 Solution

The first-order coexistence curve approaches the continuous curve as $u\uparrow 0$: both meet at $r=u=h=0$. To obtain this tricritical point, two even coefficients must be tuned to zero, while the sextic coefficient remains positive. A nonzero conjugate field removes the zero-field inversion symmetry and leads to the wing structure described in the root solution.

At $u=h=0$, the equation of state becomes $0=rM+v{M}^5$. For $r<0$,Above the transition $M=0$ and $V^{\prime \prime }(0)=r$. These are stable branches of a bounded potential, and the ordered value tends continuously to zero at $r=0$, but with $M\propto (-r{)}^{1/4}$ rather than the ordinary square-root law. Tricritical behavior requires $r=u=0$, $h=0$ and $v>0$; a generic path with fixed $u>0$ instead has ordinary critical behavior.