0
$\begingroup$

my question is quite easy(I think). I understand how to apply 4th order Runge Kutta and understand the principle of taylor series (the Mc lauren to be precise) . But I cannot fully understand how the formula of taylor series results in the algorithm/applied formulas of Runge Kutta.

What is the actual relation? I think Runge Kutta is based on Taylor series, right?

$\endgroup$
  • 1
    $\begingroup$ This is a rather complicated story, see the textbook of your choice or directly from the inventor of the systematic method math.auckland.ac.nz/~butcher/ODE-book-2008/Tutorials some slides explaining the conditions for the lower order (including 4) methods. $\endgroup$ – Lutz Lehmann May 2 '15 at 18:04
0
$\begingroup$

Some easy computations to verify the order of the method, it is by far not a proof but shows why the two steps at time increment $h/2$ are needed: Consider a linear system $y'=f(x,y)=Ay$ with an $n\times n$ matrix $A$. Then \begin{align} k_1&=f(x,y)&&=Ay\\ k_2&=f(x+h/2, y+h/2·k_1)&&=Ay+\frac h2 A^2y\\ k_3&=f(x+h/2, y+h/2·k_2)&&=Ay+\frac h2 A^2y+\frac{h^2}4 A^3y\\ k_4&=f(x+h, y+h·k_3) &&=Ay+hA^2y+\frac{h^2}2A^3y+\frac{h^3}4A^4y\\ \\ y_+&=y+\frac h6·(k_1+2k_2+2k_3+k_4)&&=y+hAy+\frac{h^2}2A^2y+\frac{h^3}6A^3y+\frac{h^4}{24}A^4y \end{align} which gives the exact partial sum of degree $4$ for the matrix exponential $e^{hA}$ of the exact solution. Which means that the local error is $O(h^5)$, as necessary for a method of order $4$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.