# [SOLVED]Laplace/Poisson Integral

#### dwsmith

##### Well-known member
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
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?