.­
9.1
•
CHAPTER 9
CONFORMAL MAPPING AND ITS APPLICATIONS
9.1
Introduction
In Chapter 2 and Chapter 5, we studied certain mappiDg properties of
analytic functions. . In this chapter, we continue the study of the maps
.
effected by analytic 'functions, and show how theY' are applied to the study
./
of fluid flow, eJ.ectro.:statics, and steady state temperatures.
Recall that in Chapter 5 we saw that when
w = f (z) = U (x, y) + i vex, y)
is an analytic function then :
I
(a)
Both
u
and
v
satisfy Laplace's equation:
\'
vxx +
We
(b)
SEq
that
u
vyy = 0 •
and
v
are harmonic functions or potential functions.
The mapping effected bY'
w =··f (z)
preserves angles and their
/
sense (its a conformai mapping) at all points where
f' (z) \: 0 •
(
(c)
c2
The curves
u (x, y) = c
1
and
v (x, y) = c
2
(where c
1
are constants), intersect at right angles at points where
f' (z) ~ 0 •
and
•
. The reader shouJ.d review sections
5.4 and 5.5 before continuing
it these items have been forgotten.
9.2
Meaning of the Laplacian operator
Every' physical problem involves three dimensional space and time.
This means that the three real space variables
coordinates and time
t
x, y and z
of Cartesian
are encountered.
now.
Imagine nov a problem of f'luid
Suppose "We are in a large channel of
-+----~y
water and that the water is flowing at
co~stant
velocity in the direction of
the y-axis.
Suppose also that a very long cylindrical pipe is placed in
the position of the z-axis.
We wouJ.d expect the stream lines of the fluid
to resemble those shown in the figure.
A stream line is the actual path
.~()
~
","---~_--.,.r
~.I'----i1D---~
~~:.-'---"""'~--­
~
~v
-"---0:lil---'""\
o
~ '~---:'J"'---~
f,~----::'l""'--_"­
y
~~/--~lI&-­ _ _
_ _~
((,,,t'~
_
,;
'. _ _ .
_.--~
"-'._"~'
'
... ,. , __ -O'9'"_'....
r-
. . . . . .• _
~
•
' . . . . . . . - - . ...
~ . ~
..
•
tak~n
by a partictQ of the fluid.
Suppose that the velocity of the :fluid
in the channel stqs the same for a very long time at each point.
During
this time we say that we are in the steady state, and the mathematical
description of the motion will not involve the variable
t.
AJ.so, suppose
the pipe is very long, and that the surface of the fluid and the walls of
the channel are very far from the origin of coordinates.
In this case, we
can simplif'y our mathematical model by assuming that the water extends to
infinity. in every. direction and that the pipe is infinite in length.
z = 0,. look exactly like the stream
case, the stream lines in the plane
•
lines in 8IJ.Y other plane
z = zO.
In this
This means that the mathematical de­
scription of the :fluid flow will not involve the variable
the motion involves only tvo s-oace dimensions.
z.
We say that
We need only look now at the
x, Y' plane to see the description of the stream lines, and their equation
y
should take the form
'Vex,
y) = c ,
/
where c
is a parameter which is con­
stant on any stream line.
In the remainder of this chapter,
•
we wiU deal with physical problems
......--1---1---+--+ X
•
involving two space dimensions ex and y variables·) and in the steactr state
(no variation with time).
When a physical phenomenon is in the stead7
state t some type of' balance or equilibrium has been obtained.
not
SOt
the lack of' balance ¥Ould cause the system to alter t thus destroying
the steactr state.
~
of' a great
(1) t
In
The equation which describes the equilibrium condition
physical problems is the eguationof Lanlace
(1) \/2 u eXt y)
•
Werethis
=
a 2 u ex, y.). + d 2 u{x, y)
~ x2
d y2
U eXt y)
= 0
is some fUnction which describes the physical
problem.
It might be the temperature at the point
example.
The operator
V2
=
~ + d2
d x2
. .}y2
eXt y) t for
is called the tvo
dimensional Laplacian operator.
The meaning of the expression \12 u
in physical problems is contained
in the following idea:
\l 2
u
eXt y)
at a point
is a measure of the difference between the
//
(2)
value of
u
at
eXt y) t and the average of the values of
infinitesimal neighborhood surrounding the point
Suppose that
•
some solid.
u
in an
eXt y) •
u eXt y) describes the temperature at each point ex, y) in
If the temperature is in the steady state, then the temperature
_._._--_..
-
---_.~_
•.
_._ .. ----' .._
-
_.. -
..
~
,.-.,_
-
~.
_..-
-
-~-
..----
.. ~
~~.----.,.-~.,
•
at the point (x, y) should equal. the aver~e ot the temperatures at points
immediately surrounding it.
It the temperature at
(x, y)
vas lover than
the average ot the temperatures surrounding it, heat would flov into the
point (x, yo) and cause the ~emperature there to rise, thus destroying the
steady state.
This. example describes vhy Laplace's equation
V 2 u =0
is the equilibrium or steady state equation.
0
We will nov suggest vby (2) is true.
Since ve must investigate
u
::r.I>JFIIUTES '''''Ai..
near the point (x, y), it is reasonable
Ne'G-t+BORHOCD
O~
to expand
•
u
in a
T~lor' s
c.x) "I)
series about
(
eXt'
(3)
yo).
u (x + h •
t
k)
=!.- t.­
n-o
o
:II:
U (x, y)
'+
+
/
u
I't'l"l:.O
+ u
X
h2
xx
2"
h3
Uxxx
hm
m I
6"""
(x, y) h + Uy (x, y)
k
+
k2
2
Uxy hk +'\ry
+
...
Perhaps the reader is seeing (:3) tor the first time.
Notice hov it is a
(
natural outgrowth of the familiar
T~lor's
series.
The difference between
/
•
u
at
(x, y) and at a nearby point
(x + h , y + k)
is
•..
. ..
•
(4)
--_.__ . -_.
---
u(x + h, y + k) - u(x, y) = ux
-
_.-
h+uy
k
­
h2
+u
~
xx:
+ uxy hk+u
7Y'
h3
+.~
~ese
The average of
rounding
(5)
...
~_.-
+
T
differences over the infinitesimal square sur­
J [U(X
2 E is
(x, y) of side length
j€
4i2
~-h.;~~)-~~::
E-
-E:
k2
2
- E
.--- .--­
-.-----.~.
72)- ~-dI1.
---,._--.--­ - -­ _.
If we replace the integrand here by the right side of (4) and integrate
we see that:
•
(a)
The first term gives
J€ JE
-E
since
h
hdkdh
=
0
-E
is an odd function.
(If you don't see this, then actually
perform the integration) .
(b)
For the same reason, the second term
JEJE
J€
-€
(c)
kdkdh
=
0
-E
The third term does not vanish since we have
IE.
Uxx
4e 2
•
-€.
~
2 E
[4J
h2
~
dhdk
=
-f­
iE
.=
~E
., I
2
6
-t
"", ;:r:",:.'-' _..--,---- .
(d)
The fourth term vanishes in the same way as the first two
t~.
(e) The fifth term resembles (e) and is
Uyy
(7)
€2
6
(f)
The remaining terms will involve
f
to a higher paver than two,
and thus they will be ignored.•
Combining
we have
[ u(x + h, Y + k) - u(x,
(8)
y)J
dlt dh
=
,.
....
•
This last relation is the mathematical statement of (2).
Thus we see that the two dimensional equation of the steady
state is Laplace's equation.
Since both the real and the imaginary parts
of an analytic function
w = f (z)
=u
(x, y) +
i
v (x, y)
/'
sat1st'y this equation
!
•
it is not surprising that analytic functions are very useful in describing
many physical problems which are at the steady state, and involve only two
9.8
•
spacoe variables.
We will explore some of these applications in the
following sections.
9.3 Stream lines in fluid flow
One important application of analytic :functions is to the study known
as hydrodynamics t aerodynamics
0 ....
fluid dynamics.
Suppose we have a two
.. dimensional flow of fluid in the steady state which has the additional
properties:
(a)
The fluid is incompressible.
(b)
The fluid is non-viscous.
(c)
The velocity of the fluid is derivable :from a potential.
•
•
The assumption that the fluid is incompressible implies that the density of
the fluid is constant.
A viscous fluid would create a tangential frictional
force on an obstacle in its path.
Since our fluid is· non-viscous, any force
it creates on the obstacle must be perpendicular to the obstacle's surface.
The assumption that the velocity of our fluid can be obtained from a potential
(harmonic) :function will be explained in the next section.
A fluid satis:f'ying the above assumptions is called an ideal fluid.
•
a fluid never really exists in nature, but is orten approximated.
I
Such
We will
9.9
•
see' that analytic f'unctions of a complex variable enable
us to visualize
these ideal nuids.
Since we have seen that the real and the imaginary parts of the
f (z) = u (x, yo) + i v. (x, yo)
analytic f'unction
satis:f'y the two dimensional
equation of the steady state, it comes as no surprise that they are useful
in examining the flow of an ideal auld.
= constant
u (x, yo)
level lines
and
Look at Figure
v (x, yo)
9.1
= constant
are sketched over
the comPlex z - plane for eight different analytic :t'un.ctions
•
The level lines for
u
v
= constant
the stream lines of some particular nuid flow?
v (x, yo) the stream :t'un.ction.
diagrams in
Figure
f (z)
are shown broken while the level lines for
Do not the solid lines
sholr.n solid.
Here the
=u
v
+ i v •
are
in each sketch look like
We call the function
The flow described bY' each of the eight
9.1· is outlined below:
Diagram (i)
A uniform horizontal flow from lef't to right.
/
Diagram (11)
A flow around a right angled corner.
(
Diagram (11i)
at
•
z
Fluid emerges from the origin.
We have a source
=o •
Diagram (iv)
..­
A vortex or whirlpool is at
z
=0
•
;
,
•
Figure. 9.•1·
Ta:ble of Stream Lines for Various Fluid
Flows
Stream lines
v(x,y) = constant - - - - - - - - - - ­ Ve10ci ty potential u(x,y) = constant - - - - - - _.- ~
""--_._--- ._-_.-.
-----------,:
(
I
I
,
-~~----s:-,
I
I
---+-----il~
.
r
I
(~)·-~ni;~rm
•
w
_
hor-i-zontBJ.
flO~·~ ---
=z
... . . - - , . _ .
I
I
I
I(i~)
I
I
--
"
Flow around a corner.
2
w =z
..
I
-
. -_._-­
1
I
I
I
I
I
I
I
'~
Source at
•
.--_ ....
z = 0 •
w = log z
:
i·-····_-- - - - . - -----------~ _.. "- - - - ..
1
. . --_.
l(iV)
1
I
I
-
I
II
I
I
Vortex at
w
=i
z=O.
.
-_._-_.. --- - - ------10
,I
log z
- -­
_._~
I
I
!
__ ..
..
_.~_.'"""~
•
•..
>.:.~.,
._,..;:.-"-..".-..z_
1._",.:.&..':::1.-'
Figure
9.1
,
•.
.
~.''"''''W:C"'':-'.'
.- . . . . _ _ ...
.... _ _ ~ • • •
_ _
__
_ . • . 0_.
C6ntinued
1
i
I
I
--------y
I
- '- - - - - - - _ .
-
I
I
I
I
I ,
-­
1..-­
--------­
(v)
•
=
Source at z
a and a
sink at z = b 0
W
.
= log
z-a
z-b
I
I (vi)
= ;
..
I.
-_._-- .--­
Dipole at
w
Ix
z
=0
•
1
----­ ---.
\
\
•
~vii)
Cylinder of radius a
in a uniform flow •
2
a uniform
to right.
9.10
•
Diagram (v) :
Fluid emerges at
We have a source at
Diagram (vi)
"a"
z
=a
and a sink at
and is absorbed at z
= b.
"b".
A dipole exists at the origin.
A dipole consists
of a large source' and sink separated by an infinitesimal distance.
Diagram (vii):
A cylindrical pipe of radius
is placed in
"a"
a uniform flow of fluid from left to right.
Di agram (viii)
A source of fluid is placed at
z
=0
into a
uniform flow from left to right.
•
We will see in the next section how the direction of the fluid flow in
each diagram, 1!:s indicated by the arrow heads on the stream. lines, is de­
termined.
We see that the imagina.ry part
considered in Figure
perfect fluid.
v (x, y)
of each of the analytic functions
9.1 defines a family of stream. lines for the flow of a
We will see in the next section how the corresponding real
,/
part
u (x, y)
is used to determine the velocity of the fluid.
Example 1
Determine the equations of the stream. lines shown in Figure
•
Solution
9 !.l
(ii) •
•
Since
z2 = (x + i y)2 = x 2 - y2 + 2 x y i , we see that the imaginary
v (x, y)
part is
=2
Thus the equations of the stream lines are
x y.
xy=c
where
c
is a parameter which is constant along any stream line.
Example 2
Determine the stream £'unction of the fluid nov in Figure
9.1 (vii).
Solution"
We must find the imaginary part of the function
•
find
f (z)
=z
+ a 2 /z •
We
a2
f (z) = x + i y +
%+1y
a 2 (x - i
x2 + y2
=x+iy+
(1)
= x +
a2 x
x 2 + y2
Thus the stream function is
+ i
rl
[y ­
a2 y
x2 + y2
J
v (x, 7) = y ­
and the family of
stream lines of fluid flow past a cylinder of radius a centered at the origin
/
is obtained by setting
v
equal to an arbitrary parameter.
Problems
Write the equations of the stream lines for the flows
•
of the ,following:
describe~
by each
.'
• 3 .. 1
The uniform flow in Figure
.3.2.
The flov described by the source at
.3.3
The flow described by the dipole in Figure
9.1 (i).
z
=0
in Figure
(iii).
9.1 (vi).
Arry stream line can be thought of as the boundary
fluid must flow.
9.1
along which the
This is possible because the assumption that the fluid is
non-viscous means that no fr:tction is created as the fluid passes over the
boundarY.
Figure
•
As an example) we coUld consider one of the hyperbolas in
9.1
(ii) as the boundary rather than the right angled corner.
Example 3
In the diagram we see two infinite plates
at right angles to each other.
Fluid continually
enters the shaded region through the slit where
the plates meet.
w'hen the steady state is
reached, desoribe the steam lines.
,/
Solution
Look at Figure
•
9.1 (iii).
If we use the two stream lines along the
positive real and imaginary axies as the boundaries of our fluid, we see at
once that the stream lines in the first quadrant of this figure satisfy the
9.13
011'"
'1
reqUirements of problem.
The stream lines are simply straight lines
y
emerging from the origin.
.
Problems
• 3. '{
.
'~--------::>~
The edges of a metal boundary are shown in the diagram.
X
Fluid
continually enters the shaded region
through the slit at
at
•
;
A and leaves
B at the same rate.
Determine
the stream lines •
cD 3.5'"' The
semi-circular hump is placed on the bottom of a flat stream bed
where the fluid had been flowing
uniformly from left to
right~,
Determine the equations of the
stream lines.
/
In some cases a stream line will divide and become two or more stream
(
'.
lines at a particular point.
Look' at the stream line entering along the
negative real axis in Figure
9.1
(vii) •
At the point
in two and travels round the cylinder to the point
z = a
z =
a it.. splits
where the two
•
stream lines once again merge and continue to infinity' along the positive
real axis.
At each point, ve can describe the velocity' of the f'luid by
means of a vector tangent to a stream line.
z = +
fined tangent a1; the points
be zero there.
a,
the velocity of the f'lu1d must
We cal.l these points stagnation points of the .fluid.
again at the stagnation points in Figure
•
Since there is no Yell de­
level lines for
u
and
that the points
z = +
v
( ) =z+
v=fz
9.1 (vii) and observe that the
do not intersect at right angles.
a
z
This means
are points vhere the analytic f'unction
a2
z
aescribing the flov fails to define a con:formal mapping.
when
Look
Thus
:f' (z)
=0
is a stagnation point.
Example 4
Demonstrate analyticeJ.ly that the :function
v
= fez) = z
+
2
a
z
/
describes a :floW' with stagnation points at
z
=
+
a.
ShoW' that the
stream line along the real B.xi.ssplits at these points to :form the circle
, z
t =a
as shown in Figure
9.1 (vii) •
.-'
•
Solution
/
-'-.".,Goo-._
~",,_
... . . __.. __
-~
~.
........__ ..
._~.~.
~'.
...
"__
4
•
•
__
.~
••• _
•
9.15
•
The stagnation points occur where
t'(z)
=1
_
a
2 =
z2
and we have
f'(z)
z =
~
a
Thus
o,
as the stagnation points.
itself' is real and thus
real axis.
f'(z) is zero.
Along the real axis,
vex, y) = 0 is a stream line along the
In Example 2 we saw that
v(x, y) = y
Thus' .v( x, y) = 0
•
in two circumstances:
=0
(a)
When
y
(real axis)
(b)
When
x 2 + y2 = a 2
(circle of' radius
a).
Problem
C. 3. (;
Find the location of' the stagnation point of' the flow in Figure
(viii) •
/
.- /
_e
9.1
9.li Velocity potential in fluid flow
The stream lines of fluid flow were found as the level lines of the
imaginary part of the appropriate analytic fUnction in the previous section.
Row can we find the velocity of the fluid at each point (x, 7) '1
The
velocity is a vector quantity which is tangent to the stream lines.
In a previous course, the student probably studied the gradient of the
fUnction
u(x, 1')
denoted by
horizontal component is
U
V u and defined as a vector whose
x and whose vertical component is . uy
We
will denote this vector by a complex number
•
(1)
V u
=
~ + i
u
7
The gradient (1) has the following properties:
(a)
. line
yo u at each point (x, 1')
is perpendicular to the level
u = constant through this point.
/
The modulus of the gradient,
+ u 2
l'
is
I
<"
•
the directional derivative
....£....J!
d s
in the direction of the
•
most rapid increase in
ot u
= u(x,
u.
To understand this statement, think
y) as!surtace in (x, y, u) space.
r
Then ( Vu
is
the slope we 'W"'Ould experience i t we were climbing up this surtace
at the point (x, "y, u) in the direction ot greatest steepness.
J vu J
IS
S(J~F=ACe
AT ()(>'t")
51-OPE
THE
T""S ""AN
AS HE
SURFI'\C.c
$-reePEsr
AT"
TME
E.><PER.II=Nc.es
WA~~S
IN
~p
THE
THE
DIR.Ec-rION
POINT
C><) 'I) t.L)
>=y
•
Now return to Figure 9.1 •
tor u(x,y)
(a)
the level lines
are perpendicular to the stream lines.
discussed above,
Vu
By property
must be tangent to the stream lines.
Since
V u
veloci ty vectors are also tangent to stream lines, could it be that
describes the velocity vector ot our ideal fluid flow?
Yes, this is the
/
case.
