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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.

Assume the scalar order parameter can take either sign. The cubic term breaks $m\mapsto -m$ symmetry, and a positive quartic coefficient bounds the Landau free energy below. Stationary points satisfyso, besides the disordered point $m=0$, the possible ordered points areThe disordered point has curvature $A^{\prime \prime }(0)=A_2$ and is locally stable when $A_2>0$. Nonzero stationary points first appear whenThat condition alone does not make them the equilibrium phase: one must compare their free energies.

At a nonzero stationary point, substitute $A_2=-A_{3}m-A_4{m}^2$ to obtainFor an ordered state to coexist with $A(0)=0$, this requiresBoth curvatures at phase coexistence are positive and equal to $2A_3^2/(9A_4)$. The barrier is at $m=-A_3/(3A_4)$. In fact the potential at phase coexistence factorizes:The two distinct minima are therefore explicit, with a finite jump in the order parameter whenever $A_3\ne 0$.

To state the equilibrium inequalities for either sign of $A_3$, put $m=-\mathrm{sgn}(A_3)\rho$ with $\rho \ge 0$. This sign always has lower energy than the opposite sign of equal magnitude. ThenThe bracket's minimum is $A_2/2-A_3^2/(9A_4)$. It follows thatThe favored ordered minimum, when it exists, isThus a first-order phase transition is possible and occurs when the coefficients cross the coexistence relation. There is no continuous transition between $m=0$ and the equilibrium ordered state while $A_3$ remains nonzero and $A_4$ remains finite and positive. The ordered state becomes globally favorable while $A_2$ is still positive, before the disordered curvature can vanish. Although a stationary solution tends to zero at $A_2=0$, it is not the global transition branch. A continuous transition would require eliminating the cubic term, outside the stated $A_3\ne 0$ condition.

The spinodal points clarify the metastable region: for $2A_3^2/(9A_4)<A_2<A_3^2/(4A_4)$ the ordered minimum is metastable; for $0<A_2<2A_3^2/(9A_4)$ the disordered minimum is metastable. At $A_2=0$ the disordered state loses local stability. These local-stability limits should not be mistaken for the first-order transition in a cubic-quartic Landau potential.