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Logarithmic capacity and single-layer scaling under homothety

Statement

Use the logarithmic energy and capacity conventions of Logarithmic capacity and the two-dimensional Laplace layer kernel. If $K \subset R^{2}$ is compact and has $0 < cap (K) < \infty$ and $s > 0$ , then $cap (s K) = s cap (K)$ . If $Γ$ is a closed piecewise-smooth curve and $λ$ is a smooth density, define $Γ_{s} = s Γ$ and $λ_{s} (s y) = s^{- 1} λ (y)$ . Then $V_{Γ_{s}} λ_{s} (s x) = V_{Γ} λ (x) - \frac{\log s}{2 π} \int_{Γ} λ d s_{y} .$ Thus changing length scale adds a constant-mode term of rank at most one (zero when $s = 1$ ); the formula is retained separately from the three-dimensional kernel $1 / (4 π ∣ x - y ∣)$ .

Facts & Assumptions

Given: The energy definition of logarithmic capacity and the two-dimensional layer kernel from Logarithmic capacity and the two-dimensional Laplace layer kernel.

[F1]

A Borel probability measure on $K$ is a nonnegative measure of total mass $1$ supported on $K$ ; dilation $y \mapsto s y$ sends such measures bijectively between $K$ and $s K$ and scales arclength by $s$ .

Proof

1.1F1givenalgebra

Let $μ$ be any probability measure on $K$ and let $s_{#} μ$ be its pushforward under $y \mapsto s y$ . Then $∣ s x - s y ∣ = s ∣ x - y ∣$ , so its logarithmic energy is $\int_{s K} \int_{s K} \log \frac{1}{∣ X - Y ∣} d (s_{#} μ) (X) d (s_{#} μ) (Y) = \int_{K} \int_{K} (\log \frac{1}{∣ x - y ∣} - \log s) d μ (x) d μ (y) = I (μ) - \log s,$ using $μ (K) = 1$ .

2.1F1step 1.1algebra

The pushforward map in [F1] is bijective, with inverse dilation by $1 / s$ . Taking the infimum over all probability measures therefore gives $I (s K) = I (K) - \log s$ . Exponentiating the definition yields $cap (s K) = e^{- I (s K)} = s e^{- I (K)} = s cap (K)$ .

3.1F1givenalgebra∎

In the curve integral, substitute $Y = s y$ . Arclength contributes $d s_{Y} = s d s_{y}$ , while $λ_{s} (s y) = s^{- 1} λ (y)$ , so their product is $λ (y) d s_{y}$ . The kernel identity $- \log ∣ s x - s y ∣ = - \log ∣ x - y ∣ - \log s$ then gives $V_{Γ_{s}} λ_{s} (s x) = - \frac{1}{2 π} \int_{Γ} (\log ∣ x - y ∣ + \log s) λ (y) d s_{y} = V_{Γ} λ (x) - \frac{\log s}{2 π} \int_{Γ} λ d s_{y},$ as asserted.

Depends on

Logarithmic capacity and the two-dimensional Laplace layer kernel

Used by

The logarithmic single layer on a circle changes by a scale term of rank at most one Example

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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)