$4^{\text{th}}$ order Runge-Kutta method

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?

• – Frenzy Li Aug 20 '18 at 16:21
• The old posts on similar topic do not have the answer of my question. – Soumen Basak Aug 20 '18 at 16:24
• 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. – Frenzy Li Aug 20 '18 at 16:26
• I suppose this particular choice of the coefficients has some advantages over other choices. I just wanted know those advantages with proper explanation. – Soumen Basak Aug 20 '18 at 16:30

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.
• 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. – Robert Israel Aug 20 '18 at 16:49
• 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. – Lutz Lehmann Aug 20 '18 at 17:00
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.
• I would imagine so then you can solve it easier if $\beta_{31} =0$ then $\beta_{32} = \alpha_{3}$ – user3417 Aug 21 '18 at 17:28