Proof that a minimum problem has no solution

Question

Please I have an exam in a few days, can you help me with the following exercise?

Let $A=\{x\in\mathbb{R}^2: 1<|x|<2\}$ and $M\geqslant 0$.

On the set $\mathcal{A}_{M}=\{u\in C(\bar{A})\cap C^1(A):u=0 \text{ on } |x| \text{ and } u=M \text{ on } |x|=1\}$ consider the area functional: \begin{equation*} F:\mathcal{A}_{M}\to[0,\infty] \end{equation*} \begin{equation*} F(u)=\int_A\sqrt{1+|\nabla u(x)|^{2}} \, dx \end{equation*}

I have to prove the following statement:

3) There exists $M_0$ such that for $M>M_0$ the minimum problem has no solution in the class $\mathcal{A}_M$

Solution: I've computed the Euler Lagrange equation in the strong form for the functional: \begin{equation*} F(\phi)= 2\pi\int_1^2 \sqrt{1+\phi'(r)}rdr \end{equation*} Obtained from the previous one using the fact that $u$ is radial and computing a change of variables with polar coordinates.

The explicit solution of the strong form of the Euler Lagrange equation is:

\begin{equation*} \phi(r)=\phi(1)-\int_1^r \dfrac{C}{\sqrt{t^2-C^2}}=M-\log\bigg({\dfrac{x+\sqrt{x^2-C^2}}{1+\sqrt{1-C^2}}}\bigg) \end{equation*}

$C$ is a real constant.

Can someone help me to find an appropriate concusion of this proof?

$\lvert X\rvert$ is the distance of the point $X=(x,y)$ from the origin? — Invisible, Apr 19 '20 at 15:46
"[...] probably the key is to prove that the functional F is invariant under rotation in the argument of u." That is correct. Do you know how to do that? And do you know how to proceed from there? — MaoWao, Apr 20 '20 at 07:20
If $O$ is an orthogonal $2\times 2$ matrix, then $F(u\circ O)=F(u)$. What does that tell us if $u$ is a minimizer? How can you use uniqueness of minimizers now? — MaoWao, Apr 20 '20 at 08:17
Why is the statement $F(u\circ O)=F(u)$ true? How can we prove it? — Pefok, Apr 20 '20 at 08:22
Use the chain rule for the derivative and the change of variables formula for the integral. — MaoWao, Apr 20 '20 at 10:38

score 3 · Answer 1 · answered Apr 21 '20 at 15:17

3

You're 95% there! You have successfully shown the following:

If a minimizer exists, it is unique
The area functional $F$ is rotationally invariant

Suppose that $u$ is a minimizer. Then, because $F$ is rotationally invariant, for any $O\in SO(2)$ the function $u\circ O$ is also a minimizer. By uniqueness, $u= u\circ O$ for all $O\in SO(2)$, i.e., $u$ is rotationally invariant, hence depends only on $|x|$.

answered Apr 21 '20 at 15:17

Neal

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thank you very much. – Pefok Apr 21 '20 at 15:57
Can you help me also with the third point? – Pefok Apr 24 '20 at 09:28
@AndreaStanghellini You should ask it as a separate question – Neal Apr 24 '20 at 17:30

Proof that a minimum problem has no solution

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