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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 \(P(r,\theta) = \frac{1}{2\pi}\sum\limits_{n = -\infty}^{\infty}r^{\lvert n\rvert}e^{in\theta}\).