0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ and i don't know why the constant is same of two different order method $\endgroup$ – seyunkim Jul 25 '19 at 16:34
  • $\begingroup$ I studied this book chapter . but i do not understand this part. so i asked $\endgroup$ – seyunkim Jul 25 '19 at 16:46
  • $\begingroup$ this book do not explain my question. $\endgroup$ – seyunkim Jul 25 '19 at 16:46
  • $\begingroup$ 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. $\endgroup$ – Lutz Lehmann Jul 25 '19 at 17:00
  • $\begingroup$ my mistake not order but step $\endgroup$ – seyunkim Jul 25 '19 at 17:04
0
$\begingroup$

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.

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.