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I made a question:

Calculus itegral on Riemann surfaces

So I thought a bit about the solution and would like some help on what I did:

Following the hint,

$\int_{\partial \Omega} i \space \partial\Phi =\int_{\partial\Omega} <grad \space\Phi,v>$.

Therefore, by Stokes' Theorem,

$\int_{\partial\Omega} <grad \space\Phi,v>= \int_\Omega div\space(\Phi_x,\Phi_y,\Phi_z)$.

Now I need to use that $\Phi$ is a real-valued function which is positive on and vanishes on the boundary of $\Omega$ to justify that $\int_\Omega div\space(\Phi_x,\Phi_y,\Phi_z)\ge 0$. And that would solve the problem. So if anyone can help me put the pieces together, I appreciate it.

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  • $\begingroup$ The divergence theorem (as you're writing it) applies only to surfaces that bound regions in $\Bbb R^3$. In the case of a 2-dimensional region, you get the double integral of the Laplacian of $\Phi$. But this won't help ... $\endgroup$ – Ted Shifrin Mar 1 '17 at 0:09
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There's no Stokes's Theorem involved in this. For one thing, you don't know anything about the Laplacian of $\Phi$ on the interior of $\Omega$. However, if you write everything out carefully in $\Bbb C$, you have \begin{align*} i\partial\Phi &= \frac i2\big(\frac{\partial\Phi}{\partial x}-i\frac{\partial\Phi}{\partial y}\big)\big(dx+i\,dy\big) \\ &= \frac12\left(i\big(\frac{\partial\Phi}{\partial x}\,dx + \frac{\partial\Phi}{\partial y}\,dy\big) - \big(\frac{\partial\Phi}{\partial x}dy - \frac{\partial\Phi}{\partial y}\,dx\big)\right) = \tfrac12\big(i\,d\Phi - \star(d\Phi)\big). \end{align*} The integral of the first term around the closed curve $\partial\Omega$ is $0$. The integral of the second term is the negative of the flux of $\text{grad}\,\Phi$ across $\partial\Omega$. Since $\Phi$ is positive on $\Omega$ and zero on $\partial\Omega$, $\text{grad}\,\Phi$ points inward everywhere (or is zero). Thus, the flux is negative, and we have our answer. (This can be extended to a Riemann surface by working with local (holomorphic) parametrizations. The geometry of the flux works out just fine.)

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  • $\begingroup$ "Thus, the flux is negative, and we have our answer". But, the flux Should not it be positive? $\endgroup$ – Manoel Mar 1 '17 at 0:28
  • $\begingroup$ Remember that we have the negative of the flux! $\endgroup$ – Ted Shifrin Mar 1 '17 at 0:29
  • $\begingroup$ Got it! OK! Thanks for your attention! $\endgroup$ – Manoel Mar 1 '17 at 0:30

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