Preliminaries

Section 10.1 Preliminaries

In most applications involving harmonic functions, a harmonic function that takes on prescribed values along certain contours must be found. In presenting the material in this chapter, we assume that you are familiar with the material covered in Sections 2.4, Section 3.3, Section 5.1, and Section 5.2. If you aren’t, please review it before proceeding.

Example 10.1.1.

Find the function \(u(x,y)\) that is harmonic in the vertical strip \(a \le \mathrm{Re}(z) \le b\) and takes on the boundary values

\begin{equation*} u(a,y) =U_1 \text{ and } u(b,y)=U_2 \end{equation*}

along the vertical lines \(x=a\) and \(x=b\text{,}\) respectively.

Solution.

Intuition suggests that we should seek a solution that takes on constant values along the vertical lines of the form \(x=x_0\) and that \(u(x,y)\) be a function of \(x\) alone; that is,

\begin{equation*} u(x,y) =P(x), \text{ for } a \le x \le b, \text{ and } \text{ for all } y \end{equation*}

Laplace’s equation, \(u_{xx}(x,y) +u_{yy}(x,y) = 0\text{,}\) implies that \(P,''(x)=0\text{,}\) which implies \(P(x)=mx+c\text{,}\) where \(m\) and \(c\) are constants. The stated boundary conditions \(u(a,y)=P(a)=U_1\) and \(u(b,y)=P(b)=U_2\) lead to the solution

\begin{equation*} u(x,y) =U_1+\frac{U_2-U_1}{b-a}(x-a) \end{equation*}

The level curves \(u(x,y) =\) constant are vertical lines as indicated in Figure 10.1.2.

Figure 10.1.2. Level curves of \(u(x,y)=U_1+\frac{U_2-U_1}{b-a}(x-a)\)

Example 10.1.3.

Find the function \(\Psi(x,y)\) that is harmonic in the sector \(0\lt \mathrm{Arg}z\lt \alpha\text{,}\) where \(\alpha \le \pi\text{,}\) and takes on the boundary values

\begin{align*} \Psi(x,0) \amp = C_1 \text{ for } x>0, \; \text{ and }\\ \Psi(x,y) \amp = C_2 \text{ at points on the ray } r>0, \; \theta =\alpha \text{.} \end{align*}

Solution.

Recalling that the function \(\mathrm{Arg}(z)\) is harmonic and takes on constant values along rays emanating from the origin, we see that a solution has the form

\begin{equation*} \Psi(x,y)=a+b\mathrm{Arg}(z)\text{,} \end{equation*}

where \(a\) and \(b\) are constants. The boundary conditions lead to

\begin{equation*} \Psi(x,y) = C_1 + \frac{C_2 - C_1}{\alpha}\mathrm{Arg}(z)\text{.} \end{equation*}

The situation is shown in Figure 10.1.4.

Figure 10.1.4. Level curves of \(\Psi(x,y) = C_1 + (C_2-C_1) \frac{1}{\alpha }\mathrm{Arg}(z)\)

Example 10.1.5.

Find the function \(\Phi(x,y)\) that is harmonic in the annulus \(1\lt |z| \lt R\) and takes on the boundary values

\begin{align*} \Phi(x,y) \amp = K_1, \text{ when } |z| = 1, \text{ and }\\ \Phi(x,y) \amp = K_2, \text{ when } |z| = R\text{.} \end{align*}

Solution.

This problem is a companion to the one in Example 10.1.3. Here we use the fact that \(\ln|z|\) is a harmonic function, for all \(z \ne 0\text{.}\) The solution is

\begin{equation*} \Phi(x,y)=K_1+\frac{K_2-K_1}{\ln R}\ln|z| \end{equation*}

, and the level curves \(\Phi(x,y)=\) constant are concentric circles, as illustrated in Figure 10.1.6.

Figure 10.1.6. Level curves of \(\Phi(x,y) = K_1 + \frac{K_2-K_1}{\ln R}\ln|z|\)

Prev Top Next