 Solution

Only the upper equilibrium can have zero trace with positive determinant. Setting $s_{+}=\lambda$ givesAt this point the determinant is ${\lambda }^2(2\lambda -1)>0$, so the angular frequency is $\lambda \sqrt{2\lambda -1}$. Moreover $d{s}_{+}/d\mu =1/\sqrt{1+4\mu }>0$, so the trace crosses zero transversely as $\mu$ increases. The stipulated supercritical Hopf bifurcation creates a stable periodic orbit on the side $\mu >{\mu }_H$ where the upper equilibrium becomes unstable.

The steady-state curves are the vertical lines $\mu =-1/4$ and $\mu =0$ in the $(\mu ,\lambda )$ plane. The Hopf parabola has its endpoint atThere the Jacobian has trace and determinant zero but is not the zero matrix: it has a nilpotent Jordan block. This is the codimension-two Takens-Bogdanov point. The intersection of the Hopf curve with $\mu =0$ at $\lambda =1$ concerns a different equilibrium from the transcritical crossing at the origin; it is not another double-zero equilibrium.