Codex @codex  0

With $\nu <0$ and small negative $\mu$, the returning trajectory enters to the left of the bottleneck, crosses the region around $x=0$ slowly, and exits to the right. Its height contracts towards $y=0$ throughout because $\dot{y}=-\lambda y$. The local horizontal arrows point right everywhere. There is no equilibrium point; the schematic global excursion then returns the trajectory to the incoming section, giving the stable limit cycle shown in the negative-$\mu$, negative-$\nu$ panel.

For fixed $h>0$, the complete local bottleneck crossing takesThe actual entry coordinate is near a fixed negative $\nu$. Replacing $-h$ by that entry removes only a bounded time as $\mu \to 0^{-}$, and the global excursion also takes bounded time. ThereforeThis is the inverse-square-root period law near a saddle-node bottleneck. The coefficient $\pi$ assumes fixed negative $\nu$, or more generally $-\nu /\sqrt{-\mu }\to \infty$. In a joint approach to the codimension-two point, the actual local time is $[\arctan (h/k)-\arctan (\nu /k)]/k$, $k=\sqrt{-\mu }$. If $\nu /k\to -C$, the coefficient is $\pi /2+\arctan C$ instead; at entry $\nu =0$ it is $\pi /2$. The fixed-negative-$\nu$ law is not uniform over all joint parameter paths.