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For the system:

$$ \dot x = r+4x^2 $$

find the value of $r$ at which the bifurcation occurs and classify it.

As far as I'm aware, you can use

$$ f(x^*)=0; f'(x^*)=0 $$

and then solve for $r$ where $x^*$ is an equillibrium point.

However this doesn't really make sense to me. I know that a bifurcation happens when an equillibrium point is created or destroyed but I don't know how to work out the bifurcation values.

Thanks

juper
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1 Answers1

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The equilibria for this differential equation are the values of $x$ where the right side $r + 4 x^2 = 0$, namely $x = \pm \sqrt{-r/4}$ if $r \le 0$, but none for $r > 0$. So the bifurcation happens at $r=0$, and it's a saddle-node bifurcation.

Robert Israel
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  • ah yes, thank you. So the bifurcation at $r=0$ is because there is none for $r>0$. If for another question there were no equllibrium points for values say $r>1$ then the bifurcation point would be $r=1$? – juper Jun 09 '17 at 17:41
  • Yes, if there are no equilibrium points for $r > 1$ but there are for $r < 1$, $r=1$ would be a bifurcation. – Robert Israel Jun 09 '17 at 19:51
  • that's perfect. thanks for y our help! – juper Jun 09 '17 at 19:53
  • One last question, how did you know it was a saddle-node bifurcation? – juper Jun 10 '17 at 10:15
  • or how would you know a system was a supercritical pitchfork? Is it upon observation of the bifurcation diagrams? – juper Jun 10 '17 at 10:24
  • In a saddle-node bifurcation, two equilibria collide and disappear. In a pitchfork bifurcation, one equilibrium becomes three. Which of these looks like this system? – Robert Israel Jun 11 '17 at 19:24