We have
tt
veloci~y
"
vector
= v u = ux
+ i u
y)
(one ot the Cauchy-Riemann equations from section
Velocity Vector
•
=
u
x
-"i v
The derivative of the function
x
fez)
and since
5.3) we have
/
•
=u
I
-'
+ i v
d fez)
d z
, can be computed
9.18
•
using
d z
dx
as
d f =
d z
(4)
u
(2)
5.1) and we have
d (u + i
(7%
=
•
Comparing
(recal.l section
v)
x + 1 v x
and
velocity vector
(3)
=
we see that
ft(z)
Thus we see that an analytic function
fez) = u + i v
is a neat
compact .package for the description of an ideal fiuid with the following
•
features:
(a) The stream function is
given by the equations
vex, y).
vex, y) = c ,
The stream lines are
where
c
is an arbitrary
constant.
(b) The velocity potential is
are given by
u( x, y) = c.
u(x, 1').
Equipotential lines
The velocity vector itself is
t t ( z) , the complex conJ ugate of the derivative of
t( z) •
/
We call
fez)
the complex potential of the fiuid f"low.
Example 1
: Show that the function
•
,
f( z)
= Vz
is the complex potential of a
.'
/
•
uniform flov rrom lett to right with
sp~ed
V.
Solution
Since
f'(z)
function is Vy
potential is
=V
, the velocity vector
t'(z)
and thus the stream lines are
Vx
and thus
th~ li3Q,u1potential
is also V.
y = constant.
lines are
x
The stream
The velocity
= constant.
Example 2
Find
(a)
the complex potential,
(b)
the velocity vector and
the stream function when a cylindrical pipe ot radius
a unitorm horizontal tlow rith veloCity
"a"
(c)
is placed in
V tar trom. the pipe •
•
•
Solution
In Figure
~a"
(5)
9.1 (vii) we see a horizontal tlow past a pipe ot radius
described by the tunction
v
=z
+
The velocity ot this flow is described by
Vi"
=1­
/
When
z
we adJ ust
far trom
•
is large (far trom the pipe) the veloeity is nearly 1.
(5)
How can
so that it ri11 have the same stream lines, but veloe!ty
z = 0 'l
It we multiply
the stream lines since the function
( 5)
by the constant
V
V we rill not alter
V vex, y) = constant
and
•
V'
(x,y)
= constant
defines the same f'amily of' stream lines.
Thus we
take
(6)
w
= V(z
2
+ ~)
z
as our new complex potential because the velocity
is nearly
V
- =
v'
V
­
V a2
-2
z
f'ar trom the origin.
Example 3
Find the complex potentia,J. describing a uniform f'low of' veJ.ocity
V
in the direction shown •
•
Solution
Since the velocity
V e i«: ,
is the complex conjugate of' the derivative
of' the complex potential we write
( ) =Ve-io<.z.
f'z
Thus
-i<z
V e
I
.
is the desired complex potentJ.al.
./
•
Problems
JI'" "...,.1
w
Determine the ve~ocit'U'
vector at each ·point (x., yo) of a fluid flow
oJ
•
described by the folloving complex potentials.
(b)
w=mlog z - a
= m log
w
,
(e)
z,
w=
Z - b
• '1_ 2..
w = m i log z ,
(c)
m
(f)
-z- ,
w=
v z +
m log z •
Find the points wher-e the speed of the flow described in
- Example 2
is a maximum. Find this maximum speed.
Suppose we have a soUrce
•
Describe the stream lines.
described by
f( z)
=m
of
n~d
at the origin with complex
poten~ial
log Z •
r
We .. call m the strength of the
source ..
What is the rate at which fluid
emerges from the origin?
i
That
is, (assuming units of feet and
seconds )what is the number of
(per foot of length perpendicular to the z
a cylinder of
cubic feet of fluid emerging per second1
/
:f'l.uid at time
t
volume =
of
A short time later,
•
r
+
dr.
Tr r 2
t + dt , the radius of the cylinder increases to
Since the velocity vector
has magnitude
Jf'(Z)
I= I ~ I
=
./
m
r
..'
/
9.22
•
and. since the speed is also described by
dr
dt
, we have
= r
m
·
dr
dt
Thus the rate at which the volume increases is
d volume
dt
=
=:
-E:
2 1'1' r
dt
m
•
21Yr
=
2f#m •
r
. We .have determined that:
•
The complex potential
m log z describes a source
of strength m at the origin in
whfch fluid emerges at the rate 27T m units of volume per' unit time.
for each unit of length perpendicular to the z plane.
Example 4
(-1, 0) at the rate of 3 cubic
Fluid 'is being created at the point
feet per second while fluid is absorbed at the rate of "2" cubic feet per
second at the point (1,
0). Determine the complex potential •
./
Solution
-
If fluid were emerging trom the origin at the rate of
per second, the complex potential
source.
_e
3
2"'"(-1,
To move the source to the point
z -(-1) to get
3
211
log
is lost at the rate of 2 cubic
(z + 1).
log
z
._~---
----_._--~_
..
­
3 cubic feet
would describe this
/
0), we replace
z
by·
A sink. at the origin in which fluid
log
reet per second is described by
sink to the point
(1~ 0)
we leP1ace
z.
To .translate this
z by z-l to get
2
log(z - 1).
2 1r'
Thus we have
t(z)
= --.L
log( z + 1) .... --.£.. log( z - 1)
21/
1
• 21r
log
2"1'1"'
(z + 1)3
(z - 1)2
as the complex potential for the flow.
Problem
• L{.,3
Fluid emerges from the point (O~ 1) at the rate of 41r cubic feet per
second~
while it is absorbed at the origin at the same rate.
.
~
complex potential. (b)
Find the velocity vector.
(c)
(a)
Find the
Find the equations
of the stream lines and the equipotential lines.
Example 5 .
Discuss the flow described by w
= m log(sin
z).
Draw the stream lines
and equipotential lines.
/
Solution
(
Near the origin sin z
v
•
~
z
= m log
~
m log
and thus
(sin z)
.r
z
/
.'
tor very small z.
!'lear
z
=1'1"',
This is a ,source ot strength
m
at the origin.
sin z ~ 11 - z and we have
v
=m log
~
mlog
•
z)
(sin
( ~ - z)
~ m log (-l)(z -
~ m
log (z -
11 )
1T' )
+m log (-1)
The term m log (-1) is simply a constant, and the addition of a constant
term to the complex potential in no way alters the physical flow being
•
described.
z
= 17'.
The term m log (z - 17') is a source of strength
Continuing in this way we find that sin z
at
has zeros at all
integral multiples of 1Y, and thus we have sources of strength
the points
z=1(n,n=O,
w
i
=m
=m
-+
-
1 , + 2 , •••
. ' ':111
at
The stream lines of
log(sin z)
-_._--..
logo/sin zl + 1m arg (sin z)
---- ----
---_.
-
._----­
---
_
..
-~
.i
/
are the curves
arg (sin z)
are seen in Figure
2.7
=c ,
where c is any constant.
of Chapter 2.
I
The equipotential lines are
constant and are seen in the same :figure •
•
These curves
sin z
/
I=
•
The Flow' Pattern described by the
-
log (sin z)
•
Re( log(sin z) ) - c
Im( log(sin z)·)
•
c~mplex
1
potential
------_ .. -­
= c2
Problems
Discuss the flov described by each of the folloving complex potentials.
Locate all sources and sinks and determine their strength.
points.
Find all stagnation:
Find the velocity vector and sketch the stream lines and the
equipotential lines •
• If.lf
w
=:=
.. l.f 5"
v
= log
(sinh 11 z)
I) tf.6
w
= log
(esc z)
.0 '1,,7
v = log (tan 17' z)
#I
•
log (cos z)·
/
9.26
•
In this and the previous sections we looked at the level lines of the
real and the imaginary parts of an analytic function
iv{x,y). We saw that the lines
:f'J.u1d :f'J.ow and that the
of
u{x, y)
vex, y) = c
vel~city
as well as from
fez)
=
u{x, y) +
resembled the stream lines of
vector could be obtained from the gradient
f'(z).
We did not derive these properties of
the complex potential from basic assumptions on the nature of the fluid.
We
simply suggested through diagrams that the complex potential can provide a
useful l;llOdel of the flow of an ideal fluid.
•
Careful derivations of these
properties are given in texts on fluid dynamics.
9.;
Conformal mapping of flow patterns
In addition to using an analytic function
fez)
as the complex potential
describing the t:low of a fluid, we can also use mapping properties of
another analytic function to transform
be more precise, imagine that
fez)
fez)
into a new flow pattern.
To
describes the flow shown, and that the
y.
z - plane on which these
stream lines are drawn is
made of rubber.
•
Now consider another analytic function
"':3
/
= M( z) , where ~ = s + i t .
This
•
tunction
:s .-
M( z) maps each point on the
plane.
z - plane to a new point on the
We can picture this mapping as a distortion of the rubber
surface to some new shape as shown.
The stream lines drawn on this
rubber sheet have also been
. distorted.
relation.
-:s = M( z)
lCl ( ~ ).
•
We can solve the
for
z
in terms of ~ and denote the result by z
=
Now the complex potential describing the new flow pattern is
f( z) = f( ~l
('S
».
Thus we see that a familiarity with mapping properties
of analytic functions'S = M( z) can generate new models of fluid flow patterns.
Example 1
In Example 2 of the previous section we saw that the complex potential
fez) = V (z + lIz) describes a
o
uniform horizontal flow which
/
encounters a cylindrical
----~~~---~x
obstacle of radius one at
Z -
the origin.
•
PLANe
Describe the
complex potential of a flow in the direction
... !
1'1
4
of velocity
V far from
9.28
•
5 with center at the
the origin about a cylindrical obstacle of radius
point
5+ 5 i •
'S -
PL.ANE
Solution
•
!n
section
5.5 we saw that the mapping function
~==
N(z)
=A
e
i oL
z ;+ ~o·
describes the following alterations of the complex
(i)
A magnification by the factor
z - plane:
A.
(ii) A rotation through the angle at- •
/
(iii) A translation by. the vector
~o.
In our problem, we wish to:
(i)
•
Magnify the pattern in the z - plane by the factor
5
(A = 5).
(11) Rotate the pattern in the z - plane through
-¥-) .
the angle
(11i) Translate the center of the cylinder to the point
5 + 5'~ (
~AUS
'0 0
= 5 + 5 i).
we see that the required mapping fUnction is
:) = M(z) = 5 ei 71'/4
Solving for
z
z +
5+ 5 i
in terms of ':) w.e get
z = ~l (~ )
•
=
e- i Tr'/4'
5
(.
.
-··".:r -. 5 -
n
5
)
i
.
Substituting this result into the complex potential
fez)
= Vo
(z +
liz)
we get
/
We must now select
For large
I~ I,
Vo
so that the speed of the now is
(1) is nearly
F(~ )~
Vo e- i
5
•
V far from ~= 0 •
and thus the velocity is nearly
11""/4(:5 - s-- S 1. )
/
------------------------------
. 9.30
•
5
Thus the speed
V is
and we sel.ect
potential is obtained. f'rom
= 5 V.
V0
Fina1.ly' our complex
(1) as
Problems
• s;, 1
A cylindrical obstacle of' radius
nov of speed
4
•
i.
5
4 is placed in a uniform horizontal
from left to right.
The center of' the cylinder is at
Find the complex potential describing the nuid motion.
• S.2:: A cylindrical obstacle of radius
f'low (upward) of speed 10.
2
is placed in uniform vertical
The center of the cylinder is at 4 + 4 i •
Find the complex potential describing the fluid motion.
The use of other analytic mapping functions
nov patterns.
:s = M( z)
produces further
It is important that the equipotential lines and the stream
/
lines intersect at right angles at points where the velocity is finite and
nonzero.
Thus we want our anal.ytic function
M(z)
to be a conformal mapping
at all points in the interior of the region in the z - plane where the flow
•
occurs.
I
We require that
M'(z) ~ 0
in the interior-of the region where we
9.31
•
draw stream. lines.
oS
I
I
,
\
z- PLAN~
~
\
\
M/(i!)
IS />JOT
FoR.
O<JTS/Oe: THE.
Z
\
Z.ERO
I
.
I
\
\
~- PL..ANE
Oes,Ac,LE
9.6
The J'oukovsId transformation
One of the most important mapping functions is
•
:s = M(z)
= ~ (z +. lIz)
known as the J'oukowski transformation. Figure
mapping properties of this function.
9.;2. shovs the important
The examples and problems of this
section illustrate their application.
Example 1
Verify the mapping illustrated in Figure
9.2
(i).
,/
Solution
We must show that:
(i) The circle
•
segment
z = ei
-e- ,Of=. -& !:: 21Y maps onto th~ line
t
=0
•
(11) The Joukowski. transormation (1) is conf'ormal for
t
1>
z
1.
This means that
M'(z) exists and is not
zero there.
z
If' we set
j
.
As
to
= ei
=
-9- 1n
~(e~-9- + e-l~)
-e- varies f'rom
0 to
along the real
1
(1) .we get
=
2 ir, cos
J -
axis.
cos
-e-
-e-- •
varies £rom
1
to
-1
and then back
Thus item (i) has been demonstrated.
From
(1) we have
•
z = 0 , and is zero for
ThuS M' (z) is undefined
for
these points are not in
Iz I>
z =
+ 1.
Since
1 , item (ii) has been demonstrated.
Remark:
We can tell that
9.2
at Figure
(i).
M' (z)
=0
for
z
=
Since the curve in the
+
1
at f'irst glance by. looking
z - plane at points
A and
C
/
is pinched at the corresponding points
that
M( z)
Example 2
•
A' and C'
in the ~ -. plane we know
is not conf'ormal (fmgle-preserving) at these points.
/
Determine the complex potential describing the flow that results when a
•
:nat plate of width
o~
velocity
Solution
2
is placed 1n a uniform vertical (upward) stream
V.
t
y
z - plane
Z -­ plane
The flow on the
z - plane is described bY' the complex potential
Vo ( z +
•
lIz).
To -JIlap this pattern onto the
Z - plane we must rotate the z - plane through
90°
Z = i z •
-Thus
by means of the mapping
z
= -
1 Z
From Figure
and the complex potential becomes
-1 V (Z - +-).
0
/
•
-:s - plane
9 .2
( i)
we know that
will map the flow pattern on the
Z - plane to that shovn on the :)--·plane.
Solving
:5
(3)
for
Z
in terms of
we get
•
Z2 _ 2 ~ Z
:
.'
+ 1
=0
(4)
denote that branch of the function that is continuous
on the ~ - plane vi th a branch cut along the real axis from - 1 to 1 and
of:
f":S 2
such that
branch is
such th~t
- 1'· is a positive
Vf'
red number when ' j 2 - 1 is a
-1
= 1
s
1
positive real number.
branch cut f r
We must take then
~2
- 1
,
•
for the minus sign in
I z I =1
(4)
would map the
':5 -
plane onto the interior of
rather than the exterior as is required.
SUbstituting
( 5)
into
(2)
(See Figure
9.2 (ii».
we get
(6)
as the complex potentie.l.
;
the velocity is
Vi.
We must now select
When
13 I is
r,::;;:,
- 2 i Vo
large t
V
0
n
so that when
1'3 Iis
2 - I' is nearly
'J
(6) becomes
F(
•
'0)
0
..' I
large t
and thus
9.35
.'
The
Thus
velocity is then, for large
2 Vo = V and
(6)
F(~
)
I':) I ,
is at last
= _'.L-.!.
·2
This is the required complex potential.
Example 3
A uniform now of speed V' in the direction
obstacle whose
•
cros~
is the ellipse shown.
45 0 encounters a cylindrical
section
Determine
the complex potential describing
the flow.
Solution
Examine the three diagrams shown •.
z~-
plane
·plane
The complex potential describing the flow in the
•
(7)
plane
z - plane is
f
I
-------------------------- --
z = e - i 1'r /4
z.
(7)
Thus
becomes
c 2 e i 1'r
v
t:':"i 1l"'/4. z +
o\e
(8)
We can map the
z
-::s -
Z - plane onto the
plane Using the JoukowsId
9.2 (11i).
formation as shown in Figure
/4)
trans­
Thus
and
•
as vas seen in
(8)
( 5)
of Example
2.
Substituting this last relation into
ve get
F(~
)
=
V
o
[.'.e- i
+
/
We must now select
For large
t:..
so that whenl~ lis large, the veloci~y is
2
I':5 I , .J -:s - 1
F(
•
Vo
i
V e
~ ~ and we have
":5 ) ~
Thus the velocity for large
2 Vo
e
- i
IJ I is
'if/4
"e
-.J •
I
i
'TT'/4
9.31
and V
=2
Vo
•
Now the complex potential
F(
V
2
'3 ) =
[
.- i
(9)
becomes
1Y/4(~ + h 2 _ 1')
+
c2 ei
'ftI/4
J+l!,2_
]
l '
Problems
• bof
Verify the mapping described in Figure
• 6.2
A flat plate of width
V e i 11'13.
•
«i 6..3
2
9.2
(iii) •
is placed in a uniform flow with velocity
Find the complex potential.
Locate the stagnation points.
A uniform flow in the direction of the real axis with velocity
encounters the elliptical obstacle of Example 3.
V
Find the complex potential
describing the flow •
• 6, Lf Find the complex potential describing the flow of a fluid which has
Wliform velocity
~
a hurdle of height
b
when it encounters
as shown.
/
••
9.38
•
.2
z-
PLA~
y
•
'.
j
(l2)
A.
C
1
-I
I
I
I
0
I
I
•
,.
.:
9.39
Figure
'Z ,,"
!
I
!
•
•
9.2 (continued)
,
((v-)
I
I
I
PL.. A AlE'
9.40
One application of the Joukowski transformation is to the study of air
flow past the wing of an airplane.
the curve in the
'!-
Look at Figure
9.2 (vi) and observe that
plane
resembles the cross-section
of an airplane wing..
supplementary problems
and
In
9.6.3
9.6.4 the flow past these "Joukowski profiles" is briefly examined •
•
,/
.. /
•
. 9.41
••••
'.
9.7 The electrostatic-field
f( z) =
We have seen in the previous sections that an analytic fUnction
u (x, y) + i
vex, y) can be thought of as a convenient "package" for the
description of certain fluid' flow problems.
The analytic function also
provides us with a description of the two dimensional electrostatic field.
Suppose we have several parallel cylindrical metalic conductors in empty
space as shown.
We assume that the cylinders are very long, so that
y
..
•
•
/
the electric field established is identical in any plane perpendicular to
these cylinders.
Suppose electric charges are placed on each conductor.
An electrostatic potential
u(x, y) is then established in the space
I
/
surrounding the conductors which has the properties:
•
(1) ~
(2)
•. .
2
u..
=0
u(x, y)
in free space,
()..
(#XX +~yy = 0)
u is a constant on the surface of each conductor.
Our knQvledge of analytic functions can help us to select appropriate
potential functions for specific problems.
We illustrate this in the
following examples.
Example 1
The positive and negative halves of the x-axis each represent infinite
•
•
" by insulation at the origin .
conducting plates separated from each other
The plate
X
.£
x
0, Y = 0
> 0,
y
=0
is kept at potential u
is kept at potential u = -2.
= 3,
.'hile the plate
Find u in the upper half plane.
Solution
We seek a harmonic function in the upper half plane
on the p.?sitive x-axis and -2 on the negative x-axis.
e
= arg(z)
e
takes on the value
pesitive real axis and 7T on the negative real axis.
and
B
The function
is the imaginary part of the function
1s thus harmonic. Furthermore,
A
y> 0 that is 3
0
and
on the
Can we find constants
"'
such that
u=A-& +
log z
B
/
."
is the desired potential?
