Nature of a fixed point in dynamical system

Question

I have the following system:

$\dot{x}=y+x(x^4+2x^2y^2-4x^2+y^4-4y^2+4)(8-(x^2+y^2)^{\frac{3}{2}})^3$ $\dot{y}=-x+y(x^4+2x^2y^2-4x^2+y^4-4y^2+4)(8-(x^2+y^2)^{\frac{3}{2}})^3$

Using cylindrical coordinates this can be rewritten as:

$\dot{r}=r(r^2-2)^2(8-r^3)^3$

$\dot{\theta}=-1$

There is a unique fixed point at the origin and I would like to investigate its nature.

Using cartesian coordinates the jacobian matrix at the point (0,0) is $\begin{bmatrix}2048&&1\\-1&&2048\end{bmatrix}$ which suggests this point is an unstable focus (clockwise).

Using cylindrical coordinates the jacobian matrix at the point (0,0) is $\begin{bmatrix}2048&&0\\0&&0\end{bmatrix}$ which gives us no information.

So according to this information I would conclude that the fixed point (0,0) is an unstable focus. However when I plot the phase portrait of this system on mathematica it shows it as a star node. Is there a problem with my calculations? Is it a star node or unstable focus? Thank you.

Hi variable, I have just added a picture of the plot to my post. Is this what yours looks like? — NormalsNotFar, Dec 17 '15 at 14:30
Ahh ok, so what is a good explanation as to why it straightens out like a star node? — NormalsNotFar, Dec 17 '15 at 14:34
What I mean is; How could I have known, without plotting, that it would appear like a star node rather than a focus? — NormalsNotFar, Dec 17 '15 at 14:36

score 5 · Accepted Answer · answered Dec 17 '15 at 14:57

The eigenvalues of your linearized system are $2048+i$ and $2048-i$. This suggests that the rate of growth in the radial direction will be much faster than the rate of rotation.

More generally, the general solution of the linear system

\begin{align} x' &= gx+y \\ y' &= -x+gy \end{align} is \begin{align} x(t) &= c_1 e^{g t} \cos (t)+c_2e^{g t} \sin(t) \\ y(t) &= c_2 e^{g t} \cos (t)-c_1e^{g t} \sin (t). \\ \end{align}

Thus, we can see that the larger $g$ is, the faster the rate of radial growth relative to the rotation. Here's an illustration of the phase portraits for various values of $g$:

I couldn't agree more; a well deserved +1 from me! :-D – Frits Veerman Dec 17 '15 at 23:34 — Frits Veerman, Dec 17 '15 at 23:34

score 4 · Answer 2 · answered Dec 17 '15 at 14:54

From the Cartesian description, you've found that the origin is an unstable focus, with eigenvalues $\lambda_{\pm} = 2048 \pm i$. The reason you don't see the turning behaviour at the origin is because the radial expansion rate is a couple of thousand times higher than the turning rate. In other words, the imaginary part is almost negligible compared with the real part of the eigenvalues. So, it looks very much like an unstable node with double real eigenvalue $\lambda = 2048$ -- which is a star.

The reason the Jacobian in cylindrical coordinates doesn't give you that much information, is that that coordinate system isn't defined at the origin. There, the entire line segment $\left\{ (r,\theta) \middle \vert\,r=0 \right\}$ corresponds with a point $(0,0)$ in the Cartesian description.

You could manually tweak the system by boosting the angular velocity, i.e. use $\dot{\theta} = - 1000$. Then, you'll see that you recover the 'focus'-type behaviour you initially expected from the analysis.

Nature of a fixed point in dynamical system

2 Answers2