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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 , with exact solution , field , and affine boundary map . Put For every admitted mesh cell with midpoint , assume the segment joining to lies in the field domain and on that segment. Let be the normalized midpoint residual on a uniform mesh and the samples of . Suppose that for all , is invertible with , and on a fixed-radius ball of radius about the maps are -Lipschitz in the block maximum norm. Assume , , and . Then has a unique zero in that ball. Every candidate in the ball satisfying obeys
where 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 , and the stated radius and mesh inequalities.
The midpoint residual has rows 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).
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 is assumed as a supplied upper bound for those discrete inverses.
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 and for with the Euclidean metric are complete, componentwise from the Cauchy criterion in , and the finite maximum of the block errors tends to zero; closedness keeps the limit in the ball.
If is differentiable with integrable then ; and a bounded derivative makes Lipschitz gives the componentwise integral of a continuous vector derivative. For and integrable when , ; for , is integrable bounds the Euclidean norm of that integral. Reversed integration limits use The integral with oriented limits: and . Apply these in each output block; taking the finite maximum preserves the same bound in the block maximum norm.
If , the perturbed operator is invertible with inverse norm at most (Neumann series and small perturbations of bounded inverses).
The chain rule differentiates a composition of totally differentiable maps (The chain rule for total derivatives: ).
The fundamental theorem of calculus applies coordinatewise to a continuously differentiable vector curve, and the norm of a Euclidean vector-valued integral is at most the integral of the norm (The second fundamental theorem: if is differentiable on with and is integrable, then , For and integrable when , ; for , is integrable).
Proof
On a mesh cell centered at , repeated vector FTC gives , where . For negative limits reverse their orientation before applying the norm inequality. Thus . Averaging over cancels the odd term and gives a difference-quotient defect at most . Applying the same integral argument to at bounds each first-order remainder by ; their average therefore bounds the endpoint-average displacement from by . Integrating on the connecting segment bounds the corresponding field difference by . The affine boundary residual is zero. Taking the maximum of the cell bounds gives .
Define . The Jacobian Lipschitz bound gives on the ball. Also , so maps the ball into itself. By [F3], it has a unique fixed point , and invertibility of makes fixed points exactly the zeros of . The fixed-point equation also gives .
The perturbation bound implies . The small-perturbation inverse bound [F5] therefore gives with norm at most .
Put and define for . The segment lies in the ball. By the chain rule [F6], , so the Lipschitz bound gives . For each Euclidean block index , [F7] gives ; taking the finite maximum over gives . Thus with this remainder bound. Since both vectors lie in the radius- ball, , and yields . Hence .
Linear interpolation is a convex combination, so . On a cell of length , repeated FTC gives , where for and for . This kernel is nonnegative and has integral , so [F4] gives the interpolation bound . The sample error of is at most by step 2.1. Combining these bounds with step 4.1 proves the displayed estimate.
Depends on
- The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$
- If $f : [a,b] \to \mathbb{R}^m$ is differentiable with integrable $f'$ then $\int_a^b f' = f(b)-f(a)$; and a bounded derivative makes $f$ Lipschitz
- Multiple-shooting and midpoint-collocation residuals for a phase-fixed periodic orbit
- The phase-bordered periodic-orbit BVP residual
- A transverse phase border makes the periodic BVP linearization boundedly invertible
- Uniform inverse bound for the sampled midpoint linearized BVP
- Normalized implicit-midpoint collocation residual for a two-point BVP
- $\mathbb{R}$ and $\mathbb{R}^n$ for $n \ge 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $\mathbb{R}$
- A contraction of a nonempty complete metric space into itself has exactly one fixed point, the limit of the iterates from any starting point
- Multivariable Taylor formula with a Lagrange remainder along a line segment
- Neumann series and small perturbations of bounded inverses
- The chain rule for total derivatives: $D(g\circ f)(a)=Dg(f(a))\circ Df(a)$
- The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G' = f$ and $f$ is integrable, then $\int_a^b f = G(b)-G(a)$
- For $a \le b$ and $f : [a,b] \to \mathbb{R}^m$ integrable when $a<b$, $\bigl\lVert\int_a^b f\bigr\rVert_2 \le \int_a^b \lVert f\rVert_2$; for $a<b$, $\lVert f\rVert_2$ is integrable
Used by
Nothing in the library uses this result yet.
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Sources
- S. M. Rump, Verification methods: Rigorous results using floating-point arithmetic, Acta Numerica 19 (2010) (standard reference, not scraped)
- E. J. Doedel and B. E. Oldeman, AUTO-07P full manual (2021) (standard reference, not scraped)