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Complex Analysis: an Open Source Textbook

Russell W. Howell, John H. Mathews

Section 11.2 The Dirichlet Problem for the Unit Disk

The Dirichlet problem for the unit disk \(D:|z|\lt 1\) is to find a real-valued function \(u(x,y)\) that is harmonic in the unit disk \(D\) and that takes on the boundary values
\begin{equation} u(\cos \theta ,\sin \theta ) = U(\theta), \text{ for } -\pi \lt \theta \le \pi\tag{11.2.1} \end{equation}
at points \(z=(\cos \theta, \sin \theta)\) on the unit circle, as shown in Figure 11.2.1.
Figure 11.2.1. The Dirichlet problem for the unit disk \(|z| \lt 1\text{.}\)
We will prove this theorem shortly, but note meanwhile that it is easy to see that the series representation in Equation (11.2.2) for \(u\) takes on the boundary values given in Equation (11.2.1) at points on the unit circle \(|z| =1\text{.}\) Since each term \(r^n\cos n\theta\) and \(r^n\sin n\theta\) in series given in Equation (11.2.2) is harmonic, it is reasonable to conclude that the infinite series representing \(u\) will also be harmonic.
Before proving Theorem 11.2.2 we establish a result that gives an integral representation for a function \(u(x,y)\) that is harmonic in a domain containing the closed unit disk. The result is the analog to Poisson’s integral formula for the upper half plane.

Proof.

Since \(u(x,y)\) is harmonic in the simply connected domain, there exists a conjugate harmonic function \(v(x,y)\) such that \(f(z)=u(x,y)+iv(x,y)\) is analytic. Let \(C\) denote the contour consisting of the unit circle; then Cauchy ’s integral formula
\begin{equation} f(z) = \frac{1}{2\pi i}\int\nolimits_c\frac{f(\xi)}{\xi -z}\,d\xi\tag{11.2.4} \end{equation}
expresses the value of \(f(z)\) at any point \(z\) inside \(C\) in terms of the values of \(f(\xi)\) at points \(\xi\) that lie on the circle \(C\text{.}\) If we set \(z^{\ast } = (\overline{z})^{-1}\) then \(z^*\) lies outside the unit circle \(C\) and the Cauchy-Goursat theorem gives
\begin{equation} 0 = \frac{1}{2\pi i}\int_c\frac{f(\xi )}{\xi -z^{\ast }}\,d\xi\text{.}\tag{11.2.5} \end{equation}
If we subtract Equation (11.2.5) from Equation (11.2.4), then use the parameterization \(\xi =e^{it}\text{,}\) \(d\xi =ie^{it}\,dt\) and the substitutions \(z=re^{i\theta}\text{,}\) \(z^{\ast }=\frac{1}{r}e^{i\theta}\text{,}\) we get
\begin{equation*} f(z) = \frac{1}{2\pi}\int_{-\pi}^{\pi}\left(\frac{e^{it}}{e^{it}-re^{i\theta}} - \frac{e^{it}}{e^{it}-\frac{1}{r}e^{i\theta}}\right)f(e^{it})\,dt\text{.} \end{equation*}
The expression inside the parentheses on the right side of this equation can be written as
\begin{align} \frac{e^{it}}{e^{it}-re^{i\theta}}-\frac{e^{it}}{e^{it}-\frac{1}{r} e^{i\theta}} \amp = \frac{1}{1-re^{i(\theta -t)}} + \frac{re^{i(t-\theta )}}{1-re^{i(t-\theta)}}\notag\\ \amp = \frac{1-r^2}{1+r^2-2r\cos (t-\theta )}\text{,}\tag{11.2.6} \end{align}
and it follows that
\begin{equation*} f(z) = \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{(1-r^2)f(e^{it})}{1+r^2-2r\cos (t-\theta)}\,dt\text{.} \end{equation*}
Since \(u(x,y)\) is the real part of \(f(z)\) and \(U(t)\) is the real part of \(f(e^{it})\text{,}\) we can equate the real parts in the latter equation to obtain Equation (11.2.3), completing the proof Theorem 11.2.3.
We now turn our attention to the proof Theorem 11.2.2.

Proof.

