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The circle $\gamma :r=1$ is a periodic orbit, since $\dot{r}=0$ there and $\dot{\theta }=1$. Its period is $2\pi$. Linearizing in $(\delta r,\delta \theta )$ along it givesThe monodromy matrix is triangular with diagonal entriesThe unit Floquet multiplier comes from translating an autonomous orbit in time: its tangent vector is a periodic solution of the variational equation, independently of $g$. A sufficient condition for asymptotic stability is ${\int }_0^{2\pi }g<2\pi$, making the normal multiplier less than one. The reverse strict inequality gives instability; equality requires nonlinear analysis.

For $g=2\sin \theta$ the normal multiplier is $e^{-2\pi }$, so the circle is asymptotically stable. That does not mean it attracts every initial point. Since $r>0$ makes $\theta$ strictly increasing, divide the radial equation by $\dot{\theta }$:For $r>1$, put $v=r-1>0$. Then $v^{\prime }=(2\sin \theta -1)v-v^2\le (2\sin \theta -1)v$, whose integrated linear coefficient decreases by $2\pi$ per revolution. Thus every exterior orbit approaches $r=1$.

For $0<r<1$ put $w=(1-r{)}^{-1}>1$. It satisfies the linear equationLet $w_c$ be its solution with $w_c(\pi )=1$. It defines the separatrix $r_c=1-1/w_c$ wherever $w_c>1$. At $\theta =\pi$, $r_c=0$, $r_c^{\prime }=0$, and differentiation gives $r_c^{\prime \prime }=2$, so $r_c\sim (\theta -\pi {)}^2$. The approach to this endpoint takes infinite physical time because $dt=d\theta /r_c$.

To locate its other endpoint, use the integrating factor $e^{-A(\theta )}$ with $A(\theta )=\theta +2\cos \theta$. It givesIn particular $w_c(0)>e^{4-\pi }>1$, whereasOn $(-\pi ,0)$, crossing $w=1$ has $w^{\prime }=-2\sin \theta >0$, so there is one crossing ${\theta }_b\in (-\pi ,0)$. There cannot be another crossing before $\pi$: on $(0,\pi )$ crossings have negative derivative, while the local quadratic form at $\pi$ is positive on its left. Thus $r_c>0$ for ${\theta }_b<\theta <\pi$, and its closure is a homoclinic orbit to the origin. At ${\theta }_b$, $r_c^{\prime }=-2\sin {\theta }_b>0$, so departure from the origin likewise takes infinite time.

Scalar-solution comparison now gives the basin picture. Initial points inside that loop have $w<w_c$ at the same angle and approach $r=0$ before reaching angle $\pi$. Points on its boundary return to the origin along the homoclinic trajectory. Points outside the loop reach angle $\pi$ with $w>1$ or lie outside the unit circle; they approach $r=1$. To check the inner exterior region quantitatively, the return map at $\theta =\pi$ is $w\mapsto e^{2\pi }(w-I)$, where $I={\int }_0^{2\pi }{e}^{-u-2+2\cos u}du<1-e^{-2\pi }$. It sends every $w>1$ increasingly to infinity, equivalently $r\to 1$.

There are no other nonconstant periodic orbits: an orbit reaching the section with $w>1$ has strictly increasing successive $w$ values, and the remaining inner trajectories tend to the origin. The origin, defined by the continuous Cartesian extension, attracts the interior of the loop but is not Lyapunov stable, since arbitrarily small lower-sector initial points outside the loop approach the unit circle. This is the homoclinic basin boundary in a sinusoidal radial flow.