Fréchet derivative of the energy functional 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...)
 A: 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 
$$DE(h)=\int_{\Omega}\langle \nabla u,\nabla h\rangle.$$
A: 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 $-\Delta u$.
