non-linear differential equation system

Hi,

I am having trouble with the following question..

dx/dt = y
dy/dt = -x + (1-x^2)y

Find the critical points and determine their stability.


Now, I found the critical point to be (0,0).
The matrix representing the linearized system is
A = [ 0, 1; -1, 0 ] and eigenvalues of this matrix are -i and i, which are complex with the real parts equal to zero, thereofore Linearization theorem can't be used here.

so.. using coordinate shift I get X=x and Y=y
thus dY/dX = -X/Y + (1-X^2).. to solve this I get an integrating factor
e^(0.5*x^2)... which can't be integrated. This is where I get stuck (assuming I have done everything else above correctly)

Does this mean the critical point is unstable and that the orbits of the non-linear system cannot be approximated by the linear approximation?

Or have I done something wrong along the way?

Thanks in advance

I get for the eigenvalues: The origin is a source but that's true only for solutions near the origin. Further qualitative analysis would show that solutions close to the origin spiral outward to the equilibrium solution which is periodic and solutions outside the equilibrium solution spiral towards it. Here's some Mathematica code to illustrate these two scenarios. Me personally, I'd always do it numerically just to check things.

Code:

---

tmax = 50;
xinit = 0.2;
yinit = 0.2;
sol = NDSolve[{Derivative[1][x][t] == y[t],
    Derivative[1][y][t] ==
    -x[t] + (1 - x[t]^2)*y[t],
    x[0] == xinit, y[0] == yinit},
  {x[t], y[t]}, {t, 0, tmax}]
p1 = ParametricPlot[Evaluate[{x[t], y[t]} /.
    sol], {t, 0, tmax}, PlotStyle -> Red]

tmax = 50;
xinit = 2.5;
yinit = 2.5;
sol = NDSolve[{Derivative[1][x][t] == y[t],
    Derivative[1][y][t] ==
    -x[t] + (1 - x[t]^2)*y[t],
    x[0] == xinit, y[0] == yinit},
  {x[t], y[t]}, {t, 0, tmax}]
p2 = ParametricPlot[Evaluate[{x[t], y[t]} /.
    sol], {t, 0, tmax}, PlotStyle -> Blue]
Show[{p1, p2}, PlotRange ->
  {{-3, 3}, {-3, 3}}]

---

Hey,

thanks shawsend. I figured out my mistake yday.. so it's all good