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Alessandrini identity for the time-dependent wave Dirichlet-to-Neumann maps

Statement

Assume Countable Choice. Let $(M, g)$ be a compact, connected, nonempty smooth Riemannian manifold with smooth boundary, let $T > 0$ , and let $q_{1}, q_{2} \in C^{\infty} (M; R)$ be independent of time. Write $Q = (0, T) \times M$ , $Σ = (0, T) \times \partial M$ , and $P = \partial_{t}^{2} - Δ_{g}$ . For $h, k \in C_{c}^{\infty} (Σ)$ , let $u$ be the zero-initial-data solution for $P + q_{1}$ with boundary value $h$ , and let $v$ be the zero-terminal-data solution for $P + q_{2}$ with boundary value $k$ , in the classical solution classes used to define the forward map $Λ_{q_{1}}^{+}$ and terminal map $Λ_{q_{2}}^{-}$ ; thus $u, v \in C^{2} ([0, T] \times M)$ . Then $\int_{0}^{T} ⁣ \int_{M} (q_{1} - q_{2}) u (t, x) v (t, x) d V_{g} (x) d t = \int_{0}^{T} ⁣ \int_{\partial M} (k Λ_{q_{1}}^{+} h - h Λ_{q_{2}}^{-} k) d S_{g} d t .$ The boundary factors are paired bilinearly, without complex conjugation, and each normal derivative is the outward classical normal trace.

Statement references

Time-dependent Dirichlet-to-Neumann map for the wave equation

Energy well-posedness for the variable-metric wave equation

Riemannian Green identity, boundary trace, and wave lifting

Riemannian volume density

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Facts & Assumptions

Given: Countable Choice, the two real smooth static potentials, smooth compactly time-supported boundary data, and the stated forward and terminal classical solutions.

[A1]

Countable Choice is the axiom base of the forward energy theorem used in the DN-map interfaces; the integral calculation itself is a choice-free application of Green's identity. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F1]

For $C^{2}$ functions $u, v$ on the time cylinder, $\int_{Q} ((\partial_{t}^{2} - Δ_{g}) u v - u (\partial_{t}^{2} - Δ_{g}) v) d V_{g} d t = \int_{Σ} (u \partial_{ν_{o u t}} v - v \partial_{ν_{o u t}} u) d S_{g} d t + \int_{M} [\partial_{t} u v - u \partial_{t} v]_{t = 0}^{t = T} d V_{g} .$ (Riemannian Green identity, boundary trace, and wave lifting)

[F2]

The forward map uses the unique zero-initial solution with Dirichlet value $h$ and returns its outward normal trace; the terminal map uses the unique zero-terminal solution with boundary value $k$ and returns its outward normal trace. (Time-dependent Dirichlet-to-Neumann map for the wave equation)

[F3]

$P + q_{j}$ means $\partial_{t}^{2} - Δ_{g} + q_{j}$ , and the Riemannian density in the interior pairing is $d V_{g}$ . (Riemannian volume density)

Proof

1.1A1F1F2F3givenalgebra

Apply [F1] to $u$ and $v$ . Since $(P + q_{1}) u = 0$ and $(P + q_{2}) v = 0$ , one has $P u = - q_{1} u$ and $P v = - q_{2} v$ . Thus the left side of [F1] is $\int_{Q} ((- q_{1} u) v - u (- q_{2} v)) d V_{g} d t = - \int_{Q} (q_{1} - q_{2}) u v d V_{g} d t .$

1.2F1F2given

The temporal endpoint term in [F1] is zero. At $t = T$ , $v (T) = v_{t} (T) = 0$ ; at $t = 0$ , $u (0) = u_{t} (0) = 0$ . These are exactly the two endpoint products in the bracket.

2.1F1F2step 1.1step 1.2algebra∎

The boundary term in [F1] equals $\int_{Σ} (h Λ_{q_{2}}^{-} k - k Λ_{q_{1}}^{+} h) d S_{g} d t$ by the prescribed traces $u ∣_{Σ} = h$ , $v ∣_{Σ} = k$ and [F2]. Combining this with steps 1.1–1.2 gives the stated identity after multiplying both sides by $- 1$ .

Depends on

Used by

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Sources

Mikko Salo, On geometric inverse problems and microlocal analysis (full author lecture notes, arXiv:2312.08123v1) (standard reference, not scraped)
Medet Nursultanov and Lauri Oksanen, Introduction to inverse problems for hyperbolic PDEs (full lecture notes, arXiv:2311.06020v2) (standard reference, not scraped)