2

I am trying to prove that $u(re^{i\theta}) = \sum\limits_{n=-\infty}^{\infty}\hat{f}(n)r^{|n|}e^{in\theta}$ is the solution to the Dirichlet problem on the unit disk if on the boundary of the unit disk, we have the Fourier series expansion of the continuous function $f(\theta) = \sum\limits_{n=-\infty}^{\infty}\hat{f}(n)e^{in\theta}$. I want to show that the solution found this way is harmonic.

I have shown that the Laplace equation $\Delta u = \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0$ may be rewritten in polar coordinates as $\Delta u = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}$.

I have two concerns. First, if I set $f(\theta) = e^{i\theta}$ or $f(\theta) = e^{-i\theta}$, then I find that $\Delta u$ is not zero.

Secondly, when I want to check generally that $\Delta u = \Delta \sum\limits_{n=-\infty}^{\infty}\hat{f}(n)r^{|n|}e^{in\theta} = 0$, is there a good way to justify that I can bring the Laplacian into the sum?

Justin
  • 23

1 Answers1

2

You have to check that $$ r^{|n|} e^{in\theta} $$ is harmonic, to include all the $r$ dependence in each term. This is easy enough, since the $r$ derivatives give $$ \frac{\partial^2}{\partial^2 r} (r^{|n|})e^{in\theta}+\frac{1}{r}\frac{\partial}{\partial r} (r^{|n|})e^{in\theta} = e^{in\theta}(|n|(|n|-1) r^{|n|-2}+ |n| r^{|n|-2}) = n^2 r^{|n|-2} e^{in\theta}, $$ and the $\theta$ derivatives give $ -n^2 r^{|n|-2} e^{in\theta} $, which cancel with it.

As far as taking the Laplacian inside the integral, this is fine provided the series is uniformly convergent in the neighbourhood of the point of evaluation. For a continuous function on $S^1$, you can check that the Fourier coefficients are bounded by the integral of $|f|$ over $S^1$. Hence you can bound the tail of the series with a geometric series with common ratio $R$, for $r<R<1$, and the sum is uniformly convergent on closed discs of radius $R<1$. Basically, you then have to use this theorem, or a slight generalisation to multiple variables.

Chappers
  • 67,606
  • Just checking a thought. Is the idea that I can use the Weierstrauss M-test to show the absolute convergence of the partial derivatives of the terms of the sum? Also, thank you for the well-written answer. – Justin Apr 15 '15 at 10:49
  • Yes, that's basically what I'm doing, although you have to check it for $f,f_r,f_{rr}$ and $f_{\theta},f_{\theta\theta}$ rather than just $f$ (see the link). – Chappers Apr 15 '15 at 10:50