With $\phi_o=0$,
$$
f(\phi)=
\begin{cases}
\frac12(a-C)\phi^2+\frac b4\phi^4,&\phi<0,\\
\frac a2\phi^2+\frac b4\phi^4,&\phi>0.
\end{cases}
$$
For $C<a$, both branches are locally convex at the origin. At
$$
\boxed{C_c=a}
$$
the negative-side curvature vanishes; for $C>a$ the interval near $0^-$ has $f''=a-C+3b\phi^2<0$, so a homogeneous composition there is unstable and the equilibrium free energy is its convex envelope. The sketch therefore has an ordinary upward quartic on the positive side and, beyond $C_c$, a negative-curvature shoulder and a minimum on the negative side.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 344 / 1 / d / i / Solution

 Solution

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