What is the Limit of the Poisson Kernel Prove for $r\to 1$?

[Dustinsfl]

Prove:
$$
\lim_{r\to 1}P(r,\theta) = \begin{cases}
\infty, & \theta = 0\\
0, & \text{otherwise}
\end{cases}
$$
For the first piece, take the summation
$$
P(1,0) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} 1^n\right).
$$
Then $\sum\limits_{n = 1}^{\infty} 1^n = \infty$.
Therefore, we have a positive number plus infinity which is infinity when $r\to 1$ and $\theta = 0$.
For the second piece, take the fractional representation of the Poisson kernel,
$$
P(1,\theta) = \frac{1}{2\pi}\frac{0}{2 - 2\cos\theta} = 0.
$$
Therefore, $P(r,\theta) = 0$ for all $\theta\neq 0$.
That is,
$$\lim_{r\to 1}P(r,\theta) = \begin{cases}
\infty, & \theta = 0\\
0, & \text{otherwise}
\end{cases}
$$

 

[tarantella]

Could you write the definition of Poisson kernel?

 

[Dustinsfl]

Could you write the definition of Poisson kernel?

$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$

 

[Sudharaka]

$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$

Can you please tell me where you found this definition?

 

[Dustinsfl]

Can you please tell me where you found this definition?

A class on Fourier series

A class on Engineering Analysis

The book Elementary Partial Differential Equations by Berg and McGregor

My Engineering Analysis book that I can't remember the name.

Separate handout notes by my Fourier Analysis professor.

 

[Sudharaka]

A class on Fourier series

A class on Engineering Analysis

The book Elementary Partial Differential Equations by Berg and McGregor

My Engineering Analysis book that I can't remember the name.

Separate handout notes by my Fourier Analysis professor.

I suggest you to check the definition again. The correct one is given >>here<<.

\begin{eqnarray}

P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}e^{in\theta}=\frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.\\

&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)+i\sum_{n=-\infty}^\infty r^{|n|}\sin(n\theta)\\

\end{eqnarray}

Since, \(r^{|n|}\sin(n\theta)\) is an odd function it is clear that the second sum is equal to zero.

\begin{eqnarray}

\therefore P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)\\

&=&1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)\\

\end{eqnarray}

Hence we finally get,

\[P_r(\theta)=1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)= \frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.\]

 

[Dustinsfl]

I suggest you to check the definition again. The correct one is given >>here<<.

\begin{eqnarray}

P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}e^{in\theta}=\frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.\\

&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)+i\sum_{n=-\infty}^\infty r^{|n|}\sin(n\theta)\\

\end{eqnarray}

Since, \(r^{|n|}\sin(n\theta)\) is an odd function it is clear that the second sum is equal to zero.

\begin{eqnarray}

\therefore P_r(\theta)&=&\sum_{n=-\infty}^\infty r^{|n|}\cos(n\theta)\\

&=&1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)\\

\end{eqnarray}

Hence we finally get,

\[P_r(\theta)=1+2\sum_{n=1}^\infty r^{n}\cos(n\theta)= \frac{1-r^2}{1-2r\cos\theta +r^2}\mbox{ where }0 \le r < 1.\]

The books are already opened to the pages. I see it clearly.

 

[Dustinsfl]

Photo of Poisson Kernel

https://www.physicsforums.com/attachments/396

Poisson Kernel -- from Wolfram MathWorld

 

[Sudharaka]

Photo of Poisson Kernel

https://www.physicsforums.com/attachments/396

Poisson Kernel -- from Wolfram MathWorld

Yes it seems that there is a slight difference in the definition of the Poisson Kernel. In some books it's defined as,

\[P(r,\theta)=\frac{1-r^2}{1-2r\cos\theta +r^2}\]

whereas in others,

\[P(r,\theta)=\frac{1}{2\pi}\frac{1-r^2}{1-2r\cos\theta +r^2}\]

You seem to be using this second definition. However notice that you are missing a \(n\) in the summation of post #3.

\[P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos({\color{red}n}\theta)\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}\]

 

[Dustinsfl]

\[P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos({\color{red}n}\theta)\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}\]

Typo.
Is my soln correct?

 

[Sudharaka]

Prove:
$$
\lim_{r\to 1}P(r,\theta) = \begin{cases}
\infty, & \theta = 0\\
0, & \text{otherwise}
\end{cases}
$$
For the first piece, take the summation
$$
P(1,0) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} 1^n\right).
$$
Then $\sum\limits_{n = 1}^{\infty} 1^n = \infty$.
Therefore, we have a positive number plus infinity which is infinity when $r\to 1$ and $\theta = 0$.
For the second piece, take the fractional representation of the Poisson kernel,
$$
P(1,\theta) = \frac{1}{2\pi}\frac{0}{2 - 2\cos\theta} = 0.
$$
Therefore, $P({\color{red}1},\theta) = 0$ for all $\theta\neq 0$.
That is,
$$\lim_{r\to 1}P(r,\theta) = \begin{cases}
\infty, & \theta = 0\\
0, & \text{otherwise}
\end{cases}
$$

Yeah it's correct. (Yes)