I have a problem to derive exactly the spatial term of the chemical potential from the free energy of the Cahn-Hillard equation

So, let us start with the free energy of a binary mixture (A and B). The free energy density in the Cahn-Hillard equation writes (with $c$ the variable related to density):

$$f(c, \nabla c) = \frac{1}{4}{\left(c^2-1 \right)^2 + \frac{\gamma}{2} \lvert \nabla c \rvert^2}\tag{1}$$

Let us define, $\mu (c) = \frac{\delta F}{\delta c}$, where $F(c) = \int d^n x f(c, \nabla c)$, with $n$ the spatial dimension.

The result is:

$$\mu (c) = c^3 -c - \gamma \nabla^2 c\tag{2}$$

How can we derive (2) from (1) ?

  • $\begingroup$ Look up how functional derivative of a gradient is evaluated en.wikipedia.org/wiki/Functional_derivative. Define grand potential $G = F - \int d^n x \mu c$ (the free energy in the Grand Canonical ensemble). Then, obtain the equation of motion from the Lagrangian density $\mathcal{L}(c,\nabla{c}) = \frac{1}{4}{\left(c^2-1 \right)^4 + \frac{\gamma}{2} \lvert \nabla c \rvert^2} - \mu c$. $\endgroup$ Commented Apr 4 at 14:11
  • $\begingroup$ Thanks for your reply. But can we not get more simply the result from (1) by applying the chain rule? $\endgroup$
    – math-int
    Commented Apr 4 at 14:18
  • $\begingroup$ No. $c$ and $\nabla c$ are to be treated as independent variables. Chain rule does not apply here. $\endgroup$ Commented Apr 4 at 14:20
  • $\begingroup$ Suppose I take the derivative of the last term: d/dc ($\gamma/2 \lvert \nabla c\rvert^2$). This would give, $\gamma/2 \lvert\nabla c\rvert d/dc (\lvert \nabla c \rvert)$. There is no way to evaluate this last term ? $\endgroup$
    – math-int
    Commented Apr 4 at 14:47
  • $\begingroup$ Derivative with respect to what? Note that you can uniformly change $c$ for all values of $x$, which will not affect the value of $\nabla c$ $\endgroup$ Commented Apr 4 at 14:48

2 Answers 2


The free energy is $F[c] = \int d^n x \frac{1}{4}{\left(c^2-1 \right)^2 + \frac{\gamma}{2} \lvert \nabla c \rvert^2}$.

In the grand canonical ensemble, the chemical potential is fixed, and the density is allowed to vary. The grand potential function is $G[c] = F[c] - \mu \int d^n x c $.

The free energy is obtained by minimizing the grand potential functional of the density, $\frac{\delta G}{\delta c(\vec{r})} = 0$.

For an infinitesimal variation $\delta c$, the change in the grand potential is

$\delta G = \int d^n x \left[\frac{1}{4}2(c^2 - 1) 2c \delta c + \gamma \nabla c \cdot \nabla \delta c - \mu \delta c\right]$

$= \int d^n x \delta c\left[c(c^2-1) - \mu \right] + \int d^n x \gamma \nabla c \cdot \nabla \delta c$.

Integrating by parts, we can rewrite the above as $\int d^n x \gamma \nabla c \cdot \nabla \delta c = -\int d^n x \gamma \delta c \nabla^2 c $. The boundary terms have to be zero because $\delta c$ has to be zero at the boundaries, to match the boundary conditions.

Combining everything,

$\delta G = \int d^n x \delta c\left[c(c^2-1) - \gamma \nabla^2 c - \mu \right]$.

Since $\delta G$ has to be zero (when $G$ is minimized) for arbitrary variations of $\delta c$, we must have $\boxed{c(c^2-1) - \gamma \nabla^2 c - \mu = 0}$.

An alternative way is to notice $G[c]$ is the "action" of a Lagrangian, $\mathcal{L}(c,\nabla{c}) = \frac{1}{4}{\left(c^2-1 \right)^4 + \frac{\gamma}{2} \lvert \nabla c \rvert^2} - \mu c$, i.e., $G[c] = \int d^n x \mathcal{L}(c,\nabla{c})$.

The equilibrium properties are given by the value of $c$ such that $G$ is minimized, which can be found from Lagrange's equation of motion, $\frac{\partial \mathcal{L}}{\partial c} = \nabla\cdot \frac{\partial \mathcal{L}}{\partial \nabla c}$, which produces the same equation for $\mu$.

Another alternative is to equate the functional derivative $\frac{\delta G}{\delta c(\vec{r})}$ with $0$.

We will use the property $\frac{\delta}{\delta c(\vec{r})} \int d^n x c(\vec{x}) = \int d^n x \delta(\vec{r} - \vec{x}) $.

Then, $0=\frac{\delta G}{\delta c(\vec{r})} = \int d^n x \left[c(\vec{x})(c^2(\vec{x})-1) -\mu\right]\delta(\vec{r} - \vec{x}) + \gamma \nabla_x c(\vec{x}) \cdot \nabla_x \delta(\vec{r}-\vec{x})$.

Integrating the last term by parts, $\frac{\delta G}{\delta c(\vec{r})} = \int d^n x \left[c(\vec{x})(c^2(\vec{x})-1) -\mu\right]\delta(\vec{r} - \vec{x}) - \gamma \nabla^2_x c(\vec{x}) \delta(\vec{r}-\vec{x})$ $ = c(\vec{r}) (c^2({r})-1)-\gamma \nabla_r^2 c(\vec{r}) - \mu$.

Since this is zero, we get back the same equation.


By following the suggestion of @ArchismanPanigrah. (2) follows straightforwardly from (1) by using the relation of functional derivative (provided in ref. 2) and by noticing that $c$ and $\nabla c$ are independent variables.

ref. 2: https://en.wikipedia.org/wiki/Functional_derivative


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