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I wonder how A-stability of a Runge-Kutta-Method implies that (asymptotic) stability is inherited from the solution of a linear initial value problem.

For a Runge-Kutta-Method $\psi^{\tau}$ there is a step size $\tau^{*}$ such that application to the Dahlquist test equation $x'=\lambda x$ yields $x_{k+1}=\psi_{\tau} x_k =R(\lambda \tau) x_k$ with rational $R$ for all $\tau < \tau^{*}$. Now $\mathbb{C}_{\_} \subset S := \left\{ x\in \mathbb{C} \mid \left|{R(z)}\right| \leq 1\right\}$ implies that (asymptotic) stability is inherited by the method $x_{k+1}R=(\tau\lambda)x_k$. I can follow the proofs so far. But how does this imply asymptotic stability is inherited by the RKM for any step size? We only know that $\psi^{\tau}=R(\tau\lambda)$ for certain small $\tau$.

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Why does $x_{k+1} = R(\lambda\tau) x_k$ only hold for small enough step size? –  Jitse Niesen Aug 6 '13 at 12:07
    
Hope I don't need to write down the whole proof, that for RKM applied to the test eq. there is a characteristic step size $\tau^{*}$ such that $\phi^{\tau}=R(\tau\lambda)$ for rational $\phi$ and $\tau \leq \tau^{*}$. I guess the crucial part is where it is shown that $\det(I-\tau\lambda A) \neq 0$ with $A$ the coefficient matrix. If it is zero for some $\tau'$ then $\frac{1}{\tau' \lambda}$ is an EV of $A$. If we assume asymptotic stability of the test eq which means $\Re{\lambda}<0$, then $A$ has an EV with positive real part. I guess this is the direction, but I don't know how to continue. –  user86638 Aug 6 '13 at 13:14
    
I get it, you are worried about poles in $R$. –  Jitse Niesen Aug 6 '13 at 13:19
    
Actually I want to derive a contraction. Something like: $R(\tau\lambda)\neq \phi^{\tau} \impl \text{The IVP is not asymptoticalle stable, i.e.} \Re{\lambda} > 0$. Well, of course only if that is true. –  user86638 Aug 6 '13 at 13:33
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The stability function $R$ is a rational function, so $R(z) = p(z) / q(z)$ with $p$ and $q$ polynomial. We have $z_{k+1} = R(\lambda\tau) z_k$ for all values of $\tau$ except for those which make the denominator $q(\lambda\tau)$ vanish. As far as I know, there is no restriction $\tau < \tau^*$, and the book of Hairer & Wanner (Solving ODEs, volume II) does not mention it (try going through the proof you are given and see whether you can get rid of it). So the only problem would be if $R$ had poles in $\mathbb{C}_{-}$. But if that were the case, then $S$ cannot contain the whole of $\mathbb{C}_{-}$, because the function goes to infinity close to a pole.

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I guess I get it now. If I'm not mistaken the determinant I mentioned is the denominator $q(z)$ you are talking about. So any step size for which $R(\tau \lambda) \neq \phi^{\tau}$ implies $R$ has a pole in $\tau \lambda$ which is not possible as you explained for an A-stable RKM. Thanks for helping! –  user86638 Aug 6 '13 at 13:49
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