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IMEX Euler has a first-order local defect and a one-step stability bound

Statement

Let d≥1 and let F=a+b on an open set U⊆Rd as in Additive autonomous ODE partitions. On a common compact convex tube T⊂U, suppose a,b are bounded and Lipschitz with constants Ma,Mb,La,Lb. For every u∈T and sufficiently small h>0, assume there is a closed ball B‾(cu,Ru)⊂T with Ru>0, centered at cu:=u+ha(u), and with hMb≤Ru; the branch lemma then gives a unique root Ψh(u)=cu+hb(Ψh(u)) in that ball when hLb≤1/2. Require that the exact solution segment from each reference state y(t)∈T stays in T and that its endpoint y(t+h) lies in B‾(cy(t),Ry(t)). Then, uniformly on the tube, ∥Ψh(y(t))−y(t+h)∥2≤Ch2,∥Ψh(u)−Ψh(v)∥2≤(1+Lh)∥u−v∥2, where one may take L=2(La+Lb) and C depends only on the stated tube bounds and branch margins. The local defect is therefore O(h2), 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 hLb≤1/2.

[F1]

One-step IMEX Euler with an implicit branch and residual status: the exact update is Y=y+ha(y)+hb(Y).

[F2]

A local IMEX Euler branch from contraction: if ∥b∥≤Mb, b is Lb-Lipschitz on the selected ball, hMb≤R, and hLb<1, the exact branch exists uniquely and satisfies the residual-to-root bound.

[F3]

Along each exact solution segment in T, s↦F(y(s)) is continuous because F is Lipschitz on T and y is continuous. Each coordinate is therefore integrable on the compact segment (A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion), so the vector derivative is integrable; the vector-valued fundamental theorem of calculus (If f:[a,b]→Rm is differentiable with integrable f′ then ∫abf′=f(b)−f(a); and a bounded derivative makes f Lipschitz) gives y(t+s)−y(t)=∫0sF(y(t+r)) dr.

[F4]
[A1]

F=a+b is bounded by MF:=Ma+Mb and Lipschitz with LF:=La+Lb 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

1.1A1F3F4algebra

By [F3], for 0≤s≤h, ϕs−y=∫0sF(ϕr) dr, so [F4] and [A1] give ∥ϕs−y∥2≤MFs. Subtracting hF(y) from the integral at h and applying [F4] again yields ∥ϕh−y−hF(y)∥2≤∫0hLF∥ϕs−y∥2ds≤12LFMFh2.

2.1A1F1F2step 1.1algebra

Set c=y+ha(y) and evaluate the implicit residual of the exact endpoint: rϕ:=ϕh−c−hb(ϕh)=(ϕh−y−hF(y))−h(b(ϕh)−b(y)). By [A1] and step 1.1, ∥rϕ∥2≤MF(12LF+Lb)h2. The exact endpoint and the branch root lie in the same admitted branch ball, so [F2] gives ∥Ψh(y)−ϕh∥2≤2MF(12LF+Lb)h2.

3.1A1F1algebra∎

For two inputs u,v with selected roots Yu,Yv, subtracting their equations in [F1] and using the Lipschitz bounds gives (1−hLb)∥Yu−Yv∥2≤(1+hLa)∥u−v∥2. Since hLb≤1/2, (1+hLa)/(1−hLb)=1+h(La+Lb)/(1−hLb)≤1+2(La+Lb)h. This is the stated one-step stability bound.

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