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Uniqueness of the outgoing exterior scalar Helmholtz solution with Dirichlet data

Statement

Assume the Axiom of Choice inherited from the modal DtN/trace suppliers, covering the Countable Choice used by the local UCP supplier. Let $O \subset R^{3}$ be a bounded open set with $C^{\infty}$ boundary (Bounded C^k domains and boundary charts) whose exterior $D = R^{3} ∖ \overline{O}$ is connected, let $k > 0$ , and let $R > 0$ with $\overline{O} \subset B_{R} (0)$ . Suppose $u$ is $C^{\infty}$ on $\overline{D}$ , solves $(Δ + k^{2}) u = 0$ on $D$ with the Euclidean Laplacian $Δ$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field), vanishes on $\partial O$ , and is outgoing in the class of Outgoing spherical trace and Dirichlet-to-Neumann map: its trace $f = u ∣_{S_{R}}$ lies in $H^{1 / 2} (S_{R})$ and its normal derivative on $S_{R}$ is $T_{R} f$ in the weak trace sense of that lemma. Then $u = 0$ on $D$ . No decay rate, asymptotic expansion or pointwise Sommerfeld formula beyond the stated outgoing class is assumed.

Facts & Assumptions

Given: The Axiom of Choice (The Axiom of Choice); a bounded $C^{\infty}$ obstacle $O$ with connected exterior $D$ ; $k > 0$ ; $R > 0$ with $\overline{O} \subset B_{R} (0)$ ; and a $C^{\infty} (\overline{D})$ solution $u$ of $(Δ + k^{2}) u = 0$ on $D$ , vanishing on $\partial O$ , outgoing in the modal class of the sphere DtN lemma, with trace $f = u ∣_{S_{R}} \in H^{1 / 2} (S_{R})$ and normal derivative $T_{R} f$ .

[L1]

First Green identity on a bounded $C^{1}$ domain: for real $w \in C^{2} (\overline{Ω})$ and $v \in C^{1} (\overline{Ω})$ , $\int_{Ω} (v Δ w + \nabla w \cdot \nabla v) d x = \int_{\partial Ω} v \partial_{ν} w d S$ (First Green identity).

[L2]

The Laplacian is $Δ f = div \nabla f = \sum_{i} \partial_{i} \partial_{i} f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[L3]

Bounded $C^{k}$ domains have the local graph property, so $D \cap B_{R} (0)$ is a bounded domain whose boundary is the disjoint union of the smooth pieces $\partial O$ and $S_{R}$ (Bounded C^k domains and boundary charts).

[L4]

Sphere DtN lemma: for $k, R > 0$ the map $T_{R} f = \sum_{l, m} t_{l} f_{l m} Y_{l m}^{R}$ is bounded $H^{1 / 2} (S_{R}) \to H^{- 1 / 2} (S_{R})$ with $Re ⟨ T_{R} f, f ⟩ \leq 0$ and $Im ⟨ T_{R} f, f ⟩ = \sum_{l, m} (Im t_{l}) ∣ f_{l m} ∣^{2} > 0$ for $f \neq 0$ ; the function $u (r n) = \sum_{l, m} f_{l m} h_{l} (k r) / h_{l} (k R) R^{- 1} Y_{l m} (n)$ , $r \geq R$ , is the unique outgoing finite-local-energy exterior solution with trace $f$ on $S_{R}$ and normal derivative $T_{R} f$ in the weak trace sense (Outgoing spherical trace and Dirichlet-to-Neumann map).

[L5]

Bounded-potential unique continuation: for nonempty open connected $Ω \subseteq R^{3}$ , $V \in L_{l o c}^{\infty} (Ω)$ and $ψ \in H_{l o c}^{2} (Ω)$ with $Δ ψ = V ψ$ distributionally, vanishing of $ψ$ on a nonempty open subset of $Ω$ forces $ψ = 0$ on $Ω$ (Weak unique continuation for bounded potentials in three dimensions).

Choice accounting. The Axiom of Choice is assumed in the Statement and is carried by the DtN/trace supplier [L4]; the local unique-continuation supplier [L5] carries Countable Choice, which AC supplies. The Green-identity, positivity and propagation steps below introduce no further choice.

