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Local and finite-time trajectory error for symplectic Euler and leapfrog

Statement

Let d≥1, Q⊆Rd be open, V∈C3(Q), and M symmetric positive definite. Set H(q,p)=12pTM−1p+V(q) and F(q,p)=(M−1p,−∇V(q)). Fix T>0 and h=T/N with N≥1. Suppose the exact solution, numerical states, all kick/drift intermediate states, and every line segment used below lie in one compact convex tube K⋐Q×Rd, with the derivatives of F through order two bounded there. For z0 in that tube, write enE for the error of iterated kick-then-drift symplectic Euler and enV for the error of iterated local leapfrog relative to the exact solution at tn=nh. Write ΨhE and ΨhV for their admitted one-step maps and φh for the exact flow. For sufficiently small h, constants CE,CV,L independent of h,N,n satisfy ∥ΨhE(z)−φh(z)∥≤CEh2,∥ΨhV(z)−φh(z)∥≤CVh3 for admitted z∈K, and whenever nh≤T, enE≤eLTe0+CETeLTh,enV≤eLTe0+CVTeLTh2. Thus symplectic Euler has global order one and leapfrog has global order two on the stated controlled tube. No unconditional domain-admissibility claim is made.

Facts & Assumptions

Given: V∈C3(Q), SPD M, an exact Hamiltonian solution and numerical trajectories contained in the stated compact convex tube, with all method stages admitted.

[A1]

The local map is the partial kick/drift composition; symplectic Euler uses kick then drift, while leapfrog uses half-kick, full drift, half-kick. Local domain restriction for kick, drift, and leapfrog maps, Local kick-then-drift symplectic Euler step

[F1]

Since M is symmetric positive definite, Mx=0 forces x=0, so M is invertible and (M−1)T=(MT)−1=M−1. Differentiating H(q,p)=12pTM−1p+V(q) therefore gives ∇H=(∇V(q),M−1p) and the coordinate field F=(M−1p,−∇V(q)).

[F2]

The exact local flow and its variational derivative exist on admitted compact trajectory segments. C¹ Euclidean maximal flows, variational dependence and the finite C² upgrade

[F4]

Apply the scalar Taylor theorem coordinatewise on the compact convex tube; finitely many coordinate remainder bounds combine into a Euclidean bound. The same argument applies to the entries of ∇2V. Multivariable Taylor formula with a Lagrange remainder along a line segment

Proof

technique · Taylor defect, one-step stability, and a finite geometric recurrence
1.1A1F1F2F3F4given

By [F1]–[F3], the exact solution satisfies z′=F(z) and z′′=DF(z)F(z). Since D2F is bounded on K, [F4] gives uniformly φh(z)=z+hF(z)+12h2DF(z)F(z)+R(z,h),∥R(z,h)∥≤C0h3.

1.2A1F1F4F5algebra

The kick derivative is I+hAK(q) with AK=(00−∇2V(q)0) and the drift derivative is I+hAD with AD=(0M−100). Their sum is DF(q,p). For kick-then-drift, the derivative product is (I+hAD)(I+hAK)=I+hDF+O(h2). For leapfrog it is (I+(h/2)AK(q1))(I+hAD)(I+(h/2)AK(q))=I+hDF+O(h2); since q1−q=O(h) and D3V is bounded, [F4] applied to each Hessian entry gives ∇2V(q1)−∇2V(q)=O(h). The remainders are uniform on K, so ∥DΨh∥≤1+Lh for a fixed L and small h. The vector-valued mean-value inequality [F5] on the stipulated convex segments gives the same one-step Lipschitz bound for pairs of compared states.

2.1A1F1step 1.1algebra

Kick-then-drift Euler gives pE=p−h∇V(q) and qE=q+hM−1pE. Thus pE=p−h∇V(q) and qE=q+hM−1p−h2M−1∇V(q). Comparing with the two coordinate components of step 1.1 gives a uniform one-step defect at most CEh2.

2.2A1F1F4step 1.1algebra

For leapfrog, the intermediate momentum is p1/2=p−(h/2)∇V(q) and the new position is q1=q+hM−1p1/2=q+hM−1p−(h2/2)M−1∇V(q). Taylor expansion of ∇V(q1), using bounded third derivatives of V, gives p1=p−h2∇V(q)−h2∇V(q1)=p−h∇V(q)−h22∇2V(q)M−1p+O(h3). These are exactly the terms through degree two in step 1.1, so the leapfrog one-step defect is at most CVh3.

3.1A1step 1.2step 2.1step 2.2algebra

For either method, if its local defect is Chs+1, steps 1.2, 2.1, and 2.2 give en+1≤(1+Lh)en+Chs+1. Iterating this inequality and using (1+Lh)j≤eLT and n≤T/h yields en≤eLTe0+CTeLThs.

4.1step 3.1algebra∎

Substitution of s=1,C=CE for symplectic Euler and s=2,C=CV for leapfrog in step 3.1 gives the two stated bounds.

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