I am a bit confused about how the Bifurcation Diagram of a parametric autonomous system $x'=f(x,μ)$ is defined.

For the one dimentional case, I think it is more obvious to me, but still not clear enough: For example, if

$$x'=μ-x^2$$ then the equilibria are:

For $μ=0$ is only the $0$ which is unstable

For $ 0\ltμ$ there are two equilibria $\sqrt{μ}, - \sqrt{μ} $ with the first one unstable and the second stable.

For $ μ\lt0$ there are no equilibria.

Now, should the bifurcation diagram be the graph of the functions? $x=0$, $x=\sqrt{μ}$ , $x= - \sqrt{μ} $ ? dotted where the $μ$ gives unstable equilibria? In my book I have a diagram like the following:enter image description here

Furthermore, what the bifurcation diagram should look like if the system $x'=f(x,μ)$ is planar? My confusion here is what the axis $x$ then should represent... Thanks.

  • $\begingroup$ No I am still looking for answer please. $\endgroup$
    – dmtri
    Commented Oct 22, 2018 at 4:32
  • 1
    $\begingroup$ To answer the second part of your question, if the system is planar, we typically still make a diagram like above, but instead of $x$, we may instead use $r=\sqrt{x^2+y^2}$ or some similar measure. Typically bifurcation diagrams only track $\mu$ and some scalar that characterizes something about our system, like a radius, for instance. $\endgroup$
    – whpowell96
    Commented Oct 22, 2018 at 18:57

1 Answer 1


Consider the equation $$x' = \mu - x^2$$

  • If $\mu > 0$, there are two equilibria: $x^* = \pm \sqrt{\mu}$.
  • The derivative of the right hand side is $Df(x, µ) = −2x$.
  • Evaluating at the fixed points we obtain the following: $Df(\sqrt{\mu},\mu) = -2\sqrt{\mu} \lt 0$, which implies that the equilibrium $x^* = \sqrt{\mu}$ is stable.
  • For $Df(−\sqrt{\mu},\mu) = 2\sqrt{\mu} \gt 0$, which means that the equilibrium $x^* = −\sqrt{\mu}$ is unstable. We can see these visually by plotting $x$ vs. $x'$ as

enter image description here

  • If $\mu \lt 0$, there are no equilibria.
  • When $\mu = 0$, the system has only one quilibrium point, $x^* = 0$. In this case, the equilibrium point is nonhyperbolic, since $Df(0, 0) = 0$, and we cannot use linearization to analyze its stability. A phase portrait, however, can help us in this case. We can see this visually by plotting $x$ vs. $x'$ as

enter image description here

  • We conclude that the equilibrium point $x^∗ = 0$ is an unstable saddle node.
  • This system has a saddle-node bifurcation at $\mu = 0$.
  • As an alternate approach, we could have also drawn a phase portrait movie or a phase line to study these stability and bifurcation results.

As for the bifurcation diagram, we can choose various approaches. For example, we can plot $x$ vs. $\mu$ and analyze the stable or unstable branch or we can do a contour plot of $x^*$ vs. $u$ based on the analysis above and arrive at

enter image description here


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