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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)↦(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∈L02(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 (ACω))

[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 Vh; 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 ∫Kdiv⁡uh=∫Kf. (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 aK(v,v)=∫KK−1v⋅v is positive definite with aK(v,v)≥κmax⁡−1∥v∥L2(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 uh a member of Vh. Summing the cell balances against any cellwise constant mean-zero qh gives the discrete second mixed equation.

1.2F1F2F3F4construct

Conversely let (uh,ph) be the unique conforming solution. Its cell balances hold by [F4]. Let Wh be the broken flux space and define the jump map J:Wh→RFhi by JF(v)=∫F(vK+⋅nK++vK−⋅nK−). Its kernel consists of fields with matching interior fluxes, hence lies in Vh by [F2]. Define ℓ(v)=∑K[aK(uh,vK)−∫Kphdiv⁡vK]. 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 WK therefore contains a basis flux with nonzero total outward flux, so DK:WK→R, DKv=∫Kdiv⁡v, is onto. The homogeneous local saddle block has equations aK(u,v)−pDKv=0 for all v∈WK and DKu=0. Taking v=u gives aK(u,u)=0, so [F6] gives u=0; then surjectivity of DK forces p=0. The square finite-dimensional block is injective and hence invertible.

2.1A1F1F2F3F7step 1.1algebra

Let c=∣D∣−1∑K∣K∣pK and set pˉK=pK−c. Then pˉ∈Qh. Sum the local first equations against a conforming test vh∈Vh. The facet-multiplier terms cancel by normal continuity; the constant pressure term contributes −c∫Ddiv⁡vh=0 by [F7]. The remaining equation is a(uh,vh)−∫Dpˉdiv⁡vh=0. Thus (uh,pˉ) 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(Jv):=ℓ(v) is well-defined and linear on RFhi. Write μ(y)=∑FμFyF and choose λF=−μF. Then ℓ(v)+∑FλFJF(v)=0 for all broken v. Taking tests supported in one cell yields each local hybrid equation; balances were already shown, so (uh,ph,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 (uK,pK) as an affine function of the multiplier vector λ, with source-dependent right sides. Substitution into JF(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 pK=c for one global constant. The local equations then read ∑F⊂∂K, Finterior(λF−c)∫FvK⋅nK=0 for all vK∈WK, since Gauss–Green and the zero physical-boundary flux give DKvK=∑Finterior∫FvK⋅nK. Using the independent facet basis [F5] gives λF=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 WK={0}; also Qh={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.

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