 Solution

A tricritical point occurs at $r=u=h=0$ with $v>0$: tuning $u$ changes a continuous transition into a first-order one. At $u=0$, the equation of state is $h=rm+v{m}^5$. It gives $m\propto (-r{)}^{1/4}$ at $h=0$, $m\propto h^{1/5}$ at $r=0$, and a minimized free energy proportional to $-|r{|}^{3/2}$ below the transition. The mean-field tricritical exponents areIn the three-dimensional parameter space $(r,u,h)$, the zero-field disordered-to-ordered continuous line $r=0,u>0$ meets the first-order line $r=3u^2/(16v),u<0$ at the tricritical point. The zero-field surface also contains coexistence of positive and negative ordered phases. For negative $u$, tilting with $h$ produces two symmetric tricritical wings of first-order coexistence, each bounded by a line of ordinary critical endpoints. Those wings and their critical edges meet at the tricritical point. The upper critical dimension of tricritical sextic theory is three, so mean-field values can acquire logarithmic corrections there.