So, lets say for the family of the explicit Runge-Kutta methods: $$y_{n+1} = y_n + \sum_{i=1}^s b_i k_i$$

where, $$k_1 = hf(t_n, y_n)$$ $$k_2 = hf(t_n+c_2h, y_n+a_{21}k_1)$$ $$\vdots$$ $$k_s = hf(t_n+c_sh, y_n+a_{s1}k_1+a_{s2}k_2+\cdots+a_{s,s-1}k_{s-1})$$

Is there any high-order Runge-Kutta scheme (preferably more than 3rd order) where either all the nodes are found on $c_i=1,$ for $i=2,...,s$ OR if $c_m\neq{1},m=2...s$ then the weight on this node should be zero $(b_m=0)$?

  • $\begingroup$ It has been suggested that the question will attract better answers at scicomp.SE, and so I am migrating... $\endgroup$ – Todd Trimble Oct 21 '14 at 11:31
  • $\begingroup$ This isn't off-topic, but it has been suggested that the question would attract better answers at scicomp.SE, and so I am migrating... $\endgroup$ – Todd Trimble Oct 21 '14 at 11:32

The following are necessary (though not sufficient) conditions for a method to have order $p$:

$$\sum_{i=1}^s b_i c_i^{k-1} = \frac{1}{k}, \ \ \ \ \ \ \ \ \ \ k=1,2,\dots,p.$$

A method with the properties you specify would have, for any $k>1$,

$$\sum_{i=1}^s b_i c_i^{k-1} = \sum_{i\notin J}^s b_i$$

where the sum on the right is independent of $k$ (here $J$ is the set of indices for which $c_i=0$). To attain order three, this sum would have to be equal to both $1/2$ and $1/3$. So such methods can have order at most two.

Note that this is really a statement about quadrature rules (more general than the statement about Runge-Kutta methods).

| cite | improve this answer | |
  • $\begingroup$ David, with what I describe it could also be $$\sum_{i}{b_i}{c_i}=1/2$$ For example if $$c_1 = 0\: and\: b_1 = anything,\: c_2 = 1/2 \: and\: b_2 = 0,\: c_3 = 1\: and\: b_3 = 1/4,\: c_4 = 1\: and\: b_4 = 1/4$$. So, I would like to derive something like this example that I stated but with the higher order the possible (preferably 4 if that is possible). $\endgroup$ – runge_kutta_questions Oct 5 '14 at 22:25
  • $\begingroup$ Sorry, I answered from my phone and didn't read carefully. I have corrected it. $\endgroup$ – David Ketcheson Oct 6 '14 at 18:43

could you give me a reference of where you can find these conditions to achieve order p? I am not sure if it's valid though for all the methods. For example, the classic RK4 method that has $$b_1=0,\,b2=0.5,\,b_3=0.5,\,b_4=1,\,and\,c_1=1/6,\,c_2=1/3,\,c_3=1/3,\,c_4=1/6$$ Then if you evaluate for $k=4$ (since its fourth-order) the above expression that you mentioned gives: $$\sum_{i=1}^s b_i c_i^{k-1} = \frac{1}{2} \neq 1/4 $$

So, there is no way for example to take $b_1=0, b_2 = b_3 = b_4 = 1.0$ and find the proper $c_i$ in order to have 4-th order accuracy?


| cite | improve this answer | |
  • $\begingroup$ That is wrong. You have switched the b's with the c's. $\endgroup$ – David Ketcheson Oct 14 '14 at 10:11

Your Answer

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