Alphabeta Math
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.

✓ 21 results · all verified · 13 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 8 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Mixed finite elements and Stokes stability

1 · Prerequisites

2 · Summary

Mixed finite elements and Stokes stability

This draft develops the real Stokes mixed formulation, a bounded divergence right inverse on bounded connected Lipschitz domains, the Babuška–Brezzi kernel-coercivity theorem, Fortin transfer, and the continuous P1+bubble/P1 mini element in dimensions two and three. Every theorem claim has a local proof carrier in the item list below. The source treatments motivate the proof routes; they do not replace the local arguments or imply independent review.

Scope and geometry

The continuous domain is a nonempty bounded connected Lipschitz open set D⊂Rd, d∈{2,3}, and the pressure gauge is Q=L02(D). The velocity space is V=H01(D;Rd) with gradient norm. We prove the finite star-shaped cover and tree-based zero-mean decomposition locally before applying the Bogovskii integral estimate to each patch. The chosen cover, overlap bumps, and their finite constants determine Cdiv(D); no uniform domain constant is claimed.

For the exact mini-element result, D is additionally polygonal/polyhedral and exactly tiled by a conforming shape-regular affine simplicial mesh. The velocity uses continuous P1 functions with zero trace plus one scalar interior bubble per cell and vector component. Pressure is continuous P1 with zero mean. No curved mesh is asserted to tile a curved domain.

Controlling plan and conflict record

The binding current plan is the NA-34 section of research/plan-numerical-spatial-track.md, read together with the explicit owner direction and its referenced mixed-foundation completion. The canonical page metadata in research/plan-spec.json and the batch task control page IDs, order, and requires.

  • The older numerical-analysis design permits “mini or Taylor–Hood.” The current plan-spec description and repaired spatial design require the exact P1+bubble/P1 mini pair in dimensions two and three, so this batch proves MINI. Taylor–Hood appears only as the requested local B-page polynomial mode calculation.
  • The current plan points to the mixed-foundation completion for the continuous divergence-lift contract on every bounded connected Lipschitz domain. This controls over an earlier star-shaped-only proposal. The finite star-shaped cover and zero-mean tree decomposition are local proved items, not assumptions imported from Durán.
  • The older spatial prerequisite prose lists separate trace/Poincare pages. The canonical task and plan-spec require finite-element-meshes-and-interpolation and lax-milgram-and-weak-elliptic-solutions; this manifest preserves those exact page edges. The simplex face trace and reference Poincare facts come from the actual FE page; the bounded-domain H0 Poincare and Korn results are proved locally.

Mathematical and axiom interfaces

The H0 definition is the Sobolev closure, not a trace definition. The local Poincare proof uses a slab, zero extension, Fubini, and closure. Korn is proved from the strain identity by integration by parts on compactly supported smooth fields and H1 closure. The Bogovskii proof uses the real Fourier convention f^(ξ)=∫e−2πix⋅ξf(x) dx and Countable Choice for L2 density and Hilbert/Lax–Milgram completions. The Stokes existence branch uses Countable Choice through the published H1 Hilbert, L2 Hilbert, Riesz, and Lax–Milgram suppliers. The local Poincare and Korn arguments use Countable Choice only through the H0 closure supplier; the finite cover and overlap-tree constructions require no choice. No full Axiom of Choice is used. No proof uses the AC-dependent general Sobolev Banach-space theorem or the nodal point-evaluation interpolation branch.

The face-average interpolant is defined by facet dual moments and never evaluates an arbitrary H1 function at a vertex. Its boundary patches preserve zero trace; its H1-to-L2 and H2-to-H1 rates are proved from cell trace, patch Poincare, Bramble–Hilbert, finite patch overlap, and affine scaling. The mini Fortin map adds cell bubbles to restore each vector cell mean, which makes the pairing exact for continuous P1 pressure. The discrete weak divergence kernel is kept distinct from the pointwise divergence-free kernel.

Reading and source disposition

The two independent full treatments are Guermond's MATH 661 course notes and Boffi–Brezzi–Fortin's textbook. Durán's complete 17-page paper was also read for the star-shaped Bogovskii formula, truncated derivative/Fourier representation, elementary ray bound, and L2 derivative estimate. Its proof does not supply the finite Lipschitz cover, tree decomposition, or our local support/divergence verification; those are proved here. Exact locators, source fetch stamps, and each harvested-result disposition are in research/numerical-analysis-remaining-71-2026-10-17-batch-31.coverage.json.

The course-note chapters on other pairs, macroelements, Taylor–Hood stability, preconditioning, and unrelated mixed families are recorded with item-specific out-of-scope reasons. They do not enter the MINI proof. The existing periodic Navier–Stokes open-problem item remains contextual only and is not a logical supplier.

Items and dependency levels

In-run levels are computed from the batch manifest and direct same-batch item dependencies; external published suppliers do not increase an item level. Suppliers are authored before consumers in the resulting order.

Level 0

Zero mean of the divergence of a zero-boundary field, A bounded-domain Poincare inequality for H1-zero fields, The homogeneous Dirichlet strain identity, Finite star-shaped cover of a bounded Lipschitz domain, The Bogovskii integral on a star-shaped set, Hilbert inf-sup gives a bounded constraint right inverse, Boundary-preserving face-average P1 interpolation

Level 1

Stokes velocity and pressure spaces and weak forms, Zero-mean L2 decomposition over a finite cover, Support, integrability, and divergence identity for the Bogovskii kernel, Babuska-Brezzi solvability by kernel coercivity and inf-sup

Level 2

Truncated Fourier formula for derivatives of the Bogovskii operator, Fortin operator for a discrete constraint, Mixed Galerkin error bound retaining beta_h, The conforming P1-bubble/P1 mini element

Level 3

Elementary Fourier ray and derivative L2 bounds, Fortin operators transfer the continuous inf-sup bound, Local and global dimensions of the mini element, Weak and pointwise divergence-free discrete velocities

Level 4

Bounded Bogovskii right inverse on a star-shaped domain

Level 5

Divergence right inverse on bounded connected Lipschitz domains

Level 6

Continuous Stokes divergence inf-sup on a Lipschitz domain

Level 7

Uniform Fortin stability of the mini element, Weak Stokes solvability on bounded connected Lipschitz domains

Level 8

Mini-element H1 velocity and L2 pressure error estimate, Discrete Helmholtz projection and pressure Schur identities

Existing page prerequisites checked

The published FE supplier page finite-element-meshes-and-interpolation defines matching affine simplicial meshes, shape regularity, continuous/broken polynomial spaces, simplex trace/Gauss–Green, reference Poincare–Wirtinger, Bramble–Hilbert, and affine scaling. Its point-evaluation interpolation branch is not used. The published Lax–Milgram supplier proves the real coercive-form result under Countable Choice; the exact CC hypothesis is retained in the Hilbert Stokes branch. The actual item bodies were read before citation.

Draft status

These are new draft items and pages for Step 1. They have not been independently audited or accepted. The coverage and proof-contract records document authoring evidence and mechanical consistency only.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Zero mean of the divergence of a zero-boundary field

Statement

Assume Countable Choice. Let D⊂Rd be bounded and open. If v∈H01(D;Rd), then div⁡v∈L2(D) and ∫Ddiv⁡v=0.

Facts & Assumptions

Given: A bounded open D and a vector field in the H1-closure of Cc∞(D;Rd).

[A1]

Countable Choice permits selecting the H0 test-function approximation sequence used to pass the integral identity to the limit. (The Axiom of Countable Choice (ACω))

[F1]

H01(D) is the closure of compactly supported smooth functions in the W1,2 norm. (Zero-boundary Sobolev space as a norm closure)

[F2]

First weak derivatives of a W1,2 field lie in L2, and its W1,2 norm controls their L2 norms. (Integer-order Sobolev spaces and their norms)

[F3]

On a finite-measure set, ∣∫Dg∣≤∣D∣1/2∥g∥2. (Holder's inequality for integrals, including the endpoint cases)

[F4]

Fubini evaluates the integral of each compactly supported coordinate derivative by iterated one-dimensional integration. (Fubini's theorem for L^1 functions on a sigma-finite product)

Proof

1.1F2algebra

By [F2], each weak derivative ∂ivi is in L2, so their finite sum div⁡v is in L2(D).

1.2givenF1F4

For v∈Cc∞(D;Rd), each component derivative has integral zero by one-dimensional integration along its coordinate, hence ∫Ddiv⁡v=0.

2.1A1F1F2F3step 1.2∎

By [A1, F1], choose vn∈Cc∞(D;Rd) converging to v in H1. Then div⁡vn→div⁡v in L2 by [F2]. Applying [F3] to the difference shows their integrals converge; the zero integrals in step 1.2 therefore give ∫Ddiv⁡v=0.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A bounded-domain Poincare inequality for H1-zero fields

Statement

Assume Countable Choice. Let D⊂Rd be bounded and open, and suppose its first coordinate projection is contained in an interval (a,b) of width L=b−a. For every real v∈H01(D), ∥v∥L2(D)≤L∥∂1v∥L2(D)≤L∥∇v∥L2(D).

Facts & Assumptions

Given: A bounded open D⊂(a,b)×Rd−1 and v∈H01(D), where H01 is the H1-closure of Cc∞(D).

[A1]

Countable Choice permits selecting the approximating test functions from the H0 closure at each n. (The Axiom of Countable Choice (ACω))

[F1]

H01(D) is the closure of compactly supported smooth functions in the W1,2 norm. (Zero-boundary Sobolev space as a norm closure)

[F2]

The W1,2 norm includes the L2 norm and the L2 norms of all first weak derivatives. (Integer-order Sobolev spaces and their norms)

[F3]

Fubini permits iterated integration of the nonnegative integrand on the bounded coordinate slab. (Fubini's theorem for L^1 functions on a sigma-finite product)

[F4]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

Proof

1.1givenF1F4

First take v∈Cc∞(D) and extend it by zero to (a,b)×Rd−1. For every x′∈Rd−1 and x1∈(a,b), the fundamental theorem of calculus gives v(x1,x′)=∫ax1∂1v(t,x′) dt. Cauchy--Schwarz yields ∣v(x1,x′)∣2≤L∫ab∣∂1v(t,x′)∣2dt.

2.1F2F3step 1.1

Integrating the last inequality over x1 gives a second factor L; integrating over x′ and using the zero extension proves ∥v∥22≤L2∥∂1v∥22.

3.1A1F1F2F3algebrastep 2.1∎

For general v∈H01(D), [A1, F1] supplies a sequence vn∈Cc∞(D) converging to v in W1,2. The two norms in step 2.1 converge by [F2], so passing to the limit gives the same inequality for v. Finally ∣∂1v∣≤∣∇v∣ almost everywhere.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

The homogeneous Dirichlet strain identity

Statement

Assume Countable Choice. For bounded open D⊂Rd and v∈H01(D;Rd), put ε(v)=(∇v+∇vT)/2. Then 2∥ε(v)∥22=∥∇v∥22+∥div⁡v∥22, so ∥∇v∥2≤2∥ε(v)∥2.

