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Given the Butcher table $$ \begin{array}{ccc} \Theta & \Big | & \Theta \\\ --&--&-- \\\ & | & 1\end{array}$$

I would like to

a) write down the corresponding evolution $\Psi$

b) find the stability function $S(z)$ of the "$\Theta$-method"

Somehow, I end up getting the Euler method but I don't know why. I would be very grateful if someone could point out my mistake to me. So here is what I did:

a) The RK formula is $$ c_i := \sum_{j=1}^s a_{ij}$$ $$ k_i := f(t_0 + c_i h, y_0 + h \sum_{j=1}^s a_{ij} k_j)$$ $$ \Psi^{t_0, t_0 + h} y_0 := y_0 + h \sum_{i=1}^s b_i k_i$$

Padding the Butcher table with zeros, I get the following Runge Kutta coefficients:

$c_1 = 0$, $c_2 = \Theta$

$a_{11}= a_{12} = a_{22} = 0$, $a_{21} = \Theta$

$b_1 = 1$, $b_2 = 0$

$k_1 = f(t + c_1 h, y_0 + h(a_{11}k_1 + a_{12} k_2) = f(t, y_0)$

$k_2 = f(t + \Theta h, y_0 + h(a_{21}k_1 + a_{22}k_2)) = f(t + h\Theta , y_0 + h \Theta f(t, y_0))$

It's a two step RK method, i.e. s = 2 and my $\Psi$ is:

$$ \Psi^{t_0 , t_0 + h} y_0 = y_0 + h(b_1 k_1 + b_2 k_2) = y_0 + h f(t_0, y_0)$$

b) The stability function of the Euler method is $S(z) = 1 + z$

I know that this is wrong because this is an old exam question and the stability function they give in their solution is $S(z) = \frac{1 + (1 - \Theta)z}{1 - \Theta z}$.

I've done it 4 times, getting the Euler method which does not depend on $\Theta$. I would be very grateful if someone could help me with this. Many thanks in advance.

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1 Answer 1

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Why do you think this is a two-step method? Take $s=1$, consider the given method as an implicit one-step method (you do not need to pad the Butcher tableau with zeroes), and you should get the stability function you want.

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  • $\begingroup$ Thank you! I padded the table because I am scared of implicit methods. Seriously: thank you so much, I have no idea why I filled the table with zeros. $\endgroup$ Commented Jan 27, 2011 at 13:32

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