Examples where the linearized system has a center at (0,0), whereas the original nonlinear system has a spiral point.
(1) An unstable spiral point at (0,0): Consider the system
x'=y, y'= -x+y3 .
Here (0,0) is the only critical point, so it is isolated. The Jacobian matrix is
0 1
-1 3y2
It has the eigenvalues ?=(3/2)y2 ±i(1/2)√(4-9y4). Hence the real part Re ? > 0 if y is nonzero. Therefore, the trajectories are "pulled away" from the origin, except on the x-axis. This forces the trajectories to be unstable spirals.
(2) An asymptotically stable spiral point at (0,0): The system
x'= y, y'= -x+(1-x2)y,
has (0,0) as its only critical point. The Jacobian at (0,0) is
0 1
-1 0
and has the eigenvalues ±i. This system is called a "van der Pol oscillator" (it is modelling an electric circuit with a vacuum tube as resistance). The unique trajectory with initial conditions x(0)=2, x'(0)=0, can be shown to be a simply closed curve (cycle) C. The other trajectories (sufficiently close to (0,0)) are spiralling fast towards C (as a limit curve) from the inside and also from the outside (see EP p.540).
REMARK. There is a lot of interest in such examples. Nevertheless, not much is known about the nature of such closed limit curves. Even for systems x'=f(x,y), y'=g(x,y) where f and g are polynomials of 2nd degree, it is unknown how many such limit cycles there can be around a given critical point. This concerns the second part of Hilbert's 16th Problem.
There exists an example with 4 cycles (but no examples with more than 4 cycles). The example was given by the Chinese mathematician Shi Song-Ling:
"A concrete example of the existence of four limit cycles for planar quadratic systems", Sci. Sinica Ser. A 23 (1980) 153-158.
Here is a link to a more modern discussion
It is known that the number of cyles must be finite whenever f and g are polynomials.
See [EP, p. 500, (5)] for a polynomial example (of 4th degree) with two limit cycles. Similar examples are readily constructed.