# Half-stable fixed point on a circle

On a line graph, it's clear that a half-stable fixed point is the limit of moving the unstable fixed point towards the stable fixed point. Some solutions go to infinity depending on the initial condition, and others asymptotically approach the fixed point.

I am having some trouble making an analogous statement on a circle graph. In the two fixed point case, some solutions will move in negative radians while others will move in positive radians towards the fixed point, like before. But unstable solutions aren't really that unstable, because they do not go to infinity. As you vary a parameter to produce the saddle node bifurcation, all of the sudden all solutions move in the same radial direction towards the fixed point.

It's clear to me that there is a distinction between "unstable" and "stable" trajectories in a line graph. I'm wondering how to interpret the difference in behavior of the saddle node bifurcation when viewed on the circle rather than the line. Apologies if this question is not clear enough.

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Stability in the local sense means that there is a neighborhood of equilibrium in which solutions are attracted to the equilibrium. This notion makes just as much sense on the circle as on the line. // Even on the line, instability does not necessarily mean that solutions escape to infinity: they may well be drawn to another equilibrium. – user53153 Jan 29 '13 at 5:09
@5PM Nevermind, I see what you're saying. – tacos_tacos_tacos Jan 29 '13 at 5:21