 Solution

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as entropy or the order parameter; a thermal first-order transition normally has latent heat. At a continuous phase transition the order parameter goes continuously to zero and there is no latent heat, although response functions and higher derivatives of the free energy can be singular.

For the symmetric scalar Landau free energy at zero conjugate field, minimizeIf $u>0$, the sixth-order term is unnecessary near $r=0$. For $r>0$ the minimum is $M=0$, whereas for $r<0$ there are two symmetry-related minima with $M^2=-r/u$ to leading order. The order parameter vanishes continuously as $r\uparrow 0$. The equilibrium free-energy density changes from zero to $-r^2/(4u)$, so its first temperature derivative is continuous and its second derivative has a jump. This is the ordinary mean-field continuous phase transition.

If $u<0$, keep $v>0$ for stability. A nonzero stationary order parameter, with $s=M^2$, obeys $r+us+v{s}^2=0$. For phase coexistence with $M=0$ its free energy must also vanish. Substituting $r=-us-v{s}^2$ into $V$ givesThe nonzero solution of $V=0$ is thereforeThe competing minima exchange stability at positive $r_t$, with a finite jump of the order parameter. This is a first-order phase transition. The metastability region is also visible: nonzero stationary points first appear at the spinodal $r=u^2/(4v)$, and the disordered local minimum loses stability at $r=0$.

For example, holding $u,v$ fixed and taking $r=a(T-T_0)$, the ordered phase has entropy lower than the disordered phase by $a{M}_t^2/2$ at coexistence. Heating through the transition absorbs latent heat per volume $T_{t}a{M}_t^2/2$. More generally the temperature dependence of all coefficients contributes to this entropy difference. With no inversion symmetry, a cubic term can also produce a first-order phase transition, as comparison of the competing minima of the Landau free energy shows.