Facts & Assumptions

Given: v∈H01(D;Rd), the closure of smooth compactly supported vector fields in H1.

[A1]

Countable Choice permits selecting the H0 test-function approximation sequence used to pass the smooth strain identity to the limit. (The Axiom of Countable Choice (ACω))

[F1]

H01(D) is the H1-closure of Cc∞(D). (Zero-boundary Sobolev space as a norm closure)

[F2]

The weak gradient has square-integrable entries and converges in L2 under H1 convergence. (Integer-order Sobolev spaces and their norms)

[F3]

For compactly supported smooth h on Rd, Fubini and the one-dimensional fundamental theorem give ∫∂jh=0 for each coordinate. (Fubini's theorem for L^1 functions on a sigma-finite product)

Proof

1.1givenalgebra

For v∈Cc∞(D;Rd), expanding the symmetric gradient gives 2∣ε(v)∣2=∣∇v∣2+∑i,j∂jvi∂ivj pointwise.

2.1F3algebrastep 1.1

For each i,j, apply [F3] to h=vi∂ivj, which is compactly supported and smooth. The product rule gives ∫D∂jvi∂ivj=−∫Dvi∂j∂ivj. Summing in j and using equality of mixed derivatives gives −∑i∫Dvi∂i(div⁡v). Apply [F3] again to h=vidiv⁡v and sum in i to obtain ∑i∫D(∂ivi)div⁡v=∥div⁡v∥22.

3.1A1F1F2algebrastep 2.1∎

By [A1, F1], approximate any v∈H01 by such smooth fields in H1. All three squared norms in the identity converge by [F2] and continuity of the finite sums, so the identity passes to the limit. Its nonnegative divergence term gives the displayed inequality.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Finite star-shaped cover of a bounded Lipschitz domain

Statement

Let D⊂Rd be nonempty, bounded, connected and Lipschitz. Here Lipschitz means that its compact boundary has a finite atlas of rigidly moved cylinders in which D is the strict epigraph xd>φ(x′) of a Lipschitz function. There are finitely many bounded open sets Ui⊂D covering D, each star-shaped with respect to a closed ball B‾i⊂Ui, and the graph joining i,j when Ui∩Uj≠∅ is connected. There are measurable sets Ei⊂Ui partitioning D and, for every edge of a spanning tree of this graph, a function ψij∈Cc∞(Ui∩Uj) with integral one.

Facts & Assumptions

Given: A nonempty bounded connected open set with the finite graph-chart definition of Lipschitz boundary stated above.

Proof

1.1givenconstruct

Fix one graph chart, with Lipschitz constant L, and a smaller horizontal ball C whose closure lies in its base. Shrink its radius r until the chart's upper height Hchart satisfies Hchart−sup⁡Cφ>4(L+1)r. Choose c with sup⁡Cφ+3(L+1)r<c<Hchart−(L+1)r and 0<ρ<r. Then the closed ball B‾((x0′,c),ρ) stays in the cylinder. Set U={(x′,xd):x′∈C, φ(x′)<xd<H} with c+ρ<H<Hchart.

2.1givenalgebrastep 1.1

If x∈U, z∈B‾((x0′,c),ρ), and w=(1−t)x+tz with 0<t<1, then z′∈C and convexity keeps w′∈C. Since ∣z′−x′∣≤2r, Lipschitz continuity gives φ(w′)≤φ(x′)+2Lrt. Also c−ρ>sup⁡Cφ+2Lr, so wd>(1−t)φ(x′)+t(c−ρ)>φ(x′)+2Lrt≥φ(w′), and wd<H by convexity of the vertical interval. Thus w∈U; the endpoints belong to U as well, proving star-shapedness with respect to the closed ball.

3.1F1givenstep 1.1step 2.1

The boundary ∂D is closed and bounded, hence compact by [F1]. At any p∈∂D, a chart containing p writes the boundary locally as xd=φ(x′), with pd=φ(p′) strictly below the chart's upper height. Shrinking a horizontal ball C about p′ makes its closure lie in the chart base and makes the height-gap inequality of step 1.1 hold; the choices there then produce a patch U={(x′,xd):x′∈C, φ(x′)<xd<H} star-shaped with respect to its selected closed ball by step 2.1. Since p lies inside the chart and inside the open set with x′∈C and xd<H, there is an ambient open ball N about p whose closure lies in that chart and open set. For every x∈N∩D, the chart description gives xd>φ(x′), so x∈U and N∩D⊂U. Let N be the family of all pairs (N,U) obtained this way from the finite chart atlas and admissible local choices; the preceding existence argument shows that its first components cover ∂D, without choosing a patch simultaneously for every boundary point. Compactness supplies a finite subfamily (Ni,Ui), 1≤i≤m, whose neighborhoods cover ∂D. Their union is an open neighborhood of the compact boundary, so some relative collar {x∈D:dist⁡(x,∂D)<δ} lies in that union and hence in ⋃i=1mUi. The remainder Kδ={x∈D‾:dist⁡(x,∂D)≥δ} is closed and bounded, hence compact by [F1], and is contained in D because every point of D‾∖D lies on ∂D. Finitely many balls with closures in D cover Kδ; each ball is convex and star-shaped with respect to a smaller concentric ball. Together these sets form a finite star-shaped cover of D.

4.1givenstep 3.1

The intersection graph is connected: if it split into two nonempty graph components, the unions of the corresponding open sets would be disjoint nonempty relatively open subsets whose union is D, contradicting connectedness. Order the cover and set Ei=Ui∖⋃j<iUj. Then the Ei are disjoint measurable sets with Ei⊂Ui and union D. A finite connected graph has a spanning tree.

5.1constructstep 4.1∎

Each tree-edge intersection is nonempty and open, so it contains a ball. Choose a nonnegative smooth bump supported in that ball. Its integral is strictly positive; dividing by that integral gives ψij∈Cc∞(Ui∩Uj) with integral one.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

The Bogovskii integral on a star-shaped set

Definition

Let U⊂Rd be bounded and open, star-shaped with respect to a closed ball B‾(a,ρ)⊂U. Choose ω∈Cc∞(B(a,ρ)) with ∫ω=1. For f∈Cc∞(U), initially define the vector-valued integral (BUf)(x)=∫Uf(y)(x−y)∫01ω ⁣(y+x−yt)dttd+1 dy. The inner integral is first interpreted for t>0; its locally integrable kernel, support and divergence identity are proved in the following lemma. All data and fields here are real. This definition makes no finite-quadrature or numerical-algorithm claim.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Hilbert inf-sup gives a bounded constraint right inverse

Statement

Assume Countable Choice. Let V,Q be real Hilbert spaces and let b:V×Q→R be a bounded bilinear form. Define B:V→Q by (Bv,q)Q=b(v,q). If Q={0}, the unique zero map is a bounded right inverse. If Q≠{0}, assume V≠{0} and define β:=inf⁡q≠0sup⁡v≠0b(v,q)∥v∥V∥q∥Q. When β>0, B is onto and has a bounded linear right inverse R:Q→V with BR=IQ and ∥R∥≤β−1.

Facts & Assumptions

Given: Real Hilbert spaces V,Q, a bounded bilinear form, and β>0 as displayed.

[A1]

Countable Choice is used by the Lax–Milgram theorem applied to the coercive form (B* q,B* r). (The Axiom of Countable Choice (ACω))

[F1]

In a real Hilbert space every bounded linear functional is inner product with a unique vector of the same norm. (Riesz representation for Hilbert spaces)

[F2]

Lax–Milgram under Countable Choice solves every bounded coercive real bilinear form on a Hilbert space, with solution norm at most the functional norm divided by the coercivity constant. (The Lax--Milgram theorem, The Axiom of Countable Choice (ACω))

[F3]

Riesz representation applied to v↦b(v,q) defines B∗q by (Bv,q)Q=(v,B∗q)V and gives ∥B∗q∥V=sup⁡v≠0∣b(v,q)∣/∥v∥V. (Riesz representation for Hilbert spaces, Hilbert space)

Proof

1.1givenF3algebra

If Q={0}, the unique map R:Q→V is a bounded right inverse and the claim is complete. Otherwise, since the scalar field is real, replacing v by −v changes the sign of b(v,q); hence the displayed supremum equals the supremum of ∣b(v,q)∣/(∥v∥∥q∥). By [F3], ∥B∗q∥V≥β∥q∥Q.

2.1A1F1F2F3step 1.1

On Q, define c(q,r):=(B∗q,B∗r)V. It is bounded by ∥B∥2∥q∥∥r∥ and coercive with constant β2 by step 1.1. For arbitrary g∈Q, the functional r↦(g,r)Q is bounded, so [F2] gives a unique q∈Q with (B∗q,B∗r)V=(g,r)Q for all r. The adjoint identity implies BB∗q=g.

3.1A1F2F3algebrastep 1.1step 2.1∎

Set Rg:=B∗q, with q from step 2.1. Uniqueness in Lax–Milgram makes this assignment linear, and the adjoint identity in [F3] gives BRg=g. Also ∥Rg∥2=(g,q)≤∥g∥∥q∥ while step 2.1 gives β∥q∥≤∥Rg∥. If Rg=0 the desired bound is immediate; otherwise combine and divide to obtain ∥Rg∥≤β−1∥g∥. Thus R is the claimed bounded right inverse.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Boundary-preserving face-average P1 interpolation

Statement

Assume Countable Choice. Let D be a bounded connected Lipschitz polygonal/polyhedral domain with a conforming shape-regular simplicial mesh, d=2 or 3, mesh size h=max⁡KhK. For each interior mesh vertex z, choose any incident facet Fz; for each boundary vertex choose an incident boundary facet. Let ψz,Fz∈P1(Fz) be the unique dual affine function satisfying ∫Fzψz,Fzφy=δzy for the facet's P1 nodal basis. Define Ihv=∑z(∫Fzψz,Fztr⁡v)φz. This is a linear face-average interpolant H1(D)→Sh, where Sh is the continuous piecewise P1 space. It maps H01(D) into Sh∩H01(D), is uniformly H1-seminorm stable, and for each element patch ωK has ∥v−Ihv∥L2(K)≤ChK∥∇v∥L2(ωK). If v∈H2(D), then ∥∇(v−Ihv)∥L2(K)≤ChK∥D2v∥L2(ωK). Constants depend only on dimension, the fixed domain, and the mesh shape-regularity bound. No point evaluation is used.

Facts & Assumptions

Given: A conforming affine mesh of a bounded connected Lipschitz polygon/polyhedron, its shape-regularity bound, and the Sobolev spaces on its cells.

