= Solution
A steady transverse <bifurcation> requires a zero <eigenvalue>, hence $\det M=0$. This gives
$$
\boxed{r_S=R^2=-\frac{|d|^2}{2Q},\qquad Q=\operatorname{Re}(\beta\overline d)<0.}
$$
The 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 $-2(d_R+\beta_Rr_S)$, is nonzero. Equivalently,
$$
\boxed{Q\ne Q_*,\qquad Q_*:=\frac{\beta_R|d|^2}{2d_R}<0.}
$$
With $d,\beta$ fixed, $\partial_r\det M=2Q\ne0$, 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 condition
$$
\boxed{Q<Q_*\quad\Longleftrightarrow\quad r_S<r_H.}
$$
For $Q_*<Q<0$, a simple steady threshold still exists, but it lies after the Hopf threshold and the pure branch is already unstable there. At $Q=Q_*$ 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 \b[phase-locked periodic solutions involving both spatial modes]: $A=a e^{i\widetilde\Omega t}$, $B=b e^{i\widetilde\Omega t}$, 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.
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