Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass
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.

Penalties below and above the Nitsche coercivity threshold

Example

Assume Countable Choice and take any nondegenerate bounded simplex with its one-cell matching mesh and degree-k space, k≥1. For either Nitsche sign θ∈{0,1}, the algebraic penalty-free boundary form, obtained by evaluating the displayed formula at zero penalty, is not coercive: the constant function has zero gradient and zero form value but positive boundary part in ∥⋅∥E,h. In contrast, any γ0>(1+θ)2C∂2/4 satisfies the sufficient coercivity threshold of Nitsche coercivity above the inverse-trace threshold. The example shows failure at zero and success above the proved bound; it does not claim that every positive penalty below that sufficient bound fails on every mesh.

Facts & Assumptions

Given: Countable Choice; one bounded simplex with all facets boundary; k≥1; a constant trial function; the form and energy seminorm of Symmetric and nonsymmetric Nitsche boundary forms; and the coercivity result Nitsche coercivity above the inverse-trace threshold.

[F1]

At γ0=0, ahθ(1,1)=0 because ∇1=0 and ∂n1=0; the energy seminorm has ∥1∥E,h2=∑FhF−1∣F∣>0 (Symmetric and nonsymmetric Nitsche boundary forms).

[F2]

If γ0>(1+θ)2C∂2/4, then ahθ(vh,vh)≥λθ∥vh∥E,h2 with λθ>0 (Nitsche coercivity above the inverse-trace threshold).

Verification

Given: The one-cell simplex setup and the data in [F1]–[F2].

1.1F1given

For vh=1, each volume-gradient and normal-flux term in the definition vanishes. At γ0=0 the penalty term vanishes too, so ahθ(1,1)=0. Since each facet has positive measure and hF>0, the finite sum ∑FhF−1∣F∣ is positive; hence no positive coercivity constant can bound this form below by ∥1∥E,h2.

2.1F2step 1.1algebra∎

For any γ0 strictly above the displayed threshold, [F2] gives the positive constant λθ and proves coercivity on the entire finite-element space, including the constant function. This establishes the stated below/above comparison without making a necessity claim for intermediate positive penalties.

Depends on

Used by

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Dependency tree · two levels

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