= Solution
Write $a=\alpha_2(T)$ and $u=\alpha_{2n}>0$. At high temperature $a>0$, so $f(m)=am^2+um^{2n}$ has one minimum at $m=0$. At low temperature $a<0$, the origin is a local maximum and two symmetry-related minima appear. The stationary equation is
$$
2m\{a+nu m^{2n-2}\}=0,
$$
so the <equilibrium magnetization> is
$$
m_{\rm eq}=0\quad(T\geq T_c),
$$
and
$$
m_{\rm eq}=\pm\left(\frac{-a}{nu}\right)^{1/(2n-2)}
\quad(T<T_c).
$$
Because $a(T)\sim T-T_c$, the two nonzero minima approach zero continuously as $T\uparrow T_c$. The <order parameter> is continuous while its response becomes singular, so this is a <continuous phase transition>.
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