= Solution
A <phase diagram> partitions a space of control parameters into regions with distinct equilibrium phases. Its boundaries mark nonanalyticities of the thermodynamic-limit <free energy>; a <first-order phase transition> has a jump in a first derivative, whereas a <continuous phase transition> has a continuously vanishing <order parameter> and singular fluctuations.
An explicit three-parameter example is the scalar <Landau free energy>
$$
V(M;r,u,h)=\frac r2M^2+\frac u4M^4+\frac v6M^6-hM,\qquad v>0,
$$
with parameter axes $(r,u,h)$. These can be realized locally by the temperature, single-ion anisotropy and magnetic field of the <Blume–Capel model>. The three dimensions here refer to control-parameter space, not necessarily to three spatial dimensions. At $h=0$, minimizing this potential gives a <continuous phase transition> at $r=0$ for $u>0$. For $u<0$, coexistence occurs at $r=3u^2/(16v)$, where
$$
V(M)=\frac v6M^2\left(M^2+\frac{3u}{4v}\right)^2.
$$
There are three <global minima>, $M=0$ and $M=\pm\sqrt{-3u/(4v)}$. Their <order parameter> jump decreases to zero as $u\uparrow0$. Thus \b[$(r,u,h)=(0,0,0)$ is a tricritical point.] The two transition curves in the $h=0$ plane meet there.
In the three-dimensional <phase diagram>, crossing $h=0$ inside the ordered region switches between opposite signs of $M$, so that region is itself a <first-order phase transition> sheet. For $u<0$, two <tricritical wings> at opposite nonzero fields separate weakly and strongly magnetized phases. Their <wing critical edges> can be obtained rather than merely sketched: require $V'=V''=V'''=0$. Since $V'''=6uM+20vM^3$, the nonzero solutions satisfy
$$
M_c^2=-\frac{3u}{10v},\qquad r_c=\frac{9u^2}{20v},\qquad h_c=\frac{6u^2M_c}{25v}.
$$
The fourth derivative there is $-12u>0$, giving ordinary continuous critical endpoints. The wings meet the zero-field coexistence sheet along the <tricritical three-phase line> and terminate at these edges. Their exact <tricritical wing coexistence factorization> supplies the surfaces in the figure: for two nonnegative minima $a\leq b$, set $s=a+b$, $p=ab$ and
$$
u=\frac{2v}{3}(-2s^2+3p),\quad r=\frac v3(s^4-s^2p+3p^2),\quad h=\frac v3s^3p.
$$
Direct expansion gives $V(M)-V(a)=(v/6)(M-a)^2(M-b)^2[(M+s)^2+p]\geq0$, proving that these surfaces describe coexistence of global, not merely local, minima.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-43-tricritical.png]
{title=Tricritical phase diagram of the sextic scalar free energy: first-order sheets, critical edges and the zero-field slice}
For the unheaded <Ising model> calculation, assume ferromagnetic exchange $J>0$ and let $M$ denote the mean <Ising spin>. On the hypercubic lattice the <coordination number of a lattice> is $z=2D$, even though the original positive-direction bond sum counts each bond only once. The <mean-field approximation> neglects products of deviations from $M$:
$$
\sigma_i\sigma_j\simeq M\sigma_i+M\sigma_j-M^2.
$$
There are $ND$ bonds, so the approximate energy is $H_{\rm MF}=-2DJM\sum_i\sigma_i+NDJM^2$. Each independent <Ising spin> therefore sees the effective field $2DJM$. With $\beta_{\rm th}=1/(k_BT)$, averaging its two Boltzmann weights gives the <self-consistency equation>
$$
\boxed{M=\frac{e^{2\beta_{\rm th}DJM}-e^{-2\beta_{\rm th}DJM}}{e^{2\beta_{\rm th}DJM}+e^{-2\beta_{\rm th}DJM}}
=\tanh\left(\frac{2DJM}{k_BT}\right).}
$$
The slope of its right side at zero is $T_c/T$, where
$$
\boxed{T_c=\frac{2DJ}{k_B}.}
$$
Above $T_c$, $\tanh x<x$ for $x>0$ excludes a nonzero solution. Below $T_c$ the initial slope exceeds one, and concavity of the hyperbolic tangent on the positive axis gives one positive solution and its negative partner. The <Ising auxiliary mean-field free energy> $DJM^2-k_BT\log[2\cosh(2\beta_{\rm th}DJM)]$ has positive quartic coefficient near $T_c$, while its quadratic coefficient changes sign there. Its stable minima therefore move continuously away from zero; this is a <continuous phase transition> in the <mean-field theory of the Ising model>.
Put $\tau=(T_c-T)/T_c>0$ and $c=T_c/T=(1-\tau)^{-1}$. The <Taylor expansion> $\tanh x=x-x^3/3+O(x^5)$ gives, on the nonzero branch,
$$
0=c-1-\frac{c^3}{3}M^2+O(M^4),\qquad
M^2=\frac{3(c-1)}{c^3}+O((c-1)^2)=3\tau+O(\tau^2).
$$
Consequently \b[$M\sim\sqrt{3}\,\tau^{1/2}$ and the mean-field order-parameter exponent is $\beta=1/2$.] The inverse-temperature symbol $\beta_{\rm th}$ used above is distinct from this <order-parameter critical exponent>.
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