Nonlinear nature of fixed point

Question

Given the system $\dot x = y-x^3$ and $\dot y = -x-y^3$ What can you say about the nonlinear nature of the fixed point?

Attempt:

I started out by employing linear analysis. The nullclines are plotted below, showing an intersection at the origin. Hence, there is one fixed point at (0,0). The Jacobian of the system at (0,0) classifies the system as a linear center. ($\tau = 0$) and ($\Delta = 1$). How do I describe the nonlinear nature of this point? Should I start by converting the system to polar coordinates?

Answer 1

Going to polar coordinates is not a bad idea. If the linearisation yields a centre, then you cannot directly conclude anything about the nonlinear nature of the fixed point, as you cannot apply the Grobman-Hartman theorem. However, you know that nonlinearly, you can have an unstable focus, a stable focus or the point can be an actual nonlinear centre. That means that at least locally, it is useful to try to separately identify the rotational and radial components of the motion.

In your case, the phase portrait suggests a stable focus, nonlinearly. When going to polar coordinates, remember that you're still interested in small values of $r$ , so you can expand the equations for $r'$ and $\theta'$ around $r=0$.

Answer 2

Applying rr' = xx' +yy'

$\dot rr = x(y-x^3)+y(-x-y^3)$

Simplifying yields:

$\dot r =-r^3 (cos^4(\theta)+sin^4(\theta))$

Similarly applying $ \dot \theta r^2 = (x\dot y -y\dot x)$ we get:

$ \dot \theta = r^2 cos(\theta)sin(\theta)-1 $

$(cos^4(\theta)+sin^4(\theta)) > {1 \over 2}$ for all $\theta$

Taking small values of r around r=0, $\dot r < {-1 \over 2} $ which indicates that the system spirals to the origin. $\dot \theta >0 $ so the system moves clockwise