The attractive 'force' between phase interfaces in the Allen-Cahn model

Question

The heuristic explanation of the behavior of phase transition in the Allen–Cahn model describes two 'forces' at play:

• the curvature of the phase interfaces—they each 'want to' minimize length;
• and an attractive 'force' between the interfaces, which decays exponentially with the distance $d$ between them. This interaction is modelled by a Toda system of particles on the real line. (Accounting for it is crucial in the construction (or exclusion) of clustered phase interfaces, starting with the work of del Pino, Kowalczyk and Wei.)

For a function $u: D \to \mathbf{R}$ that is a stationary point of the $\epsilon$-Allen–Cahn functional (working on the unit disc $D \subset \mathbf{R}^2$ for example), so a solution of the PDE \begin{equation} \epsilon^2 \Delta u + u(1-u^2) = 0, \end{equation} these two forces should be of the same order to roughly balance each other out.

How does the attractive force depend on the parameter $\epsilon > 0$? I assumed it was something like $\mathrm{e}^{-d\ln \epsilon}$ or $\mathrm{e}^{-\frac{d}{\epsilon} \ln \epsilon}$, but neither matches the asymptotics obtained by rescaling a fixed solution $u: \mathbf{R}^2 \to \mathbf{R}$ by homotheties.

Answer 1

The gradient flow $$\frac{\partial u}{\partial t}=\epsilon^2\Delta u+f(u),\;\;\text{with}\;\;f(u)=u(1-u^2),$$ of the Allen-Cahn functional has no inertia (there is no second order derivative in time), so you cannot associate it with a "force". The $\epsilon$ term governs the width of the interface between two regions where $u=+1$ and $u=-1$, for example, in one dimension $u(x)=\tanh( 2^{-1/2} x/\epsilon).$ This has no interpretation as a "balance of forces" in the Allen-Cahn equation.

An alternative way to formulate the gradient flow, where this interpretation becomes possible, is to introduce inertia with a friction term (see section 2.4 of The Calculus of Variations). Then the evolution equation becomes of second order in time, $$\frac{\partial^2 u}{\partial t^2}+a\frac{\partial u}{\partial t}=\epsilon^2\Delta u+f(u),$$ with $a$ a friction coefficient. Now you can interpret $f(u)$ as a force which attracts towards the fixed points $u=\pm 1$. For the tanh profile it decays as $\exp(-2^{1/2}|x|/\epsilon)$ away from the interface.