Showing that the Poisson Kernel for the unit disk over the interval $-\pi$ to $\pi$ is equal to 1 for $0<r<1$

Question

I know the Poisson Kernel is $$P(r,\xi)=\frac{1}{2\pi}\left(1+2\sum^{\infty}_{n=1}r^n\cos(n\xi)\right)$$ so $$U(r,\theta)=\int_{-\pi}^{\pi}P(r,\xi-\theta)f(\xi)d\xi$$ The Poisson's integral formula is$$U(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1-r^2}{1+r^2-2r\cos(\xi-\theta)}f(\xi)d\xi$$ Now, I'm not really sure how to show that the Poisson integral equals 1. I saw a proof that set $f(\xi)=1$, but I don't understand why one would have to do that.

Answer 1

It is not equal to 1 for general $f$. For each $f$, the function $U(r,\theta)$ is the solution of $$\tag{$\ast$}\begin{cases} \Delta U = 0, &\text{if } x^2+y^2<1\\ U = f, &\text{if } x^2+y^2=1 \end{cases} $$ where $\Delta$ is the Laplacian in cartesian coordinates: $$ \Delta = \frac{\partial^2}{\partial x^2}+ \frac{\partial^2}{\partial y^2}.$$

Momentarily forgetting about the Poisson kernel representation, when $f=1$, the solution to ($\ast$) can be shown to be $U=1$. This follows from the fact that $U=1$ is a solution and that solutions to ($\ast$) are unique. Returning to the Poisson kernel representation, this agrees with the fact that the integral in Poisson kernel representation can be directly computed when $f=1$ and is equal to $1$.

However, when $f$ is a general function, the solution to ($\ast$) is not $1$ (indeed, we have $U=1$ if and only if $f=1$). For example, when $f=0$ then $U=0$ (a fact that can be observed from both $(\ast)$ and the Poisson kernel representation), or when $f(x,y)=x$ then $U(r,\theta) = x=r \cos \theta $.

Answer 2

The Poisson kernel is not $1$. It is $$\operatorname{PK}(r,\theta)=\frac{1-r^2}{1+r^2-2r\cos\theta}$$ And functions of the form $$U(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^\pi \operatorname{PK}(r,\theta-\phi)f(\phi)\mathrm d\phi=\frac{1}{2\pi}\int_{-\pi}^\pi \operatorname{PK}(r,\phi)f(\theta-\phi)\mathrm d\phi$$ Satisfy the BVP $$\begin{cases}\Delta U(r,\theta)=0 & r<1 \\ U(1,\theta)=f(\theta)\end{cases}$$

As stated by @JackT When our boundary data is $f(\theta)=1$, we get $$U(r,\theta)=\frac{1}{2\pi}\int_{-\pi}^\pi\frac{1-r^2}{1+r^2-2r\cos(\phi)}\mathrm d \phi=1$$ Since $U=1$ is the only function satisfying the above boundary value problem with $f=1$. To show this we note that the Poisson kernel may be written as the real part of a meromorphic function: $$\operatorname{PK}(r,\theta)=\Re\left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right) $$ So we rewrite the above integral as $$U(r,\theta)=\frac{1}{2\pi}\Re\int_{-\pi}^\pi \frac{1+re^{i\phi}}{1-re^{i\pi}}\mathrm d\phi$$ Now let $z=re^{i\phi}\implies \mathrm dz=ire^{i\phi}\mathrm d\phi=iz\mathrm d\phi$ and switches integration to one over a circle centered at $0$ with radius $r<1$ in the complex plane: $$U(r,\theta)=\frac{1}{2\pi}\Re\int\limits_{\partial \mathbb B(0,r)}\frac{1+z}{1-z}\frac{1}{iz}\mathrm dz$$ Because $r<1$, the only singularity is the one at the origin, we can apply the residue theorem at only that point: $$U(r,\theta)=\frac{1}{2\pi}2\pi i\operatorname{Res}\left(s\mapsto \frac{1+s}{1-s}\frac{1}{is},0\right)= i\lim_{z\to 0}z\frac{1+z}{1-z}\frac{1}{iz}=i/i=1.$$

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A more useful problem is to show that the given solution agrees with the boundary data in the limit, in other words to show that $$\lim_{r\to 1}\int_{-\pi}^\pi\frac{1-r^2}{1+r^2-2r\cos\phi}f(\theta-\phi)\mathrm d\phi=f(\theta)$$ For suitable (continuous) $f$. This probably requires an application of the dominated convergence theorem or something... I'd have to take some time to think about it.