= Solution
A <tricritical point> is the meeting of a line of <continuous phase transitions> and a line of <first-order phase transitions>. In the symmetric scalar <Landau free energy>, it occurs at $r=u=0$ with $v>0$. Two independent controls are needed to tune both coefficients. The continuous transition lies at $r=0,u>0$, and the <tricritical three-phase line> found above has $r=3u^2/(16v),u<0$; both approach the <tricritical point>. At zero conjugate field the three coexisting states on this first-order line are the disordered state and the two opposite ordered states.
Setting $u=H=0$ makes the leading stabilizing interaction sextic. The nonzero minimum obeys $M^4=-r/v$, so the <order-parameter critical exponent> along this tuned direction is $\beta_{\mathrm{tri}}=1/4$. Substitution gives
$$
f_{\min}=-\frac{|r|^{3/2}}{3\sqrt v}\quad(r<0),
$$
and two temperature derivatives give the <heat-capacity critical exponent> $\alpha_{\mathrm{tri}}=1/2$. At $r=u=0$, the <equation of state> is $H=vM^5$, giving the <critical-isotherm exponent> $\delta_{\mathrm{tri}}=5$.
For <magnetic susceptibility>, differentiating $H=rM+vM^5$ gives $\chi^{-1}=r+5vM^4$. This is $r$ above the transition and $-4r$ below, hence $\gamma_{\mathrm{tri}}=1$. The quadratic fluctuation kernel still has a mass proportional to $|r|$, giving $\nu_{\mathrm{tri}}=1/2$ and $\eta_{\mathrm{tri}}=0$. Thus the <mean-field tricritical exponent calculation> yields
$$
\boxed{\beta_{\mathrm{tri}}=\frac14,\quad
\alpha_{\mathrm{tri}}=\frac12,\quad
\delta_{\mathrm{tri}}=5,\quad
\gamma_{\mathrm{tri}}=1,\quad\nu_{\mathrm{tri}}=\frac12.}
$$
Detuning $u$ gives <tricritical crossover scaling>; one must specify the approach to the <tricritical point> when quoting these <critical exponents>. The sextic interaction has <upper critical dimension> three, so sufficiently strong long-distance fluctuations change these mean-field predictions below three dimensions.
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