The integration of radially symmetric function

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I don't think this is true. Take $v'(r)\equiv 1$, $c_1=1$, $c_2=0$. In polar coordinates, the integral $(1)$ becomes $$\int_0^1 r\,dr \int_0^{2\pi} \frac{\cos \theta}{1+\sqrt{2+2\cos\theta}} d\theta $$ The inner integral evaluates to $\approx -0.6623$, so the whole thing is negative.

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Updated on August 01, 2022

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  • spatially
    spatially over 1 year

    Suppose $u\in C^\infty(\overline{B(0,1)})$ is radially symmetric. i.e., there exists a function $v$: $\mathbb R^+\to\mathbb R$ such that $u(x)=v(|x|)$. Here we take $B(0,1)\subset \mathbb R^2$.

    Notice that $|\nabla u(x)|=|v'(x)|$. And we use $x=(x_1,x_2)$ where $x_1=r\cos\theta$ for $r=|x|$.

    My question is: How we prove that $$ \int_{B(0,1)} \frac{{v'(r)\frac{x_1}{|{x}|}}}{|{v'(r)}|+\sqrt{(v'(r))^2+c_1^2+c_2^2+2[{v'(r)\frac{x_1}{|{x}|}c_1+v'(r)\frac{x_2}{|{x}|}c_2}]}}dx =0 \tag 1$$

    Here $c_1$ and $c_2$ are two arbitrary constants.

    Also, the denominate comes from $$|{v'(r)}|+\sqrt{(v'(r))^2+c_1^2+c_2^2+2[{v'(r)\frac{x_1}{|{x}|}c_1+v'(r)\frac{x_2}{|{x}|}c_2}]} =\int_{B(0,1)} \sqrt{(\partial_1u+c_1)^2+(\partial_2u+c_2)^2}\,dx $$

    Notice that it is very quick to prove that $$ \int_{B(0,1)} {v'(r)\frac{x_1}{|{x}|}}=0 $$ by doing integration with respect to $x_1$ first. But I still have trouble on how to prove $(1)$...

    Any help is really welcome!

  • spatially
    spatially over 8 years
    You are right...now I need to find out another way... Thank you!