ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-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.

A too-small SIPDG penalty destroys coercivity

Example

Use the two-cell interval specialization and piecewise affine function described in Given below. At this mesh and with penalty $γ = 1 / 4$ , the function satisfies $a_{h} (v, v) = - 1 / 2$ . Hence the form is not coercive in this case, even though its energy expression on $v$ is positive.

Facts & Assumptions

Given: $K^{-} = (0, 1 / 2)$ , $K^{+} = (1 / 2, 1)$ , unit coefficient, counting measure on all point faces, $h_{F} = 1 / 2$ at each face (the common interval-cell length), the piecewise affine function $v (x) = x$ on $K^{-}$ and $v (x) = x - 1$ on $K^{+}$ , zero endpoint values, and penalty $γ = 1 / 4$ . For functions $v, w$ on these cells, define $a_{h} (v, w) : = \sum_{K \in {K^{-}, K^{+}}} \int_{K} v^{'} w^{'} d x - \sum_{F \in {0, 1 / 2, 1}} (q_{F} (v) [w]_{F} + q_{F} (w) [v]_{F}) + γ \sum_{F \in {0, 1 / 2, 1}} h_{F}^{- 1} [v]_{F} [w]_{F},$ where $q_{1 / 2} (v) = {v^{'}}_{1 / 2}$ , $q_{0} (v) = - v^{'} (0 +)$ , and $q_{1} (v) = v^{'} (1 -)$ ; the endpoint jumps are the one-sided value traces and all face integrals use counting measure. Define the energy expression by $∥ v ∥_{D G}^{2} : = \sum_{K \in {K^{-}, K^{+}}} \int_{K} ∣ v^{'} ∣^{2} d x + γ \sum_{F \in {0, 1 / 2, 1}} h_{F}^{- 1} [v]_{F}^{2} .$

[F1]

The one-dimensional jump is $v^{-} - v^{+}$ and the interior average is the arithmetic mean of the two traces. (Jump and average conventions for symmetric interior-penalty DG)

Verification

1.1givenF1algebra

Both cell derivatives equal $1$ , so the volume term is $1$ . The endpoint terms vanish because $v (0) = v (1) = 0$ . At the interior point, [F1] gives $[v] = 1$ and ${v^{'}} = 1$ . The energy expression is $∥ v ∥_{D G}^{2} = 1 + (1 / 4) (1 / (1 / 2)) \cdot 1^{2} = 3 / 2 > 0$ .

2.1givenstep 1.1algebra∎

The two consistency terms sum to $- 2$ , while the penalty term is $(1 / 4) (1 / (1 / 2)) \cdot 1^{2} = 1 / 2$ . Therefore $a_{h} (v, v) = 1 - 2 + 1 / 2 = - 1 / 2 < 0$ , which contradicts coercivity with any positive constant in a positive energy norm.

Depends on

Jump and average conventions for symmetric interior-penalty DG

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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