existence of solution of a degenerate pde with change of variables I am looking at the pde $$u_t=x^2u_{xx},\; x\in [0,\infty) ,\; t\in (0,T], \; u(0,x)=u_0(x)$$ This is a degenerate pde with a diffusion coefficient which is not bounded from 0, so I can't apply the classic theory of existence and uniqueness since the operator is not uniformly parabolic. However, I can do change of variables as $y=ln(x)$ and arrive at another pde: $$v_t=v_{yy}-v_y,\; y \in (-\infty, +\infty),\; t\in (0,T],\; v_0(y)=u_0(e^y)$$ Now there is no degeneracy and I can state the existence of function $v=v(t,y)$. Thus, I claim that $u(t,x)$ exists as well, however only for $x\in (0,\infty)$(Question 1: What about $x=0$? Does the function exist there, can we additionally define it?). 
Question 2: Is that a right line of thought? So, as long as I can find change of variables s.t. it produces non degenerate pde I obtain the existence for the degenerate pde? What should I be careful about?
 A: $x=0$ corresponds to the limit $y \to -\infty$.  Does your solution $v(y,t)$ have a limit as $y \to -\infty$?  That's a bit of a delicate question.  However, note that if $v(t,y) = e^{y/2 - t/4} V(t,y)$ the equation for $V$ is the classical heat equation:
$\dfrac{\partial V}{\partial t} = \dfrac{\partial^2 V}{\partial x^2}$.  If, for example, $V(0,y)$ is bounded, then $V(t,y)$ has the same bound for all $t > 0$, and $v(t,y) \to 0$ as $y \to -\infty$ for any fixed $t > 0$. 
EDIT: actually it would have been better, I think, to use a different transformation:
$v(t,y) = w(t,z)$ where $y = z + t$ so you get $$ \frac{\partial w}{\partial t} = \frac{\partial^2 w}{\partial z^2}$$
and $\lim_{y \to -\infty} v(t,y) = \lim_{z \to -\infty} w(t,z)$.
But I doubt that there will always be a transformation that will act so nicely.
A: Let $u(t,x)=T(t)X(x)$ ,
Then $T'(t)X(x)=x^2T(t)X''(x)$
$\dfrac{T'(t)}{T(t)}=\dfrac{x^2X''(x)}{X(x)}=-\dfrac{4s^2+1}{4}$
$\begin{cases}\dfrac{T'(t)}{T(t)}=-\dfrac{4s^2+1}{4}\\x^2X''(x)+\dfrac{4s^2+1}{4}X(x)=0\end{cases}$
$\begin{cases}T(t)=c_3(s)e^{-\frac{t(4s^2+1)}{4}}\\X(x)=\begin{cases}c_1(s)\sqrt{x}\sin(s\ln x)+c_2(s)\sqrt{x}\cos(s\ln x)&\text{when}~s\neq0\\c_1\sqrt{x}\ln x+c_2\sqrt{x}&\text{when}~s=0\end{cases}\end{cases}$
$\therefore u(x,t)=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty C_4(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds$
$u(0,x)=u_0(x)$ :
$C_1\sqrt{x}\ln x+C_2\sqrt{x}+\int_0^\infty C_3(s)\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty C_4(s)\sqrt{x}\cos(s\ln x)~ds=u_0(x)$
$C_1\ln x+C_2+\int_0^\infty C_3(s)\sin(s\ln x)~ds+\int_0^\infty C_4(s)\cos(s\ln x)~ds=\dfrac{u_0(x)}{\sqrt{x}}$
$\int_0^\infty C_4(s)\cos(s\ln x)~ds=\dfrac{u_0(x)}{\sqrt{x}}-C_1\ln x-C_2-\int_0^\infty C_3(s)\sin(s\ln x)~ds$
$\int_0^\infty C_4(s)\cos xs~ds=e^{\frac{x}{2}}u_0(e^x)-C_1x-C_2-\int_0^\infty C_3(s)\sin xs~ds$
$\mathcal{F}_{c,s\to x}\{C_4(s)\}=e^{\frac{x}{2}}u_0(e^x)-C_1x-C_2-\mathcal{F}_{s,s\to x}\{C_3(s)\}$
$C_4(s)=\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}+C_1\delta'(s)-C_2\delta(s)-\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}$
$\therefore u(x,t)=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds+\int_0^\infty C_1\delta'(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-\int_0^\infty C_2\delta(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds=C_1e^{-\frac{t}{4}}\sqrt{x}\ln x+C_2e^{-\frac{t}{4}}\sqrt{x}+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds+\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds-C_1e^{-\frac{t}{4}}\sqrt{x}\ln x-C_2e^{-\frac{t}{4}}\sqrt{x}-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds=\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{e^{\frac{x}{2}}u_0(e^x)\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds+\int_0^\infty C_3(s)e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\sin(s\ln x)~ds-\int_0^\infty\mathcal{F}^{-1}_{c,x\to s}\{\mathcal{F}_{s,s\to x}\{C_3(s)\}\}e^{-\frac{t(4s^2+1)}{4}}\sqrt{x}\cos(s\ln x)~ds$
The function not only exist for at least $x\in(0,\infty)$ , but also has one arbitrary function left.
To determine whether the function exist at $x=0$ or not, we just need to consider $\lim\limits_{x\to0}\sqrt{x}\sin(s\ln x)$ and $\lim\limits_{x\to0}\sqrt{x}\cos(s\ln x)$ .
Since both $\sin(s\ln x)$ and $\cos(s\ln x)$ are bounded,
$\lim\limits_{x\to0}\sqrt{x}\sin(s\ln x)=0$ and $\lim\limits_{x\to0}\sqrt{x}\cos(s\ln x)=0$
So the function not only also exist at $x=0$ , but also equals to $0$ when $x=0$ .