When
of}
=1'r and
u
= -2,
When .e- = a and u
(1) now reads -2 =
=3
we have from (I) that B
tr A + 3. Thus A
= -5/11'.
= 3.
The
desired potential is then
5
-1F
of}
+
3. '.
Since -e- = tan-l
y/x, we have
(3)
u
=
_ ---L tan-l
~
y/x
+
3, 0
~
1
-t a n..- y/X
L
71
~
The electrostatic potential (2) is the real part of the function
fez)
•
=
~ i log z
+ 3.
We call t(z) the Ifcomplex electrostatic potential" or silnply the "complex
potential. t·
Example 2
Three conductors on the x-axis, separated by insulation at x
are charged as shown
•
Determine u in the upper halt plane and the complex potential.
=+
I
•
'
9.44
.
...
Solution
We assume a solution of the form
(4)
u = A -&1 + B -&2 + C
where A, 13 and C are consta."lts to be determined and
~l
and -&2 are the angles
z
shown.
- - - - - -.....--.../..----+-------#------!------~x
-1
•
We know that u is a harmonic function because -&1 is the imaginary part of
log(z -
1)
and~
2
is the imaginary part of log(z +
1).
From the given boundary conditions we see that u
~ ="6"
= 0
12'
5
Thus (4) becomes
=0
and we s.ee t.hat A
Now (4) becomes
=
·2/ rr.
-10
=
Thus (4) becoJ:les
Finally, when "6"1
2
= it2
+ 1r13 + 5
~
and we see that C
+ 0 + C
When ~l =11'- and ~2 = 0 we see that u = 7.
= 5 when
7 = TTA + 0 + 5
= 11, we have u
B
or
=
= 5.
= -10.
Now we
-17/'Tl- •
i
-
•
Since -e-l
= tan-l
17
11'-
- L and ~2
x-l
y
x - 1
-&2
+
= tan-l
17 tan-l
7r
5.
y
we have
.. /
x + 1
~
x+l
+
5, where both angles
ha.v~
are in the range [0, ft-'] •
(6) . fez) =
-
The expres sion (5) is the real part of
2i
.
1l'" In (z- 1) + l~ i
lIi (z + 1) + 5 __
and thus (6) is the desired complex potential.
Problems
• 7:, 1 Find the potential u(x, y) and the complex potential f( z) = u + i v in
the upper half plane when conductors separated by insulation on the x-axis
are charged to the potentials shown.
<) 7.2. The positive real axis and the positive
imaginary axis are conducting plates separated
by insulation at the origin and are charged
as shown.
Determine the potential u(x, y) in the first quadrant.
Also
(
find the
co~plex
potential.
/
•
.:
...
• Z. 3
Determine the potential in the
an~ar
region
sho~
VI are constants.
where
va·
and
Also find the
complex potential.
•
7. if Determine the potential between
the two charged coaxial cylinders
shown.
•
We saw previously that
a source of fluid flow.
Here we
f(z)
~gined
=m
log(z-zO)
described
an infinitely long line piercing
the complex z-plane orthogonally at the point Zo with fluid rushing out
radially in all directions.
The quantity of fluid emerging from each
linear foot of the line source was 211 m units of volume per second.
Corresponding to these line sources of fluid flow we have line sources of
/
the electrostatic potential.
to the z -plane at z=Q
Picture a straight wire conductor perpendicular
I
on which an electric charge of density q units
of charge per linear unit of distance on the wire is evenly distributed.
•
The electrostatic potential generated by this line charge is
u = -2q log r
and the complex potential is
We will not derive (7), since our principal interest is in the applications
of conformal mapping.
Corresponding to the velocity vector in fluid flow, we have the
electric field intensity vector
->
E.
~n.general,
a line charge q per unit
...:0.
length would be SUbjected to the force q E per unit length.
The electric
field intensity vector is described by the equations
•
~
\lu,
E = -
and
( 8)
Example 3
A line of charge
~
per unit length is at z
."/'
charge
~
per unit length is at z
= 1..
=0
and another of
Find the potential, the complex
potential and the electric field intensity vector.
/
Solution
•
- -
From (7) we can write down the complex potential at once as
•. _ •
~..
- -. •..
•... ... _
~
_-...
~~
'.
..... •.
~
Q"_~"'"
- .......... '.'
•.
"
.~
~.
. ............ ••
~
.,
'll_ ...... ~. . . . "
....
~
•
'," ..., - , . - . - . •
·9.48
•
fez) = -2q1 log z - 2 q2 log (z - 1) yhich is the sum of the
potentials due to the sources. at z = 0 and at z = 1.
u is given by u
= Re
u= - 2q s.ln
u= -2Qi In
u
= _q 1.
In
Since the potential
fez) ye have
1z 1
J
-· 2Q2
In
Iz
2'
x 2 +y
- 2q2
- 11
In
/(x _ 1)2 + y2
,
(x2 + y2) _ q2 In «x - 1) 2 + Y2 ).
The electric field intensity
~ector
is
...::0.
E=
-
f/(z)
"
-:0.
•
E =
-
Z
Z - ~
----_._--_. --" ------ ­
•
We have seen that the Re fez)
potential.
What does Im f( z)
the lines v
=c
=v
=u
describes the electrostatic
describe?
In our study of fluid flov
(c constant) vere called stream lines and the velocity
vectors vere tangent to these lines.
In the study of the electrostatic
//
field the lines
-.:lo.
Im
fez)
=v =c
are called
lines of force.
The vector
(
E is tangent to these lines.
/
•
'Examnle 4
Suppose that in Example 3, ql
= q2 = q.
Find the equations of the
.9.49
•
lines of force and sketch these lines and the equipotential lines.
Solution
=~
Since ql
= q, ;the complex potential (9) becomes
f(z)-: -2q log z - 2q log(z - 1)
~
fez)
= -2q log
fez)
= -2q
log
Butz(z- 1)
= x2
z(z - 1) is
arg (z(z - 1))
The
z (z - 1)
Iz
(z - 1)1
- 2q arg (z(z - 1)) i.
- y2 - x + i (2 xy - y)
= tan­ l
and therefore the angle made by
2"x:v-y
x2 - y2 - x
imaginary part of the complex potential is given by
vex, y)
= -2q
tan-l
2 xy - y
and thus the lines of force are obtained from
-2q tan
,
-1
= C •
We get
2 xy - y
2
,/x _ y2 _ x
= tan (_
~q")
= C,
where C is an arbitrary constant.
/
•
•
y,
... ....
" ...,
\
"
.\
,
,
I
,,
I
/
(
t ~_~_-i----;~~';"'-~-~---e!::--~~---l-----\.---r---~ )(
I
\
\
\
,,
•
......
/
_--­
--
---
u(x,y) = constant
equipotential lines
_
v(x,y) = constant
lines of force
---­
J'
•
I,·
/
•
Example 5
Identical line charges
z
='irn,
(n
= 0,
•
-3'-
~1, ~
• ,
-2".
2,
-IT
q
per unit length are placed at the points
... ) .
~
0
•
7T
•
2.".
•
3.".
:>-X
Find the complex potential at all points in the z-plane.
Solution
This problem. is the electrostatic analog of the fluid flow problem
considered in Example 5 of section 9.4.
•
lI1'
We need only replace the strength
of the sources of fluid flow by -2q to get our cotlplex potential
fez)
= -2q
log (sin z).
Example 6
A line of charge q per unit length is located at the point z
while the x-~~is describes a grounded conductor (has potential u
=i
= 0).
,/
Find (a) the complex potential, (b) the electrostatic potential, (c) the
lines of force, and (d) the electric intensity vector.
I
.
•
I
~
Solution
The expression -2q'log /z
il
describes the required line charge
9.51
••
at z
=i ~
but will not make the potential zero on the real axis.
we place a line of charge -q at z = -i.
Suppose
Then the potential on the real
y
u=o
----------I---;;;::.............;:;------~
X
-1.
axis will be zero.
u(x~
•
y)
c: harge - g
Thus the required electrostatic potential is
= -2q
log J z - i
I
tz
+ 2q log
+ i
I
a,nd the complex potential is
fez)
= -2q
fez)
= 2q
log
log
(z - i) + 2q log (z + i)
z + i
z - i
The charge at -i is called the image of the charge at i and' the technique
used here is called the method of images.
Since
/
z + i
z - i
=
x + i (y + 1)
x + i (y - 1)-
x
2
+ y
~i2- .
2
- 1 + i 2x
+ (y _ 1)2
we have
••
fez)
= 2q
log
= 2q log
.l!:2.L
J z- i ,
j
+ 2 q i
x 2 + ("y+1)
J X:2
21
+ (y-1 )2'
arg
/
( Z-J
Z+.i )
1
+ 2q i tan-
~_
. ':._,. .
.......
---'
-~
... - -.-_ •.. ------.
,
9.52
•
fez)
=q
log
2 + (yo + 1)2j
x
+ 2 q i tan-l
[ x2 + (yo _ 1)2
which is the complex potential.
We see at once that the electrostatic
potential is
u(x, y)
= ~e
2 + (yo + 1)2]
[
f(z) = q log
:2 + (yo _ 1)2
•
The lines of force are vex, yo)
or
.x2
+
2x
?
= CI
= 2q
tan
-1
•
-1
These can be expressed as
x
2
2
+ yo
- 1
= 2C
.
x
(C
1
=- )
c'
which reduce to
•
(x - C)
222
+ yo = C + 1
upon completing the square.
These are circles with centers (C, 0) on the
real axis passing through the line charges at
+
i.
/
---+--t---+--I+H-++--+--.--+-----t------:~
X
.­
•
/
9.53
•
~
The electric intensity vector
='(t~ +
·(~;\ 1
E
is
2q )
;i:"i
-4q-i
=
z2+1
Problems
In problems 24 through 29 find-(a) the complex potential, (b) the
electrostatic potential, (c) the lines of force and (d) the electric
•
intensity vector for each of the given distributions of line charges •
A line of charge q per unit length is at z = 1 and a line of charge
-q per unit. length is at z
• 7.6
= -1 •
Identical lines of charge q per unit length are located at
z = 1'r/2 + 1rn, where
n
is any integer.
,/
Identical lines 'of charge
q
per unit length are located at
I
z
=i
n, where
n
is any integer.
/
/
Identical lines of charge -q per unit length are located at
z
=1'Yn,
where
n
'1
is any integer.
:--~ .~.,
<11ff".'!'~·
. 9.54
Lines of charge q per unit length are located at z
of charge -q per unit length are located at z
• 7.10
=~ +
= n,
while lines
n where n is any integer •
A line of charge q:per unit length is located at z = 1 while the
imaginary axis is a grounded conductor.
A line of charge q per unit
length is located at (a, b) in the
first quadrant.
•
The positive real
and imaginary axes are grounded
conductors.
Find the complex potential and the electrostatic potential.
A useful transformation for mapping flow patterns and electrostatic
potentials is obtained through the function z
= e rJ
upper half of the z-plane onto the infinite channel
the ~ -plane.
This function maps the
0 ~ 1m (
'::J ) ~ 11 on
It is useful to visualize this mapping. in the following way.
Visualization of the lJIauning
Z
=. e~
'I
Imagine the upper half of
•
the z-plane as a rubber
sheet.
--.:.'--'--+,--J'--J.--j-.,.--I....-...I-.-.I;--..l....--!--~x
F
C
9.55
(ii) Break the rubber sheet at the
. origin.
F
(iii) Hold the edge ABC fixed and
rotate the edge DEF until
it is parallel to ABC and
the strip is of width
D
F
F
.B
......
fr.
A
(iv) Grab both the ends at A
Jr-
r m3
and D and stretch them to
11
the left to inf'inity.
The
D'
E'
F'
line segI:lent BE falls on
,
.... c'
,
Re '3
the imaginary ~ axis.
To see that this is the correct mapping the reader should demonstrate that
the real '~is on the ';;! -plane ( ';j = 8
),
t = 0, -
Co
< s<' ~ ) maps onto
the positive real axis of theI z-plane and that the line
( - Clo
<8 <:
mapping
z
G>'O
)
= e;:j
'::f =
8
+ i 1t­
maps onto the negative real axis of the z-plane under the
..
I
I
•. Also the reader should show that the line segment
~ = i t (0 $ t ~ ~) maps onto the upper half of the \1.'11t circle on the
•
9.56
z-plane.
Exampie 7
The diagram shows an inf'inite channel of width 1)'- bound on both
sides by grounded conductors.
po1nt "j = i
charge q.
At the
1"1- /2 there is a line of'
char-cae.
Determine the potential
't
III
1111/
t'uncti.on.
'U.=o
.
•
Solution
The mapping f'unction z
= etj
transf'orms the channel on the
~ -plane into the upper half of the z-plane.
~
=1
1}-/2
maps onto z = i.
The line source at
The lower boundary of the channel maps onto
the positive x-axis and the upper boundary onto the negative x-axis.
we solve the problem of having a line source of charge q at z
/
x-axis 1s at zero potential, the function z=
.
onto the desired channel in the ~ -plane.
.
e~
=i
•
= 2q
while the
will map that solution
(
In ExaI!lple 6 we saw that the
solution of this problem in the z-plane is the complex potential.
fez)
If
log
z + i
z - i
S
9.57
Thus the solution to the present problem is obtained by setting
z
=
e~
and obtaining the complex potential
(10)
= 2q
fez)
log
To find the potential u(s, t) and the family of lines of force
v(s, t)
=c
we must find the real and imaginary parts of (e ~ + i) / (e ~ - i).
e 'i + i
e'! - i
•
=
5
e eit + i
e S eit - i
-it
+ i
eS e
S
e~it
+ i
e
=
e
5
-1 + i e
e 2s +1 + i e S
(e it + e- it )
e- it )
(e it
=
e 2s _ 1 + i 2 e S cos t
2 e S sin t
e 2s + 1
25
Nowwe see that
J(e 2S ........l )2 + 4e 2s cos 2 t '
e 2s + 1 - 2e s sin t
and
/e:r
g
.(­
-! +1)
./ :
"!. -1 =
tan
-1
2e s cos t
e 2s - 1
;'
Combining these last two
exp~essions
with (10) gives us the electrostatic
potential
••
u(s, t) =q log
rce 2s........l )2 + 4e 28 cos 2 ~
t
s~
r~2S.+ 1 _ 2e 'sin ;.,
I
•
_..
.
~.'YY~
~-·--~~·r
,
---------------------
•
and the family of lines of force
v(s,
t)
=
2q tan
-1
s
2e cos t
e 2s - 1
= C' •
This last expression reduces to
eS cos t
e 2s _ 1
where
=c
C is an arbitrary constant.
Problems
• 7,,11.
Move the line of charge
q
to the point j = bi in Example 7
and
find the resulting electrostatic potential •
• 7:J5
charge
The diagram shows a line of
J..a.
1.L:: 0
at ~ = bi in a channel of
:bl.
~ charge
q
width "a" bound by two grounded con­
ducting plates.
et
III
'U.= 0
.s
Find the potential.
The diagram shows a line of
charge
/
line. of­
q
c.harge.
between two infinite
't
ffie"
L" J=e./'''''/8
grounded conducting plates making
angle 11"'/4.
\
Find the complex potential.
s
1.£.=0
~
; '..
-~'.
r .
4<~
.~
...... :-
..,.
..
...- ..-.
,_.------...
9.59
•
9.8 Heat flov
We have stud;
jed
1
problems in fluid flov and the electrostatic potential.
We nov study tvo dimensional, steady state temperature distributions.
The
figure sholl's the cross sectional
y
.boundary of a cylindrical piece of
t"emperB-t
u. reo s
on ~"e..
bOCAn da.r
I< noWn
y
?
material that extends to infinity
1n a direction perpendicular to
the z-plane.
Every plane section
"
•
of the cylinder parallel to the
z-plane looks like this.
--4-----------:;0>0 X
z - plahe.
Suppose
the temperature distribution is
known on the boundary, and that the temperatures are not changing vith
time.
What is the temperature at each point inside the cylinder?
u(x, y) denote the temperature at the point (x, y).
,/
Let
In section 9.2
Ve sav that u(x, y) satisfies Laplace's equation and is thus the real part
of some analytic function f(x)
complex temperature.
•
= u(x,
y) + i vex, y).
We call fez)
the
The problem at hand is to select the approprfate
function f( z) whose real part u(x, y) matches the given temperatures on
9.60
the boundary of the cylinder.
Exanple 1
Consider the vertical slab of
material shown.
The left face of the
slab is maintained at the constant
temperature, 30
o
x
and the right face is
at 700 (centigrade).
Find the ternp­
erature distribution throughout the
•
slab.
Solution
We must find a harmonic function u(x, y) such that
u(O, y)
= 30,
and u(l, y)
= 70
,
Of course, we also require that u(x, y) be "nice" (have no singularities
//
or discontinuities) inside the slab (0
~
x
I
~
1).
It is clear that the
function u(x, y) should be independent of the variable
boundary conditions do not chan:e with y.
•
of the form
u
= Ax
+ B, where both A and
y, since the
Suppose we try a
simpl~.
B are constants.
function
This is a
harmonic fUnction since it is the real part of the analytic function
= Az
fez)
+ B.
The boundary condition on the left edge of the slab tells
= 0,
us that when x
u
= 30
and thus
30 = A • 0 + B
The boundary condition on the right edge of the slab reads
x = 1
u
= 70
when
70 = A • 1 + B.
yielding
,
The simultaneous solution of these two equations is B
= 30,
and A
= 40.
We have then the required temperature distribution
u(x, y)
•
= 40
x + 30
and the complex temperature
fez)
= 40
z + 30.
We notice at once that this problem and its solution is exactly
equiValent to the following problem from electrostatics: "Find the potential
u(x, y) inside the channel bound by conducting plates at x
=0
and at x
=1
when the' left plate is maintained at potential 30 and the right plate is
maintained at potential 70."I We have only a change in terminology, but no
change in the required mathematical analysis.
•
If
fez)
= u(x,
y) + i v(x, y) is the complex temperature, then
the level lines of the temperature
Re f(z)
= u(x,
y)
=C
(constant)
are
c&l1ed "isotherms", and the level lines of
called "lines of nux".
Im f{z) = v{x, y) = C are
Lines of nux are lines along which heat flows.
In the previous example the isotherms are the vertical lines
the lines of flux are the horizontal lines
y = c.
x = c
while
Notice that heat novs
along the flux lines frOI!l right to left (since the right edge has the
higher temperature).
Table 9.1.
shovs the similarity in the applications con­
sidered in this chapter.
•
Remark on Example 1
A mathematical problem is called "well posed" if it has a SOlution,
and only that one solution.
the prob1em is weU
We say
pose~prob1em
a solution exists, for we found one:
fortunately, no.
Clearly
in Example 1 well posed?
u = 40 x + 30.
But is it unique?
Un­
Consider the funciton
sin'l7' z = sin ~ x cosh'11"" Y + i cos ~ x sinh 'ir y •
I
The real part of this function is harmonic, and zero for x
=0
and for x
.,
•
Thus the function
U(x, y) = 40 x + 30 + sin 1t x cosh1r y
' l " ,"""'''
,r
1 ~L
,"
.•
I
I
= 1.
