 Solution

At phase coexistence, the finite ordered global minimum has the same free energy as $m=0$. With $x=m^2>0$ and $f(0)=0$, stationarity and equal values giveSubtracting eliminates $r$:Substitute back to get $r_{\ast }=3b^2/(16c)$, soThe three degenerate equilibrium values at this temperature are $0$ and $\pm m_{\mathrm{jump}}$. A useful global check, avoiding confusion with a spinodal point, is the exact factorizationAll three zeros are therefore global minima. Their curvatures are positive: $f^{\prime \prime }(0)=3b^2/(8c)$ and $f^{\prime \prime }(\pm m_{\mathrm{jump}})=3b^2/(2c)$.

The ordered local minimum first appears on cooling at $r=b^2/(4c)$, where the nonzero stationary roots merge. It becomes globally stable only at $r_{\ast }$; the disordered minimum remains metastable down to $r=0$. These different temperatures distinguish phase coexistence from loss of local stability.