 Solution

A steady transverse bifurcation requires a zero eigenvalue, hence $\det M=0$. This givesThe inequality is required for a positive amplitude. Since $d_R>0$, the numerator cannot vanish. The zero eigenvalue is simple provided the other eigenvalue, the trace $-2(d_R+{\beta }_R{r}_S)$, is nonzero. Equivalently,With $d,\beta$ fixed, ${\partial }_r\det M=2Q\ne 0$, so the simple zero eigenvalue crosses transversely. If this is to be the first instability of the stable pure branch, its other eigenvalue must still be negative. Comparing with $r_H=-d_R/{\beta }_R$ gives the stronger conditionFor $Q_{\ast }<Q<0$, a simple steady threshold still exists, but it lies after the Hopf threshold and the pure branch is already unstable there. At $Q=Q_{\ast }$ both transverse eigenvalues vanish, so the event is a double-zero degeneracy rather than an ordinary simple steady bifurcation.

The reduced system has the reflection symmetry $C\mapsto -C$. At a generic simple steady threshold, the new mixed states therefore occur in a reflection-related pair, with small nonzero $B$ and a corresponding correction to the $A$ magnitude and common frequency. In the original variables they are phase-locked periodic solutions involving both spatial modes: $A=a{e}^{i\tilde{\Omega }t}$, $B=b{e}^{i\tilde{\Omega }t}$, with constant nonzero complex $a,b$. The odd spatial component breaks the reflection symmetry of the pure even mode; the two branches are exchanged by $b\mapsto -b$. Determining whether the branches are supercritical or subcritical, and their stability, requires the reduced nonlinear coefficient and is not determined by this linear threshold alone.