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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?

shuhalo
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  • Why do you think there is anything intuitive about RK3? See https://math.stackexchange.com/a/3362041/115115 for a visual impression. W.M.Kutta just parametrized the order equations of the general explicit 3rd order method, applied the condition that the method should reduce to the Simpson method for the quadrature case and then explored what coefficients could be set to zero. – Lutz Lehmann May 15 '20 at 12:56
  • The amount of people hoping to get explained intuition is... touching. Sorry, folks, not a chance. If you diligently do all the exercises, you may develop it. Or you may not. But intuition can't be taught or explained, by definition. –  May 16 '20 at 18:34

1 Answers1

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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.

Lutz Lehmann
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  • Well, I presume that's as good as it gets. I was hoping that the third equation can be interpreted in a straight-forward manner so that RK3 is "obviously" a reasonable approach. So that the error analysis is making rigorous what is intuitively worth a try. Apparently you can't separate the method design from the error analysis in Runge-Kutta methods, which is rather different from, say, finite difference methods, at least at the beginner level. – shuhalo May 17 '20 at 04:39
  • What you are proposing in comparing to finite difference methods vs. linear multi-step or Galerkin methods is what Heun did in 1900 just prior to Kutta in 1901. He took chains of Euler-like steps and only combined their final values in a linear combination. This gives much more flexibility to satisfy the order conditions, at the cost of more function evaluations. Kutta "condensed" that idea to get methods with a minimal number of function evaluations. – Lutz Lehmann Nov 14 '20 at 11:05
  • As to being "straight-forward", this is just not possible as the expression of the Taylor coefficients in the derivatives of the ODE function gives rather complex polynomial terms from the third order on. This is reflected in that second order methods are still "intuitive" while third order methods start to have rigid coefficient conditions that can not easily traced back to some discretization of the Picard iteration or similar. – Lutz Lehmann Nov 14 '20 at 11:09
  • If I may ask a further question, if the method is instead $a_{31}=1,a_{32}=0$ and the rest are the same, is this a valid method based on Simpson's quadrature? I could not find it in the generic third order three stage method, but I have yet to see why not. – 10understanding Nov 22 '20 at 13:54
  • Yes, it is a valid RK method, but its order will only be 2, as the third order condition can not be satisfied. – Lutz Lehmann Nov 22 '20 at 14:42