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Levin endpoint identity and conditioned residual bound

Statement

Let a<b, s≥1, ω∈R∖{0}, complex-valued f∈Cs[a,b], and real-valued g∈Cs+1[a,b] with ∣g′∣≥c>0. Use a nonsingular collocation system from Levin collocation for a nonstationary oscillatory integral, with complex-valued basis functions ψk:[a,b]→C in Cs+1[a,b]. Write Ajk=Lωψk(xj), yj=f(xj), c=A−1y, p=∑kckψk, Q=p(b)eiωg(b)−p(a)eiωg(a), and r=f−Lωp. Then Iω(f,g;[a,b])−Q=Iω(r,g;[a,b])=∑j=0s−1(−1)j(iω)j+1[Tgjr(x)g′(x)eiωg(x)]ab+(−1)s(iω)sIω(Tgsr,g;[a,b]). Put Ej=∣Tgjf(a)/g′(a)∣+∣Tgjf(b)/g′(b)∣, Vj=∥vj(a)∥2+∥vj(b)∥2 for vj,k(x)=Tgj(Lωψk)(x)/g′(x), and Ks=(∑k∥Tgs(Lωψk)∥∞2)1/2. Then ∣Iω(f,g)−Q∣≤∑j=0s−1Ej+∥A−1∥2∥y∥2Vj∣ω∣j+1+b−a∣ω∣s(∥Tgsf∥∞+∥A−1∥2∥y∥2Ks). If an approximate coefficient vector c^ has exact residual ρ=Ac^−y and endpoint vector dk=ψk(b)eiωg(b)−ψk(a)eiωg(a), its output satisfies ∣Q(c^)−Q∣≤∥d∥2∥A−1∥2∥ρ∥2=∥d∥2κ2(A)∥A∥2∥ρ∥2. If Bs denotes the right-hand side of the preceding collocation bound, then ∣Iω(f,g)−Q(c^)∣≤Bs+∥d∥2∥A−1∥2∥ρ∥2. All vector norms are Euclidean, ∥A∥2=sup⁡∥v∥2=1∥Av∥2 is the induced matrix norm, and κ2(A)=∥A∥2∥A−1∥2. Thus a small nodal residual alone does not provide an integral-error bound; the smooth residual terms and inverse solve amplification are also required.

Facts & Assumptions

Given: The stated functions, nonzero frequency, nonstationary phase, nonsingular collocation matrix, and exact arithmetic residual ρ when an approximate coefficient vector is considered.

[F1]

Levin collocation for a nonstationary oscillatory integral: Lωp=p′+iωg′p, Ajk=Lωψk(xj), and the endpoint output is p(b)eiωg(b)−p(a)eiωg(a).

[F2]

Derivative of a complex oscillatory phase: (eiωg)′=iωg′eiωg.

[F3]

If u,v are differentiable on [a,b] with u′,v′ integrable, then ∫abuv′=u(b)v(b)−u(a)v(a)−∫abu′v: for real C1 functions, ∫uv′=[uv]−∫u′v; the complex identity follows componentwise.

[F4]

Repeated integration by parts for a nonstationary oscillatory integral: the s-step expansion and its remainder bound apply to every Cs amplitude when ∣g′∣≥c>0.

[F5]

Cauchy–Schwarz: ∣⟨x,y⟩∣≤∥x∥ ∥y∥, with equality exactly for dependent pairs: for finite real or complex vectors, ∣⟨z,w⟩∣≤∥z∥2∥w∥2.

[F6]

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

[F7]

The complex modulus is subadditive, ∣z+w∣≤∣z∣+∣w∣ (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

[F8]

On Cn the Euclidean inner product is ⟨x,y⟩=∑kxkyk‾, linear in the first argument, and ∥d‾∥2=∥d∥2 by Real and complex inner-product spaces and their induced length and Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive.

Proof

1.1givenF1F2F3F6algebra

For any p∈C1[a,b], [F2] gives (peiωg)′=(Lωp)eiωg. Apply [F3] to the real and imaginary components of p and eiωg; the product rule and cancellation of the two iωg′p terms yield Iω(Lωp,g)=[peiωg]ab. With p from [F1] and r=f−Lωp, componentwise linearity [F6] gives Iω(f,g)−Q=Iω(r,g).

1.2F1givenalgebra

The coefficient equations imply A(c^−c)=ρ, so c^−c=A−1ρ. The endpoint formula is linear in the coefficient vector, hence Q(c^)−Q=d⊤A−1ρ.

2.1givenF4step 1.1

The hypotheses on f,g,ψk make r∈Cs[a,b]. Apply [F4] to r to obtain the displayed exact expansion and bound its integral remainder by (b−a)∣ω∣−s∥Tgsr∥∞.

2.2F1F5F8step 1.2algebra

With x=A−1ρ, [F8] gives d⊤x=⟨x,d‾⟩ and ∥d‾∥2=∥d∥2; Cauchy--Schwarz [F5] therefore gives ∣d⊤A−1ρ∣≤∥d∥2∥A−1∥2∥ρ∥2. The definition κ2(A)=∥A∥2∥A−1∥2 and nonsingularity give the displayed condition-number form.

3.1givenF4F5F7step 2.1algebra

Since c=A−1y, ∥c∥2≤∥A−1∥2∥y∥2. Linearity of Tg gives Tgjr=Tgjf−∑kckTgj(Lωψk). At each endpoint, [F5] bounds the corresponding divided sum by ∥c∥2∥vj(a)∥2 or ∥c∥2∥vj(b)∥2; [F7] combines it with the amplitude term. On the interval [F5] bounds ∥∑kckTgs(Lωψk)∥∞ by ∥c∥2Ks, and [F7] gives ∥Tgsr∥∞≤∥Tgsf∥∞+∥c∥2Ks. Substitution in the expansion from step 2.1 gives the stated Ej,Vj,Ks bound.

4.1F7step 3.1step 2.2algebra∎

The complex triangle inequality [F7] combines the exact-collocation bound in step 3.1 with step 2.2, giving the stated integral bound for Q(c^).

Remarks

The residual ρ in the theorem is the exact residual of the stated matrix equation. If matrix products and solves are rounded, this theorem alone does not certify that exact residual or a rounded solve.

Depends on

Used by

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Sources