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Implicit Euler Integration

  1. May 25, 2012 #1
    I'm trying to write a code to implement he backwards Euler method to integrate the equation of motion. The sticking point seems to be that the acceleration is due to drag, and thus is dependent on the new position and velocity.

    I understand the method to be:

    [itex]v_{i+1}=v_{i}+a_{i+1}δ[/itex]
    [itex]x_{i+1}=x_{i}+v_{i+1}δ[/itex]

    With only the current conditions I can’t evaluate [itex]a_{i+1}[/itex] and am stuck.

    Any help on how the implicit methods work would be really appreciated.


    I’ve considered using [itex]a_{i+1}= a_{i}+ (a_{i}-a_{i-1}){δ} [/itex] but then that’s not really an implicit method is it? – you’re simply reusing the acceleration from last time around.

    Thanks in advance,
     
  2. jcsd
  3. May 26, 2012 #2
    The backwards Euler method is iterative. The first guess is usually the forward Euler method, so the right-hand side is determined using the information at time "i". You now have a first estimate of the properties at time "i+1". You can now use these estimates in the right-hand side to get a better estimate of the properties at time "i+1".

    so for your example, first use a and v at time "i", then you have v and x at time "i+1". Now calculate a at "i+1" and recalculate v an x, but using your new estimate of a and v at "i+1":

    first iteration:
    [itex]v^1_{i+1}=v_{i}+a_{i}\Delta t[/itex]
    [itex]x^1_{i+1}=x_{i}+v_{i}\Delta t[/itex]
    calculate [itex]a_{i+1}=f(v^1_{x+1})\Delta t[/itex]
    second iteration:
    [itex]v^2_{i+1}=v_{i}+a^1_{i+1}\Delta t[/itex]
    [itex]x^2_{i+1}=x_{i}+v^1_{i+1}\Delta t[/itex]
     
  4. May 26, 2012 #3

    AlephZero

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    You take a guess at the starting value, and (with luck) the interations will converge to a better value. If they don't converge, try a smaller time step.

    That's not right, because the "dimensions" don't make sense. Multiplying ##a## by ##\delta## gives you a quantity that is a ##v##, not an ##a##.

    You could just start with ##a_{i+1} = a_i##, or if you want to extrapolate, the right fomula is ##a_{i+1} = a_i + (a_i-a_{i-1}) = 2a_i - a_{i-1}##.

    Whether it's "better" to extrapolate or not will depend on the problem you are solving and the size of the time steps.
     
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