= Solution
To determine the nonlinear <pitchfork bifurcation>, solve the four slaved steady equations exactly. Set
$$
k=\frac{h}{(4-h)\sigma^2}=\frac{\pi^2}{\alpha^2\sigma^2},\qquad u=a^2.
$$
The nonzero <equilibrium points> satisfy
$$
b=\frac{a}{1+u},\quad c=\frac{u}{1+u},\quad d=\frac{a}{\sigma(1+ku)},\quad e=\frac{ku}{1+ku},\qquad
r=(1+u)\left(1+\frac{q}{1+ku}\right).
$$
Thus
$$
r-(1+q)=[1-q(k-1)]u+qk(k-1)u^2+O(u^3).
$$
If $k\le1$, the coefficient of $u$ is positive for every physical $q\ge0$: there is \b[no change of pitchfork direction at a nonnegative rotation parameter]. If $k>1$, the change occurs at
$$
\boxed{q_D=r_\Omega^2=\frac1{k-1}=\frac{(4-h)\sigma^2}{h-(4-h)\sigma^2},\qquad r_D=1+q_D.}
$$
Below $q_D$ the nonzero branch lies on $r>r_P$; above $q_D$ it initially lies on $r<r_P$. At $q_D$, $r-r_D=ku^2+O(u^3)$, so $|a|$ grows like $(r-r_D)^{1/4}$ rather than the ordinary square-root scaling.
The <rational equilibrium curve at a degenerate pitchfork> also gives the nearby <saddle-node bifurcations> of the two reflection-related nonzero branches. For $q>q_D$,
$$
\boxed{u_{SN}=\frac{\sqrt{q(k-1)}-1}{k},\qquad
r_{SN}=\frac{k-1+q+2\sqrt{q(k-1)}}{k},}
$$
and
$$
r_P-r_{SN}=\frac{(\sqrt{q(k-1)}-1)^2}{k}.
$$
The two <saddle-node bifurcations> occur on one parameter curve by reflection symmetry; its separation from the <pitchfork bifurcation> curve is quadratic in $q-q_D$.
The stability statement needs a further condition. At $r_P$, the other coupled <eigenvalues> have polynomial $s^2+(1+2\sigma)s+\sigma[(1+\sigma)-(1-\sigma)q]$. They are stable only for $q<q_{TB}$. If $q_D<q_{TB}$, a one-dimensional <centre manifold> gives the local form $a'=\chi a[\Delta-[1-q(k-1)]a^2-ka^4+\cdots]$, where $\Delta=r-r_P$ and $\chi=\sigma/[(1+\sigma)-(1-\sigma)q_D]>0$. The quintic term stabilizes the large branch. The interval $r_{SN}<r<r_P$ then has a stable conductive state and two stable finite-amplitude states, with the two small unstable branches separating their <basins of attraction>. This is the usual subcritical convection <hysteresis>. If $q_D>q_{TB}$, the branch still reverses its direction, but the trivial state already has an additional unstable mode; this equilibrium-curve calculation alone does not imply a stable bistable wedge. At equality a two-dimensional <centre manifold> is essential.
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