1. ## Linear stability analysis

For the system
$\displaystyle \dot{x} = -y + ax(x^{2}+y^{2})$
$\displaystyle \dot{y} = x + ay(x^{2}+y^{2})$

linearise the system around the fixed point and comment on the stability properties for $\displaystyle a < 0, a = 0, a > 0.$

The system can be written in polar coordinates as
$\displaystyle \dot{r} = ar^{3}$
$\displaystyle \dot{\theta} = 1$
By solving these equations, comment on the stability properties of the fixed point for $\displaystyle a < 0, a = 0, a > 0.$

I've found the fixed point to be (0,0). I found that the fixed point is a centre, and so stable but not asymptotically stable for all a - but this seems to simple when the question specifically mentions $\displaystyle a < 0, a = 0, a > 0.$.

I've converted the equations to polar coords and solved to get $\displaystyle r = \sqrt{\frac{r_{0}}{1-2ar_{0}t}}$ but don't know where to go from here to comment on stability.

Any help would be really appreciated

2. ## Re: Linear stability analysis

Messing around with this it becomes clear that

i) if a=0, then (0,0) is a center

ii) if a>0, the trajectory is an asymptotically unstable spiral about (0,0)

iii) if a<0, the trajectory is an asymptotically stable spiral about (0,0)

I'm not seeing from the linearization why this is so yet. My first cut had $a$ dropping out of the system matrix at (0,0).

I'll continue to look at this.

3. ## Re: Linear stability analysis

Originally Posted by sophiew3
For the system
$\displaystyle \dot{x} = -y + ax(x^{2}+y^{2})$
$\displaystyle \dot{y} = x + ay(x^{2}+y^{2})$

linearise the system around the fixed point and comment on the stability properties for $\displaystyle a < 0, a = 0, a > 0.$

The system can be written in polar coordinates as
$\displaystyle \dot{r} = ar^{3}$
$\displaystyle \dot{\theta} = 1$
By solving these equations, comment on the stability properties of the fixed point for $\displaystyle a < 0, a = 0, a > 0.$

I've found the fixed point to be (0,0). I found that the fixed point is a centre, and so stable but not asymptotically stable for all a - but this seems to simple when the question specifically mentions $\displaystyle a < 0, a = 0, a > 0.$.

I've converted the equations to polar coords and solved to get $\displaystyle r = \sqrt{\frac{r_{0}}{1-2ar_{0}t}}$ but don't know where to go from here to comment on stability.

Any help would be really appreciated
You've basically completed the problem.

Just look at your solution for $r(t)$ as $t\to \infty$ under different values of $a$ and with an initial condition slightly away from the critical point, i.e. it's been perturbed.

4. ## Re: Linear stability analysis

I think you've missed the point of the question. If you linearize about the critical point, it predicts a center. The eigenvalues are $\displaystyle \pm i$. Introducing polar coordinates the stability is easily determined and depends on the sign of $\displaystyle a$ (whether positive, negative or zero).

So what does this say? That the dynamics of the actual system cannot be predicted from the linear system. Why? The critical point is non-hyperbolic, i.e. the eigenvalues of the linear system have zero real part. The Hartman–Grobman states that the behaviour of a dynamical system near a hyperbolic equilibrium point is qualitatively the same as the behaviour of its linearization near this equilibrium point. (Note: the real part of the eigenvalues have nonzero real part).