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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 mini-element bubble detects an affine pressure gradient

Statement

Assume Countable Choice. Let $K$ be a $d$ -simplex, $b_{K} > 0$ its interior barycentric bubble, and $q \in P_{1} (K)$ a nonconstant affine pressure. Define the local pairing $b_{K} (v, q) : = - \int_{K} q div v$ and let $v = b_{K} \nabla q$ . Then $b_{K} (v, q) = \int_{K} b_{K} ∣ \nabla q ∣^{2} > 0$ . This positive cell contribution does not by itself prove a global inf-sup bound.

Facts & Assumptions

Given: Countable Choice; a simplex $K$ , a nonconstant affine $q$ , and the mini-element scalar bubble $b_{K}$ .

[F1]

The bubble is positive in the cell interior and vanishes on its boundary. (The conforming P1-bubble/P1 mini element)

[F2]

The Stokes convention is $b (v, q) = - \int q div v$ ; the local pairing is its cell restriction. (Stokes velocity and pressure spaces and weak forms)

[A1]

Countable Choice is assumed by the affine simplex Gauss--Green supplier. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F3]

The divergence integral on an affine simplex equals its outward boundary flux. (Gauss–Green formula on an affine simplex)

Example

The local bubble-pressure pairing is computed by integration by parts.

Verification

technique · direct integration by parts

1.1A1F1F2F3givenconstruct

Since $\nabla q$ is constant, $div (b_{K} \nabla q) = \nabla b_{K} \cdot \nabla q$ . Apply [F3] to $b_{K} \nabla q$ ; the boundary flux vanishes because $b_{K} = 0$ on $\partial K$ . Thus $- \int_{K} q div (b_{K} \nabla q) = - \int_{\partial K} q b_{K} \nabla q \cdot n + \int_{K} b_{K} ∣ \nabla q ∣^{2} = \int_{K} b_{K} ∣ \nabla q ∣^{2}$ .

2.1F1algebradischarge-constructstep 1.1∎

Because $q$ is nonconstant, $∣ \nabla q ∣^{2} > 0$ ; because $b_{K} > 0$ in the interior, the integrand is positive on a set of positive volume. Its integral is therefore strictly positive. The calculation concerns one cell contribution only.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

21 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

Jean-Luc Guermond, MATH 661 Fall 2025, Chapter 54 (complete course notes) (standard reference, not scraped)
Daniele Boffi, Franco Brezzi, Michel Fortin, Mixed Finite Element Methods and Applications (complete textbook) (standard reference, not scraped)