'.
TABLE 9.1
•
•
COMPARISON OF TOPICS IN THE TWO DIMENSIONAL, STEADY
STATE STUDY OF FLUID FLOW. ELECTROSTATICS AND HEAT FLOW
;;
ELECTROSTATICS
FLUID FLOW
HEAT FLOW
=========================I:::=::========================::t::1========================:::::::::::======:=====
;.
velocity potential
u(x, y)
~-:,
"
I~
~
t;
..
J
~1
!
I
;1
complex electrostatic
potential
fez) = u(x, y) + i vex, y)'
complex temperature
fez) = u(x, y) + i vex, y)
stream lines
vex, y) = c
lines of force
vex, y) = c
lines of heat flux
vex, y) = c
equipotential lines
u(x, y) = c
equipotential lines
u(x, y) = c
isotherms
u(x, y) = c
I
electric intensity
v= \}u= fTTi)
E = - '1u = -
at the boundary of an
obstacle vex, y) = c
and
du(x. y) = 0
'dn
At the boundary of a
conductor
u(x, y) ='c
lleat flux
H = -K V u =- K
K = conductivity
f~,-r(z~)
alone the
material
.
bo~dary
d u(x.
dn
y)
_
f
I
(z)
of insulation
- 0
-
(n normal to the surface of the
insulation)
(n normal to the obstacle's
surface)
~
~~
:l
--_::~:~:~~:~-~--~~~~:~~~~~~--rl-~~~r:;:~=---~=======
complex potential
fez) = u(x, y) + i vex, y)
.xelocity vector
'1
temperature u(x, y)
electrostatic potential
u(x, y)
I
I
I
I
~
•0'
I\)
•.....
•
is also a solution to the problem.
fail to be unique?
Why does the solution to this problem
The reason for this is our failure to specify what is
happening to the temperature across the top and the bottom of the slab.
Since our slab goes to infinity in the
is really no top and bottom to the slab.
y
direction, we note that there
I'levertheless, we can think of
the slab as having finite, though very large, height.
. lim
cosh y =
Notice that because
OQ
y~+ 000
••
the solution U{x, y) approaches infinity for large y.
This means there
is a great heat source at the top and bottom of the slab.
If we had re­
quired that the solution u{x, y) be bounded throughout the slab, the
solution U{x, y) would have been eliminated and the solution of the problem
would then be unique!
For this reason, we will add the restriction that
the temPerature be bounded when the material under question extends to
infinity.
Besides having the temperature specified on the boundary of the
••
material whose internal temperature is to be determined, we can also specify
that a portion of the boundary is insulated.
The presence of insulation
9.64
•
means that heat cannot flow into,
or out of that portion of
the boundary.
Thus an insulated
segment of the boundary coincides
with a line of flux v(x, y)
This also means that
0, where
n
•
d
u (x. y)
(,) 'n
13'"
<
=
~"'..\
"') \" '01.­
0
is normal to the in­
sulated surface.
. the symbol
~'"
= C.
He will identify an insulated segment of a boundary by
\ \ , '" " "
Example 2
Find the bounded temperatures
inside the slab shown.
SOlution
/
Our solution
u
= 40
x + 30 to Example I gives th~ required
temperatures on the left and the right boundaries.
segment of the x-axis is a line of flux y
••
the required solution.
= O•. Thus
Also, the insulated
u
= 40
x +
39'
is
Note also that along the insulation, the normal
derivative
is zero.
Problems
• 9.1
Find the bounded temperature
distribution inside the slab shown.
Also find the complex temperature,
the lines of nux and the isotherms •
• 1f,2.
-Find the bounded temperature
inside the slab shown.
•
.
.
-----""'"'-L.---.,;;...-....:;....~-_=_-~..,x
.
• 8'.3
Find the bounded temperature
inside the region shown.
• 3. If
Find the temperature inside
the region shown.
y
Also find the
./
complex potential and the lines
of flux.
u. =70
Another useful transformation is the function
z = sin .~
z
= sin
s cosh t + i cos s sinh t .
Cl
3
/
9.66
•
It is convenient to visualize this mapping in the following way:
t;
(i)
A
Consider the semi-infinite
,
I
I
strip shown and imagine it is
I
B
made of rubber in the ~ -plane
(11 )
Fold the lines
AB
(iii)
JI\F I
I
I
I
E
I
I
I
I
I
I
I
I
I
I
I
I
I
D
....
and
A
down as shown.
ED
I
I
When these lines reach
the x-axis, the formerly hor­
•
.. izontal lines t = c should be
ellipses and the vertical lines
.should be hyperboles. s
= C.
Both these families of conics
have the points
B and D as foci and are thus called confocal ellipses
(/
and hyperbolas.
(iv)
B
Finally, move the points
and
D
sp~ctively
to -1 and +1 re­
on the z-plane.
•
y
F
s
9.67
•
The following relations derived from z
= sin 'i are
useful:
Transformations from (s, t) to (x, y):
(2)
x = sin s cosh t
(3)
y
= cos
s sinh t.
Transformations from (x, y) to (s, t):
(here. - fr/2 ~ 5in-l
CJ
~ 1172)
J (x- I) 2 + y 2 2J;1 .
•
•
The equation of the confocal ellipses is
(6)
+
+
-:,2
=
siIlh2 t
1.
The equation of the confocal hyperbolas is
= 1.
A method of deriving these relations is outlined in supplementary problem 9.8.9.
Examnle 3
/
Find the bounded temperatures
I
inside the shaded region shown
vi th the indicated boundary .
•
conditions.
•
9.68
Solution
Return to Example 2.
in that problem.
Notice that similar boundary conditions occured
Thus we need only employ the necessary transformations
to map the semi-infinite slab considered in Example 2·onto the first
quadrant to get the solution of the present problem.
f'ollowi.n~
The
two mappings ere used:
y
•
t
1
o.
,.. s
X
0
Z-plane
x
1C
2..
l.:-plane
l.:
Z·
= trZ/2
= 2l.:/tr
z-plane
z
= sin
l.:
The solution to the problem in the 'l..Plane is u = 40 X + 30 as we saw in
/
Example 2.
(Here we have replaced x by X.)
the simple magnification ~ =fT Z/2 or Z
x+
•••
,
i Y =
~ + i
1/'
2t
1t
The first transformation is
= 2 ":S /ft".
This means that
.­ /
••
and consequently
2s
x=
n-
The solution for
u
80 s
=
'n-
u
now becomes
+ 3\).
The second mapping
z = sin ~ requires the use of relation (4) to transform
this temperature into
80
u=
11­
(J
(x + 1) 2
~
y2 I
-
~ (x -
J
1) 2 + Y2 )'
+ 30
which is the desired solution •
•
Problems
Find.the bounded tenperature distribution u(x,. y)
inside each of the
shaded regions having the indicated boundary conditions.
/
I
!
•
9.70
•
•
•
3',,7
•
"t(. :::
/
••
"10
0
fr.3
•
9.9
The bilinear transformation
The mapping function
az +
w= cz
+
b
a
~here
ad - bc , 0,
1s called a bilinear transformation, a linear fractional transformation
and a M~bius transformation.
It is a simple, yet iI!lportant mapping and
we shall study its effects in this section.
The special case of (1) in which c = 0 reduces to the linear transfor­
mation
•
w = az + b
(2)
which we exacined in section 5.5.
Recall that the mapping produced by (2)
can be thought of as a magnification (by the factor t a \ ), followed by a
rotation (through the angle arg (a)), followed by a translation (by the
vector b).
Another special case of (1) is the so called "inversion transforna.tion"
,/
w=
1
z
The inversion defines a conformal mapping at all points except, of course,
Z
z
••
= o.
If we write z - re
then:we have
if;­
-i~
w= !.=!
•
z r e
.'
-e­
,I
-~-_----:':""'-----'7 X
J
-~
1/z'(,
,
9.72
•
~is
last relation tells us that if we know the polar coordinates of a
z = 2 + 2i =
complex number, for example
f'
=
2.f2
radius:
angle:
2
~ei ~/4
and ~ = ""/4, then the coordinates of lIz
I
-=
r
are easily found:
...1...- , and
2{2
-~=
_ 17­
4
An important property of the inversion transformation
This
preserves circles.
means
transformed into a circle in the
•
has coordinates
(3) is that it
that a circle in the z_plane is always
w~plane
and vice versa.
Here, however,
we must also allow circles of infinite radius, which are straight lines •
A method for proving this property is outlined in SP9.9.12.
The following
example.illustrates a geometric technique for mapping a circle under the
inversion transformation.
Example I
Find.· the mapping of
the circle shown
under
w=
I
z
••
.~
..'......~--. E -~; -.. :
o
~'":' ~
.;..;....... ,.. r. - "} .
~.,..,
.....
~
,..".,
"._
.
-4;.
x
9.73
Solution
First note that the line
0 A B D making angle
45 0 maps onto the line
V
I
/
/
DBA 0
~
lengths
-'B'
D
shown making
D'A' and
are
1/3
and
1/10 respectively.
0'
Since the mapping liz is conformal, and since the circle intersects the
•
•
line segment
AB
at right angles in the z-plane then it must intersect
AI B' at right angles in the w-plane.
Thus
our
AlB' is a diameter of
ci~cle
and we draw it
as shown.
The
di~eter
of the 'circle in the w-plane is
1/2 - 1/10
= 2/5
and thus
/
the radius is 1/5.
3/10
The center of this circle is at distance
from the origin.
1/10 + 1/5 =
The center is thus at the point
/
•
..l. e -i
10
'Tr
/4
=
3(1 - i)
10 {2'
9.74
•
and the equation of the circle is
3(1 - i)
10
IT
I=
1
5" •
Example 2
Find the mapping of the circle
Jz - 1 w
=
J3
i
I=
4
under
liz.
Solution
The center of the circle in the z-plane is
•
and the radius is 4.
A :.
2e 4 'It i/3
B
6e 11 i/3
The points
A and
B
are on a diameter passing through the origin of the z-plane and these map
A' \.-------\
onto the' points
A':
1/2 e -4 'ft 1/3
B':
1/6 e- 11" 1/3
which are shown in the v-plane •
••
Since A I BI is a diameter of the
~
.
"'I: .-~.
•
\
9.75
new circle, we can draw the image as shown.
Notice that the interior of
the circle in the z-plane maps onto the exterior of its image circle in
the w-plane.
z
=0
11
= I/O =
Hov do we know thiS? We can use a test point.
Notice that
is in the shaded region in the z-plane, and therefore its image
OQ
must be in the shaded portion of the w-plane.
To get the equation of our image circle in the w-plane we note that
its diameter is
1/2 + 1/6
•
= 2/3
and thus its radius is 1/3.
1/2 - 1/3
The center must be at a distance
= 1/6
from the origin and is thus at the point
1/6 e i2
11" /3
= 1/12
(- 1 + {3 i) •
The desired equation is then
Problem
Find the mapping of each of the following regions under the inversion
...
•
transfo!T.lation w
=
liz •
9.76
•
- 4i' ~ 2 ,
(a)
\z-21~1,
(b)
Iz
(c)
Iz
(d)
\ z + 1 - 1
(e)
'z + 1 + 1~ ~ 3
IT ,
(f)
J31
~ 2,
Re(z) ~ 2 •
loTe can now study the mapping properties of. the general bilinear function,
az + b'
c:z + d
v=
ad - be ~
O.
Multiplying by cz + d gives
czv - az + dw - b = 0,
and solving for z yields
•
(4)
z
-dw + b
cw - a
=
Relation (4) is the inverse of (1) and we sQe that it too is a bilinear
transformation.
Together, the relations (1) and (4) show that the bilinear
transformation is a one to one mapping of the extended z-plane onto the ex­
tended w-plane.
z
/
v
==
!Iotice that
dl c ~ w =4:10
a/c~z
=00
The derivative of (1) is
dw
dz
If
=
ad - be
ad - be
(cz + d)2
= 0,
./
/
then (5) is zero for all z and w must be a constant
9.77
function.
This would be a very uninteresting transformation>
w = c (constant)
J
and thus we sha.l.l always assume that
(6)
ad - bc ~
o.
With (6), (5) shows that
dw
dz
formaJ.. mapping for all z
except z = -dl c which maps into w = aJ anyway.
is never zero and thus (1) defines a con­
In previous problems and examples
the inversion transformation w
~e
= lIz.
mapped circular regions under
We now show how to map regions with
the general bilinear transformation
w=
az + b
cz + d
.
ca
Factor away
w=
to get
z + b/a
z + d/c
a
c
Next add and subtract
dl c in the numerator
z + d/c + bla - d/c
z + d/c
and simplify to get
w=
Call A
a
c
= a/c
+
, B
bc - ad
.-ae- C 2.
= (bc
1
z + dIe
- ad)~, and
... /
dIe
= D so
that we have
C2­
(7)
1
v=A+B
z +
D
Consider nov a region in the z-p1ane to be mapped to the v-plane by the
bilinear transformation (1).
Using (7) we can reduce this involved ma.pping
to a sequence of simple mappings as fo11ovs:
Translate the region by the vector D.
(i)
(ii) Apply the inversion transformation.
(11i) Magnify' the region by the factor f
B' .
(iv) Rotate the region through the ·ang1e arg (B).
Finally, translate the region through the vector A.
(v)
•
Since each of the above five steps preserves circles, ve see that the
general bilinear transformation (1), like the inversion v
= lIz,
preserves
circles.
Example 3
Find the mapping of the unit circle Izi
~
1
under .the bilinear trans­
formation.
/
(8)
w
= -i
(:
~ ~)
.
Solution
.'
•
We first convert v to the form
v=A+B
1
z + D
>
t
­
F?"""~,--
, .
.,> -.---r-
i."~::
- ..... ---:~-!"
9.79
. the procedure spec ified above.
by
(9)
Z ­
( Z +
-1
y
=
y
= -1
y
= -1
w
= -1
We have
1)
1
Z + 1 -1 -1
z + 1
+ 2i
1
z + 1
We start with
the circle shown.
•
(i) The translation z + 1
yields
(ii) The inversion yields
(iii) The magnification by .
12il = 2
gives
/
(iv)
The rotation through
~;:;:
...
the angle
arg(2i)
,
=17/2
yields
B
.'.
(v) Finally the translation
by the vector -i gives the
region shovn.
We see that (8) maps the unit circle
v
Im
~
I zI
~ 1
onto the upper half plane
o.
.
Problems
Map the regions (a)
formation
•
v
=
I zt
oS 2, and (b)
1) .
z ( z + 1
-i
1 under the trans­
..
A general mapping of the upper half plane
~ 1
<.
!m( z) c 0 under the transformation
f.f.ap the region
1vI
'Z - 11
v
=
z + 2
z - 3
Im(z) ~ 0 onto the unit circle
is given by the transformation
v = e
i~
z - a
z -
a
vhere "a" is a complex number in the upper half of the z-plane.
To see
/
this, look at the modulus of both sides of (9).
J wl=
•
) e ic><.
:
)wi = ) (10) )wl =
z -
~J
z - a
~ )
Iz - a
~
:
I
:.:
. and denominator of (10).
Iz
I \z - a: 1
- a ~
when
a.
Since
z
is in
z
y
The diagram shows the numerator
---+-----.,'-------~x
the upper half plane, and since
Ix
- al
= Ix - al
when
z
=x
(is on the real axis), we see that Im.(z)~ 0 is mapped onto' w\ ~ 1.
We can use the bilinear transformation to solve a number of problems in
fluids, electrostatics and heat flow.
The following examples illustrate
.
•
a few of these problems and their solutions •
Example 4
y
Find the temperature
distribution throughout
the region shown with
./
the givEm boundary
temperatures
u
= 20
o
for 0 c:.. ~ .(. 'fj'- and u = 0
0
for - 1} <. ~ <. O.
Find the temperature at the
I
I
•
points
z
= 0,
and z
=
i/2 •
Solution
The mapping fUnction
~
t)
= -i ( : ;
the region
maps
I z I !:
1 onto
the upper half of the
'j -
plane.
(See Example 3.)
The upper semi-circular boundary at 20
0
in the z-plane maps onto the
positive saxis while the lower boundary at 0
•
0
maps onto the negative
The bounded ter:tperature distribution in the -j -plane can be
s axis.
u = AcA.. + B, where ~ = tan-l (tIs) and 0 ~ tan-l("t:/s) ~ 7T.
round by assuming that
The boundary conditions generate the equations
= O-A
o = 'Ir·A
20
+ B
+ B
,/
which have the solution
u=
B
= 20,
A
= -201ft
and thus
t" (
20
tr tan-1 ( s) + 20.
To rind the solution in the circle on the z-plane we must write s ..and t
•
in
t~rms
of
x
and y.
The appropriate transformations are found by writing
e)
~= _
1(Z - 1)
s + 1 t
=_ 1
(x - 1 + 1
x+1+1y
s + 1 t
=_1
(Xx+1+1y
- 1· + i Y)
s + i t
=-
1 (
s + 1 t
=
2y + (1 - x 2 - y2) i
{x + 1)2 + y2
Z + 1
Y)
(
x + 1 - 1
y)
x+l-~y
x 2 + ­Y 2 _ 1 + 2 Y
(x + 1)2 + y2
i)
Equating real and imagina.""j" parts we get
(12)
•
s
=
and
(x + 1)2 + y2
t =
SUbst1tuting (12) into (11) we get
u(x, y) =
_ 20
-7T
2
tan-.1 (1 - x ..
'C;r
y:'
)
+ 20,
;s
which'" the" required temperature distribution.
u
= _ 20
~
o
~ ~a.n.(
)~ 'IT,
The temperature at (0, 0) 1s
tan-1 ('00) + 20
In the range from
0
to 1}-, tan( fI- /2) = 00
/
and thus
u(O, 0)
=-
20 (
1'l'­
~/2)
+ 20
= 100
The temperature at (0, ~) is
•
u~O, ~) = - ~ tan-1 (3/4) + 20
= 15.903o
.'
9.84
y
.G ROY tl 0
Example 5
~
ErD
a. a '" D uc:ro Po.
line of charge
q
per
unit length is located at
z
1
=~
while the circle
JzJ=­
is a grounded conductor.
Find the complex potential
inside
Izl
<
1
and the electrostatic potential.
Solution
•
• The mapping :function
1)
( zz +
1
described in Example 3 maps the conducting boundary onto the real
and the line of charge at
z = ~ onto the point
tj = i/3.
t///
LINE: of
CHARGe
~
•
't
~-axis
••
Placing an image line of charge
f
=-
f
= 2q log
-q per unit length at
-:s =
~.
-
we get
2q lo~ ( "'j - i/3) + 2 q log (~+ i/3)
'S +
~-
i/3
i/3
for the complex potential i~ the ~ -plane.
Using the bilinear transformation
we map this complex potential onto the unit circle 'z I = 1 as
fez)
= 2q
log
-i
f (z)
•
= 2q
Z-l) *.
(Z+r
(Z-l)
- :3;.
z+l
+
-i
log
2z - 1
z - 2
The electrostatic potential is
u
12Zz --
1 \
2
u(x, y) = 2q log
u(x, y)
=q
.
