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The Helmholtz single-layer and double-layer traces have the Laplace jump signs

Statement

Let $Γ = \partial Ω^{-}$ be the connected closed $C^{2}$ surface from The combined-field operator for smooth-boundary Helmholtz scattering, with outward normal $n$ . For $k > 0$ and smooth densities $λ, ϕ$ , the Helmholtz layers in that definition satisfy $γ^{-} S_{k} λ = γ^{+} S_{k} λ, \partial_{n}^{-} S_{k} λ - \partial_{n}^{+} S_{k} λ = λ,$ $γ^{-} D_{k} ϕ - γ^{+} D_{k} ϕ = - ϕ .$ The one-sided conormal limits in the single-layer formula and trace limits in the double-layer formula are taken along the fixed normal $n$ . The double-layer conormal jump is not asserted in this item.

Facts & Assumptions

Given: The C2 surface, its fixed normal, the outgoing kernel and layer definitions of The combined-field operator for smooth-boundary Helmholtz scattering, and smooth densities.

[F1]

The plane Poisson kernel has mass one, so its half-kernel has one-sided limit $\pm \frac{1}{2}$ for the planar double-layer trace (Fourier transform of the three-dimensional plane Poisson kernel).

[F2]

The outgoing kernel is $G_{k} (r) = e^{i k r} / (4 π r)$ and the Laplace kernel is $G_{0} (r) = 1 / (4 π r)$ (The combined-field operator for smooth-boundary Helmholtz scattering, The three-dimensional Laplace Newton kernel).

[F3]

The radial exponential derivative is the one established in [F91].

[F90]

Polar integration uses the published radial formula, not a linear substitution: $d x = r^{n - 1} d r d σ (ω)$ . The sphere measure agrees with graph-chart surface measure and scales by $R^{n - 1}$ . (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma, Agreement with the existing polar sphere measure)

[F91]

The complex exponential is entire with derivative itself; composing with the affine parameter $r \mapsto i k r$ gives $\frac{d}{d r} e^{i k r} = i k e^{i k r}$ (subtract the exponential difference quotient and multiply by $i k$ ). The finite-dimensional FTC in If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz applies componentwise to these smooth exponential curves. Iteration gives all higher radial derivatives, and $∣ e^{i k r} ∣ = 1$ for real $k, r$ by $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ . (The complex exponential is entire and its complex derivative is itself)

Proof

1.1F2F3givenalgebraF91

Write $H_{k} (r) = G_{k} (r) - G_{0} (r)$ . The identity $H_{k} (r) = \frac{i k}{4 π} \int_{0}^{1} e^{i k r θ} d θ$ follows by integrating the derivative of the exponential from $0$ to $r$ ; it implies $H_{k}$ and its first radial derivative are bounded for $0 < r \leq r_{0}$ . Its Cartesian first derivatives are bounded as well. Therefore the single-layer trace integrals with kernel $H_{k}$ and their first normal derivatives have identical limits from both sides, and the double-layer trace integral with kernel $\partial_{n_{y}} H_{k}$ also has identical limits: the kernels are bounded near the diagonal and the surface measure of a shrinking disk tends to zero. Thus these remainder terms do not change any of the three stated jumps.

2.1F1F2step 1.1givenalgebraF90

Fix $x_{0} \in Γ$ and rotate/translate so $x_{0} = 0$ , $n (x_{0}) = e_{3}$ , and a neighborhood of $Γ$ is $y = (z, ψ (z))$ with $ψ (0) = 0$ , $D ψ (0) = 0$ , $∣ ψ (z) ∣ \leq C ∣ z ∣^{2}$ , and $∣ D ψ (z) ∣ \leq C ∣ z ∣$ . For the Laplace kernel, the single-layer trace kernel is locally bounded by $C ∣ z ∣^{- 1}$ , which is surface-integrable; dominated convergence gives the same trace from both sides. In the normal derivative, the planar leading term is $- t (4 π (∣ z ∣^{2} + t^{2})^{3 / 2})^{- 1} λ (0)$ . Its integral over the tangent plane is $- \frac{1}{2} sgn (t) λ (0)$ by polar coordinates. Subtracting this leading term leaves an integrable remainder with a common two-sided limit: the curvature shift $ψ (z) = O (∣ z ∣^{2})$ changes $s / (∣ z ∣^{2} + s^{2})^{3 / 2}$ by at most $C ∣ z ∣^{- 1}$ because its $s$ -derivative is bounded by $C ∣ z ∣^{- 3}$ ; replacing $λ (0)$ by $λ (z, ψ (z)) J (z) = λ (0) + O (∣ z ∣)$ contributes at most $C ∣ t ∣ ∣ z ∣ / (∣ z ∣^{2} + t^{2})^{3 / 2}$ , whose disk integral is $O (∣ t ∣ ∣ \log ∣ t ∣ ∣)$ . Hence the interior-minus-exterior conormal jump is $λ (0)$ .

3.1F1F2step 1.1step 2.1givenalgebraF90∎

The double-layer trace has planar leading term $t (4 π (∣ z ∣^{2} + t^{2})^{3 / 2})^{- 1} ϕ (0)$ , whose tangent-plane integral is $\frac{1}{2} sgn (t) ϕ (0)$ . The graph normal and area factor satisfy $n (y) J (y) = (- D ψ (z), 1)$ , so its numerator is $t + O (∣ z ∣^{2})$ . As in step 2.1, subtracting the planar term leaves an integrable $C ∣ z ∣^{- 1}$ curvature remainder plus a density remainder with disk integral $O (∣ t ∣ ∣ \log ∣ t ∣ ∣)$ ; both have the same limit from either side. Thus $γ^{-} D_{0} ϕ = - \frac{1}{2} ϕ + K_{0} ϕ$ and $γ^{+} D_{0} ϕ = \frac{1}{2} ϕ + K_{0} ϕ$ for a common remainder $K_{0} ϕ$ , giving jump $- ϕ$ . Step 1.1 shows the Helmholtz-minus-Laplace term has no jump, proving the displayed double-layer relation.

Remarks

The local expansions prove the smooth-density trace and single-layer conormal jumps. The separate The Helmholtz double layer has a common weak conormal supplies all four jumps in the weak Sobolev trace convention, using Smooth-boundary trace lifts and Sobolev Helmholtz layer extensions.

Depends on

Used by

The smooth Helmholtz combined field is a compact perturbation of half the identity Theorem

Cited to discharge well-definedness by The combined-field operator for smooth-boundary Helmholtz scattering.

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Sources

Francisco-Javier Sayas, First Steps in the Boundary Element Method (2015 lecture notes) (standard reference, not scraped)
Martin Costabel, Boundary Integral Operators on Lipschitz Domains: Elementary Results (1988) (standard reference, not scraped)