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Consider the integral:

$ h(r, \theta) = \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi, r<1$

I want to show that $\Delta h=0$, but in order to do so, I need to justify this:

$\Delta \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi = \frac{1}{2\pi} \int_0^{2\pi} g(\phi) \Delta \frac{1-r^2}{1-2r \cos(\theta - \phi) + r^2} d\phi$

That way, this becomes $0$, as the Poisson kernel is harmonic.

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  • $\begingroup$ There's something not right about the integral, I think you have your letters mixed up $\endgroup$ – Ninad Munshi Jul 24 '20 at 3:30
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    $\begingroup$ probably $d\phi$ so that its a convolution integral $\endgroup$ – Calvin Khor Jul 24 '20 at 3:36
  • $\begingroup$ Whoops,... yeah it's $d\phi$ $\endgroup$ – Kay Jul 24 '20 at 3:36
  • $\begingroup$ It looks a bit like the Leibniz integral rule. It's just the $\phi$-component of $\Delta$ which seems a bit fishy to pull into the integral. $\endgroup$ – Vercassivelaunos Jul 24 '20 at 10:14
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I think I solved the problem. Seems like I can somehow relate the function to a regular analytic complex function, as shown in the last few pages of this: https://www.math.uh.edu/~shanyuji/Complex/Notes/N27.pdf

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