 Solution

A tricritical point occurs when a first-order boundary meets a continuous-transition boundary. In the spin-inversion-symmetric sextic Landau free energy, two independent parameters must tune $r=u=0$, while $v>0$ keeps the potential stable. In the $h=0$ plane, the continuous boundary is $r=0,u>0$, and the first-order boundary is $r=3u^2/(16v),u<0$. They meet at the tricritical point; the order-parameter jump on the first-order side tends to zero as $u\uparrow 0$. The first panel shows this phase diagram, including the metastability limits derived above.

Along the tuned line $u=0,h=0$, the equation of state is $M(r+v{M}^4)=0$. Thus $M=(-r/v{)}^{1/4}$ on a selected ordered branch, rather than the ordinary square-root behavior. Differentiation gives inverse magnetic susceptibility $r$ above and $4|r|$ below, while at criticality $h=v{M}^5$. Hence the mean-field tricritical exponent calculation yields $\beta =1/4$, $\gamma =1$, $\delta =5$; the minimized potential $-|r{|}^{3/2}/(3\sqrt{v})$ also gives $\alpha =1/2$. These tricritical powers require tuning the quartic coefficient, not merely varying a generic temperature path. The ordinary scalar exponents requested separately are calculated next.