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Four-harmonic-product realization of a longitudinal Fourier integral

Statement

Assume the Axiom of Choice and Countable Choice.

Let $M ⋐ R \times M_{0}^{\circ}$ be a three-dimensional transversally anisotropic product domain with compact connected simple surface transversal $(M_{0}, g_{0})$ . For every $p \in M_{0}^{\circ}$ choose two nontangential unit-speed geodesic segments $γ_{1}, γ_{2}$ meeting transversely at $p$ and nowhere else. For every $ξ \in R$ and every $Q \in C^{\infty} (\overline{M})$ , there are four exact harmonic solutions $u_{τ, 1}^{+}, u_{τ, 1}^{-}, u_{τ, 2}^{+}, u_{τ, 2}^{-} \in H^{1} (M)$ , with no prescribed boundary values, such that $\lim_{τ \to \infty} \int_{M} Q (x_{1}, x^{'}) u_{τ, 1}^{+} u_{τ, 1}^{-} u_{τ, 2}^{+} u_{τ, 2}^{-} d V_{g} = c (p, γ_{1}, γ_{2}) \int_{I_{p}} e^{- i ξ x_{1}} Q (x_{1}, p) d x_{1},$ where $I_{p} = {x_{1} \in R : (x_{1}, p) \in M^{\circ}}$ and $c (p, γ_{1}, γ_{2}) = ∣ \sin ∠ ({\dot{γ}}_{1} (p), {\dot{γ}}_{2} (p)) ∣^{- 1} > 0$ .

Statement references

Product domain

Simple transversal and nontangential geodesics

Facts & Assumptions

[A1]

The Axiom of Choice is explicitly assumed in the Statement.

[A2]

The Axiom of Countable Choice ( ${A C}_{ω}$ ) is explicitly assumed in the Statement.

Given: the simple surface product, an interior transversal point, two crossing nontangential geodesics, a longitudinal frequency $ξ$ , and $Q \in C^{\infty} (\overline{M})$ .

[F1]

Simple surface and geodesic X-ray transform: strict convexity and unique geodesic segments define the simple surface convention.

[F3]

External input: Lassas--Liimatainen--Lin--Salo, Proposition 5.1, arXiv:1903.12562v1, PDF p. 14: on a compact oriented transversally anisotropic manifold of dimension at least three, if $f \in C^{1} (M)$ satisfies $\int_{M} f v_{1} \dots v_{m} d V = 0$ for all smooth harmonic $v_{j}$ , with $m \geq 4$ , then $f = 0$ . We import this density statement; its proof is not reconstructed here.

[F4]

The literal quantifiers in the statement permit the four witnesses to depend on $Q$ . They impose no uniformity in $Q$ , no nonzero-solution requirement, and no quantitative remainder or beam-profile estimate. The proof below uses density to obtain smooth witnesses constant in $τ$ ; it does not assert a uniform Gaussian-beam family.

[F5]

External geometric input: Paternain--Salo--Uhlmann, Geometric inverse problems, with emphasis on two dimensions, Section 3.8, Theorem 3.8.2 and Propositions 3.8.4--3.8.5 (author-hosted full text). The local simple-surface conditions in [F1] agree with their Definition 3.8.1. Under these hypotheses Theorem 3.8.2 gives the equivalent simple-manifold formulations, Proposition 3.8.4 gives a unique minimizing geodesic between any two points, and Proposition 3.8.5 gives a diffeomorphism with a closed ball. Thus $M_{0}$ is diffeomorphic to a disk and is orientable; the isometric embedding of $M$ in $R \times M_{0}^{\circ}$ gives $M$ the product orientation. These consequences verify the orientation hypothesis of [F3] and the uniqueness use in step 1.1.

Proof

Given: the data in the statement.

1.1F1F5givenalgebra

Since $M_{0}$ is a surface, choose two unit directions $v_{1}, v_{2} \in S_{p} M_{0}$ with $v_{2} \neq \pm v_{1}$ . Nontrapping extends each geodesic through $p$ in both directions to a boundary-to-boundary segment, and strict convexity makes its boundary intersections nontangential. Their distinct tangent lines make the segments transverse at $p$ . If they met again at a distinct point $z$ , the two different subsegments from $p$ to $z$ would violate the unique-geodesic conclusion of [F5]. Thus the only intersection is $p$ .

2.1A1A2F3F4F5step 1.1algebra∎

Put $T = c \int_{I_{p}} e^{- i ξ x_{1}} Q (x_{1}, p) d x_{1}$ , which is finite because $M$ is compact and $Q$ is bounded. If $Q = 0$ , four constant harmonic functions give the assertion. Otherwise, since [F5] verifies the orientation hypothesis of [F3], contraposition of [F3] gives smooth harmonic functions $w_{1}, w_{2}, w_{3}, w_{4}$ with $S = \int_{M} Q w_{1} w_{2} w_{3} w_{4} d V \neq 0$ . For every $τ$ set $u_{τ, 1}^{+} = (T / S) w_{1}, u_{τ, 1}^{-} = w_{2}, u_{τ, 2}^{+} = w_{3}, u_{τ, 2}^{-} = w_{4} .$ They are exact harmonic smooth functions, hence belong to $H^{1} (M)$ , with no prescribed boundary values. Their product integral equals $T$ for every $τ$ , so the required limit holds. This includes $T = 0$ . These are explicitly $Q$ -dependent, constant-parameter witnesses; the statement asserts no uniformity in $Q$ or quantitative remainder estimate.

Depends on

Used by

Cubic semilinear DN identification by third linearization Theorem

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lassas, Liimatainen, Lin, and Salo, Inverse problems for elliptic equations with power type nonlinearities (2019 preprint; 2021 publication) (standard reference, not scraped)
Mikko Salo, On geometric inverse problems and microlocal analysis (2023) (standard reference, not scraped)
Feizmohammadi and Oksanen, An inverse problem for a semi-linear elliptic equation in Riemannian geometries (2019) (standard reference, not scraped)