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Fixed-matrix global orders and certified-action error propagation

Statement

Let T>0, N≥1, h=T/N, tn=nh, A∈Md(C), and let D⊆R×Cd be open after realification, with g:D→Cd. Let y:[0,T]→Cd solve y′(t)=Ay(t)+g(t,y(t)) Suppose the exact path, every numerical state, and every ETD2 predictor lie in a common compact time-state tube T⊂D with positive margin. Also assume the hypotheses of One-step stability after removing the exact linear flow hold on this tube: compared inputs in each time fiber and both associated ETD2 predictor states are admissible, and g is Lg-Lipschitz between the states and stages used there. Finally, let q(t):=g(t,y(t)) be C2 with ∥q′∥2≤Q1 and ∥q′′∥2≤Q2. Assume ∥esA∥2≤Meωs(0≤s≤T),M≥1,ω∈R, and put K+:=sup⁡0≤s≤T∥esA∥2,K−:=sup⁡0≤s≤T∥e−sA∥2,κ:=LgK+K−. These finite-matrix suprema are finite; in particular K+≤Memax⁡(ω,0)T. Define CE:=K+Q12,C2:=K+(Q2+LgCE2),LE:=κ,L2:=κ+Tκ22. For an exact-action exponential-Euler run use p=1,Cp=CE,Lp=LE; for exact-action ETD2 use p=2,Cp=C2,Lp=L2. If Γ(L,T):=(eLT−1)/L for L>0 and Γ(0,T):=T, then exact initial data give max⁡0≤n≤N∥yn−y(tn)∥2≤K+K−CpΓ(Lp,T)hp.

Now let a certified-action implementation return candidates y~n+1 and certified one-step action-error bounds δn as in Certified action bounds control one exponential-integrator step. Starting from y~0=y(0), its error obeys max⁡0≤n≤N∥y~n−y(tn)∥2≤K+K−eLpT(CpThp+∑j=0N−1δj). Thus local orders one and two imply fixed-A, finite-dimensional global orders one and two when exact actions are used or when ∑jδj=O(hp). The constants may depend on A, T, the tube, and the derivative bounds. The bounds are not asserted uniformly over varying matrices or growing dimensions; no unbounded-operator/PDE convergence theorem follows without the additional domain and commutator hypotheses such a theorem requires.

Facts & Assumptions

Given: The matrix, grid, solution, tube, regularity, Lipschitz and semigroup bounds, and numerical action certificates in the Statement.

[F1]

First-order systems, initial value problems, and solutions on intervals: the exact solution is differentiable and satisfies y′(t)=Ay(t)+g(t,y(t)).

[F2]

Exponential Euler has a second-order local defect: the exact-start exponential-Euler defect is at most CEh2.

[F3]

The two-stage ETD2 method has a third-order local defect: the exact-start ETD2 defect is at most C2h3.

[F4]

One-step stability after removing the exact linear flow: after the change z(t)=e−tAy(t), the exponential-Euler and ETD2 one-step maps have stability constants LE and L2 in the form 1+Lph.

[F5]

Certified action bounds control one exponential-integrator step: when the action algorithm is certified, ∥Ψ~h,n(v)−Ψh,n(v)∥2≤δn.

[F11]

A compact set and a disjoint closed set have a positive norm-distance gap: a compact subset of an open subset of a finite-dimensional normed space has positive distance from its closed complement when that complement is nonempty; if it is empty, every positive radius is a margin.

[F12]

Metric continuity characterisations, with countable choice for the sequential converse: a continuous map between metric spaces has open preimages of open sets.

[F13]

C is the real coordinate plane, with coordinate arithmetic, The standard formulas ⟨x,y⟩=∑k<nxkyk on Rn and ∑k<nxkyk‾ on Cn are inner products, and The norm ∥v∥=⟨v,v⟩ induced by a real or complex inner product: for v=(v0,…,vd−1), the componentwise map Φd(v):=(Re⁡v0,Im⁡v0,…,Re⁡vd−1,Im⁡vd−1) is real-linear, with ∥Φd(v)∥R2d2=∑j<d∣vj∣2=∥v∥22. It preserves real-time difference quotients and their limits, so the real one-step theorem applies to the realified system and its transformed state without changing the error norm.

