Serrin's overdetermined theorem and weak Bernoulli laws without Alt--Caffarelli regularity

Abstract

We study distributional Bernoulli-type conditions in geometrically irregular domains $Ω$. Here the zero extension of $u$ to $\mathbb{R}^n$ satisfies $$ Δu \;=\; c\,\mathcal{H}^{n-1}\!\lfloor_{\partial^*Ω}\;-\;f(u)\,\mathbf{1}_Ω\,dx $$ in the distributional sense. This is a weak version of the one-phase Bernoulli free boundary condition, which avoids the uniform Lipschitz/density assumptions of classical Alt--Caffarelli theory. We prove that for every $n \ge 2$ and every $f \in C^2(\mathbb{R})$ with $f(0)>0$, there exist bounded, non-spherical, finite-perimeter domains $Ω\subset \mathbb R^n$ satisfying this distributional Bernoulli law with $$ 0<\mathcal H^{n-1}(\partial^*Ω)<\infty, \qquad \mathcal H^{n-1}(\partialΩ\setminus\partial^*Ω)=0, $$ yet $$ {\rm ess}\sup_{x\in\partial^*Ω} \sup_{0<r<1} \frac{\mathcal H^{n-1}(B_r(x)\cap\partial^*Ω)}{r^{n-1}} =\infty. $$ This shows the key constraint is not absence of a reduced boundary, but failure of uniform all-scale surface density bounds. For $f \equiv 1$, these results yield counterexamples to the weak Serrin-type overdetermined problems in all dimensions, proving the distributional Bernoulli law alone cannot replace the uniform growth/density conditions core to Alt--Caffarelli theory. On the other hand, we prove a planar rigidity result: Within the Smirnov class, the associated harmonic quadrature identity forces $Ω$ to be a disk. Thus, for the constant-source Serrin/Bernoulli law, Smirnov regularity is the threshold for weak Bernoulli rigidity in $\mathbb R^2$, while uniform upper density bounds form the threshold for $n\ge 3$ according to [23].