[A1]

Countable Choice is inherited through the H0 closure interface used to identify zero boundary traces. (The Axiom of Countable Choice (ACω))

[F1]

Matching affine simplices and conforming P1 spaces have shared facet traces; continuous piecewise polynomials with zero boundary facet trace belong to H01(D). (Matching affine simplicial meshes and shape regularity, Broken and conforming polynomial spaces on a matching mesh, Zero-boundary conforming polynomials lie in W0 1p and satisfy Poincare)

[F2]

Shape regularity gives uniform affine scaling between each simplex and the reference simplex. (Shape regularity controls affine simplex geometry)

[F3]

The reference simplex has a bounded trace map and a Poincare–Wirtinger inequality; Bramble–Hilbert gives affine approximation with the stated derivative-order scaling. (Sobolev face traces and Gauss–Green on a reference simplex, Poincare–Wirtinger inequality on a reference simplex, Bramble–Hilbert approximation on a reference simplex)

[F4]

H01(D) is the closure of compactly supported smooth functions, so its trace on every boundary facet is zero by continuity of the face trace. (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms, Sobolev face traces and Gauss–Green on a reference simplex)

[F5]

On a finite-dimensional real normed space, all norms are equivalent. (Finite dimensional norm equivalence over a complete valued field)

[F6]

Affine pullback transports H1 norms and face traces between a simplex and its reference simplex; the trace bound then follows from [F3]. (Affine pullback of weak derivatives and Sobolev seminorms, Sobolev face traces and Gauss–Green on a reference simplex)

Proof

1.1A1F1F2F3F4F6construct

On each reference facet, the P1 facet mass matrix is positive definite because ∫Fp2>0 for every nonzero affine polynomial. Its inverse defines the dual functions ψ^z; affine scaling in [F2] gives ∥ψz,F∥L2(F)≤ChF−(d−1)/2. The trace map in [F3] therefore makes every coefficient of Ihv a bounded linear functional on the H1 trace of an adjacent cell. Shared coefficients define a continuous P1 function. If v∈H01, [F4] makes all selected boundary-facet averages zero, so Ihv has zero boundary nodal values and belongs to H01 by [F1].

1.2F1F2construct

The coefficients reproduce every global affine polynomial: on the selected facet its trace is P1 and the dual moment equals its value at z. Shape regularity bounds the number of cells at each vertex: each simplex's vertex solid angle has a positive lower bound, while the total angle is at most 2π in dimension two or 4π in dimension three. The vertex-star graph is connected as follows. Choose a small chart cylinder meeting only cells incident to the vertex. At an interior vertex, a small ball lies in D. At a boundary vertex, the Lipschitz chart represents this local part of D as the region above a Lipschitz graph; within a smaller base ball, vertical paths to a common height above the graph and a horizontal segment at that height connect any two points. Hence this local part is path connected. Join interior points of any two incident cells by a polygonal path there and perturb its finitely many segments to avoid the finite mesh skeleton of codimension at least two. The path then crosses only facet interiors and gives a facet-adjacency chain among cells incident to the vertex. Adjacent cell diameters are comparable by [F2]. Thus vertex-patch diameters and overlap are uniformly bounded.

2.1A1F2F3step 1.2

Let ωK be the union of cells meeting a vertex of K. Shape regularity bounds the number of cells at each vertex: each has a solid angle bounded below by the reference-cell shape bound, whereas the total angle is at most 2π in dimension two and 4π in dimension three. Each vertex-star graph is connected by step 1.2, and all stars of vertices of K contain K, so their union is connected; neighboring cell diameters are comparable by [F2]. For adjacent cells T,T′ with common facet F, let mT,mT′ be their volume means. Cell Poincare and trace in [F3] give ∥w−mT∥L2(F)≤ChK1/2∥∇w∥L2(T) and the analogous bound on T′. Since ∣F∣1/2≍hK(d−1)/2, it follows that ∣mT−mT′∣≤ChK(2−d)/2∥∇w∥L2(T∪T′). A path of bounded length through the patch graph then compares all cell means to one fixed cK and proves ∥w−cK∥L2(ωK)≤ChK∥∇w∥L2(ωK). Applying the scaled trace estimate of [F3] and ∥ψz,Fz∥L2(Fz)≤ChK−(d−1)/2 gives ∣∫Fzψz,Fz(w−cK)∣≤ChK1−d/2∥∇w∥L2(ωK) for every vertex of K.

2.2A1F2F3F5algebrastep 1.2

Suppose v∈H2(D). On each cell T⊂ωK, Bramble--Hilbert and affine scaling in [F2, F3] give pT∈P1(T) with ∥v−pT∥L2(T)≤ChT2∥D2v∥L2(T) and ∥∇(v−pT)∥L2(T)≤ChT∥D2v∥L2(T). For adjacent cells T,T′ sharing a facet F, the scaled trace estimate applied to v−pT and its first derivatives gives ∥pT−pT′∥L2(F)≤ChK3/2∥D2v∥L2(T∪T′) and ∣F∣1/2∣∇pT−∇pT′∣≤ChK1/2∥D2v∥L2(T∪T′). The second bound controls the full gradient difference, and the first then controls its constant part. Normalize both cell diameters by hK. Shape regularity and comparability across the shared facet bound the normalized affine maps and their inverses; there are only finitely many vertex-orderings. Their closure is a compact set of nondegenerate reference-pair configurations. Finite-dimensional norm equivalence [F5] varies continuously on this compact set, so its constants are uniform; affine scaling then gives ∥pT−pT′∥H1(T)≤ChK∥D2v∥L2(T∪T′). A bounded path through the connected cell-star graph lets one fixed cell polynomial p∈P1 satisfy ∥v−p∥L2(ωK)≤ChK2∥D2v∥L2(ωK) and ∥∇(v−p)∥L2(ωK)≤ChK∥D2v∥L2(ωK). This uses only local approximation and does not impose boundary values on v.

3.1A1F2algebrastep 2.1

The reference P1 basis satisfies ∥φz∥L2(K)≤ChKd/2 and ∥∇φz∥L2(K)≤ChKd/2−1. Since Ih reproduces constants, step 2.1 bounds ∥Ihw−cK∥L2(K)≤ChK∥∇w∥L2(ωK) and ∥∇Ihw∥L2(K)≤C∥∇w∥L2(ωK). Together with the patch Poincare bound for w−cK, these prove the L2 approximation and local H1 stability. Summing and using bounded overlap proves global H1-seminorm stability.

4.1A1F1algebrastep 3.1step 2.2∎

Facet duality reproduces every affine polynomial at every vertex, so Ihp=p on K. Apply the local H1 stability from step 3.1 to v−p and use step 2.2 to get ∥∇(v−Ihv)∥L2(K)≤ChK∥D2v∥L2(ωK). Summing over cells and using the bounded patch overlap from step 3.1 proves the global H1 approximation estimate.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Stokes velocity and pressure spaces and weak forms

Definition

Assume Countable Choice. Let d∈{2,3} and let D⊂Rd be a nonempty bounded connected Lipschitz open set, where Lipschitz means that ∂D has a finite rigidly moved graph-chart atlas. Here H01(D) is the H1-closure of Cc∞(D) (Zero-boundary Sobolev space as a norm closure, Integer-order Sobolev spaces and their norms). Since D has finite measure, Holder makes every L2 class integrable, so the zero-mean condition is well-defined (Holder's inequality for integrals, including the endpoint cases). Define V:=H01(D;Rd),Q:=L02(D):={q∈L2(D;R):∫Dq=0}. Equip V with ∥v∥V:=∥∇v∥L2(D) and Q with ∥q∥Q:=∥q∥L2(D). The local Poincare inequality A bounded-domain Poincare inequality for H1-zero fields makes the first a norm equivalent to the usual H1 norm. For v∈V, define ε(v):=(∇v+∇vT)/2 and b(v,q):=−∫Dq div⁡v. For viscosity ν>0, set aν(u,v):=2ν∫Dε(u):ε(v), where X:Y=∑i,jXijYij. The continuous domain class is all bounded connected Lipschitz D; the exact mini-element mesh class is the narrower polygonal/polyhedral conforming affine simplicial class.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Zero-mean L2 decomposition over a finite cover

Statement

Let D=⋃i=1NUi be the finite connected star-shaped cover from Finite star-shaped cover of a bounded Lipschitz domain, with measurable partition Ei and normalized tree-edge bumps ψe. For every q∈L02(D) there are qi∈L02(Ui), depending linearly on q, whose zero extensions satisfy q=∑iqi and ∥qi∥L2(Ui)≤Di∥q∥L2(D),Di=1+∣D∣1/2∑e incident to i∥ψe∥2.

Facts & Assumptions

Given: The finite cover, its spanning tree and unit-integral edge bumps; q∈L2(D) with zero integral.

[F1]

The cover has measurable disjoint Ei⊂Ui partitioning D and a connected intersection graph with a spanning tree; each edge bump is supported in the overlap and has integral one. (Finite star-shaped cover of a bounded Lipschitz domain)

[F2]

∥q∥L1(D)≤∣D∣1/2∥q∥L2(D). (Holder's inequality for integrals, including the endpoint cases)

Proof

1.1givenF1construct

Root the finite spanning tree. Set gi=1Eiq, mi=∫Dgi, and let Mi be the sum of mj over the subtree rooted at i. For a nonroot vertex i with parent-edge bump ψi, set qi=gi+∑j child of iMjψj−Miψi; at the root omit the final term. Every summand is supported in Ui by [F1].

2.1givenalgebrastep 1.1

Since each bump integrates to one, a nonroot patch has integral mi+∑jMj−Mi=0 by the subtree identity Mi=mi+∑jMj. At the root, the integral is mroot+∑jMj=Mroot=∑imi=∫Dq=0.

2.2F1algebrastep 1.1

In the sum of all qi, every tree-edge term −Miψi at a child cancels the +Miψi term at its parent, leaving ∑igi=q almost everywhere because the Ei partition D. All coefficients and operations are linear in q.

3.1F2algebrastep 1.1∎

For every subtree, ∣Mi∣=∣∫∪j in subtree iEjq∣≤∥q∥1≤∣D∣1/2∥q∥2 by [F2], and ∥gi∥2≤∥q∥2. The triangle inequality applied to the definition in step 1.1 gives ∥qi∥2≤∥q∥2+∣D∣1/2∥q∥2∑e incident to i∥ψe∥2=Di∥q∥2.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Support, integrability, and divergence identity for the Bogovskii kernel

Statement

Assume Countable Choice. Let U,ω and BU be as in The Bogovskii integral on a star-shaped set. For f∈Cc∞(U), write G(x,y)=(x−y)∫01ω(y+(x−y)/t)t−d−1dt. Then BUf(x)=∫UG(x,y)f(y)dy defines an L2 field supported in a compact subset of U. Writing K(x,y)=∣G(x,y)∣ and (KUg)(x):=∫UK(x,y)g(y)dy, the majorant KU∣f∣ is also in L2(U), and KU has a finite L2 bound. Moreover, div⁡BUf=f−ω∫Uf in distributions on Rd. In particular, mean-zero input has divergence f.

Facts & Assumptions

Given: Countable Choice; bounded open U star-shaped with respect to the support ball of ω∈Cc∞, ∫ω=1, and f∈Cc∞(U).

