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To determine the nonlinear pitchfork bifurcation, solve the four slaved steady equations exactly. SetThe nonzero equilibrium points satisfyThusIf $k\le 1$, the coefficient of $u$ is positive for every physical $q\ge 0$: there is no change of pitchfork direction at a nonnegative rotation parameter. If $k>1$, the change occurs atBelow $q_D$ the nonzero branch lies on $r>r_P$; above $q_D$ it initially lies on $r<r_P$. At $q_D$, $r-r_D=k{u}^2+O(u^3)$, 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 $q>q_D$,andThe 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 $a^{\prime }=\chi a[\Delta -[1-q(k-1)]a^2-k{a}^4+\cdots ]$, 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.