1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

426 CHAPTER 11 • APPLICATIONS OF HARMONIC FUNCTIONS

Figure 11.1 The harmonic function u(x,y) = U 1 + U~ - Ui (x -a).

• a

•EXAMPLE 11 .2 Find the function W (x,y) that is harmonic in the sector

0 < Arg z < a, where a ::; n, and takes on the boundary values

w(x, o) = cl>

'l!(x, y) = C2,

for x > 0 and

at points on the ray r > 0, (J = a.

Solutio-n Recalling that the function Arg z is harmonic and takes on constant
values along rays emanating from the origin, we see that a solution has the form

'11 (x, y) =a+ bArgz,

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

C2 -C1

W (x, y) = C1 + Argz.

a

The situation is shown in Figure 11.2.

y

l/f(x, y) =constant

I

Figure 11.2 T he harmonic function '11 (x, y) = C, + (C2 - Ci ) ~Arg z.