[A1]

Countable Choice is required by the affine Lebesgue change-of-variables supplier [F2]. (The Axiom of Countable Choice (ACω))

[F1]

The Bogovskii vector kernel is G(x,y)=(x−y)∫01ω(y+(x−y)/t)t−d−1dt. (The Bogovskii integral on a star-shaped set)

[F2]

Translation preserves Lebesgue measurability and measure, and the linear map z↦tz scales measure by td for t>0. Composing this dilation with translation by (1−t)y gives x=y+t(z−y) and dx=tddz. The integral substitution follows first for indicator functions, then for nonnegative simple functions and their increasing limits, and for integrable signed functions by positive and negative parts. (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, A linear map T of Rn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E) when T is invertible and T[E] Lebesgue null when it is not)

[F3]

Tonelli and Fubini allow the nonnegative/absolutely integrable changes of order in the kernel and test-function calculations. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Fubini's theorem for L^1 functions on a sigma-finite product)

[F4]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

[F6]

Distributional divergence is defined by ⟨div⁡u,η⟩=−∫u⋅∇η. (Distributional derivative)

Proof

1.1F1givenF5F7

If the integrand in [F1] is nonzero, z=y+(x−y)/t∈supp⁡ω, so x=(1−t)y+tz. Star-shapedness puts every such x in U. The sets supp⁡f, supp⁡ω and [0,1] are closed and bounded, so their product is closed and bounded in Euclidean space and compact by [F7]. Its image under the continuous map (y,z,t)↦(1−t)y+tz is compact by [F5]. Star-shapedness puts the image inside U.

1.2F1givenalgebra

Assign G(x,x)=0 on the null diagonal. If the kernel integrand is nonzero and r=∣x−y∣>0, then r/t=∣z−y∣≤R:=diam⁡(U). Thus t≥r/R and ∣G(x,y)∣≤r∥ω∥∞∫r/R1t−d−1dt≤Cr1−d. For any fixed x, decompose 0<∣x−y∣≤R into annuli 2−n−1R<∣x−y∣≤2−nR; on the nth annulus, r1−d≤(2−n−1R)1−d and its measure is at most (21−nR)d. The annular integral is therefore at most CR2−n, whose sum is finite. The same estimate with x,y interchanged proves both sup⁡x∫U∣G(x,y)∣dy<∞ and sup⁡y∫U∣G(x,y)∣dx<∞.

2.1F1F3F4step 1.1step 1.2

For the nonnegative kernel K=∣G∣, Cauchy–Schwarz gives ∣∫K(x,y)f(y)dy∣2≤(∫K(x,y)dy)(∫K(x,y)∣f(y)∣2dy). Integrating in x and using the two finite bounds in step 1.2 gives ∥KU∣f∣∥22≤MxMy∥f∥22, where Mx=sup⁡x∫UK(x,y)dy and My=sup⁡y∫UK(x,y)dx. The vector field is componentwise dominated by this majorant, so it is in L2 and has the support in step 1.1.

2.2A1F1F2F3step 1.2

Let η∈Cc∞(Rd). Using [F2], Fubini (justified by step 1.2, the compact support of f, and boundedness of ∇η), and x=y+t(z−y) gives ∫RdG(x,y)⋅∇η(x)dx=∫ω(z)(z−y)⋅∫01∇η(y+t(z−y))dt dz=∫ω(z)(η(z)−η(y))dz. The last equality is the one-variable fundamental theorem along the segment.

3.1givenF6step 2.2∎

By the definition of distributional divergence, step 2.2 says div⁡xG(⋅,y)=δy−ω; integrating against f(y) yields div⁡BUf=f−ω∫Uf. If ∫f=0, the second term vanishes.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Babuska-Brezzi solvability by kernel coercivity and inf-sup

Statement

Assume Countable Choice. Let V,Q be real Hilbert spaces, and let a:V×V→R and b:V×Q→R be bounded bilinear forms. If Q={0}, the theorem reduces to Lax–Milgram on V with Z=V and p=0. Otherwise, if Q≠{0}, assume V≠{0}. Define B:V→Q by (Bv,q)Q=b(v,q) and set β:=inf⁡q≠0sup⁡v≠0b(v,q)/(∥v∥V∥q∥Q)>0. Suppose Z:=ker⁡B is closed and a(z,z)≥α∥z∥V2 on Z, with α>0. For every F∈V′ and G∈Q′, there is a unique (u,p)∈V×Q satisfying a(u,v)+b(v,p)=F(v),b(u,q)=G(q), for all v∈V,q∈Q. If Ma=∥a∥, then one stability bound is ∥u∥V≤β−1∥G∥+α−1(∥F∥+Maβ−1∥G∥),∥p∥Q≤β−1(∥F∥+Ma∥u∥V).

Facts & Assumptions

Given: Countable Choice, real Hilbert spaces, bounded a,b, β>0, and coercivity of a on Z.

[A1]

Countable Choice is required by the Lax–Milgram step on the closed constraint kernel. (The Axiom of Countable Choice (ACω))

[F1]

The inf-sup bound yields a bounded linear right inverse R:Q→V with BR=I and ∥R∥≤β−1. (Hilbert inf-sup gives a bounded constraint right inverse)

[F2]

Lax–Milgram gives a unique solution on the Hilbert subspace Z for a bounded coercive form, and its norm is controlled by the right-hand side divided by α. (The Lax--Milgram theorem, The Axiom of Countable Choice (ACω))

[F3]

Riesz representation identifies bounded functionals on V and Q with inner products against unique vectors. (Riesz representation for Hilbert spaces)

Proof

1.1A1F1F3

If Q={0}, then G=0, Z=V, and Lax–Milgram on V gives the unique solution (u,0) with ∥u∥≤α−1∥F∥; so suppose Q≠{0}. By [F3], write G(q)=(g,q)Q for a unique g∈Q with ∥g∥=∥G∥. Let ug=Rg using [F1]. Then Bug=g, ∥ug∥≤β−1∥G∥, and the constraint equation will hold for u=ug+z with any z∈Z.

2.1A1F2step 1.1

On the closed Hilbert subspace Z, solve a(z,w)=F(w)−a(ug,w) for all w∈Z by [F2]. The restricted form is bounded and coercive by hypothesis, and the functional norm is at most ∥F∥+Ma∥ug∥. Therefore ∥z∥≤α−1(∥F∥+Maβ−1∥G∥) and u=ug+z has the stated bound. The equation a(u,w)=F(w) holds for every w∈Z.

3.1A1F1givenstep 1.1step 2.1

Define ℓ(v)=F(v)−a(u,v). It vanishes on Z by step 2.1. Set Λ(q)=ℓ(Rq). This is bounded by ∥R∥∥ℓ∥∥q∥, and for every v, v−RBv∈Z, so ℓ(v)=ℓ(RBv)=Λ(Bv).

4.1A1F1F3step 3.1step 1.1

By [F3], there is a unique p∈Q with Λ(q)=(q,p)Q. Hence ℓ(v)=(Bv,p)Q=b(v,p), giving the first mixed equation; step 1.1 gives the second. Also ∥p∥≤β−1∥ℓ∥≤β−1(∥F∥+Ma∥u∥).

5.1A1F1givenF3∎

For uniqueness, subtract two solutions. Their velocity difference w lies in Z; testing the first difference equation with w gives a(w,w)=0, so coercivity gives w=0. The pressure difference then pairs to zero with every Bv; since [F1] makes B onto, it pairs to zero with all Q, hence is zero.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Truncated Fourier formula for derivatives of the Bogovskii operator

Statement

Assume Countable Choice. Let U and ω be as in The Bogovskii integral on a star-shaped set. For ϕ∈Cc∞(Rd) and f∈Cc∞(Rd), use g^(ξ)=∫e−2πix⋅ξg(x)dx and define Tϕ,jϵf(x)=∫f(y)∫ϵ1∂xj[ϕ(y+x−yt)]dttddy. As ϵ↓0 these converge in distributions to Tϕ,jf. The part with 0<t<1/2 satisfies Tϕ,j<f^(ξ)=2πiξj∫01/2ϕ^(tξ)f^((1−t)ξ)dt. When f∈Cc∞(U), the Bogovskii formula satisfies ∂j(BUf)k=Txkω,jf−Tω,j(ykf) distributionally.

Facts & Assumptions

Given: Countable Choice; bounded star-shaped U, the normalized bump ω of The Bogovskii integral on a star-shaped set, smooth compactly supported real f,ϕ, and the displayed Fourier convention.

[A1]

Countable Choice is required by the affine Lebesgue change-of-variables supplier used in [F2] and the Fourier/Plancherel suppliers in [F1]. (The Axiom of Countable Choice (ACω))

[F1]

Plancherel and Fourier transformation agree with the integral transform on smooth integrable functions, and weak differentiation transforms to multiplication by 2πiξj. (Fourier transform agrees with l one and plancherel transforms, Distributional derivatives are polynomial Fourier multipliers)

[F2]

In the Bogovskii formula x−y=t(z−y). For t>0, dilation z↦tz scales Lebesgue measure by td and translation by (1−t)y preserves it, so their composition x=y+t(z−y) has dx=tddz. The integral identity follows for indicators, then by simple-function approximation and positive and negative parts for integrable functions. (The Bogovskii integral on a star-shaped set, A linear map T of Rn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E) when T is invertible and T[E] Lebesgue null when it is not, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation)

[F3]

Fubini applies to the compactly supported truncated integral and its Fourier transform. (Fubini's theorem for L^1 functions on a sigma-finite product)

[F4]

The kernel proof bounds its absolute integrand after integration in t by C∣x−y∣1−d for 0<∣x−y∣≤diam⁡(U); this radial bound is integrable uniformly in each variable. The Bogovskii field and all its truncations have support in the compact set of segments joining supp⁡f to supp⁡ω. (Support, integrability, and divergence identity for the Bogovskii kernel)

[F5]

An integrable absolute majorant permits passing an almost-everywhere limit through the integral. (Dominated convergence)

Proof

1.1A1givenF2F3algebra

For 0<δ<ϵ<1/2, pair Tϕ,jδf−Tϕ,jϵf with a Schwartz test function ζ. This difference integrates only over δ≤t≤ϵ, where the change of variables x=y+t(z−y) and Fubini are absolutely justified. The inner pairing contains t−1∂jϕ(z)ζ(y+t(z−y)). Integration by parts in z changes its integral to −∫ϕ(z)∂jζ(y+t(z−y))dz; the factor t−1 has canceled. Its absolute value is at most ∥ϕ∥1∥∇ζ∥∞, uniformly in y,t. Thus the paired difference is bounded by (ϵ−δ)∥f∥1∥ϕ∥1∥∇ζ∥∞, so Tϕ,jϵf has a distributional limit as ϵ↓0.

2.1A1F1F3F2F5step 1.1

For ϵ<t<1/2, Fourier transform in x and use x=y+t(z−y), whose Jacobian is td. The factor t−d cancels the Jacobian and ∂xjϕ(z)=t−1∂zjϕ(z). Integration by parts in z gives ∂jϕ^(tξ)=2πitξjϕ^(tξ) by [F1]; the remaining y integral is f^((1−t)ξ). Hence the truncated transform is 2πiξj∫ϵ1/2ϕ^(tξ)f^((1−t)ξ)dt. The integrals converge pointwise in ξ, and their absolute values are bounded uniformly in ϵ by π∣ξj∣∥ϕ∥1∥f∥1. Against any Schwartz test function this is an integrable majorant, so [F5] gives convergence in S′. Step 1.1 gives convergence in S′ as well, since its bound uses the Schwartz seminorm ∥∇ζ∥∞. Fourier transformation therefore identifies the displayed integral with the same limit.

