Adaptive Runge-Kutta-Fehlberg method constant

Question

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

Answer 1

Steps you should follow:

• Understand and implement some low-order RK methods (or copy the code) and run some example ODE.
• Understand the idea of variable step size and that control of the step size tries to obtain a mostly uniform error density, each time step contributes a part to the global error that is proportional to the time step.
• Step doubling and Richardson extrapolation allow to estimate the local error and determine an optimal step size. Find out that this requires a very high effort.
• Embedded methods re-use the same right-hand-side evaluations for two methods of different orders. These are highly tuned methods with nearly no visible, intuitive structure in the coefficients.

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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.