at which Hopf bifurcations occur. We impose the fixed point conditions $\phi' = 0$, $\rho' = 0$ as before, but now require that the Jacobian has zero trace and positive determinant—the latter two conditions are equivalent to requiring that the eigenvalues be pure imaginary.

Solving simultaneously for $\phi' = 0$, $\rho' = 0$ and trace = 0, we find $$\begin{aligned}
\label{zerotrace_conditions}
\sin \phi
   &=& -\frac{\left(K^2-4\right)
   \Omega }{F \sqrt{K-2} K
   \sqrt{K+2}}\\
\cos \phi &=& -\frac{(K-2)^{3/2}}{2
   F \sqrt{K+2}}\\
\rho &=&\sqrt{\frac{K-2}{K+2}}.\end{aligned}$$

Because $\rho$ depends only on $K$, we can go a bit further than we did in the saddle-node case. Using $\sin^2 \phi+ \cos^2 \phi = 1$ as before, $F$ can now be obtained *explicitly* in terms of $K$ and $\Omega$: $$\label{HopfcurvesK}
F_{\rm \,Hopf}=\frac{1}{2K} \sqrt{\frac{(K-2)
   \left(K^4-4 K^3+4 \left(\Omega
   ^2+1\right) K^2+16 \Omega ^2
   K+16 \Omega ^2\right)}{K+2}}$$ For the special case $K=5$ studied by Antonsen et al. \[2008\], Eq. becomes $$F_{\rm \,Hopf} = \frac{1}{10} \sqrt{\frac{3}{7}} \sqrt{225+ 196 \Omega^2}$$ Figure 1 plots the graph of $F_{\rm \,Hopf}(\Omega)$. Notice how straight it is, and that it nearly lines up with the lower branch of the saddle-node curve.

Takens-Bogdanov point
---------------------

As mentioned above, for Eq. to truly signify a Hopf bifurcation, the Jacobian determinant must be positive 