 Solution

In a first-order phase transition, two distinct global minima of the Landau free energy have equal values and exchange stability. The equilibrium order parameter jumps, and a first derivative of the equilibrium free energy is discontinuous. Depending on the direction through the phase diagram, that derivative can be magnetization, entropy or another conjugate quantity; a first-order transition need not have latent heat along every chosen path.

In a continuous phase transition, the equilibrium order parameter changes continuously while the curvature at the critical minimum vanishes. For $u>0$ and $h=0$, $M=0$ is stable for $r>0$, but for $r<0$ the stable minima have $M\simeq \pm \sqrt{-r/u}$. The order parameter therefore goes continuously to zero at $r=0$; the inverse magnetic susceptibility vanishes there.

For $u<0$, the sextic term stabilizes the Landau free energy. The nonzero minima can become favorable while $M=0$ still has positive curvature. At $r=3u^2/(16v)>0$ they coexist with $M=0$, so the order parameter jumps to magnitude $\sqrt{-3u/(4v)}$. Finding a stationary point alone does not locate the transition: equilibrium requires comparing the values of all competing minima. Spinodal boundaries describe loss of local stability, not coexistence.