I'm stuck with a problem which bothers me.
It's about the following differential equation system:
$$ x^{\prime}=y-(x^2+y^2-1)\cdot x\\ y^{\prime}=-x-(x^2+y^2-1)\cdot y $$
I have transformed the system in polar coordinates, which looks like: $$ r^\prime=-r\cdot(r^2-1)\\ \theta^\prime=-1 $$
Now to the part where I am desperate and don't know what to do:
First of all, I want to know how to determine the equilibrium points of the system in polar coordinates, since $\theta=-1$. How does that effect the equilibrium points. (0,0) is a equlibrium of the system in normal coordinates but is it also one of the system in polar coordinates?
And my bigger problem is the drawing of the phase portrait of the system in polar coordinates. I don't know how to draw it without any use of computer help, because I have to do it on my own in the exam.
I would be very grateful if someone could help me out.
The important, qualitative understanding of your differential system is that it can be expressed as:
$$\begin{cases}x'&=& \ \ \ y-ax,\\ y'&=&-x-ay \end{cases} \ \text{with} \ a=a(x,y)=x^2+y^2-1$$
i.e, along level curves:
$$a=k,\ \ k \ \text{constant}$$
which are circles with radii $\sqrt{a+1}$, the behavior of your "arrows" will be that of the linear system:
$$\binom{x'}{y'}=\begin{pmatrix}-a & 1\\-1&-a\end{pmatrix} \binom{x}{y}$$
which is a similitude matrix (with a distinction between the cases
$$a<0 \ vs. \ a>0 \ \iff \ (x,y) \ \text{ inside/outside unit circle}$$
In this way, in an exam, you can draw a thorough sketch of the phase portrait.
Fig. 1: The large view picture on the left is misleading. If you zoom on it (right picture), you will see that a spiraling effect takes place in the vicinity of the origin.