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Periodic midpoint collocation has an explicit local second-order error budget

Statement

Consider the augmented phase-fixed BVP of The phase-bordered periodic-orbit BVP residual on [0,1], with exact solution u∗∈C3, field f~, and affine boundary map B. Put M2=∥u∗′′∥∞,M3=∥u∗′′′∥∞,Lf=sup⁡∥Df~∥2,C0=M3/24+LfM2/8. For every admitted mesh cell with midpoint mi, assume the segment joining u∗(mi) to (u∗(ti)+u∗(ti+1))/2 lies in the field domain and ∥Df~∥op≤Lf on that segment. Let Gh be the normalized midpoint residual on a uniform mesh and Y∗ the samples of u∗. Suppose that for all 0<h≤h0, Jh=DGh(Y∗) is invertible with ∥Jh−1∥∞→∞≤K, and on a fixed-radius ball of radius r about Y∗ the maps DGh are LG-Lipschitz in the block maximum norm. Assume 2KLGr≤1/2, KC0h2≤r/2, and 2K2LGC0h2≤1/2. Then Gh has a unique zero Yh in that ball. Every candidate Y in the ball satisfying ∥Gh(Y)∥∞≤ε obeys

∥PhY−u∗∥∞≤4Kε+(2KC0+M28)h2,

where Ph is linear interpolation of the node values. Thus the residual and discretization constants are explicit in the supplied derivative and inverse bounds. For the augmented state, the estimate includes both the orbit coordinates and the period coordinate.

Facts & Assumptions

Given: The exact augmented solution, uniform mesh, normalized midpoint residual, bounds M2,M3,Lf,K,LG, and the stated radius and mesh inequalities.

[F1]

The midpoint residual has rows (ui+1−ui)/h−f~((ui+ui+1)/2) and the phase-periodic boundary row (Normalized implicit-midpoint collocation residual for a two-point BVP, Multiple-shooting and midpoint-collocation residuals for a phase-fixed periodic orbit).

[F2]

For a BVP with a boundedly invertible continuous linearization, its sampled midpoint Jacobian has a uniform inverse bound (A transverse phase border makes the periodic BVP linearization boundedly invertible, Uniform inverse bound for the sampled midpoint linearized BVP); here the explicit K is assumed as a supplied upper bound for those discrete inverses.

[F3]

A contraction of a nonempty complete metric space into itself has a unique fixed point (A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point). The closed ball in the node space is complete: a block-maximum Cauchy sequence is Cauchy in each of its finitely many Euclidean blocks, each block converges by R and Rn for n≥1 with the Euclidean metric are complete, componentwise from the Cauchy criterion in R, and the finite maximum of the block errors tends to zero; closedness keeps the limit in the ball.

[F5]

If ∥J−1(A−J)∥≤1/2, the perturbed operator is invertible with inverse norm at most 2∥J−1∥ (Neumann series and small perturbations of bounded inverses).

[F6]

The chain rule differentiates a composition of totally differentiable maps (The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)).

Proof

1.1F1F4F6F7givenalgebra

On a mesh cell centered at mi, repeated vector FTC gives u∗′(mi+s)=u∗′(mi)+su∗′′(mi)+r(s), where r(s)=∫0s∫0vu∗′′′(mi+w) dw dv. For negative limits reverse their orientation before applying the norm inequality. Thus ∥r(s)∥2≤M3∣s∣2/2. Averaging over [−h/2,h/2] cancels the odd term and gives a difference-quotient defect at most M3h2/24. Applying the same integral argument to u∗ at mi±h/2 bounds each first-order remainder by M2h2/8; their average therefore bounds the endpoint-average displacement from u∗(mi) by M2h2/8. Integrating Df~ on the connecting segment bounds the corresponding field difference by LfM2h2/8. The affine boundary residual is zero. Taking the maximum of the cell bounds gives ∥Gh(Y∗)∥∞≤C0h2.

2.1F2F3F4F6F7step 1.1given

Define Th(Y)=Y−Jh−1Gh(Y). The Jacobian Lipschitz bound gives ∥Th(Y)−Th(Z)∥∞≤KLGr∥Y−Z∥∞≤14∥Y−Z∥∞ on the ball. Also ∥Th(Y∗)−Y∗∥∞≤KC0h2≤r/2, so Th maps the ball into itself. By [F3], it has a unique fixed point Yh, and invertibility of Jh makes fixed points exactly the zeros of Gh. The fixed-point equation also gives ∥Yh−Y∗∥∞≤KC0h2/(1−1/4)≤2KC0h2.

3.1F5step 2.1given

The perturbation bound ∥DGh(Yh)−Jh∥≤LG∥Yh−Y∗∥ implies ∥Jh−1(DGh(Yh)−Jh)∥≤2K2LGC0h2≤1/2. The small-perturbation inverse bound [F5] therefore gives DGh(Yh)−1 with norm at most 2K.

4.1F6F7step 3.1givenalgebra

Put δ=Y−Yh and define R(t)=Gh(Yh+tδ)−Gh(Yh)−tDGh(Yh)δ for 0≤t≤1. The segment lies in the ball. By the chain rule [F6], R′(t)=(DGh(Yh+tδ)−DGh(Yh))δ, so the Lipschitz bound gives ∥R′(t)∥∞≤LGt∥δ∥∞2. For each Euclidean block index j, [F7] gives ∥Rj(1)∥2≤∫01∥Rj′(t)∥2dt≤∫01∥R′(t)∥∞dt; taking the finite maximum over j gives ∥R(1)∥∞≤∫01∥R′(t)∥∞dt≤(LG/2)∥δ∥∞2. Thus Gh(Y)=DGh(Yh)(Y−Yh)+R(1) with this remainder bound. Since both vectors lie in the radius-r ball, ∥Y−Yh∥∞≤2r, and 2KLGr≤1/2 yields ∥Y−Yh∥∞≤2Kε+12∥Y−Yh∥∞. Hence ∥Y−Yh∥∞≤4Kε.

5.1F4step 2.1step 4.1givenalgebra∎

Linear interpolation is a convex combination, so ∥Ph(Y−Yh)∥∞≤∥Y−Yh∥∞. On a cell [a,b] of length h, repeated FTC gives u∗(t)−PhY∗(t)=−∫abKt(s)u∗′′(s) ds, where Kt(s)=(b−t)(s−a)/h for s≤t and Kt(s)=(t−a)(b−s)/h for s≥t. This kernel is nonnegative and has integral (t−a)(b−t)/2≤h2/8, so [F4] gives the interpolation bound M2h2/8. The sample error of Yh is at most 2KC0h2 by step 2.1. Combining these bounds with step 4.1 proves the displayed estimate.

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