 Solution

A first-order phase transition exchanges two distinct global minima of the Landau free energy. The equilibrium free energy remains continuous, but a first derivative such as the magnetization or entropy jumps. A temperature-driven transition generically has latent heat; crossing $h=0$ between symmetry-related ordered phases instead has a magnetization jump without requiring latent heat. A continuous phase transition has no such order-parameter jump and usually has a divergent correlation length and nonanalytic higher responses.

For $u>0$, $h=0$ and $r$ crossing zero, minimizing $V=r{M}^2/2+u{M}^4/4$ gives $M=0$ for $r>0$ and $M=\pm \sqrt{-r/u}$ for $r<0$. The minima merge continuously at zero. The curvature vanishes there, so the long-distance response becomes singular.

For $u<0<v$, put $z=M^2$. A nonzero stationary minimum satisfies $r+uz+v{z}^2=0$. Equality with the $M=0$ free energy givesThe finite separation demonstrates the first-order phase transition. Metastability occurs where a nonglobal local minimum persists; its disappearance at a spinodal point is not the equilibrium coexistence criterion.