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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?

user264319
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    It may help to know that in a Hilbert space, the Frechet derivative is an actual gradient, in the sense that its Riesz representation back in the underlying space is the steepest ascent direction at the point where you compute the derivative. – Ian Aug 25 '15 at 14:52

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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 $\langle\nabla f,\cdot\rangle:C^1\to\Bbb R$.

Ellya
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    Thank you! If I understand correctly, by integration by parts we have that $dE(u) = \int \nabla f \cdot \nabla u = - \int f \Delta u$, therefore by Riesz duality we can say that $\nabla E(u) = - \Delta u$, which is exactly the claim. – user264319 Aug 25 '15 at 20:06
  • @user264319 That's right, provided you have enough regularity to write the integration by parts. Otherwise $\nabla E(u) = -\Delta u$ only in the sense of distribution. – Ian Aug 26 '15 at 13:41
  • You may also need to be careful of the fact that Riesz representation theorem holds only in Hilbert spaces – Ellya Aug 26 '15 at 15:42
  • In this case, I suppose the right space to use is $W^{2,2}$ and not $C^2$ – user264319 Aug 26 '15 at 17:26