 Solution

At the tricritical point, tune $u=0$, hold $v>0$, and let $r=at$ cross zero. The equation of state becomes $h=rM+v{M}^5$. At $h=0,r<0$, $M^4=-r/v$, while the inverse susceptibility on the ordered branch is $r+5v{M}^4=4|r|$. On the disordered side it is $r$. At $r=0$, $M=\mathrm{sgn}(h)(|h|/v{)}^{1/5}$. Therefore $\beta =1/4$, $\gamma =1$ and $\delta =5$.

The minimized singular free-energy density isTwo temperature derivatives give a singular heat capacity proportional to $|t{|}^{-1/2}$ on the ordered side. Consequently the mean-field tricritical exponent calculation givesThe displayed scaling relations for critical exponents hold numerically:These are Landau theory values with both relevant coefficients tuned: along a generic path with fixed $u>0$, the asymptotic transition is ordinary rather than tricritical. The sextic upper critical dimension is three, so fluctuations can change the powers below three dimensions, and logarithmic corrections are possible at three. The vanishing disordered mean-field singular amplitude does not invalidate the ordered-side exponent.