Intuition for the Poisson kernel

The derivation of the Poisson kernel for a disc seems to involve a trick, and I don't really understand how one would come up with it.

Let $f$ be a holomorphic function on a disc $D_{R_0}$ centered at the origin and of radius $R_0$.

We are aiming for some sort of integral representation of $f$ with a real-valued kernel, which ends up to be

$$f(z) = \frac{1}{2\pi}\int_0^{2\pi} f(Re^{i\varphi}) \text{Re}\left(\frac{Re^{\varphi} + z}{Re^{i\varphi} - z}\right) \; d\varphi$$

So the trick used is to write

$$f(z) = \frac{1}{2\pi i} \int_{D_{R_0}} f(\zeta)/(\zeta-z) - f(\zeta)/(\zeta-R^2/\bar{z}) \; d\zeta$$

I'm not sure how one would come up with adding $f(\zeta)/(\zeta-R^2/\bar{z})$, except for the fact that it magically seems to work. I've tried working backwards, but it really hasn't given me any new insight into the problem.

Solution 1:

Let $f$ be holomorphic on the closed disc $\bar{D}_R$. Let $z\in D_R$. Then $$ f(z) = \frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\text{Re}\Bigl(\frac{Re^{i\theta} + z}{Re^{i\theta} - z}\Bigr)d\theta $$

Consider for simplicity where $z = re^{i\theta}$ $$ \frac{1 - r^2}{1 - 2r\cos(\theta - \phi) + r^2} = \frac{1 - \lvert z\rvert^2}{|z - e^{i\phi}\rvert^2} = \frac{e^{i\phi}}{e^{i\phi} - z}+ \frac{e^{-i\phi}}{e^{-i\phi} - \bar{z}} - 1 $$ Also, note that the space of Harmonic functions is generated by the identity $z\mapsto z$ and $z\mapsto\bar{z}$.

Proof

Let $z = re^{i\phi}$, and let $C_R$ be the circle of radius $R$, parametrized by $\zeta = Re^{i\theta}$ so $d\zeta = iRe^{i\theta}d\theta$. By Cauchy's Theorem, \begin{align} f(z) &= \frac{1}{2\pi i}\int_{C_R}\frac{f(\zeta)}{\zeta - z}d\zeta\\ &= \frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{Re^{i\theta}}{Re^{i\theta} - re^{i\phi}}d\theta\tag{1} \end{align} Now, let $z' = R^2/\bar{z} = e^{i\phi}R^2/r$. Therefore, $\zeta\mapsto\frac{f(\zeta)}{\zeta - z'}$ is holomorphic on $\bar{D}_R$. \begin{align} 0 &= \frac{1}{2\pi i}\int_{C_R}\frac{f(\zeta)}{\zeta - z'}d\zeta\\ &= \frac{1}{2\pi}\int_0^{2\pi}f(Re^{i\theta})\frac{re^{i\theta}}{re^{i\theta} - Re^{i\phi}}d\theta\tag{2} \end{align} Now subtract (1) and (2) to get $$ \frac{R^2 - r^2}{R^2 - 2rR\cos(\theta - \phi) + r^2} = \text{Re}\Bigl(\frac{Re^{i\theta} + re^{i\phi}}{Re^{i\theta} - re^{i\phi}}\Bigr) $$