Pompeiu problem

Let n2n \ge 2, and let M(n)M(n) denote the group of rigid motions of Rn\mathbb{R}^n, i.e. the maps σ(x)=ρx+b\sigma(x) = \rho x + b with ρO(n)\rho \in O(n) and bRnb \in \mathbb{R}^n.

Let KRnK \subset \mathbb{R}^n be a bounded, simply connected Lipschitz domain, and suppose there exists a continuous function f:RnCf : \mathbb{R}^n \to \mathbb{C}, not identically zero, such that

σ(K)f(x)dx=0for every σM(n),\int_{\sigma(K)} f(x)\, dx = 0 \qquad \text{for every } \sigma \in M(n),

that is, the integral of ff vanishes over every congruent copy of KK. Then KK is a ball.

In the setting of overdetermined boundary value problems, the corresponding assertion is the following. Let ΩRn\Omega \subset \mathbb{R}^n be a bounded, simply connected domain with smooth (indeed real analytic) boundary Ω\partial \Omega, and let /n\partial/\partial n denote differentiation along the outward unit normal to Ω\partial \Omega. Suppose there exist λ>0\lambda > 0, a constant cc, and a function uC2(Ω)u \in C^2(\overline{\Omega}) not identically zero satisfying

{Δu+λu=0in Ω,un=0on Ω,u=con Ω.\begin{cases}\Delta u + \lambda u = 0 & \text{in } \Omega,\\[2pt] \dfrac{\partial u}{\partial n} = 0 & \text{on } \partial \Omega,\\[2pt] u = c & \text{on } \partial \Omega.\end{cases}

Then Ω\Omega is a ball.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

arXiv:1202.2398

Additional references

  1. arXiv:1202.2398, Pompeiu problem — the statement above follows it.
  2. Wikipedia, Pompeiu problem, the article this problem comes from.

Solutions 0

No solutions have been posted yet.