[F14]

The exponential series of a bounded operator gives ∥e−sA∥2≤es∥A∥2 for 0≤s≤T. Since s∥A∥2≤T∥A∥2, scalar monotonicity (The exponential function is strictly increasing) gives es∥A∥2≤eT∥A∥2. Thus the set defining K− contains ∥Id∥2 at s=0 and has that finite upper bound; Dedekind completeness: the least-upper-bound property supplies its finite supremum. Likewise ωs≤max⁡(ω,0)T, so the assumed semigroup bound and the same scalar monotonicity give K+≤Memax⁡(ω,0)T and a finite supremum for its nonempty defining set.

[L1]

Global error from uniform one-step defect and stability: on a common compact tube, exact-start local defect Chp+1, one-step stability 1+Lh, and exact initial values imply the fixed-time global bound CΓ(L,T)hp.

[A1]

The transformed open domain and compact tube satisfy the hypotheses stated above; the realified norm and coordinate calculus are as in [F13].

Proof

1.1givenF1F2F3F6F9F13A1algebra

By [F13], the complex state equation and the transformed equation are real finite-dimensional systems with the same Euclidean error norm. Set z(t):=e−tAy(t). The product rule and [F1], [F6] give z′(t)=e−tAg(t,y(t))=e−tAg(t,etAz(t)). Conjugating the exact numerical map and exact endpoint by e−(t+h)A cancels their common ehA linear term and multiplies their difference by e−(t+h)A. Hence [F2], [F3], and [F9] give transformed exact-start defects at most K−CEh2 and K−C2h3.

2.1givenF4F6F9F10F11F12F14L1step 1.1algebra

By [F14], both amplification suprema used here are finite. Set Dz:={(t,z):(t,etAz)∈D}, open as the preimage of D under the continuous map (t,z)↦(t,etAz) by [F6], [F9], [F10], and [F12]. The map (t,u)↦(t,e−tAu) continuously sends T to a compact tube in Dz containing the transformed exact path, states, and stages; [F10]–[F11] give positive margin. Its stability is 1+Lph by [F4], and its local constant is K−Cp by step 1.1. Thus [L1] applies with exact initial data.

2.2givenF2F3F4F5F6F9step 1.1algebra

For certified actions, put z~n:=e−tnAy~n and rn:=∥z~n−z(tn)∥2. Stability [F4], local defect [F2] or [F3] transformed by [F6] and step 1.1, and perturbation bound [F5] transformed using [F9] by e−(tn+1)A give by the triangle inequality rn+1≤(1+Lph)rn+K−Cphp+1+K−δn, with r0=0.

3.1givenF4F9L1step 2.1algebra

Applying [L1] as in step 2.1 yields max⁡n∥zn−z(tn)∥2≤K−CpΓ(Lp,T)hp. Since y(tn)=etnAz(tn) and [F9] with the definition of K+ bounds multiplication by etnA, this gives the first displayed bound.

3.2step 2.2inductionalgebra

Induction in step 2.2 gives rn≤K−∑j=0n−1(1+Lph)n−1−j(Cphp+1+δj): the base case is the empty sum, and multiplying the bound for rn by 1+Lph and adding the j=n defect and action terms gives the formula for rn+1.

4.1givenF7F8F9step 3.2algebra∎

Since 1+Lph≤eLph by [F7], each power in step 3.2 is at most eLpT by [F8] and nh≤T. Also Nh=T, so NCphp+1=CpThp. Thus rn≤K−eLpT(CpThp+∑j=0N−1δj). Multiplication by etnA and [F9] with the bound K+ give the certified-action estimate for every n, hence for their maximum.

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