1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

472 CHAP'l'ER 11 • APPLICATIONS OF HARMONIC FUNCTIONS

Figure 11. 51 Fluid flow a.round a. circle.

where A is a positive real number. The stream function for the flow in thew

plane is 1/J (u, v) =Av so that the slit lies along the streamline 1lJ (u, v) = 0.

The composite function F 2 (z} = F 1 (S(z)) determines the fluid flow in the

domain D, where the complex potential is

F 2 (z)=A(z+~),

where A > 0. We can use polar coordinates to express F 2 (z) as

F2 (z) = F2 (rei^9 ) = A (r + ~) cosO + i A (r -; ) sinO.

The streamline 1/J (r ,0) = A (r - ~) sin 0 = 0 consists of the rays

r > 1, 0 = 0 and r > 1, 0 = 1r

along the x-axis and the curve r· - !: = 0, which is the unit circle r = 1. Thus

the unit circle can be considered as a boundary curve for the fluid flow.

The approximation F2 (z) =A (z + ~)::::: Az is valid for large values of z , so

we can approximate the flow with a uniform horizontal flow having speed IVI = A
at points that are distant from the origin. The streamlines 1/J (x, y) = constant
and their images 1lJ ( u, v) = constant under the mapping w = S (z) = z +! are
illustrated in Figure 11.51.

• EXAMPLE 11.25 Find the complex potential for an ideal fluid flowing from
  left to right across the complex plane and around the segment from -i to i.

Solutio n We use the conformal mapping

w = S(z) = (z^2 + l)~ = (z +i)! (z-i)!,

where the branch of the square root of Z = z ± i in each factor is z! = Riei~,

where R = IZI, and 0 = arg-; (Z), where - 2 " < 0 $^3 ;'. The function given