 Solution

At the no-convection equilibrium $(a,b,c)=(0,0,0)$, one eigenvalue is $-\zeta$ and the other two come fromTheir trace and determinant areA zero eigenvalue occurs at $r=1+q$. If $q<\zeta /\sigma$, the other two eigenvalues are negative, and the stationary bifurcation is the pitchfork analyzed in part (c). For $q>\zeta /\sigma$ the transverse eigenvalue is positive instead.

An imaginary pair occurs at $r=1+\zeta /\sigma$, provided $q>\zeta /\sigma$, becauseThe trace crosses zero with derivative $\sigma >0$. The cubic coupling is nondegenerate here: at criticality, putting $K=\sigma \zeta q$ reduces the oscillatory variables toFor $a=R\cos ({\omega }_H\tau )$, the forced second harmonic of $c$ has sine coefficient $9R^2\zeta {\omega }_H/[2K({\zeta }^2+4{\omega }_H^2)]$. Averaging the derivative of $(a^{\prime 2}+{\omega }_H^2{a}^2)/2$ consequently gives $-9\zeta {\omega }_H^2{R}^4/[8({\zeta }^2+4{\omega }_H^2)]<0$. Thus the Hopf bifurcation is nondegenerate and supercritical.

The two thresholds meet atHere the planar block is nonzero but has trace and determinant zero, while the third eigenvalue remains negative. It is a reflection-symmetric double-zero, codimension-two point; the ordinary simple-eigenvalue reduction of part (c) is not valid there.