2
$\begingroup$

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.

Improve this question
$\endgroup$
6
  • $\begingroup$ 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 $H^1$. $\endgroup$
    – Helge
    Commented Jan 12, 2012 at 2:11
  • $\begingroup$ 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 $a \nabla f \in H^1$. $\endgroup$
    – Helge
    Commented Jan 12, 2012 at 2:13
  • $\begingroup$ 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? $\endgroup$
    – RadonNikodym
    Commented Jan 12, 2012 at 8:39
  • $\begingroup$ What's a core? A reference would suffice. $\endgroup$
    – Deane Yang
    Commented Jan 12, 2012 at 18:28
  • $\begingroup$ 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. $\endgroup$
    – RadonNikodym
    Commented Jan 12, 2012 at 21:00

0

Reset to default

You must log in to answer this question.