Lyapunov stability question from Arnold's trivium 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.
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 $\begin{pmatrix} 0 &amp; 1 \\ 0 &amp; 0 \end{pmatrix}$,
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.)
(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