Functions satisfying Neumann boundary condition

Question

I have a question about functions satisfying a condition.

• Let $D \subset \mathbb{R}^d$ be a Lipschitz domain. That is, for each $x \in \partial D$, there exists an open neighborhood $U$ of $x$ in $\mathbb{R}^d$ and a bi-Lipschitz function $\psi_{x}:B(1) \to U$ such that $\psi(0)=x$ and $\psi_{x}(B_{+}(1))=U \cap D$. Here, $B(1)=\{x \in \mathbb{R}^d \mid |x|<1\}$ and $B_{+}(1)=\{x=(x_1,\ldots,x_d) \in B(1) \mid x_d>0\}$.

• We denote $H^{1}(D)$ by the 1-st order Sobolev space on $D$ with Neumann boundary condition: $$H^{1}(D)=\{f \in L^{2}(D,dx)\mid\frac{\partial f}{\partial x_i} \in L^{2}(D,dx),\ 1\le i\le d\}.$$ Here, $\frac{\partial f}{\partial x_i}$ is the distributional derivative of $f$. We also denote $(L,\mathcal{D}(L))$ be the Neumann Laplacian on $D$. We note that \begin{equation*} D(L):=\{f \in H^{1}(D) \mid H^{1}(D) \ni g \mapsto \mathcal{E}(f,g) \text{ is continuous w.r.t. } L^{2}(D,dx)\text{-topology}\}, \end{equation*} where $\mathcal{E}(f,g)=\sum_{i=1}^{d}\int_{D}\frac{\partial f}{\partial x_i}\frac{\partial g}{\partial x_i}\,dx$.

My question

Fix $p,q \in \partial D$ with $p \neq q$. Then, can we find an $f \in D(L)\cap C(\bar{D})$ such that $f(p) \neq f(q)$ ?

My attempt

Fix an open neighborhood $U$ of $p$ in $\mathbb{R}^d$ and a bi-Lipschitz function $\psi:B(1) \to U$ such that $\psi(0)=p$ and $\psi_{p}(B_{+}(1))=U \cap D$. We may assume $p,q \in U$. Since $$\psi^{-1}(p)=0\text{ and }\psi^{-1}(q) \in \{x \in B(1)\mid x_d=0\},$$ it is easy to construct a smooth function $F$ on $B(1)$ such that $\text{supp}[F] \subset B(1)$ and $$F(\psi^{-1}(p))\neq F(\psi^{-1}(q))\text{ and }\left.\frac{\partial F}{\partial x_d}\right|_{x_d=0}=0.$$

That is, $F$ satisfies the Neumann boundary condition in the sense that $\sum_{i=1}^{d}\frac{\partial F}{\partial x_i}n_i=0$. Here, $n=(n_1,\ldots,n_d)$ denotes the outward unit normal to the boundary of upper half space.

Clearly, $\tilde{F}(x)=F(\psi^{-1}(x))$ satisfies $\tilde{F}(p) \neq \tilde{F}(q)$. Can we show the following? $$(1)\quad(\nu,\nabla \tilde{F}):=\sum_{i=1}^{d}\nu_i\frac{\partial \tilde{F}}{\partial x_i}=0\ \mathcal{H}^{d-1} \text{-a.e. on }\partial D.$$ Here, $\mathcal{H}^{d-1}$ is the $(d-1)$-dim Hausdorff measure and $\nu$ is the outward unit normal to $\partial D$. If (1) holds, I think $\tilde{F} \in D(L)$ follows by using Green's formula: \begin{equation*} -\int_{D}\frac{\partial^2 \tilde{F}}{\partial x_i^2}f\,dx=\int_{D}\frac{\partial f}{\partial x_i}\frac{\partial \tilde{F}}{\partial x_i}\,dx+\int_{\partial D}f (\nu,\nabla \tilde{F})\,d\mathcal{H}^{d-1}. \end{equation*}

Answer 1

This will only be a partial answer to your question, due to lack of time.

