Runge-Kutta Method and Stability Function

numerical methodsnumerical-calculusrunge-kutta-methodsstability-in-odes

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 $u'(t) = \lambda u(t), u(0) = 1$?

Best Answer

  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 $$1+(λh)c^T(I-λhB)^{-1}\ones=\det(I+(λh)\ones c^T(I-λhB)^{-1})=\frac{\det(I-λhB+(λh)\ones c^T)}{\det(I-λhB)}$$

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