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The harmonic affine extension minimises the Dirichlet energy

Example

Example. Assume the Axiom of Choice (The Axiom of Choice). Let Ω⊆Rn, n≥2, be a bounded C1 domain and let a(x)=c0+ℓ(x) be affine on Ω‾, so that Δa=0 and Da=ℓ is constant. Then a is the unique minimiser of the Dirichlet energy I(v)=12∫Ω∣Dv∣2 dx on the affine trace class Ka={v∈H1(Ω):Tv=Ta} (The Lp trace operator on a bounded C1 domain): for every v∈Ka, v−a∈H01(Ω) and I(v)=I(a)+12∫Ω∣D(v−a)∣2 dx ≥ I(a), with equality if and only if D(v−a)=0 almost everywhere, and then v=a almost everywhere by the Poincare inequality on H01(Ω) (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

Facts & Assumptions

Given: The Axiom of Choice; a bounded C1 domain Ω⊆Rn, an affine function a(x)=c0+ℓ(x) on Ω‾ (so Da=ℓ is a constant field and Δa=0), the Dirichlet energy I(v)=12∫Ω∣Dv∣2dx, and the affine trace class Ka={v∈H1(Ω):Tv=Ta}.

[F1]

Ka is nonempty, convex and weakly closed, and v−a∈H01(Ω) for every v∈Ka by the kernel description ker⁡T=H01(Ω) (The kernel of the trace is the closure of the test functions, The Lp trace operator on a bounded C1 domain).

[F2]

Poincare's inequality on H01(Ω): if Dη=0 almost everywhere for η∈H01(Ω), then η=0 (The Poincare inequality for zero-boundary Sobolev closures on domains bounded in one direction).

[F3]

Under the Axiom of Choice, the ultrafilter lemma, Dependent Choice and Hahn--Banach, the Dirichlet principle also identifies an affine-class minimiser with the unique weak solution (The Dirichlet principle for the Poisson equation). In the zero-forcing case here, a is weakly harmonic with trace Ta, since Da=ℓ is constant and ∫ΩDa⋅Dφ=ℓ⋅∫ΩDφ=0 for every φ∈H01(Ω) (The notation Hk and the reserved zero-boundary symbol).

Verification

technique · direct completion of the square
1.1F1algebra

Orthogonality of the cross term. For v∈Ka one has η:=v−a∈H01(Ω) by [F1], and Da=ℓ is a constant field, so ∫ΩDa⋅Dη dx=ℓ⋅∫ΩDη dx=0: each component of Dη has vanishing integral, because η is the H1-limit of functions in Cc∞(Ω) and for those the integral of each partial derivative vanishes by integration by parts against the smooth constant field (Divergence on a bounded C1 Euclidean domain). Passage to the H1 limit is valid since ∣∫Di(η−ηm)∣≤∣Ω∣1/2∥Di(η−ηm)∥2→0 by Holder (Holder's inequality for integrals, including the endpoint cases).

2.1step 1.1algebra

The energy identity. Expanding the square, ∣Dv∣2=∣Da+Dη∣2=∣Da∣2+2Da⋅Dη+∣Dη∣2, and integrating with step 1.1 gives I(v)=I(a)+12∫Ω∣Dη∣2dx≥I(a).

3.1F2step 2.1

Equality case. Equality holds exactly when Dη=0 almost everywhere, which by [F2] forces η=0, that is v=a almost everywhere.

4.1F3step 2.1step 3.1∎

a is the minimiser. By steps 2.1 and 3.1 every v∈Ka satisfies I(v)≥I(a) with equality only for v=a; since a∈Ka, it is the unique minimiser of I on Ka. This direct completion-of-the-square argument proves the example's claim. Under the additional choice hypotheses stated in [F3], the general Dirichlet principle also identifies this minimiser with the weak solution; the weak harmonicity of a was checked in [F3].

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