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[SOLVED] Laplace/Poisson Integral

dwsmith

Well-known member
Feb 1, 2012
1,673
We have a two dimensional plate whose hole has radius \(a\) and \(T(a, \theta) = f(\theta)\). Find an expression for the steady state temperature profile \(T(r, \theta)\) for \(r > a\). I am pretty sure the solution below is correct but if you want to glance it over that would be fine.

How do I find the form of Poisson Integral Formula for this problem?

Laplace's equation is
\(\frac{1}{r}\frac{\partial}{\partial r}\left(
r\frac{\partial T}{\partial r}\right) +
\frac{1}{r^2}\frac{\partial^2 T}{\partial\theta^2} = 0\).
Let \(T(r, \theta)\) be of the form \(T = R(r)\Theta(\theta)\).
\[
\frac{r}{R}\frac{\partial}{\partial r}\left(
r\frac{\partial R}{\partial r}\right) = - \frac{\Theta''}{\Theta} =
\lambda^2
\]
Since we have perfect thermal contact, our periodic boundary conditions
are
\begin{align}
\Theta(-\pi) &= \Theta(\pi)\\
\Theta'(-\pi) &= \Theta'(\pi)
\end{align}
When \(\lambda = 0\), we have \(\Theta(\theta) = b\) and \(R(r) = \alpha\ln(r) + \beta\).
Now suppose \(\lambda\neq 0\).
\[
\Theta_n(\theta) = A_n\cos(n\theta) + B_n\sin(n\theta)
\]
Let's now look at the radial equation, \(r^2R'' + rR' - n^2R = 0\), which is of the Cauchy-Euler type.
The general form of \(T(r, \theta)\) is
\[
T(r, \theta) = \alpha\ln(r) + \beta + \sum_{n = 1}^{\infty}
\left(r^n + r^{-n}\right)\left(A_n\cos(n\theta) + B_n\sin(n\theta)\right).
\]
Since \(r\) goes out to infinity, \(r^n\) and \(\ln(r)\) would blow up at
infinity.
Therefore, \(T(r, \theta)\) is of the form
\[
T(r, \theta) = A_0 + \sum_{n = 1}^{\infty}\frac{A_n}{r^n}\cos(n\theta) +
\frac{B_n}{r^n}\sin(n\theta).
\]
To solve for the Fourier coefficients, we need to use the boundary
condition on the hole of radius \(a\).
\begin{alignat*}{2}
T(a, \theta) &= A_0 + \sum_{n = 1}^{\infty}\frac{A_n}{a^n}\cos(n\theta)
+ \frac{B_n}{a^n}\sin(n\theta) &&{} =f(\theta)\\
A_0 &= \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\theta)d\theta\\
A_n &= \frac{a_n}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)d\theta\\
B_n &= \frac{a_n}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)d\theta
\end{alignat*}
[HR][/HR][HR][/HR]We can re-write the solution as
\[
T(r,\theta) = \sum_{n = 0}^{\infty}\frac{C_n}{r^n}e^{in\theta}.
\]
The Poisson kernel is \(P(r,\theta) = \frac{1}{2\pi}\sum\limits_{n = -\infty}^{\infty}r^{\lvert n\rvert}e^{in\theta}\).
 

dwsmith

Well-known member
Feb 1, 2012
1,673
We can write \(T(r, \theta) = \sum\limits_{n = -\infty}^{\infty}\left(\frac{a}{r}\right)^nc_n\exp(in\theta)\). Then \(c_n = \frac{a^n}{2\pi}\int_{-\pi}^{\pi}f(\varphi)\exp(-in\varphi)d\varphi\).
\[
\sum\limits_{n = -\infty}^{\infty}\left(\frac{a}{r}\right)^n \left(\frac{1}{2\pi} \int_{-\pi}^{\pi}f(\varphi)\exp(-in\varphi)d\varphi\right)\exp(in\theta) =
\int_{-\pi}^{\pi}f(\varphi)\left[\frac{1}{2\pi}\sum_{-\infty}^{\infty}r^{-n} \exp(in(\theta - \varphi))\right]d\varphi
\]
Poisson's kernel is \(P(r, \theta) = \frac{1}{2pi}\sum\limits_{n = -\infty}^{\infty}r^{|n|}e^{in\theta}\).
In our case \(r > a\), we have
\[
P(r, \theta - \varphi) = \frac{1}{2\pi}\sum_{-\infty}^{\infty}r^{-n} \exp(in(\theta - \varphi)).
\]
We can re-write \(P(r, \theta)\) as
\[
P(r, \theta) = \frac{1}{2\pi}\left[\sum_{n = 0}^{\infty}r^{-n}e^{in\theta} + \sum_{n = 1}^{\infty}r^{-n}e^{-in\theta}\right]
\]
Let \(z = \frac{1}{r}\exp(i\theta)\).
Then we have two geometric series.
\[
P(r, \theta) = \frac{1}{2\pi}\left[\frac{1}{1 - z} + \frac{\bar{z}}{1 - \bar{z}}\right]
\]

At the moment, I have to go so I can't finish it yet.

However, is this the correct idea?