Core of divergence form operator with unbounded coefficient

Consider the unbounded operator $L$ on $L^2(\mathbb{R^d})$ to be the self-adjoint extension of $$Lf := \nabla \cdot \left(a(x) \nabla f(x) \right)$$ on $C^2_c(\mathbb{R^d})$.

I also assume that $a(x) > 0$ for all $x$ and $a(x)$ is differentiable. However, I make no assumptions on the boundedness of $a$.

Does this operator have a core? If so, can it be identified explicitly?

Thanks.

-
I don't think your operator is defined on $C^2_c$ without assuming more on $a$ like differentiability. Writing down the core for the quadratic form $q(f) = \int a(x) |\nabla f(x)|^2 dx is easy, since it's all$f$such that$a^{1/2} |\nabla f| \in L^2(\mathbb{R}^d)$. Something similar works for$L$(it's the obvious condition of everything being defined, so$a \nabla f$being in$H^1$. – Helge Jan 12 '12 at 2:11 I can't delete the previous comment for some reasons. Here's the scrambled part. since it's all$f$such that$a^{1/2} |\nabla f| \in L^2(\mathbb{R}^d)$. Something similar works for$L$. It should just be the$f$such that$a \nabla f \in H^1$. – Helge Jan 12 '12 at 2:13 Yes, you're right,$a(x)$should be differentiable. I'll edit the question. Another question: So if a set C is a core for the Dirichlet form wouldn't this imply C is a core for L? – RadonNikodym Jan 12 '12 at 8:39 What's a core? A reference would suffice. – Deane Yang Jan 12 '12 at 18:28 Let$L$be a closed operator on$L^2(D)$. Then$\mathcal{D} \subset D(L)$is said to be a core of$L$if$\mathcal{D}$is dense in$D(L)$with respect to the graph norm$||u||_{L^2} + ||Au||_{L^2}\$. Ethier and Kurtz is a good reference and contains various sufficient condition for a set of functions to be a core. –  RadonNikodym Jan 12 '12 at 21:00