Talk:Symplectic integrator

message: It says in the article:
"Symplectic integrators are designed for the numerical solution of Hamilton's equations"
What happens if you try to use a symplectic integrator whose time evolution cannot be derived from a hamiltonian (eg. if the system also includes a dissipative force)? I think this should be mentioned in the article. Ossi 12:48, 25 October 2007 (UTC)
replies: text: A symplectic integrator is defined in terms of a Hamiltonian, H.  If there is no H (as you say, the system cannot be derived from a Hamiltonian) then there is simply no integrator to iterate. Does that make sense? Jcandy 22:16, 27 October 2007 (UTC)
thread_title:  A question 

message: "Symplectic integrators possess as a conserved quantity a Hamiltonian which is slightly perturbed from the original one. By virtue of these advantages".
I am a mere pusher of 1s and 0s, and don't really know what the advantage is of mostly conserving the original Hamiltonian. In addition to the statement above, I would love to see a more "practical" statement of the virtues. Maybe something like: "the error is therefore of the order (dt^n)" or "numeric stability is improved significantly over non-SI methods", or whatever will help me compare this usefully to say Runge-Kutta or other non-SI methods.  — Preceding unsigned comment added by 76.104.205.238 (talk) 18:43, 31 August 2011 (UTC)
thread_title:  A Request for more practical info 

message: Substituted at 07:29, 30 April 2016 (UTC)
thread_title: Assessment comment

message: As stated in the page, the proposed second-order example should match the Verlet method. However, by doing the math, it seems that this is only true with c1 = 1 and c2 = 0, and not the other way around (d1 and d2 are OK). I tried to use c1 = 0 and c2 = 1 (as it is currently stated in the article), but this results in integration steps that I can clearly see that do not hold for constant accelerations. Am I interpreting something wrong?
: I was also confused by this However, a look at previous equations reveals that the higher order indices are applied first! Since they are operators, they act from right to left. Therefore, Verlet's method is recovered by first setting c=1, d=1/2, then again with c=0, d=1/2. Therefore, the coefficients are ok, but the "simplified equations" are, in my opinion, wrong. They should regressive equations, giving e.g. v_{i-1} in terms of x_i and v_i, to be run from i=k to i=1. --Daniel (talk) 11:29, 17 September 2021 (UTC)
replies: text: This seems an important **open issue** in this page. I ran across it when implementing the fourth-order version, and running into trouble. I then saw this discussion: https://scicomp.stackexchange.com/questions/20533/test-of-3rd-order-vs-4th-order-symplectic-integrator-with-strange-result  ...which agrees with your statement that the order should be reversed. How to best do this without changing too much at the same time?  — Preceding unsigned comment added by 92.225.65.73 (talk) 11:41, 15 November 2023 (UTC)

text: Note: I now at least inserted a remark in the 'Examples' section stating that the index is always traversed in decreasing order. It would be, however, much better, if the operator statements were rephrased so they have the ordering with the lowest index value to the far right (applied first), because that would then make everything below very easy to follow. It would also require a change in the order of the listed coefficients further below.  — Preceding unsigned comment added by 92.225.65.73 (talk) 11:55, 15 November 2023 (UTC)
thread_title: problem in "A second-order example" section?