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.

  • $\begingroup$ 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

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.)

  • $\begingroup$ "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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.