I have to solve the following exercise:
Let $X=\{u\in C^{1}(-1,1)\cap C[-1,1]:u(\pm 1)=0\}$ and consider the length and the area functionals $F,G:X\rightarrow\mathbb{R}$: $$ F(u)=\int_{-1}^{1}\sqrt{1+u'(x)^{2}}dx $$ $$ G(u)=\int_{-1}^{1}u(x)dx $$ For a fixed constant v>0 consider the minimum problem: $$ \min\{F(u):u\in X \ \text{such that} \ G(u)=v\} $$ i) Given $\varphi,\phi\in C_c^\infty(-1,1)$ define $H:\mathbb {R}^2\to \mathbb {R}$ as $H(\epsilon,\tau)=G(u+\epsilon\varphi+\tau\psi)$.
Show that there exist $\psi$ and a function $\tau(\epsilon)$ such that $H(\epsilon, \tau (\epsilon)) = v$ for all $\epsilon$ .
This function depends on $\varphi$.
ii) Let u be a minimizer for the above problem.
Prove that there exists $\lambda \in \mathbb{R}$ (the Lagrange multiplier, a constant curvature!) such that $$\frac{d}{dx}\left(\frac{u'}{\sqrt{1+u'^2}}\right)=\lambda $$
Here the advice is to use the fact that $$\frac{d}{d\epsilon}|_{\epsilon=0}F(u+\epsilon\varphi+\tau\psi)=0$$ which is nothing but the EL equation in the weak form.
iii) Find the solution of the minimum problem
I have done parts 1 and 3 (1 as following) but I miss how to prove ii).
Any help would be greatly appreciated
i) $$v=H(\epsilon,\tau(\epsilon))=G(u+\epsilon\varphi+\tau\psi)=$$$$\int_{-1}^1udx+\int_{-1}^1\epsilon\varphi dx+\int_{-1}^1\tau(\epsilon)\psi dx$$ $$=v+\epsilon \int_{-1}^1 \varphi dx+\tau(\epsilon)\int_{-1}^1\psi dx $$ and we can now impose $\int_{-1}^1 \psi dx=1$ to get $$\tau=-\epsilon\int_{-1}^1\varphi dx $$
I know the problem is the same as Exercise in calculus of variation but in the latter the last two points are missing.
Choose $\tau=-\epsilon\int_{-1}^1\varphi dx,\ \int_{-1}^1\psi=1 dx$ as (i), then $H(\epsilon,\tau(\epsilon))=v$ for all $\epsilon$,
\begin{align*}0=\frac{d}{d\epsilon}|_{\epsilon=0}H(\epsilon,\tau(\epsilon)) &=\int_{-1}^1\frac{(u'+\epsilon \varphi'+\tau(\epsilon)\psi')(\varphi'-\psi'\int_{-1}^1\varphi dt)}{\sqrt{1+(u'+\epsilon \varphi'+\tau(\epsilon)\psi')^2}}dx|_{\epsilon=0}\\ &=\int_{-1}^1\frac{u'}{\sqrt{1+u'^2}}(\varphi'-\psi'\int_{-1}^1\varphi dt)dx,\ \forall\varphi,\psi\in C_c^\infty(-1,1). \end{align*} which implies $\frac{d}{dx}(\frac{u'}{\sqrt{1+u'^2}})=\lambda$ for some constant $\lambda$ as following ($g=\varphi-\psi\int_{-1}^1\varphi dt$).
Lemma: If $f(x)\in C^1(-1,1)$ satisfies $\int_{-1}^1f(x)g'(x)dx=0,\ \forall g(x)\in C_c^\infty(-1,1)$ $s.t.\int_{-1}^1g(x)dx=0$, then we have $f'(x)=constant$.
(since $0=\int_{-1}^1f(x)g'(x)dx=f(x)g(x)|_{-1}^1-\int_{-1}^1f'(x)g(x)dx=-\int_{-1}^1f'(x)g(x)dx$)