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The three-dimensional Newton kernel is a fundamental solution

Statement

For every $y \in R^{3}$ , the locally integrable kernel $G_{3} (x, y) = (4 π ∣ x - y ∣)^{- 1}$ of The three-dimensional Laplace Newton kernel satisfies $- Δ_{x} G_{3} (x, y) = δ_{y}$ in the sense of distributions: for every $φ \in C_{c}^{\infty} (R^{3})$ , $\int_{R^{3}} G_{3} (x, y) (- Δ φ (x)) d x = φ (y) .$

Facts & Assumptions

Given: The Axiom of Choice, a point $y \in R^{3}$ , the kernel in The three-dimensional Laplace Newton kernel, and a test function $φ \in C_{c}^{\infty} (R^{3})$ .

[A1]

The Axiom of Choice is assumed; it implies the Countable Choice required by the cited divergence and polar-measure theorems (The Axiom of Choice).

[F1]

The divergence theorem on a bounded smooth domain identifies the integral of $div F$ with the outward boundary flux for $C^{1}$ vector fields (Divergence on a bounded C1 Euclidean domain).

[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)

[F2]

The sphere parametrization $(\sin θ \cos φ, \sin θ \sin φ, \cos θ)$ has surface Jacobian $\sin θ$ for $0 < θ < π$ , $0 < φ < 2 π$ . Integrating gives $∣ S^{2} ∣ = 2 π \int_{0}^{π} \sin θ d θ = 4 π$ ; the omitted poles and meridian are surface-null in charts. Thus [F90] gives $∣ \partial B_{ϵ} ∣ = 4 π ϵ^{2}$ . (The derivatives of sine and cosine are cosine and minus sine, The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ )

Proof

1.1A1F1givenchoose

Translation reduces to $y = 0$ . For $x \neq 0$ , direct radial differentiation gives $Δ (1 / ∣ x ∣) = 0$ . Fix $R$ so that $φ$ vanishes in a neighborhood of $\partial B_{R}$ , and apply Green's identity, obtained by applying [F1] to $G_{3} \nabla φ - φ \nabla G_{3}$ , on the annulus $B_{R} ∖ \overline{B_{ϵ}}$ . Its inner outward normal is $- e_{r}$ , and $\partial_{- e_{r}} G_{3} = 1 / (4 π ϵ^{2})$ ; therefore $\int_{B_{R} ∖ B_{ϵ}} G_{3} (- Δ φ) d x = \int_{\partial B_{ϵ}} (φ \partial_{- e_{r}} G_{3} - G_{3} \partial_{- e_{r}} φ) d S .$ The outer flux vanishes by the choice of $R$ .

2.1F2F90step 1.1givenalgebra∎

On $\partial B_{ϵ}$ , the first inner term is the spherical average of $φ$ because $\partial_{- e_{r}} G_{3} = 1 / (4 π ϵ^{2})$ , and so tends to $φ (0)$ by continuity. The second inner term has absolute value at most $\sup ∣ \nabla φ ∣ (4 π ϵ^{2}) / (4 π ϵ) = O (ϵ)$ and tends to zero. The omitted ball integral of $G_{3} Δ φ$ is also $O (ϵ^{2})$ since $Δ φ$ is bounded and $\int_{0}^{ϵ} r^{- 1} r^{2} d r = ϵ^{2} / 2$ . Letting $ϵ ↓ 0$ proves the identity at $y = 0$ ; translating coordinates proves it for every $y$ .

Depends on

Used by

Cited to discharge well-definedness by The three-dimensional Laplace Newton kernel.

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Sources

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