The real-valued function
\begin{equation*} P(r,t-\theta ) = \frac{1-r^2}{1+r^2-2r\cos (t-\theta)} \end{equation*}
is known as the Poisson kernel. Expanding the left side of Equation (11.2.6) in a geometric series gives
\begin{align*} P(r,t-\theta) \amp = \frac{1}{1-re^{i(\theta -t)}} + \frac{re^{i(t-\theta )}}{1-re^{i(t-\theta)}}\\ \amp = \sum_{n=0}^{\infty}r^ne^{in(\theta -t)}+\sum\limits_{n=1}^{\infty}r^ne^{in(t-\theta)}\\ \amp = 1+\sum\limits_{n=1}^{\infty}r^n[e^{in(\theta -t)}+e^{in(t-\theta)}] =1+2\sum\limits_{n=1}^{\infty}r^n\cos[n(\theta -t)]\\ \amp = 1+2\sum\limits_{n=1}^{\infty}r^n(\cos n\theta \cos nt+\sin n\theta \sin nt)\\ \amp = 1+2\sum\limits_{n=1}^{\infty}r^n\cos n\theta \cos nt+2\sum\limits_{n=1}^{\infty}r^n\sin n\theta \sin nt\text{.} \end{align*}
We now use the above result in Equation (11.2.3) to obtain
\begin{align*} u(r\cos \theta, r\sin \theta) \amp = \frac{1}{2\pi}\int_{-\pi}^{\pi}P(r, t-\theta )U(t)\,dt\\ \amp = \frac{1}{2\pi}\int_{-\pi}^{\pi}U(t)\,dt +\frac{1}{\pi} \int_{-\pi}^{\pi}\sum_{n=1}^{\infty}r^n\cos n\theta \cos nt \, U(t) \,dt\\ \amp \qquad \qquad \qquad + \frac{1}{\pi}\int_{-\pi}^{\pi}\sum\limits_{n=1}^{\infty}r^n\sin n\theta \cos nt\,U(t)\,dt\\ \amp = \frac{1}{2\pi}\int_{-\pi}^{\pi}U(t)\,dt +\sum\limits_{n=1}^{\infty}\frac{r^n}{\pi}\cos n\theta \int_{-\pi}^{\pi}\cos nt\,U(t)\,dt\\ \amp \qquad \qquad \qquad + \sum\limits_{n=1}^{\infty}\frac{r^n}{\pi}\sin n\theta \int_{-\pi}^{\pi}\sin nt\,U(t)\,dt\\ \amp = \frac{a_0}{2}+\sum\limits_{n=1}^{\infty}a_nr^n\cos n\theta +\sum\limits_{n=1}^{\infty}b_nr^n\sin n\theta \text{.} \end{align*}
where \(\{a_n\}\) and \(\{b_n\}\) are the Fourier series coefficients for \(U(t)\text{.}\) This observation establishes the representation for \(u(r\cos \theta ,r\sin \theta )\) in Equation (11.2.2).

Example 11.2.4.

Find the function \(u(x,y)\) that is harmonic in the unit disk \(|z|\lt 1\) and takes on the boundary values
\begin{equation*} u(\cos \theta, \sin \theta ) = U(\theta) = \frac{\theta}{2}, \text{ for } -\pi \lt \theta \lt \pi \end{equation*}
Solution.
Using Example 10.1.1, we begin with the Fourier series for \(U(\theta)\text{:}\)
\begin{equation*} U(t) = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin nt\text{.} \end{equation*}
Using formula (11.2.2) for the solution of the Dirichlet problem, we obtain
\begin{equation*} u(r\cos \theta ,r\sin \theta ) =\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}r^n\sin n\theta\text{.} \end{equation*}
Note that that the above series representation for \(u(r\cos \theta, r\sin \theta)\) takes on the prescribed boundary values at points where \(U(\theta)\) is continuous. The boundary function \(U(\theta)\) is discontinuous at \(z=-1\text{,}\) which corresponds to \(\theta =\pm \pi\text{;}\) and \(U(\theta )\) was not prescribed at these points. Graphs of the approximations \(U_7(t)\) and \(u_7(x,y)=u_7(r\cos \theta ,r\sin \theta)\text{,}\) which involve the first seven terms in equations are shown in Figure 11.2.5.
Figure 11.2.5. Functions \(U_7(t)\) and \(u_7(r\cos \theta, r\sin \theta)\) for Example 11.2.4

Exercises Exercises

For problems 1–6, find the solution to the given Dirichlet problem in the unit disk \(D\) by using the Fourier series representations for the boundary functions that were derived in the examples and exercises of Section 11.1.

1.

\(U(\theta) = \begin{cases}\;\;\, 1 \amp \text{ for } 0\lt \theta\lt \pi, \\ -1 \amp \text{ for } -\pi \lt \theta \lt 0. \end{cases}\)
Solution.
\(u(r\cos\theta, r\sin\theta) = \frac{4}{\pi}\sum\limits_{n=1}^{\infty}\frac{1}{2n-1}r^{2n-1}\sin[(2n-1)\theta]\text{.}\)

2.

