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Complex Variables Chapter 20 Conformal Mappings

May 21, 2013 Lecturer: Shih-Yuan Chen

1

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Contents Complex functions as mappings Conformal mappings Linear fractional transformations Schwarz-Christoffel transformations

2

Complex Functions as Mappings

For w = f (z), a point z = x + iy in the domain of f (z-plane) is said to map to the point w = f (z) in the range of f (w-plane). The complex function w = f (z) = u(x, y) + iv(x, y) is considered as the planar transformation u = u(x,y) & v = v(x, y). And w = f (z) is the image of z under f.

3



Images of curves Note that if z(t) = x(t) + iy(t), a ≤ t ≤ b, describes a

curve C in the region, then w = f (z(t)), a ≤ t ≤ b, is a parametric representation of the corresponding curve C′ in the w-plane.

A point z on the level curve u(x, y) = a is mapped to a point w on the vertical line u = a, & a point z on the level curve v(x, y) = b will be mapped to a point w lying on the horizontal line v = b.

4

Ex. Horizontal strip 0 ≤ y ≤ π lies in the domain of f (z) = ez. Vertical line segment x = a in this region can be described by z(t) = a + it, 0 ≤ t ≤ π ⇒ w = f (z(t)) = eaeit. Thus the image is a semi-circle with its center at w = 0 & r = ea. Similarly, a horizontal line y = b can be described by z(t) = t + ib, −∞ < t < ∞ ⇒ w = f (z(t)) = eteib. Since Arg(w) = b & |w| = et, the image is a ray emanating from the origin. Since 0 ≤ Arg(w) ≤ π, the image of the entire horizontal strip is the upper half- plane v ≥ 0.





Ex. f (z) = 1/z has domain z ≠ 0 and real & imaginary parts u(x, y) = x/(x2 + y2) & v(x, y) = −y/(x2 + y2). Level curve u(x, y) = a ≠ 0 can be written as

Point z on this circle other than zero is mapped to a point w on the line u = a. Similarly, v(x, y) = b ≠ 0 can

be written as

Point z on this circle is mapped to a point w on the line v = b.

22

222

21

21or 01

=+

−=+−

ay

axyx

ax

222

21

21

=

++

bbyx

6

Since w = 1/z, we have z = 1/w. Thus f −1(w) = 1/w & f = f −1. We conclude that f maps the horizontal line y = b to the circle u2 + (v+1/2b)2 = (1/2b)2 & maps the vertical line x = a to the circle (u−1/2a)2 + v2 = (1/2a)2.


7




Translation & rotation Linear function f (z) = z + z0 can be interpreted as a

translation in the z-plane. Let z = x + iy & z0 = h + ik. Since w = f (z) = (x + h) +

i(y + k), point (x, y) has been translated h units in the horizontal direction & k units in the vertical direction to the new position at (x+h, y+k).

can be interpreted as a rotation through θ0 degrees.

⇒ If z = reiθ, then

( ) zezg i 0θ=

( ) ( )0θθ +== irezgw8

( ) 00 zzezh i += θ


If the complex mapping is applied to a region R centered at the origin, the image region R′ may be obtained by first rotating R through θ0 degrees & then translating the center to the new position z0.

9




Ex. Find a complex function that maps the horizontal strip −1 ≤ y ≤ 1 onto the vertical strip 2 ≤ x ≤ 4. (sol) Rotating the horizontal strip −1 ≤ y ≤ 1 by 90° results in the vertical strip −1 ≤ x ≤ 1. And the vertical strip 2 ≤ x ≤ 4 can be obtained by shifting this vertical strip 3 units to the right.

( )3

32

+=+=∴

izzezh

iπ




Magnification A magnification is a complex function of the form

w = f (z) = αz, where α is a positive real constant. Note that |w| = |αz| = α|z| ⇒ f changes the length of

the complex number z by a factor α. If g(z) = az + b & , then the vector z is

rotated through θ0 degrees, magnified by a factor r0 & then translated by b. Ex. Map the disk |z| ≤ 1 onto the disk |w − (1+ i)| ≤ 1/2. First halve the radius of the disk & then translate its center to the point 1+ i ⇒ w = f (z) = z/2 + (1+ i)

00

θiera =

11


Power functions A complex function of the form f (z) = zα, where α is

a positive real constant, is called a real power function.

For z = reiθ ⇒ w = f (z) = rαeiαθ. Thus, for 0 ≤ Arg(z) ≤ θ0 ⇒ 0 ≤ Arg(w) ≤ αθ0.

12



Ex. Map the upper-plane y ≥ 0 onto the wedge 0 ≤ Arg(w) ≤ π/4. 0 ≤ Arg(z) ≤ π ⇒ 0 ≤ Arg(w) ≤ π/4

Successive mappings If ζ = f (z) maps R onto R″ & w = g(ζ) maps R″ onto R′, then w = g(f (z)) maps R onto R′.