= Re
fez) and thus
log (2x - 1)2 +4y 2
(x - 2)2 + y2
Example 6
Find the temperatures inside the
/
semi-circular region with the given
boundary values.
x
l
Solution
•
- - _... .-- ..· ..
z -
The mapping function
,,'~_
...._.. •· ..' __ . .
·-,·_.···_..,..,....·,~.,··
...
·.·""·.,._1~·_
......
r:3=-i
p~A.I'JE
z - 1
z + 1
.,..-__ -.. -............__ .-.. '"_._........ _.•
.. _.__ ......
~,.,._
> .....
_.~'
. . . . .~ ~ .
~
...
_ _ ~__ ._. __ ~
.• ~
r
.., _ _
.....
9.86
•
described in Example 3 maps this region
onto the first quadrant of th~
"'3 -plane.
The lower boundary on the z-plane maps
onto the positive t-axis and the upper
boundary maps onto the positive
s-axis.
The solution in the oj -plane is
u=
200
1)-
t
an-1 (~)
S··
':$ _
Using the relations (12) we have
u=
where
200
-n­
PL A rJ r:;­
.
tan- l
0 ~ tan- l (
)
tt!..
11"/2,
for our solution.
Problems
Find the temperature distribution throughout each of the shaded regions
due to the
temperat~es
maintained on the boundaries.
//
.. /
••
9.87
.1.s­
•
t
1
s
U=30°
1
/
·'
-e
The boundary of the disc
A line of charge
q
lz1
~
1 repres~nts a grounded conductor.
per unit length is placed at
z
= 1/4.
Find the
complex potential that results.
9.10 More properties of the bilinear transformation
In this section we introduce additional properties of the bilinear
transformation
v=
•
Property I
az + b
cz + d
ad - bc
~
O.
Mapping three given points into three given points.
y
• Wz..
z·
x
2.
• W,3
.•..
\ole can always find a bilinear function (1) which maps any three given
distinct points zl' z2' z3 onto three specified distinct
•
.
correspond~ng
.
.
!""";"'lI'''14~--''-
, -
~
.~"
~
-v~~'?'-_
........... !"
f
Invariance or the cross ratio
Property II
If zl' z2'
z3"~~
are distinct points in the z-plane vhich map onto the
corresponding distinct points vI' v 2 ' v ' v4 in the v-plane by the bilinear
3
transformation
(1), then
(2) (z4 - zl) (Z2 - Z3)
=
(v4 - vI) (v2 - v3)
Zl) (z4 - z3)
(V2 - VI) ("4 ... v3)
The left side of (6) is called the cross ratio of
(Z2
If ve
v1sh to compute the bilinear trans fo:rn.ation mentioned in Property I, ve
simply replace z4 by z and v4 by v in (2).
•
A method for deriving Property II is outlined in supplementary problem
Sf9.10.6 •.
Example 1
Find a bilinear transformation mapping the points 1, i, -1 in the z-plane
into the points 0, 1, 2
in the v-plane, respectively.
/
Solution
I
We use the cross ratio (2) and set zl
VI
•
= 0,
v2
= 1,
v3
=2
and v4
= 1,
z2
= v.
We get
(z - 1) (1 + 1)
(i - 1) (z + 1)
=
(v .;. 0) (1 - 2)
(1 - 0) (v - 2)
= i,
z3
= -1,
z4
= z,
.'
:
I
9.90
•
i + 1
Since
(1 + i)
(1 - i)
=
i - 1
(1 + i) = - 2i
2
(1 + i)
= -1 ,
the. above express'ion reduces to
1)
- 1 ( z z + 1
=
v
w - 2
.
Solving for v ve get
i(w - 2) (z - 1). = v(z + 1)
which ,simplifies to
«1 - i) z + 1 + i)
•
v
=-
2 i (z - 1)
and finally
- 2i (z - 1)
.. w =
«1 - i) z + 1 + 1
Example 2
Find a bilinear transformation mapping the points 0, 1,co on the z-p1ane
into
()o ,
i, 1 on the v-plane" respective1y .
,/
Solution
v
•
3
= 1 and v4 = v.
(z - 0) (1-00)
(1
0) (z - 00 )
We get
=
(v
-~
) (i -'1)
(i - 00) (v - 1)
"'"., -::'-!. .• , • -
"-;'-',~:-'::
. I . t" .••
1;". • ' -
•
9.91
•
We 1nterpret the terms
1 - 06
and
z-OO
$imply as "1"because
w-OO
1-co
a - tj
11m
-::f0+00
b - ~
=1
•
We nov have
z
=
1 - 1
w-1
and solving for w we get
z - 1 +
·w=
•
i
z
Pr.oblem
• 1.0.1
Find the bilinear transformation mapping the three given points in
the z-plane onto the specified points in the v-plane.
(a)
w
w
lt
m
-2
o
i
1
2
-1
o
,/
.'
Definition of symmetric points
(
Two points P and P2 are said
l
to be symmetric with respect to
.e
the circle C having center at
if and only if
0
(10
(c)
z
v
o
00
1
1
00
o
-1
0
1
1
9.92
(i)
A radical line starting from the center
contains both Pl and P2 •
The center
(OP2 ) = R2
(iii)
0
0 and extending to infinity
cannot be between P and P •
2
l
, where R is the radius of the circle C.
If C is a straight line,
F1
~
\
the points P1 and P2 are on
\
\
opposite sides of C.
segment
-Pl
P2
\'\
The line
is bisected by
C and is perpendicular to C.
•
Property III
Preservation of symmetric points
Suppose we have a circle C with two.points Pl and P symmetric with
2
respect to it in the z-plane.
The bilinear
trans~ormation
circle onto C/ and the points Pl and P2 onto P~
the v-plane.
The
pi
and
p£
respectively in
are symmetric with respect to the circle
C ".
v
/
Y
••
and P2I
(1) makes the
/
9.93
.'
A method for proving'Property III is outlined in supplementary problem
SP 9.10.8.
Example 3
Two points 1
= 2. ,
+ i and "q" 'are symmetric with respect to the cirCl:le Izi
/
nd"~.
/
Solution
11
Since
+ il
=~
then
-2­
a
Iql !!lust satisfy the relation
•
.r2 I q I
Jql
= 22
= 4
which yields
//2. Because arg (1
+ i) is
il'j't,
we
have
q= 2(1 + i).
Thus the points 1 + i and 2 + 2i are symmetric with respect to the circle
/
Izl
= 2.
Example 4
.'
•
Let
p
and
q
be complex numbers, and suppose that the points in the
z-plane described by
p
and
q
are symmetric with respect to the circle
/~
1z -
= R.
al
(3)
q
=a
Show that
+
p - a
Solution
The i'irgure shows the
circle 1 z - a\
= R and
the two symmetric points
p
•
and
q.
The condition
speci fying the fact that
"lp - al
Iz- al=~
. Iq - al
p
a.nd
q
are symmetric is
a)
(q
= R2 •
Since
-ie>l.
p - a = \p - a\ e
and
iO(.
q - a = \q - a \ e
we see that
2
R =
Ip -
and thus
a'
\q -
a\
=
R2
=( (p _ ii)
(p -
q = a +
-a)
(q - a)
-
p - a
•
¥
',.::,'
•
-
~
'r",·
.'!"'~~.
-~'~.':-.~'~"
],.
..
_...........
~
'.
9.95
Problems
• 10.,2
I + {3 i and.
Let
Iz' = 4.
•
10,,~
lz -
be points sym:!.etric with respect to the circle
Find q •
7 + i
Let
q
6' = 3.
and
q
be points symmetric. wi th respect to the circle
Find q.
The fact that a bilinear transformation maps a circle C and tyO symmetric
points PI and P into a circle C I Yitn corresponding symmetric points
2
P/
•
and P: can be used to find specific mapping functions as the following
example illustrates •
Exarn.ple 4
Find ~ bilinear transformation mapping the upper half plane
onto the circle JwI ~ I and such that
z
= 2i
maps onto Y
Re
= O.
Solution
The points 2i and - 2i are
symmetric with respect to the
x-axis.
.e
These two points must
map into points symmetric yith
- 2. i
fz}
~ 0
respect to the unit circle
with respect to
lv'
= 1,
I vi
= 1.
and since z
z = -2i must map onto v = Oa.
~e points
= 2i
0
and 0.. are sYI:JIlletric
must map onto
v
= 0,
then
The ratio
w= z - 2i
z + 2i
satisfies these requirements.
:!ultiplication by a unit vector
e i1To
vill
not alter the size of the image circle and thus
v = ei'eo
(~
- 2i)
+
2i
maps as required •
•
koblems
• 10.Lf
onto the circle I vi ~
e1(),s
He( z) ~ 0
Find a bilinear transformation mapping the half plane
2
and the point
z = 1 onto
v = 0 •.
Find a,bilinear transformation mapping the half plane
the unit circle I vi:: 1
and the point
/
9.11 Poisson's integral formulas
Imagine the region shown represents
some uniform material and that the
tempe,rature at each point
Q on the
z = i onto
v = O.
•
y> x
onto
9.97
boundary is known.
Suppose we wish to determine the steady state
temperature at each point
P
inside the region.
general problem were solved in previous sections.
Special cases of this
This problem is called
t" e.
Dirichlet problem.
The Dirichlet Problem:
Let the values of a harmonic function be given on the boundary of some
simply connected region.
Determine the values .of the function inside the
region.
•
It can be shown that a unique solution to the Dirichlet problem exists
under very loose restrictions on the shape of the region and the boundarJ
values.
There are two cases in which a simple formula can be written for the
solution
of the Dirichlet problem.
is the .unit circle
Jzl
the upper half plane
The first is the case where the region
~ 1, and the second is the case where the region is
~ O.
Im( z)
(
In the first case we have
•
u(r, -e-) =
lj
2f)-
o
-.;u~(1=.;,'----!:...0.:,..;),.;;;d~0~_
21r.....:.::(1=---_....::r:.-,.2..:..)
1 - 2r cos (-e- - 0) + r 2
.-
!
9.98
.",
which is called Poisson's fOrJ!1ula for a circle.
coordinates of a point
in the unit circle
I zl
Here (r, -e-) are the ~olar
~ 1 and (1,
llJ) is a general
point on the boundary.
The fUnction u(l, llJ) is
Figure 1
y
assumed known, and the
right hand side of (1)
allows us to
com~ute
the
harmonic function u(r, -tt)
at the point (r, -e-) inside
-i---------~-----L----'---~~~
the circle by simply knowing
its values u(l, ~) on the
boundary."
J~
Our second formula is
(2)
u(x, y)
/
=;
00
u(
,
(~-x)
g)
"d~
Y
+y2
-00
which is called Poisson's formula for a half
the- harmonic function
u
•
....
"
'-~-,
y)
Here the values of
are assumed known on the. real axis (u (
and the right hand side of (2)
(x,
~lane.
in the upper half plane
15,0»;
.
allO\lS
y > 0
"":"' ':r:..-o:' ,
~ ~ ,~-
us to compute u(x, y) at any point
by si~ply knowing the values of
.. ~~.~-.
u
X
9.99
•
·on the x-axis.
'/
P ('X') y)
TiL--eY
I
I
I
------
•
X
-+-_...J....-~----~~--;::::_
Figure 2
It is useful to determine the geometric significance of the terms that
appear in the integrands of (1) and (2).
from the law of cosines.
Looking at Figure 1 we see that
Thus (1) can be written as
/
u(r, -&) =
In (3), the point P at (r,
-&)
is fixed, while the point
Q at (1, ¢) I:lOves
over the boundary of the circle to perform the integration.
••
To see the geometric meaning in (2), we
i
ex~ne
Figure 2.
Here the point
-------------
-
-
-
9.100
.'
P
at (x, y) is fixed, while the point
the x-axis to perform the integration.
constants while '( is variable.
( ~d
_ ~
x)2 + y2
.
=
Q moves from left to right over
Thus
x
and
y
are thought of as
Manipulating the integrand we have
tl
y
=
From Figure 2 we see that
•
Recalling from the
elenent~J
calculus that
¥; x)
we see that (u =
=d-&.
/
J
'It­
Thus we can write (2) as
(4)
u(x, y) =
~
-6-="2
u(
~
,0)
d-e­ "il-­
~~-y
~
Before tryins to derive the Poisson integrals, it is instructive to use
-e
them 'to actually calculate the numerical values of a harmonic function.
i
"rO,
.
........
,.. ­
;
~.
.'
9.101
Example 1
Suppose the temperature is measured at points on the boundary of the
unit circle as shown in the table.
Estimate the temperature
at
z =
~
using Poisson's integral (1).
angle ~
in degrees
•
Solution
00
0
3.00
10
temperature
60
0
20
will aI;lproximate the integral (3) by a
90
0
30
-
..
temperature
u(l, ~) on
the boundary
To estimate the value of the
.
Riemann sum.
u
at the point
Since there
(~,
0), we
are twelve
120 0
20
points on the unit circle where temperatures
1500
10
are measured, it is convenient to use twelve
1800
0
··
·
330
··
0·
0
terms for the Riemann sum.
we see the boundary of the unit circle
,
We substitute for cI~
PQ
In figure 3,
subdivided into twelve equal parts.
30
o
ori 7T/6 radians.
We can measure the distances
from Figure 3.
I
I
9.102
We have then
:U'
u(r, -Go) =
J
t~1 2
o
u(~,
VT
o
u(~, o)~
U(1, (6) d ~
J
0) =
.lPQJ 2
.12..
-[U(1, 0)
2ft
l!5Ql 2
•
¢=o
+
u(~, O)~
,t'Q
~) d @
u(1,
u(l, IX
60°)
(11:-)
0
+
"
(1i'"6')
+
¢:c.~oo
r:
.~
a
2~" [(.5)2
+
10
<'62 )2
--,-.....:0:'-""2
(1.5)
•
u<~, o)~
u(~, o)·~
+
+
+
a
+
a . . .
1
~
+ 0
[26.01 + 26.42 + 23.92 + 11.14 + 4.63]
.0625
5.16.
Of course, in computing this approximation of the temperature at (~, 0)
we have assumed that the tenperature distribution is not wildly varying
betwee~
the points
~ = 0°, 30°, 60°,
... .
For example, we might assume
that u(1, ~) varies according to the graph. shown.
30
U(l)
¢)
I
'2.0
.r
•
I
10
a"
fO
O
IS-O o
2.70 0
3(0 0
>- y5
9.103
ei
o
'Y
II
~
"U.. = 0 - fC<;;-----
i"so
-----f'~---_::
-:u.-=o
e
""t<. = 0
I
"U..= 0
•
'U. =0
/
Figure 3
I
e
I······':
. ':
.~,.
.... ~
:.
~ '
·-:-i Y'TJT""S.-"--·
---:-:-:"!"···---7.~·~·
•
--
--~-------------------
Problems
• 11_1
The values of the temperature on the boundary of the unit circle are
measured at
be
20°
~
= 5°,
15°, 25°, 35°, •.• , 175°
centigrade.
¢
Howeyer, for
= -5
and found in each case to
~15 °
0·,
, -25 ° , ••• , -175°
temperature is found to be zero degrees centigrade in each case.
the temperature at the point
solution to Example
• 11_2
•
4 of
x
= 0,
section
y ::
~.
Estimate
Compare your answer vi. th the
9.9 •
Use the Poisson integral (3) to estimate the temperature at (0, 0)
required
in problem 46 of this chapter.
Measure the temperature on the
¢ = 0°,
boundary u(l, ¢) at 360 evenly spaced.points
Compare your estimate
• 11_3
the
~ith
1°, 2°, ••• , 359 0
•
the exact solution obtained previously .
Prove that the value of a harmonic function at the point
the average of its values on the circular boundary
Iz r =
z
=0
equals
1.
Uext we turn our attention to Poisson I s formula for the half plane (2)
and its equivalent (4).
..
•
;
F.xample 2
Estimate the tenperature at the point
z
= i,
when the temperatures
9.105
on the x-axis are found to be
o for x<. 1
u(x, 0)=
f
lOx for 1 ~ x
°
for 3 <. x.
Solution
(4).
We will use Poisson's integral
77/2­
u(x, yo).:
i,.
J
u(
~
-&-:-1r/%'
•
-1
~ ~J<>l~T:/(a;
u(O, 1)
,,".-
d~
,0)
-&-
f>-3.
J10'4
= ;.
e--::
:s =
. "7('/'1
+-
7r/
-J
::0 (..Q"l .
3
d(tan-1 )( )
¥
This last integral is true because u( ~ , 0) =
° for values of
-e-
outside
y
•
1.
/
•
~
O
-A.:::.....
-...r.-----­
"U.=o
~
~
~
~
~
~
G'
~
...
~
~
~
()
~
¢..,
-...
- - - - - - - ­ ........ _ - - - - - ­
y
-u..=10X
~
~ I
~---
-u..=o
----------------
-
-
... ;._..:.. ...
__
. ...-..-. -
9.106
For convenience, 'We subdivide the interval
intervals of length 0.2.
1
~
x
~
3
into 10 equal sub­
With this subdivision 'We generate an approximation
by means of a Riemann sum for the last integral.· We see that the end points
of the intervals are 1, 1.2,
u
1.4, ••• ,
at the midpoint of each interval.
2.8, 3.0.
~1e select the value of
Thus 'We vil1 select
u
at
x = 1.1,
1. 3, 1. 5, ••• , 2.9, for the v'alues of the temperature on the boundary.
)0
We get
•
u(O, 1)
~
L
·10~
A ~ (angle in radians)
1)­
~c.J
anet 'Where
A '6"k (radians)
17'""
=
=
b ~k (degrees)
180
tan -1 ( xk + 0.1 ) - tan -1( xk - 0.1 ) .
(degrees)
180
x
----+----~---+--~~r
>(K+ o ,1
We summarize our calculations in tabular form:
,,
..
•
/
I
­
9.107
•
•
u(~
,
0)
=
tre;k
It
xk
1
1.1
11
5.2
0
57.2
2
1.3
13
4.30
55.9
3
1.5
15
3.50
52.5
4
1.7
17
3.00
51
5
1.9
19
2.50
41.5
6
2.1
21
2.1 0
44.1
7
2.3
..
23
1.8
0
41.4
8
2.5
25
1.60
40
9
2.7
27
1.40
37.8
10
2.9
29
1.20
34.8
10xlt
(degrees)
10:l1t
A-&k
(degrees)
462.2
Thus we have
u(O,
l)~
L
10
10>it
A'6k ( degrees)
180
k = 1
=
462.2
180
=
2.57
Example 3
Find an approximate value for the temperature considered in the previous
example by approximating
Poisson's integral with equal segments of angle
.-. .
d -&
= 10o •
9.108
Solution
In the figure we see radial lines emerging from z = i making angles
-&
= 00 ,
.:!:. 100
,
.:!:. 200
,
•••
,
.:!:. 90 0
with the negative y-axis.
only the rays making angle~:-&- = 50
interval
1~xs.3
0
600 and 700
,
on which the temperature
u
We see that
meet' the x-axis in the
is not zero.
'I
-­
-­ -­
-­ --­
e
-- -­
90"
-
8'Oc)
//
..
/
/
2.,7
/
/
_.soo
I
I
I
-30 0
I
I
/
/
- '{0°
/
/
/
,
,
__
I
I
-20"
I.
