$$\dot x = \mu −(x^2)/2+(x^4)/4$$
a)Determine the types of bifurcations of equilibria that may occur, and their location in the $\mu$ − $x$ plane.
i tried subbing in values of $µ$ and sketching the $µ$-$x$ diagram. at the point $\mu=0$ and µ=1/4 bifurcations happen and i beleive i have drawn the graph correctly. what i want to know is how do i classify the bifurcations with 3 different bifurcation points. should i treat each one indivdually? or classify the whole thing.
Let $G(x,\mu)=\mu-\frac{x^2}2 + \frac{x^4}4$. By studying $G(x,\mu)=0$ we can easily check that
for $\mu<0$ we have two equilibrium points, $x_{1,2}=\pm\sqrt{1+\sqrt{1-4\mu}}$;
for $\mu=0$ we have three equilibrium points, $x_{1,2}=\pm 1$ and $x_3=0$;
for $0<\mu<\frac 14$ we have four equilibrium points, $x_{1,2}=\pm\sqrt{1+\sqrt{1-4\mu}}$ and $x_{3,4}=\pm\sqrt{1-\sqrt{1-4\mu}}$;
for $\mu = \frac 14$ we have $x_{1,2}=\pm 1$;
for $\mu>\frac 14$ there are no equilibrium points.
To classify the bifurcations and the stability of the equilibrium points you have to use the local theory of bifurcations. I essentially follows the local bifurcation analysis of Glendinning, Chapter 8. Start considering $(x,\mu)=(0,0)$: since $G(0,0)=G_x(0,0)=0$, $G_\mu(0,0)=1\neq 0$ and $G_{xx}(0,0)=-1<0$ we have a saddlenode bifurcation from $x=0$ when $\mu=0$, and since $G_\mu(0,0) G_{xx}(0,0)<0$ we have that the upper stationary point is stable and the lower is unstable. An analogous analysis in $(x,\mu)=\left(1,\frac 14\right)$ and in $(x,\mu) = \left(-1,\frac 14\right)$ shows that also these two points are saddlenode bifurcations and gives the stability classification.
Remark. In this case you can also plot the graph of $G(x,\mu)$ in the $(x,\dot x)$ plane and infer the stability of the equilibrium points from here.
The resulting bifurcation diagram is