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Symmetry cancels the first kernel moment
Statement
For the kernel W and its scaled family W_ε of Scaled symmetric compactly supported kernel family, the first moment vanishes: ∫zW(z) dz=0 and ∫zW_ε(z) dz=0 for every ε>0.
Facts & Assumptions
Given: The normalized radial compactly supported kernel W and ε>0.
Scaled symmetric compactly supported kernel family: W(−z)=W(z), W is continuous and compactly supported, and W_ε(z)=ε^(−d)W(z/ε).
A compactly supported Riemann integrand admits the global change-of-variables formula from a diffeomorphism near the relevant compact preimage: let U⊂R^d be open and T:U→R^d be injective C¹ with invertible derivative. If f:R^d→R is compactly supported Riemann integrable with supp(f)⊂T(U), then ∫{R^d}f(y)dy=∫{R^d}h(x)dx, where h(x)=f(T(x))|det DT(x)| on U and h=0 outside U. Here U=R^d and the reflection and positive scaling maps are onto R^d.
A continuous real function on a compact Jordan measurable set is Riemann integrable over that set: a continuous real function on a compact Jordan box is Riemann integrable there.
Proof
For each coordinate k, g_k(z)=z_kW(z) is continuous and compactly supported by [F1], hence Riemann integrable on a bounding box by [F3]. It satisfies g_k(−z)=−g_k(z). Apply [F2] to T(z)=−z, whose determinant has absolute value one: ∫g_k(z)dz=∫g_k(−z)dz=−∫g_k(z)dz. Hence ∫z_kW(z)dz=0.
For each k, z_kW_ε(z) is continuous and compactly supported, so [F3] makes it integrable on a bounding box. Apply [F2] to T(u)=εu: ∫z_kW_ε(z)dz=ε∫u_kW(u)du=0 by step 1.1. This holds for every k, so both vector first moments vanish.
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Sources
- Volker Springel, Smoothed Particle Hydrodynamics in Astrophysics, Annual Review of Astronomy and Astrophysics 48 (2010), author full text (standard reference, not scraped)
- Benjamin Seibold, MIT OpenCourseWare 18.336, Numerical Methods for Partial Differential Equations, Lecture 25: Particle Methods (Spring 2009) (standard reference, not scraped)