show than this PDE can be reduced to heat equation How to reduce this PDE to heat equation
$$x^2G_{xx}=G_t$$   
($G_{xx}$ is the 2nd order derivative on $x$, $G_t$ is the 1st derivative on $t$)
We wish to obtain a form such that $G(x,t)=F(U(x,t))$, 
when substituted into the original equation we have
$$U_{xx}=U_t$$

Let $\begin{cases}x_1=\ln x\\t_1=t\end{cases}$ ,
Then $G_x=G_{x_1}(x_1)_x+G_{t_1}(t_1)_x=\dfrac{G_{x_1}}{x}=e^{-x_1}G_{x_1}$
$G_{xx}=(e^{-x_1}G_{x_1})_x=(e^{-x_1}G_{x_1})_{x_1}(x_1)_x+(e^{-x_1}G_{x_1})_{t_1}(t_1)_x=(e^{-x_1}G_{x_1x_1}-e^{-x_1}G_{x_1})e^{-x_1}=e^{-2x_1}G_{x_1x_1}-e^{-2x_1}G_{x_1}$
$G_t=G_{x_1}(x_1)_t+G_{t_1}(t_1)_t=G_{t_1}$
$\therefore e^{2x_1}(e^{-2x_1}G_{x_1x_1}-e^{-2x_1}G_{x_1})=G_{t_1}$
$G_{x_1x_1}-G_{x_1}=G_{t_1}$
Let $\begin{cases}x_2=x_1-t_1\\t_2=t_1\end{cases}$ ,
Then $G_{x_1}=G_{x_2}(x_2)_{x_1}+G_{t_2}(t_2)_{x_1}=G_{x_2}$
$G_{x_1x_1}=(G_{x_2})_{x_1}=(G_{x_2})_{x_2}(x_2)_{x_1}+(G_{x_2})_{t_2}(t_2)_{x_1}=G_{x_2x_2}$
$G_{t_1}=G_{x_2}(x_2)_{t_1}+G_{t_2}(t_2)_{t_1}=G_{t_2}-G_{x_2}$
$\therefore G_{x_2x_2}-G_{x_2}=G_{t_2}-G_{x_2}$
$G_{x_2x_2}=G_{t_2}$
