# Solution Cahn-Hilliard 1d for a domain of finite size

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 like this one :

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 thus looking for that kind of profile :

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.

What do you think about it ? If this proposition were to be right, what could be a formalism with whom we could build a solution for a finite domain ?

I don't know if such a concentration profile is possible physically. So I shall make remarks about math. From the initial equation $$f'(c(x)) - \tilde{\mu}- \epsilon^2 c''(x) = 0,$$ follows equation $$\frac{d}{dx}\left(f(c(x)) - \tilde{\mu}c(x) - \frac{\epsilon^2}2 (c'(x))^2 \right) = 0,$$ from which also follows $$f(c(x)) - \tilde{\mu}c(x) - \frac{\epsilon^2}2 (c'(x))^2 = C_1. \qquad (1)$$ Here $$C_1$$ is a constant. I don't understand why are you equating this constant with $$-\tilde{\mu}C$$. Further, the equation (1) has two solutions $$c'(x) = \pm \sqrt{\frac{2}{\epsilon^2}(f(c(x))-\tilde{\mu}c(x)-C_1)}\quad (2)$$ Definite sign in expression (2) implies that $$c(x)$$ is a monotonic function of $$x$$ because of $$c'(x)$$ is nonnegative or nonpositive. The $$\tanh$$ solution corresponds to this case, as I understand.

In case when $$c'(0) = 0$$ and $$c(x)$$ is a nonmonotonic function one should choose in (2) plus sign for $$x < 0$$ and minus sign for $$x > 0$$.

Update. Equations here strongly resemble those of classical mechanics. The direct correspondence is obvious if we say that $$c$$ is a position of a particle on a line, $$x$$ is time, $$\epsilon^2$$ is a mass of the particle, $$-C_1$$ is energy of particle and $$U(c) = \tilde{\mu}c - f(c)$$ is potential energy. Then equation (1) looks like the energy of the particle equal to the sum of kinetic and potential energies. Hence the solution of the equation for $$c(x)$$ is equivalent to the solution of the mechanical problem about the motion of a particle in a potential well. In my opinion, the photo below shows the situation when the solution of the equation for $$c(x)$$ has properties you might be interested in. The correspondent solution will have no plateau $$c(x) = const$$ in the center.

• I edited the question and added images maybe it's gonna be clearer. I'd like a solution that would take into account the fact that the amount of c is fixed. – J.A Aug 1 '19 at 20:53
• @J.A, I edited my answer. – Gec Aug 2 '19 at 12:15
• You can continue the calculation. Given the condition : $c'(0)=c'(\infty)=0$, we obtain $C_1=f(c(\infty))-\tilde{\mu}c(\infty)=f(c(0))-\tilde{\mu}c(0)$. From which : $\tilde{\mu}=(f(c(\infty))-f(c(0)))/(c(\infty)-c(0))$ – J.A Aug 2 '19 at 14:03
• But I have a doubt if this changes something to the fact that if you have $c'(0)=0$, you just have a flat solution... but I don't know how to prove it, or to prove the opposite. – J.A Aug 2 '19 at 14:24
• @J.A I added to my answer about the solution of the equation for $c(x)$. – Gec Aug 2 '19 at 16:35

I have at least a part of the answer. In one phrase, the boundary conditions play a role in the case where the space is finite.

If the boundary coundition on the wall in the finite space is a Neuman boundary condition, meaning : $$\mathbf{n}.∇((f′(c)−μ)−ϵ^2Δc)=0$$ which is in 1D : $$\mathbf{e_x}.\frac{d((f′(c)−μ)−ϵ^2Δc)}{dx}=0$$ the solution which is true for the infinite space will be correct in the finite space too. And the conservation of matter will just give us the argument $$x_0$$ of $$c(x_0)=0$$.

In the case of an infinite space and a finite quantity of matter, even if we start with a nucleus, it will evaporate to a stable steady state.