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 $j(x) = -\frac{d}{dx}\left(c^3 - c - \gamma\frac{d^2c}{dx^2}\right)$

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, $j(x) = \frac{dc}{dx}$, so this is actually equivalant to boundary condition 1, not 2.

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.