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Alessandrini identities for scalar conductivity and Schrödinger DN maps

Statement

Assume the Axiom of Choice and Countable Choice.

Let (M,g) be a nonempty compact connected smooth Riemannian domain with nonempty smooth boundary. If γ1,γ2∈C∞(M,R) are strictly positive and uj solves div⁡g(γj∇guj)=0 with smooth boundary value fj, then ⟨(Λγ1−Λγ2)f1,f2⟩∂M=∫M(γ1−γ2)g(∇gu1,∇gu2‾) dVg. If instead q1,q2∈C∞(M,R) have trivial Dirichlet kernels for −Δg+qj and uj solves (−Δg+qj)uj=0 with smooth boundary value fj, then ⟨(Λq1−Λq2)f1,f2⟩∂M=∫M(q1−q2)u1u2‾ dVg.

Facts & Assumptions

[A1]

The Axiom of Choice is inherited from the weak DN-map well-posedness suppliers [F2]--[F3].

[A2]

The Axiom of Countable Choice (ACω) is inherited from the weak DN-map well-posedness suppliers [F2]--[F3].

Given: the compact smooth Riemannian domain, two positive conductivities or two real nonresonant potentials, and arbitrary smooth complex boundary values f1,f2.

[F1]

Scalar conductivity and Schrödinger Dirichlet-to-Neumann maps: the conductivity DN map is the weak boundary form of div⁡g(γ∇g), and the Schrödinger DN map is the weak boundary form of −Δg+q; their smooth-data fluxes use γ∂νout and ∂νout, respectively.

[F2]

Weak solvability of the positive scalar conductivity equation: positive smooth conductivity gives a unique weak solution for each H1/2 boundary value.

[F3]

Smooth-manifold elliptic nonresonant solvability and regularity: a trivial Schrödinger Dirichlet kernel gives a unique weak solution for each H1/2 boundary value, with smooth solutions for smooth boundary data.

[F4]

Riemannian volume density: in each chart the Riemannian volume is a positive density, so the integration-by-parts identities can be patched without assuming orientability.

Proof

Given: the data in the statement.

1.1A1A2F1F2F4givenalgebra

Write aγj(v,w)=∫Mγjg(∇v,∇wˉ) dVg. By [F1], ⟨Λγ1f1,f2⟩=aγ1(u1,u2), since u2 is an extension of f2. Let u~1 solve the γ2 equation with boundary value f1, which exists by [F2]. Then ⟨Λγ2f1,f2⟩=aγ2(u~1,u2). Since u1−u~1∈H01(M) and u2 solves the γ2 weak equation, Hermitian symmetry gives aγ2(u1−u~1,u2)=aγ2(u2,u1−u~1)‾=0. Hence aγ2(u~1,u2)=aγ2(u1,u2). Subtracting the two equalities gives the conductivity identity.

1.2A1A2F1F3F4givenalgebra

Write aqj(v,w)=∫M(g(∇v,∇wˉ)+qjvwˉ) dVg. By [F1], ⟨Λq1f1,f2⟩=aq1(u1,u2) because u2 extends f2. Let u~1 be the q2 solution with boundary value f1, which exists by [F3]. Then ⟨Λq2f1,f2⟩=aq2(u~1,u2). Since u1−u~1∈H01(M) and u2 solves the q2 weak equation, real-valuedness of q2 gives aq2(u1−u~1,u2)=aq2(u2,u1−u~1)‾=0. Thus ⟨Λq2f1,f2⟩=aq2(u1,u2).

2.1step 1.2algebra∎

Subtract the identities in step 1.2. The gradient terms cancel and the remaining volume term is ∫M(q1−q2)u1u2‾ dVg, which proves the Schrödinger identity.

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