How to show $\int_{-\pi}^{\pi} P(r,\theta-t)P(s,t)dt=P(rs,\theta)$

Question

I've posted the same question before. Although some user's hint, I cannot found solution. In fact, I just know vary basic concept of complex analysis.(ex, Euler formula) So If you need to use the concept of complex analysis. Could you explain some detail about solution. Thank you...

Prove that $\int_{-\pi}^{\pi} P(r,\theta-t)P(s,t)dt=P(rs,\theta)$ where $P(r,t)=\frac{1}{2\pi}\frac{1-r^2}{1-2rcos(t)+r^2}$is Poisson's kernel. HINT. Use the series expansion.

Answer 1

Use the fact that $P_r {x}$ is the sum of $r^{|n|} e^{i n x}$ over all integers n, the series converging uniformly in x for fixed $r<1$. Term by term integration is permitted and you get the equation immediately.

Answer 2

The Poisson kernel comes from solving the Laplace equation in the unit disk with prescribed boundary values on the unit circle. The solutions of $\nabla^2 F=0$ obtained by separation of variables in polar coordinates have the form $$ r^{|m|}e^{im\theta},\;\; m=0,\pm 1,\pm 2,\cdots. $$ The general such solution is $u(r,\theta)=\sum_{m=-\infty}^{\infty}A_mr^{|m|}e^{im\theta}$, where the constants $A_m$ are chosen to satisfy a prescribed boundary condition $u(1,\theta)=f(\theta)$, which takes the form of a Fourier series expansion: $$ f(\theta)=\sum_{m=-\infty}^{\infty}A_me^{im\theta}. $$ This gives $$ A_m = \frac{1}{2\pi}\int_{0}^{2\pi}f(\theta')e^{-im\theta'}d\theta',\;\; m=0,\pm 1,\pm 2,\cdots. $$ Therefore, $$ F(r,\theta)=\frac{1}{2\pi}\int_{0}^{2\pi}f(\theta')\sum_{m=-\infty}^{\infty}r^{|m|}e^{im(\theta-\theta')}d\theta' $$ The Poisson kernel is $$ P(r,\theta)=\frac{1}{2\pi}\sum_{m=-\infty}^{\infty}r^{|m|}e^{im\theta}. $$ Using this form, for $r,s < 1$, \begin{align} &\int_{0}^{2\pi}P(r,\theta-\theta')P(s,\theta')d\theta' \\ & = \frac{1}{(2\pi)^2}\int_{0}^{2\pi}\sum_{m=-\infty}^{\infty}r^{|m|}e^{im(\theta-\theta')}\sum_{n=-\infty}^{\infty}s^{|n|}e^{in\theta'}d\theta' \\ & =\frac{1}{(2\pi)^2}\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}r^{|m|}s^{|n|}e^{im\theta}\int_{0}^{2\pi}e^{i(n-m)\theta'}d\theta' \\ & =\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}\sum_{m=-\infty}^{\infty}r^{|m|}s^{|n|}e^{im\theta}\delta_{n,m} \\ & =\frac{1}{2\pi}\sum_{n=-\infty}^{\infty}(rs)^{|n|}e^{in\theta}=P(rs,\theta). \end{align}

