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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.

A three-dimensional spherical pulse leaves a quiet interior

Example

Assume the Axiom of Countable Choice. Let c>0, x0∈R3, r>0, and let (u0,u1)∈Cc3×Cc2 be supported in B‾r(x0) (The support of a function on Rn and its compactly supported Riemann integral). Then the Kirchhoff solution u of Kirchhoff's formula in three dimensions satisfies u(x,t)=0 whenever ∣x−x0∣>ct+r or (for ct>r) ∣x−x0∣<ct−r; at time t the pulse is carried by the spherical shell

ct−r≤∣x−x0∣≤ct+r,

and the interior Bct−r(x0) behind the front is quiet. This is the concrete illustration of the strong Huygens principle in three dimensions (The strong Huygens principle in odd spatial dimensions(b)), and it is the three-dimensional side of the contrast with A two-dimensional pulse has a tail inside the cone.

Facts & Assumptions

Given: ACω; c>0, x0∈R3, r>0, compactly supported data (u0,u1) with support in B‾r(x0); the Kirchhoff solution u.

[F1]

Kirchhoff's formula defines the solution and evaluates it from u0, its radial derivative, and u1 on the sphere ∂Bct(x), for t>0. (Kirchhoff's formula in three dimensions)

[F2]

Shell form of strong Huygens in odd dimensions: if the data are supported in a compact K and ∂Bct(x)∩K=∅, then u(x,t)=0. (The strong Huygens principle in odd spatial dimensions, The strong Huygens principle in the homogeneous Cauchy setting)

Verification

1.1givenalgebra

The support ball lies inside the sphere: if ∣x−x0∣<ct−r (with ct>r) and ∣y−x0∣≤r, then ∣y−x∣≤∣y−x0∣+∣x0−x∣<r+(ct−r)=ct, so B‾r(x0) is disjoint from ∂Bct(x) with positive distance; if ∣x−x0∣>ct+r and ∣y−x0∣≤r, then ∣y−x∣≥∣x−x0∣−∣y−x0∣>ct+r−r=ct, so again the support ball is disjoint from the sphere with positive distance.

2.1givenstep 1.1F1F2algebra∎

Vanishing: in either case of step 1.1 the data are supported in a compact set disjoint from ∂Bct(x), so [F2] gives u(x,t)=0; hence u(⋅,t) vanishes both outside the outer sphere ∣x−x0∣=ct+r and inside the inner sphere ∣x−x0∣=ct−r, so its support is contained in the closed shell ct−r≤∣x−x0∣≤ct+r; the quiet interior behind the front is the case ∣x−x0∣<ct−r, and the statement is exactly the three-dimensional instance of the shell form [F2], evaluated from the data on ∂Bct(x) as [F1] prescribes.

Depends on

Used by

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Sources