Alphabeta Math
TheoremStatement: 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.

FE-HMM macro, micro, and sampling error separation

Statement

Let UH be the FE-HMM solution from Periodic FE-HMM with simplicial macro and periodic P1 micro meshes on a matching macro triangulation of Q=(0,1)2, and let u0∈H01(Q) solve the homogenized equation with coefficient A∗. For each macro cell let AKex be the exact periodic-coupling sampled tensor and AKh its discrete micro tensor. Define Δmicro=max⁡K∥AKh−AKex∥op,Δres=max⁡K∥AKex−A∗∥op,Δ=Δmicro+Δres, and ηH=inf⁡V∈VH∥∇(u0−V)∥L2(Q). If Δ<λ, then ∥∇(UH−u0)∥2≤λ+Λλ−Δ ηH+Δλ−Δ ∥∇u0∥2. The first term is macro best approximation; the second is explicitly split by Δmicro and Δres. By Energy identity for FE-HMM micro tensor error, Δmicro≤max⁡KηK2. If every micro ratio is an integer, No sampling resonance for period-aligned periodic micro cells gives Δres=0. The condition Δ<λ is a sufficient perturbation condition; no arbitrary-scale accuracy is asserted.

Facts & Assumptions

Given: A finite conforming macro mesh of the unit square, its continuous zero-boundary P1 space, positive cell weights summing to area, uniformly elliptic symmetric local micro tensors, a homogenized solution, and the exact sampled tensors.

[F1]

Periodic FE-HMM with simplicial macro and periodic P1 micro meshes: BH(U,V)=∑K∣K∣∇U∣K⋅AKh∇V∣K and the macro equation has right side ∫QfV.

[F2]

Well-posedness of the finite FE-HMM micro solves: each AKh is symmetric and λ-elliptic; the macro system has a unique solution.

[F3]

Energy identity for FE-HMM micro tensor error: ∥AKh−AKex∥op≤ηK2.

[F4]

No sampling resonance for period-aligned periodic micro cells: integer period ratios give AKex=A∗.

[F5]

Symmetry and uniform ellipticity of the effective tensor: A∗ is symmetric with λI≤A∗≤ΛI.

[F6]

Compatible 2D newest-vertex-bisection meshes and Matching affine simplicial meshes and shape regularity: the selected macro triangulation is finite, conforming, and covers Q‾ with disjoint cell interiors.

Proof

1.1F2F5F1F6F7algebra

By [F2] the macro solution and all local tensors are well defined. Write DK=AKh−A∗ and D0=max⁡K∥DK∥op. The triangle inequality gives D0≤Δ. For macro functions V,W, piecewise constancy of their P1 gradients and Cauchy--Schwarz over the finite cells imply ∣BH(V,W)−∫QA∗∇V⋅∇W∣≤D0∥∇V∥2∥∇W∥2. By [F5] the reference form is at least λ∥∇V∥22, so [F1] and D0≤Δ<λ also give BH(V,V)≥(λ−Δ)∥∇V∥22.

2.1F5step 1.1F1algebra

Fix any VH∈VH and put eH=UH−VH. Since both macro and homogenized equations have the same load, for every test in VH, BH(eH,eH)=∫QA∗∇(u0−VH)⋅∇eH+(∫QA∗∇VH⋅∇eH−BH(VH,eH)). The right side has magnitude at most (Λ∥∇(u0−VH)∥2+D0∥∇VH∥2)∥∇eH∥2 by [F5] and step 1.1. Combining with coercivity in step 1.1 and ∥∇VH∥2≤∥∇u0∥2+∥∇(u0−VH)∥2 gives ∥∇(UH−u0)∥2≤λ+Λλ−D0∥∇(u0−VH)∥2+D0λ−D0∥∇u0∥2. This also holds if eH=0, when division is unnecessary.

3.1F3F4step 2.1algebra∎

Take a sequence VH whose approximation errors decrease to ηH; finite-dimensionality ensures the infimum is finite, and the displayed inequality passes to the limit. The two coefficient functions are increasing for 0≤D0≤Δ<λ, so replace D0 by Δ to obtain the stated estimate. Finally, the exact decomposition AKh−A∗=(AKh−AKex)+(AKex−A∗) gives the separate micro and resonance terms, with their micro-space and aligned-sampling interpretations from [F3]–[F4].

Depends on

Used by

Dependency tree · two levels

45 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