Answer 3

\begin{align} P(r,\phi) &= {\bf Re}\frac{1+re^{i\phi}}{1-re^{i\phi}} \\ {\bf Re}\frac{1+re^{i\phi}}{1-re^{i\phi}} &= {\bf Re}\left((1+re^{i\phi})\sum_{k\geq0}(re^{i\phi})^k\right) \\ &= {\bf Re}\left(1+2\sum_{k\geq1}(re^{i\phi})^k\right) \\ \dfrac{1-r^2}{1-2r\cos\phi+r^2} &= 1+2\sum_{k=1}^\infty r^k\cos k\phi \\ 4\pi^2\int_{-\pi}^{\pi} P(r,\theta-t)P(s,t)dt &= \int_{-\pi}^{\pi}\left(1+2\sum_{m=1}^\infty r^m\cos m(\theta-t)\right) \left(1+2\sum_{n=1}^\infty s^n\cos nt\right) dt \\ &= \int_{-\pi}^{\pi}dt+2\sum_{m=1}^\infty r^m\int_{-\pi}^{\pi}\cos m(\theta-t)+2\sum_{n=1}^\infty s^n\int_{-\pi}^{\pi}\cos nt dt \\ &+4\sum_{m=1}^\infty\sum_{n=1}^\infty r^ms^n\cos nt\cos m(\theta-t) dt \\ &= 2\pi+4\sum_{n=1}^\infty(rs)^n\pi\cos n\theta \\ &= 2\pi P(rs,\theta) \end{align} because \begin{align} &\int_{-\pi }^{\pi } 1 \, dt =2\pi\\ &\int_{-\pi }^{\pi } \cos n t \, dt =0\\ &\int_{-\pi }^{\pi } \cos n (\theta -t) \, dt =0\\ &\int_{-\pi }^{\pi } \cos n t \cos m(\theta -t) \, dt =0\hspace{2cm}m\neq n\\ &\int_{-\pi }^{\pi } \cos n t \cos n(\theta -t) \, dt =\pi\cos n t \end{align}

Answer 4

This solution uses existence and uniqueness of solutions of the Laplace equation to deduce the desired property of the Poisson kernel.

Let $f(\theta)$ be a continuous periodic function of $\theta$ on $[0,2\pi]$. Let $F(x,y)$ be the unique solution of the Laplace equation on $0 \le x^2+y^2 < 1$ such that the radial limit of $F$ on $x^2+y^2=1$ is $f$. Define $S(r)f$ to be the continuous periodic function on $[0,2\pi]$ that is given by $F(r\cos\theta,r\sin\theta)$. Then it follows that $$ S(r)S(s)f=S(rs)f $$ because any solution of $\nabla^2F=0$ on $0 \le x^2+y^2 < 1$ satisfies $$ \nabla^2F(rx,ry)=r^2(\nabla^2F)(rx,ry)=0,\;\; 0 \le r \le 1. $$ In other words, the solution operator $S$ for the radial Laplace equation is a multiplicative semigroup on the continuous periodic functions $C_p[0,2\pi]$. It is a $C_0$ semigroup, meaning that $\lim_{r\uparrow 1}P(r)f=f$, and this limit occurs uniformly for a continuous periodic function $f$ on $[0,2\pi]$. Solution operators for linear ODEs and PDEs have such properties because of uniqueness of solutions, if they have some type of symmetry operation that leaves the equation invariant such as scaling in the radial component.

The Poisson integral is a representation of this solution semigroup: \begin{align} (P(r)f)(\theta) & = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}f(\theta')d\theta' \\ & = \int_{0}^{2\pi}P(r,\theta-\theta')f(\theta')d\theta'. \end{align} Because of this, $$ P(rs)f=P(r)P(s)f. $$ This general property forces the Poisson kernel to have a related property: $$ \int_{0}^{2\pi}P(rs,\theta-\theta'')f(\theta'')d\theta'' \\ = \int_{0}^{2\pi}P(s,\theta-\theta')\int_{0}^{2\pi}P(r,\theta'-\theta'')f(\theta'')d\theta''d\theta' \\ = \int_{0}^{2\pi}\left(\int_{0}^{2\pi}P(s,\theta-\theta')P(r,\theta'-\theta'')d\theta'\right) f(\theta'')d\theta'' $$ Because this is true for all continuous functions, then $$ P(rs,\theta-\theta'') = \int_{0}^{2\pi}P(r,\theta-\theta')P(s,\theta'-\theta'')d\theta' \\ P(rs,\theta) = \int_{0}^{2\pi}P(r,\theta+\theta''-\theta')P(s,\theta'-\theta'')d\theta' \\ P(rs,\theta) = \int_{0}^{2\pi}P(r,\theta-t)P(s,t)dt. $$