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✓ 20 results · all verified · 10 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 10 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Mixed H(div) fluxes, Darcy flow and local conservation

1 · Prerequisites

2 · Summary

This page develops real $H (div)$ fluxes with the Green-duality normal trace, then defines Raviart–Thomas and Brezzi–Douglas–Marini polynomial families on conforming affine simplices. The canonical RT0 facet interpolation is defined only for $H^{1}$ inputs, where the trace moments are bounded. For arbitrary zero-normal $H (div)$ fields, the bounded commuting projection is constructed from the stable discrete divergence lift and an $L^{2}$ projection onto the finite-dimensional divergence-free kernel; no rough-field facet moment is used.

For a bounded connected Lipschitz polyhedral domain in dimension two or three, the page fixes the no-flow Darcy sign convention, the zero-mean pressure gauge, and measurable uniformly positive symmetric permeability. It proves continuous and RT0/P0 solvability, a mesh-uniform divergence lift, a bounded commuting projection, an explicit mixed error estimate with its permeability-contrast dependence, and exact cellwise mass balance with normal continuity.

The hybrid system uses one constant multiplier on each interior facet. It is equivalent to the conforming mixed method after quotienting the common pressure/multiplier shift, and local flux-pressure blocks can be eliminated to a facet system with only that gauge null direction. One-cell meshes are handled separately because their discrete flux and mean-zero pressure spaces vanish. The four finite examples are on the companion page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

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

Contravariant Piola map

Definition

Let $\hat{K}, K \subset R^{d}$ be nondegenerate open simplices and let $F_{K} : \hat{K} \to K$ be the orientation-preserving affine bijection $F_{K} (\hat{x}) = A_{K} \hat{x} + b_{K}$ , where $J_{K} : = \det A_{K} > 0$ . For a measurable vector field $\hat{v}$ on $\hat{K}$ , define its contravariant Piola push-forward by

(P_{K} \hat{v}) (x) : = J_{K}^{- 1} A_{K} \hat{v} (F_{K}^{- 1} (x)), x \in K .

The inverse pull-back is $\hat{v} (\hat{x}) = J_{K} A_{K}^{- 1} v (F_{K} (\hat{x}))$ . Facet normals are oriented consistently with the chosen orientation of $F_{K}$ ; the associated oriented surface vector transforms by $n d s = J_{K} A_{K}^{- T} \hat{n} d \hat{s}$ .

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

The real H(div) space

Definition

Assume Countable Choice for the published Sobolev interfaces cited below. Let $D \subseteq R^{d}$ be open and use real scalars. $L^{2} (D; R^{d})$ means the $d$ -tuple of real $L^{2}$ equivalence classes, with norm $∥ v ∥_{2}^{2} = \sum_{j = 1}^{d} ∥ v_{j} ∥_{2}^{2}$ . The test space $C_{c}^{\infty} (D; R)$ consists of real smooth functions compactly supported in $D$ .

For $v \in L^{2} (D; R^{d})$ , say that $g \in L^{2} (D; R)$ is its distributional divergence when

\int_{D} v \cdot \nabla ϕ d x = - \int_{D} g ϕ d x for every ϕ \in C_{c}^{\infty} (D; R) .

The $L^{2}$ class $g$ is unique when it exists. Indeed, if $g_{1}, g_{2}$ both satisfy the identity, put $h = g_{1} - g_{2} \in L^{2} (D)$ . Every compact $C \subset D$ has finite Lebesgue measure by Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, and Cauchy-Schwarz inequality for $L^{2}$ gives $\int_{C} ∣ h ∣ \leq ∣ C ∣^{1 / 2} ∥ h ∥_{L^{2} (D)} < \infty$ . Thus $h \in L_{l o c}^{1} (D)$ . Subtracting the two divergence identities gives $\int_{D} h ϕ = 0$ for every real test $ϕ$ ; splitting a complex test into real and imaginary parts gives the same identity for complex tests. Injectivity of the regular-distribution map in Locally integrable functions embed in distributions therefore yields $h = 0$ almost everywhere, hence $g_{1} = g_{2}$ as $L^{2}$ classes. Define

H (div; D) : = {v \in L^{2} (D; R^{d}) : there is g \in L^{2} (D; R) with g = div v distributionally},

and equip it with the graph norm

∥ v ∥_{H (div; D)} : = (∥ v ∥_{2}^{2} + ∥ div v ∥_{2}^{2})^{1 / 2} .

The Sobolev notation $H^{1} (D) = W^{1, 2} (D)$ and $H_{0}^{1} (D) = W_{0}^{1, 2} (D)$ uses the norm and zero-boundary closure conventions of Integer-order Sobolev spaces and their norms and Zero-boundary Sobolev space as a norm closure. No pointwise boundary value is part of this definition.

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The real zero-mean $L^{2}$ space is Hilbert

Statement

Assume Countable Choice. Let $D \subset R^{d}$ be a measurable set of finite measure, and set $L_{0}^{2} (D) : = {q \in L^{2} (D; R) : \int_{D} q d x = 0} .$ With the real $L^{2}$ inner product $(q, r) = \int_{D} q r d x$ , $L_{0}^{2} (D)$ is a closed real Hilbert subspace of $L^{2} (D; R)$ .

Facts & Assumptions

Given: Countable Choice, a finite-measure set $D$ , and the real Hilbert space $L^{2} (D; R)$ with its integral inner product.

[A1]

The published real $L^{2}$ Hilbert-space result assumes Countable Choice. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The integral inner product is well-defined, induces the quotient $L^{2}$ norm, and makes real $L^{2} (D)$ complete. ( $L^{2}$ with the integral pairing is a Hilbert space)

[F2]

Holder's inequality gives $∣ \int_{D} q d x ∣ \leq ∣ D ∣^{1 / 2} ∥ q ∥_{2}$ for $q \in L^{2} (D)$ . (Holder's inequality for integrals, including the endpoint cases)

[F3]

A closed linear subspace of a Banach space is complete with the inherited norm. (A closed subspace of a Banach space is Banach)

Proof

1.1F2given

Define $ℓ (q) : = \int_{D} q d x$ . By [F2], $∣ ℓ (q) ∣ \leq ∣ D ∣^{1 / 2} ∥ q ∥_{2}$ , and linearity of the integral makes $ℓ$ a bounded real linear functional. Its kernel is exactly $L_{0}^{2} (D)$ .

2.1step 1.1algebra

The kernel of a continuous functional is closed: if $q_{n} \in L_{0}^{2} (D)$ and $q_{n} \to q$ in $L^{2}$ , then $∣ ℓ (q) ∣ = ∣ ℓ (q - q_{n}) ∣ \leq ∣ D ∣^{1 / 2} ∥ q - q_{n} ∥_{2} \to 0$ , so $ℓ (q) = 0$ . Thus $L_{0}^{2} (D)$ is a closed linear subspace of $L^{2} (D)$ .

3.1A1F1F3step 2.1∎

The inner product on $L^{2} (D)$ restricts to an inner product on $L_{0}^{2} (D)$ and induces the inherited $L^{2}$ norm. By [F1], $L^{2} (D)$ is complete; by [F3] and step 2.1, its closed subspace $L_{0}^{2} (D)$ is complete. Therefore it is a real Hilbert space with the restricted inner product.

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

Smooth-neighborhood density in H1 of a simplex

Statement

Assume Countable Choice. Let $K \subset R^{d}$ be a nondegenerate open simplex. For every $u \in H^{1} (K; R)$ and every $ε > 0$ , there is a function $u_{ε}$ smooth on an open neighborhood of $\overline{K}$ with $∥ u_{ε} - u ∥_{H^{1} (K)} < ε$ .

Facts & Assumptions

Given: Countable Choice, a nondegenerate simplex $K$ , a real Sobolev class $u \in W^{1, 2} (K)$ , and $ε > 0$ .

[A1]

Countable Choice is inherited through the published Sobolev, density, change-of-variables, and mollification interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$W^{1, 2} (K)$ consists of real $L^{2}$ classes with first weak derivatives in $L^{2} (K)$ . (Integer-order Sobolev spaces and their norms)

[F2]

For every $g \in L^{2} (R^{d})$ and $η > 0$ , some $ϕ \in C_{c}^{\infty} (R^{d})$ satisfies $∥ g - ϕ ∥_{2} < η$ . (Smooth compactly supported functions of an open set are dense in $L^{2}$ )

[F3]

An invertible linear map scales Lebesgue measure by its absolute determinant, and translations preserve measurability and measure. Thus an affine map $T (x) = A x + b$ satisfies $∣ T (E) ∣ = ∣ \det A ∣ ∣ E ∣$ . Applying this identity to indicator functions, then nonnegative simple functions and their increasing approximations, and finally positive and negative parts gives $\int_{T (E)} f (y) d y = ∣ \det A ∣ \int_{E} f (T (x)) d x$ for integrable real $f$ (and for nonnegative measurable $f$ ). Null sets are preserved in both directions, so composition also respects almost-everywhere classes. (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ 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)

[F4]

The smooth-bump lemma gives $χ \in C_{c}^{\infty} (B_{1 / 2} (0))$ with $0 \leq χ \leq 1$ and $χ = 1$ on ${\overline{B}}_{1 / 4} (0)$ . Its support has finite measure by Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, while $B_{1 / 4} (0)$ contains the cube $(- 1 / (8 \sqrt{d}), 1 / (8 \sqrt{d}))^{d}$ , whose measure is positive by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included. Thus $m : = \int χ$ is finite and positive. The normalized bump $ρ : = χ / m$ has unit mass and support in $B_{1} (0)$ ; its rescalings are an $L^{1}$ approximate identity by A unit-mass smooth bump generates an $L^{1}$ approximate identity. (A smooth bump between concentric Euclidean balls, The mollifier family generated by a unit-mass smooth bump)

[F5]

For every $g \in L^{2} (R^{d})$ , $∥ ρ_{η} * g - g ∥_{2} \to 0$ as $η \to 0^{+}$ . (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ )

[F6]

Interior mollification commutes with first weak derivatives: if $w \in H^{1} (U)$ and the $η$ -neighborhood of $K$ lies inside $U$ , then $D_{i} (ρ_{η} * w) = ρ_{η} * (D_{i} w)$ on $K$ . (Interior mollification commutes with weak derivatives)

[F7]

Convolution of a locally integrable function with a smooth compactly supported mollifier is smooth and derivatives pass under the integral sign. (Convolution with a mollifier is smooth, and derivatives pass under the integral sign)

[F8]

An $L^{2}$ function on a bounded set is locally integrable by Holder's inequality and finiteness of Lebesgue measure on bounded sets. (Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure)

Proof

1.1A1F1F3constructalgebra

Choose $c \in K$ and $s \in (0, 1)$ , and put $T_{s} (x) = c + s (x - c)$ and $U_{s} = T_{s}^{- 1} (K)$ . Since $K$ is convex and $c$ is interior, $T_{s} (\overline{K})$ is a compact subset of $K$ ; hence $U_{s}$ is an open simplex containing $\overline{K}$ with positive distance from $\overline{K}$ to $U_{s}^{c}$ . Define $u_{s} = u \circ T_{s}$ on $U_{s}$ . For $ψ \in C_{c}^{\infty} (U_{s})$ , change variables $y = T_{s} (x)$ in the weak derivative identity of $u$ ; with $\tilde{ψ} (y) = ψ (T_{s}^{- 1} y)$ one has $D_{y_{i}} \tilde{ψ} = s^{- 1} (D_{x_{i}} ψ) \circ T_{s}^{- 1}$ , so the identity gives $D_{i} u_{s} = s (D_{i} u) \circ T_{s}$ distributionally.

