Visualise the bifurcation diagram

Question

Can anyone help me visualise the bifurcation diagram that would be produced by

1. $\dot x = (x−μ)(1+μ−x^2)$

and

2. $\dot x = (μ^2−1)(μ−2)−x$

I know for 2. there is only one equilibrium point so should this give me a wave like graph?

I know for 1. I have 3 solutions so is this a pitchfork bifurcation plot?

Answer 1

First take the second dynamical system, which is simpler than the first. We have $$ \dot x = (\mu^2-1)(\mu-2) -x,\quad \mu\in \mathbb{R}. $$ For every $\mu\in \mathbb{R}$ we have the stationary point $x_\star = (\mu^2-1)(\mu-2)$, that is also (aymptotically) stable. In this case there is no bifurcation.

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Concerning $$ \dot x = (x-\mu)(1+\mu-x^2) $$ I essentially follow the local bifurcation analysis presented in Glendinning, Chapter 8. Let $G(x,\mu) = (x-\mu)(1+\mu-x^2)$. For every $\mu\in \mathbb{R}$ we have the stationary point $x_\star = \mu$, whereas the quadratic part gives $x_{\pm} = \pm \sqrt{1+\mu}$ for $\mu\geq -1$. Note that for $\mu=-1$ we have a single stationary point $x=0$.

To undesrtand the local bifurcation, start considering $(x,\mu)=(0,-1)$: since $G(0,-1)=G_x(0,-1)=0$, $G_\mu(0,-1)=1\neq 0$ and $G_{xx}(0,-1)=-2<0$ we have a saddlenode bifurcation from $x=0$ when $\mu=-1$, and since $G_\mu(0,-1) G_{xx}(0,-1)<0$ we have that the upper stationary point is stable and the lower is unstable.

Now note that if $\mu=\frac{1-\sqrt{5}}{2}=: \mu_-$ then $x_-=\mu_-$, and that if $\mu=\frac{1+\sqrt{5}}{2}=: \mu_+$ then $x_+=\mu_+$. In these two cases one of the two stationary points given by the quadratic part coincides with $x_\star$, thus we have two other bifurcations to study.

• If $(x,\mu)=(x_+,\mu_+)$ then $G(x_+,\mu_+)=G_x(x_+,\mu_+)=G_\mu(x_+,\mu_+)=0$, $G_{xx}(x_+,\mu_+)=-4x_->0$ and $G_{\mu x}(x_+,\mu_+)^2-G_{xx}(x_+,\mu_+)G_{\mu\mu}(x_+,\mu_+)>0$, thus we have a transcritical bifurcation for $\mu=\mu_-$. Furthermore, since $G_{xx}(x_+,\mu_+)=(2x_--1)^2>0$ we have that the upper stationary point is unstable and the lower one is stable.

• An analogous computation shows that there is a transcritical bifurcation for $\mu=\mu_+$, with the upper stationary points that is stable and the lower one that is unstable.

Remark. You can also plot the graph of $G(x,\mu)$ in the (,˙) plane and infer the stability of the equilibrium points from here.

The resulting bifurcation diagram turns out to be

.