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Gaussian and Simpson mass assembly give distinct exact and approximate checks

Example

Assume Countable Choice for the Lebesgue measure construction. For the degree-one monic Legendre basis P0=1,P1=x on [−1,1], the two-node Gaussian mass matrix is diag⁡(2,2/3). For the Chebyshev arcsine pairing, composite Simpson with two panels gives MS=(π00π/3),G=(π00π/2). The Simpson error in the (1,1) entry is π/6 and satisfies the itemwise error estimate.

Facts & Assumptions

Given: Countable Choice, Lebesgue measure on [−1,1], its two-node Gauss–Legendre rule, and the Chebyshev pairing sampled by two-panel composite Simpson.

[A1]

The Axiom of Countable Choice (ACω): Countable Choice is assumed for the Lebesgue measure construction and Riemann–Lebesgue agreement used in the local two-node Gaussian calculation.

[F1]

Positive-measure Gaussian quadrature moment data, Monic orthogonal polynomials exist uniquely through the quadrature order, The m-node Gaussian quadrature rule for a positive measure, and Gaussian quadrature is exact through degree 2m−1 and has positive weights: valid order-three moment data for m=2 give the unique monic orthogonal polynomial, its two root nodes, cardinal-integral weights, and exactness through degree three.

[F2]

Legendre polynomials have explicit norms and weighted derivative orthogonality: ∫−11P02=2, ∫−11P0P1=0, and ∫−11P12=2/3.

[F3]

Assemble a weighted Galerkin mass matrix from Gaussian nodes and Gaussian quadrature exactly assembles the orthogonal Galerkin mass matrix: the mass entries are Gaussian sums and equal the corresponding exact polynomial integrals.

[F4]

Arcsine-weighted inner product on the Chebyshev interval: Gij=∫0πTi(cos⁡θ)Tj(cos⁡θ) dθ for i,j∈{0,1}.

[F5]

Composite-Simpson assembly of the Chebyshev weighted mass matrix: for two panels, MijS=π6(gij(0)+4gij(π/2)+gij(π)).

[F6]

Composite Simpson error bound for Chebyshev weighted mass entries: ∣Gij−MijS∣≤πh4(i+j)4/180 for even panel count and h=π/M.

[F7]

Chebyshev polynomials are orthogonal for the arcsine weight: the exact mass matrix for T0,T1 is diag⁡(π,π/2).

[F8]

Quarter-turn values and shifts by pi/2 and pi: cos⁡0=1, cos⁡(π/2)=0, and cos⁡π=−1.

[F12]

The Lebesgue integral is linear on L1(μ): finite linear combinations of integrable polynomial functions may be integrated termwise.

[F14]

Square roots exist: a unique a≥0 with (a)2=a; the positives are {x2:x≠0} supplies the positive root 3; Basic properties of the absolute value, A function differentiable at c is continuous at c, and Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function give continuity of ∣x∣k and the polynomial integrands, as used with [F11]. Measure monotonicity in step 1.1 follows from the measure axioms in [F10]: if A⊆B, then B=A⊔(B∖A) and μ(B)=μ(A)+μ(B∖A)≥μ(A).

Proof

1.1A1F1F10F11F14algebra

Under [A1], let μ be Lebesgue measure restricted to the relative Borel sets of [−1,1]. It is a positive measure with total mass 2 by [F10]. Every relative open neighborhood of a point of [−1,1] contains a positive-length interval, including a one-sided interval at either endpoint, so [F10] and [F14] give that neighborhood positive measure. Thus the relative support is all of [−1,1], containing at least three points. Every ∣x∣k for 0≤k≤3 is continuous and bounded on [−1,1] by [F14], so [F11] gives its integrability. These are exactly the order-three, m=2 moment-data conditions in [F1]. The primitives x,x2/2,x3/3,x4/4 give moments μ0=2,μ1=0,μ2=2/3,μ3=0 by [F11].

1.2F4F5F7F8F9algebra

By [F9], g00=1, g01=cos⁡θ, and g11=cos⁡2θ. Substitution of the quarter-turn values [F8] into [F5] gives M00S=π, M01S=0, and M11S=π/3. The exact mass matrix is G=diag⁡(π,π/2) by [F7].

2.1F1F12F14step 1.1algebra

The monic polynomial p2(x)=x2−1/3 is orthogonal to 1 and x, since μ2−μ0/3=0 and μ3−μ1/3=0 by step 1.1 and [F12]. Finite linearity extends this to every polynomial of degree below two, so uniqueness in [F1] identifies p2 with the Gaussian orthogonal polynomial. Its roots satisfy x2=1/3, hence are x1=−1/3 and x2=1/3 by [F14]. Their cardinal polynomials are ℓ1(x)=(1−3x)/2 and ℓ2(x)=(1+3x)/2, so [F12] and the first two moments give w1=w2=1. The rule Definition and exactness theorem in [F1] therefore give Q2(f)=f(−1/3)+f(1/3), exact for degrees through three.

3.1A1F1F2F3step 2.1algebra

The Gaussian sum for the basis 1,x is (1−1/311/3) ⁣T(1001)(1−1/311/3)=(2002/3). The entries agree with exact integrals by [F2], as also asserted by [F3].

4.1F6F13step 1.2algebra∎

The only nonzero entry error is ∣G11−M11S∣=π/6. The Simpson theorem [F6] applies with M=2, h=π/2, and i=j=1, and gives its bound π5/180. This bound is at least π/6, since [F13] gives π4>81>30, equivalently π5/180>π/6. Therefore the displayed finite calculation matches the general entrywise estimate.

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