Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Boundary determination of all scalar conductivity jets

Statement

Assume the Axiom of Choice and Countable Choice.

Let Ω⊂Rn, n≥2, be a bounded smooth domain, and let γ1,γ2∈C∞(Ω‾,R) be strictly positive. If their full physical conductivity Dirichlet-to-Neumann maps agree on all boundary data, then every inward boundary-normal derivative agrees: ∂xnkγ1∣∂Ω=∂xnkγ2∣∂Ω(k=0,1,2,…), where xn=dist⁡(x,∂Ω) in a boundary normal collar.

Connectedness of Ω is not required. Write its finitely many connected components as Ω1,…,ΩJ. On a possibly disconnected domain, the full physical DN map means the finite direct sum Λγ=⨁a=1JΛγ∣Ω‾a on H1/2(∂Ω)=⨁a=1JH1/2(∂Ωa), using the connected-domain DN definition on each component.

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the DN identity and exact correction suppliers [F1]--[F3].

[A2]

The Axiom of Countable Choice (ACω) is inherited from the DN identity and exact correction suppliers [F1]--[F3].

Given: the smooth bounded domain, two strictly positive smooth conductivities with equal full-data DN maps, and the fixed-atlas trace convention.

[F1]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: Λγ is the weak full-boundary map of div⁡(γ∇) with physical flux γ∂νout.

[F2]

Alessandrini identities for scalar conductivity and Schrödinger DN maps: for conductivity solutions uj with boundary values fj, ⟨(Λγ1−Λγ2)f1,f2⟩=∫Ω(γ1−γ2)∇u1⋅∇u2‾ dx.

[F3]

Boundary quasimode jets and exact-solution correction: for every boundary point, unit tangential covector, cutoff χ=1 near that point, integer k≥0, and residual order N>k, the approximate solutions have residual and exact-correction H1 errors O(λ−N); their weighted gradient product satisfies λk∫dist⁡(x,∂Ω)kF∇v1⋅∇v2‾ dx→ck∫∂ΩF∣χ∣2 dS with ck>0.

Proof

Given: the data in the statement.

1.1A1A2F1F2F3givenconstructalgebra

We first make explicit the full physical DN map and the identity on a possibly disconnected Ω. The compact smooth manifold Ω‾ has finitely many connected components Ma=Ω‾a, 1≤a≤J: connected coordinate neighborhoods give an open cover, and a finite subcover meets every component. Each Ma has nonempty boundary; otherwise its same-dimensional inclusion in Rn would make it both open and compact, impossible for a nonempty subset of Rn. Thus the connected-domain suppliers [F1] and [F2] apply on every Ma. Fixed finite atlases identify H1(Ω)=⨁aH1(Ωa) and H1/2(∂Ω)=⨁aH1/2(∂Ωa), with the corresponding finite sum for the boundary dual. The full physical map is Λγ=⨁aΛγ∣Ma: solve separately on each component, and its weak boundary pairing is the sum of the component weak pairings. This is exactly the global physical flux pairing, not an additional assumption on the conductivities or data. Equality of the global maps implies equality of the component maps by inputs and tests supported on one component. Applying [F2] on each component and summing gives the global Alessandrini identity used below. The quasimode supplier [F3], Proof 5.1, already constructs the exact solutions componentwise. Boundary normal collars and derivatives are componentwise as well. Consequently all later localization and normal-jet steps apply without a connectedness restriction.

1.2A1A2F1F2F3givenalgebra

Fix x0∈∂Ω and a sufficiently small patch. Choose χ∈Cc∞(∂Ω) supported in the patch and equal to one near x0, and use [F3] to construct approximate beams for γ1,γ2 with the same boundary trace. Let uλ,j be their exact conductivity solutions. Equality of the DN maps and [F2] give ∫Ω(γ1−γ2)∇uλ,1⋅∇uλ,2‾ dx=0. Replacing each exact solution by its approximate beam changes this integral by O(λ−N): the coefficient is bounded, the beam H1 norms are uniformly bounded, and each exact correction is O(λ−N) in H1 by [F3].

2.1F3step 1.2contradiction

Set k=0 in [F3] and let λ→∞ in step 1.2. Since c0>0, ∫∂Ω(γ1−γ2)∣χ∣2 dS=0. If γ1−γ2 were positive or negative at any boundary point, continuity would give a patch on which it has that strict sign; choosing χ supported in that patch and equal to one near the point would make this integral nonzero. Thus γ1=γ2 on ∂Ω.

3.1A1A2F2F3step 1.2step 2.1algebra

Fix an integer k≥1 and suppose inductively that ∂xnj(γ1−γ2)=0 on a boundary patch for every j<k. Taylor's formula with integral remainder in the inward normal coordinate gives γ1(y,xn)−γ2(y,xn)=xnkFk(y,xn),Fk(y,0)=∂xnk(γ1−γ2)(y,0)k!, for a smooth Fk. Apply [F2] to the exact beams from [F3], replace them by their approximate beams, and multiply the resulting integral identity by λk. The replacement error is O(λk−N), since the weight xnk is bounded and the corrections are O(λ−N) in H1. Choose N>k and apply the weighted limit in [F3]; then ck∫∂ΩFk(y,0)∣χ(y)∣2 dS(y)=0. Since ck>0, the same localized sign argument as in step 2.1 gives Fk(y,0)=0 throughout the patch. Therefore the kth inward normal derivatives agree there.

4.1step 2.1step 3.1induction∎

Step 2.1 is the base case for the induction. Step 3.1 proves each subsequent order on a patch that may be chosen around any prescribed boundary point. Thus every boundary point has a neighborhood on which the order-k normal derivative difference is zero for every integer k≥0, proving the full boundary-jet assertion.

Depends on

Used by

Dependency tree · two levels

34 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources