# Adaptive Runge - Kutta - Fehlberg method constant

in text book

Runge - Kutta - Fehlberg method

$$\tilde{w}_{i+1} = w_i + \frac{16}{135}k_1 + \frac{6656}{12825}k_3 + \frac{28561}{56430}k_4 - \frac{9}{50}k_5 + \frac{2}{55}k_6\tag1$$

is 6order Runge-Kutta method and

$$w_{i+1} = w_i + \frac{25}{216}k_1 + \frac{1408}{2565}k_3 + \frac{2197}{4104}k_4 - \frac{1}{5}k_5 \tag2$$ Runge-Kutta 5 order

then why coefficient is below?

$$k_1 = hf(t_i,w_i)$$

$$k_2 = hf(t_i+\frac{h}{4},w_i+\frac{1}{4}k_1)$$

$$k_3 = hf(t_i+\frac{3}{8}h,w_i+\frac{3}{32}k_1+\frac{9}{32}k_2)$$

and $$k_4,k_5,k_6$$ and so on.

and why (1) formula has $$w_i$$ instead of $$\tilde{w}_i$$

and I don't know why this method don't use Runge - Kutta method order4 and 5 or 4 and 6 instead of order 5 and 6

• and i don't know why the constant is same of two different order method – seyunkim Jul 25 '19 at 16:34
• I studied this book chapter . but i do not understand this part. so i asked – seyunkim Jul 25 '19 at 16:46
• this book do not explain my question. – seyunkim Jul 25 '19 at 16:46
• How sure are you about your order assignments? 6 stage Runge-Kutta methods have maximally order 5, there is no 5-stage order 5 method, so the method in (2) has order 4 which makes this the 4(5) embedded RKF method. – Lutz Lehmann Jul 25 '19 at 17:00
• my mistake not order but step – seyunkim Jul 25 '19 at 17:04

As for the cited embedded system, order 4 with an embedded order 5 step for the error estimator: From the construction of the method you get that $$w_{i+1}=\tilde w_{i+1}+Ch^5+O(h^6)$$ and the local error of $$\tilde w_{i+1}$$ is $$O(h^6)$$, so that $$w_{i+1}-\tilde w_{i+1}=Ch^5+O(h^6)$$ is a valid estimate of the dominant term $$Ch^5$$ of the step error of the 4th order method and can be used to adapt the step size to the desired global error level.