1.2F2F3algebraalgebra

For any $g \in L^{2} (K)$ , extend it by zero to $R^{d}$ . Given $η > 0$ , choose $ϕ \in C_{c}^{\infty} (R^{d})$ with $∥ g - ϕ ∥_{L^{2} (K)} < η$ by [F2]. Because $T_{s} (K) \subset K$ , change of variables [F3] gives $∥ (g - ϕ) \circ T_{s} ∥_{L^{2} (K)} \leq s^{- d / 2} η$ , while uniform continuity of $ϕ$ on $\overline{K}$ gives $∥ ϕ \circ T_{s} - ϕ ∥_{L^{2} (K)} \to 0$ as $s \to 1^{-}$ . The triangle inequality and arbitrary $η$ therefore show $∥ g \circ T_{s} - g ∥_{L^{2} (K)} \to 0$ .

2.1F1step 1.1step 1.2algebra

Apply step 1.2 to $g = u$ and to every $g = D_{i} u$ . By step 1.1, $D_{i} u_{s} = s (D_{i} u) \circ T_{s}$ , and $s \to 1$ ; hence $u_{s} ∣_{K} \to u$ in the full $H^{1} (K)$ norm as $s \to 1^{-}$ .

3.1A1F4F5F6F7F8step 2.1choosealgebra∎

Choose $s$ close enough to $1$ that $∥ u_{s} - u ∥_{H^{1} (K)} < ε / 2$ , and let $r = dist (\overline{K}, U_{s}^{c}) > 0$ . Choose the bump from [F4] and then $0 < δ < r$ . The zero extension of $u_{s}$ is locally integrable by [F8], so $w_{δ} = (ρ_{δ} * E_{0} u_{s})$ is smooth by [F7]. On $K$ , [F6] gives $D_{i} w_{δ} = ρ_{δ} * E_{0} (D_{i} u_{s})$ . Each zero extension is in $L^{2} (R^{d})$ ; [F5] therefore gives $∥ w_{δ} - u_{s} ∥_{H^{1} (K)} \to 0$ as $δ \to 0^{+}$ . Choose $δ < r$ small enough to make this norm less than $ε / 2$ . Then $w_{δ}$ is smooth on a neighborhood of $\overline{K}$ , and the triangle inequality with step 2.1 gives $∥ w_{δ} - u ∥_{H^{1} (K)} < ε$ .

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

Weak normal trace by Green duality

Statement

Assume Countable Choice for the Sobolev-space interfaces below. Let $D \subseteq R^{d}$ be open and $v \in H (div; D)$ . Define the trace quotient $T_{D} : = H^{1} (D) / H_{0}^{1} (D), ∥ [w] ∥_{T_{D}} : = \inf_{z \in H_{0}^{1} (D)} ∥ w + z ∥_{H^{1} (D)} .$ Then the formula

⟨ γ_{n} v, [w] ⟩ : = \int_{D} (v \cdot \nabla w + (div v) w) d x

is independent of the representative $w$ , defines a bounded real linear functional $γ_{n} v \in T_{D}^{'}$ , and satisfies $∥ γ_{n} v ∥_{T_{D}^{'}} \leq ∥ v ∥_{H (div; D)}$ . This functional is the weak normal trace; no pointwise boundary value is asserted.

Facts & Assumptions

Given: Countable Choice; an open set $D$ ; a real $v \in H (div; D)$ ; $H^{1} (D) = W^{1, 2} (D)$ and $H_{0}^{1} (D) = W_{0}^{1, 2} (D)$ .

[A1]

Countable Choice is assumed by the cited Sobolev class and zero-boundary closure interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Integer-order Sobolev spaces and their norms, Zero-boundary Sobolev space as a norm closure)

[F1]

The distributional divergence satisfies $\int_{D} v \cdot \nabla ϕ + (div v) ϕ = 0$ for every $ϕ \in C_{c}^{\infty} (D; R)$ . (The real H(div) space)

[F2]

$H_{0}^{1} (D)$ is the $H^{1}$ -norm closure of $C_{c}^{\infty} (D; R)$ . (Zero-boundary Sobolev space as a norm closure)

[F3]

The real $L^{2}$ Cauchy–Schwarz inequality controls both scalar pairings and the componentwise vector pairing. (Cauchy-Schwarz inequality for $L^{2}$ )

Proof

1.1F1F3algebra

For $w \in H^{1} (D)$ put $L_{v} (w) : = \int_{D} (v \cdot \nabla w + (div v) w)$ . The two integrals are finite by [F3], and Cauchy–Schwarz in $R^{2}$ gives $∣ L_{v} (w) ∣ \leq ∥ v ∥_{2} ∥ \nabla w ∥_{2} + ∥ div v ∥_{2} ∥ w ∥_{2} \leq ∥ v ∥_{H (div; D)} ∥ w ∥_{H^{1} (D)} .$ Thus $L_{v}$ is a bounded real linear functional on $H^{1} (D)$ .

2.1A1F1F2step 1.1choosealgebra

For every $ϕ \in C_{c}^{\infty} (D; R)$ , [F1] gives $L_{v} (ϕ) = 0$ . If $z \in H_{0}^{1} (D)$ , then for each $ε > 0$ , [F2] gives such a $ϕ$ with $∥ z - ϕ ∥_{H^{1}} < ε$ ; step 1.1 yields $∣ L_{v} (z) ∣ = ∣ L_{v} (z - ϕ) ∣ \leq ∥ v ∥_{H (div)} ε$ . Since this holds for every $ε > 0$ , $L_{v} (z) = 0$ . Therefore $L_{v}$ is constant on each coset of $H_{0}^{1} (D)$ and defines $γ_{n} v$ on $T_{D}$ .

3.1step 1.1step 2.1algebra∎

For any $[w] \in T_{D}$ , step 2.1 makes $L_{v} (w)$ the same for every representative. Applying step 1.1 to every representative $w + z$ , $z \in H_{0}^{1} (D)$ , and taking the infimum of their $H^{1}$ norms gives $∣ ⟨ γ_{n} v, [w] ⟩ ∣ \leq ∥ v ∥_{H (div; D)} ∥ [w] ∥_{T_{D}}$ . Hence $γ_{n} v$ is bounded with the claimed norm.

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

Piola divergence and facet-flux identities

Statement

Let $F_{K} (\hat{x}) = A_{K} \hat{x} + b_{K}$ be as in Contravariant Piola map, and let $\hat{v} \in C^{1} (\overline{\hat{K}}; R^{d})$ . Set $v = P_{K} \hat{v}$ . Then

({div}_{x} v) (F_{K} (\hat{x})) = J_{K}^{- 1} {div}_{\hat{x}} \hat{v} (\hat{x}) .

For each facet $\hat{F}$ mapped to $F$ , with the corresponding oriented normals, the integrated flux is preserved:

\int_{F} v \cdot n_{F} d s = \int_{\hat{F}} \hat{v} \cdot {\hat{n}}_{\hat{F}} d \hat{s} .

Facts & Assumptions

Given: An orientation-preserving affine bijection $F_{K}$ with matrix $A_{K}$ , determinant $J_{K} > 0$ , a smooth reference vector field $\hat{v}$ , and its contravariant Piola push-forward.

[F1]

The push-forward is $v (F_{K} (\hat{x})) = J_{K}^{- 1} A_{K} \hat{v} (\hat{x})$ . (Contravariant Piola map)

[F2]

The chain rule for total derivatives states $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ for differentiable maps between finite-dimensional real spaces. (The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ )

Proof

1.1F1F2algebra

Write $v_{i} (F_{K} (\hat{x})) = J_{K}^{- 1} \sum_{j} (A_{K})_{i j} {\hat{v}}_{j} (\hat{x})$ . The chain rule [F2] for $F_{K}^{- 1}$ gives $\partial_{x_{i}} = \sum_{k} (A_{K}^{- 1})_{k i} \partial_{{\hat{x}}_{k}}$ ; since $A_{K}$ is constant, summing the derivatives gives $({div}_{x} v) \circ F_{K} = J_{K}^{- 1} \sum_{i, k, j} (A_{K}^{- 1})_{k i} (A_{K})_{i j} \partial_{{\hat{x}}_{k}} {\hat{v}}_{j} = J_{K}^{- 1} \sum_{j} \partial_{{\hat{x}}_{j}} {\hat{v}}_{j} .$ The middle equality uses $A_{K}^{- 1} A_{K} = I$ .

2.1F1F2constructalgebra∎

Choose oriented tangent vectors ${\hat{t}}_{1}, \dots, {\hat{t}}_{d - 1}$ on $\hat{F}$ . Under $F_{K}$ the tangent vectors are $A_{K} {\hat{t}}_{i}$ ; multilinearity of the determinant gives the oriented area-normal transformation $n_{F} d s = cof (A_{K}) {\hat{n}}_{\hat{F}} d \hat{s}$ . Since $A_{K}$ is invertible, $cof (A_{K}) = J_{K} A_{K}^{- T}$ . Therefore the Piola flux element satisfies $v \cdot n_{F} d s = (J_{K}^{- 1} A_{K} \hat{v}) \cdot (J_{K} A_{K}^{- T} {\hat{n}}_{\hat{F}}) d \hat{s} = \hat{v} \cdot {\hat{n}}_{\hat{F}} d \hat{s} .$ Integrating over the corresponding facets proves the flux identity.

Commuting diagram

For smooth fields, the proved identity is represented by

diagram failed to render:
C^1(\overline{\widehat K};\mathbb R^d) \arrow[r,"\operatorname{div}"] \arrow[d,"\mathcal P_K"'] & C(\widehat K) \arrow[d,"J_K^{-1}(\cdot)\circ F_K^{-1}"] \\
C^1(\overline K;\mathbb R^d) \arrow[r,"\operatorname{div}"'] & C(K)

The top and bottom arrows are divergence, the left arrow is the contravariant Piola push-forward, and the right arrow is scalar pull-forward with the Jacobian factor. The proof above establishes the commutation algebraically; the diagram is only a record of the identity.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Zero-normal-trace H(div) space

Definition

Assume Countable Choice for the Sobolev and weak normal trace interfaces. For an open set $D \subseteq R^{d}$ , let $T_{D} = H^{1} (D) / H_{0}^{1} (D)$ be the trace quotient of Weak normal trace by Green duality, and let $γ_{n} : H (div; D) \to T_{D}^{'}$ be its weak normal trace. Define

H_{0} (div; D) : = {v \in H (div; D) : γ_{n} v = 0 in T_{D}^{'}} .

Thus zero normal trace means that the Green-duality functional vanishes on every H1 trace class; no pointwise normal value is required.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Affine simplicial meshes and RT/BDM spaces

Definition

Assume Countable Choice. Let $d \in {2, 3}$ and let $D \subset R^{d}$ be a bounded, connected polyhedral Lipschitz domain. A matching affine simplicial mesh and its shape regularity have the exact meaning of Matching affine simplicial meshes and shape regularity; additionally, call the mesh facet-connected when its dual graph (one vertex per cell and one edge per shared $(d - 1)$ -facet) is connected. Write $h = \max_{K \in T_{h}} diam K$ .

For $k \geq 0$ , $P_{k} (K)$ is the real scalar polynomial space of total degree at most $k$ , and ${\tilde{P}}_{k}$ is the real homogeneous scalar polynomial space of degree exactly $k$ (for $k = 0$ , the constants). Define ${RT}_{k} (K) = P_{k} (K; R^{d}) + x {\tilde{P}}_{k} (K), {BDM}_{k} (K) = P_{k} (K; R^{d}) (k \geq 1) .$ For either local family $X_{k}$ , its global H(div)-conforming space is

The zero-normal RT0 flux space and mean-zero P0 pressure space are $V_{h} : = {RT}_{0} (T_{h}) \cap H_{0} (div; D), Q_{h} : = {q \in L^{2} (D) : q ∣_{K} is constant for every K, \int_{D} q = 0} .$ For each mesh facet $F$ , fix one unit normal $n_{F}$ ; on a cell boundary use the outward unit normal $n_{K}$ , which equals either $n_{F}$ or $- n_{F}$ . The notation $P_{0} (T_{h})$ means cellwise constants. No existence of a mesh for every $D$ is asserted by this definition.

