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SIPDG coercivity above the discrete inverse-trace threshold

Statement

Assume Countable Choice. Use the mesh, coefficient field, energy norm, and form of Symmetric interior-penalty DG form and energy seminorm, and let Cflux be the constant of Discrete inverse-trace bound for SIPDG coefficient fluxes. If γ≥4Cflux, then ah(vh,vh)≥12∥vh∥DG2(vh∈Vhbr). Thus the discrete bilinear form is coercive with constant 1/2 in the broken energy norm. For every f∈L2(Ω), the finite-dimensional SIPDG system has a unique solution. The sufficient threshold depends on the polynomial degree, shape regularity, and coefficient ellipticity bounds; no universal penalty value is asserted.

Facts & Assumptions

Given: Countable Choice; k≥1; symmetric cellwise constant AK with ellipticity bounds 0<λ≤Λ; a finite matching shape-regular mesh; γ≥4Cflux; and the SIPDG form and energy norm.

[F1]

The form is the cell energy minus the two symmetric flux-jump pairings plus γ∑FhF−1([v]F,[w]F)L2(F). (Symmetric interior-penalty DG form and energy seminorm)

[F2]

The squared coefficient-flux trace satisfies ∑FhF∥qF(vh)∥L2(F)2≤Cflux∑K∫KAK∇vh⋅∇vh. (Discrete inverse-trace bound for SIPDG coefficient fluxes)

[F3]

The broken energy expression is a norm on the finite-dimensional discontinuous polynomial space. (The SIPDG broken energy expression is definite)

[F4]

Countable Choice is the explicit hypothesis in the inverse-trace and energy-norm suppliers [F2]–[F3]. (The Axiom of Countable Choice (ACω))

[F5]

The L2 case of Holder's inequality gives ∣∫Ffg dS∣≤∥f∥L2(F)∥g∥L2(F). (Holder's inequality for integrals, including the endpoint cases)

Proof

1.1F1F2F4F5F6algebra

Put X2:=∑K∫KAK∇vh⋅∇vh and J2:=∑FhF−1∥[vh]F∥L2(F)2. The integral Cauchy–Schwarz inequality [F5] on each facet, followed by [F6] on the finite weighted face sum and [F2] give ∣∑F∫FqF(vh)[vh]F∣≤(∑FhF∥qF(vh)∥F2)1/2J≤CfluxXJ.

2.1F1step 1.1givenalgebra

By [F1], ah(vh,vh)=X2−2∑F∫FqF(vh)[vh]F+γJ2. Young's inequality gives 2CfluxXJ≤12X2+2CfluxJ2, hence ah(vh,vh)≥12X2+(γ−2Cflux)J2≥12(X2+γJ2)=12∥vh∥DG2.

3.1givenF1F3step 2.1algebra∎

The coefficient matrix of ah in any basis of the finite-dimensional space Vhbr is symmetric by [F1]. If its matrix kernel contained a nonzero vector, the corresponding vh≠0 would satisfy ah(vh,vh)=0, contradicting [F3] and step 2.1. Therefore the square matrix is injective, hence invertible in finite dimension, so every load has exactly one solution.

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