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Tensor composite Simpson error for a two-dimensional force integral

Statement

Let $a < b$ , $c < d$ , let $n_{x}, n_{y} \geq 2$ be even integers, and set $R = [a, b] \times [c, d]$ , $h_{x} = (b - a) / n_{x}$ , $h_{y} = (d - c) / n_{y}$ . If $F \in C^{4} (R, R^{2})$ and $M_{40} = \max_{R} ∣ \partial_{x}^{4} F ∣, M_{04} = \max_{R} ∣ \partial_{y}^{4} F ∣,$ then the tensor evaluator of Tensor composite Simpson force evaluator on a rectangle satisfies $∣ \int_{R} F - Q_{y} Q_{x} F ∣ \leq \frac{\sqrt{2} ∣ R ∣}{180} (h_{x}^{4} M_{40} + h_{y}^{4} M_{04}), ∣ R ∣ = (b - a) (d - c) .$ Here and in applications of this estimate, $F \in C^{4} (R, R^{2})$ means that each coordinate of $F$ is the restriction to $R$ of a scalar $C^{4}$ function on an open neighborhood of $R$ , in the sense of $C^{k}$ maps and multi-index derivative notation in Euclidean space. The displayed derivatives are taken coordinatewise from these extensions; their values on $R$ are independent of the extensions, since the extensions agree on the interior and their derivatives are continuous up to the boundary.

On every declared input, if each oracle query returns a value or evaluation-failure, the evaluator terminates with its specified status. For the valid inputs above, with the exact oracle returning $F$ at every grid point, the evaluator terminates and returns a vector after exactly $(n_{x} + 1) (n_{y} + 1)$ oracle queries. This is an exact-arithmetic quadrature bound, not a roundoff estimate.

Facts & Assumptions

Given: The rectangle, even panel counts, $F \in C^{4} (R, R^{2})$ , and the derivative bounds in the Statement.

[F1]

The tensor evaluator applies the one-dimensional composite Simpson rule successively in each coordinate (Tensor composite Simpson force evaluator on a rectangle).

[F2]

For a scalar $C^{4}$ function on an interval of length $B - A$ with even panel count and width $h$ , Simpson error is at most $(B - A) h^{4} \max ∣ f^{(4)} ∣ / 180$ (Composite Simpson error bound for a four-times continuously differentiable function).

[F3]

The composite Simpson weights are nonnegative and sum to the interval length (Composite Simpson rule on an even uniform grid).

[F4]

Continuous fourth derivatives attain finite maxima on the compact rectangle ( $C^{k}$ maps and multi-index derivative notation in Euclidean space, Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[F5]

The Euclidean norm on $R^{2}$ has the triangle inequality and the displayed coordinate formula (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation).

Proof

1.1givenF1F2F4F5algebra

Invalid endpoints or panel counts return immediately without queries. Otherwise, the grid is finite, and any failed query returns immediately. If no query fails, each supplier call uses an evaluable interpolant, admissible panel counts and a finite sum, so the evaluator returns a vector. For the valid inputs in the error bound, there are exactly $(n_{x} + 1) (n_{y} + 1)$ grid points; the exact oracle returns $F$ at each, so none of the valid-input queries fails. The remaining supplier calls use evaluable interpolants, even positive panel counts and finite sums. Thus the evaluator terminates with a vector after the stated number of oracle queries. Fix a component $F_{k}$ , $k = 0, 1$ , and write $I_{x}, I_{y}$ for exact integration and $S_{x}, S_{y}$ for the composite Simpson operators. Add and subtract $I_{y} S_{x} F_{k}$ : $I_{y} I_{x} F_{k} - S_{y} S_{x} F_{k} = I_{y} (I_{x} - S_{x}) F_{k} + (I_{y} - S_{y}) S_{x} F_{k} .$ The maxima in the Statement are finite by [F4]. By [F5], each coordinate of a vector has absolute value at most its Euclidean norm, so $∣ \partial_{x}^{4} F_{k} ∣ \leq M_{40}$ and $∣ \partial_{y}^{4} F_{k} ∣ \leq M_{04}$ on $R$ . For each fixed $y$ , [F2] bounds the first one-dimensional error by $(b - a) h_{x}^{4} M_{40} / 180$ ; integrating in $y$ gives $(b - a) (d - c) h_{x}^{4} M_{40} / 180$ .

2.1F2F3step 1.1

The $y$ derivative commutes with the finite $x$ -weighted sum. Applying [F2] to each fixed $x$ and then summing the nonnegative Simpson weights from [F3] bounds the second term by $(b - a) (d - c) h_{y}^{4} M_{04} / 180$ . Thus each coordinate error is at most $E = ∣ R ∣ (h_{x}^{4} M_{40} + h_{y}^{4} M_{04}) / 180$ .

3.1F5step 2.1algebra∎

If both coordinate errors have absolute value at most $E$ , their Euclidean norm is at most $(E^{2} + E^{2})^{1 / 2} = \sqrt{2} E$ . Substitution of the value of $E$ from step 2.1 proves the Statement.

Depends on

Used by

Cited to discharge well-definedness by Tensor composite Simpson force evaluator on a rectangle.

Dependency tree · two levels

67 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Tobin Driscoll and Richard Braun, Fundamentals of Numerical Computation, Section 5.6 (standard reference, not scraped)
MIT OpenCourseWare 18.330, Introduction to Numerical Analysis, Chapter 5 (standard reference, not scraped)