1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

466 CHAPTER. 11 • APPLJCATIONS OF HARMONIC FUNCTIONS

(b) Find the electrostatic potential cf> (x, y) in the domain D that satisfies

the boundary values shown in Figure 11.46.

ef>(x, y) = 100

cf>(x, y) = 200

y

Figure 11.46

when lzl = 5;

when lz - 21 = 2.

w=--z-^10

2z-5

u

11.7 Two-Dimensional Fluid Flow

Suppose that a fluid flows over the complex plane and that the velocity at the

point z = x + iy is given by the velocity vector

V (x, y) = p(x, y) +iq(x, y). ( 11 -30)

We also require that the velocity does not depend on time and the compo-
nents p (x, y) and q(x, y) have continuous partial derivatives. The divergence of
t he vector field in Equation (11-30) is given by

div V (x, y) = p., (x, y) + qy (x, y)

and is a measure of the extent to which the velocity field diverges near the point.
We consider only fluid flows for which the divergence is zero. This condition is
more precisely characterized by the requirement that the net flow through any
simply closed contour be identically zero.
If we consider the flow out of the small rectangle shown in Figure 11.47, then
the rate of outward flow equals the line integral of the exterior normal component
of V ( x, y) taken over the sides of the rectangle. The exterior normal component
is given by -q on the bottom edge, p on t he right edge, q on the top edge, and

• p on the left edge. Integrating and setting the resulting net flow to zero y ields

f,

11+t.11 l'"+t.z

!I (p(x+t.x,t)- p(x,t)]dt+"' [q(t,y+t.y)-q(t,y))dt=O.

(11-31)