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In the text "Function Theory of a Complex Variable" the intuition behind the Poisson Integral Formula is given as follows: The Poisson Integral formula shows how to calculate a harmonic function on the disc from it's "boundary values" that is it's values on the circle that bounds the disc.

Theorem (7.3.3)

Let $u \rightarrow \mathbb{R}$ be a harmonic function on a neighborhood of the closed disk $D(0,1)$. Then for any point $ a \in D(0,1)$ $u(a)=\frac{1}{2 \pi}\int_{0}^{2 \pi}u(e^{i \phi} \cdot \frac{1 - |a|^{2}}{|a-e^{i \phi}|^{2}}d \phi$

My initial attack on deepening the intuition for the poisson integral formula that there exists Open Set $U$ such that there is a Closed Disk where $u$ is harmonic within the neighborhood of that closed disk, then for any point within the open disk on can calculate values of the harmonic function $u$ but bounded by the closed disk D(0,1).

Note: Basically it looks like the closed disk bounds where our function is harmonic within the open set U, so when one calculates the harmonic function (i.e $u(a)=\frac{1}{2 \pi}\int_{0}^{2 \pi}u(e^{i \phi} \cdot \frac{1 - |a|^{2}}{|a-e^{i \phi}|^{2}}d \phi$ everything is bounded by the closed disk D(0,1)

Is my intuition correct and rigour correct or do I need to be corrected also I noticed that the poisson integral can be expressed in a more convenient form.

Also is there a better way to establish/define there's a bound when one calculates harmonic functions via the Poisson Integral Formula ?

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  • $\begingroup$ To produce Latex-formatted stuff, enclose it in dollar signs, like $x^2$, which produces $x^2$. Double-dollar-signs around an equation make it a displayed equation. $\endgroup$ – John Hughes Jul 28 '17 at 19:48
  • $\begingroup$ ohh sorry I copied this from /r/math :(, i'll have to edit. $\endgroup$ – Zophikel Jul 28 '17 at 19:56
  • $\begingroup$ I don't really understand your question, but in case it helps: The value at a point $x$ of the hyperbolic plane of the harmonic extension of a (say) continuous function $u$ on the circle at infinity, is the average of $u$ as you see it from $x$. (You see it as a function on the unit circle of the tangent plane at $x$.) $\endgroup$ – Pierre-Yves Gaillard Jul 28 '17 at 20:11
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I'm not exactly sure what you are asking, but I'm going to give a, at times, hand-wavy "proof" of Poisson integral formula which demystifies it and makes it apparent why it works.

OK let me first write the statement of the problem (since there are many issues with the text of the question). Given $0\leq a<1$, the Poisson kernel $P_a:S^1\to \mathbb{C}$ ($S^1$ being the unit circle parametrized by $0\leq \theta<2\pi$) is defined as $$ P_a(\theta)=\sum_{n=-\infty}^\infty r^{|n|}e^{in\theta}=\frac{1-r^2}{1-2r\cos\theta +r^2}=\frac{1-r^2}{|1-re^{i\theta}|} $$ Now if $u:D(0,1)\to \mathbb{R}$ is harmonic, then $$ u(ae^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi P_a(\theta-\phi)u(e^{i\phi})d\phi $$ The question of is, I think, what is the intuition behind this?


First, we need to ask what does a harmonic function in $D(0,1)$ look like? Writing $u(r,\theta)$ instead of $u(re^{i\theta})$ and working with polar coordinates of $\mathbb{R}^2$, we find that $$ 0=\Delta u=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)+\frac{1}{r^2}\frac{\partial^2 u}{\partial \theta^2} $$ A special class of harmonic functions are separated, i.e. $u(r,\theta)=R(r)\Theta(\theta)$. With this assumption the PDE becomes two distinct ODEs, depending on a parameter $\lambda$ $$ r^2R''+rR'+\lambda R=0, \qquad \Theta''=\lambda\Theta $$ Now since $u(r,\theta)$ is periodic for $\theta\to \theta+2\pi$, we must have $\lambda=-n^2$, and $$\Theta_n(\theta) = a_n e^{in\theta}+a_{-n} e^{-in\theta},\qquad \Theta_0(\theta)=a_0$$ The radial equations $$r^2R''+rR'-n^2 R=0$$ has solution (which you can check easily), $c_n r^{|n|} +d_{n}r^{-|n|}$ for $n\neq 0$ and $c_0+d_0\ln r$ for $n=0$. However we do not want the solutions to blow up at the origin, so we put $d_n=0$. This gives the full solution as $$ u(re^{i\theta})=\sum_{n=-\infty}^\infty a_n r^{|n|}e^{in\theta} $$


Note how similar this separable solution is to Poisson kernel $P_r(\theta)$! Finding $u$, is now, basically reduced to finding the coefficients $a_n$. But $$ a_n=\frac{1}{2\pi}\int_{-\pi}^\pi u(e^{i\phi}) e^{-in\phi}d\phi $$ Therefore $$ u(re^{i\theta})= \sum_{n=-\infty}^\infty \int_{-\pi}^\pi r^{|n|}e^{in(\theta-\phi)}d\phi $$ Since convergence is not an issue (I'm being sloppy here. You asked for intuition, so I guess sloppy is fine), we can swap the integral and the infinite sum, which gives you $$ u(re^{i\theta})=\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta-\phi)u(e^{i\phi})d\phi $$ So this is a (sorta) proof. But it also gives some insight: I assumed $u(re^{i\theta})$ is a separable harmonic function, and I also hand-waved my way around why the sum and integral commute. But other than that, as you can see, there is no mystery in Poisson integral formula. The reason it works is due to the special form the harmonic functions on a unit disk have to take. Also note that if the disk is not unit, then the Poisson kernel (the sum) does not converge, and this whole method becomes nonsensical.

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  • $\begingroup$ Hmmm I'll have to clarify my question essentially what I was trying to get was if my intuition was correct on understanding how the Poisson Integral Formula calculates values of a harmonic function on it's boundary values. One thing i'll have to figure out is how to derative bounds for the poisson integral and the possion integral formula for a region in $R^{1}$ $\endgroup$ – Zophikel Jul 28 '17 at 21:35

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