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Hyperbolic inverse problems and finite-time visibility — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Exact-solution correction errors vanish in the Gaussian-beam ray limit

Example

Let (M,g) be a compact Riemannian manifold, let T>0, and let Q=(0,T)×M carry product measure dVgdt. Fix a unit-speed geodesic segment γ:[0,L]→M and t0 with 0<t0<T−L, and set Γ(s)=(t0+s,γ(s)), 0≤s≤L; ds denotes its arclength parameter. Let a∈C∞([0,T]×M), and for each λ≥1 let uj,λapp∈L2(Q) and Rj,λ∈L2(Q) for j=1,2. Suppose ∥uj,λapp∥L2(Q)≤C0, ∥Rj,λ∥L2(Q)≤C1λ−1/2, and ∫Qa u1,λappu2,λapp dVgdt⟶∫0La(t0+s,γ(s)) ds. Set uj,λ=uj,λapp+Rj,λ. Then ∣∫Qa u1,λu2,λ dVgdt−∫Qa u1,λappu2,λapp dVgdt∣=O(λ−1/2), and hence the exact-solution product has the same ray-integral limit. For the real self-adjoint wave application, assume additionally Countable Choice, that M is connected and nonempty with nonempty smooth boundary, and that the two operators are Pj=∂t2−Δg+qj(x) with qj∈C∞(M;R) independent of time. The beams must satisfy the following additional supplier conditions: the observed ray has a compact embedded phase-space extension with immersed base projection; its two endpoint cutoff neighborhoods lie outside the measurement cylinder; and the approximate forward beam vanishes near t=0, while the approximate terminal beam vanishes near t=T. These temporal vanishing conditions are separate from the location of the spatial endpoint residuals. Use physical time along the unit-speed spatial ray and the normalization ∣∂tϕ∣=1, with initial transported mass one. Then the Gaussian-beam lemma gives the unweighted ray measure and residual O(λ−N) throughout Q. Apply the wave energy wellposedness theorem with zero boundary data and zero initial data for the forward correction, or zero terminal data for the terminal correction, to the negative residual. Since ∥Pjuj,λapp∥Lt1Lx2≤T∥Pjuj,λapp∥L2(Q), its energy estimate gives an L2(Q) correction O(λ−N), hence the assumed O(λ−1/2) order after choosing N≥1. The exact waves retain the actual approximate boundary traces. This application asserts no temporal vanishing merely from an extension outside the measurement domain; the generic error estimate above only needs its explicitly stated norm and product-limit hypotheses.

Facts & Assumptions

Given: A smooth coefficient a on the compact spacetime cylinder; two uniformly L2-bounded approximate solutions; and two L2 correction remainders of order λ−1/2.

[F1]

For real-valued square-integrable v,w, Holder's inequality for integrals, including the endpoint cases gives ∫∣vw∣≤∥v∥2∥w∥2. For complex-valued v,w, apply that inequality to the real functions ∣v∣ and ∣w∣; ∣vw∣=∣v∣∣w∣ and their L2 norms are unchanged.

Proof

1.1F1givenalgebra

Expand the exact product: u1,λu2,λ−u1,λappu2,λapp=R1,λu2,λapp+u1,λappR2,λ+R1,λR2,λ. Multiplying by a, integrating, and applying [F1] to each term gives an absolute value at most ∥a∥L∞(Q)(∥R1,λ∥2∥u2,λapp∥2+∥u1,λapp∥2∥R2,λ∥2+∥R1,λ∥2∥R2,λ∥2).

2.1step 1.1givenalgebra∎

By the hypotheses, the preceding expression is bounded by ∥a∥∞(2C0C1λ−1/2+C12λ−1), which is O(λ−1/2) for λ≥1. Since this difference tends to zero, the assumed approximate-product limit passes unchanged to the exact product.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

A potential perturbation beyond the two-trip distance is invisible before time T

Example

Assume Countable Choice. Let (M,g) be a compact, connected, nonempty smooth Riemannian manifold with nonempty smooth boundary, let T>0, and let q1,q2∈C∞(M;R) be time independent. Put Q0=supp⁡(q1−q2), the closure of the set on which q1−q2 is nonzero, and suppose Q0≠∅ and dg(Q0,∂M)>T/2, where dg is the intrinsic Riemannian distance. For every smooth Dirichlet input h∈Cc∞((0,T)×∂M), the full time-dependent forward Dirichlet-to-Neumann maps satisfy Λq1+h=Λq2+hin L2((0,T)×∂M,dSgdt). Thus a perturbation supported farther than half the observation time from the boundary cannot be detected by boundary measurements made before time T.

