 Solution

A Hopf crossing of the transverse stability matrix requires zero trace and strictly positive determinant. Since ${\beta }_R<0$, the trace condition gives the positive thresholdThe determinant there isThus the required coefficient condition and the rotating-frame Hopf frequency areThe trace derivative $-2{\beta }_R$ is strictly positive, so with $d,\beta$ fixed the real parts cross zero with nonzero speed. Under this condition the determinant stays positive from $r=0$ to $r_H$; the pure branch first loses linear stability through this Hopf bifurcation. When $Q<Q_{\ast }$ its first instability is instead the steady one of part (a), and $D_H<0$ prevents a Hopf pair at the trace-zero point. Equality gives the already noted double-zero degeneracy.

Generically the new branch is periodic in the symmetry-reduced rotating frame, with a small oscillating odd-mode amplitude and a modulation of the even-mode amplitude. Reconstructing the common temporal phase combines the modulation frequency with the carrier frequency. The original amplitudes therefore describe modulated periodic oscillations, generically a two-frequency quasiperiodic state. If the two frequencies are rationally related, the reconstructed motion can be periodic instead. The spatial pattern no longer has a fixed mixture of the even and odd modes: their relative amplitudes and phase oscillate. As with the steady case, nonlinear nondegeneracy is needed for an ordinary Hopf branch, and its criticality or stability cannot be inferred from the trace and determinant alone.