= Solution
Consider the $(u,r)$ plane at $h=0$ for $V=rM^2/2+uM^4/4+vM^6/6$, with fixed $v>0$. For $u>0$, the continuous boundary is $r=0$: above it the <global minimum> is zero, and below it there are two symmetry-related ordered <pure thermodynamic phases>. For $u<0$, the equilibrium boundary is the <tricritical three-phase line>
$$
\boxed{r=\frac{3u^2}{16v}.}
$$
Above this curve the disordered minimum wins; below it the two ordered minima win. On the curve,
$$
V(M)=\frac v6M^2\left(M^2+\frac{3u}{4v}\right)^2\geq0.
$$
Its three zeros are exactly $M=0$ and $M=\pm\sqrt{-3u/(4v)}$, proving equality of three <global minimum> values. <Spin inversion symmetry> enforces the equality of the two ordered phases automatically, so tuning a single remaining coefficient can produce three-phase coexistence on a line in this symmetry plane. This does not contradict the generic phase-rule counting without symmetry restrictions.
The <tricritical point> is $(u,r)=(0,0)$, where the discontinuity vanishes, the first-order boundary meets the continuous boundary, and the positive sextic term becomes the leading stabilizer. The local metastability limits $r=0$ and $r=u^2/(4v)$, $u<0$, bracket the first-order curve but are not phase boundaries. In the full $(r,u,h)$ parameter space, the zero-field line is attached to <tricritical wings>; a nonzero field splits the equality of the two ordered phases.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-49-tricritical.png]
{title=Zero-field sextic Landau phase diagram, showing the continuous boundary and three-phase coexistence line}
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