Facts & Assumptions

Given: Countable Choice; the compact Riemannian manifold, static real potentials with dg(Q0,∂M)>T/2, and smooth boundary input h.

[A1]

Countable Choice is used through the DN-map and energy-solution interfaces and the uniqueness of the locally integrable representative of a regular distribution. (The Axiom of Countable Choice (ACω))

[F1]

For any forward solution for q1 with boundary value h and terminal solution for q2 with boundary value k, the interior product pairing is zero when dg(Q0,∂M)>T/2. (Finite propagation for the wave equation on a Riemannian manifold)

[F2]

The forward DN map is the outward normal trace of the zero-initial solution; the terminal map is the outward normal trace of the zero-terminal solution. Both are in L2 for the stated classical solution class. (Time-dependent Dirichlet-to-Neumann map for the wave equation)

[F3]

For these forward and terminal solutions, ∫Q(q1−q2)uv dVgdt=∫Σ(kΛq1+h−hΛq2−k) dSgdt. (Alessandrini identity for the time-dependent wave Dirichlet-to-Neumann maps)

[F4]

If a locally integrable function on the smooth boundary cylinder pairs to zero with every compactly supported smooth test, then it vanishes almost everywhere, by chart localization and regular-distribution uniqueness. (Smooth boundary tests separate locally integrable traces)

Proof

1.1A1F1F3given

Fix arbitrary k∈Cc∞((0,T)×∂M). Let u1 be the forward solution for q1 with boundary input h, and let v2 be the terminal solution for q2 with boundary input k. By [F1], ∫Q(q1−q2)u1v2 dVgdt=0. Using [F3] therefore gives ∫Σ(kΛq1+h−hΛq2−k) dSgdt=0.

1.2F3given

Apply [F3] again with both potentials equal to q2. Its interior term is then zero, so ∫Σ(kΛq2+h−hΛq2−k) dSgdt=0.

2.1A1F2F4step 1.1step 1.2given∎

Subtract step 1.2 from step 1.1. The terminal terms cancel, yielding ∫Σk(Λq1+h−Λq2+h) dSgdt=0 for every smooth compactly supported k. The difference of the two traces belongs to L2(Σ) by [F2], hence is locally integrable. Its pairing with every k is zero; [F4] implies that its L2 class vanishes. This is the asserted equality of forward DN data.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

Adequate observation time on a simple surface

Example

Assume Countable Choice. Let (M,g) be a compact, connected, nonempty simple two-dimensional Riemannian manifold with smooth boundary. Suppose L∗<∞ is an upper bound for the lengths of all maximal unit-speed geodesics in M. Let T>L∗ and let q1,q2∈Cc∞(M∘;R) be static interior potentials. If their full time-dependent Dirichlet-to-Neumann maps agree for every smooth boundary input compactly supported in (0,T)×∂M, then q1=q2 throughout M.

This gives a concrete observation-time condition: observing longer than a uniform upper bound on the maximal geodesic lengths suffices for coefficient uniqueness in the simple-surface case.

Facts & Assumptions

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

[A1]

Countable Choice is the axiom base of the finite-time coefficient-uniqueness and two-dimensional X-ray injectivity suppliers. (The Axiom of Countable Choice (ACω))

[F1]

Under Countable Choice, on a compact connected nonempty simple surface, if L∗<∞ bounds all maximal unit-speed geodesic lengths and T>L∗, equality of the full forward DN maps on smooth time-compact boundary inputs implies q1=q2 for real static q1,q2∈Cc∞(M∘;R). The vanishing ray integrals are proved inside this corollary, not required as an additional premise. (Static wave coefficients are determined on a simple surface with adequate observation time)

[F2]

The unweighted scalar geodesic X-ray transform is injective on a compact simple surface. (Unweighted scalar geodesic X-ray injectivity on simple surfaces)

Proof

1.1given

By hypothesis, every maximal unit-speed geodesic has length at most L∗<T. Thus all maximal rays relevant to the finite-time coefficient-uniqueness corollary lie within the observation interval; the strict inequality also covers a ray attaining the upper bound.

2.1A1F1F2step 1.1given∎

Equal DN maps and the simple-surface hypotheses satisfy [F1]; since T>L∗, the corollary yields q1−q2=0 on M. Its X-ray injectivity step is justified by [F2], so q1=q2 as claimed.

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