Let $D\subset\mathbb R^n$ be a smoothly bounded open domain and $0\in D$. For any $x\in\partial D$ there holds \begin{eqnarray*} 2 \,a'(\vert x\vert)\,(x\cdot\nu(x))+(n1)\,a(\vert x\vert) \, H(x) = \text{const.} \end{eqnarray*} Here $a:(0,\infty)\to (0,\infty)$ is a positive smooth function, $a'$ denotes its derivative, $\nu(x)$ is the outer unit normal vector in $x\in\partial D$ and $(n1)H(x)$ denotes the mean curvature in $x\in\partial D$. Does this imply that $D$ is a ball?
1 Answer
If you let $\partial D$ be a hypersurface of revolution whose generatrix $\gamma: I \to [0,\infty) \times \mathbb{R}$ is such that $\gamma$ is monotonic, you then have $H(x)$ and $x\cdot \nu(x)$ can both be written as functions of $x$ only. One can further arrange that $x\cdot \nu(x) > \epsilon > 0$ for all positions.
Let $G(x) = H(x) / x \cdot \nu(x)$.
If you choose $a(s)$ to be any solution of
$$ (\ln a(s))' = \frac{1n}{2} G(s) $$
your desired equation is satisfied with constant 0.

$\begingroup$ Many thanks! Would the answer be also affirmative, if the monotonicity assumption on $\vert\gamma\vert$ fails? $\endgroup$– guest61Jul 6, 2021 at 15:44

$\begingroup$ @guest61: I don't understand your comment. My answer provides infinitely many counterexamples to your claim. $\endgroup$ Jul 6, 2021 at 17:03

$\begingroup$ I don't see why a surface of revolution has a mean curvature that only depends on the modulus of the position vector? The same question for $x\cdot\nu(x)$? $\endgroup$– guest61Jul 10, 2021 at 16:35

$\begingroup$ the monotonicity assumption forces there to be exactly one circle for each value of $x$, and the mean curvature is obviously constant on that one circle by rotational symmetry. $\endgroup$ Jul 11, 2021 at 3:15