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For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot{y}-y\dot{x})/r^2$. The polar coordinates satisfyHence the rotation is clockwise. The origin is an unstable equilibrium, while $r=1$ is a limit cycle of period $2\pi$. Writing $u=r^2$ gives the logistic differential equation $\dot{u}=2u(1-u)$. For $r(t_0)=r_0>0$,All nonzero solutions tend to the unit-circle attracting limit cycle as $t\to \infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to -\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes atThis finite-time blow-up of an ordinary differential equation occurs backward in time. Such exterior solutions have no $t\to -\infty$ behavior: their maximal time interval is $(t_{\ast },\infty )$. The zero solution exists for all time and has no defined polar angle.

The planar system $\dot{x}=y+(1-x^2-y^2)x$, $\dot{y}=-x+(1-x^2-y^2)y$ has polar coordinates $\dot{r}=r(1-r^2)$, $\dot{\theta }=-1$. Every nonzero orbit approaches the clockwise unit-circle limit cycle, of period $2\pi$. For $r(t_0)=r_0>0$,Interior orbits approach the origin backward in time, but exterior orbits have backward finite-time blow-up of an ordinary differential equation at $t=t_0+\frac{1}{2}\log (1-r_0^{-2})$. Thus forward convergence to a limit cycle does not imply existence for all negative times.