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

Zero mean of the divergence under zero normal trace

Statement

Assume Countable Choice. Let $D \subset R^{d}$ be bounded and open. If $v \in H_{0} (div; D)$ , then $div v \in L^{1} (D)$ and $\int_{D} div v d x = 0.$

Facts & Assumptions

Given: Countable Choice, a bounded open set $D$ , and $v$ with zero weak normal trace.

[A1]

Countable Choice is inherited from the Sobolev and trace-quotient interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$H_{0} (div; D)$ is the kernel of the weak normal trace. (Zero-normal-trace H(div) space)

[F2]

Green duality defines $⟨ γ_{n} v, [w] ⟩ = \int_{D} (v \cdot \nabla w + (div v) w)$ . (Weak normal trace by Green duality)

[F3]

Bounded sets have finite Lebesgue measure, so the constant function $1$ belongs to $L^{2} (D)$ and has zero weak gradient in $H^{1} (D)$ . (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, Integer-order Sobolev spaces and their norms)

[F4]

Holder's inequality implies $∥ g ∥_{L^{1} (D)} \leq ∣ D ∣^{1 / 2} ∥ g ∥_{L^{2} (D)}$ for $g \in L^{2} (D)$ . (Holder's inequality for integrals, including the endpoint cases)

[F5]

The divergence of an H(div) field belongs to $L^{2} (D)$ . (The real H(div) space)

Proof

1.1A1F3F4F5construct

Since $div v \in L^{2} (D)$ by [F5], and $∣ D ∣ < \infty$ by [F3], [F4] shows $div v \in L^{1} (D)$ . The constant class $1$ is an admissible element of $H^{1} (D)$ by [F3] and the definition of the weak Sobolev norm, so its trace class $[1]$ lies in $T_{D}$ .

2.1F2step 1.1algebra

Apply [F2] with $w = 1$ . Since $\nabla 1 = 0$ , Green duality gives $⟨ γ_{n} v, [1] ⟩ = \int_{D} div v d x$ .

3.1F1step 2.1∎

By [F1], $γ_{n} v = 0$ as an element of $T_{D}^{'}$ acting on $T_{D}$ , so its value at $[1] \in T_{D}$ is zero. The identity in step 2.1 therefore gives $\int_{D} div v = 0$ .

LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passOpen item page →

H1-zero fields have zero weak H(div) normal trace

Statement

Assume Countable Choice and let $D \subset R^{d}$ be open. Every real vector field $w \in H_{0}^{1} (D; R^{d})$ belongs to $H_{0} (div; D)$ , where the latter is the kernel of the weak normal trace. In particular, $div w \in L^{2} (D)$ and its weak normal trace vanishes on the whole boundary trace quotient.

Facts & Assumptions

Given: Countable Choice, an open set $D$ , and $w \in H_{0}^{1} (D; R^{d})$ .

[A1]

Countable Choice selects the compactly supported smooth approximants in the H1-zero closure. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

Each component of $H^{1} (D; R^{d})$ has its first weak derivatives in $L^{2} (D)$ ; their finite sum is the distributional divergence. (Integer-order Sobolev spaces and their norms, The real H(div) space)

[F2]

$H_{0}^{1} (D)$ is the $H^{1}$ -norm closure of $C_{c}^{\infty} (D)$ . (Zero-boundary Sobolev space as a norm closure)

[F3]

Green duality defines the weak normal trace by $⟨ γ_{n} w, [ϕ] ⟩ = \int_{D} (w \cdot \nabla ϕ + (div w) ϕ)$ . (Weak normal trace by Green duality)

[F4]

$H_{0} (div; D)$ is exactly the kernel of this weak normal trace. (Zero-normal-trace H(div) space)

[F5]

Cauchy–Schwarz controls each volume pairing in Green duality. (Cauchy-Schwarz inequality for $L^{2}$ )

Proof

1.1F1constructalgebra

Since each component of $w$ and each first weak derivative is in $L^{2} (D)$ by [F1], the finite sum $div w = \sum_{j = 1}^{d} \partial_{j} w_{j}$ lies in $L^{2} (D)$ . Thus $w \in H (div; D)$ by its definition.

2.1A1F2F3F4F5choosealgebra∎

By [A1,F2], choose vector fields $w_{n} \in C_{c}^{\infty} (D; R^{d})$ with $w_{n} \to w$ in $H^{1}$ . For arbitrary $ϕ \in H^{1} (D)$ , the weak-derivative identity for $ϕ$ , tested against each compactly supported component $(w_{n})_{j}$ , gives $\int_{D} w_{n} \cdot \nabla ϕ + \int_{D} (div w_{n}) ϕ = 0.$ The H1 convergence implies $w_{n} \to w$ in $L^{2}$ and $div w_{n} \to div w$ in $L^{2}$ . Specifically, [F5] bounds the difference of the Green pairings by $∥ w_{n} - w ∥_{2} ∥ \nabla ϕ ∥_{2} + ∥ div w_{n} - div w ∥_{2} ∥ ϕ ∥_{2} ⟶ 0.$ Thus $\int_{D} w \cdot \nabla ϕ + \int_{D} (div w) ϕ = 0$ . This holds for every $ϕ \in H^{1} (D)$ ; hence [F3] gives $γ_{n} w = 0$ , and [F4] implies $w \in H_{0} (div; D)$ .

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H(div) is Hilbert and H0(div) is closed

Statement

Assume Countable Choice. For every open $D \subseteq R^{d}$ , the real space $H (div; D)$ with the graph inner product

(v, w)_{H (div)} : = (v, w)_{L^{2} (D; R^{d})} + (div v, div w)_{L^{2} (D)}

is a Hilbert space. Its subspace $H_{0} (div; D)$ of Zero-normal-trace H(div) space is closed, hence is Hilbert with the restricted inner product.

Facts & Assumptions

Given: Countable Choice and an open set $D \subseteq R^{d}$ .

[A1]

Countable Choice is inherited from the real $L^{2}$ Hilbert-space interface used for graph completeness. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

Real $L^{2}$ is complete for its integral inner product and quotient norm. ( $L^{2}$ with the integral pairing is a Hilbert space)

[F2]

The graph norm is $∥ v ∥_{H (div)}^{2} = ∥ v ∥_{2}^{2} + ∥ div v ∥_{2}^{2}$ . (The real H(div) space)

[F3]

The weak normal trace is bounded by the H(div) graph norm. (Weak normal trace by Green duality)

[F4]

Holder's inequality passes the two $L^{2}$ pairings to their limits. (Holder's inequality for integrals, including the endpoint cases)

[F5]

A closed subspace of a Banach space is complete in the restricted norm. (A closed subspace of a Banach space is Banach)

Proof

1.1A1F1F2F4algebraalgebra

Let $(v_{n})$ be Cauchy in the graph norm. By [F1] applied componentwise, $v_{n} \to v$ in $L^{2} (D; R^{d})$ and $div v_{n} \to g$ in $L^{2} (D)$ for some $v, g$ . For every $ϕ \in C_{c}^{\infty} (D; R)$ , the distributional identity gives $\int_{D} v_{n} \cdot \nabla ϕ = - \int_{D} (div v_{n}) ϕ$ . Holder's inequality [F4] passes both pairings to their $L^{2}$ limits, so $\int_{D} v \cdot \nabla ϕ = - \int_{D} g ϕ$ . Hence $g = div v$ distributionally, $v \in H (div; D)$ , and [F2] gives graph-norm convergence $v_{n} \to v$ .

2.1F1F2step 1.1algebra

The displayed graph inner product is the sum of the real $L^{2}$ inner products of $v$ and its divergence, so it induces exactly the graph norm. By step 1.1 that norm is complete; therefore H(div) is Hilbert.

3.1F3F5step 2.1algebra∎

By [F3], if $v_{n} \in H_{0} (div; D)$ and $v_{n} \to v$ in H(div), then $γ_{n} v_{n} \to γ_{n} v$ in $T_{D}^{'}$ . Since each $γ_{n} v_{n} = 0$ , the limit trace is zero, so $v \in H_{0} (div; D)$ . This proves closedness; [F5] and step 2.1 show that the restricted inner-product space is complete.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

No-flow mixed Darcy problem and coefficient norms

Definition

Assume Countable Choice. Let $D \subset R^{d}$ be a bounded connected Lipschitz polyhedral domain, $d \in {2, 3}$ , and set $V : = H_{0} (div; D)$ and $Q : = L_{0}^{2} (D)$ with the real graph and integral inner products from The real H(div) space and The real zero-mean $L^{2}$ space is Hilbert. A permeability tensor $K (x)$ is a measurable real symmetric $d \times d$ matrix field for which there are constants $0 < κ_{\min} \leq κ_{\max} < \infty$ satisfying $κ_{\min} ∣ ξ ∣^{2} \leq ξ \cdot K (x) ξ \leq κ_{\max} ∣ ξ ∣^{2}$ for almost every $x \in D$ and every $ξ \in R^{d}$ . It is positive definite almost everywhere, so $K^{- 1}$ is measurable and has eigenvalues between $κ_{\max}^{}$ and $κ_{\min}^{}$ . Define $a (u, v) : = \int_{D} K^{- 1} u \cdot v, b (v, q) : = - \int_{D} q div v,$ and the weighted flux norm $∥ v ∥_{K^{- 1}} : = a (v, v)^{1 / 2}$ . For data $F \in V^{'}$ and $f \in Q$ , the no-flow mixed Darcy problem is to find $(u, p) \in V \times Q$ such that $a (u, v) + b (v, p) = F (v) (v \in V), b (u, q) = - \int_{D} f q (q \in Q) .$ The pressure space's zero-mean constraint fixes the additive constant (Neumann) gauge; $u \in H_{0} (div; D)$ encodes zero weak normal flux on the full boundary. In the standard source-only Darcy problem used for hybridization, $F = 0$ .

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Cell averaging is the $L^{2}$ projection onto P0

Statement

For a finite conforming simplicial mesh of bounded $D$ , define $P_{0} : L^{2} (D) \to P_{0} (T_{h})$ cellwise by $(P_{0} g) ∣_{K} = ∣ K ∣^{- 1} \int_{K} g .$ Then $P_{0}$ is linear, equals the identity on piecewise constants, is the $L^{2}$ -orthogonal projection onto that space, satisfies $∥ P_{0} g ∥_{2} \leq ∥ g ∥_{2}$ , and preserves the integral $\int_{D} P_{0} g = \int_{D} g$ .

Facts & Assumptions

Given: A finite conforming mesh of bounded $D$ and $g \in L^{2} (D)$ .

[F1]

The mesh cells have positive finite measure, disjoint interiors, and cover $D$ up to their measure-zero affine faces. (Matching affine simplicial meshes and shape regularity, Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null)

[F2]

Cauchy–Schwarz gives $∣ \int_{K} g ∣ \leq ∣ K ∣^{1 / 2} ∥ g ∥_{L^{2} (K)}$ . (Cauchy-Schwarz inequality for $L^{2}$ )

Proof

1.1F1givenalgebra

The cell integral is linear, so the formula defines a linear piecewise-constant map. If $g$ is constant on each cell, its cell average is the same constant, so $P_{0} g = g$ . For any piecewise constant $q_{h}$ , $\int_{D} (g - P_{0} g) q_{h} = \sum_{K} q_{K} (\int_{K} g - ∣ K ∣ ∣ K ∣^{- 1} \int_{K} g) = 0.$ Thus the range is orthogonal to the residual and $P_{0}$ is the orthogonal projection onto $P_{0} (T_{h})$ .

