# How to solve Cahn-Hilliard free energy extremization for a domain of finite size ?

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First I have to say I asked this question in physicsSE but afterwards somebody advised me to ask it here. Do I have to remove it from SE ?

I'm trying to get the solution of the Cahn-Hilliard equation in 1d with a certain mass $C$. We have two components, and let's assume we have the relation $c_1+c_2=1$.Hence we take only the variable $c=c_1$.

The total energy with the Lagrange parameter $\tilde{\mu}$ (which is a sort of non-local chemical potential) writes :

$$F[c(\mathbf{r})]=\int \{f(c(\mathbf{r}))+\frac{\epsilon^2}{2} (\nabla c)^2 \}d\Omega -\tilde{\mu}\int (c(\mathbf{r}) -C) d\Omega$$
In 1 dimension :
$$\frac{\delta F}{\delta c}=0\implies \frac{df}{dc}-\tilde{\mu}-\epsilon^2 \frac{d^2c}{dx^2}=0$$
Multiplying with $dc/dx$ leads to :
$$\frac{\epsilon}{\sqrt{2}}\frac{dc}{\sqrt{f-\tilde{\mu}(c-C)}}=dx$$
Symmetry imposes $$c'(0)=0\implies f(c(0))-\tilde{\mu}(c(0)-C)=0$$
At infinity, we also have $c'(\infty)=0 \; ;\;c(\infty)=-1$ (or $0$ depending on the potential you're using).

This equation is solvable for the classical Cahn-Hilliard with  $f-\tilde{\mu}(c-C)=(c^2-c_0^2)^2$. The classical way is to get $x(c)$ and then invert it. You find a $\tanh$ solution. But this solution does not respect the symmetry condition $c'(0)=0$ (right you can make it very very close to $0$ by building manually a solution with tanh functions... but I'm looking for an exact solution of the equation). Meaning it only gives the profile of an interface between 2 semi-infinite media.

What I don't understand is how to get a profile respecting the symmetry condition, meaning with a nucleus/aggregate of one phase into the other phase. Meaning a phase of finite size (for example $c=1$) into the other phase ($c=-1$).

I'm wondering wether my problem is overconstrained since the equation $\frac{\epsilon}{\sqrt{2}}\frac{dc}{\sqrt{f-\tilde{\mu}(c-C)}}=dx$ admits only one new constant and there are 3 constraints : $c'(0)=c'(\pm \infty)=0$ and $\int_{\mathbb{R}}c dx=C$ (about this one I have a doubt since $C$ enters the potential).

REMARK :  I was wondering maybe there was something missing in the equations. But actually no, since the dynamical equation used in simulations is :$\partial_t c = \nabla.(M(c)\nabla((f'(c)-\tilde{\mu)}-\epsilon^2\Delta c))$, so it's logical that the static picture is given by $(f'(c)-\tilde{\mu)}-\epsilon^2\Delta c=0$.
However what could be is that indeed the system is overconstrained and there is no stable solution. Fortunately the $\tanh$ function provides a landscape that is "quasi-stable" (very very slowly unstable) in the sense that beyond the size of the interface it's as if we had a semi-infinite domain since we are very close to it and that's why we use this model in simulations.
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