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Finite-order modified energy control on a compact tube

Statement

Let Q⊆Rd be open and star-shaped about a point q∗∈Q, meaning q∗+t(q−q∗)∈Q for every q∈Q and t∈[0,1]. Let M be real symmetric positive-definite, U=Q×Rd, and H(q,p)=12pTM−1p+V(q) with V∈C∞(Q). Choose one of this page's fixed-step methods: kick-then-drift symplectic Euler with r=1, or local leapfrog or an admitted implicit midpoint branch with r=2. Fix a finite truncation order N≥r and choose h0∈(0,1] small enough for the selected map and comparison flow branches. For one step size 0<h≤h0, let zk+1=Φh(zk) for k=0,…,n−1, with n∈N0. Assume every numerical endpoint, every method stage, every truncated modified-flow comparison segment, and every segment used to compare H with Hh[N] lies in one compact tube K⋐U. Then there are constants C0,C1, independent of h,n, such that ∣H(zn)−H(z0)∣≤2C0hr+C1nhN+1. The bound holds only while the stated compact-tube and domain hypotheses hold; if nh≤T, its accumulated defect term is at most C1ThN. It asserts neither exact conservation of H by the numerical method nor convergence of an infinite modified series.

Facts & Assumptions

Given: A smooth Hamiltonian, one selected page method and its order r, a fixed N≥r, a fixed step 0<h≤h0≤1, and a trajectory satisfying the compact-tube hypotheses.

[A1]

The finite modified Hamiltonian exists globally on the star-shaped phase domain. A finite formal logarithm has Hamiltonian coefficients on a star-shaped domain

[F1]

On the compact tube, ∣Hh[N]−H∣≤C0hr. A fixed finite modified Hamiltonian is close to the original energy

[F2]

Each numerical step changes the finite modified Hamiltonian by at most C1hN+1. The finite modified Hamiltonian has a bounded one-step defect

[F3]

Symplectic Euler has order one, while leapfrog and the admitted implicit midpoint branch have order two under their respective tube and branch conditions. Local and finite-time trajectory error for symplectic Euler and leapfrog, The admitted implicit midpoint branch is symplectic and second order

Proof

technique · telescope the modified energy and compare its endpoints with the original energy
1.1A1F2F3givenalgebra

For the selected method, [F3] fixes the order r of the finite modification. For each k<n, [A1] and [F2] then give ∣Hh[N](zk+1)−Hh[N](zk)∣≤C1hN+1. The triangle inequality over the n consecutive increments yields ∣Hh[N](zn)−Hh[N](z0)∣≤C1nhN+1.

2.1F1step 1.1algebra∎

Add and subtract the modified Hamiltonian at both endpoints: H(zn)−H(z0)=(H−Hh[N])(zn)+(Hh[N](zn)−Hh[N](z0))+(Hh[N]−H)(z0). Applying [F1] to the first and third terms and step 1.1 to the middle term gives the stated bound.

Remarks

Symplecticity, reversibility, exact invariants, and modified-energy control are separate properties. This result concerns only the unconstrained maps specified above. Lie and Strang splitting are treated at the exact split-flow compositions ↗, while constrained integration is treated at the classical RATTLE step ↗; no claim here transfers to those methods.

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