A coarse Euler variational step gives a spurious unstable multiplier

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 $\dot{\vartheta }=\omega$. Hence $r=1$ is a periodic orbit with least period $2\pi \mathrm{/}\omega =3$. Differentiating $r(1-r)$ at $r=1$ gives $\dot{\eta }=-\eta$, whose exact solution is $\eta (3)=e^{-3}\eta (0)$; this transverse multiplier lies strictly between zero and one.

By the forward-Euler update specified in the Statement, with $a=-1$ and $h=3$ we have ${\eta }_1=(1-3){\eta }_0=-2{\eta }_0$. Thus its one-step amplification has modulus $2>1$, despite the exact transverse multiplier being stable. Since this calculation provides no enclosure of the truncation defect or exact variational flow, it cannot serve as a certified multiplier claim under [F1].