Solving Laplace Equation over a Disk
More often than not, we may have to solve the two-dimensional Laplace equation over a circular disk instead of a rectangular domain. Then it will be more natural to employ the polar coordinates, where the Laplace equation now reads
\begin{align}
\nabla^2 u(r,\theta) = \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0 \tag{1}
\end{align}
where the Laplacian in polar coordinates is \( \nabla^2 = \frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial}{\partial r}) + \frac{1}{r^2} \frac{\partial^2 }{\partial \theta^2} \). Given that the disk has a radius of \(a\), the B.C. will be in the general form of
\begin{align}
u(r=a,\theta) = f(\theta), -\pi \leq \theta \leq \pi \tag{2}
\end{align}
The technique of separation of variables is still usable in this scenario, where we now let
\begin{align}
u(r,\theta) = R(r) \Theta(\theta) \tag{3}
\end{align}
with \( R(r) \) being a function of \(r\) and \(\Theta(\theta)\) a function of the polar angle \(\theta\) only, respectively. Plugging (3) into (1) then produces
\begin{align}
R^{\prime\prime}(r)\Theta(\theta) + \frac{1}{r}R'(r)\Theta(\theta) + \frac{1}{r^2} R(r) \Theta^{\prime\prime}(\theta) &= 0 \\
(r^2 \frac{R^{\prime\prime}(r)}{R(r)} + r\frac{R'(r)}{R(r)}) + \frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)} &= 0 \tag{4}
\end{align}
The first two terms depend on \(r\) only, while the last term depends on \(\theta\) only. Hence, we may assume the separation that
\begin{align}
r^2 \frac{R^{\prime\prime}(r)}{R(r)} + r\frac{R'(r)}{R(r)} &= \lambda \tag{5} \\
\frac{\Theta^{\prime\prime}(\theta)}{\Theta(\theta)} &= -\lambda \tag{6}
\end{align}
for some constant \( \lambda \). (5) can be rearranged into an Euler form second-order ODE (see this tutorial):
\begin{align}
r^2 R^{\prime\prime} + rR’ -\lambda R &= 0 \tag{7}
\end{align}
which has the solution of
\begin{align}
R = r^\sqrt{\lambda} = r^n, \sqrt{\lambda} = n = 0,1,2,\ldots \tag{8}
\end{align}
We will quickly see the reason behind the requirement that \( n \) is a non-negative integer. Then, (6) may be solved as
\begin{align}
\Theta^{\prime\prime} + \lambda \Theta &= 0 \\
\Theta^{\prime\prime} + n^2\Theta &= 0 \\
\Rightarrow \Theta &= \cos(n\theta), \sin(n\theta) \tag{9}
\end{align}
This confirms that \( n \) has to be an integer for single-valuedness. The full solution will then be the superposition of the three basis functions \( R\Theta = 1, r^n\cos(n\theta), r^n\sin(n\theta) \):
\begin{align}
u(r,\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty} (A_nr^n\cos(n\theta) + B_nr^n\sin(n\theta)) \tag{10}
\end{align}
Enforcing the B.C., we have
\begin{align}
u(a,\theta) = \frac{A_0}{2} + \sum_{n=1}^{\infty} (A_na^n\cos(n\theta) + B_na^n\sin(n\theta)) = f(\theta) \tag{11}
\end{align}
To obtain the coefficients \( A_n, B_n \), we simply project (11) onto the Fourier basis as before to obtain
\begin{align}
A_0 &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\xi) d\xi \tag{12} \\
A_n &= \frac{1}{\pi a^n} \int_{-\pi}^{\pi} f(\xi) \cos(n\xi)d\xi \tag{13} \\
B_n &= \frac{1}{\pi a^n} \int_{-\pi}^{\pi} f(\xi) \sin(n\xi)d\xi \tag{14}
\end{align}
Putting these back to (10) then produces the desired solution:
\begin{align}
u(r,\theta) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\xi) d\xi \\
&\quad+ \frac{1}{\pi} \sum_{n=1}^{\infty} (\frac{r}{a})^n \left[(\int_{-\pi}^{\pi} f(\xi) \cos(n\xi)d\xi) \cos(n\theta) + (\int_{-\pi}^{\pi} f(\xi) \sin(n\xi)d\xi) \sin(n\theta)\right] \tag{15}
