A problem to understand the stability analysis in the Cahn-Hillard equation

Question

Let us suppose the general diffusion equation (Cahn-Hillard equation):

$$\frac{\partial c}{\partial t} = M \nabla^2 \mu, \tag{1}$$

where $c (\underline{r},t)$ is the concentration of a given species that changes in space ($\underline{r} = (x,y)$, in two dimensions, e.g.) and time, $t$. $M$ is a transport coefficient (called the mobility coefficient) and $\mu$ the chemical potential. The latter can be written as,

$$\mu (c) = \frac{\delta F}{\delta c} = \left(\frac{\partial f}{\partial c}\right)_{c=c_0} -\gamma \nabla^2c,$$ where $F$ is the system free energy and $f$ the free energy density, $\gamma$ a coefficient, and $c_0$ the average composition concentration.

In the traditional treatment of linear stability analysis, we typically assume a perturbation of the homogenous state, $c_0$, as

$$c (\underline{r},t) = c_0+\delta c (\underline{r},t), \tag{2}$$ $$\mathrm{where} \\ \delta c (\underline{r},t) = a\ \mathrm{exp}(wt)\ \mathrm{sin}(\underline{q}.\underline{r}),$$

where $w$ is the growth rate and $\underline{q}$ the wavevector. Then, we replace Equation (2) into Equation (1) and obtain an equation for the perturbation field, $\delta c (\underline{r},t)$.

$\textbf{My questions:}$

1. What mathematically justifies to write Equation (2) and then replaces it into Equation (1) ? I mean to me it seems like we assume the solution of Equation (1) can be written as Equation (2), but we do not know that a priori, no?
2. Let us assume a diffusion equation that does not admit any homogenous solution, e.g., in Equation (1), you add a term like $$\frac{\partial c}{\partial t} = M \nabla^2 (\mu + f(\underline{r},t)),$$

where $f(\underline{r},t))$ is a given function of space and time. How would the linear analysis look like ? Can we still write, Equation (2) ?

Answer 1

Writing $c$ as a general perturbation of the homogeneous state $c_0$ does not introduce any additional assumptions on the general solution, as if $c = c_0 + \delta c$, then $\delta c = f - c_0$ can reproduce any function $f$. The second equation specifying the exact form of $\delta c$ does introduce an assumption, but it is well justified by the following argument:

Rewrite the PDE as an abstract evolution equation on some reasonable function space $\frac{\partial c}{\partial t} = F(c)$. Since $c_0$ is a steady state, we have that $F(c_0)=0$. We now expand in a Taylor series about $c_0$ to obtain $$ \begin{aligned} \frac{\partial c}{\partial t} = F(c) &= F(c_0) + F'(c_0)\delta c + \mathcal{O}(\delta c ^2) \\ &=F'(c_0)\delta c + \mathcal{O}(\delta c ^2). \end{aligned} $$ However, notice that $\frac{d}{dt}(c_0+\delta c) = \frac{d \ \delta c}{dt}$, so if we take $\delta c$ small enough to neglect the higher order terms in the Taylor series, we obtain $$ \frac{d \ \delta c}{dt} = F'(c_0)\delta c. $$ We now look at the form of the operator $F'(c_0)$. It contains terms involving combinations of the Laplacian $-\nabla ^2$ and the spatially homogeneous (and hence scalar!) steady state $c_0$. This operater then has the same eigenfunctions as the standard Laplacian, given here by $\sin (q\cdot r)$ with associated eigenvector $\lambda_q$. The eigenvalues of the linearized CH equation will be some function of $\lambda_q$, call them $\lambda_q^F$ and we will have $$ \frac{d \ \delta c}{dt} = F'(c_0)\delta c = \lambda_q^F \delta c, $$ which has exact solution $\delta c = e^{\lambda_q^F t} \sin(q\cdot r)$. Since eigenfunctions are only unique up to a constant, we can multiply by some constant $a$, and we could also rewrite $\lambda_q^F = w$ and we have obtained the desired form.

As for the nonhomogeneous, it becomes more difficult for two reasons:

1. If $f$ depends on $t$, then the equation is not longer autonomous and standard linear stability analysis no longer applies without additional information on $f$, as a "stable" critical point at one time may not exist or change stability at future times.

2. If $f$ depends on $r$, then steady states $c_0$ may not be homogeneous in space and the eigenfunctions of $F'(c_0)$ will no longer be simple scalar multiples of Laplacian eigenfunctions, complicating analysis further.