transformation from pde to a simpler heat equation

Question

Please give me hints to transform the following PDE to a simpler heat equation:

$$\dfrac{\partial u(t,x)}{\partial t}+k(\rho-\ln x)x\dfrac{\partial u(t,x)}{\partial x}+\dfrac{1}{2}\sigma^2x^2\dfrac{\partial^2u(t,x)}{\partial x^2}-\rho u(t,x)=0$$

Answer 1

Since this PDE itself already belongs to a Heat equation, you only can ask whether this PDE can transform to a Heat equation of simpler form.

In fact the answer is "yes" .

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

Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial x_1}\dfrac{\partial x_1}{\partial x}+\dfrac{\partial u}{\partial t_1}\dfrac{\partial t_1}{\partial x}=\dfrac{1}{x}\dfrac{\partial u}{\partial x_1}$

$$\frac{\partial^2u}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{1}{x}\frac{\partial u}{\partial x_1}\right)=\frac{1}{x}\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x_1}\right)-\frac{1}{x^2}\frac{\partial u}{\partial x_1}=\frac{1}{x}\left(\frac{\partial}{\partial x_1}\left(\frac{\partial u}{\partial x_1}\right)\frac{\partial x_1}{\partial x}+\frac{\partial}{\partial t_1}\left(\frac{\partial u}{\partial x_1}\right)\frac{\partial t_1}{\partial x}\right)-\frac{1}{x^2}\frac{\partial u}{\partial x_1}=\frac{1}{x^2}\frac{\partial^2u}{\partial x_1^2}-\frac{1}{x^2}\frac{\partial u}{\partial x_1}$$

$\dfrac{\partial u}{\partial t}=\dfrac{\partial u}{\partial x_1}\dfrac{\partial x_1}{\partial t}+\dfrac{\partial u}{\partial t_1}\dfrac{\partial t_1}{\partial t}=\dfrac{\partial u}{\partial t_1}$

$\therefore\dfrac{\partial u}{\partial t_1}+k(\rho-x_1)x\dfrac{1}{x}\dfrac{\partial u}{\partial x_1}+\dfrac{1}{2}\sigma^2x^2\left(\dfrac{1}{x^2}\dfrac{\partial^2u}{\partial x_1^2}-\dfrac{1}{x^2}\dfrac{\partial u}{\partial x_1}\right)-\rho u=0$

$\dfrac{\partial u}{\partial t_1}+k(\rho-x_1)\dfrac{\partial u}{\partial x_1}+\dfrac{\sigma^2}{2}\dfrac{\partial^2u}{\partial x_1^2}-\dfrac{\sigma^2}{2}\dfrac{\partial u}{\partial x_1}-\rho u=0$

$\dfrac{\partial u}{\partial t_1}+\dfrac{\sigma^2}{2}\dfrac{\partial^2u}{\partial x_1^2}+\left(k\rho-\dfrac{\sigma^2}{2}-kx_1\right)\dfrac{\partial u}{\partial x_1}-\rho u=0$

Let $\begin{cases}x_2=k\rho-\dfrac{\sigma^2}{2}-kx_1,\\t_2=t_1,\end{cases}$

Then $\dfrac{\partial u}{\partial x_1}=\dfrac{\partial u}{\partial x_2}\dfrac{\partial x_2}{\partial x_1}+\dfrac{\partial u}{\partial t_2}\dfrac{\partial t_2}{\partial x_1}=-k\dfrac{\partial u}{\partial x_2}$

$\dfrac{\partial^2u}{\partial x_1^2}=\dfrac{\partial}{\partial x_1}\left(-k\dfrac{\partial u}{\partial x_2}\right)=-k\dfrac{\partial}{\partial x_2}\left(\dfrac{\partial u}{\partial x_2}\right)\dfrac{\partial x_2}{\partial x_1}-k\dfrac{\partial}{\partial t_2}\left(\dfrac{\partial u}{\partial x_2}\right)\dfrac{\partial t_2}{\partial x_1}=k^2\dfrac{\partial^2u}{\partial x_2^2}$

