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CorollaryStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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Static wave coefficients are determined on a simple surface with adequate observation time

Statement

Assume Countable Choice. Let (M,g) be a compact, connected, nonempty simple two-dimensional Riemannian background with smooth boundary. Suppose L∗<∞ bounds the lengths of all maximal unit-speed geodesics in M, and let T>L∗. For real static potentials q1,q2∈Cc∞(M∘;R), if the full time-dependent Dirichlet-to-Neumann maps for ∂t2−Δg+q1 and ∂t2−Δg+q2 agree for every smooth boundary input compactly supported in (0,T)×∂M, then q1=q2 on M. This equality statement asserts no stability modulus or recovery in the presence of noisy data.

Facts & Assumptions

Given: Countable Choice; a simple compact connected nonempty Riemannian surface; a finite upper bound L∗ on all maximal unit-speed geodesic lengths; T>L∗; real smooth interior potentials with equal full time-dependent DN maps.

[A1]

Countable Choice is the axiom base of the wave-data and two-dimensional X-ray suppliers. (The Axiom of Countable Choice (ACω))

[F1]

Equal wave DN maps on a simple background imply that the integral of q1−q2 over every maximal unit-speed geodesic of length less than T is zero. (Wave boundary data determine integrals along maximal geodesics shorter than T)

[F2]

The scalar geodesic X-ray transform integrates a real interior function over each incoming maximal unit-speed boundary geodesic. (Simple surface and geodesic X-ray transform)

[F3]

On a compact simple surface, zero unweighted scalar geodesic X-ray data for a real smooth interior function imply that the function vanishes. (Unweighted scalar geodesic X-ray injectivity on simple surfaces)

Proof

1.1A1F1F2given

Every maximal unit-speed geodesic has length at most L∗<T. Therefore [F1] applies to each such ray and gives I0(q1−q2)=0 on all incoming boundary unit vectors, using the transform convention in [F2].

2.1A1F3step 1.1given∎

The real function q1−q2 lies in the class covered by [F3]. Its transform vanishes by step 1.1, so [F3] gives q1−q2=0 throughout M. Hence q1=q2.

Depends on

Used by

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