^{\int^v\xi(v)~dv}\dfrac{\partial g}{\partial z_1}\right)\dfrac{\partial z_1}{\partial z}+\dfrac{\partial}{\partial v_1}\left(e^{\int^v\xi(v)~dv}\dfrac{\partial g}{\partial z_1}\right)\dfrac{\partial v_1}{\partial z}=e^{2\int^v\xi(v)~dv}\dfrac{\partial^2g}{\partial z_1^2}$

$\dfrac{\partial g}{\partial v}=\dfrac{\partial g}{\partial z_1}\dfrac{\partial z_1}{\partial v}+\dfrac{\partial g}{\partial v_1}\dfrac{\partial v_1}{\partial v}=\xi(v)e^{\int^v\xi(v)~dv}z\dfrac{\partial g}{\partial z_1}+\dfrac{\partial g}{\partial v_1}$

$\therefore\xi(v)e^{\int^v\xi(v)~dv}z\dfrac{\partial g}{\partial z_1}+\dfrac{\partial g}{\partial v_1}=\eta(v)e^{2\int^v\xi(v)~dv}\dfrac{\partial^2g}{\partial z_1^2}+(\xi(v)z+\nu(v))e^{\int^v\xi(v)~dv}\dfrac{\partial g}{\partial z_1}+(\chi(v)z+\zeta(v))g$

$\xi(v)e^{\int^v\xi(v)~dv}z\dfrac{\partial g}{\partial z_1}+\dfrac{\partial g}{\partial v}=\eta(v)e^{2\int^v\xi(v)~dv}\dfrac{\partial^2g}{\partial z_1^2}+\xi(v)e^{\int^v\xi(v)~dv}z\dfrac{\partial g}{\partial z_1}+\nu(v)e^{\int^v\xi(v)~dv}\dfrac{\partial g}{\partial z_1}+(\chi(v)z+\zeta(v))g$

$\dfrac{\partial g}{\partial v}=\eta(v)e^{2\int^v\xi(v)~dv}\dfrac{\partial^2g}{\partial z_1^2}+\nu(v)e^{\int^v\xi(v)~dv}\dfrac{\partial g}{\partial z_1}+(\chi(v)e^{-\int^v\xi(v)~dv}z_1+\zeta(v))g$

With reference the substitution concept in Solving a PDE arising from physics,

Let $g=e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h$ ,

Then $\dfrac{\partial g}{\partial v}=e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial v}+z_1\chi(v)e^{-\int^v\xi(v)~dv}e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h$

$\dfrac{\partial g}{\partial z_1}=e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial z_1}+\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h$

$\dfrac{\partial^2g}{\partial z_1^2}=e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial^2h}{\partial z_1^2}+\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial z_1}+\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial z_1}+(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h=e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial^2h}{\partial z_1^2}+2\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial z_1}+(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h$

$\therefore e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial v}+z_1\chi(v)e^{-\int^v\xi(v)~dv}e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h=\eta(v)e^{2\int^v\xi(v)~dv}\left(e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial^2h}{\partial z_1^2}+2\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial z_1}+(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h\right)+\nu(v)e^{\int^v\xi(v)~dv}\left(e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}\dfrac{\partial h}{\partial z_1}+\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h\right)+(\chi(v)e^{-\int^v\xi(v)~dv}z_1+\zeta(v))e^{z_1\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv}h$

$\dfrac{\partial h}{\partial v}+z_1\chi(v)e^{-\int^v\xi(v)~dv}h=\eta(v)e^{2\int^v\xi(v)~dv}\left(\dfrac{\partial^2h}{\partial z_1^2}+2\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~\dfrac{\partial h}{\partial z_1}+(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2h\right)+\nu(v)e^{\int^v\xi(v)~dv}\left(\dfrac{\partial h}{\partial z_1}+\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~h\right)+(\chi(v)e^{-\int^v\xi(v)~dv}z_1+\zeta(v))h$

$\dfrac{\partial h}{\partial v}+\chi(v)e^{-\int^v\xi(v)~dv}z_1h=\eta(v)e^{2\int^v\xi(v)~dv}\dfrac{\partial^2h}{\partial z_1^2}+2\eta(v)e^{2\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~\dfrac{\partial h}{\partial z_1}+\eta(v)e^{2\int^v\xi(v)~dv}(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2h+\nu(v)e^{\int^v\xi(v)~dv}\dfrac{\partial h}{\partial z_1}+\nu(v)e^{\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~h+\chi(v)e^{-\int^v\xi(v)~dv}z_1h+\zeta(v)h$

$\dfrac{\partial h}{\partial v}=\eta(v)e^{2\int^v\xi(v)~dv}\dfrac{\partial^2h}{\partial z_1^2}+\left(2\eta(v)e^{2\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv+\nu(v)e^{\int^v\xi(v)~dv}\right)\dfrac{\partial h}{\partial z_1}+\left(\eta(v)e^{2\int^v\xi(v)~dv}(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2+\nu(v)e^{\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv+\zeta(v)\right)h$

Let $\begin{cases}z_2=z_1+2\int^v\eta(v)e^{2\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv~dv+\int^v\nu(v)e^{\int^v\xi(v)~dv}~dv\\v_1=v\end{cases}$ ,

Then $\dfrac{\partial h}{\partial z_1}=\dfrac{\partial h}{\partial z_2}\dfrac{\partial z_2}{\partial z_1}+\dfrac{\partial h}{\partial v_1}\dfrac{\partial v_1}{\partial z_1}=\dfrac{\partial h}{\partial z_2}$

$\dfrac{\partial^2h}{\partial z_1^2}=\dfrac{\partial}{\partial z_1}\left(\dfrac{\partial h}{\partial z_2}\right)=\dfrac{\partial}{\partial z_2}\left(\dfrac{\partial h}{\partial z_2}\right)\dfrac{\partial z_2}{\partial z_1}+\dfrac{\partial}{\partial v_1}\left(\dfrac{\partial h}{\partial z_2}\right)\dfrac{\partial v_1}{\partial z_1}=\dfrac{\partial^2h}{\partial z_2^2}$

$\dfrac{\partial h}{\partial v}=\dfrac{\partial h}{\partial z_2}\dfrac{\partial z_2}{\partial v}+\dfrac{\partial h}{\partial v_1}\dfrac{\partial v_1}{\partial v}=\left(2\eta(v)e^{2\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv+\nu(v)e^{\int^v\xi(v)~dv}\right)\dfrac{\partial h}{\partial z_2}+\dfrac{\partial h}{\partial v_1}$

$\therefore\left(2\eta(v)e^{2\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv+\nu(v)e^{\int^v\xi(v)~dv}\right)\dfrac{\partial h}{\partial z_2}+\dfrac{\partial h}{\partial v_1}=\eta(v)e^{2\int^v\xi(v)~dv}\dfrac{\partial^2h}{\partial z_2^2}+\left(2\eta(v)e^{2\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv+\nu(v)e^{\int^v\xi(v)~dv}\right)\dfrac{\partial h}{\partial z_2}+\left(\eta(v)e^{2\int^v\xi(v)~dv}(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2+\nu(v)e^{\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv+\zeta(v)\right)h$

$\dfrac{\partial h}{\partial v}-\left(\eta(v)e^{2\int^v\xi(v)~dv}(\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv)^2+\nu(v)e^{\int^v\xi(v)~dv}\int^v\chi(v)e^{-\int^v\xi(v)~dv}~dv+\zeta(v)\right)h=\eta(v)e^{2\int^v\xi(v)~dv}\dfrac{\partial^2h}{\partial z_2^2}$

Which is separable.