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Equal additive stage perturbations attain quadratic RKC1 amplification

Example

Let the state space be Euclidean, take $A = 0$ , choose a unit vector $v$ , and after each of the $s$ RKC1 stage updates add the same perturbation $ϵ v$ with $ϵ \geq 0$ . This is a deterministic additive-perturbation model; it does not assert that floating-point errors have this form.

Facts & Assumptions

Given: The RKC1 recurrence, the zero self-adjoint matrix, and the equal stage perturbations in the Example.

[F1]

First-order stabilized explicit Runge–Kutta–Chebyshev step: the linear stages satisfy $Y_{0} = y$ , $Y_{1} = (I + h A / s^{2}) Y_{0}$ , and $Y_{j} = 2 (I + h A / s^{2}) Y_{j - 1} - Y_{j - 2}$ for $j \geq 2$ .

[F2]

Chebyshev stage perturbations have an explicit quadratic bound: for a self-adjoint nonpositive operator in the certified interval, stage perturbations of norm at most $ϵ$ yield final error at most $s (s + 1) ϵ / 2$ .

Verification

1.1F1givenalgebra

Since $A = 0$ , the unperturbed stages all equal $y$ . If $e_{j}$ is the perturbed-minus-unperturbed stage difference, then $e_{0} = 0$ , $e_{1} = ϵ v$ , and [F1] gives $e_{j} = 2 e_{j - 1} - e_{j - 2} + ϵ v$ for $2 \leq j \leq s$ .

2.1step 1.1inductionalgebra

Induction gives $e_{j} = ϵ j (j + 1) v / 2$ : it holds for $j = 0, 1$ , and substituting the formulas at $j - 1, j - 2$ into the recurrence yields $ϵ {(j - 1) j - (j - 2) (j - 1) / 2 + 1} v = ϵ j (j + 1) v / 2$ .

3.1F2step 2.1algebra∎

At the output $j = s$ , $∥ e_{s} ∥_{2} = ϵ s (s + 1) / 2$ because $∥ v ∥_{2} = 1$ . The assumptions of [F2] hold for $A = 0$ , so the lemma supplies the matching upper bound; this perturbation pattern attains equality in its quadratic stage-count estimate.

Depends on

Used by

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Sources

Ibrahim Almuslimani and Gilles Vilmart, Explicit Stabilized Integrators for Stiff Optimal Control Problems, §2.2.1 (standard reference, not scraped)
B.P. Sommeijer, L.F. Shampine, and J.G. Verwer, RKC: An explicit solver for parabolic PDEs, §2 (standard reference, not scraped)