I
,I0
_10
I
I
r
I
I
I
I
0"
\
\
.\ \
\
,
\
\
toO
'" "'­
'\
""-
\,
ZOO
\
­
"-
1,0
l
30°
3,0
"
""
"-
"­
"­
"­
"­
'COo
F<:,o
"­
'10°
These three rays meet the x-axis at the points where
x
is 1.1, 1.7 and 2.7
and thus the temperatures (lOx) there are 11, 17 and 27 respectively.
I
d -& is 10
•
o
we have
d -& (radians)
1)-
=
100
1800
X
=
1
18
.- /
Since
-7~
9.109
•
Thus our approximation is
u(O, 1)
J~
=
u(
'i,
0)
fr
0=-
2
~
L
u( ¥k' 0)
k
- ~ 11(1/18) + 17(1/18) + 27(1/18)
~
3.06.
This answer is 20% larger than the approximation found in Example 2.
Since
.
the previous approximation involved 10 selected values of the temperature
•
and the present only three, the previous answer, 2.57, is probably more
accurate.
Example 4
Use Poisson's integral (4) to find the exact bounded temperature at any
point (x, y) in the upper half plane when the values on the x-axis are given
by
u(x, 0)
=
for
0 I
<. x
for
x < 0 •
...
•
-
-------
,I
. ,
••
-----'.-.
, - - - - ..... - .....
- . , . .......
~.-
. . . . . _ _ . . . . . . . , ... - ... - . . . . . . . _ - . . . ._ " ' . ,
......
_._-~
...
,.
' _ - - , , ,. . - - . . . - • • - •
.-.--..--
_ . . . . . _ _ ••
0<
•••••
_
... ,---..
9.110
•
. (X,)')
Solution:
t:
yl
.We have
L--60
!l
u(x, y) =
u(x, y)
•
where .e­
y
<r­
~
J
-e-s-Ji
t
u(
11,
_
-X
oj d-&
=
o
= tan-1
-x
._=
-1
-tan
Thus we have
u(x, y) = ~ + 0/n-) tan-1
Since tan-1 -yx
u(x, y)
=3
=
x/y
'iT - tan-1
-2
v
....
x
-(5/2)tan-1 y/x •
(
This same problem was solved by another method in Example 1 of section
•
9.7 •
9.111
Problems
#I 11.'1 Consider the temperature distribution given in Example 2. Use the
z = 1 + i.
method of that example to approximate the temperature at the point
Use the same values of
tI 11.5
~.
Set a second approximation to the temperature at
previous problem.
Use the method of Ex8J:lp1e 3 with d
z
=1
-e- =
+ i found in the
100
•
In problems 61 and 62, use Poisson's integral formula to find the values
of a bounded harmonic function u(x, y) in the upper half of the z-p1ane
•
.
.
when the values u(x, 0) are given on the x-axis.
5 for
u(x, 0)
=
7 for
-10 for
1
<
x
-1 < x < 1
x
<
-1 •
Compare your answer with Example 2 of section 9.7.
/
" 11.7
u(x, 0)
r:
for
5 <:x
for
0 <+ .(,.5
for
x -< 0
/
(See problem
•
20(c) of this chapter.)
We will now derive Poisson's half plane formula (2).
Consider the contour
9.112
Let f( z)
C"
r
shown.
be a fUnction analytic
for
z
in the upper
-p..
half
p1~e.
R
We have
from Cauchy I s integral fornula
fez) =
1.
[
211-
i
t1J
for
z
inside Cr.
o
•
J'
f
~-
d ~
z
Since -; is outside this contour we have
~ -2~i
"B )d~
f(
~- Z
i
Adding these last two expressions we get
fez) - 0
fez)
/
~
1
2fl"i
"2Ir-J
fez) =
,
I f(z)\
1
2 1i'i
'j-z
f(;S)
[
1
J:
j
1
~ -:- z
~-z
f(:5) d ~
~ - z
d~
f(
(z - z) d ~
( ~ - z)(
z)
1-
0
fez) =
and let
f(:S)d~
j . :s )
2~i
Now write
•
:oj )
f(
R
~ 00 •
~1
21l'i
j'
-,
.~
We assume that
/'
+
2i>i~.
fez) is such
tha~
'lim
r
=
K.~ooJA
o.
."
I
If
is bounded for Im(z) ~ 0 this limit will vanish as required.
(See
9.113
Theorem 1 of section 6. Tl.
On the integral
)--LJ rl<tl
integration is on the real axis and .thus
f( z
-
fez) =
211 i _.,..
-Lj
.~
t -
GO.
S
'! =
t.
the variable
>\ole
1 of
have
dt
2y i
x - iy)( t - x + i y 1
f(t)
(t - x)
~ +dty2
_0'­
Next write
fez)
= u(x,
y) + i vex, y) and get
u(x, y) + i v (x, y)
=
1
~
JOO
00
+
u(t, 0) y dt
(t - x)2 + y2
i (vct,c) Yd
7T
je"t:-x).. . + y
-co
_00
Taking the real part of this last expression we get the required Poisson
.
integral
•
Co
1
ti(x, y) = -;;;:-
1
u(t, 0)
y dt
(t - x)2 + y2
_0..
A method for deriving Poisson's formula (1) for the circle is outlined
in supplementary problem
9.11.7.
The Poisson formulas solve the Dirichlet problem when the region is a
circle or a half plane.
regions?
of interest
How can we solve the Dirichlet problem for other
If we can find an analytic function
which maps the region
R onto the unit circle Iwl ~ 1, then we can map the boundary
values of a harmonic function on
•
f( z)
R onto the boundary lw\
=1
/
also.
Solve
the resulting Dirichlet problem on the circle with Poisson's integral and then
'"
L
9.114
•
fez)
y
v
1.
R
~ -
w-
PLANE
map the solution back to
PLAN£
R in the z-plane with thefUnction
z
= f-l(w).
Thus we see that the Dirichlet problem can be solved for the region
•
1wl
~ 1.
R if
we can find a conformal mapping
w = f( z) of
:function
Unfortunately there is no general method
f( z) always be found?
for constructing
f(z) given
Rier.Ja.nn Happin!'j '.:'heorem
f(z)
exists.
~~ile
construct a specific
R.
R onto
Can the
However, there is a result known as the
which states that if
R is simply connected, then
we might be frustrated in our attempts to find or
f(z), we can take some comfort, at least, in knowing
that it does exist and that effort might result in finding it.
9.12
The Schwarz - Christoffel transformation
In this section we examine a formula which enables us to map the upper
•
half of the z-plane onto the interior of a eeneralpolygon in the w-plane •
9.115
•
- Z -
PL-ANE'
'11-
PLANE
Of-course, the boundary of the polygon will map onto the x-axis.
descr~bing
the formula which describes this mapping we must specify con­
ventions regarding the vertices, angles,
•
Select a poin#t
w~
on the
boundary of the polygon which
will map onto the point
z
= 00
This point may, or may not be
,/
a vertex of the polygon.
(ii)
The positive orientation
of the boundary of the polygon
•
~and
Notations and Conventions Regarding Polygons
(i)
Before
is the 'same as the positive
orientation of the polygon.
9.116
•
orientation for contours of
integration.
In the case at
hand it is clockwise.
In
more complicated cases we
imagine walking on the
boundary so that the interior
(shaded) is always on our left.
(iii)
Starting from
,
traverse the boundary of the polygon in the
positive sense naming the vertices in succession wI' w2'
•
will map onto points
I
be shown since
We..
x
5
, wn '
These
xl' x 2 ' ••• , Xn", respectively, of the x-axis in the
z-plane which are in increasing order
In this case, if
•
Wc:;o
xl
~
x2 < ••• <.
~.
= w5 ' then only the four points Xl' x 2 ' x3' x4
would
is now at infinity •
•
-.
-- ........., ..... ~.'
".-
;.
. . . ~.-.
_.
~ .. ~rtr1W'-~
'--~
4
J
9.117
•
(iv)
In this example, as we
--
move from w 00 through . wl
to w2
we make an abrupt
W•
l
change in direction at
This change in direction,
called the
at wl "
It
exterior angle
is denoted by
"0<
'i}"
It is measured, as usual, in the positive
o~
counterclockwise sense.
\
•
•
\
\
The exterior angles at
\
W5'
each vertex are
-- --
measured and denoted
byQC. 1 ~, 0(2 1f",
,
<:A. n 1r .
Remarks:
/
""
....
1.
In the case of the pc;>lygon illustrated, all the
cJ,..
i 1?' are positive.
However, negative exterior angles
.­
can occur as at the corner illustrated
where w·e %'lust neasure in the clockwise
..... ,..............'.,. ..'
'1";r'CII_
9.118
•
sense.
2.
The
of the exterior angles of a closed polygon must equal
sum
~1'ir + '""'2'ff+ •••
+
04
n
2
rr-.
1"t=21l'-.
Thus we have
+
(1)
0(
2
+
... +" n = 2.
We can now write the formula which maps the upper half of the z-plane
onto the polygon in the w-plane known as the Schwarz - Christoffel transformation:
Case 2:
Now x
•
n
=~.
In this case we simply drop the factor (z - x)
n
the above formula and get only
n - 1
factors in the integrand:
....o<.h
."
. from
9.119
•
dz
Remarks (continued)
3.
We can picture the mapping
effected by (2) or (3) by
imagining the polygon in
the w-plane as a sheet of
•
rubber •
We break the rubber at w
and begin to stretch the
polygon open.
Continue opening the polygon
until
the boundary is a
/
straight line.
The points
Pl , P2 , ••• , P5 , must finally correspond to
xl) x2' ••• , x5
on the x-axis, while the ragged edge resulting from the initial break at
•
Woo is stretched to infinity.
The rubber sheet is now the upper half of
+B
•
9.120
the z-plane.
4.
The constants
A and
B are in general complex numbers.
By appropriately
selecting these numbers we can effect a translation (B), magnification (
'A\ )
and rotation (arg A) of the'polygon.
5. When the vertices wl ' v 2 '
their images xl'
arbitrary manner.
~,
••• ,
~
••• , v n
on the x-axis cannot be selected in an
We can arbitrarily select three of these xi's, but the
remainder must then be calculated.
•
are given points in the v-plane,
In general, this calculation can be
.
very difficult.
A method for deriving (2) is outlined in supplementary problen
9.12.21.
1
EX~le
I
Without specifying the numbers A, B or the Xi s, write both forms of the
Schwarz-Christoffel transformation, (2) and (3), for the mapping of 'the upper
/
half of the z-plane onto a rectangle in
Solution
v
'II...
)
"
/
/,
"
"
•
z-
FLI\AJE:;
w-
PL.ANt:
,
.......
9.121
Notice that all the exterior angles are 'it /2, and thus
do i
=~
ror i
=
1, 2, 3, 4.
w=A
Ir
Wee
JZ
=,w4' then the point.x4 is at infinity and we use (3) to get
w=A
+
B .
In many probleos of interest, one or more vertices of the polygon are
at infinity as the next example illustrates.
Example 2
•
Without specifying
or the Xi ' s,
A, B,
·te b0 th f orms
wr~
of the Schwarz-Christoffel
transformation mapping the
upper half of the z-plane
/
onto the polygon shown.
Solution
Now there are three vertices at infinity and two that are finite for a
•
total of five.
If we select our starting point
w~
on the real w-axis we
. t . - . _ - : :.. ~.:;...... ~ . . .
l--u..l_....lI .... ,·:~
-#
••••••••
_~
.........
..:.<
••,:": • • ~ . . . . . ~.,.;,:.... '.:.
_,.:..~._... ~ .. ,.
t.
_.~.~~...
',·.._~ .....~w-__
,.
~~~•. ,''':.
-~.;,
__ ....
.
",_~~J_._._.__~
9.122
can denote the vertices as shown:
Of course, the vertices wI' w3 and w
5
~e
at infinity.
We must now determine the exterior angles at each vertex.
•
The exterior
angle is defined as our change in direction as we pass through the vertex •
At WI then, we see that we must make a 180
0
turn in the counterclockwise
(positive) sense.
E>(IE~IOp'"
AIJG-I-E:
7r
( Ye<..roR5
'tJ·THE
ROTItTe'
POSITIVE!
The exterior angle at each vertex is shown below:
l
•
/80
seN5e:
0
)
.
9.123
•
Notice that the sum of the exterior angles
equals
2
~
(as is always the case when the polygon is simply connected).
The following table summar! zes our findings.
vertex
vi
exterior angle
«'i1>"'
1
--
31'1""
v2
~
v3
- 3/4
fr'
1
1Y
-7:[
v4
•
i
'if'
v1
V
"
- 1/4
1r-
s
. JZ
1
We can nov write the transformation (2) as
(4)
v
= A.
If we select
w
1/4
(z - xh)
dz
(z - Xl) (z - x3) (z - x )
5
(z - X2)
= v
5
t
3/4
then x
5
+ B
is at infinity and ve see from (3) that the
I
Notice that any factor in the integrand of (4) would be de1eated by selecting
the starting point
•
Woo at the corresponding vertex on the polygon.
Problems:
Without specifying the constants
I
A, B or the Xi s, write the two forms
•
of the Schwarz-Christoffel transformation (2) and (3) for the mapping of the
upper half of the z-plane onto the polygon shown in the w-plane .
• /2.1
-.
.%,
-------.­
-
W4<L
,
------- - --- - ------- - --------------
-
-- -
---------- ------
--- -
\"
-,
u
!
~
\
'.
- -.
­
.
\, "\
\.,,---\>\ \.\~
\:~,,, \<0: \'<\
:
~~~L:......:..-:-.::.L...:::L-.:.%~~~.r&~~
- .----- -
~~,~~
<\",
----'------~----------'-----------------
cp 12. .. 7
"-'-0
CL
9.125
•
•
J2.1
Suppose the points xl t x 2 t
(2) are selected such that xl > x
•••
t
x
n
in the Schwarz-Christoffel formula
> 2 > ••• )
now maps onto the interior of the polygon
~.
What ree;ion in the z-plane
in the w-plane?
In the previous examples t certain constants
At
Schwarz-Christoffel formula were left undetermined.
"
. the x. s
B and
J.
OT t:he
In the folloving eXa.t:lples
we illustrate how specific values for these constants can be selected.
Example 3
Use the Schwarz-Christoffel transformation to map Im( z) ~ 0
~
~~gion
shown.
w-
Solution
This region has three
vertices w1
e·
=-
1r /2 t
If we select our starting
PLANE
onto the
,,"
"' • • ,J.
J..'.
_ .... _
-;~:
101__
~
.
'••
""'-~-"';'
•
..
.....
~_.
r.:.- •• r_"
'"- ..
~
_........... ,..~ • .
.•• _ •. ~
;.--~-
- -..- . _ - - - . 9.126
•
Woo = w
point at
v
3
=
AI
= 0:» we have from (3) with the exterior angles shown
z
J
(z - ~)- 1/2(Z - x )- 1/2 dz + B.
2
4 above stated that any three of the xi s are arbitrary.
Remark
ve have only three vertices, each of Xl' x 2 ' x 3
We have already set x3 =
can be selected at will.
Suppose we select Xl = -1
Cb
Since
and x 2 = 1.
We
now have
•
f:
ve can vrite the last expression as
v
C
A
11
C
A
(1 + z)- 1/2 (1 - z)- 1/2 dz + B
J h~:2
7
+ B
I
where we have simply changed A to A because of the influence of the factor
.
(_1)1/2
.
/
Since
J~l
--
dz
2'
- z
we have
v'=
A sin-l z + B.
(
=
sin-l
z
/
9.127
•
We must select A and B so that w =
- 1t72 when
z =
-1
'fr= A sin-1 (- 1) + B
_
2
-
~
=A
(- 1)- /2) + B.
1'1"/2 when z = 1
lie must have w =
~
= A sin-1
(- 1) + B
'jJ-
= 2A1t
.
'2
We see that
w
•
= sin-1
+ B
A
=1
and
B
=D
satisfies these two equations and thus
z
•
is the appropriate mapping function.
Exam.ple 4
r,Iap the region
a--~
__
Im(z) ~ 0
onto the region shown.
----,~...£-....t.........t........-£..._""::""-""':"'...L..-+-
~"U.
o
•
...
.::.~'
.....
..
~ - ~ ~ , ~.~.~
..
9.128
•
Solution
As in the previous example there are only three vertices
W3.
wl' w2
It we select the
starting point at
Woo
= w3 we have
trom (3)
(See Remark 4.)
Now three of the xi's are always "arbitrary.
selected x
3
v = AI
=
Oc ,
J7.
.and nov we take
~
z- 1/2 (z - 1)1/2
=0
and x 2
= 1.
We have
We have
dz + B.
Writing
A I (z _ 1)1/2 = A(l _ z)1/2
we have
=.J
v
Z
J~
1
;.' dz + B.
To evaluate this integral we set
z
= sin2
-&
, dz
=2
sin
-e-
cos
-e-
d-e­
/
•
and get
'JZJo/
dz
=
2J~ J 1
sin -e- cos
~
d -&
and
9.129
21: co.2~ d~
J co.
s
(1 +
R
= -& + sin 2
2
2 -fr) d-e­
-e­
== -&" + sin -& cos
Since sin -&
=.fZ
~.
we have
';1- Z
,
and our mapping :function is thus
w
=A
(sin-l.r; + Jz(l - z) ) + B.
We previously made the correspondences
and
l
The f'irst of' these yields
0
= A( sin-1
0
+ 0) + B and theref'ore" B
= O.
9.130
~
The second yields
1 = A (s1n-l (1) + 0)
1
= fl'/2
and thus
A
= 2i/ 'fr
~
v =
(6)
A
•
We have nov our mapping :function
r;:
(sin-l
+ Jz(l - z)
) •
Example 5
(a) Sholl' that the point
z = i
maps onto the point v~ - 0.77 + 1.06 i
under the function (6) of the previous example.
•
\
(b) Picture .the region shaded on the v-plane in the previous example as a
large ocean of fluid.
The ocean floor is the step of height one shovn.
If
the fluid is moving vith velocity V in a horizontal direction from left
to right far from the origin v
the point v
Solution
=-
= 0,
determine the velocity of the fluid at
0.77 + 1.06 i
/
(a) We must first calculate (6) vith
+ riel - i)
z
=i
)
.r
•
v=
~
2i
1t"
. , :!
~J •
~
,
.~S·
.
-.r
.-.~.
/
9.131
•
The inverse sine can be approximated easily f'rom the contour map of' the sine
function in Figure 2.7.
f'
=1
= f'r/4
~
and
We see there that the contour lines
0.57 + 0.75
intersect at
1.
Thus
I
sin-l(ei 1.r/4) ~
0.57·+ 0.75-
i
Bov
=
~
1.10 + .46
1.
Simple calculation with (7) now yields
2i
'j)-
(.57 + .75
- .77 + 1.06
i
+ 1.10 + .46
i)
i
:
y
-----./ i
p
>
-
-
•
PI..ANE
pi
. p'.~
--"
l-!'---------
~~
AI
Z -
>
w=-.77 + 1.0<; 1.• J
B'
w-
PLAN£<
"U.
-;>
.
!
I
•
The horizontal stream lines in the z-plane map onto the bentstream
lines of the v-pl~e under the mapping function (6) of Example 4.