2.1F1F2algebra∎

By [F2], $∣ K ∣ ∣ (P_{0} g) ∣_{K} ∣^{2} = ∣ K ∣^{- 1} ∣ \int_{K} g ∣^{2} \leq \int_{K} ∣ g ∣^{2}$ . Summing over cells gives $∥ P_{0} g ∥_{2}^{2} \leq ∥ g ∥_{2}^{2}$ . Summing the defining averages multiplied by cell measures gives $\int_{D} P_{0} g = \sum_{K} \int_{K} g = \int_{D} g$ .

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Piola maps RT and BDM polynomial families onto the physical cell

Statement

For any $d \geq 1$ , write $P_{k}$ for real scalar polynomials of total degree at most $k$ and ${\tilde{P}}_{k}$ for homogeneous scalar polynomials of degree $k$ , including zero, and define the local families ${RT}_{k} = P_{k}^{d} + x {\tilde{P}}_{k}$ and ${BDM}_{k} = P_{k}^{d}$ .

Let $F (\hat{x}) = A \hat{x} + b$ be an orientation-preserving affine bijection between nondegenerate $d$ -simplices, with $J = \det A > 0$ , and use the contravariant Piola map of Contravariant Piola map. For every $k \geq 0$ , it maps ${RT}_{k} (\hat{K})$ bijectively onto ${RT}_{k} (K)$ , and for every $k \geq 1$ it maps ${BDM}_{k} (\hat{K})$ bijectively onto ${BDM}_{k} (K)$ .

Facts & Assumptions

Given: The affine map $F$ , its matrix $A$ , determinant $J > 0$ , and the polynomial families in Affine simplicial meshes and RT/BDM spaces.

[F1]

The Piola push-forward is $(P \hat{v}) (x) = J^{- 1} A \hat{v} (A^{- 1} (x - b))$ , and its inverse is $\hat{v} (\hat{x}) = J A^{- 1} v (A \hat{x} + b)$ . (Contravariant Piola map)

[F2]

The local polynomial formulas in the Statement agree, in dimensions two and three, with the local families in Affine simplicial meshes and RT/BDM spaces. Here they define the notation for all $d \geq 1$ ; no global mesh or domain is needed for this cellwise assertion.

Proof

1.1F1F2algebra

For BDM, if $\hat{v} \in P_{k} (\hat{K})^{d}$ , then $\hat{v} \circ F^{- 1}$ is a vector polynomial of degree at most $k$ in $x$ because $F^{- 1}$ is affine. Multiplication by the constant matrix $J^{- 1} A$ preserves this degree, so [F1] maps BDM into BDM. Applying the same argument to the inverse affine map proves onto-ness.

1.2F1F2algebra

Write an RT field as $\hat{v} (\hat{x}) = \hat{p} (\hat{x}) + \hat{x} \hat{q} (\hat{x})$ with $\hat{p} \in P_{k}^{d}$ and homogeneous $\hat{q}$ of degree $k$ . The push-forward of the first term is in $P_{k}^{d}$ . For the second term, use $\hat{x} = A^{- 1} (x - b)$ : its image is $J^{- 1} (x - b) \hat{q} (A^{- 1} (x - b))$ . The degree- $k$ homogeneous part of $\hat{q} (A^{- 1} (x - b))$ is $q_{k} (x) : = \hat{q} (A^{- 1} x)$ , with remainder of degree at most $k - 1$ (zero when $k = 0$ ). Thus $(x - b) \hat{q} (A^{- 1} (x - b)) = x q_{k} (x) + p_{k} (x)$ for some vector polynomial $p_{k}$ of degree at most $k$ : all terms from $x$ times the remainder and from $b$ times the full scalar polynomial have degree at most $k$ . Therefore the image lies in ${RT}_{k} (K)$ .

2.1F1step 1.2algebra∎

The inverse Piola map is the contravariant Piola map for the orientation-preserving affine bijection $F^{- 1}$ , up to the displayed inverse formula in [F1]. The same polynomial decomposition from step 1.2 applied to $F^{- 1}$ maps ${RT}_{k} (K)$ into ${RT}_{k} (\hat{K})$ . Hence the forward inclusion is onto.

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The RT0/P0 discrete kernel is exactly divergence-free

Statement

Assume Countable Choice and the mesh conventions of Affine simplicial meshes and RT/BDM spaces. With $V_{h} = {RT}_{0} (T_{h}) \cap H_{0} (div; D)$ , $Q_{h} = P_{0} (T_{h}) \cap L_{0}^{2} (D)$ and $b (v_{h}, q_{h}) = - \int_{D} q_{h} div v_{h}$ , the discrete kernel $Z_{h} : = {v_{h} \in V_{h} : b (v_{h}, q_{h}) = 0 for every q_{h} \in Q_{h}}$ is exactly ${v_{h} \in V_{h} : div v_{h} = 0}$ .

Facts & Assumptions

Given: A conforming mesh, the spaces $V_{h}, Q_{h}$ , and $v_{h} \in V_{h}$ .

[A1]

Countable Choice is inherited from the zero-normal trace and Sobolev interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The divergence of an H(div) field with zero normal trace has zero integral. (Zero mean of the divergence under zero normal trace)

[F2]

By definition, $V_{h}$ consists of global H(div) fields whose cell restrictions are RT0; their cellwise divergence is constant, and membership in $V_{h}$ includes zero normal trace. (Affine simplicial meshes and RT/BDM spaces, Zero-normal-trace H(div) space)

Proof

1.1A1givenF1F2

By [F2], $div v_{h}$ is cellwise constant; by [F1], its global mean is zero. Thus $div v_{h} \in Q_{h}$ .

2.1givenstep 1.1algebracases∎

If $v_{h} \in Z_{h}$ , test its defining orthogonality with the admissible choice $q_{h} = div v_{h}$ from step 1.1. Then $0 = b (v_{h}, q_{h}) = - ∥ div v_{h} ∥_{L^{2} (D)}^{2}$ , so its divergence vanishes in $L^{2}$ . Conversely, if $div v_{h} = 0$ , the defining integral $b (v_{h}, q_{h})$ is zero for every $q_{h} \in Q_{h}$ . This proves the equality, including the case $Q_{h} = {0}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Hybridized no-flow RT0/P0 Darcy system

Definition

Assume Countable Choice.

Assume the no-flow Darcy data of No-flow mixed Darcy problem and coefficient norms and a facet-connected conforming mesh from Affine simplicial meshes and RT/BDM spaces. Let $F_{h}^{i}$ be its interior facets. On each cell $K$ , let $W_{K} : = {v \in {RT}_{0} (K) : v \cdot n_{K} = 0 on every boundary facet of D contained in \partial K} .$ The broken flux space is $W_{h} : = \prod_{K \in T_{h}} W_{K}$ , the cell pressure space is $M_{h} : = \prod_{K \in T_{h}} R$ , and the skeleton multiplier space is $Λ_{h} : = \prod_{F \in F_{h}^{i}} R$ . The scalar $λ_{F}$ is constant on facet $F$ and uses the fixed orientation only to label that facet; local equations use the outward normal of each cell.

For the source-only problem $F = 0$ , a hybrid solution $(u_{h}, p_{h}, λ_{h}) \in W_{h} \times M_{h} \times Λ_{h}$ for $f \in Q$ satisfies, for every cell $K$ , every $v_{K} \in W_{K}$ , and every interior facet $F = K^{+} \cap K^{-}$ , $\int_{K} K^{- 1} u_{K} \cdot v_{K} - \int_{K} p_{K} div v_{K} + \sum_{F \subset \partial K, F \in F_{h}^{i}} λ_{F} \int_{F} v_{K} \cdot n_{K} = 0,$ $\int_{K} div u_{K} = \int_{K} f,$ $\int_{F} (u_{K^{+}} \cdot n_{K^{+}} + u_{K^{-}} \cdot n_{K^{-}}) = 0.$ Because each RT0 normal component is constant on a facet, the last equation is pointwise normal-flux matching. With physical-boundary flux zero, the transformation $p_{K} \mapsto p_{K} + c$ for all cells and $λ_{F} \mapsto λ_{F} + c$ leaves the hybrid equations unchanged: in the local first equation the change is $- c \int_{K} div v_{K} + c \sum_{F interior} \int_{F} v_{K} \cdot n_{K} = 0$ by Gauss–Green and the zero physical-boundary flux; the mass and jump equations do not use pressure or multipliers. This is the common pressure gauge.

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H1 traces agree across interior mesh facets

Statement

Assume Countable Choice. Let $D$ be a polyhedral open set with a conforming affine simplicial mesh, let $K_{+}$ and $K_{-}$ share an interior facet $F$ , and let $w \in H^{1} (D)$ . The elementwise $L^{2} (F)$ traces of $w ∣_{K_{+}}$ and $w ∣_{K_{-}}$ agree almost everywhere on $F$ .

Facts & Assumptions

Given: Countable Choice, adjacent mesh cells $K_{+}, K_{-}$ , their shared interior facet $F$ , and one global Sobolev class $w \in H^{1} (D)$ .

[A1]

Countable Choice is inherited from the Sobolev, smooth-density and mollification interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

Each element restriction has a unique bounded $L^{2}$ facet trace, obtained by continuous extension from smooth-neighborhood functions. (H1 facet trace inequality on shape-regular simplices)

[F2]

The smooth-bump lemma gives $χ \in C_{c}^{\infty} (B_{1 / 2} (0))$ with $0 \leq χ \leq 1$ and $χ = 1$ on ${\overline{B}}_{1 / 4} (0)$ . Its support is bounded, and $B_{1 / 4} (0)$ contains the cube $(- 1 / (8 \sqrt{d}), 1 / (8 \sqrt{d}))^{d}$ of positive measure by A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included; hence $0 < m : = \int χ < \infty$ by Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure. The normalized bump $ρ : = χ / m$ is a nonnegative unit-mass bump supported in $B_{1} (0)$ . Its rescalings are an $L^{1}$ approximate identity by A unit-mass smooth bump generates an $L^{1}$ approximate identity. The same smooth-bump lemma supplies cutoffs equal to one near a compact subset of an open set by a finite ball cover. (The mollifier family generated by a unit-mass smooth bump, A smooth bump between concentric Euclidean balls, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included)

[F3]

Approximate identities converge in $L^{2} (R^{d})$ , and mollification commutes with first weak derivatives in an interior neighborhood. (Every $L^{1}$ approximate identity converges to the identity in $L^{p}$ for $1 \leq p < \infty$ , Interior mollification commutes with weak derivatives)

[F4]

A Sobolev $H^{1} (R^{d})$ function is in $L^{2}$ ; on each compact set, Holder's inequality and finite measure make it $L^{1}$ . Thus $W \in H^{1} (R^{d})$ is locally integrable, and convolution with its smooth compactly supported mollifier is smooth. (Integer-order Sobolev spaces and their norms, Holder's inequality for integrals, including the endpoint cases, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure, Convolution with a mollifier is smooth, and derivatives pass under the integral sign)

Proof

1.1A1F2givenconstructalgebra

Fix a compact $(d - 1)$ -simplex $G$ contained in the relative interior of $F$ . A sufficiently small open tube around $G$ has closure in $D$ . Cover its closure by finitely many balls whose larger concentric balls still have closure in $D$ ; [F2] supplies a smooth cutoff equal to one on each smaller ball and supported in its larger ball. The function $χ = 1 - \prod_{i} (1 - χ_{i})$ is compactly supported in $D$ and equals one near $G$ . The zero extension $W$ of $χ w$ belongs to $H^{1} (R^{d})$ : for a test function $ψ$ , the weak product rule follows directly from $\int_{D} χ w \partial_{i} ψ = \int_{D} w \partial_{i} (χ ψ) - \int_{D} w (\partial_{i} χ) ψ = - \int_{D} (χ \partial_{i} w + w \partial_{i} χ) ψ,$ and compact support of $χ$ removes any boundary term on extension.

