 Solution

Fix $c>0$. Stationary points satisfy $m(a+b{m}^2+c{m}^4)=0$. With $x=m^2$, the nonzero candidates areKeep only real positive roots. At such a point,so $x_{+}$ gives local minima and $x_{-}$ gives local maxima when both exist. The solution $m=0$ is a local minimum for $a>0$. To choose the phase, compare free energies: using the stationary point relation $a=-bx-c{x}^2$ givesFor $b>0$, this expression is negative whenever a nonzero stationary minimum exists. The paramagnetism phase is stable for $a>0$, and the ferromagnetism phase with $m=\pm \sqrt{x_{+}}$ is stable for $a<0$. The phase transition at $a=0$, $b>0$, is continuous and second order.

For $b<0$, equality with $F(0)$ requires $x=-3b/(4c)$. Substitution into the stationary equation gives the first-order coexistence curveThe disordered phase is the global minimum above this curve, and the ordered phase is the global minimum below it. At coexistence the factorizationexhibits three equally low minima, $m=0$ and $m=\pm \sqrt{-3b/(4c)}$, proving the discontinuous jump.

The ordered spinodal is $a=b^2/(4c)$, $b<0$, where its local minimum meets the intervening local maximum. The disordered spinodal is $a=0$, $b<0$. Thus the ordered phase is metastable for $a_{\mathrm{t}}<a<b^2/(4c)$; the disordered phase is metastable for $0<a<a_{\mathrm{t}}$. These spinodals are limits of metastability, not equilibrium transition curves.

The first-order curve meets the continuous transition line at the tricritical point $(a,b)=(0,0)$. On the specially tuned path $b=0$, the equilibrium ordered solution is $m=\pm (-a/c{)}^{1/4}$ for $a<0$ and tends continuously to zero at the origin.