 Solution

A continuous phase transition has a continuously vanishing order parameter and no latent heat. A first-order phase transition exchanges globally stable phases with discontinuous first derivatives of a thermodynamic potential; the order parameter generally jumps, and an entropy jump gives latent heat. In Landau theory the distinction is determined by global minima, not just by the appearance of additional stationary points.

For the uniform scalar potential $U(M)=r{M}^2/2+u{M}^4/4$ with $u>0$ and $h=0$, its stationary equation is $M(r+u{M}^2)=0$. For $r>0$, $M=0$ is the minimum; for $r<0$, the minima are $M=\pm \sqrt{-r/u}$. They approach zero continuously as $r\uparrow 0$, giving an ordinary continuous transition.

For $u<0$ and $v>0$, use the sextic potential. Write $y=M^2$. Stationarity of a nonzero phase requires $r+uy+v{y}^2=0$, while equality of its potential with $U(0)=0$ requires $ry/2+u{y}^2/4+v{y}^3/6=0$. Eliminating $r$ yieldsAt this value the potential factors as $U(M)=\frac{v}{6}{M}^2(M^2+3u/(4v){)}^2\ge 0$, proving that the three stationary zeros really are degenerate global minima. Crossing this line makes the order parameter jump: the transition is first order. The disordered and ordered spinodal points instead occur at $r=0$ and $r=u^2/(4v)$ respectively, delimiting metastability; these are not the equilibrium transition. If the symmetry allows a cubic term, that term can also produce a first-order transition.