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Global divergence-integral error separates state and quadrature error

Statement

Let T>0, N∈N with N≥1, h=T/N>0, and tn=nh. Let K⊆Rd be compact, let x:[0,T]→K be an exact path, and suppose every numerical state xn also belongs to K. Let q:[0,T]×K→R be a continuous scalar integrand. Define In:=∫tntn+1q(s,x(s)) ds,ℓn:=ℓ0+∑j=0n−1Ij. Suppose the one-step state method satisfies the hypotheses of Global error from uniform one-step defect and stability at its positive integer order p, and write Kx for its resulting constant, so max⁡n∥xn−x(tn)∥2≤Kxhp. Let Qh(tn,⋅):K→R be the increment map in the numerical scalar update ℓ^n+1=ℓ^n+Qh(tn,xn), with ∣ℓ^0−ℓ0∣≤Cℓ,0hr for r∈R with r≥0, using the real powers of Real powers for positive bases, with the zero-base positive-exponent convention. If constants Kq,D≥0 and an integer m≥1 satisfy, uniformly for n<N and u,v∈K, ∣Qh(tn,u)−Qh(tn,v)∣≤hKq∥u−v∥2,∣Qh(tn,x(tn))−In∣≤Dhm+1, then max⁡0≤n≤N∣ℓ^n−ℓn∣≤Cℓ,0hr+TKqKxhp+DThm.

Facts & Assumptions

Given: The exact path, continuous integrand, numerical state and scalar updates, and the uniform state-solver, sensitivity, and quadrature-defect hypotheses in the Statement.

[F1]

Under the full hypotheses of Global error from uniform one-step defect and stability, the state method has a fixed-time bound Kxhp with the explicit L>0 and L=0 constants. These hypotheses require a common compact time-state tube with positive margin in the open ODE domain, containing the entire exact path, all numerical grid states, and every method stage; the one-step map is defined on each grid-time fiber, with local-defect bound Chp+1, fiberwise stability factor 1+Lh, and initial error bound C0hp, for constants independent of N and n.

[A1]

For each step, the scalar update is Lipschitz in its starting state with constant hKq and has exact-start quadrature defect at most Dhm+1; the initial scalar error is at most Cℓ,0hr.

[F3]

For h>0 and r∈R, hr is the real power defined by hr=exp⁡(rlog⁡h) (Real powers for positive bases, with the zero-base positive-exponent convention).

Proof

1.1F1A1algebra

Let en:=∣ℓ^n−ℓn∣. By [F1], ∥xn−x(tn)∥2≤Kxhp. The exact and numerical scalar updates, the triangle inequality, and [A1] give en+1≤en+∣Qh(tn,xn)−Qh(tn,x(tn))∣+∣Qh(tn,x(tn))−In∣≤en+hKqKxhp+Dhm+1.

2.1A1F3step 1.1algebra∎

Iterating the recurrence from step 1.1 for n steps yields en≤e0+n(hKqKxhp+Dhm+1). Since n≤N=T/h and e0≤Cℓ,0hr, this is at most Cℓ,0hr+TKqKxhp+DThm. Taking the maximum proves the claim.

Remarks

For a continuous normalizing flow, q(t,x)=−div⁡xf(t,x). This theorem bounds that divergence integral without identifying it as a probability log density; the separate identification is proved in A continuous flow changes log density by minus its divergence. The bound makes a higher-order state solver insufficient by itself: the chosen divergence quadrature must also have the required order m and state sensitivity bound.

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