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We have been asked to prove the following Runge-Kutta method is of order 3

$y_{i+1} = y_i + \frac{K_1}{4} + \frac{3K_3}{4}$

with

$K_1 = h \times f(x_i,y_i)$

$K_2 = h \times f(x_i + \tfrac{h}{3}, y_i + \tfrac{K1}{3})$

$K_3 = h \times f(x_i + \tfrac{2h}{3}, y_i + \tfrac{2K_2}{3})$

We have ideas from this answer -Determining Runge-Kutta Order - but are struggling to apply our system to this method.

Thanks

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    This can not be right as the coefficients in the first equation of the $K$ have to sum up to $1$ for order-1 consistency. Just put in a constant $f$. – 2017-02-25

1 Answers 1

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According to the man the Butcher tables are named after (see these slides), the conditions for order 3 are \begin{align} b_1+b_2+b_3&=1\\ b_2c_2+b_3c_3&=\frac12\\ b_2c_2^2+b_3c_3^2&=\frac13\\ \text{and}\quad b_3a_{32}c_2=\frac16 \end{align} which gives $$b_2c_2(c_3-c_2)=\frac12c_3-\frac13$$ which for $c_2=\frac13$ and $c_3=\frac23$ results in $b_2=0$ and $b_3=\frac34$ which finally has $b_1=\frac14$, $a_{32}=\frac23$, $a_{31}=0$.