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

[ATY]

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 f_1,f_2,f_3,f_4. 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 f_2,f_3,f_4 ? 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

 

[FactChecker]

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.

 

[BvU]

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

Never mind, FC was faster

 

[ATY]

Thank you for the answer :)

 

[ATY]

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
t=t^n ?
I wrote an email to the author of the text and got this reply:

But this does not really help me. I do not know what exactly this means. \Delta t 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

 

[BvU]

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

 

[ATY]

Ok. So I will try to go along the Taylor series notation of wikipedia right now:

so f is the same f as from the appendix.
\Delta t is the step size and is the same as (x-a).
So does this mean, that (f)_{t=t^n} is the same as f(a) ?
This would mean, that t=t^n is the position at which I evaluate my function.

 

[BvU]

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.

in a manner that isn't fully consistent with his convention

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

 

[ATY]

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 f_4 = (f)_{t=t^n}+\Delta t (\frac{\partial f}{\partial t}+f_3 \frac{\partial f}{\partial y})_{t=t^n}.
If we just look at the beginning now; it means that I will take the value of f at the position t^n.
(while the function f_4 get's evaluated at the point t+\Delta t).
Is this now finally correct ?

 

[BvU]

I think you've got it !
Perhaps the animation of slide 56 here helps picture what happens in an RK4 routine ?

 

[ATY]

Thank you so much :)