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Explicit cellwise error for kernel value and gradient quadrature

Statement

Assume the cell rule of Meshfree cell quadrature and fill radius, h≤ε≤1, x is interior, and f∈C¹(R^d) is bounded with bounded first derivative. Define

Qε,h0f(x)=∑jVjf(xj)Wε(x−xj),Qε,h1f(x)=∑jVjf(xj)∇Wε(x−xj).

Let S_εf(x)=∫f(y)W_ε(x−y)dy and D_εf(x)=∫f(y)∇W_ε(x−y)dy. There is C=C(d,R,C_L,W), independent of ε,h,x, such that

∣Qε,h0f(x)−Sεf(x)∣≤Ch(∥∇f∥∞+ε−1∥f∥∞),

∣Qε,h1f(x)−Dεf(x)∣≤Ch(ε−1∥∇f∥∞+ε−2∥f∥∞).

Facts & Assumptions

Given: An interior point, a cell rule with the fill-radius and local-volume bounds of Meshfree cell quadrature and fill radius, a fixed admissible kernel, ε,h with h≤ε≤1, and f∈C¹ with bounded f and ∇f.

[F1]

Scaled symmetric compactly supported kernel family: W is supported in the closed ball \overline B(0,R), and M₀,M₁,M₂ bound W, ∇W and D²W.

[F2]

Meshfree cell quadrature and fill radius: 0<h<∞, cells meeting B(x,r) have total volume at most C_L(r+h)^d, and an interior point satisfies B(x,Rε+2h)⊂Ω. Hence supp(g₀),supp(g₁)⊂\overline B(x,Rε)⊂B(x,Rε+2h)⊂Ω, so both integrands meet the cellwise lemma's support hypothesis and the full-space integrals equal the corresponding Ω integrals.

[F3]

Scaling identities for kernel moments and derivatives: ||W_ε||∞=ε^(−d)M₀, ||∇W_ε||∞=ε^(−d−1)M₁, and ||D²W_ε||∞=ε^(−d−2)M₂.

[F4]

Cellwise oscillation bound for particle quadrature: cell quadrature error is bounded by 2h times the Lipschitz constant of the integrand times the total active cell volume.

[F5]

Proof

1.1F1F2F3F4F5F6F7

Put g₀(y)=f(y)W_ε(x−y). On any segment y(t)=y₀+t(y₁−y₀), [F6] gives the chain and product rules, and Cauchy-Schwarz [F7] gives |(g₀∘y)'(t)|≤(ε^(−d)M₀||∇f||∞+ε^(−d−1)M₁||f||∞)|y₁−y₀| by [F3]. Applying [F5] on that segment gives the same Lipschitz constant for g₀. Its support is contained in the closed ball \overline B(x,Rε), hence in B(x,(R+1)ε). By [F2], the total volume of cells meeting that support is at most C_L((R+1)ε+h)^d≤C_L(R+2)^dε^d. Apply [F4] and cancel ε^d; the factor 2 is absorbed into C. This gives the first inequality.

2.1F1F2F3F4F5F6F7step 1.1∎

Put g₁(y)=f(y)∇W_ε(x−y). On any segment y(t)=y₀+t(y₁−y₀), [F6] gives componentwise chain and product rules, and Cauchy-Schwarz [F7] gives ||(g₁∘y)'(t)||₂≤(ε^(−d−1)M₁||∇f||∞+ε^(−d−2)M₂||f||∞)|y₁−y₀| by [F3]. Applying the vector mean value inequality [F5] bounds its Lipschitz constant by this expression. Its support is contained in the same closed ball, so [F2] gives the active-volume bound from step 1.1. Apply [F4] to each vector component; taking the Euclidean norm costs at most a factor √d, and [F4] contributes a factor 2; both are absorbed into C. Cancelling ε^d proves the second inequality.

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