2.1A1F1F2F3F4step 1.1algebra

Let $W_{ε} = ρ_{ε} * W$ with the mollifier of [F2]. By [F3] applied to $W$ and its weak derivatives, and by derivative commutation on the tube where $χ = 1$ , $W_{ε} \to w$ in $H^{1}$ on every smaller tube around $G$ . Each $W_{ε}$ is smooth by [F4]. Cover $G$ by finitely many smaller simplices on each side whose shared facet portions lie in that tube. The trace bound [F1] on each smaller simplex shows the traces of $W_{ε}$ converge in $L^{2}$ there to the local traces of $w$ . By locality and uniqueness of the bounded trace extension, these are the restrictions of the elementwise traces defined on $K_{+}$ and $K_{-}$ . The smooth $W_{ε}$ has identical restrictions from both sides, so the two traces are equal; their limits are equal as well.

3.1A1step 2.1algebra∎

Countably many compact facet simplices $G$ exhaust the relative interior of $F$ ; its boundary has $(d - 1)$ -dimensional measure zero. Step 2.1 gives equality on each such patch, hence equality almost everywhere on all of $F$ .

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RT0 facet-flux unisolvence and global H(div) conformity

Statement

Assume Countable Choice and the mesh conventions of Affine simplicial meshes and RT/BDM spaces. On each $d$ -simplex $K$ , ${RT}_{0} (K) = {a + c x : a \in R^{d}, c \in R}$ has dimension $d + 1$ , and the $d + 1$ integrated outward facet-normal fluxes determine its element uniquely. A piecewise RT0 field belongs to $H (div; D)$ exactly when its oriented integrated fluxes match across every shared facet; such a field belongs to $H_{0} (div; D)$ exactly when its normal component (equivalently its integrated outward flux) also vanishes on every boundary facet.

Facts & Assumptions

Given: Countable Choice, a conforming affine simplicial mesh, and piecewise RT0 vector polynomials with the stated facet flux conditions.

[A1]

Countable Choice is inherited through the Sobolev trace and density interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

RT0 on a cell is $P_{0} (K)^{d} + x P_{0} (K)$ , and each simplex has $d + 1$ facets. (Affine simplicial meshes and RT/BDM spaces)

[F2]

Piola preserves RT0 and its divergence and oriented facet flux. (Piola maps RT and BDM polynomial families onto the physical cell, Piola divergence and facet-flux identities)

[F3]

On a smooth field on a simplex, the integral of its divergence equals the sum of its outward facet fluxes. (Gauss–Green formula on an affine simplex)

[F4]

H1 functions on a cell have $L^{2}$ facet traces; the traces from adjacent cells agree for a global H1 function. (H1 facet trace inequality on shape-regular simplices, H1 traces agree across interior mesh facets)

[F5]

Smooth-neighborhood functions are dense in H1 of each cell. (Smooth-neighborhood density in H1 of a simplex)

[F6]

The weak normal trace on the global domain is Green duality, and H0(div) is its zero kernel. (Weak normal trace by Green duality, Zero-normal-trace H(div) space)

[F7]

The regular-distribution map is injective on $L_{l o c}^{1}$ modulo almost-everywhere equality under Countable Choice. An $L^{2}$ function is locally integrable: on compact $C$ , finite Lebesgue measure and Cauchy–Schwarz give $\int_{C} ∣ h ∣ \leq ∣ C ∣^{1 / 2} ∥ h ∥_{2}$ . (Locally integrable functions embed in distributions, Cauchy-Schwarz inequality for $L^{2}$ , Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure)

Proof

1.1A1F1F3algebra

On the reference simplex ${\hat{S}}_{d} = {x_{i} > 0, \sum_{i} x_{i} < 1}$ , each coordinate facet has outward normal $- e_{i}$ and area $1 / (d - 1)!$ , while the top facet has outward normal $(1, \dots, 1) / \sqrt{d}$ and area $\sqrt{d} / (d - 1)!$ . For $\hat{v} = a + c x$ , the constant-vector facet fluxes sum to zero by [F3], and the flux sum of $c x$ is $c / (d - 1)! = d c ∣ {\hat{S}}_{d} ∣$ . If all facet fluxes vanish then $c = 0$ ; the coordinate-facet fluxes then force each $a_{i} = 0$ . The flux map is injective between spaces of dimension $d + 1$ , hence bijective.

2.1F1F2step 1.1algebra

For any physical simplex, [F2] transports the reference flux degrees of freedom bijectively, so the same unisolvence holds. On each facet, $(a + c x) \cdot n$ is constant because $x \cdot n$ is constant on its supporting hyperplane; equal integrated fluxes therefore give equal normal components.

3.1F1F3step 2.1algebra

Let $v$ be a piecewise RT0 field with matching shared-facet fluxes. For $ϕ \in C_{c}^{\infty} (D)$ , apply [F3] to $ϕ v$ cellwise and sum: interior facet terms cancel by step 2.1 and the common value of $ϕ$ , and boundary terms vanish because $ϕ$ is compactly supported. Thus $\int_{D} v \cdot \nabla ϕ = - \sum_{K} \int_{K} (div v) ϕ$ . The cellwise divergence is constant and in $L^{2}$ , so $v \in H (div; D)$ by definition.

3.2A1F3F7step 2.1constructcontradiction

Conversely, let $v$ be a piecewise RT0 field in H(div), with global $L^{2}$ divergence $g$ . For every test supported in one cell, the global distributional identity and cellwise Gauss–Green give zero pairing with $g$ minus that cell's polynomial divergence. This difference is in $L^{2} (K)$ and hence in $L_{l o c}^{1} (K)$ by [F7]; extending from real to complex tests by linearity, injectivity [F7] makes the difference zero almost everywhere on the cell. For arbitrary $ϕ \in C_{c}^{\infty} (D)$ , summing cellwise Green formulas and comparing with the global distributional identity yields $\sum_{F = K^{+} \cap K^{-} i n t e r i o r} \int_{F} ϕ (v_{K^{+}} \cdot n_{K^{+}} + v_{K^{-}} \cdot n_{K^{-}}) = 0$ . Each normal sum is constant on its facet. If one were nonzero, choose a nonnegative smooth test supported in a small ball around a relative-interior point of that facet, meeting no other facet, and positive on a subpatch; the sum would be a nonzero constant times a positive surface integral, a contradiction. Hence all interior fluxes match.

4.1A1F4F5F6step 2.1step 3.1constructalgebra

Suppose the interior fluxes match. For $w \in H^{1} (D)$ , approximate each restriction $w ∣_{K}$ in H1 by smooth-neighborhood functions using [F5]. The trace bound [F4] and volume convergence pass cellwise Gauss–Green to each restriction. Interior traces of $w$ agree by [F4], and the normal components of $v$ agree by step 2.1, so interior terms cancel. If $c_{F}$ denotes the constant outward normal component on a boundary facet $F$ , the resulting identity is $\int_{D} (v \cdot \nabla w + div v w) = \sum_{F \subset \partial D} c_{F} \int_{F} γ_{F} w$ . By [F6] this is the weak normal-trace functional; in particular, if every boundary flux is zero then $v \in H_{0} (div; D)$ .

5.1A1F4F6step 3.2step 4.1constructcontradiction∎

Conversely, if $v \in H_{0} (div; D)$ , its zero trace and the identity in step 4.1 give $\sum_{F \subset \partial D} c_{F} \int_{F} γ_{F} w = 0$ for every $w \in H^{1} (D)$ . For any selected boundary facet, choose a smooth function on $R^{d}$ supported in a small ball about a relative-interior point of that facet, positive on a subpatch, and meeting no other boundary facet; its restriction lies in $H^{1} (D)$ . The sum then reduces to the selected $c_{F}$ times a positive surface integral, so $c_{F} = 0$ . Thus every boundary facet flux vanishes.

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Continuous mixed Darcy well-posedness with pressure gauge

Statement

Assume Countable Choice and the hypotheses in No-flow mixed Darcy problem and coefficient norms on a bounded connected Lipschitz domain in dimension two or three. For every $F \in V^{'}$ and $f \in Q$ , there is a unique $(u, p) \in V \times Q$ solving $a (u, v) + b (v, p) = F (v), b (u, q) = - \int_{D} f q$ for all $v \in V, q \in Q$ . The graph norm on the flux and zero-mean pressure norm make these equations well-posed; the mean-zero pressure convention fixes the Neumann gauge.

Facts & Assumptions

Given: Countable Choice, $D, K, V, Q, a, b, F, f$ as in the no-flow Darcy definition.

[A1]

Countable Choice is inherited from the right-inverse and Babuška–Brezzi suppliers. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$V = H_{0} (div; D)$ and $Q = L_{0}^{2} (D)$ are real Hilbert spaces in their graph and integral norms. (H(div) is Hilbert and H0(div) is closed, The real zero-mean $L^{2}$ space is Hilbert)

[F2]

Every $v \in V$ has $\int_{D} div v = 0$ , hence $div v \in Q$ . (Zero mean of the divergence under zero normal trace)

[F3]

The divergence has a bounded right inverse $B_{D} : Q \to H_{0}^{1} (D; R^{d})$ with $div B_{D} q = q$ and $∥ \nabla B_{D} q ∥_{2} \leq C_{d i v} ∥ q ∥_{2}$ . (Divergence right inverse on bounded connected Lipschitz domains)

[F4]

For a bounded domain in a coordinate slab of width $L$ , $∥ w ∥_{2} \leq L ∥ \nabla w ∥_{2}$ for $w \in H_{0}^{1} (D)$ . (A bounded-domain Poincare inequality for H1-zero fields)

[F5]

Bounded forms with positive inf-sup and coercivity on the closed constraint kernel give a unique mixed solution under Countable Choice. (Babuska-Brezzi solvability by kernel coercivity and inf-sup)

[F6]

The real $L^{2}$ Cauchy–Schwarz inequality bounds the data functional and the bilinear form pairings. (Cauchy-Schwarz inequality for $L^{2}$ )

[F7]

Every H1-zero vector field belongs to the zero-normal H(div) space. (H1-zero fields have zero weak H(div) normal trace)

Proof

1.1givenF1algebra

The form bounds follow from the permeability eigenvalue bounds: $∣ a (u, v) ∣ \leq κ_{\min}^{} ∥ u ∥_{L^{2}} ∥ v ∥_{L^{2}} \leq κ_{\min}^{} ∥ u ∥_{V} ∥ v ∥_{V}$ , and $∣ b (v, q) ∣ \leq ∥ div v ∥_{2} ∥ q ∥_{2} \leq ∥ v ∥_{V} ∥ q ∥_{Q}$ . Thus $M_{a} \leq κ_{\min}^{}$ and $M_{b} \leq 1$ .

1.2A1F3F7construct

For $q \in Q$ , the right inverse [F3] gives $B_{D} q \in H_{0}^{1} (D; R^{d})$ , so [F7] places $B_{D} q$ in $V$ . Its graph norm is finite.

1.3F1F2givenalgebra

Define $B : V \to Q$ by $(B v, q)_{Q} = b (v, q)$ . Since $b (v, q) = - \int_{D} q div v$ and [F2] places the divergence in $Q$ , $B v = - div v$ . Hence $Z = \ker B = {v \in V : div v = 0}$ ; it is closed because divergence is continuous in the graph norm. For $v \in Z$ , the graph norm equals $∥ v ∥_{2}$ , and $a (v, v) \geq κ_{\max}^{} ∥ v ∥_{2}^{2} = κ_{\max}^{} ∥ v ∥_{V}^{2}$ .

