= 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 $V'(M)=0$ and $V(M)=V(0)$ with $M\ne0$. Setting $q=M^2$ gives $r+uq+vq^2=0$ and $r/2+uq/4+vq^2/6=0$. Eliminating $r$ gives
$$
\boxed{M^2=-\frac{3u}{4v},\qquad r=\frac{3u^2}{16v}.}
$$
Both 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 $u,v$, $\partial_TV_{\min}=r'(T)M^2/2$ 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.
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