$\dfrac{\partial u}{\partial t_1}=\dfrac{\partial u}{\partial x_2}\dfrac{\partial x_2}{\partial t_1}+\dfrac{\partial u}{\partial t_2}\dfrac{\partial t_2}{\partial t_1}=\dfrac{\partial u}{\partial t_2}$

$\therefore\dfrac{\partial u}{\partial t_2}+\dfrac{k^2\sigma^2}{2}\dfrac{\partial^2u}{\partial x_2^2}-kx_2\dfrac{\partial u}{\partial x_2}-\rho u=0$

With reference to Change variables into Fokker-Planck PDE,

Let $\begin{cases}x_3=x_2e^{kt_2},\\t_3=t_2,\end{cases}$

Then $\dfrac{\partial u}{\partial x_2}=\dfrac{\partial u}{\partial x_3}\dfrac{\partial x_3}{\partial x_2}+\dfrac{\partial u}{\partial t_3}\dfrac{\partial t_3}{\partial x_2}=e^{kt_2}\dfrac{\partial u}{\partial x_3}=e^{kt_3}\dfrac{\partial u}{\partial x_3}$

$\dfrac{\partial^2u}{\partial x_2^2}=\dfrac{\partial u}{\partial x_2}\left(e^{kt_3}\dfrac{\partial u}{\partial x_3}\right)=\dfrac{\partial u}{\partial x_3}\left(e^{kt_3}\dfrac{\partial u}{\partial x_3}\right)\dfrac{\partial x_3}{\partial x_2}+\dfrac{\partial u}{\partial t_3}\left(e^{kt_3}\dfrac{\partial u}{\partial x_3}\right)\dfrac{\partial t_3}{\partial x_2}=e^{2kt_3}\dfrac{\partial^2u}{\partial x_3^2}$

$\dfrac{\partial u}{\partial t_2}=\dfrac{\partial u}{\partial x_3}\dfrac{\partial x_3}{\partial t_2}+\dfrac{\partial u}{\partial t_3}\dfrac{\partial t_3}{\partial t_2}=kx_2e^{kt_2}\dfrac{\partial u}{\partial x_3}+\dfrac{\partial u}{\partial t_3}$

$\therefore kx_2e^{kt_2}\dfrac{\partial u}{\partial x_3}+\dfrac{\partial u}{\partial t_3}+\dfrac{k^2\sigma^2e^{2kt_3}}{2}\dfrac{\partial^2u}{\partial x_3^2}-kx_2e^{kt_2}\dfrac{\partial u}{\partial x_3}-\rho u=0$

$\dfrac{\partial u}{\partial t_3}-\rho u=-\dfrac{k^2\sigma^2e^{2kt_3}}{2}\dfrac{\partial^2u}{\partial x_3^2}$

Let $u=e^{\rho t_3}v$ ,

Then $\dfrac{\partial u}{\partial x_3}=\dfrac{\partial(e^{\rho t_3}v)}{\partial x_3}=e^{\rho t_3}\dfrac{\partial v}{\partial x_3}$

$\dfrac{\partial^2u}{\partial x_3^2}=\dfrac{\partial}{\partial x_3}\left(e^{\rho t_3}\dfrac{\partial v}{\partial x_3}\right)=e^{\rho t_3}\dfrac{\partial^2v}{\partial x_3^2}$

$\dfrac{\partial u}{\partial t_3}=\dfrac{\partial(e^{\rho t_3}v)}{\partial t_3}=e^{\rho t_3}\dfrac{\partial v}{\partial t_3}+\rho e^{\rho t_3}v$

$\therefore e^{\rho t_3}\dfrac{\partial v}{\partial t_3}+\rho e^{\rho t_3}v-\rho e^{\rho t_3}v=-\dfrac{k^2\sigma^2e^{(2k+\rho)t_3}}{2}\dfrac{\partial^2v}{\partial x_3^2}$

$e^{\rho t_3}\dfrac{\partial v}{\partial t_3}=-\dfrac{k^2\sigma^2e^{(2k+\rho)t_3}}{2}\dfrac{\partial^2v}{\partial x_3^2}$

$\dfrac{\partial v}{\partial t_3}=-\dfrac{k^2\sigma^2e^{2kt_3}}{2}\dfrac{\partial^2v}{\partial x_3^2}$