2.1F3F4step 1.2algebra

For nonzero $q \in Q$ , choose $v = - B_{D} q$ . Then $b (v, q) = ∥ q ∥_{2}^{2}$ by [F3]. The Poincare bound [F4] gives $∥ B_{D} q ∥_{V}^{2} = ∥ B_{D} q ∥_{2}^{2} + ∥ q ∥_{2}^{2} \leq (1 + L^{2} C_{d i v}^{2}) ∥ q ∥_{2}^{2}$ . Therefore the inf-sup constant for $b$ on $V \times Q$ is at least $(1 + L^{2} C_{d i v}^{2})^{- 1 / 2} > 0$ .

3.1A1F1F2F3F5F6step 1.1step 1.2step 1.3step 2.1algebracases∎

If $Q = {0}$ , [F2] makes $Z = V$ , so the Countable Choice Lax–Milgram branch of [F5] applies using coercivity from step 1.3. If $Q \neq {0}$ , [F3] supplies a nonzero $q$ ; then $B_{D} q \in V$ by step 1.2 and is nonzero since its divergence is $q$ . The positive inf-sup from step 2.1 and the nonzero Hilbert spaces [F1] allow the mixed branch of [F5]. In both cases boundedness from step 1.1, closedness and coercivity from step 1.3, and data functionals $F$ and $G (q) = - \int_{D} f q$ satisfy [F5]; [F6] makes $G$ bounded. The theorem gives the asserted unique pair, and the pressure mean-zero condition removes the additive constant.

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Canonical RT0 interpolation on H1 fields

Definition

Assume Countable Choice, a conforming affine simplicial mesh $T_{h}$ from Affine simplicial meshes and RT/BDM spaces, and $v \in H^{1} (D; R^{d})$ . The facet trace inequality of H1 facet trace inequality on shape-regular simplices makes each moment $m_{F} (v) : = \int_{F} γ_{F} v \cdot n_{F}$ finite. On each cell $K$ , use its outward normals to prescribe the $d + 1$ facet moments (with sign reversed when $n_{F}$ points inward to $K$ ); by RT0 flux unisolvence in RT0 facet-flux unisolvence and global H(div) conformity, there is a unique $I_{K} v \in {RT}_{0} (K)$ with those moments. Define $I_{h} v$ by $(I_{h} v) ∣_{K} = I_{K} v$ . The elementwise traces of the global H1 field agree almost everywhere across each shared facet by H1 traces agree across interior mesh facets, so the shared moment is single-valued in the fixed orientation; the same unisolvence result gives $I_{h} v \in {RT}_{0} (T_{h}) \subset H (div; D)$ . No canonical facet-moment definition on rough H(div)-only input is made here.

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Exact RT0/P0 element mass balance and normal continuity

Statement

Assume Countable Choice, the mesh conventions of Affine simplicial meshes and RT/BDM spaces, and let $(u_{h}, p_{h}) \in V_{h} \times Q_{h}$ solve the discrete Darcy equations with source $f \in L_{0}^{2} (D)$ . Then on every cell $K$ , $div u_{h} ∣_{K} = ∣ K ∣^{- 1} \int_{K} f, \int_{K} div u_{h} = \int_{K} f .$ The normal flux is single-valued on every interior facet and zero on every boundary facet.

Facts & Assumptions

Given: A source $f \in Q$ , a conforming mesh, and a discrete solution $(u_{h}, p_{h})$ .

[A1]

Countable Choice is inherited through the weak normal-trace and divergence-mean suppliers. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The discrete second equation is $- \int_{D} q_{h} div u_{h} = - \int_{D} f q_{h}$ for every $q_{h} \in Q_{h}$ . (No-flow mixed Darcy problem and coefficient norms)

[F2]

$div u_{h}$ is cellwise constant and has zero mean. (Affine simplicial meshes and RT/BDM spaces, Zero mean of the divergence under zero normal trace)

[F3]

Cell averaging maps $f$ to a mean-zero piecewise constant and is the $L^{2}$ projection onto cell constants. (Cell averaging is the $L^{2}$ projection onto P0)

[F4]

Global RT0 conformity means matching interior normal flux and zero boundary normal flux. (RT0 facet-flux unisolvence and global H(div) conformity)

Proof

1.1A1F1F2F3constructalgebra

Let $P_{0} f$ be the cellwise average of $f$ . By [F2] and [F3], both $div u_{h}$ and $P_{0} f$ are in $Q_{h}$ . Put $g_{h} = div u_{h} - P_{0} f \in Q_{h}$ . For every $q_{h} \in Q_{h}$ , [F1] and the orthogonality property in [F3] give $\int_{D} g_{h} q_{h} = 0$ . Taking $q_{h} = g_{h}$ yields $∥ g_{h} ∥_{2}^{2} = 0$ .

2.1F2F4step 1.1algebra∎

On each cell the equality $div u_{h} = P_{0} f$ says the constant divergence equals the cell average of $f$ ; multiplying by $∣ K ∣$ gives the exact element balance. The field $u_{h}$ is globally RT0 and in H0(div), so [F4] supplies matching interior normal values and zero boundary normal values. The H(div) conformity result equates the flux with the weak normal trace, hence the asserted normal conditions hold.

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Boundedness and divergence commutation of RT0 interpolation

Statement

Assume Countable Choice, a shape-regular mesh with $h = \max_{K} h_{K}$ , and $I_{h}$ from Canonical RT0 interpolation on H1 fields. There is $C$ depending only on dimension and the shape bound such that every $v \in H^{1} (D; R^{d})$ satisfies $∥ I_{h} v ∥_{L^{2} (D)} \leq C (∥ v ∥_{L^{2} (D)} + h ∥ \nabla v ∥_{L^{2} (D)}), (div I_{h} v) ∣_{K} = ∣ K ∣^{- 1} \int_{K} div v .$ If $v \in H_{0}^{1} (D; R^{d})$ , then $I_{h} v \in V_{h}$ .

Facts & Assumptions

Given: Countable Choice, a shape-regular conforming mesh, and a vector field $v \in H^{1} (D; R^{d})$ .

[A1]

Countable Choice is inherited through the Sobolev trace and interpolation interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$I_{K} v$ is the unique RT0 field with the facet moments of $v$ ; matching moments give a global H(div) field. (Canonical RT0 interpolation on H1 fields, RT0 facet-flux unisolvence and global H(div) conformity)

[F2]

The simplex trace inequality is $∥ γ_{F} v ∥_{F} \leq C (h_{K}^{- 1 / 2} ∥ v ∥_{K} + h_{K}^{1 / 2} ∥ \nabla v ∥_{K})$ . (H1 facet trace inequality on shape-regular simplices)

[F3]

Shape regularity gives $∣ K ∣ ≍ h_{K}^{d}$ , $∣ F ∣ ≍ h_{K}^{d - 1}$ , and each altitude comparable to $h_{K}$ . (Shape regularity controls affine simplex geometry)

[F4]

The Gauss–Green formula holds on smooth fields on each affine simplex. (Gauss–Green formula on an affine simplex)

[F5]

Smooth-neighborhood functions are dense in H1 on every cell, and H1_0 is the H1 closure of compactly supported smooth functions. (Smooth-neighborhood density in H1 of a simplex, Zero-boundary Sobolev space as a norm closure)

[F6]

$V_{h}$ is the intersection of the global RT0 space with $H_{0} (div; D)$ . (Affine simplicial meshes and RT/BDM spaces)

[F7]

The facet moment is bounded by the product of the $L^{2}$ facet norms. (Cauchy-Schwarz inequality for $L^{2}$ )

Proof

1.1A1F1F2F3F7algebra

Let $a_{F}$ be the vertex opposite facet $F$ in $K$ and define $ϕ_{K, F} (x) = (x - a_{F}) / (d ∣ K ∣)$ . On $F$ , $(x - a_{F}) \cdot n_{K}$ is the altitude $H_{F}$ , so $\int_{F} ϕ_{K, F} \cdot n_{K} = H_{F} ∣ F ∣ / (d ∣ K ∣) = 1$ by the simplex pyramid identity $d ∣ K ∣ = H_{F} ∣ F ∣$ . On any other facet, $a_{F}$ and $x$ lie in its hyperplane, hence $ϕ_{K, F} \cdot n_{K} = 0$ . Thus $I_{K} v = \sum_{F \subset \partial K} m_{K, F} (v) ϕ_{K, F}$ , where $m_{K, F}$ is the outward moment. By [F2], [F7], and Cauchy–Schwarz, $∣ m_{K, F} (v) ∣ \leq C (h_{K}^{(d - 2) / 2} ∥ v ∥_{K} + h_{K}^{d / 2} ∥ \nabla v ∥_{K})$ . Also [F3] gives $∥ ϕ_{K, F} ∥_{K} \leq C h_{K} ∣ K ∣^{- 1 / 2} \leq C h_{K}^{1 - d / 2}$ . Multiplying and summing the fixed $d + 1$ terms proves $∥ I_{K} v ∥_{K} \leq C (∥ v ∥_{K} + h_{K} ∥ \nabla v ∥_{K})$ . Squaring and summing cells yields the global stability estimate since $h_{K} \leq h$ .

2.1A1F2F4F5step 1.1algebra

The divergence of $I_{K} v \in {RT}_{0} (K)$ is constant. For smooth-neighborhood approximants to $v ∣_{K}$ from [F5], Gauss–Green [F4] says the sum of their outward facet moments equals the integral of their divergence. The trace bound [F2] and H1 convergence let both sides pass to the limit, so $\int_{K} div I_{K} v = \sum_{F \subset \partial K} m_{K, F} (v) = \int_{K} div v$ . Dividing by $∣ K ∣$ gives the stated cell average.

3.1A1F1F2F5F6step 2.1algebra∎

If $v \in H_{0}^{1} (D; R^{d})$ , choose compactly supported smooth vector fields converging to it in H1 by [F5]. Their boundary facet moments are zero; the trace bound [F2] passes this zero moment to $v$ . Hence $I_{h} v$ has zero boundary facet flux, and [F1] gives membership in $H_{0} (div; D)$ ; [F6] then gives $I_{h} v \in V_{h}$ .

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Uniform RT0/P0 divergence lift and inf-sup bound

Statement

Assume Countable Choice, a bounded connected Lipschitz polyhedral $D$ in dimension two or three, and a shape-regular conforming mesh. Let $C_{I}$ be the mesh-independent constant in the RT0 interpolation estimate and let $L$ be a coordinate-slab width for $D$ . If $B_{D}$ is the actual right inverse of Divergence right inverse on bounded connected Lipschitz domains, define $S_{h} q_{h} : = I_{h} B_{D} q_{h}, C_{R} : = C_{I} C_{d i v} (L + diam D) .$ Then $S_{h} : Q_{h} \to V_{h}$ is a linear right inverse of divergence and $∥ S_{h} q_{h} ∥_{L^{2} (D)} \leq C_{R} ∥ q_{h} ∥_{L^{2} (D)}$ . Consequently $∥ S_{h} q_{h} ∥_{H (div)} \leq \sqrt{1 + C_{R}^{2}} ∥ q_{h} ∥_{2}$ . When $Q_{h} \neq {0}$ , the RT0/P0 inf-sup constant satisfies $β_{h} \geq (1 + C_{R}^{2})^{- 1 / 2}$ uniformly in mesh size.

Facts & Assumptions

Given: Countable Choice, $D$ , a shape-regular mesh, and $q_{h} \in Q_{h}$ .