\(U(\theta) = \begin{cases}\frac{\pi}{2}-\theta \amp \text{ for } 0 \le \theta \lt \pi, \\ \frac{\pi}{2}+\theta \amp \text{ for } -\pi \lt \theta \lt 0. \end{cases}\) (See Figure 11.2.6)
Figure 11.2.6. Approximations for \(U_5(\theta)\) and \(u_5(r\cos \theta ,r\sin \theta)\) in Exercise 2

3.

\(U(\theta) = \begin{cases}-1 \amp \text{ for } \frac{\pi}{2}\lt \theta \lt \pi, \\ \;\;\, 1 \amp \text{ for } -\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2},\\ -1 \amp \text{ for } -\pi \lt \theta \lt -\frac{\pi}{2}. \end{cases}\)
Solution.
\(u(r\cos\theta, r\sin\theta) = \frac{4}{\pi}\sum\limits_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}r^{2n-1}\cos[(2n-1)\theta]\text{.}\)

4.

\(U(\theta) = \begin{cases}\pi -\theta \amp \text{ for } \frac{\pi}{2} \le \theta \le \pi, \\ \;\;\;\; \theta \amp \text{ for } -\frac{\pi}{2} \le \theta \lt \frac{\pi}{2}, \\ -\pi -\theta \amp \text{ for } -\pi \lt \theta \lt -\frac{\pi}{2}. \end{cases}\)

5.

\(U(\theta ) = \begin{cases}\pi -\theta \amp \text{ for } \frac{\pi}{2} \le \theta \le \pi, \\ \;\;\; \frac{\pi}{2} \amp \text{ for } -\frac{\pi}{2} \le \theta \lt \frac{\pi}{2}, \\ \pi +\theta \amp \text{ for } -\pi \lt \theta \lt -\frac{\pi}{2}. \end{cases}\) (See Figure 11.2.7)
Solution.
\(u(r\cos\theta, r\sin\theta) = \frac{3\pi }{8}+\frac{2}{\pi}\sum\limits_{n=1}^{\infty }\frac{1}{(2n-1)^2}r^{2n-1}\cos[(2n-1)\theta]\) \(-\frac{4}{\pi}\sum\limits_{n=1}^{\infty}\frac{1}{2^2(2n-1)^2}r^{4n-2}\cos[2(2n-1)\theta]\text{.}\)
Figure 11.2.7. Approximations for \(U_5(\theta )\) and \(u_5(r\cos \theta ,r\sin \theta )\) in Exercise 5

6.

\(U(\theta) = \begin{cases}\;\;\, 1 \amp \text{ for } \frac{\pi}{2}\lt \theta\lt \pi, \\ \;\;\, 0 \amp \text{ for } -\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2}, \\ -1 \amp \text{ for } -\pi \lt \theta \lt -\frac{\pi}{2}. \end{cases}\) (See Figure 11.2.8)
Figure 11.2.8. Approximations for \(U_7(\theta)\) and \(u_7(r\cos \theta, r\sin \theta)\) in Exercise 11.2.6

7.

\(U(\theta ) = \begin{cases}\;\,\, 0 \amp \text{ for } \frac{\pi}{2} \le \theta \le \pi,\\ \frac{\pi -\theta}{2} \amp \text{ for } 0 \le \theta \lt \frac{\pi}{2},\\ \frac{\pi +\theta}{2} \amp \text{ for } -\frac{\pi}{2} \le \theta \lt 0, \\ \;\,\, 0 \amp \text{ for } -\pi \lt \theta \lt -\frac{\pi}{2}. \end{cases}\)
Solution.
\(u(r\cos\theta, r\sin\theta) = \frac{\pi}{8} + \frac{2}{\pi}\sum\limits_{n=1}^{\infty }\frac{1}{(2n-1)^2}r^{2n-1}\cos[(2n-1)\theta]\) \(+\frac{4}{\pi }\sum\limits_{n=1}^{\infty }\frac{1}{2^2(2n-1)^2}r^{4n-2}\cos[2(2n-1)\theta]\text{.}\)

8.

\(U(\theta ) = \begin{cases}\;\;\, 0 \amp \text{ for } \frac{\pi}{2}\lt \theta \lt \pi,\\ -1 \amp \text{ for } 0\lt \theta \lt \frac{\pi}{2},\\ \;\;\, 1 \amp \text{ for } -\frac{\pi}{2}\lt \theta \lt 0,\\ \;\;\, 0 \amp \text{ for } -\pi \lt \theta \lt -\frac{\pi}{2}. \end{cases}\)

9.

Write a report on the Dirichlet problem and include some applications.

10.

Look up the article on the Poisson integral formula and discuss what you found.
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