# Closed surfaces of prescribed mean curvature

Let $$D\subset\mathbb R^n$$ be a smoothly bounded open domain and $$0\in D$$. For any $$x\in\partial D$$ there holds $$\begin{eqnarray*} 2 \,a'(\vert x\vert)\,(x\cdot\nu(x))+(n-1)\,a(\vert x\vert) \, H(x) = \text{const.} \end{eqnarray*}$$ 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 $$(n-1)H(x)$$ 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

$$(\ln a(s))' = \frac{1-n}{2} G(s)$$

your desired equation is satisfied with constant 0.

• Many thanks! Would the answer be also affirmative, if the monotonicity assumption on $\vert\gamma\vert$ fails? Jul 6, 2021 at 15:44
• @guest61: I don't understand your comment. My answer provides infinitely many counterexamples to your claim. Jul 6, 2021 at 17:03
• 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 $x\cdot\nu(x)$? Jul 10, 2021 at 16:35
• 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. Jul 11, 2021 at 3:15