3.1A1F2F3F4F5step 1.1∎

For ϵ>0, let BUϵf be the defining integral truncated to t≥ϵ. For fixed x≠y, the kernel integrand is zero for all sufficiently small t, because nonzero values require ∣x−y∣/t=∣z−y∣≤R:=diam⁡(U). By [F4], the absolute t integral is bounded by C∣x−y∣1−d10<∣x−y∣≤R. Its product with ∣f(y)∣ is integrable in (x,y) by the annular bound and Fubini. Applying [F5] to this absolute majorant proves convergence of the truncated fields to BUf in L1(Rd) and hence in distributions; no sign condition on ω is required. Rewrite the k-th component using xk−yk=t(zk−yk) from [F2] as the integral of [zkω(z)−ykω(z)]f(y) against dt/td. Differentiating the truncated fields in xj, then using step 1.1 for the distributional limit, gives Txkω,jf−Tω,j(ykf); the minus sign comes from the second term.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Fortin operator for a discrete constraint

Definition

Let V,Q,b be as in Stokes velocity and pressure spaces and weak forms, and let Vh⊂V, Qh⊂Q be finite-dimensional subspaces. A Fortin operator is a bounded linear map Πh:V→Vh for which there is a constant CF such that ∥Πhv∥V≤CF∥v∥V(v∈V),b(Πhv,qh)=b(v,qh)(v∈V, qh∈Qh). The second condition is exact constraint preservation. It does not assert div⁡Πhv=div⁡v pointwise.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Mixed Galerkin error bound retaining beta_h

Statement

Assume Countable Choice. Let V,Q be real Hilbert spaces and let F∈V′, G∈Q′ be bounded data. Suppose bounded bilinear forms a,b and (u,p)∈V×Q satisfy a(u,v)+b(v,p)=F(v),b(u,q)=G(q) for all v∈V,q∈Q. Let Vh⊂V, Qh⊂Q be finite-dimensional conforming subspaces, with Vh≠{0}, Qh≠{0}, and discrete inf-sup constant βh:=inf⁡qh≠0sup⁡vh≠0b(vh,qh)/(∥vh∥V∥qh∥Q)>0. Assume the discrete kernel coercivity a(vh,vh)≥α∗∥vh∥V2 for every vh∈Zh:={wh:b(wh,qh)=0 for all qh∈Qh}, where α∗>0. Put Ma=∥a∥, Mb=∥b∥. The Galerkin system with the same F,G has a unique solution (uh,ph). For any wh∈Vh, rh∈Qh, define δu=∥u−wh∥V, δp=∥p−rh∥Q, Rh=βh−1Mbδu, and Uh=δu+α∗−1{Ma(δu+Rh)+Mbδp}+Rh, Ph=δp+βh−1{Ma(Uh+δu)+Mbδp}. Then ∥u−uh∥V≤Uh and ∥p−ph∥Q≤Ph. In particular the pressure bound retains βh−1 explicitly; taking infima over wh,rh gives the corresponding best-approximation bound.

Facts & Assumptions

Given: A continuous solution, nonzero conforming finite-dimensional subspaces Vh,Qh, positive discrete inf-sup βh, boundedness constants Ma,Mb, and coercivity on Zh with constant α∗. The discrete equations use the same F,G.

[A1]

Countable Choice is used through the Babuška–Brezzi/Lax–Milgram solvability theorem for the discrete pair. (The Axiom of Countable Choice (ACω))

[F1]

On any real Hilbert pair, coercivity on the constraint kernel and a positive inf-sup constant give unique mixed solvability. (Babuska-Brezzi solvability by kernel coercivity and inf-sup)

[F2]

The discrete inf-sup bound gives a right inverse of the discrete constraint operator with norm at most βh−1. (Hilbert inf-sup gives a bounded constraint right inverse)

Proof

1.1A1F1given

Apply [F1] to Vh,Qh with the assumed Zh-coercivity and discrete inf-sup constant βh. Since the restricted forms are bounded, the Galerkin system with the same F,G has a unique solution (uh,ph).

2.1A1givenstep 1.1

Fix wh,rh and set θu=wh−uh, θp=rh−ph, ηu=u−wh, ηp=p−rh. Subtracting the discrete equations from the continuous equations gives a(θu,vh)+b(vh,θp)=−a(ηu,vh)−b(vh,ηp) and b(θu,qh)=−b(ηu,qh).

3.1A1F2step 2.1

The second identity implies ∥Bhθu∥Q≤Mbδu. By [F2] choose ru∈Vh with Bhru=Bhθu and ∥ru∥V≤Rh. Then zh=θu−ru∈Zh.

4.1givenstep 2.1step 3.1

Test the first identity of step 2.1 with zh. Since b(zh,θp)=0, a(zh,zh)=−a(ηu,zh)−b(zh,ηp)−a(ru,zh). Discrete-kernel coercivity and boundedness imply, when zh≠0, ∥zh∥≤α∗−1{Ma(δu+Rh)+Mbδp}; if zh=0 this bound is immediate. Therefore ∥u−uh∥≤∥ηu∥+∥zh∥+∥ru∥≤Uh.

5.1A1F2algebrastep 2.1step 4.1∎

Since θu+ηu=u−uh, the first identity of step 2.1 gives b(vh,θp)=−a(u−uh,vh)−b(vh,ηp). Therefore ∣b(vh,θp)∣≤[MaUh+Mbδp]∥vh∥ by step 4.1. Taking the supremum and applying the discrete inf-sup yields ∥θp∥≤βh−1(MaUh+Mbδp)≤βh−1{Ma(Uh+δu)+Mbδp}. Hence ∥p−ph∥≤δp+∥θp∥≤Ph.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

The conforming P1-bubble/P1 mini element

Definition

Assume Countable Choice. Let D⊂Rd, d=2 or 3, be a bounded connected Lipschitz polygonal/polyhedral domain with a conforming shape-regular affine simplicial mesh Th. On a simplex K with barycentric coordinates λ0,…,λd, define the scalar interior bubble bK=(d+1)d+1∏i=0dλi. Let Sh1={w∈C0(D‾):w∣K∈P1(K)},Sh,01=Sh1∩H01(D). The mini-element spaces are Vh=[Sh,01]d⊕⨁K∈ThbKRd⊂H01(D;Rd),Qh=Sh1∩L02(D). Each bubble is extended by zero off its cell. The zero-boundary conformity of the piecewise affine part is supplied by Zero-boundary conforming polynomials lie in W0 1p and satisfy Poincare. The local/global dimensions are proved in the next dimension lemma.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Elementary Fourier ray and derivative L2 bounds

Statement

Assume Countable Choice. Let U be bounded and open, ϕ∈Cc∞(Rd), and ρ>0. For every ξ∈Rd and coordinate j, ∫0∞2π∣ξj∣∣ϕ^(tξ)∣dt≤ρ−1∥ϕ∥1+ρ∥∂j2ϕ∥1. If Tϕ,j is the distributional operator of Truncated Fourier formula for derivatives of the Bogovskii operator, then for every f∈Cc∞(U) its restriction to U is in L2(U) and ∥Tϕ,jf∥L2(U)≤Aj(ϕ)∥f∥L2(U),Aj(ϕ)=2(d−1)/2(ρ−1∥ϕ∥1+ρ∥∂j2ϕ∥1)+2d∣U∣∥∂jϕ∥∞. By density, this restriction has a unique bounded extension L2(U)→L2(U) with the same bound.

Facts & Assumptions

Given: Countable Choice; bounded U, ϕ∈Cc∞, ρ>0, and the truncated Fourier formula for Tϕ,j.

[A1]

Countable Choice is assumed by the Plancherel/Fourier-multiplier interfaces and is used for the L2 density extension. (The Axiom of Countable Choice (ACω))

[F1]

For 0<t<1/2, Tϕ,j<f^(ξ)=2πiξj∫01/2ϕ^(tξ)f^((1−t)ξ)dt. (Truncated Fourier formula for derivatives of the Bogovskii operator)

[F2]

Plancherel identifies the Fourier and L2 norms, and ∂j2ϕ^(η)=(2πiηj)2ϕ^(η). (Fourier transform agrees with l one and plancherel transforms, Distributional derivatives are polynomial Fourier multipliers)

[F3]

Tonelli permits integrating the nonnegative Fourier bound in (t,ξ), and Countable Choice gives density of Cc∞(U) in L2(U). (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Smooth compactly supported functions of an open set are dense in L2, The Axiom of Countable Choice (ACω))

[F4]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

Proof

1.1givenalgebra

If ξj=0, the left side of the ray estimate is zero. Otherwise put t0=(2πρ∣ξj∣)−1. On [0,t0], use ∣ϕ^∣≤∥ϕ∥1 to bound the integral by 2π∣ξj∣t0∥ϕ∥1=ρ−1∥ϕ∥1.

2.1A1F2algebrastep 1.1

On [t0,∞), [F2] gives ∣ϕ^(tξ)∣≤(4π2t2ξj2)−1∥∂j2ϕ∥1. Integrating 2π∣ξj∣t−2/(4π2ξj2) from t0 to infinity gives ρ∥∂j2ϕ∥1. Adding the intervals proves the ray bound.

3.1A1F1F2F3F4step 2.1

Write m(t,ξ)=2π∣ξj∣∣ϕ^(tξ)∣. Cauchy–Schwarz in t and step 2.1 yield ∣T<f^(ξ)∣2≤Rj(ϕ)∫01/2m(t,ξ)∣f^((1−t)ξ)∣2dt, where Rj(ϕ)=ρ−1∥ϕ∥1+ρ∥∂j2ϕ∥1. Integrate in ξ, set η=(1−t)ξ, and then s=t/(1−t). Since (1−t)−(d+1)dt=(1+s)d−1ds≤2d−1ds for 0≤s≤1, the inner integral is at most 2d−1Rj(ϕ)∥f∥22 by the ray bound and [F2]. Plancherel [F2] then gives ∥Tϕ,j<f∥2≤2(d−1)/2Rj(ϕ)∥f∥2.

4.1A1F1givenF4step 3.1

For 1/2≤t≤1, the kernel t−d∂xj[ϕ(y+(x−y)/t)] has absolute value at most 2d+1∥∂jϕ∥∞; integrating the sharper factor t−d−1 over this interval gives at most 2d∥∂jϕ∥∞. On U×U its integral operator has L2(U) norm at most 2d∣U∣∥∂jϕ∥∞ by Cauchy–Schwarz in y and then x. Add this to the short-range bound of step 3.1, which was proved on all of Rd and hence also holds after restriction to U.

