= Solution
Take
$$
F[\phi]=\int\left[\frac a2\phi^2+\frac b4\phi^4+\frac\kappa2|\nabla\phi|^2\right]d\mathbf r.
$$
Its variational chemical potential is
$$
\mu=\frac{\delta F}{\delta\phi}=a\phi+b\phi^3-\kappa\nabla^2\phi.
$$
Conservation gives $\dot\phi=-\nabla\mathbin{\cdot}\mathbf J$. Assuming an isotropic constant mobility $M>0$, local linear irreversible thermodynamics gives $\mathbf J=-M\nabla\mu$. Hence the noiseless <conserved order-parameter dynamics> is
$$
\boxed{\dot\phi=-\nabla\mathbin{\cdot}\mathbf J},
\qquad
\boxed{\mathbf J=-M\nabla(a\phi+b\phi^3-\kappa\nabla^2\phi)}.
$$
It decreases the free energy because $\dot F=-\int M|\nabla\mu|^2d\mathbf r\leq0$ under closed or no-flux boundary conditions.
Back to article page
