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IMEX Euler has a first-order local defect and a one-step stability bound
Statement
Let and let on an open set as in Additive autonomous ODE partitions. On a common compact convex tube , suppose are bounded and Lipschitz with constants . For every and sufficiently small , assume there is a closed ball with , centered at , and with ; the branch lemma then gives a unique root in that ball when . Require that the exact solution segment from each reference state stays in and that its endpoint lies in . Then, uniformly on the tube, where one may take and depends only on the stated tube bounds and branch margins. The local defect is therefore , and the exact-root method is first order locally.
Facts & Assumptions
Given: The additive IVP, a common tube with uniform field bounds, the positive-radius branch balls and branch hypotheses for every starting state in that tube, exact path segments and endpoints admitted in their branch balls, and .
One-step IMEX Euler with an implicit branch and residual status: the exact update is .
A local IMEX Euler branch from contraction: if , is -Lipschitz on the selected ball, , and , the exact branch exists uniquely and satisfies the residual-to-root bound.
Along each exact solution segment in , is continuous because is Lipschitz on and is continuous. Each coordinate is therefore integrable on the compact segment (A continuous function on is Riemann integrable, by Heine-Cantor and Riemann's criterion), so the vector derivative is integrable; the vector-valued fundamental theorem of calculus (If is differentiable with integrable then ; and a bounded derivative makes Lipschitz) gives .
For and integrable when , ; for , is integrable: for an integrable vector function on an increasing interval, .
is bounded by and Lipschitz with on the tube; the exact path and its endpoints remain in the stated convex tube, and each selected root belongs to its stated branch ball in that tube.
Proof
By [F3], for , , so [F4] and [A1] give . Subtracting from the integral at and applying [F4] again yields .
Set and evaluate the implicit residual of the exact endpoint: . By [A1] and step 1.1, . The exact endpoint and the branch root lie in the same admitted branch ball, so [F2] gives .
For two inputs with selected roots , subtracting their equations in [F1] and using the Lipschitz bounds gives . Since , . This is the stated one-step stability bound.
Depends on
- One-step IMEX Euler with an implicit branch and residual status
- A local IMEX Euler branch from contraction
- 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
- A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion
- 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
Dependency tree · two levels
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Sources
- Uri M. Ascher, Steven J. Ruuth, and Raymond J. Spiteri, Implicit-explicit Runge-Kutta methods for time-dependent partial differential equations (standard reference, not scraped)
- Lorenzo Pareschi and Giovanni Russo, Implicit-explicit Runge-Kutta schemes and applications to hyperbolic systems with relaxation (standard reference, not scraped)