Complex Analysis - Maths KU

292 Chapter 5 Integration in the Complex Plane
C
T
Figure 5.54 Flow of fluid tends to
turn C.
F
�
F · T ds < 0 mean that the fluid tends to rotate C in the counterclockwise
C
and clockwise directions, respectively. See Figure 5.54.
Now if N =(dy/ds) i − (dx/ds) j denotes the normal unit vector to a
positively oriented, simple closed contour C, and if F(x, y) =P (x, y)i +
Q(x, y)j again represents a velocity field of a two-dimensional fluid flow, we
define the net flux of F as the real line integral of the normal component of
the velocity vector F:
�
net flux = F · N ds. (19)
Specifically, (19) defines a net rate at which the fluid is crossing the curve C
in the direction of the normal N and is measured in units of area per unit
time. In other words, the net flux across C is the difference between the rate
at which fluid enters and the rate at which fluid leaves the region bounded by
C. A nonzero value of �
C
C
F · N ds indicates the presence of sources or sinks
for the fluid inside the curve C.
Now if f(z) =P (x, y) +iQ(x, y) is the complex representation of the
velocity field F of a fluid, the line integrals (18) and (19) can be computed
simultaneously by evaluating the single contour integral �
f(z) dz. To see
C
how this is done, first observe:
�
�
�
dx dy
F · T ds = (P i + Q j) · i +
C
C
ds ds j
� �
ds = Pdx+ Qdy
C
�
�
�
dy dx
F · N ds = (P i + Q j) · i −
ds ds j
� �
ds = Pdy−Qdx, C
and then:
�
C
C
�
f(z) dz = (P − iQ)(dx + idy)
C
��
� ��
�
= Pdx+ Qdy + i Pdy−Qdx C
�� � ��
C
�
= F · T ds + i F · N ds . (20)
C
Equation (20) shows that (18) and (19) can be found by computing �
f(z) dz
C
and identifying the real and imaginary parts of the result. That is,
�� �
�� �
circulation = Re f(z) dz and net flux = Im f(z) dz . (21)
C
EXAMPLE 5 Circulation and Net Flux
Use (21) to compute the circulation and the net flux for the flow and curve C
in part (a) of Example 4.
Solution Frompart (a) of Example 4, the flow is f(z) =(z − i) 2 and C is
the circle |z| = 1. Then f(z) = � z − i � 2 =(¯z + i) 2 =¯z 2 +2i¯z − 1, and C is
C
C
C