The com­
plex potential in the z-plane is then
=U z
fez)
where
U
is the velocity at infinity.
z = g(v) ve could then write
of v in the form
..
f(g(v»
If ve could solve (6) for z in terms
= U g(v)
as the complex potential in the v-plane.
•
~i
no easy yay to write
g(v) explicitly.
Unfortunately, there seems to be
Nevertheless, ve can still find the
velocity, for the velocity in the v-plane is given by
velocity
(8) velocity
­
df
= a.v
=u
=
dz
dv
dv
dz from (6), but it is much easier to find this deriva.tive.'
We can compute
dv
from (5) as
,
U
dg(v)
dv
(i7;'
/
=
2i
'1r
Therefore
~
dv
•
and
=
.it...
2i
(8) becomes
velocity
= 11"
- 2-i
7
---..;z~·
[
1 - z
U
.
9.133
•
Nov we must select
When w is near
!im
r
~
U so that when
w is near
z is also near
00 ,
_' = N
0-
00 ,
the velocity is
V.
and we have
= i.
%.~e- ..J 1 - z
Thus we have near infinity'
velocity
and we take
=
..1l:.­
= -
U
2
U
2V
U=
- T
Thus we have
velocity
=- Vi
J~
1
i
I
•
Finally we find the velocity at
W'
~
-
.77 + 1.06 i by sUbstituting the
image of this point (Z= i) into this last expression to get
= _ Vi
velo.city
=
,---:i'
J~
,
_ViP-i
2
=v~
.J7"
V (1.18 + .86 i)
This vector has magnitude
1.46 V and makes angle 3~ with the u-axis.
Example 6
Use the Schwarz-Christoffel transformation to map
Im(z)
~
0
onto the
;
region shown •
-~.......
- ...
~:
. 1·;
'.....
••.. too.
';'"r
-.­
~
..
~-,
.......,...­
_.. ;
.........
..,.,.~~~
. -' - -,
,
'W -
PLA NC;
Solution
This region in the v-plane has four
•
~~ w4 =
alw~s
at>.
v~rtices,
vl
= 0,
v 2 = i, v
3
=0
We are
free to select
three of the points
at will..
Suppose we
take xl. =' -l, x
number
Xl
•
x
2
3
= 1 and x4 =
00.
We cannot arbitrarily speci!'y the
since it is a fourth value.
and x3 ud thus
-l. <
"2 <
All we know is that it is between
l.
."
9.135
:.
The following table summarizes our selections.
vertices
vi
images
xi
"'l == 0
-1
v
i
2 =
~
-1r
- 1
v3 == 0
1
11"/2
~
v4
Oa
211
2
=Co
exterior
angles eI- i 7T
1l' /2
S7. "2)
::It i
~
Thus our mapping function is trom (3)
Ce
+ B
~ z2 _ l'
..
J Z z - "2 dz
v=A
J1
v=A
dz
(z ­
/
v=A
+ B
- z2'
z dz
S~
. j1 _ z2
J
,
v=
-A /1 - z2
- x2
We must nov determine
(9)
o
/
=
TJ"La A
vhen
(10) i
=-
fl1
Since v
=0
(11) , 0
= _ 1r x 2
A
Since
v = 0 vhen . z = - 1
ve have
ve have
I
1
- x 22
z = 1
A
2
+ B.
A, B and x2.
z = x2
vhen
A sin-1 z
+ B
+ B
2
Since v= i
-"21~:
dz
- z2'
-
X
-1 x2 + B.
sin
2 A
ve have
+ B.
.'
9.136
Adding (10) and (11) we see that
This means that either
the constant function
have from (10), A
Y
I:
= -i.
A
Y
=0
=0
B
= O.
or x2
= O.
With
B
It A
which is impossible.
= 0,
= 0,
(9) becomes
~ A
= O.
the mapping becomes
Thus
x
2
=0
and we
Thus we get
i
Problems
Use the Schwarz - Christoffel transformation to determine a function
.­
•
which maps
-1i
Im.{ z) 2: 0 onto the polygon shown in the w-plane •
•
- 21 .....
' ..L.--"'::"'-~""""':;'---,,~
1
I
9.137
r
.
•
•
12.1 G
u(x~
Determine the temperature
y) at each point ot the
shaded region when the temperatures
on the boundary are as shown.
6) f 2..17
Determine the temperature
at the, point (-
0.77, 1.06)
w1th the given boundary temperatures.
(See problem 46 and Examples 4 and 5.)
i
' . (!)
12..~/5
Describe the region on the w-plane which is the image ot
Im( z)
under the mapping function
w
=
J:z.
z- 2/3(1 _ z)- 2/3 dz.
t>
/
/
~
0
P9.1
APPENDIX I
SOLUTIONS TO PROBLEMS
Problems from Chapter 9
i/
~
"d--=- c
w- -= 2 : :
~
C
X+~~) ~ ~ ~':'qL"
~ ~~~,
-I
~
~
"t-=c.x)
=
zLZ::z£u ~ ~ . ~ -¥=:
Xz.+ "'f 7..
..
X
-+ "'r
ah..JL.,
'2..
+
,
~ ~
+
J
'teL
-
_11' =.
x
C,
cYt"
•
...."
0/
/).f~~
(AI-)
~
+1.
i ?:-1.
12-+1/ - 4
=:
'''-1/
7h..- ~ ~
~'
V
b=.1.
)~ ~ ~ ~ .A'..-.- ~
.7k ~
tl__
4
~
C1J:-l
if-
F~ 7,1
~~~
-- '1~_1 •• ....... ~,
'-*
4(·=+1)_~(~-1)=
+l
cVUJ,
X+!
~
_
t IV>-1f'
( H i ) - ""r
(~-1) 3; ,
~
~'
it
-= Co
x-!
7.£- ~ ~ ~ ~
-f/("",")=O,
-f(~)=:; V2:+-~~-C
-tIer)
d~
~=
=:
V + ~ ::::.
- -;-
;";d&-
0
~~,
•
P9.3
(it
7/
(ru)
-FI(~)
-t(~):::: ~ ~2..
=
21'1M,
~
)
~ ~ .~ ~ ~
== 2./11(..,(x-~ir)
Dd~~
I
~~~~
-f( e)::: ~.z; 2.::: /h7... C><' +.-e: "'t' ) 1..
=
><~
(c)
:::
nn-
(>< 2.._ 11" 2.)
+
--
•
2. ""'"--
-
>< if" -<..
..
Co
Y~.­
-~"'­
­
_ ;m..6t: +~
x)
><2.+7-2..
(01..)
Jr~ -=
c.v-/ =
-~
~-6
-+­
/rn.-_
~-a...
~ ~ ea..12R- ~ Ca..) ::: a..., ~ ~ Ca.) ==
CU-r f&-Cb)~b, ~ ~Ch)-=' b:1-, 7.-£.­
cu-~-4/;=';: I +..:[~(;:-,,-)
Q...2.,
-ah.y-(r-I.)] .....
P9.4
7/CrL) C~)
ur=_tJ~- \ ~-=-: I = ~:'[~I
~r
-k' ~-b2.J
6,
1t-a...z..
X-
Q..I
)<'-
(e)
(-p)
w- ~
Y ~ == u.J-7
V-e +_ ~
+ ~.J4y
V-r +~ ~-I ~
2-
~ -tJA- zL;c
~/
V + ~
t
=
=
V)(
h.
=
+-)
~
+':(V-r
<!.
~~~,
Jb,..
~
,
~A-w-~ ~ ~
2-
.~~
W-I
-=
V(i-
a. 2.
2:
)::::.
2..
V(i-
(L'
.
/z. 2. e-'" 1.&
)
•
=
V
(1 - "-~~:''')
~V[(i- :>c<t-2.<*) 1-;~:~2t>1
'
7~~~':+.z£-~~
\ C:Z;:' ('- = V
(1- ~:
+ :: ~~2"
2.- [
<:cd-
2-<>- ) ' -
]
•
~J2-~~~~~
~~~~~
·-Tr~.(;;-~ 7T
.h-~a.....
~~~
I
~
~2.~ -=-1.
~a-eL~~~~~
a....
.L .-A 'n:~'
~_r,-, ~,
M
e:r-:- LJ I
O"~ ~
-e-:::; + ~ ·
•
~/v~~~~J~~
/v
~ ~ ~~.J
a..-,
~
~.~~~~~~
~ aX ±,(...~..) Zl&.- ~ ~~
r+::t4~,
~~
Y
7--4~ oX~
~V,
d~ ~ ~
~
cv
~
aX"
24 (c-l.)
+~~
(0)1)
~ ~ -2.~c ~a~ c;i:C~~ o-f~~~, ~
~~~~
-F ( 2-) =
'=
2.
4
(-e -~) - 2..
2-
4
-e~--G'
..
4
-c
,
~.
-f (i':)
-= 2.
~ \ ~~/
I + 2.L [;t;.,:'
'>r;' - z:,' f]
+ zl [~..::'
"J-
~' -z..::' ! ]
J
•
'"' +11' a. -z-:I--r.'
V 2..
(/_c!)()(2.+~;l.)
-
(!.X
-2
'Z.,
_<!..~z...=
if-' +1 -:
0
0
,
d~~~~~~~
~ ~~
ah.P­
C J'
•
P9.7
'0/
F~ ~
5.; ~~~
4('"-L~)
~a.;~~~~
_,
2.".
~ r~ ~~ ~
~ c-a-.e
-= f'
r
+
7Th.,..J
a:;C
o-I-~
~ ~-~
~:"n~
'7~ ~ ~·eX
=0,
~
CI"-~
h... '::. 0.)
± t) ±
2.)
IN,
d'­
~ ~ ~~ -r'(...)"" °
f(i!)
=.
f/(~)=
4
(~=)
t
_~-e
~ - ~ ~
~e
~ ~ r
~ ~
I'L-~~~)
J
e:::tLlT,
f/(r)
~~~~
7.....e-~ ~~
=
-
~~
y
•
~ ~-=n...7T)~
,
P9.10
••
,~
'y
jf-~ ~
;t;:.
~;z-£.. ~ ~
-F(c)=-
~
(e
5
+
~)
~ ;Z ~
~:=. M(~)
=
-r.+4l
F(--:S)=
..
~
I
'I ~,
2=
~-4...(...
5(~-lf':- +-
"
:S -
d~
F~
~
.
if",'
)
,
~ I -s I ) Fe -S)?<:-. £ '.}' ~"....- ~
~.~ I-';d..-- ~ ~~-
F"""
if~~~~~~
Iy
~(c)~ IOCr:- + ~)
~
C.i)
~ ~;ae.. ~
't,,") a.-l.-
•
Gi) ~;d... ~.z:.
~
= fvl (-e) =
e1,lr/2.
~= -l~ -'1-",:"
F(~)-= I (J
F<YI..- ~ I~I)
F 1("'5)
~
'f+lf":, '7~
(~ + '1+ 'IA..) = ~"C r + if +-(/A-') ~
I
7.h.-
~ ~~
(--<.' '"3
- tf _ '-/.:.
F ("'5)
~ -,:
(0
+
J
')
Lf
-:- -S -" - 'I~
),
~
10.l.
•
\.
(--...
P9.9
12/C~)
~h~r.:....-, 7~~~~
~~~~~~~.s­
~~~~d-~'
(.
I;Y 7,.g.~~~
-F(2) = ~ (~77"~) -= 4
(:':;: )
_ 4 (~7T~) - 4 (~7rr)
~
7~~~~a.Z r=h- ~ ~
oa:
2 = ~ + n..
e-f- ~ .:.-,,-, d~ ~
~~
+- fer)
-
•
P9.12:
Vo (-2. +
~(s)
-+
~)
\ /0
-=
V,
(e-l. "../~ Z +
':'
z
e
""/3
).
~'L~pX~~~
I
,
-::::!:~
(
_- +
1­
2..
e
,l, 7{"/,3
+e..-.l.
/,3) = ..
1r
7T
~3
--1_
-
•
,
'7k~ cv=.-~ ~ aX: -s±l'
~;z;&- ~~. CZ<-~ ~ 2-) ~
~
v. ~ ~
aL-
VO=:-'f,
IY7~~4~u~3
~ d'.,;;t; ~
,
Z
~~~
=.,
-==
=
a--r/-//
~
~~
•
P9.111
'y
on.-:d.- .;;-
~J ~ ~
~ ~=ce./.9"J ~
7.A..~
I<c..
of~~~
"'3 -:::.
;,.
~
-A..
(~-+- ~
')
-t=
~~
I
= 2:
(
<!e
i-e-
.u.. ~
Ite DF
+ c
-e
e..
)
~(c.--k)~~
a-v-;d...
~
oj-;z:4
~
~~~~, Q,..-~i)
;Z
~ ~,;d..,,;c ;z;&.. ~
~ ~ ~ I~I ~c.,
17/
/
r;I~.~~ ~z-, ~
e­
2...
~~~7li.z~~~
~
a..
~ir
L
•
-r.-
Z
~
-=
e
1.
ze-;t:J2..
7T"/3
r.
z- ~4 ~
~ ~
""
~
-.:. 7r/3
e
Z
P9.14
Irl
(~)
~~
.a- ~
j-R-r,
e.'"
I
"S ==
•
-----
7r
T..,e.,~~
0::
b r. == -J. b -r.
~
e/
•
SP9.~
SUPPLEMENTARY PROBLEMS
e
9.2.~
Let
time t .
u (x, '1', t)
denote the temperature at the point
(x, '1')
at
Interpret each term in the "heat equation"
k V'2 u =
vhe~e
k
is a positive constant
substance.
the thermal properties of the
Review the discussion of the one dimensional heat equation in
8.6.
section
9.2.2
•
ref~ecting
Let
u (x, '1', t) denote the vertical dispacement of the skin of a
drum heaSi at point
(x, '1')
and at time
t.
Interpret each term. in the
"wave equation ll
where
k
2
is a constant reflecting the density and tension in the skin of
the drum head.
Why i~
the wave equation an application of Newton r s force
.•/
equal.s mass times acceleration law.
Review section
8.7 in which the one
dimensional. vaveequation is examined.
I
e.
9.2.3
Using Taylor's theorem in three variables
SP9.2
\.
(----
=
u (x + h, Y + k, z + 1)
LLL
·m:: 0
h
""m + n + p
a
u (x, Y, z)
mIn I
p= 0
=0
p
1
show that the three dimensional Laplacian
V
2
u=
u
.
xx +uyy +uzz
measures the difference between
u
at the point (x, y, z)
and the average
ot u in an infinitesimal neighborhood of the point.
9.3.1
•
diagrams (i v) , (v)
(i
Answers
v)
v=
and (viii) •
-tan-1
(y/x) ;
(v)
v = tan
-1
(y_a
2
x - a
-tan-1 (
Y - b2 )
where
a = al + i
(~)
.
x ­ b1
(vi)
9.3.2
9.1
Find the stream functions for the flows described in Figure
v
= V
Y
+ m tan
-1
~'
1
b = bl + i b 2 ;
Find the stream functions, sketch the stream lines, and locate
stagnation points for fluid motion described by:
(a)
•
(c)
e
z
or
/
9.4.1. Find the velocity vector for each of the flows in problem 9.3.2
SP9.3
(it
9.4.2
Fluid enters the region shown in Example 3 of section "9.3 at the
rate of
and
(b)
Answers:
8 cubic feet per second.
Determine
the velocity vector.
( a)
16
'11
16
log z ,
7ri
is absorbed at (0, 2) at the rate of 4
•
the complex potential
Fluid emerges from the point (2, 0) at the rate of 8 Trft 3jsec.
9.4.3
•
(a)
1r ft3jsec •
(a)
Find the complex potential.
(b)
Find the velocity vector.
(c)
Find the equations of the stream lines and the equipotential lines.
9.4.4
and
Discuss the flow described by each of the following complex potentials.
Locate all sources and sinks.
Find all stagnation points.
Find the velocity
vector and sketch the stream lines and the equipotential lines.
(b)
Y
= log (cosh z)
(c)
w
= log
(cot z)
(d)
w
= log
(sec z)
Spg.4
9.4.5
Consider the flov described by the complex potential
y =
(a)
VG+ :2)
+ mi
log z
Shov that ve can consider the cylinder
II
z
= a to be an obstacle in
the fluid.
(b)
Find the stagnation points and show that there are three cases:
(i)
m
(ii)
<2
a V in which case there are tvo stBgnation points on the cylinder;
m = 2 a V and there is a stagnation point at
(iii)
m) 2 a V
z =
- a i ;
and the stBgnation point is belOW' the cylinder.
(c)
Find the equations of the stream lines and the equipotential lines.
(d)
Find the velocity vector.
(e)
Sketch the stream lines.
Answers:
(c)
v
=V ~
-
;2)
sin -&+ m log r
= c1..'
2
+ -a-)
r
(d) V
~
m i
z;
or
.~
I
l
SP9.5
•
9.4.6
•
A source of stre~h
strength
(a)
~
is· located at (:.
is located at ( E. , 0) and a sink of
E. , 0).
Determine the complex potential.
answer
(b)
~
m
E.
log
z-E.
z+E.
Show that as the source and the sink near each other, and their strengths
increase
/ (Eo -7' 0),
the complex potential just found approaches
which is a dipole at the origin.
(See Figure
-2."""".
.. \.""Z"
i
9.1 (vii) ).
I
•
SP9.6
(,.
A log of radiu~ "a"
9.4.7
is placed
at the bottom of a deep stream bed in
which water nows from left to right
with uniform velocity
V.
-­
tn this
problem we will show that the complex
potential describing this flow is
. 'f( z) =7r a
'(a)
•
Show that if f
bed (y
(b)
= 0)
V coth
=u
(~
a
~
X
+ i v, then the stream line
and the log (
Iz - i
Show that for large z,
al
v
=0
is the stream,
= a) •
f' (z) ~ V.
This means that far from the
origin, the velocity is uniform from left to right with speed
V.
SP9.1
9.5.1
A cylindrical obstacle of radius one is placed in a uniform flow of
V making the angle et.. with the real axis.
velocity
).
potential is
9.5.2
Show that the complex
A cylindrical obstacle of radius
3
has its center placed at the
point (3, - 3) of a uniform flow with velocity
8 e i '11/3.
Find the complex
potentiil describing the flow.
9.6.1
•
Vi.
9.6.2
A flat plate of width
is placed in a uniform flow with velocity.
Find the complex potential describing the fluid flow.
A uniform flow with velocity
of Example 3, section
9.6.3
2b
9.6.
V i
encounters the elliptical obstacle
Find the complex potential describing the flow.
It is possible to construct the mapping effected by the Joukowski
,/
transformation geometrically. Suppose
in the figure.
Draw the line segment
z
is located at the -point
0 P
P
and measure its length
shown
r
and
I
angle -&.
as shown.
From the point
Calculate
l/r
P
draw the line segment
PQ
and layoff this distance from
making angle
P
along
PQ
-
~
to
SP9.8
o
locate the point
•
segment
0 R
R.
This point
R is at
-.
z + 1z
and bisect it to determine the point
Demonstrate the mappings shown in Figures
Now draw the line
~ =~
(z
+
i) .
9.2 (iv), (v) and (vi), by
drawing the appropriate circles on the z - plane.
