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RK4 is:

$$\begin{aligned}k_1&=hf(x_n, y_n)\\ k_2&=hf(x_n+h/2. y_n+k_1/2)\\ k_3&=hf(x_n+h/2, y_n+k_2/2)\\ k_4&=hf(x_n+h, y_n+k_3)\\ y_{n+1}&=y_n+(k_1+2k_2+2k_3+k_4)/6\end{aligned}$$

I want to know the reason of the coefficients $1/2, 1/6, 1/3$.

Is this the solution to some optimization problem? What is the general rule?

  • Kutta constructed this method explicitly as extension of the Simpson quadrature rule, see https://math.stackexchange.com/questions/2527302/whats-the-motivation, https://math.stackexchange.com/questions/3436705/how-to-derive, https://math.stackexchange.com/questions/3167436/why-use-classic-fourth-order – Lutz Lehmann Apr 20 '22 at 16:45
  • See also [https://math.stackexchange.com/questions/528856/explanation-and-proof-of-the-4th-order-runge-kutta-method] ( Explanation and proof of the 4th order Runge-Kutta method ) – Winther Nov 08 '22 at 11:21

1 Answers1

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Short answer: you can think of $v:=\frac{1}{6}\left(k_1+2 k_2+2 k_3+k_4\right)$ as a weighted average slope, so $$ y_{n+1}=y_n+h v $$ is nothing but a kind of Euler method but with an improved slope. Instead of taking the slope at $(x_n,y_n)$ you take several measures ($k_i$) in the interval $(x_n,x_{n+1})$. Why these weights exactly? This requires a long answer.

Long answer: see this great answer. To summarize, the exact value of $y_{n+1}$ is $y_n+\int_{x_n}^{x_{n+1}}f(x,y)dx$ so you should approximate this integral numerically (because you don't know $y(x)$). When you use Simpson's rule you obtain the formula you have presented.

A. J. Pan-Collantes
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