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A I need some help with the derivation of fourth order Runge Kutta

  1. Jan 27, 2017 #1

    ATY

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    Hey guys, I need your help regarding the derivation of the fourth runge kutta scheme.
    So, I found http://www.ss.ncu.edu.tw/~lyu/lecture_files_en/lyu_NSSP_Notes/Lyu_NSSP_AppendixC.pdf
    this derivation. Maybe you have a clue what tehy are doing in C.54.
    So before this they are calculating the taylor expansion series of [tex]f_1,f_2,f_3,f_4[/tex]. This makes somehow sense. But I get confused when they throw everything together in equation C.54.
    Why is therre just the "ordinary" f everywhere instead of [tex] f_2,f_3,f_4 [/tex] ? I thougt that they just insert everything into equation C.46, but this seems not to be the case, or they are skipping some steps (like getting f_4 back to f_1 or so...)
    I hope that somebody takes a look at the derivation and can tell me what exactly they are doing there
     
  2. jcsd
  3. Jan 27, 2017 #2

    FactChecker

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    C.51 .. C.53 show how to replace f2, f3, f4 with terms expressions of partial derivatives of f. It says that is what they did.
     
  4. Jan 27, 2017 #3

    BvU

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    On the righthand side of 47, 51, 53 there are only ##f##, no ##f_2## etc. ?!

    Never mind, FC was faster
     
  5. Jan 27, 2017 #4

    ATY

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    Thank you for the answer :)
     
  6. Jan 31, 2017 #5

    ATY

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    Hey guys,
    I got one more question. I hope you do not mind it.
    I spend some time with the equations but, there is still one thing I do not understand. What do they mean with
    [tex]t=t^n[/tex] ?
    I wrote an email to the author of the text and got this reply:
    fd2366d65d.jpg

    But this does not really help me. I do not know what exactly this means. [tex]\Delta t[/tex] is from what I know the step size "h" or (x-a) (this is the notation the english wikipedia uses in the article about the taylor expansion).
    I hope that you want to help me one last time.
    Have a nice day
     
  7. Jan 31, 2017 #6

    BvU

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    It's just a notation convenience to write ##t^n## instead of ##t=n\Delta t##. He doesn't use it in lecture 2, but in table 3.1 he does -- in a manner that isn't fully consistent with his convention. But it's readable. Try to master 2nd order first by writing it out in full - if stuck, grab another textbook or pdf.

    For you it might be confusing that in section in appendix C.1, ##x## plays the role of ##t## in the Runge Kutta section C.2
     
    Last edited: Jan 31, 2017
  8. Jan 31, 2017 #7

    ATY

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    Ok. So I will try to go along the Taylor series notation of wikipedia right now:
    dadd9a8618.png
    so [itex]f[/itex] is the same [itex]f[/itex] as from the appendix.
    [itex]\Delta t[/itex] is the step size and is the same as [itex](x-a)[/itex].
    So does this mean, that [itex](f)_{t=t^n}[/itex] is the same as [itex]f(a)[/itex] ?
    This would mean, that [itex]t=t^n[/itex] is the position at which I evaluate my function.
     
  9. Jan 31, 2017 #8

    BvU

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    Higher order Runge Kutta evaluates the function at intermediate points. E.g. for 4th order in the starting point (C.47), halfway (##\Delta t\over 2##) using ##f_1## (C.48), halfway (##\Delta t\over 2##) using ##f_2## (C.49) and at the end of the step (##\Delta t##) using ##f_3## (C.50). So: yes, in this case ##\Delta t## is the step size. But: no, the function is evaluated not just at ##t^n## but also at ##t^{n+{1\over 2}}##

    Compare these to the notation in table 3.1 -- post if there is any doubt whatsoever left over.

    I didn't really do him justice there: it's pretty consistent, but he doesn't use it all over, probably for reasons of clarity :smile:
     
  10. Jan 31, 2017 #9

    ATY

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    Sorry for the confusion by my bad choice of words.
    I think I got it, but I just want to be sure:
    When we take a look at e.g. C.53
    we have [itex]f_4 = (f)_{t=t^n}+\Delta t (\frac{\partial f}{\partial t}+f_3 \frac{\partial f}{\partial y})_{t=t^n}[/itex].
    If we just look at the beginning now; it means that I will take the value of [itex]f[/itex] at the position [itex]t^n[/itex].
    (while the function [itex]f_4[/itex] get's evaluated at the point [itex]t+\Delta t[/itex]).
    Is this now finally correct ?
     
  11. Jan 31, 2017 #10

    BvU

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    I think you've got it !
    Perhaps the animation of slide 56 here helps picture what happens in an RK4 routine ?
     
  12. Jan 31, 2017 #11

    ATY

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    Thank you so much :)
     
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