Proof

technique · direct

Given: The data and solution of the Given paragraph above.

1.1L1L2L3

Green's identity on the truncated exterior and vanishing of the imaginary boundary pairing. The truncated region $A : = D \cap B_{R} (0)$ is a bounded open set whose boundary is the disjoint union of $\partial O$ and $S_{R}$ , with the outward normal of $A$ equal to the outer normal of $S_{R}$ and to the inward normal of $\partial O$ [L3]. Write $u = a + i b$ with real smooth $a, b$ . Applying [L1] to $(w, v) = (a, a)$ and $(b, b)$ and adding gives $\int_{A} (∣ \nabla u ∣^{2} - k^{2} ∣ u ∣^{2}) d x = \int_{\partial A} Re (\overline{u} \partial_{ν} u) d S$ , because $Δ a = - k^{2} a$ and $Δ b = - k^{2} b$ . For the imaginary part, apply [L1] to $(w, v) = (b, a)$ and $(a, b)$ and subtract: the volume integrand is $a Δ b - b Δ a + \nabla b \cdot \nabla a - \nabla a \cdot \nabla b = 0$ , and the boundary integrand is $a \partial_{ν} b - b \partial_{ν} a = Im (\overline{u} \partial_{ν} u)$ . Thus the FULL complex identity is $\int_{A} (∣ \nabla u ∣^{2} - k^{2} ∣ u ∣^{2}) d x = \int_{\partial A} \overline{u} \partial_{ν} u d S$ . The inner boundary contributes zero because $u = 0$ on $\partial O$ , so $\int_{S_{R}} Im (\overline{u} \partial_{ν} u) d S = 0$ . Since $\partial_{ν} u = T_{R} f$ on $S_{R}$ in the weak trace sense of [L4] and $f = u ∣_{S_{R}}$ , the last identity is the vanishing of the imaginary part of the dual pairing, $Im ⟨ T_{R} f, f ⟩ = 0$ .

2.1L4step 1.1

The boundary trace vanishes. By [L4] the imaginary pairing of the DtN operator is $Im ⟨ T_{R} f, f ⟩ = \sum_{l, m} (Im t_{l}) ∣ f_{l m} ∣^{2}$ with every coefficient $Im t_{l} > 0$ . Since the left side is zero by step 1.1 and every term of the sum is nonnegative, every coefficient $f_{l m}$ vanishes, so $f = 0$ ; consequently also $T_{R} f = 0$ and, by the trace identity of [L4], $∥ f ∥_{H^{1 / 2} (S_{R})} = 0$ .

3.1L4step 2.1

Vanishing on the exterior of the large sphere. Restricted to ${r > R}$ , the field $u$ is an outgoing finite-local-energy solution of the Helmholtz equation with trace $f = 0$ on $S_{R}$ and normal derivative $T_{R} f = 0$ : $u$ is $C^{\infty}$ on the closed region ${r \geq R}$ , hence in every $H^{1} ({R < ∣ x ∣ < R_{1}})$ , and its trace and weak normal derivative on $S_{R}$ are the ones named in the statement. By the uniqueness clause of [L4] there is exactly one such solution, namely the outgoing series with coefficients $f_{l m} = 0$ ; hence $u = 0$ on ${r > R}$ . The set $D \cap {r > R}$ is a nonempty open subset of $D$ because $O$ is bounded, so $u$ vanishes on the nonempty open set $D \cap {r > R}$ .

4.1L5step 3.1∎

Unique continuation on $D$ . The real and imaginary parts of $u$ are $C^{\infty}$ on $\overline{D}$ , hence in $H_{l o c}^{2} (D)$ , and satisfy $Δ w = - k^{2} w$ on $D$ because $Δ u = - k^{2} u$ ; the coefficient $V \equiv - k^{2}$ lies in $L_{l o c}^{\infty} (D)$ and $D$ is nonempty, open and connected. By step 3.1 each of them vanishes on the nonempty open subset $D \cap {r > R}$ , so [L5] applied to $w = Re u$ and $w = Im u$ on $Ω = D$ gives $Re u = Im u = 0$ on $D$ , that is, $u = 0$ on $D$ , as asserted.

Uniqueness of the outgoing exterior scalar Helmholtz solution with Dirichlet data

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources