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I would like to know the motivation behind the choice of numbers or coefficients in front of $k_1$, $k_2$, $k_3$ and $k_4$ in $4^{\text{th}}$ order Runge-Kutta method. There are many choices of the coefficients one can make. However, $\frac16$, $\frac13$, $\frac13$, $\frac16$ are the most popular set. Can anyone explain this point?

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  • $\begingroup$ The old posts on similar topic do not have the answer of my question. $\endgroup$ – Soumen Basak Aug 20 '18 at 16:24
  • $\begingroup$ Well, I did not start my comment with "Duplicate:", just "Relevant." In particular, the second post probably points in the correct direction but did not lay everything out explicitly. $\endgroup$ – Frenzy Li Aug 20 '18 at 16:26
  • $\begingroup$ I suppose this particular choice of the coefficients has some advantages over other choices. I just wanted know those advantages with proper explanation. $\endgroup$ – Soumen Basak Aug 20 '18 at 16:30
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Consider a differential equation $y' = f(x,y)$ with initial condition $y(0)=0$, and write out the series solution to order $x^5$ (in terms of the coefficients of the bivariate Taylor series of $f$ at $(0,0))$. Then compare with what you get from the Runge-Kutta scheme with coefficients $k_1, \ldots, k_4$. In order for the Runge-Kutta to agree with the series solution to order $x^5$, there will be a set of equations to solve. It will turn out that $k_1 = 1/6$, $k_2 = 1/3$, $k_3 = 1/3$, $k_4 = 1/6$ are the solution.

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  • $\begingroup$ Thats not true. In case of 4th order "The system involves 11 equations in 13 unknowns, so two of them could be chosen arbitrary." cfm.brown.edu/people/dobrush/am33/Matlab/RK4.html $\endgroup$ – Soumen Basak Aug 20 '18 at 16:40
  • $\begingroup$ That's if you want to choose not just $k_1, \ldots, k_4$, but the points at which to evaluate the $f$'s. Once you make the classical choices for those, the $k_1,...k_4$ are determined. $\endgroup$ – Robert Israel Aug 20 '18 at 16:49
  • $\begingroup$ For the specific sequence of Euler predictor, midpoint corrector twice and evaluation at the end-point, it is sufficient to compute the Taylor expansions for the special case $y'=f(x,y)=y$ to get the uniqueness of the coefficients. $\endgroup$ – Lutz Lehmann Aug 20 '18 at 17:00
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Runge Kutte forms an infinite family of ODE solvers. The coefficients come from something called a Butcher Tableau. The most basic form of Runge Kutte is Eulers method. Later there were better methods made with the mid-point method then Heun's method. Then Kutta gave an explanation of $4th$ order methods. The evaluation of the stages give the tableau. There is a lecture here on derivation for RK4 which is typically done in numerical analysis.

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  • $\begingroup$ Thanks for sharing two useful references on numerical analyis. In the 2nd reference on numerical analysis, it is mentioned that alpha2=1/2 and beta31=0 is the most useful (see the section above table 2) choice. Why this particular choice is most useful ? Dow you know the reason ? This can not be just human preference. $\endgroup$ – Soumen Basak Aug 21 '18 at 16:34
  • $\begingroup$ I would imagine so then you can solve it easier if $\beta_{31} =0$ then $ \beta_{32} = \alpha_{3}$ $\endgroup$ – user3417 Aug 21 '18 at 17:28
  • $\begingroup$ For other order 4 methods see math.auckland.ac.nz/~butcher/ODE-book-2008/Tutorials, the slides on low order methods contain the tableaus and their order conditions for the usual and some unusual methods up to order 4. $\endgroup$ – Lutz Lehmann Oct 15 '18 at 13:37

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