 Solution

Consider the $(u,r)$ plane at $h=0$ for $V=r{M}^2/2+u{M}^4/4+v{M}^6/6$, 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 lineAbove this curve the disordered minimum wins; below it the two ordered minima win. On the curve,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.