[A1]

Countable Choice is inherited through the continuous divergence-lift and interpolation suppliers. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$B_{D} q_{h} \in H_{0}^{1} (D; R^{d})$ , $div B_{D} q_{h} = q_{h}$ , and $∥ \nabla B_{D} q_{h} ∥_{2} \leq C_{d i v} ∥ q_{h} ∥_{2}$ . (Divergence right inverse on bounded connected Lipschitz domains)

[F2]

$I_{h}$ maps H1 fields to RT0, commutes with divergence cell averages, is zero-normal on H1_0 inputs, and obeys $∥ I_{h} w ∥_{2} \leq C_{I} (∥ w ∥_{2} + h ∥ \nabla w ∥_{2})$ . (Boundedness and divergence commutation of RT0 interpolation)

[F3]

$∥ B_{D} q_{h} ∥_{2} \leq L ∥ \nabla B_{D} q_{h} ∥_{2}$ for the chosen slab width. (A bounded-domain Poincare inequality for H1-zero fields)

Proof

1.1A1F1F2construct

Since $q_{h}$ is cellwise constant, its cell average is itself. By [F1] and the commuting identity [F2], $div S_{h} q_{h} = P_{0} (div B_{D} q_{h}) = q_{h}$ . The input $B_{D} q_{h}$ is in H1_0, so [F2] gives $S_{h} q_{h} \in V_{h}$ . Linearity follows from the linear divergence lift and interpolation.

2.1F1F2F3step 1.1algebra

The stability estimate [F2] and [F1], [F3] imply $∥ S_{h} q_{h} ∥_{2} \leq C_{I} C_{d i v} (L + h) ∥ q_{h} ∥_{2} \leq C_{R} ∥ q_{h} ∥_{2},$ because $h \leq diam D$ . The right-inverse identity from step 1.1 then yields $∥ S_{h} q_{h} ∥_{H (div)}^{2} = ∥ S_{h} q_{h} ∥_{2}^{2} + ∥ q_{h} ∥_{2}^{2} \leq (1 + C_{R}^{2}) ∥ q_{h} ∥_{2}^{2}$ .

3.1F2step 1.1step 2.1algebra∎

For nonzero $q_{h} \in Q_{h}$ , take $v_{h} = - S_{h} q_{h}$ . Then $b (v_{h}, q_{h}) = ∥ q_{h} ∥_{2}^{2}$ and $∥ v_{h} ∥_{H (div)} \leq \sqrt{1 + C_{R}^{2}} ∥ q_{h} ∥_{2}$ by step 2.1. This admissible test gives the lower bound on the supremum for each $q_{h}$ ; taking the infimum over nonzero $q_{h}$ proves the stated $β_{h}$ bound. The constants depend on $D$ , its chosen continuous lift, and the shape bound, but not on $h$ .

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Bounded commuting projection onto the zero-normal RT0 space

Statement

Assume Countable Choice and a shape-regular mesh. Let $P_{0}$ be the cellwise $L^{2}$ -orthogonal projection onto piecewise constants, let $S_{h} : Q_{h} \to V_{h}$ be the stable right inverse above, and put $Z_{h} = {z_{h} \in V_{h} : div z_{h} = 0}$ . There is a linear projection $Π_{h} : H_{0} (div; D) \to V_{h}$ such that $div Π_{h} v = P_{0} div v, ∥ Π_{h} v ∥_{H (div)} \leq C ∥ v ∥_{H (div)},$ where $C$ depends on the domain and mesh shape bound, not mesh size. The construction uses the finite-dimensional $L^{2}$ -orthogonal projection onto $Z_{h}$ and never applies facet moments to rough H(div)-only fields.

Facts & Assumptions

Given: Countable Choice, a shape-regular mesh, and $v \in H_{0} (div; D)$ .

[A1]

Countable Choice is inherited through the H(div) and real $L^{2}$ interfaces; the finite-dimensional projection is constructed below. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

$div v$ has zero mean, so $q_{h} : = P_{0} div v$ belongs to $Q_{h}$ ; $P_{0}$ is an $L^{2}$ contraction and preserves the integral. (Zero mean of the divergence under zero normal trace, Cell averaging is the $L^{2}$ projection onto P0)

[F2]

The right inverse obeys $div S_{h} q_{h} = q_{h}$ , $∥ S_{h} q_{h} ∥_{2} \leq C_{R} ∥ q_{h} ∥_{2}$ , and $∥ S_{h} q_{h} ∥_{H (div)} \leq C_{S} ∥ q_{h} ∥_{2}$ , where $C_{S} = \sqrt{1 + C_{R}^{2}}$ . (Uniform RT0/P0 divergence lift and inf-sup bound)

[F3]

$Z_{h}$ is finite-dimensional, since the mesh is finite and each local RT0 space has dimension $d + 1$ . Real $L^{2}$ has the integral inner product used below. ( $L^{2}$ with the integral pairing is a Hilbert space)

[F4]

The divergencefree members of $V_{h}$ are exactly the discrete kernel $Z_{h}$ . (The RT0/P0 discrete kernel is exactly divergence-free)

Proof

1.1A1F3constructalgebra

If $Z_{h} = {0}$ , set $P_{Z_{h}} = 0$ . Otherwise take a basis $z_{1}, \dots, z_{m}$ of $Z_{h}$ and its Gram matrix $G_{i j} = (z_{j}, z_{i})_{L^{2}}$ . For nonzero $c \in R^{m}$ , $c^{T} G c = ∥ \sum_{j} c_{j} z_{j} ∥_{2}^{2} > 0$ , so $G$ is invertible. For any $r \in L^{2} (D; R^{d})$ , solve $G c = ((r, z_{i})_{L^{2}})_{i = 1}^{m}$ and set $P_{Z_{h}} r = \sum_{j} c_{j} z_{j}$ . This is linear, and $(r - P_{Z_{h}} r, z_{i})_{L^{2}} = 0$ for every basis vector. Thus $r - P_{Z_{h}} r$ is orthogonal to $Z_{h}$ , $P_{Z_{h}} z = z$ for $z \in Z_{h}$ , and $∥ r ∥_{2}^{2} = ∥ P_{Z_{h}} r ∥_{2}^{2} + ∥ r - P_{Z_{h}} r ∥_{2}^{2}$ gives $∥ P_{Z_{h}} r ∥_{2} \leq ∥ r ∥_{2}$ . These conclusions hold also for $Z_{h} = {0}$ .

2.1A1F1F2F4step 1.1constructalgebra

Define $q_{h} = P_{0} div v$ , $r = v - S_{h} q_{h}$ , and let $P_{Z_{h}}$ be the orthogonal projection constructed in step 1.1. Set $Π_{h} v : = S_{h} q_{h} + P_{Z_{h}} r .$ Both terms are in $V_{h}$ , and the second lies in $Z_{h}$ . Thus [F2] and [F4] give $div Π_{h} v = q_{h} = P_{0} div v$ . The cell-average map, $S_{h}$ , and the orthogonal projection are linear, so $Π_{h}$ is linear.

3.1F2F4step 1.1step 2.1algebra

If $v_{h} \in V_{h}$ , then $div v_{h}$ is cellwise constant, so $P_{0} div v_{h} = div v_{h}$ . Consequently $v_{h} - S_{h} (div v_{h}) \in V_{h}$ has zero divergence and lies in $Z_{h}$ by [F4]. Its orthogonal projection is itself; substituting in the formula of step 2.1 gives $Π_{h} v_{h} = v_{h}$ . Hence $Π_{h}$ is a projection.

4.1F1F2step 1.1step 2.1algebra∎

Let $C_{S} = \sqrt{1 + C_{R}^{2}}$ . Since $∥ q_{h} ∥_{2} \leq ∥ div v ∥_{2}$ , [F2] gives $∥ S_{h} q_{h} ∥_{2} \leq C_{R} ∥ q_{h} ∥_{2}$ , hence $∥ r ∥_{2} \leq ∥ v ∥_{2} + C_{R} ∥ q_{h} ∥_{2}$ . Also $∥ P_{Z_{h}} r ∥_{2} \leq ∥ r ∥_{2}$ and its divergence is zero. Therefore [F2] gives $∥ Π_{h} v ∥_{H (div)} \leq C_{S} ∥ q_{h} ∥_{2} + ∥ r ∥_{2} \leq C_{S} ∥ q_{h} ∥_{2} + ∥ v ∥_{2} + C_{R} ∥ q_{h} ∥_{2} \leq (1 + C_{S} + C_{R}) ∥ v ∥_{H (div)} .$ This bound is uniform in mesh size. The formula uses only the $L^{2}$ divergence, the discrete right inverse, and a finite-dimensional $L^{2}$ projection, so no facet trace of $v$ is used.

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RT0/P0 Darcy stability and flux/pressure error bound

Statement

Assume Countable Choice, the Darcy data in No-flow mixed Darcy problem and coefficient norms, and a facet-connected shape-regular mesh. For at least two cells, set $C_{R}$ as in the stable lift theorem and define $\hat{β} = (1 + C_{R}^{2})^{- 1 / 2}, M_{a} = κ_{\min}^{}, M_{b} = 1, α_{*} = κ_{\max}^{} .$ Let $(u, p)$ solve the continuous source-only Darcy problem ( $F = 0$ ) and $(u_{h}, p_{h})$ the RT0/P0 Galerkin system. For any $w_{h} \in V_{h}, r_{h} \in Q_{h}$ , put $δ_{u} = ∥ u - w_{h} ∥_{H (div)}$ , $δ_{p} = ∥ p - r_{h} ∥_{2}$ , and $R_{h} = {\hat{β}}^{- 1} M_{b} δ_{u},$ $U_{h} = δ_{u} + α_{*}^{- 1} {M_{a} (δ_{u} + R_{h}) + M_{b} δ_{p}} + R_{h},$ $P_{h} = δ_{p} + {\hat{β}}^{- 1} {M_{a} (U_{h} + δ_{u}) + M_{b} δ_{p}} .$ Then the discrete pair exists uniquely and $∥ u - u_{h} ∥_{H (div)} \leq U_{h}, ∥ p - p_{h} ∥_{2} \leq P_{h}, ∥ u - u_{h} ∥_{K^{- 1}} \leq κ_{\min}^{} U_{h} .$ For a one-cell mesh $V_{h} = Q_{h} = {0}$ , so the unique discrete pair is $(0, 0)$ ; the nonzero- $Q_{h}$ error theorem is not invoked. The constants display permeability contrast and do not assert contrast-independent robustness.

Facts & Assumptions

Given: Countable Choice, a continuous source-only Darcy solution, a facet-connected shape-regular mesh, and the stated permeability bounds.

[A1]

Countable Choice is required by the continuous mixed and Galerkin solvability suppliers. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The continuous Darcy pair exists uniquely in $V \times Q$ . (Continuous mixed Darcy well-posedness with pressure gauge)

[F2]

For nonzero $Q_{h}$ , the stable lift gives $β_{h} \geq \hat{β} > 0$ . (Uniform RT0/P0 divergence lift and inf-sup bound)

[F3]

The discrete mixed kernel is exactly divergencefree. (The RT0/P0 discrete kernel is exactly divergence-free)

[F4]

The Darcy form obeys $∣ a (v, w) ∣ \leq κ_{\min}^{} ∥ v ∥_{H (div)} ∥ w ∥_{H (div)}$ and $∣ b (v, q) ∣ \leq ∥ v ∥_{H (div)} ∥ q ∥_{2}$ ; on the divergencefree kernel, $a (v, v) \geq κ_{\max}^{} ∥ v ∥_{H (div)}^{2}$ . (No-flow mixed Darcy problem and coefficient norms)

[F5]

For nonzero conforming finite-dimensional subspaces, positive $β_{h}$ , and discrete-kernel coercivity, the actual mixed Galerkin solution is unique and satisfies the displayed explicit $U_{h}, P_{h}$ bound with $β_{h}$ retained and with the exact bilinear-form norms as continuity constants. (Mixed Galerkin error bound retaining beta_h)

[F6]

A one-cell RT0 field with zero flux on every facet is zero, by facet-flux unisolvence; a one-cell mean-zero P0 space is zero. (RT0 facet-flux unisolvence and global H(div) conformity, Affine simplicial meshes and RT/BDM spaces)

Proof

1.1A1F2F3F4given

Suppose first there are at least two cells. Their independent cell constants, subject to one mean constraint, give $Q_{h} \neq {0}$ . The stable lift [F2] then gives $V_{h} \neq {0}$ and a positive discrete inf-sup constant. If $z_{h}$ belongs to the discrete kernel, [F3] gives $div z_{h} = 0$ ; hence its graph norm is its $L^{2}$ norm and the coefficient lower bound [F4] gives discrete coercivity with $α_{*} = κ_{\max}^{}$ . Write $m_{a} = ∥ a ∥$ and $m_{b} = ∥ b ∥$ for the exact norms on $V \times V$ and $V \times Q$ . By [F4], $m_{a} \leq M_{a} = κ_{\min}^{}$ and $m_{b} \leq M_{b} = 1$ .

