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So in dynamical system class, I ran into this equation.

$\dot{x} = x [r -(1 - x^2)] [r - (2x^3 - 2x)]$

How can one possibly sketch the bifurcation diagram and locate and identify all bifurcations in this family?

I know if I solve $0 = x [r -(1 - x^2)] [r - (2x^3 - 2x)]$ I can find some of the fixed (equilibrium) points for this family. There is one at $x = 0$, $x = \pm\sqrt{1-r}$, and there should be a few more. But I don't even know how to start to sketch a bifurcation diagram for this family! Please help!

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  • $\begingroup$ Your fixed points seem to be wrong. $x=0$ $x=\pm\sqrt{1-r}$ (for $0<r<1$) and then you have to solve $2x^3-2x=r$. Are you sure you copied the right question? $\endgroup$
    – MrYouMath
    Commented Oct 25, 2017 at 10:40
  • $\begingroup$ Oh I just solved for the fixed points wrong that's why. It is the right problem. $\endgroup$
    – Itsnhantransitive
    Commented Oct 25, 2017 at 14:58
  • $\begingroup$ The solution to $2x^3-2x=r$ is not so nice that is why I asked if you copied the correct equation. $\endgroup$
    – MrYouMath
    Commented Oct 25, 2017 at 15:00
  • $\begingroup$ Yeah I tried to graph all three parts of the equation together on a graph of $\dot{x}$ as a function of $x$ then do the phase line for all three parts on the same graph to see whether the fixed points are stable or unstable or semistable.... and it led to something really messy... $\endgroup$
    – Itsnhantransitive
    Commented Oct 25, 2017 at 15:03

1 Answer 1

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Not sure if this is what you are looking for but this is a plot of the equilibrium solutions $x$ as functions of $r$. We can see from the graph that the number of equilibria changes moving from left to right from 4 to 5 to 6 to 3(on the $x$ axis) to 6 to 5 to 4 and then finally 2 for $r\geq1$. The Mathematica code I used is:

sol1 = Solve[x^2 + r - 1 == 0, x];

sol2 = Solve[2 x^3 - 2 x - r == 0, x];

Plot[{0, sol1[[1, 1, 2]], sol1[[2, 1, 2]], sol2[[1, 1, 2]], sol2[[2, 1, 2]], sol2[[3, 1, 2]]}, {r, -5, 5}]

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  • $\begingroup$ Oh man this looks right! This looks like it's from mathematica! If you please give me the code that would be awesome! $\endgroup$
    – Itsnhantransitive
    Commented Oct 25, 2017 at 3:35
  • $\begingroup$ Nevermind I figured it out! Woot you are awesome! Thank you so much @JEM $\endgroup$
    – Itsnhantransitive
    Commented Oct 25, 2017 at 3:45

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