Physical meaning of Cahn-Hilliard boundary conditions

Question

Consider the 1D Cahn-Hilliard equation for a two-component mixture, on an interval $x\in[a,b]$:

$\frac{dc}{dt} = -\frac{d}{dx}j(x)$ where the flux $j(x) = -D\frac{d}{dx}\left(c^3 - c - \gamma\frac{d^2c}{dx^2}\right)$

The typical boundary conditions are:

1. $j(x)=0$ at $x=a$ and $x=b$, meaning no flux across the boundaries.

2. $\frac{dc}{dx}=0$ at $x=a$ and $x=b$.

What is the physical rationale for boundary condition 2?

It is tempting to compare boundary condition 2 to the zero-flux boundary condition for the diffusion equation, which has the same mathematical form $\frac{dc}{dx}=0$ at $x=a$ and $x=b$. However, for the diffusion equation, $j(x) = -D\frac{dc}{dx}$, so the condition $\frac{dc}{dx}=0$ is equivalent to $j(x)=0$, which is analogous to boundary condition 1, not 2.

Answer 1

You are absolutely correct that the first condition is the no-flux condition. The second condition is a bit more subtle and actually determines which value of $c$ is favored at the boundary! The normal derivative of $c$ directly determines the contact angle of droplets at the wall! The standard condition $\partial_x c=0$ implies a neutral condition where the contact angle is 90 degrees. This boundary condition can be directly derived by minimizing the total free energy including a contact potential. We discuss this in some detail in one of our recent publications: https://doi.org/10.1063/5.0207761 (see particularly Eqs. 1 and 3).

Answer 2

However, for the diffusion equation, $j(x) = -D\frac{dc}{dx}$, so the condition $\frac{dc}{dx}=0$ is equivalent to $j(x)=0$, which is analogous to boundary condition 1, not 2.

You are correct if everything was purely diffusion, but your overall flux is not diffusive, so $\text dc/\text dx$ does not imply that $j(x)=0$ (just plug that in and see for yourself).

You are correct that $\text dc/\text dx=0$ means that there is no diffusive flux though.