hello $$$$ I am trying to find explanation how to derive cahn hilliard equation:
$$ u_t =\Delta (w'(u)-\epsilon ^2 \Delta u)$$ as gradient flow of energy functional $$E[u]=\int w(u)+\epsilon ^2 |\nabla u|^2.$$ I tried to follow the definition of gradient flow from this blog but I got stuck.
I read that it is $ H^{-1} $ gradient flow of the functional. can anyone tell me what is $ H^{-1} $ gradient flow? thanks.
Let $H$ be a Hilbert space and $F :H \to \mathbb R$. Suppose that $F$ is Gâteaux differentiable at $u \in H$, then gradient of $F$ at $u$, denoted by $\text{grad}F(u)$, is given by the unique $w \in H$ which satisfies $$\langle F'(u), v \rangle_H = \langle w, v \rangle_H \quad \forall v \in H.$$ When we restrict $v$ to some set $X \subset H$, we have the so-called constraint gradient. Then we usually write $\text{grad}_H^X F(u)$ to denote the unique element of the following set $$\left\{ f \in H : \quad \lim_{t \to 0} \frac 1t \big( F(u+tv) - F(u) \big) = \langle f, v \rangle_H \quad \forall v \in X\right\}$$ least norm. Hence, for each $X$ and $H$, we have certain flows which mainly depend on how we calculate $\text{grad}_H^X F(u)$.
For example, let say $$F(u) = \frac 12 \int_\Omega |du|^2 dx.$$ Clearly, with $X = C_0^\infty (\Omega)$, we have $$\text{grad}_{L^2}^X F(u) = -\Delta u$$ while we have $$\text{grad}_{H^1}^X F(u) = u$$ and $$\text{grad}_{H^{-1}}^X F(u) = \Delta^2 u.$$ In other words, the equation $$u_t = -\Delta u$$ will be a $L^2$-gradient flow associated to the energy functional $F$ above. However, it is no longer the $H^{-1}$-gradient flow associated to the energy functional $F$ but the following equation $$u_t = \Delta^2 u.$$
to get first a Cahn-Hilliard system first we write the mass conservation i.e. $\dfrac{\partial u}{\partial t}=-h_x$, where $h$ is the mass flux which is related to the chemical potential $\mu$ by a constitutive relation $h=-\mu_x$, and that the chemical potential $\mu$ is a variational derivative of $\Psi$ with respect to $u$, we end up with the sixth-order Cahn-Hilliard system:
\begin{eqnarray} \dfrac{\partial u}{\partial t}=\Delta \mu \end{eqnarray} \begin{eqnarray} \mu=\dfrac{\delta E}{\delta u} \end{eqnarray}
now we will find the variational derivative of $E$ w.r.t $u$
\begin{eqnarray*} \delta E&=& \int \left[w'(u) \delta u+2\epsilon^2 \nabla u \cdot \delta(\nabla u)\right]\\ &=&\int \left[w'(u) \delta u+2\epsilon^2 \nabla u\nabla(\delta u)\right]\\ &=& \int \left[w'(u) \delta u-2\epsilon^2 \Delta u \delta u\right]\\ &=&\int \left[w'(u)-2\epsilon^2 \Delta u\right]\delta u \end{eqnarray*}
so we get $$\mu=w'(u)-2\epsilon^2 \Delta u$$
i.e. $$u_t=\Delta(w'(u)-2\epsilon^2 \Delta u)$$
