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.


1 Answer 1


(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.

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

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