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Small-step collocation maps are Lipschitz stable

Statement

Let $f$ be $L$ -Lipschitz in state on a common invariant stage tube, and let the collocation stage branches for initial states $u, v$ lie in that tube. If $h L ∥ Q ∥_{\infty} \leq 1 / 2$ , then $∥ U (u) - U (v) ∥_{\infty, 2} \leq 2 ∥ u - v ∥_{2} .$ Consequently the collocation step map $Ψ_{h} (u) = u + h \sum_{j} b_{j} f (t + c_{j} h, U_{j} (u))$ satisfies $∥ Ψ_{h} (u) - Ψ_{h} (v) ∥_{2} \leq (1 + 2 h L ∥ b ∥_{1}) ∥ u - v ∥_{2}, ∥ b ∥_{1} : = \sum_{j} ∣ b_{j} ∣ .$

Facts & Assumptions

Given: A collocation tableau, two admissible initial states and their selected collocation stages in one common tube, a state Lipschitz constant $L$ , and $h L ∥ Q ∥_{\infty} \leq 1 / 2$ .

[F1]

Collocation tableau from distinct quadrature nodes: $U (u) = 1 u + h Q F (U (u))$ and $Ψ_{h} (u) = u + h \sum_{j} b_{j} F_{j} (U (u))$ .

[F2]

Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation: the Euclidean norm satisfies the triangle inequality and is homogeneous.

[F3]

Collocation tableau from distinct quadrature nodes defines $∥ Q ∥_{\infty} = \max_{i} \sum_{j} ∣ Q_{i j} ∣$ ; Collocation quadrature data are well defined proves its entries are finite.

[A1]

Both assumed stage branches exist in the same set where the state-Lipschitz bound in the Statement applies, and all stage values are admissible.

Proof

1.1A1F1F2F3algebra

Subtract the stage equations in [F1] for $u$ and $v$ . The branches are the ones assumed in the Statement; the block triangle inequality [F2], state Lipschitz bound [A1], and row-sum definition [F3] give $∥ U (u) - U (v) ∥_{\infty, 2} \leq ∥ u - v ∥_{2} + h L ∥ Q ∥_{\infty} ∥ U (u) - U (v) ∥_{\infty, 2} .$ Since $h L ∥ Q ∥_{\infty} \leq 1 / 2$ , moving the last term to the left yields the stated factor $2$ .

2.1A1F1F2step 1.1algebra∎

Subtract the endpoint formulas in [F1]. By [F2] and the same state Lipschitz bound, $∥ Ψ_{h} (u) - Ψ_{h} (v) ∥_{2} \leq ∥ u - v ∥_{2} + h L \sum_{j} ∣ b_{j} ∣ ∥ U_{j} (u) - U_{j} (v) ∥_{2} .$ Apply step 1.1 and the definition of $∥ b ∥_{1}$ to obtain the asserted one-step bound.

Depends on

Used by

One-node Gauss collocation is implicit midpoint of order two Proposition
A stage residual tolerance preserves collocation order Theorem
M-node collocation has global order M Theorem
M-node Gauss collocation has global order 2M Theorem

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ernst Hairer, Syvert P. Nørsett, and Gerhard Wanner, Solving Ordinary Differential Equations I, 2nd ed., Chapter II §II.7 (standard reference, not scraped)
Mathew F. Causley and David C. Seal, On the convergence of spectral deferred correction methods, Communications in Applied Mathematics and Computational Science 14 (2019), §3 (standard reference, not scraped)