Application of Poincaré-Bendixon Theorem 

Consider the planar system in polar corrdinates:
    $$
\frac{dr}{dt}=r-r^3+r^2\sin\phi,\qquad\frac{d\phi}{dt}=1+\frac{1}{2}r\cos\phi.
$$
    Show that it has at least one periodic orbit.


As far as I know, the theorem of Poincaré-Bendixon says that if the $\omega$-limit set $\Omega(r_0,\phi_0)$ of some point $(r_0,\phi_0)$ is finite and does not contain a critical point, then $\Omega(r_0,\phi_0)$ is a periodic orbit.
The solution says that the $\omega$-limit set of any non-zero point must be finite and must lie at non-zero $r$ since for small $r>0$, we have $\dot{r}>0$ and for all large $r>0$, we have $\dot{r}<0$. I do not understand this. Why does this follow?
Moreover, it says that there can't be an equilibrium at $r\neq 0$: If $\dot{\phi}=0$, then, it is said, we have $r\geq 2$ and hence $\dot{r}\leq -2$. I cannot see this, since if $\dot{\phi}=0$ we have $r=-\frac{2}{\cos\phi}$. Why is this $\geq 2$?
 A: It follows from the fact that ${dr\over{dt}}= r-r^3+r^2sin(\phi)$, you can write ${dr\over{dt}} =r(1-r^2+rsin(\phi))$, so there exists an interval $[0,u]$ on which $dr/dt>0$ since $lim_{r\rightarrow 0}1-r^2+rsin(\phi)=1$. This implies that if the norm $r(t)$ of an orbit is contained in $(0,u]$ it increases, so the $\omega$-limit of a non zero point can't be zero.
on the other hand, you can also write
  ${dr\over{dt}} =r^3(1/r^2-1+(sin(\phi))/r)$ this shows that if the length of $r(t)$ is bigger than a given constant $M$, it decreases since $lim_{r\rightarrow +\infty}1/r^2-1+(sin(\phi))/r=-1$, so the $\omega$-limit can't  go to the infinity.
The $\omega$-limit has finitely fixed points since they are finitely many fixed points. A fixed point is characterised by $dr/dt=0$, and $d\phi/dt=0$. If you write $dr/dt=0$, you can express $sin(\phi)$ has a rational function of $r$, you obtain $cos(\phi)$ has a rational function of $r$ if you write $d\phi/dt=0$. By writing $cos^2\phi+sin^2\phi=1$, you obtain a polynomial equation $P(r)$ which has only finitely many solutions this implies also that $sin\phi$ and $cos\phi$ are elements of a finite set (they are obtain from the solutions of $P(r)=0$ with $dr/dt=0, d\phi/dt=0$). this implies that the point such that $dr/dt=0$ and $d\phi/dt=0$ are finite.
