Let us assume I have an A-stable numerical scheme. I believe that given any linear equation $y' = Ay$, it means that the numerical scheme applied to this equation is stable (and therefore convergent since it is consistent) if the eigenvalues of A have a negative real part.

My question is then, does this result extend to the non-linear case ? I am interested in particular in a system $y' = f(y,t)$ where the Jacobian of $f$ has negative real eigenvalues $\forall y$.

edit: Also I guess I could give a particular example here (but my question is not limited to that case): for example let us say the ODE describes a 2nd order chemical reaction network, i.e. $f \in R^n \rightarrow R^n$ is a polynomial function of the $y$ of order 2, where $n$ is the number of species. Considering the network I was able to prove that the eigenvalues of the Jacobian are real negative for any $y$, however I don't know if I can make the further argument that any A-stable method will converge (by the way, what about L-stable methods, B-stable methods, ... ?)

  • $\begingroup$ Usually, yes. But there are pathological counterexamples. $\endgroup$ Mar 8, 2013 at 12:23
  • $\begingroup$ Thanks. Could you give me a counter example I could take a look at ? I also edited my question for further details $\endgroup$
    – Tibo
    Mar 8, 2013 at 17:20
  • $\begingroup$ regarding your edit: If you are really dealing with a fixed-size system of ODEs (and not a PDE semi-discretization), then any one-step method is guaranteed to converge. You don't need $A$-stability for that. You only need to show that $f$ is Lipschitz continuous in $y$. $\endgroup$ Mar 10, 2013 at 10:06

1 Answer 1


I'm going to answer a more general question than the one you asked: do the eigenvalues of an initial value ODE determine the stability of the solution? Here I'm referring to mathematical stability, not numerical stability. Of course, a "yes" to this question is a necessary condition for a "yes" to your question. And unfortunately, the answer is "no".

In general, for nonlinear and/or non-autonomous problems, studying the eigenvalues of a frozen linearization, i.e. the Jacobian with $t$ fixed, gives very useful insight. However, there are two major caveats:

  1. For non-normal matrices, eigenvalues don't tell the whole story, and one must consider pseudospectra instead. This caveat applies even to linear, constant-coefficient problems. See the book by Trefethen & Embree.

  2. There exist pathological examples of systems where the solution behavior is completely unrelated to the eigenvalues. I will give an example that is attributed to Vinograd:

$$y'(t) = A(t) y(t)$$


$$A(t) = \begin{pmatrix} -1-9\cos^2(6t)+6\sin(12t) & 12\cos^2(6t) + \frac{9}{2}\sin(12t) \\ -12\sin^2(6t) + \frac{9}{2}\sin(12t) & -1 - 9 \sin^2(6t)-6\sin(12t) \end{pmatrix}$$

The eigenvalues of $A(t)$ are $\lambda=-1, -10$ -- independent of $t$. But the solution is

$$y(t) = C_1 \exp(2t) \begin{pmatrix} \cos(6t) + 2\sin(6t) \\ 2\cos(6t)-\sin(6t) \end{pmatrix} + C_2 \exp(-13t) \begin{pmatrix} \sin(6t) - 2\cos(6t) \\ 2\sin(6t)+\cos(6t) \end{pmatrix}$$

So in this case, the eigenvalues suggest stability, but the solution exhibits unbounded exponential growth. For a nice discussion of generalizations of this example, see this excellent book. It is out of print, but used copies can sometimes be found -- for a hefty price -- on Amazon.

Some further comments: I don't know of a nonlinear, autonomous example exhibiting this kind of behavior, and it may well be that they don't exist. I also don't know of an example arising in applications. People do generally use the eigenvalues of the frozen jacobian as an indicator of stability in practice.

  • $\begingroup$ thanks a lot, that is what I wanted to know. I will try to get a hand on these two books too. In the meantime, I guess I will just make the argument using the eigenvalues of the Jacobian and mention that some counterexamples exist. $\endgroup$
    – Tibo
    Mar 10, 2013 at 17:27
  • $\begingroup$ @Tibo did you see my last comment above (on your question)? $\endgroup$ Mar 10, 2013 at 18:26
  • $\begingroup$ just saw it thanks. You are right I may have not done a correct distinction in my question between convergence (which means that the numerical scheme will give me the right answer when $\Delta t \rightarrow 0$) and stability (which is that the solution of the numerical scheme is bounded). $\endgroup$
    – Tibo
    Mar 10, 2013 at 23:06
  • $\begingroup$ For the convergence case of the pure ODE system as in a chemical reaction network, the Lipschitz argument is a bit tricky though: the function f being a polynomial of order 2, it is Lipschitz continous on any interval, but the size of the Lipschitz constant grows with the size of the interval, so I am not sure I can apply the convergence theorem for any proper one step method... $\endgroup$
    – Tibo
    Mar 10, 2013 at 23:15
  • $\begingroup$ I mentioned A-stability because I wanted to be sure there was convergence for any $\Delta t$ due to the fact that the eigenvalues of the Jacobian are real negative but unbounded (same reason that the Lipschitz constant $L$ is unbounded) $\endgroup$
    – Tibo
    Mar 10, 2013 at 23:20

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