On Wikipedia, the page about symplectic integrators talks about how the semi-implicit Euler method is a first-order symplectic integrator.
But when I read the page about the semi-implicit Euler method, they do not show the same algorithm.

On the first page, it says

Update the position of the particle by adding to it its velocity multiplied by c1
Update the velocity of the particle by adding to it its acceleration (at the updated position) multiplied by d1
c1 = d1 = 1

Which can be translated to:

$\text{Let g(t, x) = acceleration of the particle at t, x}$ \begin{align} x_{n+1} &= x_n + c_1 * v_n \, \Delta t\\[0.3em] v_{n+1} &= v_n + d_1 * g(t_{n+1}, x_{n+1}) \, \Delta t \end{align}

However, on the second page it is:

\begin{align} v_{n+1} &= v_n + g(t_n, x_n) \, \Delta t\\[0.3em] x_{n+1} &= x_n + f(t_n, v_{n+1}) \, \Delta t \end{align}

Am I failing to see the similarity between the two?

Edit: I've done numerical testing of the two algorithms and they are not equivalent, the first one is not time-reversible. Did I misinterpret something?


There are two variants of the symplectic Euler method. They are time-reverse to each other. This is actually clearly mentioned in the Wikipedia page. (While it is questionable if the general non-autonomous formulation is appropriate. It will still be an order-1 method, however the symplecticity depends on the system actually being Hamiltonian.)

Alternating both variants is equivalent to the (velocity) Verlet method of double step size. However, also one of the variants with modified initial conditions will give you an instance of the Verlet method.

The best short introduction for these symplectic integrators is Hairer, Ernst; Lubich, Christian; Wanner, Gerhard (2003). "Geometric numerical integration illustrated by the Störmer/Verlet method", see also http://www.unige.ch/~hairer/ under preprints and polycopies.

  • $\begingroup$ What do you mean by time-reverse to each other? Do I have to alternate between the two when changing the sign of the time? $\endgroup$ – Bloc97 Mar 17 '17 at 18:01
  • $\begingroup$ And doesn't symplectic mean that the integration is time-reversible? Why is the first method not time-reversible? $\endgroup$ – Bloc97 Mar 17 '17 at 18:09
  • 1
    $\begingroup$ No, symplectic means that it preserves some (symplectic) 2-form on the phase space, which implies volume preservation on the phase space which translates into energy preservation for some Hamiltionian function that is related to the Hamiltonian function of the ODE system. See en.wikipedia.org/wiki/Hamiltonian_mechanics $\endgroup$ – LutzL Mar 18 '17 at 10:15

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