Verlet integration

Remarks

The two-step formula comes from adding the Taylor expansions of $x(t + h)$ and $x(t - h)$: the odd powers cancel, and $x(t+h) - 2x(t) + x(t-h) = h^2 x''(t) + O(h^4)$. So the local error is $O(h^4)$. In a two-step recurrence of this kind, a local error made at one step is carried forward with a factor of size at most $O(n)$. Summing over $n \le T/h$ steps gives a global error $O(h^2)$. For clause 1, the leapfrog positions satisfy $x_{n+1} - x_n = h v_{n+1/2}$ and $v_{n+1/2} = v_{n-1/2} + h a(x_n)$, so $x_{n+1} - x_n = x_n - x_{n-1} + h^2 a(x_n)$, and both start from the same $x_1$. Velocities do not appear in the two-step form; they can be recovered by $v_n \approx (x_{n+1} - x_{n-1}) / (2h)$, with error $O(h^2)$. Example: for $x'' = 0$ the recurrence gives $x_n = x_0 + n h v_0$, the exact solution.

Going deeper

For a student

The two-step form needs only positions, which halves the storage. Its drawbacks are that velocities are only implicit and that round-off is larger, because a small term $h^2 a$ is added to large positions. The velocity Verlet form fixes both points and is the same method. Its good long-time behaviour comes from it being symmetric and symplectic.