0
$\begingroup$

consider a Runge-Kutta method with $R$ stages and coefficients given by Butcher-Tableau:

\begin{array}{c|c} a & B \\ \hline & c^T \\ \end{array}

Now consider the step $t \rightarrow t + h, y_0 \rightarrow y_1$ and let $g_i = y_0 + h \sum_{i=0}^R B_{i,j}k_j$. So the $g_i$ are approximations of $u(t+a_ih)$ and the $k_i$ are approximations of $u'(t+a_ih)$.

I have some questions concerning this construction:

  1. How do I get (probably implicit) formulas for $g = (g_1, ..., g_R)^T$ and $y_1$.

  2. Can I then use them to get $\omega(z) = y_1 = \frac{det(Id-zB-zec^T)}{det(Id-zB)}$ ($\omega$ is the stability function)? Maybe by applying the formulas of 1. to $u'(t) = \lambda u(t), u(0) = 1$?

$\endgroup$
2
  • $\begingroup$ To say nothing against the variability of variable-letter assignment, but the usual convention for Butcher tableaux has them as $\begin{array}{c|c}c&A\\\hline&b^T\end{array}$. $\endgroup$ Dec 3, 2019 at 15:22
  • $\begingroup$ @Dr.LutzLehmann Yes that's true, but unfortunately we do it differently in our lecture. $\endgroup$
    – Chaser01
    Dec 3, 2019 at 15:27

1 Answer 1

1
$\begingroup$
  1. Yes, in implicit methods you need to solve the implicit non-linear system. Preferably using a super-linear method.

  2. Yes, the stability function covers the behavior for linear problems, with a focus on non-expanding dynamics. You get to solve $\newcommand{\ones}{{\bf 1}}$ $$k=λ(y\ones+hBk)\implies k=λ(I-λhB)^{-1}\ones y$$ and $$y_{+1}=y+hc^Tλ(I-λhB)^{-1}y=(1+(λh)c^T(I-λhB)^{-1}\ones)y$$.

    Then apply the determinant identity $\det(I+uv^T)=1+v^Tu$ in reverse $$1+(λh)c^T(I-λhB)^{-1}\ones=\det(I+(λh)\ones c^T(I-λhB)^{-1})=\frac{\det(I-λhB+(λh)\ones c^T)}{\det(I-λhB)}$$

$\endgroup$
2
  • $\begingroup$ Thank you! But what do you mean by a super-linear method? I just don't know how to start with the first question. $\endgroup$
    – Chaser01
    Dec 3, 2019 at 16:29
  • 1
    $\begingroup$ Newton's method or similar. In practice, one would prefer to evaluate the Jacobian at most once per step or use the same over several steps. With the corresponding simplified Newton method you get linear convergence with a rate $O(h^2)$ which is still faster than the "naive" fixed-point iteration of the stage equations which has a rate $O(h)$. $\endgroup$ Dec 3, 2019 at 16:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .