The Poisson Integral Formula

The Poisson integral formula provides an explicit solution to the Dirichlet problem on the disk and reveals the reproducing kernel structure of harmonic functions.

Statement

Theorem8.12Poisson integral formula

Let $u$ be harmonic on the open disk $D(0, R)$ and continuous on $\overline{D}(0, R)$ . Then for every $z = re^{i\phi}$ with $r < R$ :

$u(re^{i\phi}) = \frac{1}{2\pi}\int_0^{2\pi} P_r(\phi - \theta)\,u(Re^{i\theta})\,d\theta$

where the Poisson kernel is

$P_r(\psi) = \frac{R^2 - r^2}{R^2 - 2Rr\cos\psi + r^2} = \text{Re}\left(\frac{Re^{i\psi} + r}{Re^{i\psi} - r}\right).$

Proof

Step 1. By the Cauchy integral formula, for $|z| < R$ and $f = u + iv$ holomorphic:

$f(z) = \frac{1}{2\pi i}\oint_{|\zeta|=R}\frac{f(\zeta)}{\zeta - z}\,d\zeta.$

With $\zeta = Re^{i\theta}$ , $d\zeta = iRe^{i\theta}\,d\theta$ :

$f(z) = \frac{1}{2\pi}\int_0^{2\pi}\frac{Re^{i\theta}}{Re^{i\theta} - z}\,f(Re^{i\theta})\,d\theta.$

Step 2. The point $R^2/\bar{z}$ lies outside $|\zeta| = R$ , so by Cauchy's theorem:

$0 = \frac{1}{2\pi}\int_0^{2\pi}\frac{Re^{i\theta}}{Re^{i\theta} - R^2/\bar{z}}\,f(Re^{i\theta})\,d\theta = \frac{1}{2\pi}\int_0^{2\pi}\frac{\bar{z}e^{i\theta}}{\bar{z}e^{i\theta} - R}\,f(Re^{i\theta})\,d\theta.$

Step 3. Subtract the conjugate of Step 2 from Step 1. After simplification:

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

since only the real part of $f$ contributes (by taking real parts of both sides). The kernel is

$\text{Re}\frac{Re^{i\theta}+z}{Re^{i\theta}-z} = \frac{R^2 - |z|^2}{|Re^{i\theta}-z|^2} = P_r(\phi - \theta). \qquad \blacksquare$

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Properties of the Poisson Kernel

RemarkKey properties

The Poisson kernel $P_r(\psi)$ satisfies:

Positivity: $P_r(\psi) > 0$ for $r < R$ .
Normalization: $\frac{1}{2\pi}\int_0^{2\pi} P_r(\psi)\,d\psi = 1$ .
Approximate identity: $P_r(\psi) \to 0$ uniformly on $[\delta, 2\pi-\delta]$ as $r \to R^-$ for any $\delta > 0$ .
Fourier series: $P_r(\psi) = 1 + 2\sum_{n=1}^\infty (r/R)^n\cos(n\psi)$ .

Properties 1--3 make $P_r$ an approximate identity, ensuring that $u(re^{i\phi}) \to g(Re^{i\phi})$ as $r \to R^-$ at points of continuity of $g$ .

ExamplePoisson formula and Fourier series

If $u(Re^{i\theta}) = \sum_{n=-\infty}^\infty c_n e^{in\theta}$ is the Fourier series of the boundary data, then

$u(re^{i\phi}) = \sum_{n=-\infty}^\infty c_n\left(\frac{r}{R}\right)^{|n|} e^{in\phi}.$

Each Fourier mode is damped by the factor $(r/R)^{|n|}$ as we move inward. This explains why harmonic functions are smooth: high-frequency oscillations on the boundary are exponentially suppressed in the interior.

RemarkConnection to probability

The Poisson kernel can be interpreted probabilistically: $u(z)$ is the expected value of $g(B_\tau)$ where $B_t$ is a Brownian motion starting at $z$ and $\tau$ is the first exit time from the disk. This connection to stochastic processes extends to general domains.