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Let $r \in [0,1)$ and $\theta \in [-\pi,\pi]$ and define, $$P_r(\theta) = \frac{1-r^2}{2\pi(1+r^2-2r\cos(\theta))} = \frac{1}{2\pi}\sum_{n=-\infty}^{\infty}r^{|n|}e^{in\theta}.$$ Then I want to show that $\Delta P_r(\theta) = 0.$ Should I just use the fact that, $$\Delta f = {\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial f}{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}\\= {\frac {\partial ^{2}f}{\partial r^{2}}}+{\frac {1}{r}}{\frac {\partial f}{\partial r}}+{\frac {1}{r^{2}}}{\frac {\partial ^{2}f}{\partial \theta ^{2}}}$$

and just compute the relevant partial derivatives using the analytic expression of $P_r(\theta)$ as follows:

$$\frac{\partial^2}{\partial^2 r}\left(\frac{1-r^2}{2\pi \left(1+r^2-2r\cos \left(\theta \right)\right)}\right)+\frac{1}{r}\cdot \frac{\partial}{\partial r}\left(\frac{1-r^2}{2\pi \left(1+r^2-2r\cos \left(\theta \right)\right)}\right)\\+ \frac{1}{r^2}\cdot \frac{\partial^2}{\partial^2 \theta }\left(\frac{1-r^2}{2\pi \left(1+r^2-2r\cos \left(\theta \right)\right)}\right) \\=\frac{4\left(\cos ^2\left(θ\right)+\sin ^2\left(θ\right)-r^2\cos ^2\left(θ\right)-r^2\sin ^2\left(θ\right)+r^2-1\right)}{\pi \left(r^2-2r\cos \left(θ\right)+1\right)^3}\\=0?$$

Also, should I consider the case $r=0$ separately?

Student
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2 Answers2

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That will work eventually, but it's easier to apply $\Delta$ to each $r^{\left|n\right|}\exp{in\theta}$ term. Note in particular the $r$ derivatives multiply by $n^2$ (the first expression for $\Delta$ gets this faster), while the $\theta$ sector multiplies by $-n^2$.

J.G.
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  • But you have to justify differentiating the infinite series. How do you propose to do that? – Student May 12 '19 at 06:24
  • @model_checker $\partial_z f(z)$ has the same radius of convergence as $f(z)$. The proof is very technical. You can use the same techniques to address Laplacians of complex functions. This is worth knowing; but if such material is currently unfamiliar, you should probably just augment your existing strategy with user657324's observation. – J.G. May 12 '19 at 09:42
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Acctually, your calculation is fine :

use $$\cos^2(\theta )+\sin^2(\theta )=1$$ $$-r^2\cos^2(\theta )-r^2\sin^2(\theta )=-r^2,$$ and conclude.

user657324
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