= Solution
The first-order coexistence curve approaches the continuous curve as $u\uparrow0$: 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 $0=rM+vM^5$. For $r<0$,
$$
M^4=-r/v,\qquad V_{\min}=-\frac{|r|^{3/2}}{3\sqrt v},\qquad V''(M)=4|r|.
$$
Above the transition $M=0$ and $V''(0)=r$. 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. \b[Tricritical behavior requires $r=u=0$, $h=0$ and $v>0$; a generic path with fixed $u>0$ instead has ordinary critical behavior.]
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