# Free energy in Allen-Cahn PDE

I am a mathematician and I am taking a mathematical physics course. In the part of reaction-diffusion equations, there is something that I do not understand. I have been defined the Allen-Cahn equation as $$u_t=u_{xx}+u-u^3.$$ Then the professor wrote the equation as $$u_t=u_{xx}-W'(u),$$ where $$W(u)=1/4-u^2/2+u^4/4.$$ The free energy was defined as $$H[u]=\int_\Omega \left(\frac12 u_x^2+W(u)\right)\,\mathrm{d}x.$$ I do not understand what this free energy tries to model. I read on Wikipedia that the part $1/2\int_\Omega u_x^2\,\mathrm{d}x$ is called Dirichlet energy, but I cannot find any physical interpretation of this. Could you provide any simple explanation of $H[u]$ for a theoretical mathematician?

• $H$ is minimised when $u$ satisfies the AC equation. Thus, it measures how far $u$ is from the solution to the PDE. The first term penalises the derivative, and the second one the value of $u$ itself. Physicists would say "kinetic energy" and "potential energy", although I am not sure how useful these terms are for you. Is this what you are looking for? can you clarify what exactly are you expecting answers to look like? are you perhaps asking for an explicit "real-life" situation the AC models? – AccidentalFourierTransform Jan 5 '18 at 17:23
• @AccidentalFourierTransform Thank you for your comment. I computed $\frac{d}{dt}H[u]=-\int_\Omega u_t^2\,dx$, when $u$ satisfies AC equation. I also computed the first variation of $H[u](t)$ for each $t$. In both cases, I obtained equilibria if $u_{xx}-W'(u)=0$. What do you mean by minimizing $H$? Also, the kinetic energy here has something to do with $1/2\,mv^2$? What is the mass and velocity in the Dirichlet energy? I learned that minus gradient of the potential is the force. Is this concept applicable here? – user39756 Jan 5 '18 at 18:06
• Sorry, I misread the functional $H$. My comment above is mostly wrong. – AccidentalFourierTransform Jan 5 '18 at 18:21

$H$ is like a potential and the system will always "roll" to smaller values of $H$ (unlike in Hamiltonian mechanics but much rather as in "gradient flows"), i.e. $H$ is a Lyapunov function of the Allen-Cahn dynamics: it decreases in time unless a stationary condition is reached. In this case, the time-evolution will try to smoothen inhomogeneities (i.e. decrease $(\partial_x u)^2$) and there will be an attraction towards the values $u=\pm 1$ (since those are the minima of $W$). Depending on the initial conditions, a minimum of $H$ may or may not be reached (If the initial data are e.g. $u(t=0,x\geq0)=1=-u(t=0,x<0)$, then the resulting solution can not reach a minimum of $H$.
• Thank you for your answer. I have a question. As I said in a previous comment, equilibrium is obtained if $0=u_{xx}-W'(u)=u_{xx}+u-u^3$. We may view this as an autonomous system of ODEs $\dot{X}=Y$, $\dot{Y}=X^3-X$, and plot its dynamics on a plane with axes $X=u$ and $Y=u_x$. Let $\tilde{H}(X,Y)=Y^2/2+W(X)$ be the function inside the integral of $H$. As $X_0=\pm1$ is a minimum of $W$, then $(\pm 1,0)$ is a stable fixed point (the attraction you mentioned towards $\pm1$), by considering the Lyapunov function $V=\tilde{H}-\tilde{H}(\pm 1,0)$. My question is... – user39756 Jan 6 '18 at 16:26
• ... why $H$ is called a Lyapunov function, when actually $\tilde{H}$ (well, $V$) is the function being a Lyapunov function for a system of ODEs. Is there any concept of Lyapunov function for PDEs? – user39756 Jan 6 '18 at 16:26
• And finally, on another subject, why is the part $1/2\int_\Omega u_x^2 dx$ usually called Dirichlet energy? Where is the mass or velocity so that it could be interpreted as a kinetic energy? And why is $W$ called a potential there? Is its gradient some sort of force? – user39756 Jan 6 '18 at 16:31
• I think that in the physics community the nomer "Lyapunov function" is indeed used more broadly: In some dynamical system (in this case density profiles evolving according to Allen-Cahn) a function $f$ (domain=state space, codomain=R) is Lyapunov if it never increases in time (weaker version) or if it even has to decrease unless the system has arrived in a stationary state (=what you call equilibrium. This is a stronger version of the Lyapunov property). The Dirichlet energy is only formally analogous to kinetic energy (in that derivatives $\partial_x$ act on the field in this term). – Thibaut Demaerel Jan 6 '18 at 22:11
• in the term $W$ no derivatives act on the field $u$. That's why people will think and speak of this as a potential. Anyway, these are just issues of terminology and -a priori- perhaps not so important. – Thibaut Demaerel Jan 6 '18 at 22:13