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I have given two formulations of a general Runge Kutta Method with $s$ stages

  1. $$ y(x+h) = y_0 + h\sum_{i=1}^{s}b_ik_i$$ $$ k_i = f\left(x_0+c_ih;\,y_0+h\sum_{i=1}^{s}a_{ij}k_j\right)$$
  2. $$ y(x+h) = y_0 + h\sum_{i=1}^{s}b_if\left(x_0+c_ih;\,y_i\right) $$ $$ y_i = y_0 + \sum_{i=1}^{s}a_{ij}f\left(x_o+c_ih;\, y_i\right)$$

Those forumalas produce the same approximation, do they? But how can one show this?

Just putting in the definition of $k_i$ or respectivly $y_i$ leads to an "endless cycle".

Osch Bert
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1 Answers1

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Set $y(x+h)$ from the first definition equal to $y(x+h)$ from the second, and show that they result in the same thing (which they should, because the definitions are the same).

Emily
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  • I tried this approach, but I end up with $\sum_{i=1}^{s}k_i = \sum_{i=1}^{s}f\left(x_0+c_ih;,y_i\right)$ and do not know how to show this equality. – Osch Bert Nov 27 '13 at 21:20
  • @OschBert Well, use the definitions of $k_i$ and $y_i$. So, $$\sum k_i = \sum f\left(x_0+c_ih; y_0 + h\sum a_{ij}k_j\right) = \sum f(x_0+c_ih; y_i)$$ and so on. – Emily Nov 27 '13 at 21:26
  • I'm sorry but I do not see the second equality. – Osch Bert Nov 27 '13 at 21:36
  • So, if $y_i = y_0 +h\sum_{i=1}^{s}a_{ij}k_j$ then $k_i = f\left(x_0+c_ih;,y_i\right)$ and the result follows. Or vice versa if $k_i = f\left(x_0+c_ih;,y_i\right)$ then $y_i = y_0 +h\sum_{i=1}^{s}a_{ij}k_j$ and the result follows. But I go round in circles, don't I?! – Osch Bert Nov 27 '13 at 21:48
  • Start stepwise. Note that you have an error in your definition of $k_i$. It should read:

    $$k_i = f\left(x_0 + c_ih; y_0 + h \color{red}{\sum_{j=1}^i} a_{ij}k_j\right).$$

     – Emily Nov 27 '13 at 21:51
  • Unless you're looking at an implicit RK formula, in which case the upper bound can be $s$. But even still you can still look at it stepwise. – Emily Nov 27 '13 at 21:54
  • It is a general RK method (so the upper bound s should be correct). But now I end up at the point where i have to show that $y_i = y_0+h\sum a_{ij}k_j$. This equality holds if $k_j=f\left(x_0+c_ih;,y_i\right)$ but this equality holds if $y_i = y_0+h\sum a_{ij}k_j$ and so on... – Osch Bert Nov 27 '13 at 22:03
  • @OschBert In the explicit case, you can start from $i=1$ and build up. In the implicit case, you get $y_i = y_0+h\sum a_{ij}k_j = y_0+h\sum a_{ij} f(x_0+c_ih; y_j)$. There is no circular reasoning. You end up with an implicit equation. This is why it's called an implicit method. – Emily Nov 27 '13 at 22:16
  • I got my error in reasoning. The trick is to set $y_i = y_0 + h\sum a_{ij}k_j$ and derive the second formulation from the first one. TY for your assistance. – Osch Bert Nov 27 '13 at 23:21