# Bifurcation Diagram-confusion

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:

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.

• No I am still looking for answer please. Commented Oct 22, 2018 at 4:32
• 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. Commented Oct 22, 2018 at 18:57

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

• 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

• 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