1

I have a Runge-Kutta method given by the Butcher tableau: $$ \begin{array}{c|ccc} 0 & & & \\ 1/2 & 1/2 & & \\ 1/3 & 0 & 1/3 & \\\hline & -1/3& 1/3 &1 \end{array} $$ I would like to prove that hasn't order $3$. However, I don't know how to attack the problem, so I appreciate any hint or solution. Thanks in advance.

glS
  • 6,818
  • 2
    Are you allowed to use order conditions for RK methods? I.e. $\sum_i b_i = 1$ for the first order, $\sum_i b_i c_i = \frac{1}{2}$ for the second order and a pair of $\sum_i b_i c_i^2 = \frac{1}{3}, \sum_{i,j} b_i a_{ij} c_j = \frac{1}{6}$ for the third order. – uranix May 20 '15 at 14:31
  • I shouldn't use the conditions –  May 20 '15 at 14:53

1 Answers1

3

Any Runge-Kutta method can be used as a quadrature formula. Just apply it to $$ \frac{dy}{dx} = f(x) $$ so $$ y_{n+1} = y_{n} + h \sum b_i f(c_i). $$ Let's assume that the method is of third order. Thus evaluating $y_{n+1} - y_n$ as $$y_{n+1} - y_n = \int_{x_n}^{x_{n+1}} f(x) dx \approx h \sum b_i f(c_i)$$ shoud have an error of $O(h^4)$ magnitude (local error of RK method of third order). Consider $f(x) = x^2$. $$ \int_0^h x^2 dx = \frac{h^3}{3}\\ h\sum_i b_i c_i^2 = \frac{h}{3}\frac{h^2}{4} + h\frac{h^2}{9} = \frac{7}{36}h^3 $$ which is $O(h^3)$ far from the right answer, so no third order.

uranix
  • 7,503