I am trying to get an intuitive understanding of the third-order Runge method. That method defines
$y_{k+1} = y_{k} + \left( \frac h 6 k_1 + \frac {4h} 6 k_2 + \frac h 6 k_3 \right)$
where
$k_1 = f(x_k,y_k)$
$k_2 = f( x_k + \frac h 2, y_k + \frac h 2 k_1 )$
$k_3 = f( x_k + h, y_k - h k_1 + 2h k_2 )$
The first three of those four equations make intuitively sense to me:
- The definition of $y_{k+1}$ is from Simpson's rule (Newton-Cotes degree $2$)
- The definition of $k_1$ is clear
- $k_2$ uses Euler with half the step size
However, I don't understand the definition of $k_3$. This does not look like it's taken from a quadrature formula. It seems like it is trying to approximate $y(x_k+h)$
$y_{k} - h y'(x_k) + 2h y'(x_k+h/2)$.
What is the reason for that choice?
Runge's third order method (1895) is in Kutta's notation a 4-stage method with Butcher tableau \begin{array}{c|cccc} 0&\\ \frac12&\frac12\\ 1&1\\ 1&0&0&1\\ \hline &\frac16&\frac46&0&\frac16 \end{array} See What's the motivation for Runge-Kutta methods? for references to the historical sources
The RK3 method you ask about was presented by Kutta (1901) in the course of systematically exploring the space of all 3rd order 3-stage explicit one-step methods. \begin{array}{c|ccc} c_1&\\ c_2&a_{21}\\ c_3&a_{31}&a_{32}\\ \hline &b_1&b_2&b_3 \end{array} Basing it on the Simpson quadrature method $(b_1,b_2,b_3)=\frac16(1,4,1)$ requires $c_1=0$, $c_2=\frac12$, $c_3=1$. This then also fixes $a_{21}=\frac12$. \begin{array}{c|ccc} 0&\\ \frac12&\frac12\\ 1&a_{31}&a_{32}\\ \hline &\frac16&\frac23&\frac16 \end{array} The one remaining condition is then the second 3rd order condition $$ b_3a_{32}c_2=\frac16\implies a_{32}=2 $$ and thus to balance $c_3=a_{31}+a_{32}$ then also $a_{31}=-1$ follows.