Let $d \ge 2$ be a positive integer. Let $\Phi\colon \mathbb{R}^{d-1} \to \mathbb{R}$ is $C^3$-function with $\Phi(0)=0.$ We define a unbounded domain $D \subset \mathbb{R}^d$ by
\begin{align*}
D=\{x=(x_1,\ldots,x_{d-1},x_d) \in \mathbb{R}^d\mid x_d>\Phi(x_1,\ldots,x_{d-1})\}.
\end{align*}
Let $a \in \overline{D}$ and $r \in (0,1]$ such that $B(a,r) \cap \partial D\neq \emptyset.$ Here, $B(a,r)$ denotes a open ball centered at $a$ with radius $r$.
I am looking for a function $f\colon \mathbb{R}^d \to \mathbb{R}$ with the following property:
- $f$ is nonnegative and $0< f \le 1$ on $D \cap B(a,r)$,
- $f$ is compactly supported and smooth (more precisely, $f \in C^3(\overline{ D})$),
- $f$ satisfies the Neumann boundary. That is, for any $x \in \partial D$, we have
$
\langle \nabla f(x),\nu(x) \rangle=0.
$ Here, $\nu(x)$ denotes the inward unit normal vector at $x \in \partial D.$
Can we construct such a function $f$?
If $D$ is the half space, I can construct such a function. Indeed, we take a smooth function $g\colon \mathbb{R}^d \to \mathbb{R}$ such that $g=1$ on $B(a,r)$ and $g=0$ outside $B(a,2r)$. For $x=(x_1,\ldots,x_{d-1},x_d)$, we let $\hat{x}=(x_1,\ldots,x_{d-1},-x_d)$. Then,
\begin{align*}
f(x)=\{g(x)+g(\hat{x})\}/2, \quad x \in \mathbb{R}^d
\end{align*}
satisfies the all requirements.
Best Answer
In my opinion the difficult component of this problem is interpreting the 3rd condition $\vec\nabla f \cdot\vec\nu=0$ visually as the requirement of $f(\vec{y})$ for $y\in\mathbb{R}^d$ to be increasing only in the direction normal to $\vec\nu$, hence tangent to the surface defined by $\Phi(\vec x)$ for $x\in\mathbb{R}^{d-1}$
We can do this very simply by defining a cylindrical frame that varies with $\vec y$ such that $\vec \nu$ defines the polar/vertical direction. We see that the tangent plane for $\Phi$ in $\mathbb{R}^d$ is actually the $\mathbb{R}^{d-1}$ subspace defined by: $$\vec\rho_x=\left(\frac{\vec\nabla_x\Phi}{\|\vec\nabla_x\Phi\|},\|\vec\nabla_x\Phi\|\right)$$
where $\vec\nabla$ is the usual gradient operator on Euclidean space. When normalized, this frame becomes
$$\hat\rho_x=\frac{1}{\|\vec\nabla\Phi\|\sqrt{1+\|\vec\nabla\Phi\|^2}}\;\left(\vec\nabla\Phi,\,\|\vec\nabla\Phi\|^2\right)$$ $$\hat z_x=\vec\nu(x)$$
Technically, this is not a full frame, since we would need another set of $d-2$ vectors that are orthonormal with both $\hat\rho$ and $\hat\nu$. However this does not prevent us from writing the usual test function, normalized by the maximum diameter of the ball, on $\mathbb{R}^d$ that merely depends on $\rho$ and not the other $d-2$ directions:
$$f_x(y)=\exp\left(-\frac{1}{1-(\rho_x/2)^2}\right)$$
where the covector $\rho_x$ is dual to $\hat\rho_x$ such that $\rho_x(\hat\rho_x)=1$ and $\rho_x(\hat z_x)=0$