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Centered Neumann voltage sequences on the three-dimensional ball

Statement

Assume the Axiom of Choice, Countable Choice and Dependent Choice. Let Ω=B1(0)⊂R3, fix 0<ρ<1 and 0<m<1<M, and let E be the closed real subspace of L∞(Ω) consisting of functions vanishing almost everywhere outside Bρ(0). Set A={a∈E:ess inf⁡(1+a)>m, ess sup⁡(1+a)<M} and σ=1+a. Choose an orthonormal spherical-harmonic basis Yℓk of mean-zero L2(S2), with ℓ≥1 and 1≤k≤2ℓ+1. For each basis current there is a unique weak solution uσ,ℓk∈H1(Ω) with boundary mean zero such that ∫Ωσ∇uσ,ℓk⋅∇v‾=∫S2YℓkTr⁡v‾(v∈H1(Ω)). The centered sequence G(a):=(Tr⁡(uσ,ℓk−u1,ℓk))ℓ,k belongs to Y=ℓ2(L2(S2)). The map G:A→Y is Fréchet differentiable, and both G and G′ are Lipschitz on A, with constants depending only on m,M,ρ and the ball. Its derivative in direction h∈E is the trace sequence of the unique boundary-mean-zero solutions ∫Ωσ∇vh,ℓk⋅∇w‾=−∫Ωh∇uσ,ℓk⋅∇w‾. The uncentered sequence (Tr⁡u1,ℓk)ℓ,k does not belong to Y. Thus absolute voltage data in this convention lie in an affine translate of Y, whereas data differences belong to Y. All complex spaces here may be regarded as real Banach spaces when taking derivatives with respect to real conductivities.

Facts & Assumptions

Given: The choices, fixed collar radius, uniformly elliptic conductivities and mean-zero current basis in the Statement.

[F1]

The degree-ℓ harmonic space has dimension 2ℓ+1, and these spaces form a complete orthogonal expansion on S2. Spherical harmonics form a complete basis with the addition formula

[F2]

The ball is an extension domain, satisfies the mean-zero Poincaré inequality, and has a bounded trace H1(Ω)→W1/2,2(S2)⊂L2(S2). Bounded C^k domains admit integer-order Sobolev extension, Poincare-Wirtinger on bounded connected extension domains by Rellich compactness, Boundary trace quotient and chart norm equivalence

[F3]

Bounded coercive forms have unique solutions for bounded conjugate-linear functionals. The Lax--Milgram theorem

[F4]

The divergence theorem on a bounded C1 domain gives Green's identity for smooth functions; the complex form follows by real and imaginary parts. Divergence on a bounded C1 Euclidean domain

[F5]

Restrictions of ambient smooth functions are dense in H1(B1), so identities proved for smooth tests pass to all H1 tests. Ambient smooth restrictions are dense on bounded C^k domains

[A1]

Choice assumptions are used at [F1]–[F5]; the Axiom of Choice selects the countable family of finite-dimensional bases. The Axiom of Choice, The Axiom of Countable Choice (ACω), The axiom of dependent choice: a relation in which every element is related to something admits an N-indexed chain

Proof

1.1A1F2F3givenalgebra

Let V={w∈H1(Ω):∫S2Tr⁡w=0}. This is closed by [F2]. For w∈V, write wˉ for its volume mean. Poincaré gives ∥w−wˉ∥H1≤C∥∇w∥2, and the trace bound and boundary-mean condition give ∣wˉ∣ ∣S2∣≤∣S2∣1/2∥Tr⁡(w−wˉ)∥2≤C∥∇w∥2. Thus ∥w∥H1≤C∥∇w∥2 on V. The form Aσ(w,v)=∫σ∇w⋅∇v‾ is bounded and coercive there, uniformly for m<σ<M. By [F3], any functional v↦−∫H⋅∇v‾ with H∈L2(Ω)3 has a unique solution in V, of H1 norm at most C∥H∥2.

2.1A1F1F2F4F5step 1.1givenalgebra

The background solution is u1,ℓk(r,ω)=rℓYℓk(ω)/ℓ. It is a harmonic polynomial with normal derivative Yℓk and boundary mean zero. For smooth test functions, the divergence theorem [F4] gives its weak equation. The density [F5] and continuity of the trace pass the identity to every H1 test. The same integration by parts on Bρ gives ∥∇u1,ℓk∥L2(Bρ)2=ρ2ℓ+1/ℓ. In particular Aρ:=∑ℓ≥1(2ℓ+1)ρ2ℓ+1/ℓ<∞, since (2ℓ+1)/ℓ≤3.

3.1F2step 1.1step 2.1algebra

Apply step 1.1 with H=a∇u1,ℓk, supported in Bρ, to obtain wa,ℓk∈V satisfying Aσ(wa,ℓk,v)=−∫a∇u1,ℓk⋅∇v‾. The sum u1,ℓk+wa,ℓk solves the required Neumann equation, first for v∈V and then for all v∈H1 by subtracting its boundary mean. The current annihilates constants. Uniqueness follows by testing the difference with itself and using step 1.1. Summing the energy estimates gives ∑∥wa,ℓk∥H12≤C∥a∥∞2Aρ. The trace bound proves G(a)∈Y. Moreover ∑∥∇uσ,ℓk∥L2(Bρ)2≤2Aρ+2∑∥wa,ℓk∥H12≤C, uniformly over A, since ∥a∥∞≤max⁡{1−m,M−1}.

4.1F2step 1.1step 3.1algebra

For a,b∈A, subtract the weak equations to obtain A1+a(u1+a,ℓk−u1+b,ℓk,v)=−∫(a−b)∇u1+b,ℓk⋅∇v‾. By step 1.1 and the summed interior bound in step 3.1, the sequence of differences has ℓ2(H1) norm at most C∥a−b∥∞. For h∈E, the same estimate constructs the asserted derivative sequence va,h with norm at most C∥h∥∞. Subtracting its equation from the difference equation for a+h,a shows that the remainder has forcing −h∇(u1+a+h−u1+a) and hence ℓ2(H1) norm at most C∥h∥∞2. Taking traces proves Fréchet differentiability. Finally, subtract the derivative equations at a,b: A1+a(va,h−vb,h,w)=−∫h∇(u1+a−u1+b)⋅∇w‾−∫(a−b)∇vb,h⋅∇w‾. The established estimates bound its sequence norm by C∥a−b∥∞∥h∥∞, proving the operator-norm Lipschitz derivative bound and the asserted Lipschitz bound for G.

5.1F1step 2.1step 3.1algebra∎

By step 2.1, Tr⁡u1,ℓk=Yℓk/ℓ. Its squared sequence norm is ∑ℓ≥1(2ℓ+1)/ℓ2, which diverges since each summand is at least 2/ℓ. Thus the absolute background voltage sequence is not in Y, while every centered difference is in Y by step 3.1. This establishes the affine-data distinction without changing the set of measured currents.

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