 Solution

The divergence of the Cartesian vector field isOn a periodic orbit with $r^2=s$ andthis becomes the constantSince the period is $2\pi$, the divergence integral isIt is positive on the inner branch $s_{-}$, negative on the outer branch $s_{+}$, and zero where the branches meet at $s=\lambda /2$. The divergence test therefore classifies the inner orbit as unstable, the outer orbit as stable, and the double orbit as nonhyperbolic, exactly as the radial phase line does.

There is also no conflict with the exclusion form of the divergence criterion. For a circular orbit of radius $R$,so the divergence necessarily changes sign or vanishes somewhere in its interior; it cannot have the strict one-sign property that would exclude the orbit.