 Solution

Write $F(x)={\mu }^2-(x^2-a{)}^2$. Its fixed points satisfy $x^2=a\pm |\mu |$, retaining only nonnegative right sides. At a simple fixed point,Thus on the outer branches $x^2=a+|\mu |$, the positive point is stable and the negative point unstable. On the inner branches $x^2=a-|\mu |$, the positive point is unstable and the negative point stable.

For $a>0$, there are four simple points for $0<|\mu |<a$, two for $|\mu |>a$, and a saddle-node at each $(\mu ,x)=(\pm a,0)$, where the two inner points coalesce. At $\mu =0$, there are also bifurcations at $x=\pm \sqrt{a}$. The smooth branches $x=\pm \sqrt{a+\mu }$ and $x=\pm \sqrt{a-\mu }$ cross and exchange stability. Locally $F={\mu }^2-4a{\xi }^2+\text{higher terms}$, where $\xi =x-x_0$. After a parameter-dependent shift this is a transcritical bifurcation. It is not an ordinary saddle-node: two nearby branches exist on both sides of zero. At zero itself each crossing point is semistable, attracting from the right and repelling from the left.

For $a<0$, the only possible branch is $x=\pm \sqrt{a+|\mu |}$, existing for $|\mu |\ge |a|$. Each point $(\pm |a|,0)$ is a saddle-node bifurcation. There are no fixed points for $|\mu |<|a|$; outside that interval the positive point is stable and the negative point unstable. At either threshold $F\sim 2a{x}^2<0$, so the zero point is semistable. The first row of the diagram displays these branches; solid curves are stable and dashed curves unstable.