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[SOLVED] Limit of Poisson kernel

dwsmith

Well-known member
Feb 1, 2012
1,673
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,
$$\lim_{r\to 1}P(r,\theta) = \begin{cases}
\infty, & \theta = 0\\
0, & \text{otherwise}
\end{cases}
$$
 

girdav

Member
Feb 1, 2012
96
Could you write the definition of Poisson kernel?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
$$
P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos\theta\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}
$$
Can you please tell me where you found this definition?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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,

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

dwsmith

Well-known member
Feb 1, 2012
1,673
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.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.

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

dwsmith

Well-known member
Feb 1, 2012
1,673
\[P(r,\theta) = \frac{1}{\pi}\left(\frac{1}{2} + \sum_{n = 1}^{\infty} r^n\cos({\color{red}n}\theta)\right) = \frac{1}{2\pi}\frac{1 - r^2}{1 - 2r\cos\theta + r^2}\]

Typo.
Is my soln correct?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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)