Existence of periodic solutions to non-linear system of ODEs. (Polar form) This is an example problem in my lecture notes and I simply want to check whether I'm following or not.

Consider the system \begin{align}  x'&=-y-x(x^2+y^2-3x-1)\\ 
 y'&=x-y(x^2+y^2-3x-1) \end{align}
where $r=\sqrt{x^2+y^2}$. We will find an explicit expression for the
  corresponding flow introducing polar coordinates $x=r\cos{\theta}, \
 y=r\sin{\theta}.$ 
\begin{align} rr'&=xx'+yy'=-(x^2+y^2)(x^2+y^2-3x-1)\\
 &=-r^2(r^2-3r\cos{\theta}-1) \end{align}   thus
   $r'=-r(r^2-3r\cos{\theta}-1)$. We observe that
   $(r^2-3r\cos{\theta}-1)<0$ for $r<\epsilon$ with some $\epsilon$ small
  enough and that $(r^2-3r\cos{\theta}-1)>0$ for $r>R$ for some $R>0$
  large enough. Thus solutions starting in the ring shaped domain
  $\epsilon < r < R$ do not leave it.
A more precise analysis is the following: The sign of $r'$ only
  depends on the sign of $x^2+y^2-3x-1$ having level sets as circles
  with center in the points $(3/2,0)$. Now
  $x^2+y^2-3x-1=(x-3/2)^2+y^2-13/4$. It means that the circle $C$ with
  equation $(x-3/2)^2+y^2=(\sqrt{13}/2)^2$ separates domains where $r'$
  is positive and negative. Choosing a circle with centre on the origin
  that is contained inside the circle $C$ and another that contains $C$
  gives an annulus domain that is positively invariant set for our
  equation:

To conclude about the number of equilibrium points inside this
  positively invariant set we derive an expression for the polar angle
  $\theta'$:
\begin{align}
 (\tan{\theta})'&=\frac{1}{\cos^2{\theta}}\theta'=\left(\frac{y}{x}\right)'=\frac{y'x-x'y}{x^2}\\
 &=\frac{x^2+y^2}{x^2}=\frac{r^2}{r^2\cos^2{\theta}}=\frac{1}{\cos^2{\theta}}.
 \end{align}
so $\theta' = 1$ and never zero. This implies that the system has no
  equilibrium points other than the origin, which is not in the
  positively invariant annulus. Thus by Pioncare-Benedixsons theorem the
  annulus $\epsilon < r < R$ must contain at least one periodic orbit.

I have 3 questions: 


*

*How can one conclude that $C$ separates ddomains where $r'$ is positive and negative without trying different values for $x$ and $y$?

*Now, inside the black circle $C$ we have $r'<0$ and outside of it $r'>0$. To be a positively invariant set, once trajectories enter the domain, it can never leave. But if we start at the point $(-1,0)$ then $r'>0$ so eventually we will leave the circle $r=R.$ So this annular region does not make sense to me.

*Why does $\theta'=1$ imply no equilibrium points? Is it because the angle constantly changes and for it to be an equilibrium point the angle needs to be zero at some point?

 A: I prefer to think in terms of the (equivalent) Lyapunov function
$V((x,y)) = { 1\over 2} (x^2+y^2)$.
The advantage of this function is its simplicity and geometric appeal, the disadvantage is that it does not match up exactly with the underlying dynamics in terms of
geometric interpretations. (Hence the contained and containing circles.)
Let $\phi(t) = V((x(t),y(t)))$, then we
see that
$\phi'(t) = -(x^2+y^2)((x-{3 \over 2})^2+y^2-{13 \over 4})$. In particular, with
$C=\{((x,y)| (x-{3 \over 2})^2+y^2 ={13 \over 4} \}$ we see that
$\phi'(t) \ge 0$ when $(x(t),y(t)) $ is 'inside' $C$ and $\phi'(t) \le 0$ when
'outside'.
Note that $C$ 'contains' a small circle $C_0$ centred at the origin and $\phi'(t) \ge 0$ if $(x(t),y(t)) \in C_0$.
Also, there is a large circle $C_1$ centered at the origin, that contains
$C$ and we see that $\phi'(t) \le 0$ if $(x(t),y(t)) \in C_1$.
In particular, if $A$ is the (compact) annulus 'between' $C_0,C_1$ then we see that $A$ is invariant.
To show that $A$ contains no equilibrium points, we need to show the dynamics are
not zero in $A$, since $0 \notin A$ it is sufficient to show that $\phi' \neq 0$ in $A$.
