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Exact RATTLE has second order global accuracy on compact tubes

Statement

Let the constrained vector field and exact RATTLE map be those of Regular holonomic phase space and constrained dynamics and Scaled classical RATTLE step map. Suppose the exact solution on [0,T] stays in a compact subset of a convex coordinate chart whose closure lies inside a regular chart, with A(q)−1 uniformly bounded. Let h=T/N>0, let z0h∈M satisfy ∥z0h−z(0)∥≤C0h2 in these coordinates, and let zn+1h=Ψh(znh) use the selected exact branches. For all sufficiently large N, the numerical path remains in the regular chart and max⁡0≤n≤N∥znh−z(tn)∥≤CTh2, where CT is independent of N.

Facts & Assumptions

Given: A smooth constrained vector field, its exact local flow, an exact RATTLE branch, the compact-tube hypothesis, and the initial error bound in the Statement.

[F1]

The constrained Hamiltonian vector field has a unique local flow: the vector field is X(q,p)=(M−1p,−∇V(q)−J(q)TΛ(q,p)) and has a unique local flow.

[F2]

Scaled RATTLE equations have unique local multiplier branches: the selected multipliers are smooth in local phase coordinates, have common step ranges and bounded derivatives on compact phase tubes, and at h=0 both equal Λ(q,p)/2.

[F3]

Exact RATTLE preserves both constraints and is symmetric: Ψ−h=Ψh−1 on the local branches.

[F4]

Multivariable Taylor formula with a Lagrange remainder along a line segment: a C3 map has a quadratic Taylor polynomial with a cubic remainder on a compact convex chart.

Proof

1.1F1F2algebra

By [F1] and [F2], differentiating the RATTLE equations at h=0 gives DhΨ0(z)=X(z): the position derivative is M−1p, and the momentum derivative is −∇V−JT(Λ/2+Λ/2). The selected map is C3 in (h,z) by [F2].

1.2F2F3algebra

In chart coordinates choose a slightly larger compact phase neighborhood around the exact path. The revised compact-subset conclusion of [F2] gives one step range and uniform C3 bounds there. Write Ψh(z)=z+hX(z)+h2a(z)+O(h3) uniformly on this neighborhood. Substitution into the inverse equation gives Ψh−1(z)=z−hX(z)+h2(DX(z)X(z)−a(z))+O(h3). By [F3], this equals Ψ−h(z)=z−hX(z)+h2a(z)+O(h3). Comparing quadratic coefficients yields a(z)=12DX(z)X(z).

1.3F2F5

Since Ψ0 is the identity and Ψh is C3, its state derivative is DzΨh=I+O(h) uniformly on the larger compact chart. Thus ∥DzΨh∥≤1+Lh for some L≥0 and small h>0. Applying [F5] in the convex chart gives ∥Ψh(x)−Ψh(y)∥≤(1+Lh)∥x−y∥ there.

2.1F1F2F4step 1.2

The exact flow satisfies z˙=X(z) and hence z¨=DX(z)X(z). Its third time derivative is bounded on the compact tube. Taylor's theorem [F4] therefore gives z(t+h)=z(t)+hX(z(t))+12h2DX(z(t))X(z(t))+O(h3). Comparing with step 1.2 proves the local defect ∥Ψh(z(t))−z(t+h)∥≤Ch3 uniformly in t.

3.1givenstep 2.1step 1.3algebra∎

Until a numerical state leaves the larger chart, steps 2.1 and 1.3 give en+1≤(1+Lh)en+Ch3, where en=∥znh−z(tn)∥. Induction yields en≤(1+Lh)ne0+Ch3∑j=0n−1(1+Lh)j≤eLT(C0+CT)h2. For sufficiently small h, this bound is less than the positive distance from the exact compact path to the boundary of the larger chart, so a first-exit index cannot exist. The estimate therefore holds for every n≤N.

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