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Mean-zero weak Neumann problem for uniformly elliptic conductivities

Statement

Assume the Axiom of Choice and Countable Choice. Let d≥2, let Ω⊂(0,1)d be nonempty, bounded, connected and C1, and let γ∈L∞(Ω;R) satisfy 0<m≤γ≤M<∞ almost everywhere. Let D(Ω)=H1(Ω;C)/H01(Ω;C) be the complex trace quotient and write D˙=D(Ω)/C1. Let D˙∗ denote the continuous conjugate-linear dual of D˙. For each F∈D˙∗ there is a unique weak solution class [u]∈H1(Ω;C)/C1 satisfying ∫Ωγ∇u⋅∇v‾ dx=F([Tr⁡v])(v∈H1(Ω)). The resulting solution operator Rγ:D˙∗→D˙ is bounded. Its unique representative whose boundary trace has surface mean zero solves the same weak equation. In particular, for f∈L2(∂Ω) with ∫∂Ωf dS=0, the functional Ff([g])=∫∂Ωfg‾ dS is admissible and defines the mean-zero Neumann-to-Dirichlet map on mean-zero L2 fluxes.

Facts & Assumptions

Given: the Axiom of Choice and Countable Choice, Ω, γ, and F∈D˙∗ as in the statement.

[F1]

The hypotheses stated above give a real coefficient with m≤γ≤M almost everywhere; these bounds make the weak form bounded and coercive. No narrower conductivity class is used for this estimate.

[F2]

On a bounded connected extension domain, subtracting the volume mean gives ∥v−vΩ∥L2≤C∥∇v∥L2. (Poincare-Wirtinger on bounded connected extension domains by Rellich compactness)

[F3]

The trace is bounded from H1(Ω) to H1/2(∂Ω) and the trace quotient norm is equivalent to the boundary norm; constants correspond under the quotient identification. (Boundary trace quotient and chart norm equivalence)

[F4]

A bounded coercive sesquilinear form on a Hilbert space and a bounded conjugate-linear functional have a unique weak solution with norm at most the functional norm divided by the coercivity constant. (The Lax--Milgram theorem)

[F5]

Every bounded C1 domain is a Sobolev extension domain, and its compact boundary has finite positive surface measure. (Bounded C^k domains admit integer-order Sobolev extension, Surface integration on compact C1 hypersurfaces, Chart and partition independence of surface measure)

[A1]

The Axiom of Choice is used at the Poincaré and trace interfaces. (The Axiom of Choice)

[A2]

Countable Choice is used at the Lax–Milgram interface. (The Axiom of Countable Choice (ACω))

Proof

Given: The data in the statement.

1.1A1F1F2givenalgebra

Let V={v∈H1(Ω):∫Ωv dx=0}, a closed Hilbert subspace of H1(Ω). By [F2], ∥v∥H1≤C0∥∇v∥2 on V, where C0=(1+C2)1/2. By [F1], the form aγ(u,v)=∫Ωγ∇u⋅∇v‾ is bounded on V and aγ(v,v)≥m∥∇v∥22≥mC0−2∥v∥H12. Thus it is bounded and coercive on V.

2.1A2F3F4step 1.1given

Define ℓF(v)=F([Tr⁡v]) for v∈V. By [F3], the quotient trace norm of [Tr⁡v] is at most C1∥v∥H1, so ∣ℓF(v)∣≤C1∥F∥∥v∥H1; hence ℓF is bounded on V. Applying [F4] to step 1.1 gives a unique u0∈V with aγ(u0,v)=ℓF(v) for every v∈V.

3.1F1step 1.1step 2.1algebra

For arbitrary v∈H1(Ω) write v=v0+c with v0=v−vΩ∈V and c=vΩ. Gradients are unchanged by adding c, and F acts on the trace quotient by constants, so the equation for u0 on V implies it for every v∈H1. If u1 represents another solution class, put w=(u1−u0)−(u1−u0)Ω∈V. Subtracting the two weak equations gives aγ(w,v)=0 for all v∈V; setting v=w and using coercivity from step 1.1 gives w=0. Thus the solution is unique modulo constants.

4.1F3F4F5step 2.1step 3.1givenalgebra∎

The estimate in [F4] and step 2.1 give ∥u0∥H1≤C02C1m−1∥F∥. The bounded trace map [F3] therefore makes the solution class in D˙ a bounded linear function of F. By [F5], the compact smooth boundary has finite positive surface measure. Subtracting the surface mean of Tr⁡u0 selects one representative with boundary mean zero; this constant shift preserves the weak equation and is the only such shift. For mean-zero f∈L2(∂Ω), Cauchy–Schwarz and the trace bound show Ff is continuous, and its zero mean makes it independent of the representative modulo constants. This proves the final assertion.

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