2.1A1F1F4F5step 1.1algebra

Apply [F5] to the conforming spaces $V_{h} \subset V$ , $Q_{h} \subset Q$ , continuous solution [F1], and the exact norms $m_{a}, m_{b}$ from step 1.1. This gives a unique Galerkin pair and the error bounds with $β_{h}, m_{a}, m_{b}$ . The supplied expressions are nondecreasing in the nonnegative quantities $β_{h}^{- 1}, m_{a}, m_{b}$ : $R_{h}$ is nondecreasing first, then $U_{h}$ , then $P_{h}$ . Since $β_{h}^{- 1} \leq {\hat{β}}^{- 1}$ , $m_{a} \leq M_{a}$ , and $m_{b} \leq M_{b}$ , these replacements give exactly the stated upper bounds $U_{h}, P_{h}$ . For $e = u - u_{h}$ , the weighted bound follows from [F4] and $∥ e ∥_{2} \leq ∥ e ∥_{H (div)}$ , since $∥ e ∥_{K^{- 1}}^{2} \leq κ_{\min}^{} ∥ e ∥_{2}^{2}$ .

3.1F4F6step 2.1algebracases∎

If the mesh has one cell, every facet is a physical-boundary facet. The boundary-zero facet moments and [F6] force $V_{h} = {0}$ , while a constant with zero global mean is zero, so $Q_{h} = {0}$ . The zero pair is the sole discrete pair. Finally $M_{a} / α_{*} = κ_{\max} / κ_{\min}$ , making the permeability-contrast dependence explicit in the formulas rather than uniform in that ratio.

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Hybridization, static condensation, and pressure gauge

Statement

Assume Countable Choice and the source-only Darcy system and hybrid equations of Hybridized no-flow RT0/P0 Darcy system on a facet-connected mesh. The hybrid solutions modulo the common pressure shift $(p, λ) \mapsto (p + c, λ + c)$ are in one-to-one correspondence with the conforming RT0/P0 Darcy solution. If the mesh has at least two cells, every local flux-pressure saddle block is invertible; eliminating these local variables gives a linear system on the interior-facet multipliers whose nullspace is exactly the one-dimensional constant gauge. For one cell, the conforming flux and mean-zero pressure spaces are zero and the hybrid pressure is entirely gauge.

Facts & Assumptions

Given: A source $f \in L_{0}^{2} (D)$ , a facet-connected conforming mesh, fixed positive permeability bounds, and the hybrid spaces and equations in Hybridized no-flow RT0/P0 Darcy system.

[A1]

Countable Choice is inherited through the discrete mixed solvability and normal-trace interfaces. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

The hybrid method has broken RT0 cell fluxes with zero exterior normal flux, cell pressures, facet multipliers, local equations, cell balances, and flux-jump constraints; shifting all pressures and multipliers by the same constant is a gauge. (Hybridized no-flow RT0/P0 Darcy system)

[F2]

Piecewise RT0 fields with matching oriented facet fluxes and zero boundary flux lie in $V_{h}$ ; each facet flux is a unisolvent local degree of freedom. (RT0 facet-flux unisolvence and global H(div) conformity)

[F3]

The conforming RT0/P0 source problem has a unique solution for the mean-zero pressure. (RT0/P0 Darcy stability and flux/pressure error bound)

[F4]

Every conforming solution satisfies the cell balances $\int_{K} div u_{h} = \int_{K} f$ . (Exact RT0/P0 element mass balance and normal continuity)

[F5]

On each cell, a local RT0 basis function has unit outward flux on one selected facet and zero flux on all other facets. (RT0 facet-flux unisolvence and global H(div) conformity)

[F6]

The local quadratic form $a_{K} (v, v) = \int_{K} K^{- 1} v \cdot v$ is positive definite with $a_{K} (v, v) \geq κ_{\max}^{} ∥ v ∥_{L^{2} (K)}^{2}$ . (No-flow mixed Darcy problem and coefficient norms)

[F7]

Every no-flow H(div) field has zero total divergence integral. (Zero mean of the divergence under zero normal trace)

Proof

1.1F1F2givenalgebra

Start with a hybrid solution. Its jump constraints and zero boundary flux, together with [F2], make $u_{h}$ a member of $V_{h}$ . Summing the cell balances against any cellwise constant mean-zero $q_{h}$ gives the discrete second mixed equation.

1.2F1F2F3F4construct

Conversely let $(u_{h}, p_{h})$ be the unique conforming solution. Its cell balances hold by [F4]. Let $W_{h}$ be the broken flux space and define the jump map $J : W_{h} \to R^{F_{h}^{i}}$ by $J_{F} (v) = \int_{F} (v_{K^{+}} \cdot n_{K^{+}} + v_{K^{-}} \cdot n_{K^{-}})$ . Its kernel consists of fields with matching interior fluxes, hence lies in $V_{h}$ by [F2]. Define $ℓ (v) = \sum_{K} [a_{K} (u_{h}, v_{K}) - \int_{K} p_{h} div v_{K}]$ . The conforming first equation gives $ℓ = 0$ on $\ker J$ .

1.3F5F6algebra

Suppose there are at least two cells. Every cell is incident to an interior facet because the dual graph is connected and has more than one vertex. The local space $W_{K}$ therefore contains a basis flux with nonzero total outward flux, so $D_{K} : W_{K} \to R$ , $D_{K} v = \int_{K} div v$ , is onto. The homogeneous local saddle block has equations $a_{K} (u, v) - p D_{K} v = 0$ for all $v \in W_{K}$ and $D_{K} u = 0$ . Taking $v = u$ gives $a_{K} (u, u) = 0$ , so [F6] gives $u = 0$ ; then surjectivity of $D_{K}$ forces $p = 0$ . The square finite-dimensional block is injective and hence invertible.

2.1A1F1F2F3F7step 1.1algebra

Let $c = ∣ D ∣^{- 1} \sum_{K} ∣ K ∣ p_{K}$ and set ${\overset{ˉ}{p}}_{K} = p_{K} - c$ . Then $\overset{ˉ}{p} \in Q_{h}$ . Sum the local first equations against a conforming test $v_{h} \in V_{h}$ . The facet-multiplier terms cancel by normal continuity; the constant pressure term contributes $- c \int_{D} div v_{h} = 0$ by [F7]. The remaining equation is $a (u_{h}, v_{h}) - \int_{D} \overset{ˉ}{p} div v_{h} = 0$ . Thus $(u_{h}, \overset{ˉ}{p})$ is the conforming mixed solution, which is unique by [F3].

2.2F1F3F5step 1.2constructalgebra

The map $J$ is onto: for each interior facet $F$ , use [F5] on one adjacent cell with zero flux on its other facets and zero field on all other cells; its image is the coordinate vector for $F$ . Since $ℓ$ vanishes on $\ker J$ , the formula $μ (J v) : = ℓ (v)$ is well-defined and linear on $R^{F_{h}^{i}}$ . Write $μ (y) = \sum_{F} μ_{F} y_{F}$ and choose $λ_{F} = - μ_{F}$ . Then $ℓ (v) + \sum_{F} λ_{F} J_{F} (v) = 0$ for all broken $v$ . Taking tests supported in one cell yields each local hybrid equation; balances were already shown, so $(u_{h}, p_{h}, λ)$ is hybrid. This proves the correspondence before quotienting the pressure gauge.

3.1A1F1F3F5step 1.1step 2.1step 2.2step 1.3algebra

Invert the local blocks to express each $(u_{K}, p_{K})$ as an affine function of the multiplier vector $λ$ , with source-dependent right sides. Substitution into $J_{F} (u) = 0$ gives a linear system solely on the finitely many interior-facet multipliers; this is static condensation. Existence and correspondence follow from steps 1.1–2.2. For its homogeneous nullspace, any multiplier vector yields a homogeneous hybrid solution, hence by step 2.1 a conforming solution with zero source. Uniqueness [F3] forces $u = 0$ and $p_{K} = c$ for one global constant. The local equations then read $\sum_{F \subset \partial K, F i n t e r i o r} (λ_{F} - c) \int_{F} v_{K} \cdot n_{K} = 0$ for all $v_{K} \in W_{K}$ , since Gauss–Green and the zero physical-boundary flux give $D_{K} v_{K} = \sum_{F i n t e r i o r} \int_{F} v_{K} \cdot n_{K}$ . Using the independent facet basis [F5] gives $λ_{F} = c$ on every interior facet. Conversely a common constant vector is a null vector by the gauge identity in [F1]. Thus the kernel is exactly the constant gauge.

4.1F1F2F3algebracases∎

For one cell there are no interior multipliers. All its facets are physical boundary facets, so zero boundary flux and local unisolvence [F2] force $W_{K} = {0}$ ; also $Q_{h} = {0}$ . The hybrid cell pressure is arbitrary, but every two such values differ by the common gauge, so the quotient contains one solution, corresponding to the zero conforming pair.

5 · Examples, counterexamples and false statements

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Mixed H(div) fluxes, Darcy flow and local conservation

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Contravariant Piola map

Definition

The real H(div) space

Definition

The real zero-mean L2L2 space is Hilbert

Statement

Facts & Assumptions

Proof

Smooth-neighborhood density in H1 of a simplex

Statement

Facts & Assumptions

Proof

Weak normal trace by Green duality

Statement

Facts & Assumptions

Proof

Piola divergence and facet-flux identities

Statement

Facts & Assumptions

Proof

Commuting diagram

Zero-normal-trace H(div) space

Definition

Affine simplicial meshes and RT/BDM spaces

Definition

Zero mean of the divergence under zero normal trace

Statement

Facts & Assumptions

Proof

H1-zero fields have zero weak H(div) normal trace

Statement

Facts & Assumptions

Proof

H(div) is Hilbert and H0(div) is closed

Statement

Facts & Assumptions

Proof

No-flow mixed Darcy problem and coefficient norms

Definition

Cell averaging is the L2L2 projection onto P0

Statement

Facts & Assumptions

Proof

Piola maps RT and BDM polynomial families onto the physical cell

Statement

Facts & Assumptions

Proof

The RT0/P0 discrete kernel is exactly divergence-free

Statement

Facts & Assumptions

Proof

H1 facet trace inequality on shape-regular simplices

Statement

Facts & Assumptions

Proof

Hybridized no-flow RT0/P0 Darcy system

Definition

H1 traces agree across interior mesh facets

Statement

Facts & Assumptions

Proof

RT0 facet-flux unisolvence and global H(div) conformity

Statement

Facts & Assumptions

Proof

Continuous mixed Darcy well-posedness with pressure gauge

Statement

Facts & Assumptions

Proof

Canonical RT0 interpolation on H1 fields

Definition

Exact RT0/P0 element mass balance and normal continuity

Statement

Facts & Assumptions

Proof

The real zero-mean $L^{2}$ space is Hilbert

Cell averaging is the $L^{2}$ projection onto P0