I have heard that the gradient flow of the Dirichlet energy gives a solution of the heat equation, i.e. if $u(t,x) \in C^1( [0,\infty) \times \mathbb R^d)$ solves
$$ u_t(t,x) = – dE(u(t,x)), $$
where $$ E(u) := \dfrac{1}{2} \int \|\nabla_x u\|^2, $$
then
$$ \partial_t u = \Delta_x u := \sum_i \partial_{x_i,x_i}^2 u. $$
As I am trying to prove this fact, I have some problems in calculating the Frechet derivative of the Dirichlet energy $dE$.
If I understand correctly, fixed $f$, $dE$ is a linear operator from $C^1$ into $\mathbb R$ which should satisfy
$$ \lim_{\|h\|_{C^1} \to 0} \dfrac{E(f+h) – E(f) – dE(h) }{\|h\|_{C^1}} =
\lim_{\|h\|_{C^1} \to 0} \dfrac{(\int 2 \nabla f \cdot \nabla h + \nabla h \cdot \nabla h) – dE(h)}{\|h\|_{C^1}} = 0 ,$$
but how to find it?
Or should I not compute the differential of $E$ at all?
Best Answer
Perhaps it would be wiser to use the following expansion $$E(f+h)=E(f)+dE(h)+o(h),$$ from what you have calculated, we see that $$\int \nabla f\cdot\nabla h+\frac{1}{2}\nabla h\cdot\nabla h=E(f+h)-E(f)=dE(h)+o(h).$$ Since the the last term on the LHS is $o(h)$, we obtain $dE(h)=\int\nabla f\cdot\nabla h$. Note that this is the Frechet derivative at the function $f$, we then attribute the Frechet derivative to be the action $\langle\nabla f,\cdot\rangle:C^1\to\Bbb R$.