As far as I know, all the symplectic Runge-Kutta methods are implicit which need to solve non-linear equations during the calculation. Is there any explicit method? If not, why?


1 Answer 1


There are explicit, symplectic methods for certain types of Hamiltonian problems. For example, the symplectic Euler method

\begin{align} p_{n+1} &= p_n - h H_q(p_{n+1}, q_n) \\ q_{n+1} &= q_{n} + h H_p(p_{n+1}, q_n) \end{align}

is symplectic, see e.g. Theorem 3.3 on p. 189 in the book by Hairer, Wanner and Lubich (see full reference below). For simple functions $H$ like $H(p, q) = \frac{1}{2} \left( p^2 + q^2 \right)$, this becomes

\begin{align} p_{n+1} &= p_n - h q_n \\ q_{n+1} &= q_n + h p_{n+1} \end{align}

which is explicit. More generally, this method is explicit for separable Hamiltonians (see comments after the Theorem mentioned above).

In VIII.6 on p. 325, Hairer et al state that

Symplectic methods for general Hamiltonian equations are implicit, and so are symmetric methods for general reversible systems.

Therefore, there are no sympletic methods that are explicit for general Hamiltonian functions. I couldn't find the specific Theorem that states this, though.

Hairer, Ernst; Lubich, Christian; Wanner, Gerhard, Geometric numerical integration. Structure-preserving algorithms for ordinary differential equations, Springer Series in Computational Mathematics 31. Berlin: Springer (ISBN 3-540-30663-3/hbk). xvii, 644 p. (2006). ZBL1094.65125.

  • $\begingroup$ Under most definitions I think the trivial "do nothing" method $p^{n+1}=p^n$, $q^{n+1}=q^n$ would count as an explicit symplectic method. It's a fairly awful one though $\endgroup$
    – origimbo
    Commented May 24, 2019 at 16:22
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    $\begingroup$ Given that this is not even consistent with the solved differential equation, I would doubt whether it can be called a "method" in any reasonable sense of the word. $\endgroup$
    – Daniel
    Commented May 24, 2019 at 16:23
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    $\begingroup$ I couldn't find the specific Theorem that states this, though. I believe the theorem you're looking for is in section 2 of this paper by Sanz-Serna, see also these course notes, theorem 6 $\endgroup$
    – GoHokies
    Commented May 24, 2019 at 18:03
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    $\begingroup$ @GoHokies To be clear, the proof is sketched in Sanz-Serna's paper, where he also mentions that Lasagni gave a proof but didn't publish it. $\endgroup$ Commented May 26, 2019 at 6:21

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