5.1A1F1F3step 3.1step 4.1∎

The estimates hold first for f∈Cc∞(U). Density in L2(U) extends the restriction of the distributional operator uniquely to all L2(U) inputs and preserves the same bound. Convergence in L2(U) implies convergence against every test function in Cc∞(U), so this extension agrees on U with the distributional limit from the definition.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Fortin operators transfer the continuous inf-sup bound

Statement

Let Vh⊂V and Qh⊂Q, and suppose the continuous inf-sup constant of b is β>0. If Qh={0} the constraint assertion is vacuous. Otherwise, for a Fortin operator as in Fortin operator for a discrete constraint with bound CF, define βh:=inf⁡qh≠0sup⁡vh≠0b(vh,qh)/(∥vh∥V∥qh∥Q). Then βh≥β/CF.

Facts & Assumptions

Given: Conforming subspaces, β>0, and a bounded linear Fortin operator preserving b against Qh.

[F1]

For all v∈V and qh∈Qh, b(Πhv,qh)=b(v,qh) and ∥Πhv∥V≤CF∥v∥V. (Fortin operator for a discrete constraint)

Proof

1.1F1givenalgebra

The case Qh={0} is vacuous, so fix qh≠0. In these real spaces, replacing a nonzero velocity by its negative shows that the supremum of b(v,qh)/∥v∥V equals the supremum with ∣b(v,qh)∣ in the numerator, both on V and on any nonzero subspace Vh. The continuous inf-sup bound gives sup⁡v≠0∣b(v,qh)∣/∥v∥V≥β∥qh∥Q>0, so some v has b(v,qh)≠0. Constraint preservation [F1] then implies Πhv≠0, and its norm bound forces CF>0; in particular Vh≠{0}. Define s(qh):=sup⁡vh≠0∣b(vh,qh)∣/∥vh∥V. For every nonzero v∈V with Πhv≠0, [F1] and the definition of this fixed-qh supremum give ∣b(v,qh)∣=∣b(Πhv,qh)∣≤s(qh)∥Πhv∥V≤CFs(qh)∥v∥V. If Πhv=0, preservation gives b(v,qh)=0, so the same bound holds.

2.1givenalgebrastep 1.1∎

Taking the supremum over nonzero v in step 1.1 and using the continuous bound yields β∥qh∥Q≤CFs(qh). Divide by the positive CF∥qh∥Q, and only then take the infimum over qh≠0. The real-sign equivalence in step 1.1 identifies inf⁡qh≠0s(qh)/∥qh∥Q with the stated βh, giving βh≥β/CF.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Local and global dimensions of the mini element

Statement

Assume Countable Choice. For a nonempty connected conforming simplicial mesh in dimension d=2 or 3, the local scalar velocity polynomial space P1(K)⊕span⁡{bK} has dimension d+2, the vector velocity dimension is d(d+2), and the local pressure space P1(K) has dimension d+1. If the mesh has Nv vertices, Nv,0 interior vertices and NK cells, then dim⁡Vh=dNv,0+dNK,dim⁡Qh=Nv−1.

Facts & Assumptions

Given: The mini-element spaces on a finite connected conforming simplicial mesh, with one barycentric bubble per cell and per vector component.

[A1]

Countable Choice is inherited from the mini-element/H0 space interface. (The Axiom of Countable Choice (ACω))

[F1]

Vh is the zero-boundary continuous piecewise P1 velocity space plus independent cell-supported barycentric bubbles, and Qh is continuous P1 with zero mean; the zero-boundary P1 space is defined as Sh1∩H01(D). (The conforming P1-bubble/P1 mini element)

[F2]

For a linear map with finite-dimensional domain, domain dimension equals kernel dimension plus image dimension. (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T)

[F3]

H01(D) is the H1-closure of compactly supported smooth functions; affine pullback transfers H1 convergence to a reference simplex, where the facet trace is continuous and agrees with classical restriction on affine functions. (Zero-boundary Sobolev space as a norm closure, Affine pullback of weak derivatives and Sobolev seminorms, Sobolev face traces and Gauss–Green on a reference simplex)

[F4]

A conforming piecewise polynomial with zero trace on every boundary facet belongs to H01(D). (Zero-boundary conforming polynomials lie in W0 1p and satisfy Poincare)

Proof

1.1A1F1algebra

On one d-simplex, P1(K) has the d+1 barycentric coordinate basis. The bubble bK is nonzero in the interior and vanishes at every vertex; because its polynomial degree is d+1>1, it is not in P1(K). Thus the scalar local velocity sum is direct with dimension d+2, and its d vector components give d(d+2). The local pressure dimension is d+1.

1.2A1F1F3F4algebra

A continuous piecewise P1 scalar function is uniquely determined by its Nv vertex values: affine uniqueness on each simplex proves injectivity, and assigning shared vertex values defines such a function by face-to-face conformity. If w∈Sh1∩H01(D), choose wn∈Cc∞(D) with wn→w in H1(D) by [F3]. On each boundary cell, affine pullback gives convergence in H1 on the reference simplex; the reference traces of wn vanish there, so continuity of the trace gives zero trace for the pullback of w. Since that pullback is affine, its trace is its classical restriction, and the nodal values of w vanish on every boundary facet. Conversely, if all boundary vertex values vanish, the affine trace of w vanishes on each boundary facet, so [F4] gives w∈H01(D). Hence exactly the Nv−Nv,0 boundary nodal values are forced to zero, leaving Nv,0 independent velocity nodal values per component.

2.1A1F1F2algebrastep 1.1step 1.2∎

Each cell bubble is linearly independent of the nodal functions because its support interior is disjoint from every other cell bubble and it vanishes on the cell boundary, while a P1 polynomial vanishing on all vertices is zero. Hence the velocity dimension is dNv,0+dNK. The integral functional on Sh1 is nonzero (it takes the positive value ∣D∣ on 1), so its zero-mean kernel has codimension one; therefore dim⁡Qh=Nv−1.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Weak and pointwise divergence-free discrete velocities

Definition

For the conforming mini-element spaces Vh,Qh, define the weak discrete divergence kernel by Zh:={vh∈Vh:b(vh,qh)=0 for every qh∈Qh}, and the exact discrete divergence kernel by Zh0:={vh∈Vh:div⁡vh=0 almost everywhere on D}. Since b(vh,qh)=−∫Dqhdiv⁡vh, one has Zh0⊆Zh. Equality is not part of either definition; the companion example proves strictness for a concrete mesh.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Bounded Bogovskii right inverse on a star-shaped domain

Statement

Let U⊂Rd be bounded and open, star-shaped with respect to a ball, and let ω and BU be as in The Bogovskii integral on a star-shaped set. Under Countable Choice there is a bounded linear extension BU:L02(U)→H01(U;Rd) such that div⁡BUf=f,∥∇BUf∥2≤CU∥f∥2. After translating the ball center to 0, choose RU≥sup⁡y∈U∣y∣ and define Aj(ϕ)=2(d−1)/2(ρ−1∥ϕ∥1+ρ∥∂j2ϕ∥1)+2d∣U∣∥∂jϕ∥∞. One valid constant is CU=(∑k,j=1d[Aj(xkω)+RUAj(ω)]2)1/2.

Facts & Assumptions

Given: Countable Choice; bounded star-shaped U, unit-integral ω∈Cc∞ supported in the ball of radius ρ, and the Bogovskii formula.

[A1]

Countable Choice is used to select the L2 approximation sequence and take its H1 limit. (The Axiom of Countable Choice (ACω))

[F1]

The formula has compact support in U for compactly supported smooth input and satisfies div⁡BUf=f−ω∫Uf distributionally. (Support, integrability, and divergence identity for the Bogovskii kernel)

[F2]

Each derivative of the formula is the difference Txkω,jf−Tω,j(ykf) in the distributional sense. (Truncated Fourier formula for derivatives of the Bogovskii operator)

[F3]

For the restriction to U, the Fourier ray estimate and bounded-kernel tail give ∥Tϕ,jg∥L2(U)≤Aj(ϕ)∥g∥L2(U) when g∈Cc∞(U). (Elementary Fourier ray and derivative L2 bounds)

[F4]

On bounded U, zero-boundary functions satisfy ∥w∥2≤CP(U)∥∇w∥2. (A bounded-domain Poincare inequality for H1-zero fields)

[F5]
[F6]

Every open ball contains a nonnegative smooth compactly supported function with strictly positive integral; normalizing its integral gives a unit-mass bump. (A smooth bump between concentric Euclidean balls)

[F7]

Scaling a unit-mass smooth bump gives a mollifier family with unit mass and support radius ϵ. (The mollifier family generated by a unit-mass smooth bump)

[F8]

Convolution of a locally integrable function with a smooth compactly supported mollifier is smooth, and its support is contained in the closure of the support sum. (Convolution with a mollifier is smooth, and derivatives pass under the integral sign, The support of a convolution lies in the closure of the support sumset)

[F9]

Mollification commutes with weak derivatives and an L2 approximate identity converges to the identity in L2, under Countable Choice. (Interior mollification commutes with weak derivatives, Every L1 approximate identity converges to the identity in Lp for 1≤p<∞, The Axiom of Countable Choice (ACω))

[F10]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

Proof

1.1A1F2F3givenalgebra

First take f∈Cc∞(U) with ∫f=0. After translating the ball center to 0, both ω and xkω are supported in B(0,ρ)⊂U. For either ϕ=ω or ϕ=xkω and any g∈Cc∞(U), every point in the support of a truncated Tϕ,jg has the form x=(1−t)y+tz with y∈supp⁡g, z∈supp⁡ϕ, and 0≤t≤1. Star-shapedness puts every such segment in U; compactness of supp⁡g×supp⁡ϕ×[0,1] shows the resulting support is contained in a compact subset of U. Hence the distributional operators in [F2] are supported in U and their ambient L2 norms equal their L2(U) norms. By [F2]–[F3], for each k,j, ∥∂j(BUf)k∥2≤[Aj(xkω)+RUAj(ω)]∥f∥2, since ∣yk∣≤RU on U. Squaring and summing the finitely many derivative components gives ∥∇BUf∥2≤CU∥f∥2.

2.1A1F1F6F7F8F9choosestep 1.1

By [F1], this field is supported in a compact subset K⋐U, and step 1.1 puts it in H1(Rd). Choose the normalized bump from [F6] and its scaled mollifier ρϵ from [F7]. For ϵ<12dist⁡(K,Rd∖U), [F8] makes ρϵ∗BUf smooth with compact support in U. Its first derivatives are ρϵ∗∂j(BUf) and converge in L2 together with the field by [F9], so the approximants converge to BUf in H1. Hence BUf∈H01(U). The divergence identity in [F1] reduces to div⁡BUf=f.

2.2A1F5givenF10step 1.1

For arbitrary f∈L02(U), choose hn∈Cc∞(U) with hn→f in L2 using [F5], and put fn=hn−ω∫Uhn. Then fn∈Cc∞(U), ∫fn=0, and ∥fn−f∥2≤(1+∣U∣1/2∥ω∥2)∥hn−f∥2→0 by Cauchy–Schwarz and ∫f=0.

