# Steps in minimising free-Interfacial thickness

I am trying to obtain the interfacial thickness of a mixture of two fluids in the context of the Cahn-Hilliard equation. In the simplest case and in one dimension the free energy has the form $$F=\int dx \left[H(\psi)+\frac{1}{2}D \left(\frac{\partial \psi}{\partial x} \right)^2\right]$$ where, in first approximation $H(\psi)\approx \frac{1}{2}\tau \psi^2$ with $\tau<0$.

I know that the solution of this equation is of the form $\psi \sim \tanh(x/\xi)$ and $\xi$ represents the interfacial thickness, but if I put $\psi=\tanh(x/\xi)$ in the free energy and minimize with respect to $\xi$ I get $$\xi^2=\frac{C_2 D}{C_1 \tau}$$ with $C_1,C_2$ positive constants of the order of $1$. This is basically correct ($\xi^2$ should be $D/|\tau|$ ) except that since $\tau<0$ the sign does not make any sense.

Can somebody tell me if the procedure makes sense? My idea is: plug a 'test' function into the free energy equation and minimize it with respect to the parameter that is free in my test function.

• You assume a functional form and minimize wrt to its parameters? You are not guaranteed to get the minimal surface, but you will get an upper bound for the free energy. Why are you doing this? 1D C-H (and the functional form) is easy to solve analytically and you should be able to devise rather straightforward numerical algorithms for more complex problems. Commented Mar 11, 2017 at 19:35
• Thanks for your answer. I just want to be sure of the proceeding for the easiest case (where I know the analytical solution) because the next step is using it with a distortion ( a particle in the interface) for which I don't. Commented Mar 11, 2017 at 19:44

You could try and solve $\frac{\delta F}{\delta \phi} = 0$ with your approximated potential and using the assumption that $\phi \rightarrow \phi_c$ for $x\rightarrow \pm \infty$. For small enough $x$ i get the solution $\phi(x) = \phi_c\sin(\sqrt{\frac{-t}{D}}x )$ which seems to have the correct width. The calculation is however only marginally simpler than just solving the problem with a double well potential.