Can someone explain to me what Bifurcation Points and Bifurcation Values are? I'm looking over a problem my professor gave out regarding the Supercritical Pitchfork Bifurcation: $\frac{dx}{dt} = \mu x - x^3, \mu \in \Bbb R, x \in \mathbb R$
Let $f(x,\mu) = \mu x - x^3$. She said by definition $\overline x$ is a bifurcation point and $\overline \mu$ is a bifurcation value if $f(\overline x, \overline \mu) = 0 \text{ and } f_x(\overline x, \overline \mu) = 0$
So I understand that $f(\overline x, \mu) = 0 \text{ if } \overline x =0, \pm \sqrt{\mu} \text{, and that } f_x(x,\mu) = \mu - 3x^2$. This means that:
$f_x(\overline x, \mu) = \begin{cases} -2\mu; & \overline x = -\sqrt{\mu} \\ \mu; & \overline x = 0 \\ -2 \mu; & \overline x = \sqrt{\mu} \end{cases}$
But then she writes that bifurcation occurs at the point $(\overline x, \overline \mu) = (0,0)$. Where does this come from? Why is $\overline x = 0$ the only bifurcation point when $\pm \sqrt{\mu}$ were also equilibrium points?
First of all, $f_x(0,\mu)=0$ if and only if $\mu=0$, and $f_x(\pm\sqrt{\mu},\mu)=-2\sqrt{\mu}\neq 0$ for $\mu>0$. Thus by definition, $(\bar{x},\bar{\mu})=(0,0)$ is a bifurcation point. Second, note, for small $x$ $$ \dot{x}=\mu x-x^3\approx \mu x. $$ The solution of $\dot{x}=\mu x$ is $x(t)=Ce^{\mu t}$, and hence the behavior of $\dot{x}=\mu x$ depends on the sign of $\mu$, i.e., if $\mu>0$, $x(t)$ is increasing and if $\mu<0$, $x(t)$ is decreasing (assume $C>0$). Thus $(\bar{x},\bar{\mu})=(0,0)$ is a bifurcation point of $\dot{x}=\mu x-x^3$.