( ) 41zzf =∴

13



Ex. Map the horizontal strip 0 ≤ y ≤ π onto the wedge 0 ≤ Arg(w) ≤ π/4.

Ex. Map the wedge π/4 ≤ Arg(z) ≤ 3π/4 onto v ≥ 0.

( )

( ) ( ) ( )

( )( ) ( ) 4

Arg04Arg0

041

zz

g

ezf

eegzfgww

yz

===∴

≤≤ ←≤≤ →≤≤

=

=

πζπ

πζζ

( ) ( )

( ) ( ) ( )

( )( ) ( ) 224

2Arg0Arg0

43Arg42

4

izzezfgw

wz

i

g

zezf i

−===∴

≤≤ ←≤≤ →≤≤

−

=

= −

π

ζζπζ

πππ

π

14

Angle-preserving mappings Complex mapping w = f (z) defined on a domain D

is called conformal at z = z0 in D if, for C1 & C2 intersect in D at z0 and C1′ & C2′ are the corresponding images in the w-plane, the angle θ between C1 & C2 is equal to the angle φ between C1′ & C2′.

Conformal Mappings

15



Conformal Mappings If z1′ & z2′ denote tangent vectors to curves C1 & C2,

respectively, then, applying the law of cosines to the triangle determined by z1′ & z2′, we have

Likewise, if w1′ & w2′ denote tangent vectors to curves C1′ & C2′, respectively, then

( )1 2

cos

cos2

21

221

22

211

212

22

12

21

′′′−′−′+′

=⇒

′′−′+′=′−′

−

zzzzzz

zzzzzz

θ

θ

( )2 2

cos21

221

22

211

′′′−′−′+′

= −

wwwwww

φ16

Conformal Mappings Conformal mapping

Thm. If f (z) is analytic in the domain D & f '(z0) ≠ 0, then f is conformal at z = z0. (proof) If a curve C in D is parameterized by z = z(t), then

w = f (z(t)) describes the image curve in w-plane. Apply Chain rule to w = f (z(t)) ⇒ w′ = f ′(z(t)) z′(t). If curves C1 & C2 intersect in D at z0, w1′ = f ′(z0) z1′ & w2′ = f ′(z0) z2′. Since f '(z0) ≠ 0, using (2) gives

( ) ( ) ( ) ( )( ) ( ) θφ =

′′′′′′−′′−′′+′′

= −

2010

22010

220

2101

2cos

zzfzzfzzfzzfzzfzzf

Conformal Mappings Ex. (a) Analytic function f (z) = ez is conformal at all points in the z-plane, since f ′(z) = ez is never zero. (b) Analytic function g(z) = z2 is conformal at all points except z = 0 since g′(z) = 2z ≠ 0 for z ≠ 0.

Ex. Vertical strip −π/2 ≤ x ≤ π/2 is the fundamental region of w = sin z. Vertical line x = a within this region can be described by z(t) = a + it, −∞ < t < ∞. From (21) of Ch.17,

Also,

( ) taitaitaivuyxiyxzsinhcoscoshsinsin

sinhcoscoshsinsin +=+=+∴

+=⇒

1cossin

1sinhcosh 2

2

2

222 =−∴=−

av

autt 18

Conformal Mappings The image of x = a is therefore a hyperbola with

sin a as u-intercepts, and since −π/2 ≤ a ≤ π/2, the hyperbola crosses the u-axis between u = ±1. Note that if a = −π/2, then w = −cosh t, & so x = −π/2 is mapped onto (−∞,−1] on the −u axis. Likewise, x = π/2 is mapped onto [1,∞) on the +u axis.

Similarly, the horizontal line segment described by z(t) = t + ib, −π/2 < t < π/2, is mapped onto the upper (lower) portion of the ellipse for b > 0 (b < 0)

1sinhcosh 2

2

2

2

=+b

vb

u

19

Conformal Mappings Mapping of f (z) = sin z. Since f ′(z) = cos z, f is conformal at

all points in the region except z = ±π/2.

The hyperbolas & ellipses are therefore orthogonal since they are images of the orthogonal families of horizontal segments & vertical lines.