As you effectively noted, any $f\in C^2(\overline{D})$ with compact support and such that $(\nu, \nabla f) = 0 \ \mathcal{H}^{d-1}$-a.e. on $\partial D$ will belong to $D(L)$. The problem with your argument is that, given a differentiability point $p \in \partial D$ and a general $U \subseteq \mathbb{R}^d$ and $\psi : B(1) \to U$ as you describe, it is not true that $$(\nu, \nabla f) = 0 \ \text{at} \ p \Leftrightarrow \partial_d(f \circ \psi)\big|_0 = 0 .$$ To see why, suppose you had a different pair $(U',\psi')$ satisfying the same requirements. Then, by the chain rule, a.e. on $\psi'^{-1}(U \cap U')$ it holds that $$\partial_d(f \circ \psi')\big|_{x} = \partial_d(f \circ \psi \circ \psi^{-1} \circ \psi')\big|_{x} = \sum_{i=1}^d \partial_i(f \circ \psi)\big|_{y(x)}\partial_d y_i\big|_{x}$$ where $y := \psi^{-1} \circ \psi' : \psi'^{-1}(U \cap U') \to \psi^{-1}(U \cap U')$, i.e. the change of variables map. Note that $y(x) = 0$, hence you have (a.e.) $$\partial_d(f \circ \psi')\big|_0 = \sum_{i=1}^d \partial_i(f \circ \psi)\big|_0 \partial_d y_i\big|_0.$$ Since it is certainly possible for the two sides of this equation to be different, it is meaningless to attempt to describe the vanishing of the normal derivative of $f$ in terms of the vanishing of the $d$-th partial derivative in generic boundary co-ordinates.

What is true is that, in each co-ordinate system $(U,\psi)$, the (unit) normal derivative of $f$ will correspond to a certain linear combination of partial derivatives of $f \circ \psi$. Notice that this linear combination will always feature a non-zero multiple of the $d$-th partial derivative.

If your boundary were better than Lipschitz (for instance if it were smooth) then you could also set up special coordinate systems adapted to $\partial D$ which, in addition to your requirements, also have the property that the $d$-th partial derivative of $f \circ \psi$ on $\{ x_d=0\}$ really does equal the (unit) normal derivative of $f$ relative to $\partial D$. These are known as geodesic (or Gaussian) normal co-ordinates (adapted to the hypersurface $\partial D$). I don't know what the theory is for these co-ordinate systems in such a low regularity setting as Lipschitz.

In order to answer your question you thus need some way of generating $C^2(\overline{D})$ functions $f$ with prescribed boundary data, i.e. with (a.e.) vanishing normal derivative together with the condition $f|_{\partial D} = g$ for some non-constant $g$. You should be able to do so using the celebrated Whitney extension theorem. I expect that the argument will go broadly speaking as follows: The vanishing of the normal derivative is essentially a constraint on a particular linear combination of partial derivatives of $f$. Assume WLOG that locally around $p \in \partial D$, the a.e. defined vector field of normal derivatives $\nu = (\nu_1, \ldots, \nu_d)$ is such that $\nu_d \neq 0$. Hence $$(\nu, \nabla f) = \left(\sum_{i=1}^{d-1} \nu_i \partial_i f\right) + \nu_d \partial_d f, \quad \text{and} \quad \partial_d f = -(\nu_d)^{-1}\sum_{i=1}^{d-1} \nu_i \partial_i f \ \text{if} \ (\nu, \nabla f) = 0.$$ Hence, you can choose anything you like for the partial derivatives $\partial_i f$, $i=1,\ldots,d-1$ on $\partial D$, and provided the $d$-th partial derivative is then given by the formula above, you have ensured that $f$ satisfies the Neumann boundary condition. Going the other way around, suppose you have the following collection of functions on $\partial D$: (i) a (non-constant, to solve your problem!) $g$ on $\partial D$; (ii) for $i=1, \ldots, d-1$, functions $f_i$; (iii) $f_d := -(\nu_d)^{-1} \sum_{i=1}^{d-1}\nu_i f_i$. Then (provided you can check that the hypotheses of the Whitney extension theorem are satisfied) there locally exists a function $f$, $C^2$ up to and including $\partial D$, which restricts to $g$ on $\partial D$ and which, by construction, satisfies the Neumann boundary condition there. From here, globalize to get the desired result.