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The mollified Biot-Savart kernel is smooth and globally Lipschitz

Statement

Assume Countable Choice and the mollifier of Smooth mollified Biot-Savart kernel. For every $ε > 0$ , $G_{ε}$ and $K_{ε}$ are $C^{\infty}$ on $R^{2}$ , and there is $C_{ρ} < \infty$ independent of $ε$ such that $\sup_{x \in R^{2}} ∥ D K_{ε} (x) ∥_{o p} \leq C_{ρ} ε^{- 2} .$ Consequently $K_{ε}$ is globally Lipschitz with constant at most $C_{ρ} ε^{- 2}$ .

Facts & Assumptions

Given: Countable Choice, the singular kernel $K (z) = z^{⊥} / (2 π ∣ z ∣^{2})$ , a smooth radial compactly supported unit-mass mollifier $ρ_{ε}$ , and $ε > 0$ .

[A1]

Assume Countable Choice for the smooth convolution and polar-coordinate suppliers (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Convolution of a locally integrable function with a smooth compactly supported mollifier is smooth, and derivatives pass to the mollifier (Convolution with a mollifier is smooth, and derivatives pass under the integral sign).

[F2]

Polar coordinates give the integral of a nonnegative radial function on $R^{2}$ (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma).

[F3]

The linear scaling map $x \mapsto ε x$ scales Lebesgue measure by its determinant (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F4]

A globally bounded derivative on the convex domain $R^{2}$ gives a global Lipschitz bound (On a convex open set, a uniform bound $∥ D f (z) v ∥_{2} \leq M ∥ v ∥_{2}$ implies $∥ f (y) - f (x) ∥_{2} \leq M ∥ y - x ∥_{2}$ ).

Proof

1.1A1F1F2givenalgebra

By [F1], $G_{ε}$ and $K_{ε}$ are smooth and $D K_{ε} (x) = \int_{R^{2}} K (x - z) D ρ_{ε} (z) d z,$ componentwise. The gradient $D ρ_{ε}$ is supported in $∣ z ∣ \leq ε$ and satisfies $∥ D ρ_{ε} ∥_{\infty} = ε^{- 3} ∥ D ρ ∥_{\infty}$ . If $∣ x ∣ \leq 2 ε$ , then $∣ x - z ∣ \leq 3 ε$ . Since $∣ K (w) ∣ \leq (2 π ∣ w ∣)^{- 1}$ , polar integration [F2] gives $\int_{∣ w ∣ \leq 3 ε} ∣ K (w) ∣ d w \leq C ε$ . Thus $∥ D K_{ε} (x) ∥ \leq C_{ρ} ε^{- 2}$ in this region.

2.1F1F3step 1.1algebra

If $∣ x ∣ > 2 ε$ and $∣ z ∣ \leq ε$ , then $∣ x - z ∣ > ∣ x ∣ / 2$ and $∣ K (x - z) ∣ \leq C / ∣ x ∣$ . Scaling [F3] gives $∥ D ρ_{ε} ∥_{1} = ε^{- 1} ∥ D ρ ∥_{1}$ , so $∥ D K_{ε} (x) ∥ \leq \frac{C}{∣ x ∣} ∥ D ρ_{ε} ∥_{1} \leq C_{ρ} ε^{- 2} .$ Combining this with step 1.1 yields the global derivative bound.

3.1F4step 2.1∎

Apply [F4] to the smooth map $K_{ε}$ along the segment between any $x, y \in R^{2}$ . The derivative bound in step 2.1 gives $∣ K_{ε} (x) - K_{ε} (y) ∣ \leq C_{ρ} ε^{- 2} ∣ x - y ∣ .$ This is the claimed global Lipschitz property.

Depends on

Used by

Cited to discharge well-definedness by Smooth mollified Biot-Savart kernel.

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Sources

J. Liu and X. Xin, Convergence of Vortex Methods for Weak Solutions to the 2-D Euler Equations with Vortex Sheet Data (standard reference, not scraped)
Gregory R. Baker and Lan D. Pham, A Comparison of Blob Methods for Vortex Sheet Roll Up (standard reference, not scraped)