 Solution

A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as the entropy or order parameter. It can have latent heat when the entropy jumps. A continuous phase transition has a continuously vanishing order parameter and no latent heat, with singular higher derivatives and a diverging correlation length. The LG theory compares global minima, not merely the points where a local minimum loses stability.

For the uniform quartic Landau free energythe equation of state is $rm+u{m}^3=h$. At zero field the stable minimum is $m=0$ for $r>0$, and $m=\pm \sqrt{-r/u}$ for $r<0$. Thus tuning $r$ through zero gives a continuous transition. For a fixed $r<0$, varying $h$ through zero instead switches between the two ordered minima and makes $m$ jump, giving a first-order phase transition in the conjugate field.

A temperature-like first-order transition can occur at zero field when $u<0$ and a positive sextic term $v{m}^6/6$ stabilizes the potential. Write $q=m^2$. Stationarity of a nonzero phase gives $r+uq+v{q}^2=0$, while equality with $V(0)=0$ gives $rq/2+u{q}^2/4+v{q}^3/6=0$. Solving these two conditions yieldsThe order parameter jumps from zero to $\pm \sqrt{q_{\mathrm{coex}}}$. At this pointso the competing stationary points are genuinely global minima. This phase coexistence condition differs from the spinodal points $r=0$ and $r=u^2/(4v)$, which mark loss or creation of local stability, not equilibrium coexistence.

If the symmetry permits a cubic term $w{m}^3/3$, a positive quartic coefficient does not preclude a first-order transition. For $V=r{m}^2/2+w{m}^3/3+u{m}^4/4$ with $w\ne 0,u>0$, stationarity and coexistence give $m=-2w/(3u)$ and $r=2w^2/(9u)$. This is the first-order transition in a cubic-quartic Landau potential; the symmetry restriction on the expansion is therefore part of the prediction.