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Finite-time global error of the velocity-Verlet method: ∥zn−Φnh(z0)∥≤C(eLT−1)h2/L

Statement

Use the data, finite field and derivative bounds, and constants h1,C1 of The velocity-Verlet step matches the exact flow through order two: local truncation error O(h3) on a convex tube on an open bounded convex tube W, with compact K1 and positive tube margin. Let T>0, 0<∣h∣≤h1, and n∈N satisfy n∣h∣≤T. Suppose zj=Vhj(z0) and yj=Φjh(z0) both lie in K1 for 0≤j≤n. With L=L1=sup⁡W∥DXH∥, the following holds:

∥zn−yn∥≤C1h2eLT−1L(L>0),∥zn−yn∥≤C1Th2(L=0).

The constants are independent of n,h,T, with the displayed data fixed.

Facts & Assumptions

Given: The stated data, including finite bounds on the convex tube and retention of both sequences in K1.

[F1]

The velocity-Verlet step matches the exact flow through order two: local truncation error O(h3) on a convex tube supplies existence of the short exact trajectories from each point of K1 in W, and ∥Vh(z)−Φh(z)∥≤C1∣h∣3.

[F2]

On the convex W, the derivative bound and the chain rule give ∥XH(x)−XH(y)∥≤L∥x−y∥; the exact flow obeys its integral equation (The local flow of an autonomous first-order system, The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a)).

[F4]

The field XH=(M−1p,−∇V(q)) is C1 on its phase domain and hence locally Lipschitz. Clause 2 of The local flow is a partial one-parameter group: identity, composition and inverse-time laws applies to its unique maximal autonomous solutions: if Φs(z) and Φh(Φs(z)) are defined, then Φs+h(z)=Φh(Φs(z)), for signed real times as well.

Proof

1.1givenF1F2F3algebra

Exact-flow stability. Write a=∣h∣. For short trajectories starting at x,y∈K1, their distance u(t), with time oriented by the sign of h, satisfies u(t)≤u(0)+L∫0tu(s)ds on 0≤t≤a by [F2]. If v(t)=u(0)+L∫0tu(s)ds, then v′≤Lv; differentiation of e−Ltv(t) proves v(t)≤eLtu(0). Thus ∥Φh(x)−Φh(y)∥≤eLa∥x−y∥.

2.1givenF1F4step 1.1algebra

Recurrence. Set ej=∥zj−yj∥. For j<n, the datum yj=Φjh(z0) lies in K1, so [F1] supplies Φh(yj); [F4] with s=jh gives Φh(yj)=Φ(j+1)h(z0)=yj+1. Thus by [F1], ej+1≤∥Vh(zj)−Φh(zj)∥+∥Φh(zj)−Φh(yj)∥≤C1a3+eLaej. Since e0=0, induction gives en≤C1a3∑k=0n−1eLak.

3.1step 2.1F3algebra∎

Evaluation. For L>0 the sum is (eLna−1)/(eLa−1), and eLa−1≥La, obtained by integrating LeLt≥L on [0,a]. Hence en≤C1a2(eLT−1)/L. For L=0 the sum equals n and na≤T gives en≤C1Ta2.

Remarks

This is a conditional finite-time trajectory estimate. Retention in the compact tube is assumed. The analytic energy theorem does not give exponential-time accuracy of the original trajectory. Negative steps obey the same bound with elapsed time n∣h∣.

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