\end{align}
Poisson Integral Formula
We can rewrite (15) in a more condensed way as an integral. First, we apply the trigonometric identity to get
\begin{align}
u(r,\theta) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} (1+2 \sum_{n=1}^\infty (\frac{r}{a})^n \cos(n(\xi-\theta))) f(\xi)d\xi \tag{16}
\end{align}
Then, define the Poisson kernel as
\begin{equation}
P(r,\xi) = \frac{1}{2\pi} (1+2 \sum_{n=1}^\infty r^n \cos(n\xi)) \tag{17}
\end{equation}
so that (16) becomes
\begin{equation}
u(r,\theta) = \int_{-\pi}^{\pi} P(\frac{r}{a},\xi-\theta)f(\xi) d\xi \tag{18}
\end{equation}
Using Euler’s Formula, we can let \( z = re^{i\xi} = r(\cos(\xi) + i\sin(\xi))\) and by De Moivre’s formula, we have
\begin{align}
z^n = r^ne^{in\xi} = r^n(\cos(n\xi) + i\sin(n\xi)) \tag{19}
\end{align}
So (17) can be expressed in terms of (19) as
\begin{align}
1+2 \sum_{n=1}^\infty r^n \cos(n\xi) &= \Re(1+2 \sum_{n=1}^\infty z^n) \\
&= \Re(1+2\frac{z}{1-z}) \\
&= \Re(\frac{1+z}{1-z}) = \Re(\frac{1+re^{i\xi}}{1-re^{i\xi}}) \tag{20}
\end{align}
The required real part may be found as
\begin{align}
\frac{1+re^{i\xi}}{1-re^{i\xi}} &= \frac{1+re^{i\xi}}{1-re^{i\xi}} (\frac{1-re^{-i\xi}}{1-re^{-i\xi}}) \\
&= \frac{1-r^2+r(e^{i\xi} -e^{-i\xi})}{1+r^2-r(e^{i\xi} + e^{-i\xi})} \\
&= \frac{1-r^2+2ir\sin\xi}{1+r^2-2r\cos\xi} \\
&= \frac{1-r^2}{1+r^2-2r\cos\xi} + i\frac{2r\sin\xi}{1+r^2-2r\cos\xi} \tag{21}
\end{align}
Therefore, (18) becomes the desired Poisson Integral Formula:
\begin{align}
u(r,\theta) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{1-(\frac{r}{a})^2}{1+(\frac{r}{a})^2-2(\frac{r}{a})\cos(\xi-\theta)} f(\xi) d\xi \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{a^2-r^2}{a^2+r^2-2ra\cos(\xi-\theta)} f(\xi) d\xi \tag{22}
\end{align}
Notably, at \( r = 0 \), (22) becomes just
\begin{align}
u(r,\theta) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(\xi) d\xi \tag{23}
\end{align}
Hence, the value at the center in the solution of a Laplace equation is always equal to the average value along the boundary.
Exercise
Solve the Laplace equation on the disk \( x^2 + y^2 \leq 4 \) with the B.C. \( u(x,y) = xy, x^2 + y^2 = 4 \).
Answer
In polar coordinates, \(x = r\cos\theta\), \(y = r\sin\theta\), \( a = 2 \), and the B.C. becomes
\begin{align}
u(2,\theta) = 4\cos\theta\sin\theta = f(\theta)
\end{align}
By (15), the solution is
\begin{align}
u(r,\theta) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} 4\cos\xi\sin\xi d\xi \\
&\quad+ \frac{1}{\pi} \sum_{n=1}^{\infty} (\frac{r}{2})^n \left[(\int_{-\pi}^{\pi} 4\cos\xi\sin\xi \cos(n\xi)d\xi) \cos(n\theta) + (\int_{-\pi}^{\pi} 4\cos\xi\sin\xi \sin(n\xi)d\xi) \sin(n\theta)\right] \\
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} 2\sin(2\xi) d\xi \\
&\quad+ \frac{1}{\pi} \sum_{n=1}^{\infty} (\frac{r}{2})^n \left[(\int_{-\pi}^{\pi} 2\sin(2\xi) \cos(n\xi)d\xi) \cos(n\theta) + (\int_{-\pi}^{\pi} 2\sin(2\xi) \sin(n\xi)d\xi) \sin(n\theta)\right]
\end{align}
By the orthogonality of the Fourier basis, we have
\begin{align}
\int_{-\pi}^{\pi} 2\sin(2\xi) d\xi &= 0 \\
\int_{-\pi}^{\pi} 2\sin(2\xi) \cos(n\xi)d\xi &= 0 \\
\int_{-\pi}^{\pi} 2\sin(2\xi) \sin(n\xi)d\xi &= \begin{cases}
0 & n\neq 2 \\
2\pi & n=2
\end{cases}
\end{align}
Hence the final answer is
\begin{align}
u(r,\theta) &= 2(\frac{r}{2})^2 \sin(2\theta)
\end{align}
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