3.1A1F4F5algebrastep 1.1step 2.1∎

By step 1.1 applied to differences, BUfn is Cauchy in the gradient norm. [F4] makes it Cauchy in H1; completeness and closedness in [F5] give a limit in H01(U). The limit is independent of the approximating sequence by the same bound, so it defines a linear operator and retains the gradient bound. Continuity of divergence H1→L2 and step 2.1 give div⁡BUf=f.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Divergence right inverse on bounded connected Lipschitz domains

Statement

Let D⊂Rd be nonempty, bounded, connected and Lipschitz, with d∈{2,3}. Under Countable Choice there is a bounded linear map BD:L02(D)→H01(D;Rd) such that div⁡BDq=q,∥∇BDq∥2≤Cdiv(D)∥q∥2. For a chosen finite cover and tree decomposition from Finite star-shaped cover of a bounded Lipschitz domain and Zero-mean L2 decomposition over a finite cover, one valid bound is Cdiv(D)=∑iCUiDi; it is finite and independent of any finite-element mesh.

Facts & Assumptions

Given: Countable Choice; a bounded connected Lipschitz domain, the finite star-shaped cover, its zero-mean decomposition constants, and the local Bogovskii right inverses.

[A1]

Countable Choice is inherited through the local L2 divergence right-inverse extension. (The Axiom of Countable Choice (ACω))

[F1]

The cover has finitely many star-shaped Ui and every q∈L02(D) decomposes linearly into qi∈L02(Ui) with q=∑iqi and ∥qi∥2≤Di∥q∥2. (Finite star-shaped cover of a bounded Lipschitz domain, Zero-mean L2 decomposition over a finite cover)

[F2]

Each local operator maps L02(Ui) boundedly into H01(Ui)d with divergence qi and gradient bound CUi. (Bounded Bogovskii right inverse on a star-shaped domain)

[F3]

Zero-boundary fields on bounded open sets have zero-mean divergence. (Zero mean of the divergence of a zero-boundary field)

Proof

1.1A1F1F2construct

For q∈L02(D), form qi by [F1], apply the local theorem [F2], extend each BUiqi by zero to D, and define BDq=∑iBUiqi~. The extension belongs to H01(D)d: approximate in H1(Ui) by compactly supported smooth fields and extend those by zero.

2.1A1F1F2step 1.1

Zero extension preserves weak derivatives because the approximating functions are compactly supported inside Ui; hence div⁡BUiqi~ is the zero extension of qi. Summing and using [F1] gives div⁡BDq=q. Linearity follows from the linear decomposition and local operators.

2.2A1F1F2algebrastep 1.1

The triangle inequality and [F1]–[F2] yield ∥∇BDq∥2≤∑iCUi∥qi∥2≤(∑iCUiDi)∥q∥2. Each sum is finite, and its ingredients depend only on the chosen domain cover, not on a mesh parameter.

3.1A1F3step 2.1∎

Conversely, any v∈H01(D)d has zero-mean divergence by [F3], so L02(D) is the compatible pressure range. The identity in step 2.1 proves that every such datum is attained by the constructed right inverse.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Continuous Stokes divergence inf-sup on a Lipschitz domain

Statement

Assume Countable Choice. On the spaces V=H01(D;Rd) and Q=L02(D) of Stokes velocity and pressure spaces and weak forms, equipped with ∥v∥V=∥∇v∥2, the form b(v,q)=−∫Dq div⁡v is bounded by d. Its inf-sup constant satisfies inf⁡q≠0sup⁡v≠0b(v,q)∥v∥V∥q∥2≥Cdiv(D)−1.

Facts & Assumptions

Given: Bounded connected Lipschitz D, real V,Q, the stated b, and the bounded right inverse BD.

[A1]

Countable Choice is inherited from the bounded Lipschitz divergence right-inverse theorem. (The Axiom of Countable Choice (ACω))

[F1]

BDq∈H01(D)d, div⁡BDq=q, and ∥∇BDq∥2≤Cdiv(D)∥q∥2. (Divergence right inverse on bounded connected Lipschitz domains)

[F2]

The conventions are b(v,q)=−∫Dqdiv⁡v and ∥v∥V=∥∇v∥2. (Stokes velocity and pressure spaces and weak forms)

[F3]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

Proof

1.1F2F3

For any matrix G∈Rd×d, ∣tr⁡G∣≤d∣G∣F by Cauchy–Schwarz on its diagonal. Hence ∣b(v,q)∣≤∥q∥2∥div⁡v∥2≤d∥q∥2∥∇v∥2.

2.1A1F1F2algebrastep 1.1∎

For q≠0, take v=−BDq. By [F1]–[F2], b(v,q)=∥q∥22 and ∥v∥V≤Cdiv(D)∥q∥2. Thus sup⁡v≠0b(v,q)/∥v∥V≥Cdiv(D)−1∥q∥2. Dividing by ∥q∥2>0 and taking the infimum over nonzero q proves the bound.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

Uniform Fortin stability of the mini element

Statement

On the mini-element spaces of The conforming P1-bubble/P1 mini element, there is a Fortin operator Πh:V→Vh preserving b(v,qh) for every qh∈Qh, with ∥Πhv∥V≤CF∥v∥V. The constant depends only on d and the mesh shape-regularity bound, not on h. Consequently βh≥(CFCdiv(D))−1.

Facts & Assumptions

Given: Countable Choice; a bounded connected Lipschitz polyhedral domain with shape-regular conforming simplices; the continuous Stokes spaces, the face-average interpolant Ih, cell bubbles bK, and the continuous inf-sup bound.

[A1]

Countable Choice is inherited through the face-average H0 interpolation input. (The Axiom of Countable Choice (ACω))

[F1]

Ih maps H01 to continuous zero-boundary P1, is H1 stable, and satisfies the local L2 estimate ∥v−Ihv∥L2(K)≤ChK∥∇v∥L2(ωK). (Boundary-preserving face-average P1 interpolation)

[F2]

b(v,q)=−∫Dqdiv⁡v and Qh is continuous piecewise affine with zero mean. (Stokes velocity and pressure spaces and weak forms, The conforming P1-bubble/P1 mini element)

[F3]

The continuous inf-sup constant is at least Cdiv(D)−1, and a Fortin map of norm CF transfers it to βh≥β/CF. (Continuous Stokes divergence inf-sup on a Lipschitz domain, Fortin operators transfer the continuous inf-sup bound)

[F4]

A Fortin operator is a bounded linear map into Vh that preserves b(v,qh) for every qh. (Fortin operator for a discrete constraint)

[F5]

A continuous piecewise P1 pressure belongs to H1(D), with weak gradient equal to its cellwise polynomial gradient. (Continuous matching piecewise polynomials belong to W1p)

[F6]

Affine scaling transfers the fixed reference-simplex trace and Poincare inequalities to each mesh cell; the scaling constants depend only on d and the shape-regularity bound. (Shape regularity controls affine simplex geometry, Affine pullback of weak derivatives and Sobolev seminorms, Sobolev face traces and Gauss–Green on a reference simplex, Poincare–Wirtinger inequality on a reference simplex)

Proof

1.1A1F1F2construct

For each cell put cK=(∫KbK)−1∫K(v−Ihv)∈Rd, and define Πhv=Ihv+∑KbKcK. The denominator is positive because bK>0 in the cell interior. The added functions vanish on cell boundaries, so Πhv∈Vh.

1.2A1F1F6algebra

We track the constants in the construction of [F1]. Shape regularity bounds each vertex solid angle below, so at most N(d,σ) cells meet a vertex, where σ is the shape-regularity bound. Each vertex star is facet connected: a sufficiently small interior ball or boundary Lipschitz-epigraph neighborhood is path connected, and paths avoiding the codimension-two mesh skeleton cross only facets. Thus cells in ωK are linked by uniformly bounded facet chains and have comparable diameters by [F6]. For adjacent cells T,T′ with means mT,mT′, cell Poincare and trace give ∣mT−mT′∣≤ChK1−d/2∥∇v∥L2(T∪T′). Summing along those chains and choosing c to be one cell mean gives ∥v−c∥L2(ωK)≤ChK∥∇v∥L2(ωK). The facet dual functions of [F1] have norm at most ChK−(d−1)/2; the scaled trace inequality therefore gives ∣∫Fzψz,Fz(v−c)∣≤ChK1−d/2∥∇v∥L2(ωK) for each vertex of K. Since Ih reproduces constants, the local P1 basis bounds yield ∥∇Ihv∥L2(K)+hK−1∥v−Ihv∥L2(K)≤C∥∇v∥L2(ωK). All these constants and the patch-overlap bound depend only on d,σ; the domain charts were used only for connectivity.

2.1A1F1F2F5step 1.1

Put w=Πhv−v∈H01(D;Rd). By [F5], qh∈H1(D); approximating w in H1 by compactly supported smooth fields and using the weak-derivative definition of qh gives b(w,qh)=∫Dw⋅∇qh. Since ∇qh is constant on each cell and step 1.1 gives ∫Kw=0, this equals ∑K∇(qh∣K)⋅∫Kw=0. Therefore b(Πhv,qh)=b(v,qh).

2.2A1F6step 1.1step 1.2

Reference scaling in [F6] gives ∫KbK≥cd∣K∣ and ∥∇bK∥L2(K)≤ChKd/2−1. Thus ∥∇(bKcK)∥L2(K)≤ChK−1∥v−Ihv∥L2(K)≤C∥∇v∥L2(ωK) by step 1.2. Squaring, summing cells and using bounded patch overlap gives a uniform bound for the bubble correction; adding the H1 stability of Ih from step 1.2 proves ∥Πhv∥V≤CF∥v∥V.

3.1A1F3F4step 2.1step 2.2∎

The construction is linear in v and has the preservation and norm properties in steps 2.1 and 2.2, so [F3] applies with β≥Cdiv(D)−1. It yields βh≥(CFCdiv(D))−1 uniformly over the mesh family.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Weak Stokes solvability on bounded connected Lipschitz domains

Statement

Assume Countable Choice. Let D⊂Rd be nonempty, bounded, connected and Lipschitz, d=2 or 3, and let ν>0. For every bounded linear functional F on V=H01(D;Rd) equipped with ∥v∥V=∥∇v∥2, there is a unique (u,p)∈V×Q, Q=L02(D), satisfying 2ν∫Dε(u):ε(v)−∫Dpdiv⁡v=F(v),−∫Dqdiv⁡u=0 for every v∈V,q∈Q. The pressure is normalized to have mean zero. One stability bound is ∥u∥V≤ν−1∥F∥V′ and ∥p∥2≤3Cdiv(D)∥F∥V′.

Facts & Assumptions

Given: Countable Choice; a bounded connected Lipschitz D; the real Sobolev spaces; ν>0 and F∈V′.

