 Solution

Set $h=0$ in the even sextic Landau free energy. For $u>0$, near $r=0$ the nonzero stationary branch satisfies $M^2=-r/u+O(r^2)$ for $r<0$. The minimum value is $V_{\min }=-r^2/(4u)+O(r^3)$ below the transition and zero above. The order parameter is continuous, and if $r$ crosses zero linearly with temperature, the first temperature derivative of the minimized free-energy density is continuous too. There is no latent heat; a second derivative has a finite jump at this level.

For $u<0$, solve the two conditions $V^{\prime }(M)=0$ and $V(M)=V(0)$ with $M\ne 0$. Setting $q=M^2$ gives $r+uq+v{q}^2=0$ and $r/2+uq/4+v{q}^2/6=0$. Eliminating $r$ givesBoth the disordered and ordered minima have positive curvature there, but their values coincide. Crossing this phase coexistence boundary changes the global minimum discontinuously. The order parameter jumps; for $r=r(T)$ at fixed $u,v$, ${\partial }_T{V}_{\min }=r^{\prime }(T)M^2/2$ jumps as well, generically yielding a nonzero latent heat. Local extrema can persist past coexistence as metastable states, which explains hysteresis in a dynamically delayed transition. The spinodal points, where local curvature vanishes, do not locate the equilibrium transition.