TITLE: Proof that a minimum problem has no solution
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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 $|x|$.