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Suppose we have a minimization problem $F(u(x))=\int_A \sqrt{1+|\nabla u|^2}dA$, where $A$ is a ring such that $1<|x|<2$. After going to the polar coordinates we obtained: $$ F(\varphi(r,\theta))=\int_A \sqrt{1+\varphi_r^2+\frac{1}{r^2}\varphi_\theta^2}\ rdrd\theta $$ and now because of the symmetry the solution is independent of the angle $\theta$, but I do not know how to prove it rigorously to get: $$ F(\varphi(r))=2\pi\int_1^2\sqrt{1+\varphi_r^2}\ rdr $$

I suppose that the idea is to show that if we add some variation of the angle it will make things worse:

$$ F(\varphi(r))\leq F(\varphi(r)+d(\varphi(\theta))) $$ But I do not know if it is a way to do this and how to proceed.

Upd: To clarify. I understand that if $\varphi$ is independent of $\theta$, then we have $F(\varphi(r))=2\pi\int_1^2\sqrt{1+\varphi_r^2}\ rdr$, but I do not know how to prove that it is indeed independent, even though it is kind of obvious

KeepKolmogorov
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  • Shouldn't it be $\mathrm{d}A = \mathrm{d}x,\mathrm{d}y$ in the fisrt integral before using polar coordinates ? Otherwise $F[u(x)]$ is a line integral instead of a surface integral. – Abezhiko Mar 26 '23 at 11:14
  • If $\varphi$ is independent of $\theta$ thanks to symmetry, then $\varphi_\theta = 0$ and $\int_0^{2\pi}\mathrm{d}\theta = 2\pi$. – Abezhiko Mar 26 '23 at 11:16
  • yes, you're right, it is $dxdy$, I shorted the notation as $x=(x_1,x_2)$ – KeepKolmogorov Mar 26 '23 at 11:17
  • I don't understand how rigorous is "thanks to symmetry", I also have this intuition but my question was -- how to prove it without "thanks to symmetry" in a more precise way – KeepKolmogorov Mar 26 '23 at 11:19
  • To clarify. I understand that if $\varphi$ is independent of $\theta$, then we have $F(\varphi(r))=2\pi\int_1^2\sqrt{1+\varphi_r^2}\ rdr$, but I can not prove that it is indeed independent, even though it is kind of obvious – KeepKolmogorov Mar 26 '23 at 11:22
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    Little warning : it should also be $\frac{\phi_\theta^2}{r^2}$ instead of $\frac{\phi_\theta}{r^2}$. Then, since the integrand is positive, assuming $\phi_\theta = 0$ to suppress this term permits to lower the final result. A more systematic approach would use the Euler-Lagrange equations associated to the Lagrangian $L = r\sqrt{1+\phi_r^2+\frac{\phi_\theta^2}{r^2}}$. – Abezhiko Mar 26 '23 at 11:44
  • You are indeed right, I made a typo, thanks! – KeepKolmogorov Mar 26 '23 at 11:55

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