solution for the degenerate parabolic PDE Look at $u_t=a(x)u_{xx}$ if I have $a(x)\geq a_0>0$ then I can see in all books that $C^{2,1}$ solution exist and it is unique. However, if $a(x)\geq0$, that is degenerate, I see in Friedman's book the construction of $K_{\epsilon}$: sequence convergent uniformly to $K$ which is the solution of the degenerate equation. But he doesn't state the property of the latter. What are the problems of those equations? Looks like I have problem constructing a weak solution because I lack coercivity property, so I have little hope to have it $C^{2,1}$. But what class a solution of the degenerate equation belongs to? thnaks!
 A: This is not a complete answer, but is too long to be a comment. It is an analysis of the pure Cauchy problem in the case $a(x)=x^2$. Using semigroup theory, it can be proved that the problem is well posed on the space
$$
X=\Bigr\{u\colon\mathbb{R}\to\mathbb{R}:\int_{-\infty}^\infty|u(x)|^2\frac{dx}{x^2}<\infty\Bigl\}.
$$
The fact that the initial value $u_0(x)=u(x,0)\in X$ implies that $u_0$ must vanish to a certain order at $x=0$.
The equation $u_t-x^2u_{xx}=0$ can be transformed into the heat equation by means of a change of variable. Consider first the region $x>0$. 
The change $x=e^{-z}$ transforms the equation into $u_t-u_z-u_{zz}=0$, $u(z,0)=u_0(e^{-z})$. Now if $v(z,t)=u(z-t,t)$, then $v_t-v_{zz}=0$ and $v(z,0)=u_0(e^{-z})$. A similar computation can be caried out for $x<0$. This allows to prove for instance that if
$$
\lim_{x\to0}x\,u_0'(x)=\lim_{x\to0}x^2\,u_0''(x)=0,\tag{1}
$$
then there exists a unique classical solution.
Some remarks are in order. We have $u(0,t)=u_0(0)$ for all $t>0$. Moreover, $x=0$ acts as a barrier to diffusion. If $u_0$ has compact support contained in $(0,\infty)$, then $u(x,t)=0$ for all $x<0$ and $t>0$. This is in sharp contrast with the heat equation. The equation is not regularizing at $x=0$. The solution is $C^{2,1}$ 0n $(\mathbb{R}\setminus\{0\})\times(0,\infty)$, but only as regular as $u_o$ at $x=0$. What happens if $u_0$ does not satisfy (1)? The solution can still be constructed, but in general it will be a solution only in a weak sense. For instance, if $u_0(x)=\sin(-\log|x|)$, then $u(x,t)=e^{-t} \sin(-\log|x|-t)$, which is discontinuous at $x=0$ for all $t>0$.
Finally, can something be said for the more general equation $u_t-|x|^\alpha u_{xx}=0$, $\alpha\ge0$? If $0\le\alpha<2$, then the operator $|x|^\alpha u''$ has some compactnes properties, and it is possible to obtain some results. If $\alpha>2$, very little is known.
