Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)
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.

Element and edge residual representation for P1 Galerkin errors

Statement

Assume Countable Choice, as in the Sobolev and simplex trace suppliers.

Let T be a matching triangular mesh, U∈V(T), and u solve the weak model from Residual indicators and data oscillation for P1 AFEM. For every v∈H01(Ω), ∫Ω∇(u−U)⋅∇v+(b⋅∇(u−U))v=∑K∈T∫KRK(U)v−∑E interior∫EJE(U)v.

Facts & Assumptions

[A1]

Countable Choice is assumed in the cited Sobolev and reference-simplex suppliers (The Axiom of Countable Choice (ACω)).

Given: Countable Choice, the weak equation, matching affine triangles, continuous P1 function U, and v∈H01(Ω).

[F1]

On an affine simplex, Green's identity holds for the polynomial vector field ∇U and Sobolev test functions; opposite outward normals cancel on a shared matching facet (Sobolev face traces and Gauss–Green on a reference simplex, Interior facet orientations cancel in a matching mesh).

[F2]

H01 is the closure of smooth compactly supported functions, and continuous matching piecewise-affine functions with zero boundary values belong to this space (Zero-boundary Sobolev space as a norm closure, Continuous matching piecewise polynomials belong to W1p, Zero-boundary conforming polynomials lie in W0 1p and satisfy Poincare).

Proof

1.1A1F1given

For smooth compactly supported v, integrate ∫K∇U⋅∇v=−∫KΔU v+∫∂K(∇U⋅nK)v on every cell using [F1]. Since U∣K is affine, ΔU=0. The sum of boundary integrals on each interior edge is ∫EJE(U)v, and boundary-edge terms vanish because v is compactly supported.

2.1step 1.1givenalgebra

The weak residual is (f,v)−∫∇U⋅∇v−∫(b⋅∇U)v. Substituting step 1.1 gives cell integrands f−b⋅∇U=RK(U) and the negative of the summed outward flux jumps. This is the displayed formula for smooth compactly supported v.

3.1A1F1F2step 2.1∎

Smooth compactly supported functions are dense in H01 by its definition. Every term in the identity is continuous in the H1 norm: volume terms use f,∇U∈L2, and edge terms use the trace inequality from [F1] on the finite mesh. Passing to the limit extends step 2.1 to all v∈H01.

Depends on

Used by

Dependency tree · two levels

39 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