Complex Analysis - Maths KU

136 Chapter 2 Complex Functions and Mappings
–4 –4 –2
2 4
4
y
2
–2
–4
Figure 2.62 Streamlines in the planar
flow assicated with f(z) =z
x
These differential equations are independent of each other and so each can be
solved byseparation of variables. This gives the general solutions x(t) =c1e t
and y(t) =c2e −t where c1 and c2 are real constants. In order to plot the
curve z(t) =x(t)+iy(t), we eliminate the parameter t to obtain a Cartesian
equation in x and y. This is easilydone bymultiplying the two solutions to
obtain xy = c1c2. Because c1 and c2 can be anyreal constants, this familyof
curves can be given by xy = c where c is a real constant. In conclusion, we
have shown that particles in the planar flow associated with f(z) =¯z move
along curves in the familyof hyperbolas xy = c. In Figure 2.62, we have used
Mathematica to plot the streamlines corresponding to c = ±1, ±4, and ±9 for
this flow. These streamlines are shown in black superimposed over the plot
of the normalized vector field of f(z) =¯z.
EXAMPLE 3 Streamlines
Find the streamlines of the planar flow associated with f(z) =¯z 2 .
Solution We proceed as in Example 2. Since the function f can be expressed
as f(z) =¯z 2 = x 2 − y 2 − 2xyi, we identify P (x, y) =x 2 − y 2 and Q(x, y) =
−2xy. Thus, the streamlines of this flow satisfythe system of differential
equations:
dx
dt = x2 − y 2
dy
dt
= −2xy.
The chain rule of elementarycalculus states that (dy/dx) · (dx/dt) =dy/dt,
and so after solving for dy/dx we have that
� ��� �
dy dx
=
dt dt
dy
dx .
Therefore, bydividing dy/dt = −2xy by dx/dt = x 2 − y 2 , we find that the
system in (4) is equivalent to the first-order differential equation:
dy −2xy
=
dx x2 − y2 or 2xy dx + � x 2 − y 2� dy =0. (5)
Recall that a differential equation of the form M(x, y)dx + N(x, y)dy =0
is called exact if ∂M/∂y = ∂N/∂x. Given an exact differential equation, if
we can find a function F (x, y) for which ∂F/∂x = M and ∂F/∂y = N, then
F (x, y) =c is an implicit solution to the differential equation. Identifying
M(x, y) =2xy and N(x, y) =x 2 − y 2 , we see that our differential equation
in (5) is exact since
∂M
∂y
∂
∂ � 2 2
= (2xy) =2x = x − y
∂y ∂x
� = ∂N
∂x .
(4)