Numerical integration - verlet algorithm - accuracy

  • Thread starter QPingy
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Main Question or Discussion Point

In my computational physics textbook, three different velocity estimators are derived for a problem with equation of motion: [itex]\ddot x = F(x)[/itex] where the positions are found by using the Verlet algorithm:
[itex]x(t+h) = 2 x(t) - x(t-h) + h^2 F[x(t)][/itex]

The three velocity estimators are:
[itex]
v(t) = \frac{x(t+h) - x(t-h)}{2h} + \mathcal{O}(h^2)[/itex]
[itex]
v_{improved}(t) = \frac{x(t+h) - x(t-h)}{2h} - \frac{h}{12}\left( F[x(t+h)] - F[x(t-h)] \right) + \mathcal{O}(h^3)
[/itex]
[itex]
v_{leapfrog}(t + h/2) = \frac{x(t+h) - x(t)}{h} + \mathcal{O}(h^2)
[/itex]

I have no problems deriving these equations, so far everything is clear.
But, in the textbook they apply the methods for the 1D harmonic oscillator and they conclude:
The leap-frog energy estimator is an order of magnitude worse than the other two. This is not surprising since the fact that the velocity is not calculated at the same time instants as the position results in deviation of the energy from the continuum value of order h instead of h^2.

So, just because the time instants are different, the leapfrog's results are 1 orde worse than the other two? I can't find an explanation/reasoning for this...

Can someone help me?

Regards,
Jan
 

Answers and Replies

  • #2
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So, just because the time instants are different, the leapfrog's results are 1 orde worse than the other two?
Wrong time also means wrong position and therefore wrong potential. As the potential is monotonically increasing / decreasing for many steps at a time, you get a consistent direction of the error there.
 
  • #3
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Okay, but I still don't see why the leapfrog is one order worse than the first estimator. Both use 2 positions, which are calculated using the verlet algorithm. I understand what you're saying about the potential, but I don't get why this would result in such a difference between the 2 estimators...
 

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