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The Duffy weighted product rule has an explicit bound using pure coordinate derivatives

Statement

Assume Countable Choice, let α<2, ρ=1−α, n≥1, and let G∈C2n in each coordinate on [0,1]2. Let p~n,ρ and p~n,0 be the monic node polynomials of the pushed-forward Gauss–Jacobi rules for rρdr and ds, respectively. Define Mρ=∫01rρdr=1ρ+1,Kρ=1(2n)!R∫01p~n,ρ(r)2rρdr,K0=1(2n)!R∫01p~n,0(s)2ds. Then the rule of Weighted product Gauss quadrature after the Duffy transform satisfies ∣∫T∥(x,y)∥2−αq(x,y) dx dy−Qn,αD(G)∣≤Kρ∥∂r2nG∥∞+MρK0∥∂s2nG∥∞, where G is the transformed amplitude from The Duffy map turns a vertex singularity into an integrable radial weight.

Facts & Assumptions

Given: Countable Choice, α<2, n≥1, continuous q on the reference triangle, the Duffy rule, and the stated coordinate smoothness of its transformed amplitude G.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice is used to supply the Lebesgue-density positive measures underlying the two Gaussian rules.

[F1]

The Duffy map turns a vertex singularity into an integrable radial weight: the singular triangle integral equals ∫01∫01rρG(r,s)dsdr.

[F2]

Weighted product Gauss quadrature after the Duffy transform: its finite sum is the tensor product of the n-node rules for rρdr and ds.

[F3]

Gauss–Jacobi exactness and the derivative remainder: each pushed-forward one-dimensional rule has error constant (2n)!−1∫p~n2dμ and integrates the measure mass exactly.

[F4]

A tensor product of two positive Gaussian rules has a derivative error bound: the tensor error is bounded by the first coordinate constant times the second mass and the first mass times the second coordinate constant.

[F5]

The Jacobi weight defines a finite positive measure with full support: the weights (1+t)ρ for ρ>−1 and 1 define finite positive full-support measures with moments of every finite order.

[F6]

The improper integral of a power at zero: ∫01rρdr=1/(ρ+1) for ρ>−1.

Proof

1.1A1F2F5givenalgebra

The exponent ρ=1−α satisfies ρ>−1. By [F5] and [A1], the Jacobi weights (1+t)ρ and 1 define finite positive measures with full support; the affine node/weight scalings in [F2] give their Gaussian rules for rρdr and ds.

1.2F2F3F6F7algebra

The first measure has mass Mρ=1/(ρ+1) by [F6], and the second has mass 1. For the first affine pushforward, r=(1+t)/2 gives rρdr=2−(ρ+1)(1+t)ρdt and its monic node polynomial is p~n,ρ(r)=2−npn,ρ(2r−1). Hence its squared-node-polynomial integral is 2−(ρ+1+2n)∫−11pn,ρ(t)2(1+t)ρdt. The derivative of H((1+t)/2) contributes 2−2n, so [F3] gives exactly Kρ as defined. The same calculation with ρ=0 gives K0. Positivity of the factorial denominator follows from [F7].

2.1F1F2F4step 1.2∎

Apply [F4] to the transformed integral and tensor rule identified by [F1]–[F2]. Substituting the masses and constants from step 1.2 yields the displayed error bound.

Depends on

Used by

Cited to discharge well-definedness by Weighted product Gauss quadrature after the Duffy transform.

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