Langford–Laugesen two-disk Neumann eigenvalue conjecture

Let Ω⊂C\Omega\subset\mathbb{C} be a bounded simply connected Lipschitz domain, let ω∈C2(Ω)∩C(Ω‾)\omega\in C^2(\Omega)\cap C(\overline{\Omega}) be positive on Ω‾\overline{\Omega}, and equip Ω\Omega with the metric g=ω∣dz∣2g=\omega|dz|^2. Suppose that the Gaussian curvature satisfies Kg≤KK_g\le K, let A=∫Ωω dx>0A=\int_\Omega\omega\,dx>0, and assume KA<4πKA<4\pi whenever K>0K>0. If the Neumann eigenvalues are enumerated as 0=λ0(Ω,g)<λ1(Ω,g)≤λ2(Ω,g)≤⋯0=\lambda_0(\Omega,g)<\lambda_1(\Omega,g)\le\lambda_2(\Omega,g)\le\cdots, then the conjecture asserts that λ2(Ω,g)<λ1 ⁣(DK(A/2))\lambda_2(\Omega,g)<\lambda_1\!\left(D_K(A/2)\right), where DK(A/2)D_K(A/2) is the geodesic disk of area A/2A/2 in the constant-curvature-KK model. The bound is sharp in the class of connected disk-type membranes, with extremizing sequences degenerating to the disjoint union of two copies of DK(A/2)D_K(A/2).

References

Primary source

arXiv

Additional references

Progress summary

Refreshed
Claimed solved

A September 2026 preprint claims to prove the two-disk conjecture for much rougher surfaces, but the proof has not been independently verified.

The Langford–Laugesen conjecture, posed in 2023, predicts that two equal constant-curvature disks maximize the relevant second Neumann eigenvalue among connected disk-type membranes. It concerns the strict bound identified as Conjecture 1.41.4 in their 2023 paper.

September 2026 preprint claim

Meiqi Liu, Zhouyu Long, and Wenming Zou claim the conjecture for bounded Lipschitz domains, without boundary differentiability or a simplicity assumption. They prove the stronger reciprocal inequality

1λ2+1λ3>2λ1(DK(A/2)),\frac{1}{\lambda_2}+\frac{1}{\lambda_3}>\frac{2}{\lambda_1(D_K(A/2))},

and claim sharpness via connected domains degenerating to two disks; the preprint is unrefereed and no independent verification or refutation was found.

Current status (as of September 2026): the conjecture has a claimed proof under the stated Lipschitz hypotheses, but the result remains unverified.

Sources

Solutions 0

No solutions have been posted yet.