# 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...)

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 $$-\Delta u$$.

• 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? Apr 7 '19 at 21:54
• 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 Apr 7 '19 at 22:03
• Sorry, I can’t understand... Apr 7 '19 at 23:42
• Ok i will write answer in a little bit Apr 8 '19 at 15:24