After selecting various
points on these circles, determine their mapping by the above construction
.,
and obtain the profiles in the ~ - plane.
/
9.6.4
A cylindrical obstacle whose cross-sectional area is the Joukowski
profile shown in Figure
•
velocity
V e i ~.
stagnation points.
9.2
(vi) is placed in a uniform stream with
Find the (a) complex potential and determine the (b)
SP9.9
•
e =
~{l
[. e
-ie<
V
(a)
Answers:
+ a)2 + b 2
[+
2'
,
and
e
e
iot.
z
+
C
= i'S+"'3 2
/
-.
• eia<.
_1
-a+ib+
.
•
2
J
where
+ a - i b.
,
SP9.10
Find the potential
=
t( z)
u(x, y) and the complex potential
u(x, y) + i vex, 1') in the upper halt plane when conducting plates,
separated by insulation on the x-axis are charged to the potentials shown.
'tC.
(a)
"U.. -:. 0
(b)
'U..
(c)
·f
=- 2­
"t.L.
0
-t
= 3
0
=V
>><
~=o
0
~X
1­
""C<-:: 2.
-s9.1.2
1.t:=C
~
"'t.(. ::
i.
"'«.
0
=
0
~X
D
O
Find the potential in the
first quadrant when conducting plates
•
on the positive real and imaginary
~
"
;f
':5 - axies
Hint:
are charged as shown.
through
1.( ':.
0
~.s
1
9.1.1 (b)
Map the solution of problem
In problemS 9.1.3
u.= 1
onto this region.
9.1.8 find (a) the complex potential, (b)
the electrostatic potential, (c) the lines of force and (d) the electric
intensity vector for each ot the given distributions of line charges.
A line ot charge
q per Unit length is at
z
=a
and a line of
I
charge - q per unit length is at
z
=- a.
,I
•
Identical lines ot charge
1( 11'/2 + 71 n) where
n
q per unit length are located at
is any integer.
SP9.11
Identical lines of charge
z
=
n/2
where
n
-q
and b
9.7.8
1r n
q
are located at
are located at 'it n
Lines of charge
8.
per uni ~ length are located at
is any integer.
Lines of charge
charge
q
q
where
n
is
1)" /2
+ ir n
while lines of
e.rr:r integer.
are located at the points
z = a + b n ~ where
are real numbers and n is any integer.
Lines of charge
+ i, where
n
q
per unit .length are located at the points
is any integer, while the
z
=
x-axis is a grounded con­
ducting plate.
The diagram shows two
lines ot charge
'5 = 7r/3
length at
1=
2 11" /3 t.
u=o
per unit
q
i
and
Both boundaries
of the channel are grounded con­
ducting plates.
9.7.10
•
Find the potential in the channel.
The diagram shows a line of
.­
charge q
per unit length located at
~ ='a e iB
between two infinite
/
SP9.1:2..
t
__- - - - - _
--~--------------­
~.-----_.--.
grounded conducting plates
o
making angle c/.... •
Find the
C::"&"
<l.
---==----­
OF
~-i' @ 1.~
.
._.--­
':!= ae
----_...,.;;..;;..----L~
~_-:-S---~
::
potential.
(~)
~~/;;
----
Find the potential
1n the infinite channel shown when
the lower boundary is a grounded conducting plate and the upper boundary
1s a conducting plate at potential
=1•
u
•
------------1------------~x
Z-
PL.ANE:
(b) Sho'W' that the mapping function
z
=
log
"3-1
i +1
maps the above channel
onto the lens shaped region
shown in the
1-
. s:z. +"t;
:1-_
2
-r c..or k.
plane.
(c) Find the potential within
•
A/
this lens shaped region when
S
-'"*"....:;......~--..::---..:-"--<:;.._.::::----<:~~.....:::...~~-~;>_
I
1
'j
,
the circular arc ABC
-1 D
I
is a
F
~
i
SP9./:3
'.
grounded conducting plate while the lover conducting plate
, maintained at potential
D/E/r I'is
u = 1.
Find the bounded temperature inside each ot the shaded regions shown
in problems
9.8.1 to 9.8.8. Describe the isotherms and the lines ot tlux.
ANSWER. :
-
'loX
+
a.
5"0
.
.•
9.• 8.2
/'
o
o
"­
;t
"
-u. -:::
•
-!
.§:<2- 1: a h.
7T
."
o·
~
II
/
-
Y
X
+
£>0
•
Y
,......,
SP9.1 ~
I
AN $
'NER., ~
.
9.8.4
tL=
30
'~-o(
./
-!
+ _ 2.0/1
\
Y
ta.n..­
X
-
,,8-0<..
\
\
,
\
x
t
3
6
9.8.5
•
9.8.6
o
-o
/1
.......
.' 1/
-;s
./
/
•
5"00(
In this problem we examine the connection between relations (1)
~hrough
(7) of section 9.8.
(a) Derive relations (2) and (3) from (1).
•
(b) Derive relations (6) and (7) from (2) and (3) .
fe) Show that the foci of the ellipses and hyperbolas described by (6) and
(7) are at the points
(d) Derive (4).
(~l,
0).
The simplest way to see this is to use the defining
relation for a hyperbola in the form:
that
Fl P - F2P = d, where F
l
and
"The locus of all points
F2
are the foci and
d
P
such
is the distance
between the vertic es."
(e) Derive (5).
that
•
­
Note that an ellipse is the locus of all points
Fl P + F2 P
=rn,
P
such
where Fl and F 2 are the foci and m is the major axis •
.
'
Find the mapping of each of the following regions under the in­
SP9.I'
~
version transformation y;:
2l
(a)
Iz
+
(e)
Iz
+ 2\ ~ 2,
(e)
tz
- 1 +
(g)
Re(z)
1, ,
<.
~
i; ~ 5.[2,
lIz.
r~
(b)
rz
(d)
'z + IT - if ~
(f.l
+ 3 i
1,
2,
Im(z) ~ 1,
o.
Use the transformation
y;:
_
to map the regions
i
(Zz +- 1)
1
(a)
J zl ~ 4 and
(b)
J z - il ~
1.
Map the region Im( z) ~ 1 under the transformation y;:
Map the first quadrant in the
v=
AnsYer:
z-p1ane under the transformation
z - i
z + i
The loyer half of the circle
In problems
~ ~ ~
9.9.5
through 9.9.11
I y}
~ 1.
find the temperature distribution
/
throughout each of the shaded regions resulting from the temperatures on
the boundaries.
•
/
./
SP9.17
ce
•
.-
,I
SP9.1 S
In this problem we prove that the inversion transformation
v
= lIz
(a)
preserves circles.
Let
A, B, C, D be real numbers and let
relation
~
(b)
If A = 0
O.
Shov that the
Azz+bz+bz+D=O
defines a circle with center at
A
b = B + i C.
z = b
and radius
Jibl 2
7
-
DA
IA
if
it is a straight line (circle of infinte radius).
Shov that under the inversion w
= lIz
the above circle maps onto a
a circle in the v-plane.
\e
Find the bilinear transformation mapping the three given points
9,.10.1
in the z-plane
(a)
(b)
2
1
00
2
00
ClQ
0
4
i
-1
1
2i
0
2
-2i
3
0
v-plane.
v
v
w
th~
z
z
z
-3
onto the corresponding points in
(c)
;
/
9.10.2
Let
J3
- i
and
/
q be points symnetric vith respect to the
/
circle
e
I z I = 6.
Let
circle
Iz
4
/
Find q.
+ 3i
- 3 - 2i
and
I = 5.
q be points symmetric vith respect to the
Find
q.
SP9.1cr
9.10.4
Find a bilinear transformation mapping the half plane
onto the circle
Jw J =:
3
and the point
z
=1
+ i
onto
w
= o.
Find a bilinear transformation mapping the half plane
onto the unit circle
w
Iv I ~
i·
and mapping the point . z
Re(z) ~ 0
= .(3
Im{z) ~ 1
+ i
onto
o.
=
9.10.6·
In this problem we will derive the expression for the cross ratio
(2) of section 9.10.
We wish to find the bilinear transformation that
maps the three distinct points
distinct corresponding points
(a)
zl' z2' z3
vl' w2' v3
in the z-plane'onto the three
in the v-plane respectively.
Show that if
az + b
cz + d '
w=
then
(ad - bc) (z - Zk)
(cz + d) (czk + d)
where
(b)
k
Use
= 1,
2, 3.
(1) above to show that
_ W3 )
(v2 - wl)(w - w3)
. (w - WI)
(W2
=
(z _
ZI) (Z2 _ z3)
(z2 - zl)(z - Z3)
l
.'
51'9 •.2..0
•
(Hint:
Start with the left side of this expression and convert it into
z's using (1) fovr times.)
9.10.7
Prove that the cross ratio of four distinct points
zl' :2' z3'
(Z4 _ Zl) (Z2 _ Z3)
(Z2 - zl)(z4 - z3 J
is real if' and only if they are on a circle.
9.10.8
•
In this problem we will prove that a bilinear transformation
maps a circle
to
C and two points
z = p
C in the z-plane onto a circle
symmetric with respect to
/
and
z
=q
symmetric with respect
C/ and two points
C in the w-plane.
v
=p /
and w
= q/
We abreviate this statement
by saying "a bilinear transformation preserves circles and their symmetric
points".
(a)
Argue that a simple translation preserves circles and their symmetric
~/
points.
(b)
•
Argue that a simple rotation preserves circles and their symmetric
points •
(e) ,Show that a magnification preserves circles and their symmetric points.
SP9 •. 2. ,
(d)
Show that the inversion w
= lIz
preserves circles and their symmetric
points by examining the following:
A circle with center
z
=
blA and radius
II b12 -
7
IA is
DA
given by
c:
A z
Z + 'bz
z-plane.
+ bz + D
(Recall
=0
(where
Spg.9.l2).
A and-' D are real) in the
Show that under inversion this
becomes
C/ :
•
D w v + by + bw + A
=0
w = bID
Jib /2
and radius
(ii) Show that
p
and
to circle
p/
-
DA'
ID •
are symmetric with respect to circle
C
A pq - bp - bq + D = O.
if and only it'
(iii) Show that
q
which is a circle having center
= IIp
and
ql = l/q
are symmetric with respect
CI •
(e) Using (a), (b), (c) and (d) above, prove that the general bilinear
,/
transformation preserves circles and their symmetric points.
9.10.9
•
Let
C be a circle in the z-plane and let z
two points symmetric with respect to C.
both p
and
q.
Show that -rr intersects
=p
and z
= q be
Let -r be a circle passing through
C at right angles.
{Hint:
Map
SP9 •. 22
•
C onto a straight line.)
9.10.10
Let
side
C.
Construct two
lines
~rom
to circle
T/ •
C be a circle with center at
o.
Let
Q be a point out­
T
Q tangent
C at
T and
Let P be the
-,
point at which TT
intersects
with respect to circle
OQ.
Prove that
P and
Q are symmetric
C.
"
•
Let ~l be the circle Izi
9..10.11
Iz
+
01
inside
(a)
=R
where
=1
and let ~
C is real and positive.
2
be the circle
Assume also that""'-l
is
r--;.
Find two points,
z
=p
and
z
= q,
symmetric with respect to both
I
/
.'
•
(b)
Find a bilinear transformation that maps
r
1 and
P2
on the z-plane
I'.
SP9.2-3
onto tva circles ,....,; and
"j -plane.
answers:
I
"'1
J
~2
(c)
r;' having 'their cen'ters a't 'the origin in 'the
I
/
Wri'te 'the equa'tions of \""'1
and)72'
~ =
, ~f
'~I
z - p
z - q
= p
=
·p(R - C) - p
2
p(R-c) - 1
Suppose T'l is main'tained a't 'tempera'ture
T in 'the z-plane.
2
Tl
and j7 2 is at 'tempera'ture
De'termine 'the 'tempera'ture dis'tribution
u(x, y) be'tween
'these 'two circles.
Answer:
(T2 _ Tl) log
-l-
+ T
Use Poisson's in'tegral formula for 'the circle 'to es'tima'te 'the
9.11.1
'tempera'ture a't 'the poin't
~
p) + Y
(px _ 1)2 + ~2
p(R - c) - p ~
p(R - c) - 1 )
~
2 log
47 of
{ ex - 2 2]
~~is chap'ter.
= 50,
z
= -
1/2 for 'the disc described in problem
Use values of 'the 'tempera'ture a't 'the discre'te poin'ts
15°, 25 0 , ••• , 3550 as given on 'the boundary.
Compare your
resul't wi'th 'the exac't answer given by 'the solu'tion 'to problem 47.
9.11.2
this, use 'the mapping
vJ
~ R.
To do
w = R z in conjunc't10n wi'th (1) of sec'tion
9.11.
Find Poisson '5 in'tegral formula for 'the circle J
SP9.2,. Lj
•
Use the variables
'W'
( P....a.)
Answer:
u,
,--.;;r
J
=pei
=
z
-&,
=r
e:!:e', z = re i -& •
~n-
1
2'11'"
.
~}
~
(R2 -p 2) U(R,
d
-­
R2 - 2R ~ cos(-& - 0) +(52. .
"
Prove that the value of a harmonic function at a point
P equals
the average of its values on the boundary of any circle vi th center at
P.
(Of course, we assume the function is harmonic at all points inside this
circle.)
Hint:
Use the result of the previous problem.
Consider the temperature distribution examined in Example
•
section 9.11.
Use the method of that example to approximate the temperature
at z = 2 + i. Use the values of
~
selected in that example.
Get a second approximation to the temperature at
9.11.5
found in the previous problem.
z
=2
+ i
Use the method of Example 3 'With
de-
/
harmonic function
u(x, y} in
Im(z»
0 when the values of u(x, 0) are
given on the x-axis as fo110'W's:
(a) u(x, 0) =
•
= 100 •
Use Poisson's integral formula to find the values of a bounded
9.11.6
(b)
2 of
u(x, 0) =
[
(
2~ for
for
30
for
o for
1~
x
x<. 1 •
2< x
2~x<.2
-30 for x < -2 •
.­
I
SP9.2.5
(c)
u(x, 0)
where
0 tor
=
2n~x<:2n+l
[ 1 tor
2n+l<x<2n
is any integer.
n
9.11.7
In this problem ve will derive Poisson's formula tor the unit circle
«1) ot section 9.11).
z = re:i.& be a point inside' z\
Let
ei~ be a point on the circle , zr =
I z(
be analytic tor
(a) Explain
t(z)
•
..
when
'zl
=
<:
(e)
t(z) = u(r, -ET) + i
v(r,~)
;!; 1.
·1
1
1-
2 11'"i
]
l/z
d'j
1.
1 .
[
Let
why
Set Cj = ei~
(b)
1.
Let ~=
= 1.
and
:s 1- l/iJ1
'j - z
z = re1-& and shov that
d
~
=
Mu1.tiply numerator and denominator of this last expression by
2 r cos (¢ - ~) - r 2 - 1
(d)
•
Combine this last result vith (a) to get
J
~1l"
u(r,-&)
=
1
~
()
u(l, ¢) (1 - r 2 ) d ¢
1 + r 2 ~ 2r cos (¢ --&)
.­ I
'"'•
Without determining the constants
A, B or the Xj, IS, write. the two forms
of the Schwarz-Christoffel transformation ((2) and (3) of section 9.12)
tor the mapping of
Im( z) ~ 0 onto the polygon shown in the v-plane.
9.12.1
9.12.2
V
1.
't(. .
9.12.4
-;L=::==;;\L..._.....L_--=;.'U..
/
•
.r
I
SP9.27
9.12.8
v
.' In this problem we vill demonstrate that the function
y=A
.
s~
n
i
'.
maps the unit circle
'.
The vertices
Vi (i
1'r'
=1
d z·+ B
I zI
= 1,
<..
1
onto the polygon shown in the v-plane.
2, ••• , n) of the polygon map onto the points
Zi on the boundary of the circle.
s,
(a)
•
S:l,
s~
-t-----------?>
.5t'1.
S
Write the Schwarz-Christoffel transformation for the mapping of the
polygon in the v-plane onto
1m (
1 )~
O.
""t{.
(b)
~ - i
z = "4
~ + i
Show that
maps
Im( "j ) ~ 0
I z, ~ 1.
onto
(c). Using the bilinear transformation in (b) rewrite the answer for (a)
in terms of. z.
(d)
Call zi the image of
si' (i = 1, 2, ••• , n).
Simplify the result in" (c) ~sing the fact that
E
o(..i = 2
to get
the desired mapping function.
Answers:
(a)
z + 1
[ z - 1
(c)
_
1]
70i +
70i - 1
-c:<
i
dz
(z - 1)
2 + B •
Use the Schwarz-Christoffel transformation to determine a function which
maps
lm( z)
constants
9.12.10
•
2
0
onto the polygon shown in the w-plane.
Determine all
A, B and the xi's which appear in (2) and (3) of section 9.12
9.12.11
SP9.'2..f
•
Ansver:
= bi (z - 1) 2
V
•
11 -
J
z2
9.12.16
9.12.17
..
?II
•
Ansver:
v=
1 - b
2
log (z - 1) +
.
where
b
2
9.12.18
•
+ 1 - b 16g(1-b)
2
1".
+
2
1
.." •
_cg
\ ..
l+b log(l+b)
. r
j
..
.
~
= a1
2
Find the temperature at the point
with the given boundary values.
log (z + 1)
i .
satisfies the equation .
111'(·1 - b)
1 + b
{Hint:
z
=
2 ~i in the region shovn .
Use the result of problem
9.12.14.
SP9.W
z = 2
The point
IT
i
will map
onto "i" in the upper half' plane.)
50.
Answer:
9.12.19
Y
5
2
=
Determine the region in the w-planeinto which
-ot..
z
(1 - z)
-p
dz
\
o
Im(z) ~ O.
maps
\
Answer:
\
r?t'
The triangle shown where
~=
'.
and
\
2-co(-{3
a=
r
(1 - ~ )
r
f7 ( ¥
(1
)
-13 )
/
with k > 1
maps
Im(z)
~
0
v
A
B
-k
C
-1
c'
o
1
k
-a.
onto the rectangle shown.
a=
~
1
Shov that
J
dx
where F denotes the so calJ.ed complete elliptic integral on the left.
Also
shov that
b =
9.12.21
1.
k
In this problem we will show that the Schwarz-Christoffel trans­
formation ((2) of section 9.12) maps
polygon.
Im(z)
~
0 onto the interior of the
We use the notation employed in the first part of section
Which describes the polygon, its vertices, etc.
(a)
Show that
. . . (z _
(b)
Show that
...
(e)
•
We vill nov begin
traversing the x-axis
from ,right to left and
_0(.
~)
n
dz.
9.2
we will watch the direction
of dw.
Show that when
x,
arg dw
(d)
= arg
Sholl' that when
changes by
(e)
A - 2 1)-.
As
of arg
x
0<.
x moves to a point between
xl
and
x2' arg dw
1 7"r-.
continues to pass overx2 , x3'
...
, describe the behavior
dw.
z
(f)
Describe the curve formed in the w-plane as1moved from -
00
to +
00
along the x-axis.
(g)
Why does the region
Im(z)
> a
map onto the interior of this polygon?
,/
/
/
'.