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An inexact midpoint stage solve has a controlled update error

Example

Assume Countable Choice for the Lebesgue interval-measure formula. Use one-node Gauss collocation on $y^{'} = - y$ , $y (0) = 1$ , with $h = 1 / 10$ . Let $U^{*}$ be its exact midpoint stage and take the approximate stage $V = U^{*} + 1 / 2000$ .

Facts & Assumptions

Given: The scalar IVP, one-node Gauss tableau, step size and approximate stage in the Example; use the explicit-left-Euler SDC sweep when comparing sweep and stage residuals.

[F1]

One-node Gauss collocation is implicit midpoint of order two: the exact collocation stage is the implicit midpoint stage and its method has order two under a common-tube smoothness assumption.

[F3]

Small-step implicit Runge–Kutta stages have a unique branch and controlled residual error: on a closed product ball, an invariant contraction map with factor $q < 1$ has a unique stage and an update residual $G (V)$ bounds the stage error by $∥ G (V) ∥ / (1 - q)$ .

[F4]

First-order systems, initial value problems, and solutions on intervals: for this scalar IVP the stage field is $f (t, Y) = - Y$ .

[F5]

Finite explicit-left-Euler spectral deferred-correction sweep: its explicit-left-Euler stage block $Q_{Δ}$ is strictly lower triangular, so for one node $Q_{Δ} = 0$ and the formal sweep is $S (V) = 1 - (h / 2) V$ for this field.

[A1]

Countable Choice, inherited from the one-node Gaussian moment data used in [F1].

[A2]

For [F3], use center $c = 1$ , radius $R = 1$ , matrix $A = (1 / 2)$ , and $F (Y) = - Y$ on the closed ball $B = [0, 2]$ . The bound parameters are $M_{F} = 2$ and $L_{F} = 1$ ; they are denoted with subscripts to distinguish $M_{F}$ from the number of nodes.

Verification

1.1A1A2F1F3F4F5givenalgebra

Here $Q_{11} = 1 / 2$ , $b_{1} = 1$ , and by [F4] the exact stage equation is $U^{*} = 1 - (h / 2) U^{*}$ , so $U^{*} = 20 / 21 \in [0, 2]$ . For [F3], the ball is $[0, 2]$ , and its map is $Φ (Y) = 1 - (h / 2) Y = 1 - 0.05 Y$ . It maps $[0, 2]$ into $[0.9, 1] \subset [0, 2]$ ; its contraction factor is $q = h ∣ A ∣ L_{F} = 0.1 (1 / 2) (1) = 0.05$ , and the invariant-ball bound is $h ∣ A ∣ M_{F} = 0.1 (1 / 2) (2) = 0.1 \leq R = 1$ . Thus the closed-ball and contraction hypotheses of [F3] hold. The stage update residual map is $G (V) = V - 1 + (h / 2) V = (1 + h / 2) (V - U^{*})$ ; at $V - U^{*} = 1 / 2000$ it has magnitude $21 / 40000 = 0.000525$ . For the explicit-left-Euler sweep in [F5], $Q_{Δ} = 0$ , so $∣ S (V) - V ∣ = ∣ G (V) ∣$ and the SDC contraction factor is also $h / 2 = 1 / 20$ .

2.1F3step 1.1algebra

The specified $V = 20 / 21 + 1 / 2000$ lies in $[0, 2]$ , so [F3] applies. With $q = 1 / 20$ , the numerical residual threshold $(1 - q) h^{3} = 19 / 20000 = 0.00095$ exceeds $∣ G (V) ∣ = 21 / 40000$ from step 1.1. Directly, [F3] gives $∣ V - U^{*} ∣ \leq ∣ G (V) ∣ / (1 - q) = 21 / 38000 < 1 / 1000 = h^{3}$ .

3.1F1F4step 2.1algebra∎

For this field $f (t, Y) = - Y$ and endpoint weight $b_{1} = 1$ , the approximate endpoint differs from the exact collocation endpoint by $h ∣ V - U^{*} ∣ = 0.1 / 2000 = 1 / 20000 = 0.00005$ . The stage bound in step 2.1 also gives the upper bound $h h^{3} = h^{4} = 0.0001$ . The example verifies a single stage solve; a global order-two conclusion would additionally require uniform per-step defect, stability, and stage-error estimates on the common tube.

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Sources

Ernst Hairer and Gerhard Wanner, Introduction à l’Analyse Numérique, Chapter III: Équations Différentielles Ordinaires (standard reference, not scraped)
John C. Butcher, Implicit Runge–Kutta Methods, tutorial accompanying Numerical Methods for Ordinary Differential Equations (standard reference, not scraped)