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Additive per-step errors in a Lipschitzian one-step scheme accumulate geometrically: η((1+Lh)n−1)/(Lh)

Statement

Let (X,d) be a metric space (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric), let V:X→X map X into itself, and let L≥0 and h>0. Assume

d(V(x),V(y))≤(1+Lh) d(x,y)for all x,y∈X,

which is a Lipschitz hypothesis on V with constant 1+Lh (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction). Fix z0∈X, put zj+1:=V(zj) for j∈N, and let x0:=z0 together with points xj+1∈X satisfying

d(xj+1,V(xj))≤η(j∈N)

for a number η≥0 ("computational error per step"). Then for every n∈N:

  1. if L>0, d(xn,zn)≤η (1+Lh)n−1Lh, and if L=0, d(xn,zn)≤nη;
  2. if moreover F:X→R satisfies the Lipschitz bound ∣F(x)−F(y)∣≤Λ d(x,y) for all x,y∈X and some 0≤Λ<∞, then ∣F(xn)−F(zn)∣≤Λη (1+Lh)n−1Lh(L>0),∣F(xn)−F(zn)∣≤Λnη(L=0);
  3. if nh≤T for some T>0 and L>0, then (1+Lh)n≤eLT and therefore d(xn,zn)≤η eLT−1Lh≤η eLTTh.

Facts & Assumptions

Given: The metric space (X,d), the map V, the constants L≥0, h>0, η≥0, the exact iterates zj=Vj(z0) and the computed points xj with d(xj+1,V(xj))≤η.

[F1]

Metric axioms and the triangle inequality in the forms d(a,c)≤d(a,b)+d(b,c) (Metric space: d(x,y)=0 iff x=y, symmetry, and the triangle inequality; pseudometric and ultrametric).

[F2]

The Lipschitz hypothesis on V, and the Lipschitz bound on F in clause 2 (Lipschitz map, α-Hölder map for rational 0<α≤1, and contraction).

Proof

technique · direct induction on $n$ with a finite geometric sum
1.1givenF1F2

One-step recursion. By [F1] and [F2], d(xj+1,zj+1)≤d(xj+1,V(xj))+d(V(xj),V(zj))≤η+(1+Lh) d(xj,zj).

2.1step 1.1algebra

Induction. We show d(xj,zj)≤η∑i=0j−1(1+Lh)i for all j. For j=0 this is d(x0,x0)=0. If it holds for j, then step 1.1 gives d(xj+1,zj+1)≤η+(1+Lh)η∑i=0j−1(1+Lh)i=η∑i=0j(1+Lh)i.

3.1step 2.1F2algebra

Geometric evaluation. For L>0 the finite geometric sum is ∑i=0n−1(1+Lh)i=(1+Lh)n−1Lh, and for L=0 it is n; this proves clause 1 in both cases. Clause 2 follows by applying the Lipschitz bound on F to the pair (xn,zn) and substituting clause 1.

4.1step 3.1F3algebra∎

Long-time form. If nh≤T then by [F3], (1+Lh)n≤eLhn≤eLT, so (1+Lh)n−1≤eLT−1 and d(xn,zn)≤η(eLT−1)/(Lh). Finally (eLT−1)/(Lh)=Th⋅eLT−1LT≤TheLT because eLT−1≤LT eLT for LT≥0; this proves clause 3.

Remarks

  • The lemma is a general accumulation statement about a one-step scheme with an external per-step perturbation; it does not model any concrete floating-point arithmetic, and η is a hypothesis on the computed points, not a derived roundoff bound. For velocity Verlet, its application requires a metric state set (X,d) on which all stages are defined, Vh:X→X, computed points remaining in X, and a positive step h>0. Establish separately a numerical-map bound d(Vh(x),Vh(y))≤(1+Lmaph)d(x,y) for every x,y∈X. Then, for nh≤T and Lmap>0, this lemma contributes η(eLmapT−1)/(Lmaph) to the finite-time error comparison, or nη when Lmap=0. The exact-field derivative constant in Finite-time global error of the velocity-Verlet method: ∥zn−Φnh(z0)∥≤C(eLT−1)h2/L bounds the exact flow and does not establish this map bound or invariance of X. Combining the estimates requires their hypotheses on a common admissible set; exact-arithmetic symplecticity in The drift, kick and velocity-Verlet maps are canonical: (DC)TJ DC=J and C∗ω=ω does not itself provide these error hypotheses.
  • No choice principle is used; the induction is finite and the geometric sums are finite sums.

Depends on

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