Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

V.I. Arnold put the following question in his Mathematical trivium:

Can an asymptotically stable equilibrium position become unstable in the Lyapunov sense under linearization?

It puzzled me for a while, since my experience doesn't include such a pathological case - but I'm still not sure about this one.

share|cite|improve this question
Just a guess: what if there are singularities in the potentials or forces? Even if the force is generally attractive, this could lead to instabilities. – Raskolnikov Dec 31 '10 at 11:05
up vote 7 down vote accepted

For the system $dx/dt=y-x^3$, $dy/dt=-y^3$, the origin is an asymptotically stable equilibrium. (It is even globally attracting, as can be seen by sketching the phase portrait with the help of the nullclines $y=x^3$ and $y=0$.)

The linearized system at the origin is $dx/dt=y$, $dy/dt=0$, which is not Lyapunov stable: the solution starting at $(x_0,y_0)=(0,\epsilon)$ is $(x(t),y(t)) = (\epsilon t, \epsilon)$, which goes to infinity if $\epsilon \neq 0$.

The matrix corresponding to the linearized system is $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$, which has zero as a double eigenvalue. If both eigenvalues have nonzero real part, then the linearized system determines the stability of the original system. If the eigenvalues are $\pm c i$ (for some real $c \neq 0$), then the linearized system is a neutral center (hence Lyapunov stable). So the phenomenon in question can only occur if at least one eigenvalue is zero. (I haven't really thought about what can or cannot happen in the case with one zero eigenvalue and one real nonzero eigenvalue.)

share|cite|improve this answer
Eigenvalues of the linearized system (around the equilibrium state $(0,0)$) are $\pm i$, and that is on the edge of stability - can we do better and take the eigenvalues to the right semiplane? Or am I missing something? Roughly the same example is used in my Nonlinear Systems textbook to show a possibility of linearization method to fail - but it just shows that you can't make any conclusions about the original system if the eigenvalues of the linearized one are on the imaginary axis. I thought Arnold had something even more 'pathological' in mind with this question. – H. M. Šiljak Dec 31 '10 at 21:55
The eigenvalues are zero, not $\pm i$ (see my updated answer). – Hans Lundmark Jan 1 '11 at 10:35
My bad, $0\cdot1=1$ in my head while writing that comment. Thank you very much! – H. M. Šiljak Jan 1 '11 at 12:25
One more thing: As should be clear from the context, the statement "can only occur" refers to the two-dimensional case. For higher dimensions, I don't know. – Hans Lundmark Jan 1 '11 at 18:25

(without proof)

  • Linearized system has an eigenvalue at the imaginary axis => no conclusion can be made.
  • Linearized system only eigenvalues at open left complex halfplane => nonlinear system asympt. stable.
  • Linearized system at least one eigenvalue in the open right complex
    half plane => nonlinear system unstable

therefore you can conclude: an asymptotical nonlin. sys. after linearisation can be unstable if you get eigenvalues at the imaginary axis with algebraic multiplicity > geometric multiplicity

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.