[A1]

Countable Choice is assumed by the Hilbert-space completeness and Lax–Milgram branch used below; no full Axiom of Choice is used by the local Korn or Poincare arguments. (The Axiom of Countable Choice (ACω))

[F1]

H1(D) is Hilbert under Countable Choice, H01(D) is its closed subspace, and bounded-domain Poincare makes the gradient norm equivalent to the H1 norm. (The Sobolev space H1 is a Hilbert space, Zero-boundary Sobolev space as a norm closure, A bounded-domain Poincare inequality for H1-zero fields, The Axiom of Countable Choice (ACω))

[F2]

Real L2(D) is Hilbert under Countable Choice, and the integral functional is bounded by ∣D∣1/2∥q∥2, so its zero-mean kernel Q is closed and Hilbert. (L2 with the integral pairing is a Hilbert space, Holder's inequality for integrals, including the endpoint cases, The Axiom of Countable Choice (ACω))

[F3]

2∥ε(v)∥22=∥∇v∥22+∥div⁡v∥22 for every zero-boundary field. (The homogeneous Dirichlet strain identity)

[F4]

The divergence form is bounded and has continuous inf-sup constant at least Cdiv(D)−1. (Continuous Stokes divergence inf-sup on a Lipschitz domain)

[F5]

On real Hilbert spaces, bounded a,b, coercivity on the constraint kernel and positive inf-sup imply unique mixed solvability with the stated norm bounds. (Babuska-Brezzi solvability by kernel coercivity and inf-sup)

[F6]

Every zero-boundary velocity has zero-mean divergence. (Zero mean of the divergence of a zero-boundary field)

[F7]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

Proof

1.1A1F1F2

By [F1], V is complete in the H1 norm. If (vn) is Cauchy in ∥∇⋅∥2, [F1] and Poincare make it Cauchy in H1; its H1 limit is in the closed subspace H01 and its gradients converge. Thus V is Hilbert in the gradient norm. By [F2], Q is Hilbert.

2.1F3F7step 1.1

Cauchy–Schwarz and ∣ε(v)∣≤∣∇v∣F give ∣aν(u,v)∣≤2ν∥u∥V∥v∥V. By [F3], aν(v,v)=2ν∥ε(v)∥22≥ν∥v∥V2 on all of V, hence on ker⁡B; take Ma=2ν and α=ν.

3.1A1F4F5step 1.1step 2.1

The form b has Mb≤d and inf-sup β≥Cdiv(D)−1 by [F4]. Apply [F5] with G=0. It yields a unique velocity and a unique mean-zero pressure satisfying the two displayed equations. The bound on velocity is ∥u∥V≤α−1∥F∥=ν−1∥F∥.

4.1A1F4F5F6step 3.1∎

By [F6], div⁡u∈Q. The constraint equation tested with q=div⁡u therefore gives ∥div⁡u∥22=0, so the weak constraint is exact incompressibility. Also [F5] gives ∥p∥2≤β−1(∥F∥+Ma∥u∥V). Use β−1≤Cdiv(D), Ma=2ν and the velocity estimate of step 3.1 to obtain ∥p∥2≤3Cdiv(D)∥F∥. Mean-zero gauge is part of Q, so the pressure is unique in that normalization.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Mini-element H1 velocity and L2 pressure error estimate

Statement

Assume Countable Choice. Let D be a bounded connected Lipschitz polygonal/polyhedral domain in dimension d=2 or 3, with a conforming shape-regular simplicial mesh family of mesh size h. Put V=H01(D;Rd) and Q=L02(D). For ν>0 and F∈V′, suppose (u,p)∈V×Q satisfies 2ν∫Dε(u):ε(v)−∫Dp div⁡v=F(v),−∫Dq div⁡u=0 for every v∈V,q∈Q, with u∈H2(D;Rd) and p∈H1(D). Let (uh,ph)∈Vh×Qh solve the same equations for every vh∈Vh,qh∈Qh, where Vh,Qh are the P1-bubble/P1 mini spaces. Then ∥u−uh∥H1(D)+∥p−ph∥L2(D)≤Ch(∥u∥H2(D)+∥p∥H1(D)). The constant is independent of h and depends on d, D, shape regularity, ν, Cdiv(D), and the Poincare constant. The mixed error formula retains Ma=2ν, Mb=d, discrete-kernel coercivity α∗=ν, and βh≥(CFCdiv(D))−1 explicitly.

Facts & Assumptions

Given: Countable Choice; the continuous Stokes solution with the stated regularity, the mini-element spaces, a conforming shape-regular mesh, and positive viscosity ν.

[A1]

Countable Choice is inherited through the Sobolev Hilbert and interpolation inputs. (The Axiom of Countable Choice (ACω))

[F1]

The mini pair is Vh=[Sh,01]d⊕∑KbKRd, Qh=Sh1∩L02, with continuous piecewise P1 pressures. (The conforming P1-bubble/P1 mini element)

[F2]

The face-average map is defined on all of H1(D) with values in continuous piecewise P1, has the H1-to-L2 and H2-to-H1 estimates, and maps H01(D) into the zero-boundary subspace. (Boundary-preserving face-average P1 interpolation)

[F3]

The mini Fortin map gives βh≥(CFCdiv(D))−1, and the mixed Galerkin theorem gives the explicit velocity/pressure bounds Uh,Ph. (Uniform Fortin stability of the mini element, Mixed Galerkin error bound retaining beta_h)

[F4]

The Stokes conventions are aν(u,v)=2ν∫ε(u):ε(v), b(v,q)=−∫qdiv⁡v, and the velocity gradient norm; the zero-boundary Poincare inequality controls the full H1 norm. (Stokes velocity and pressure spaces and weak forms, A bounded-domain Poincare inequality for H1-zero fields)

[F5]

The homogeneous Korn identity gives 2∥ε(v)∥22=∥∇v∥22+∥div⁡v∥22. (The homogeneous Dirichlet strain identity)

[F6]

Holder bounds the mean of an L2 error by ∣D∣1/2 times its L2 norm. (Holder's inequality for integrals, including the endpoint cases)

[F7]

The L2 Cauchy--Schwarz inequality gives ∫∣fg∣≤∥f∥2∥g∥2. (Cauchy-Schwarz inequality for L2)

Proof

1.1A1F1F2F6given

Set wh=Ihu. By [F2], wh∈Vh and δu:=∥u−wh∥V≤CIh∥u∥H2. Define p~h=Ihp and subtract its mean, rh=p~h−∣D∣−1∫Dp~h. Then rh∈Qh. Since p has zero mean, ∣D∣1/2∣D∣−1∣∫D(p~h−p)∣≤∥p~h−p∥2, and [F2] gives δp:=∥p−rh∥2≤2CIh∥p∥H1.

2.1A1F3F4F5F7step 1.1

The strain bound ∣ε(v)∣≤∣∇v∣F and Cauchy–Schwarz give Ma≤2ν; ∣div⁡v∣≤d∣∇v∣F gives Mb≤d. By [F5], aν(v,v)=2ν∥ε(v)∥22≥ν∥∇v∥22 for every v∈Vh, so α∗=ν. The uniform Fortin theorem in [F3] supplies the stated βh bound.

3.1A1F2F3step 1.1step 2.1

Apply [F3] with the approximants of step 1.1 and these constants. The explicit quantities Rh,Uh,Ph are linear combinations of δu,δp with coefficients depending only on ν,d,βh; the lower bound on βh makes those coefficients uniform in h. Thus ∥u−uh∥V+∥p−ph∥2≤Ch(∥u∥H2+∥p∥H1).

4.1A1F4step 3.1∎

Since u−uh∈H01(D)d, [F4] gives ∥u−uh∥H1≤(1+CP(D)2)1/2∥u−uh∥V. Absorb this fixed domain factor into C; it is independent of h. This proves the asserted full H1 velocity estimate without asserting additional regularity of the Stokes solution.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Discrete Helmholtz projection and pressure Schur identities

Statement

Assume Countable Choice. Let Vh,Qh be the stable mini-element spaces on a bounded connected Lipschitz polyhedral domain with a conforming mesh, with viscosity ν>0, and let Zh={vh:b(vh,qh)=0 ∀qh∈Qh}. Choose bases of Vh,Qh, let M be the velocity mass matrix, A the Stokes stiffness matrix, and B the constraint matrix with entries Bℓj=b(ϕj,χℓ). The mass-orthogonal projector onto Zh is PM=I−M−1BT(BM−1BT)−1B. On the mean-zero pressure coordinates, both the mass pressure Schur matrix BM−1BT and stiffness pressure Schur matrix BA−1BT are positive definite. Before imposing the pressure gauge on the full continuous P1 space, the constant pressure is exactly the one-dimensional nullspace of BT.

Facts & Assumptions

Given: Countable Choice; the bounded connected Lipschitz polyhedral mesh, the Stokes bilinear forms and pressure gauge, and the uniform discrete inf-sup bound.

[A1]

Countable Choice is inherited through the mini-element and Fortin inputs. (The Axiom of Countable Choice (ACω))

[F1]

The discrete weak kernel is the nullspace of the mini constraint pairing; the mini Fortin theorem gives βh>0. (Weak and pointwise divergence-free discrete velocities, Uniform Fortin stability of the mini element)

[F2]

The zero-boundary divergence has zero integral, so constants pair to zero against all velocity fields; the pressure space is gauged by zero mean. (Zero mean of the divergence of a zero-boundary field, Stokes velocity and pressure spaces and weak forms)

[F3]

The homogeneous Korn identity makes the Stokes stiffness form positive definite on Vh for ν>0. (The homogeneous Dirichlet strain identity)

Proof

1.1A1F1given

The mass matrix M is symmetric positive definite because for a nonzero coefficient vector x, xTMx=∥vh(x)∥L22>0. The inf-sup condition in [F1] says that BTy=0 for a mean-zero pressure coefficient vector y implies y=0; hence B has full row rank on the gauged pressure coordinates.

2.1algebrastep 1.1

Therefore SM:=BM−1BT is symmetric positive definite: for y≠0, yTSMy=(BTy)TM−1(BTy)>0. The formula for PM is defined. Direct multiplication gives BPM=B−BM−1BTSM−1B=0, and if Bx=0 then PMx=x. Thus its range is exactly ker⁡B.

3.1algebrastep 2.1

For every x, x−PMx=M−1BTSM−1Bx. If z∈ker⁡B, then zTM(x−PMx)=(Bz)TSM−1Bx=0. Hence x−PMx is mass-orthogonal to the kernel and PM is the mass-orthogonal projector.

3.2A1F3algebrastep 2.1

By [F3], A is symmetric positive definite. Full row rank of B from step 2.1 then makes SA:=BA−1BT positive definite by the same quadratic-form argument as in step 2.1. Eliminating velocity from Au+BTp=f, Bu=g gives SAp=BA−1f−g, which is the pressure Schur identity.

4.1A1F1F2algebrastep 1.1∎

If pressure coordinates are instead taken in the full continuous P1 space, [F2] puts constants in ker⁡BT. Conversely, if BTy=0, the corresponding qh∈Sh1 pairs to zero with every velocity. Its mean-zero part qh−∣D∣−1∫Dqh also pairs to zero by [F2]; the stable inf-sup in [F1] forces that part to vanish. Thus the nullspace consists exactly of constants, and restricting to Qh=L02∩Sh1 removes that gauge mode.

5 · Examples, counterexamples and false statements

None yet.

Sources