Existance (of lack thereof) of minimizers of the area functional on $A_M=\{u\in C(\bar{A})\cap C^1(A):u=0 \; if\; x=2 \; u=M \;if \; x=1\}$

Question

I was trying to find to find the minimizer of the area functional $$F(u)=\int_A \sqrt{1+|\nabla u(x)|^2}dx $$ where $A=\{A=\{x\in\mathbb{R}^2: 1<|x|<2\}\}$ with the solution belonging to the set $\mathcal{A}_{M}=\{u\in C(\bar{A})\cap C^1(A):u=0 \text{ on } |x| \text{ and } u=M \text{ on } |x|=1\}$.

I actually want to prove that for $M>M_0$ the problem has no solution.

It all came to prove that for the minimizer u we need to have $u(x)=\varphi|x|=\varphi(r)$, id est the minimizer is radial, and the EL equation associated with the problem is $$\frac{d}{dr}\left(\frac{r\varphi'}{\sqrt{1+\varphi'^2}}\right)=0 $$ but now I'm stuck since I don't know how to solve this.

Any help is appreciated.

Answer 1

We can explicitate the function $\varphi(r)$ $$ \begin{split} \frac{d}{dr}\left(\frac{r\varphi'}{\sqrt{1+\varphi'^2}}\right)&=0\\ \implies \frac{r\varphi'}{\sqrt{1+\varphi'^2}} &=c\\ \implies r^2\varphi'^2 & =c^2+c^2\varphi'^2\\ \implies \varphi'&=\frac{c}{\sqrt{r^2-c^2}}\\ \implies \varphi &=c \ln\left(\left|\frac{\sqrt{r^2-c^2}+r}{c}\right|\right)+d. \end{split}$$ We can now impose the boundary condition $$ \begin{split} 0= \varphi(2)&=c \ln\left(\left|\frac{\sqrt{4-c^2}+2}{c}\right|\right)+d\\ \implies d&=-c ln\left(\left|\frac{\sqrt{4-c^2}+2}{c}\right|\right)\\ M =\varphi(1) &=c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{c}\right|\right)-c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{c}\right|\right)\\ \implies M &=c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{\sqrt{4-c^2}+2}\right|\right) \end{split} $$ Now we notice that for the function to be real valued we need $|c|\le 1$ and so we can conclude that for the solution to exist the following inequality has to be satisfied $$ \inf_{c\in[-1,1]}c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{\sqrt{4-c^2}+2}\right|\right)\le M\le \sup_{c\in[-1,1]}c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{\sqrt{4-c^2}+2}\right|\right)$$ but actually since $c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{\sqrt{4-c^2}+2}\right|\right)$ in regards of the variable $c$ is in $C([-1,1])$ we have that the minimum and maximum are attained (and so they are finite) and so we can write $$ \min_{[-1,1]}c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{\sqrt{4-c^2}+2}\right|\right)\le M\le \max_{[-1,1]}c \ln\left(\left|\frac{\sqrt{1-c^2}+1}{\sqrt{4-c^2}+2}\right|\right).$$

So if $M$ is outside this interval then the minimum problem for the area functional has no solution.