Solution
= Solution
For
$$
\mathcal F=\int\left[
f(\phi)+\frac{\kappa(\phi)}2|\nabla\phi|^2
\right]d\mathbf r,
$$
the <functional derivative> is
$$
\boxed{
\mu=f'(\phi)-\nabla\mathbin{\cdot}
[\kappa(\phi)\nabla\phi]
+\frac{\kappa'(\phi)}2|\nabla\phi|^2
=f'(\phi)-\kappa(\phi)\nabla^2\phi
-\frac{\kappa'(\phi)}2|\nabla\phi|^2
}.
$$
The constrained minimum again makes $\mu$ constant. Evaluating it in either uniform bulk phase gives $\mu=f'(\pm\phi_B)=0$, so $\mu=0$ everywhere.