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Nonstationary error bound for endpoint Filon interpolation

Statement

Let $a < b$ , $ω \in R ∖ {0}$ , $f \in C^{2} [a, b]$ be complex-valued, and $g \in C^{3} [a, b]$ be real-valued with $∣ g^{'} ∣ \geq c > 0$ . Use the endpoint Lagrange rule from Filon quadrature from nodal interpolation and phase moments and put $r = f - p_{n}$ . If $μ_{j} = \int_{a}^{b} ℓ_{j} e^{i ω g}$ and the rule uses ${\hat{μ}}_{j}$ with $∣ {\hat{μ}}_{j} - μ_{j} ∣ \leq ϵ_{j}$ , then $∣ I_{ω} (f, g; [a, b]) - Q_{n} ∣ \leq \frac{∣ T_{g} r (a) / g^{'} (a) ∣ + ∣ T_{g} r (b) / g^{'} (b) ∣ + (b - a) ∥ T_{g}^{2} r ∥_{\infty}}{∣ ω ∣^{2}} + η_{n} .$ With exact moments $η_{n} = 0$ , this is an $O (∣ ω ∣^{- 2})$ endpoint-interpolation error bound. The constant depends on the fixed amplitude, phase, nodes, and derivative norms.

Facts & Assumptions

Given: The hypotheses in the Statement, the integral convention on Finite-interval oscillatory integral and phase convention, and the endpoint nodes $c_{0} = a, c_{n} = b$ .

[F1]

Lagrange interpolation at distinct real nodes: $p_{n} = \sum_{j = 0}^{n} f (c_{j}) ℓ_{j}$ and $p_{n} (c_{j}) = f (c_{j})$ .

[F2]

Filon quadrature from nodal interpolation and phase moments: $Q_{n} = \sum_{j} f (c_{j}) {\hat{μ}}_{j}$ and $η_{n} = \sum_{j} ∣ f (c_{j}) ∣ ϵ_{j}$ .

[F3]

Repeated integration by parts for a nonstationary oscillatory integral: the two-step identity for $I_{ω} (h, g)$ has first boundary term $[(h / g^{'}) e^{i ω g}]_{a}^{b} / (i ω)$ , second term $- [(T_{g} h / g^{'}) e^{i ω g}]_{a}^{b} / (i ω)^{2}$ , and remainder $I_{ω} (T_{g}^{2} h, g) / (i ω)^{2}$ .

[F4]

The complex modulus is subadditive, $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F5]

Real Riemann integrals on an interval are linear by Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ with $m = 1$ ; the componentwise complex definition Finite-interval oscillatory integral and phase convention extends this to complex integrands.

Proof

1.1givenF1F2F4F5algebra

Let $Q_{n}^{*} = \sum_{j} f (c_{j}) μ_{j}$ . By [F1] and [F5], linearity of the integral gives $Q_{n}^{*} = I_{ω} (p_{n}, g)$ ; by [F2] and the finite-sum triangle inequality [F4], $∣ Q_{n}^{*} - Q_{n} ∣ \leq η_{n}$ . The same linearity gives $I_{ω} (f, g) - Q_{n} = I_{ω} (r, g) + (Q_{n}^{*} - Q_{n})$ .

1.2givenF1F3algebra

The endpoint nodes and [F1] give $r (a) = r (b) = 0$ , so the first boundary term in [F3] vanishes. Applying [F3] to $r$ leaves $I_{ω} (r, g) = - \frac{[(T_{g} r / g^{'}) e^{i ω g}]_{a}^{b}}{(i ω)^{2}} + \frac{I_{ω} (T_{g}^{2} r, g)}{(i ω)^{2}} .$

2.1givenF2F3F4step 1.1step 1.2algebra∎

Since $∣ e^{i ω g} ∣ = 1$ , the complex triangle inequality [F4] bounds the displayed boundary term by $∣ T_{g} r (a) / g^{'} (a) ∣ + ∣ T_{g} r (b) / g^{'} (b) ∣$ , and the integral term is at most $(b - a) ∥ T_{g}^{2} r ∥_{\infty}$ by [F3]. Combine these bounds with step 1.1 and $∣ (i ω)^{2} ∣ = ∣ ω ∣^{2}$ using [F4] to obtain the assertion.

Depends on

Used by

A Fourier integral missed by one sample but captured by Filon moments Example

Dependency tree · two levels

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Sources

NIST Digital Library of Mathematical Functions, §2.3(i) (standard reference, not scraped)
A. Iserles, S. P. Nørsett, and S. Olver, Highly oscillatory quadrature: The story so far, §3.2 (standard reference, not scraped)
D. Levin, Procedures for computing one- and two-dimensional integrals of functions with rapid irregular oscillations, §3 (standard reference, not scraped)