I am reading a chapter of Evan's book on weak convergence methods for nonlinear PDE's p.49 and it states that the Euler Lagrange equation for the functional \begin{equation} I[w]:=\int_U|Dw|^2 \end{equation} (for some bounded, open domain, $U\subset \mathbb{R^n}$ for $n\ge 2$, with smooth boundary) on the set $$\mathcal{A}:=\{w\in W^{1,2}(U,\mathbb{R^m}) : |w|=1 \text{ a.e. and } w=g \text{ on } \partial U\}$$ is given by the solution to $$ \begin{cases} -\Delta u=|Du|^2u \\ |u|=1 \text{ a.e. } \end{cases} $$ in $U$.
My question is: how did they get this? I haven't really done much calculus of variations before but I think when you ignore that you are restricting the possible $w$'s the minimising function is just one with $\Delta u =0$ but I don't know how to use E-L for cases when the possible functions are restricted.
Also, is there a general way of dealing with these questions?
Thanks!
We want to prove that $$\int_U Du:Dv=\int_U|Du|^2u\cdot v,\ \forall\ v\in W_0^{1,2}(U;\mathbb{R}^m)\cap L^\infty(U,\mathbb{R}^m)$$
Assume that $u\in\mathcal{A}$ is a critical point of $I$. Take $v\in W_0^{1,2}(U;\mathbb{R}^m)\cap L^\infty(U,\mathbb{R}^m)$ and note that for small $t\in\mathbb{R}$, we have that $u+tv\neq 0$ a.e. in $U$. Define $$u_t=\frac{u+tv}{|u+tv|}$$
I - Show that for small $t\in\mathbb{R}$, $u_t\in W^{1,2}(U,\mathbb{R}^m)$,
II - Compute $$\frac{d}{dt}I[u_t]_{t=0}$$
To conclude, note that $\displaystyle\frac{d}{dt}I[u_t]_{|t=0}=0$.