Let $\Omega \subset\mathbb{R}^n$ be an open set and $$E(u)=\frac{1}{2}\int_{\Omega} | \nabla u|^2 \quad (u \in H_0^1 (\Omega)). $$ Then, what is the Fréchet derivative of the functional $E$? And why? (I want to show it directly...)


The Frechet derivative $DE$, if it exists, is unique and satisfies

$$E(u+h)=E(u)+DE(h)+r(h),\ $$ where $r(h)$ is $o(h).$ So, if we can find a candidate that satisfies the equation, we are done.

Claim (admittedly with the foreknowledge that the claim is true):

$$DE(h)=\int_{\Omega}\langle \nabla u,\nabla h\rangle$$

The proof is a calculation:

$$E(u+h)-E(u)=\frac{1}{2}\left (\int_{\Omega} | \nabla (u+h)|^2-\int_{\Omega} | \nabla (u)|^2\right )=\frac{1}{2}\left (\int_{\Omega} \langle\nabla (u+h),\nabla (u+h)\rangle-\int_{\Omega} | \nabla (u)|^2\right )=\int_{\Omega}\langle \nabla u,\nabla h\rangle+\frac{1}{2}\int_{\Omega}\langle \nabla h,\nabla h\rangle,$$

from which we see that, setting $r(h)=\frac{1}{2}\int_{\Omega}\langle \nabla h,\nabla h\rangle$ and noting that is is $o(h)$, we have

$$DE(h)=\int_{\Omega}\langle \nabla u,\nabla h\rangle.$$


it's $-\Delta u$ (as a functional...). Why, you may ask...

We require that $E(u+h)-E(u) = \langle \nabla E(u) , h \rangle$ for any $h \in H_0^1 (\Omega)$.

$E(u+h)-E(u) = \frac{1}{2}\int_\Omega 2 \nabla u \nabla h$. Now use integration by parts on this expression to get the answer.

We get that $\langle \nabla E(u) , h \rangle = \langle -\Delta u , h\rangle\>$. This tells us we can associate the functional $\nabla E(u)$ acting on a function $h$ with integrating $h$ times $-\Delta u$.

  • $\begingroup$ Thank you. I have a question. We need to show the limit of $\frac{E(h)}{||h||_{H_0^1}}$ as h approaches 0 in $H_0^1$ is 0. How is it showed? $\endgroup$
    – Ichiko
    Apr 7 '19 at 21:54
  • 1
    $\begingroup$ So we need to show that $E(h)$ decays to $0$ faster than $\int_\Omega h^2$ does. Can you think of how to do that? Hint: IBP and Cauchy-Schwarz $\endgroup$
    – fGDu94
    Apr 7 '19 at 22:03
  • $\begingroup$ Sorry, I can’t understand... $\endgroup$
    – Ichiko
    Apr 7 '19 at 23:42
  • $\begingroup$ Ok i will write answer in a little bit $\endgroup$
    – fGDu94
    Apr 8 '19 at 15:24

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