# On the domain of the Neumann Laplacian

Let $$U$$ be a bounded domain of $$\mathbb{R}^d$$, and write $$m$$ for the Lebesgue measure on $$U$$. For $$k=1,2$$, we denote by $$H^k(U)$$ the set of all locally $$m$$-integrable functions $$u\colon U \to \mathbb{R}$$ such that for any multi-index $$\alpha$$ with $$|\alpha|\le k$$, the weak derivative $$D^\alpha u$$ exists and belongs to $$L^2(U,m)$$.

Define the Neumann Laplacian $$(L,\text{Dom}(L))$$ on $$U$$ by $$\begin{eqnarray*} \text{Dom}(L) & = & \{u \in H^1(U) : H^1(U) \ni v \mapsto \int_{U}\nabla u\cdot \nabla v\,dm \\ & & \text{ is continuous on L^2(U,m)}\} \\ -\int_{U}v Lu\,dm &= &\int_{U} \nabla u\cdot \nabla v\,dm,\quad u \in \text{Dom}(L),\,v \in H^1(U). \end{eqnarray*}$$

I know that if $$U$$ is a bounded $$C^2$$ domain, $$\text{Dom}(L) \subset H^2(U)$$ (see Section~10.6.2 in [1] ). Even if $$U$$ is a bounded Lipschitz domain, does this inclusion hold ? I don't think this is correct in general.

For example, if $$U$$ is a $$C^{1,1}$$ domain or convex domain, does this holds?

I would like to know various conditions for $$U$$ such that the inclusion $$\text{Dom}(L) \subset H^2(U)$$ holds.

• This question is related to yours, and it has some information in the comments: mathoverflow.net/questions/154710/… Commented Nov 5, 2022 at 18:06
• @ChristianRemling Thank you for your helpful comment! Commented Nov 5, 2022 at 18:38
• Sections 2.4 ($C^{1,1}$ domain) and 3.2 (convex domain) in Grisvard's 'Elliptic problems in nonsmooth domains' could be of interest to you. (The latter follows the approach outlined by Giorgio Metafune as I understand it.) Commented Nov 14, 2022 at 10:05
• @Hannes Thank you for your comment. I found a positive result in the Grisvard's book. Commented Nov 16, 2022 at 4:41

This is a partial (positive) answer for the convex case only but not every detail has been worked out. Let first $$U$$ be convex and smooth and all functions be in $$C^3$$ up to the boundary. Integrating by parts one obtains $$\int_U |\Delta u|^2=\int_U |D^2 u|^2+\int_{\partial U}\sum_{i,j}(D_{jj}uD_i u-D_{ij}uD_j u)\nu_i d\sigma.$$ Here $$|D^2 u|^2 =\sum_{i,j}|D_{ij}u|^2$$. The first term in the boundary integral vanishes because $$\langle \nabla u, \nu \rangle =0$$. Differentiating this equality along any tangent vector $$h$$ to $$\partial U$$ one gets $$\langle D^2 u\, h, \nu \rangle +\langle \nabla u , D_h \nu \rangle=0$$.
If $$h=\nabla u$$, the convexity of $$U$$ gives $$\langle h, D_h \nu \rangle \geq 0$$ and then $$\langle D^2 u\, \nabla u, \nu\rangle \leq 0$$ and finally by the equality displayed $$\|D^2 u\|_2 \leq \|\Delta u\|_2$$. This is an a priori estimates for smooth functions and smooth convex sets, with a constant independent on the smoothness of $$U$$ and an approximation argument should give the result.
• Thank you for your very helpful comment. I see. I understood the estimate of the norm of the Hessian norm of $u$. Commented Nov 16, 2022 at 4:45