0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It may be obvious but I find it interesting to point out that in this and many others problems in calculus of variation is worth reflecting on if there are some obvious solutions to the problem. As an example we know by exercience that circles (and in general spheres) are the optimal shape in minimizing the boundary after that we fixed the area/volume/n-volume $\endgroup$ – Alain Ngalani Jun 5 '20 at 20:02
1
$\begingroup$

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 $0=\int_{-1}^1f(x)g'(x)dx=f(x)g(x)|_{-1}^1-\int_{-1}^1f'(x)g(x)dx=-\int_{-1}^1f'(x)g(x)dx$)

$\endgroup$
  • $\begingroup$ Thank you very much $\endgroup$ – Frankie123 Apr 25 '20 at 15:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.