A stable planar cycle can have a singular phase slice

The identities $\dot{r}=(x{x}^{\mathrm{\prime }}+y{y}^{\mathrm{\prime }})\mathrm{/}r$ and $\dot{\vartheta }=(x{y}^{\mathrm{\prime }}-y{x}^{\mathrm{\prime }})\mathrm{/}{r}^2$ give $\dot{r}=r(1-r)$ and $\dot{\vartheta }=1$. Thus $r=1$ is invariant and $\vartheta (t)=\vartheta (0)+t$, so the least period is $2\pi$. Differentiating $r(1-r)$ at $r=1$ gives the radial variational equation $\dot{\rho }=-\rho$, whose multiplier is $e^{-2\pi }<1$; the angular variational equation is $\dot{\alpha }=0$, whose multiplier is $1$. Because the polar-coordinate Jacobian at times $0$ and $2\pi$ is the same invertible matrix, the Cartesian monodromy is conjugate to $\mathrm{diag}(e^{-2\pi },1)$. For $r<1$ the radial derivative is positive, and for $r>1$ it is negative, so each nearby radius moves monotonically toward $1$ and remains between its initial value and $1$. It has a limit $\bar{r}$; if $\bar{r}(1-\bar{r})\mathrm{\ne }0$, then $\dot{r}=r(1-r)$ is eventually bounded away from zero, contradicting convergence. Hence $\bar{r}$ is an equilibrium, and the interval between the initial radius and $1$ contains only the equilibrium $1$, so $r(t)\to 1$. These bounds also give Lyapunov stability of the unit circle. Thus the orbit is orbitally asymptotically stable and its neutral multiplier is simple.

At $(1,0)$ the vector field is $(0,1)$. For ${\psi }_b=x-1$, $c_b=(1,0)$ and $c_b^{\mathrm{\top }}f(x_{\ast })=0$, so [F2] gives a singular shooting Jacobian. For ${\psi }_g=y$, $c_g=(0,1)$ and $c_g^{\mathrm{\top }}f(x_{\ast })=1$; the phase multiplier is simple by step 1.1, so [F2] gives an invertible Jacobian.