The question: consider a hole of radius a in a two-dimensional plane. We let the temperature on the hole's boundary be given by $f(\theta)$, where $\theta$ is a polar angle. From the temperature profile $T(r,\theta)$ for $r>a$, find the Poisson Integral Formula.

My solution: We know the general solution to Laplace's Equation in plane polars is


To make sure it doesn't blow up as $r\rightarrow \infty$, we make $C_0=0$ and $C_n=0$. Combining $A_n$ and $B_n$ with $D_n$ to get $Q_n$ and $W_n$, we get

$$T(r,\theta)=D_0+\sum_{n=1}^{\infty}r^{-n}(Q_n\cos(n\theta)+W_n\sin(n\theta))$$ We ntoe that these constants can be found using the boundary conditions, but are not important for my question.

Our formula above can be rewritten as $$T(r,\theta)=\sum_{n=0}^{\infty}\frac{C_n}{r^n}e^{in\theta}$$

For some constant $C_n$. Now, from here, how can I find the Poisson Integral Formula? I know the integral kernal is given by


But I don't know how to go from there. Can anyone help me understand this?

Thank you in advance.


Here is how I did it:

We know that on the boundary of the hole $T(a, \theta) = f(\theta)$. 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)\tag{1}\\ \Theta'(-\pi) &= \Theta'(\pi)\tag{2} \end{align} Suppose $\lambda = 0$. Then $\Theta(\theta) = a\theta + b$. Using equation (1), we have that \begin{gather*} \Theta(-\pi) = -a\pi + b = a\pi + b = \Theta(\pi)\\ 2a\pi = 0\Rightarrow a = 0 \end{gather*} When $\lambda = 0$, we have $\Theta(\theta) = b$. Additionally, we then have $\frac{r}{R}\frac{\partial}{\partial r}\left(r\frac{\partial R}{\partial r}\right) = 0$. Then $R(r) = \alpha\ln(r) + \beta$. Now suppose $\lambda\neq 0$. Our $\Theta$ equation is then that of a simple harmonic oscillator. $$ \Theta(\theta) = A\cos(\lambda\theta) + B\sin(\lambda\theta) \tag{3} $$ We can now use equations (1) and (2) on equation (3) to determine the eigenfunction of the solution to the Laplace equation. \begin{align*} \Theta(-\pi) &= \Theta(\pi)\\ 2B\sin(\lambda\pi) &= 0 \end{align*} Therefore, $B = 0$ or $\lambda = n\in\mathbb{Z} - \{0\}$. \begin{align*} \Theta'(-\pi) &= \Theta'(\pi)\\ 2A\lambda\sin(\lambda\pi) &= 0 \end{align*} Now we have that $A = 0$ or $\lambda = n\in\mathbb{Z} - \{0\}$. Since we aren't looking for the trivial solution, $\lambda = n\in\mathbb{Z} - \{0\}$. $$ \Theta_n(\theta) = A_n\cos(n\theta) + B_n\sin(n\theta) \tag{4} $$ Let's now look at the radial equation, $r^2R'' + rR' - n^2R = 0$, which is of the Cauchy-Euler type. Let $R = r^m$. Then we have $r^m\left[m(m - 1) + m - n^2\right] = 0$ so $m = \lvert n\rvert$. 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 $$ T(r, \theta) = A_0 + \sum_{n = 1}^{\infty}\frac{A_n}{r^n}\cos(n\theta) + \frac{B_n}{r^n}\sin(n\theta). \tag{5} $$ 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 \tag{6}\\ A_n &= \frac{a^n}{\pi}\int_{-\pi}^{\pi}f(\theta)\cos(n\theta)d\theta \tag{7}\\ B_n &= \frac{a^n}{\pi}\int_{-\pi}^{\pi}f(\theta)\sin(n\theta)d\theta \tag{8} \end{alignat*} To find the steady state temperature for $r > a$, we have equation (5) and equation (6) to (8) for the Fourier coefficients. We can also write equation (5) in terms of exponentials. $$ T(r, \theta) = \sum_{n = -\infty}^{\infty}\left(\frac{a}{r}\right)^n c_n\exp(in\theta) \tag{9} $$ where the Fourier coefficient is $$ c_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\varphi)\exp(-in\varphi)d\varphi. $$ Let's substitute $c_n$ in to equation (9). $$ \sum_{n = -\infty}^{\infty}\left(\frac{a}{r}\right)^n\left[ \frac{1}{2\pi}\int_{-\pi}^{\pi}f(\varphi)\exp(-in\varphi)d\varphi \right]\exp(in\theta) = \int_{-\pi}^{\pi}f(\varphi)\left[\frac{1}{2\pi} \sum_{n = -\infty}^{\infty}\left(\frac{a}{r}\right)^n \exp(in(\theta - \varphi))\right]d\varphi $$ where $\mathcal{P}(r, \theta - \varphi) = \frac{1}{2\pi}\sum\limits_{n = -\infty}^{\infty}\left(\frac{a}{r}\right)^n\exp(in(\theta - \varphi))$ is the Poisson kernel for $r > a$. We can write the Poisson kernel as $$ \mathcal{P}(r, \theta - \varphi) = \frac{1}{2\pi}\left[ \sum_{n = 0}^{\infty}\left(\frac{a}{r}\right)^n \exp(in(\theta - \varphi)) + \sum_{n = 1}^{\infty} \left(\frac{a}{r}\right)^n\exp(-in(\theta - \varphi))\right] $$ Let $z = \frac{a}{r}\exp(i(\theta - \varphi))$. Then the summations in our Poisson kernel are geometric series. That is, we can write the Poisson kernel as \begin{align*} \mathcal{P}(r, \theta - \varphi) &= \frac{1}{2\pi}\left[ \frac{1}{1 - z} + \frac{\bar{z}}{1 - \bar{z}}\right]\\ &= \frac{1}{2\pi}\frac{1 - z\bar{z}}{1 - (z + \bar{z}) + z\bar{z}} \end{align*} Note that $z\bar{z} = \frac{a^2}{r^2}$ and $z + \bar{z} = 2\frac{a}{r}\cos(\theta - \varphi)$. Let's multiple through by $r^2$ as we make the substitution. $$ = \frac{1}{2\pi}\frac{r^2 - a^2}{r^2 - 2ar\cos(\theta - \varphi) + a^2} $$ The solution to the Laplace steady state diffusion equation can be represented as $$ T(r, \theta) = \int_{-\pi}^{\pi}f(\varphi)\mathcal{P}(r, \theta)d\varphi. $$ Therefore, the Poisson Integral Formula for equation (5) is \begin{align*} T(r, \theta) &= \int_{-\pi}^{\pi}f(\varphi) \mathcal{P}(r, \theta - \varphi)d\varphi\\ &= \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{r^2 - a^2}{r^2 - 2ar\cos(\theta - \varphi) + a^2}f(\varphi)d\varphi \end{align*}


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