0
$\begingroup$

I have a question about a Dirichlet form.

Let $D$ be a open subset of $\mathbb{R}^d$. Then, we can define $H^{1}(D)$ by \begin{equation*} H^{1}(D)=\{f \in L^{2}(D,dx):\frac{\partial f}{\partial x_i} \in L^{2}(D,dx),\,1\le i\le d\}. \end{equation*} It is well known that $H^{1}(D)$ becomes a Hilbert space with inner norm \begin{equation*} (f,g)_{H}:=\mathcal{E}(f,g)+\int_{D}fg\,dx, \end{equation*} where $\mathcal{E}(f,g):=\frac{1}{2}\int_{D}\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial x_i}\,dx$. Moreover, $(\mathcal{E}, H^{1}(D))$ becomes a Dirichlet form on $L^{2}(D,dx)$. Hence, from a general theory of Dirichlet form, there exists a unique (non-positive) closed linear operator $(L,\text{Dom}(L))$ such that \begin{equation*} (-Lf,g)=\mathcal{E}(f,g),\quad f \in \text{Dom}(L),\ g\in H^{1}(D). \end{equation*}

My question

If $D =\mathbb{R}^d$, I know $\text{Dom}(L)=W^{2,2}(\mathbb{R}^d)$.

  • For a general open subset $D$, $\text{Dom}(L)=W^{2,2}(D)$?
  • Is there a sufficient condition for $f \in \text{Dom}(L)$?

If you know related results, please let me know.

$\endgroup$

1 Answer 1

3
$\begingroup$

(More an extended comment than a full answer)

Your $L$ is the Neumann Laplacian and so functions in its domain have to satisfy Neumann boundary conditions, informally speaking. The domain certainly won't be all of $W^{2,2}(D)$ in general. For instance, you might try working it out with $D = (0,1)$ in $\mathbb{R}^1$. When you integrate by parts, you find that in order to get the boundary terms to vanish, you have to have $f'(0)=f'(1)=0$.

In higher dimensions, the condition becomes that at the boundary, the normal derivative of $f$ should vanish (i.e. the directional derivative in the direction normal to the boundary). If $\partial D$ is not smooth enough to have a well-defined normal direction, then things get a lot more complicated.

In some cases the boundary may be too small to have any effect. If you take $\mathbb{R}^2$ minus a point, I think you should find that the domain is all of $W^{2,2}$. But for $\mathbb{R}^2$ minus a slit, it isn't.

This should be discussed, at least in simple cases (e.g. smooth boundary), in standard PDE books, but I don't have a specific reference for you at the moment.

$\endgroup$
1
  • $\begingroup$ It is not likely to know $\text{Dom}(L)$ as you say. $\endgroup$
    – sharpe
    Jan 27, 2018 at 16:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.