I'm looking at this following equation

$\dot{x} = \frac{dx}{dt} = \frac{x^2}{x^2 + 1} - rx$,

and trying to classify the bifurcations that appear when the parameter $r$ is varied. I'm pretty much following the terminology of Strogatz. Questions:

1) For $r = 0$, there's an unstable fixed point at $x = 0$, but this becomes stable as r is increased. Am I right in saying that this is a transcritical bifurcation, ie. one with the normal form $\dot{x} = rx - x^2$?

2) For $0 < r < 1/2$ there are two fixed points - one stable, one unstable - that come closer together as $r$ increases and meet at $x = 1$ when $r = 1/2$. Am I right in saying this is a saddle-node bifurcation, ie. normal form $\dot{x} = r - x^2$?

I would like to show the two above statements, but I can't quite get it to work. I'm thinking something along the lines of a Taylor expansion around $x = 0$ and $x = 1$.

Cheers! \T

EDIT: OK, the first question is actually pretty straight forward when you look at the Taylor expansion...

The second one is trickier though. Around $x = 1$, my Taylor expansion of $\dot{x}$ looks like this:

$\dot{x} \approx 1/2 + \frac{x-1}{2} - 1/4(x-1)^2 -r(x-1) - r$.

Does this resemble the normal form $\dot{x} = r - x^2$ ? (Signs and factors don't matter).


1 Answer 1


When a stable and an unstable fixed point collide and annihilate, that's not a pitchfork, it's a saddle-node bifurcation. A pitchfork has three fixed points coming together on one side of the critical $r$ and one on the other side.

EDIT: If $r = s + 1/2$ and $x = y - 2 s + 1 $, your Taylor expansion becomes $$ \dot{y} = \dot{x} = - y^2/4 - s + s^2$$ A further transformation can make $-s + s^2$ into $-s$ near $0$.

  • $\begingroup$ That's what I meant, edited! $\endgroup$
    – trolle3000
    Jul 15, 2012 at 22:00

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