20



Conformal Mappings Ex. f (z) = z + 1/z is conformal at all z except at z = ±1 & 0. The function is conformal in the upper half- plane of |z| > 1. If z = reiθ, then w = reiθ + (1/r)e−iθ,

Note that if r = 1, then u = 2cosθ & v = 0. Therefore, the semicircle z = eit, 0 ≤ t ≤ π, is mapped to the segment [−2,2] on the u-axis. It follows from (3) that if r > 1, then the semicircle z = reit, 0 ≤ t ≤ π, is mapped onto the upper half of the ellipse:

( )3 sin1 ,cos1θθ

−=

+=⇒

rrv

rru

rrb

rra

bv

au 1 and 1 where,12

2

2

2

−=+==+ 21

For a fixed θ, the ray z = teiθ (t ≥ 1) is mapped to the portion of the hyperbola

in the upper half-plane v ≥ 0. Since f is conformal for |z| > 1 & a ray θ = θ0 intersects a circle |z| = r at a right angle, the hyperbolas & ellipses in the w-plane are orthogonal.

Conformal Mappings

411sincos

22

2

2

2

2

=

−−

+=−

tt

ttvu

θθ



Conformal Mappings Conformal mappings are categorized as

elementary mappings, mappings to half-planes, mappings to circular regions, & miscellaneous mappings.

Ex. Use the table to find a conformal mapping between the strip 0 ≤ y ≤ 2 & the upper plane v ≥ 0. What is the image of the negative x-axis? (sol) Let a = 2 ⇒ f (z) = eπz/2



Ex. Find a conformal mapping between the strip 0 ≤ y ≤ 2 & the disk |w| ≤ 1. What is the image of the negative x-axis? The strip can be mapped by f (z) = eπz/2 onto the upper half-plane. The complex mapping maps the upper half-plane to the disk |w| ≤ 1.

maps the strip 0 ≤ y ≤ 2 onto the disk |w| ≤ 1.

( )( ) 2

2

z

z

eieizfgw π

π

+−

==∴

Conformal Mappings

ζζ

+−

=iiw



Conformal Mappings Dirichlet problem & Harmonic functions A bounded harmonic function u = u(x, y) that takes

on prescribed values on the entire boundary of a region R is called a solution to a Dirichlet problem on R.

Recall from ch.17 that the real & imaginary parts of an analytic function are both harmonic. Since there are lots of analytic functions, we can find closed-form solutions to many Dirichlet problems.

25

Conformal Mappings Transformation theorem for harmonic fxs

Thm. Let f be an analytic function that maps domain D onto domain D′. If U is harmonic in D′, then the real-valued function u(x, y) = U(f (z)) is harmonic in D. (proof) If U has a harmonic conjugate V in D′ ⇒ H = U + iV is analytic in D′, & so H(f (z)) = U(f (z)) + iV(f (z)) is analytic in D. It follows from ch.17 that the real part U(f (z)) is harmonic in D. For U to have a harmonic conjugate, let

( )vUi

uUwg

∂∂

−∂∂

=

Conformal Mappings The 1st Cauchy-Riemann equation

is equivalent to Laplace eq., which is satisfied because U is harmonic in D′. The 2nd Cauchy-Riemann equation

is equivalent to the equality of the 2nd-order mixed partial derivatives.

∂∂

−∂∂

=

∂∂

∂∂

vU

vuU

u

∂∂

−∂∂

−=

∂∂

∂∂

vU

uuU

v

27

Conformal Mappings Therefore, g(w) is analytic in the simply connected

domain D′ & has an antiderivative G(w) from ch.18.

If G(w) = U1 + iV1, then

Since , U & U1 have equal 1st

partial derivatives. Therefore, H = U + iV1 is analytic in D′, and so U has a harmonic conjugate in D′.

( ) ( )v

Uiu

UwGwg∂∂

−∂∂

=′= 11

( )vUi

uUwg

∂∂

−∂∂

=

28

Conformal Mappings Solving Dirichlet problems Find a conformal mapping w = f (z) that transforms

the region R onto R′. Transfer the boundary conditions from the

boundary of R to the boundary of R′.

29

R R′


Solve the corresponding Dirichlet problem in R′. The solution to the original Dirichlet problem is

u(x, y) = U(f (z)).

Ex. U(u, v) = (1/π)Arg w is harmonic in the upper half- plane v > 0 since it is the imaginary part of the analytic function g(w) = (1/π)Ln w. Use this function to solve the Dirichlet problem in the LHS figure.

Conformal Mappings

30



Conformal Mappings The analytic function f (z) = sin z maps the original

region to v ≥ 0 & maps the boundary segments to those shown in the RHS figure.

The harmonic function U(u,v) = (1/π)Arg w satisfies the transferred BC U(u,0) = 0 for u > 0 & U(u,0) = 1 for u < 0. Therefore, u(x,y) = U(sin z) = (1/π)Arg(sin z) is the solution to the original problem. If tan−1(v/u) is chosen to lie between 0 & π, the solution can also be written as

( )

= −

yxyxyxu

coshsinsinhcostan1, 1

π31

Conformal Mappings Ex. For mapping in the figure, the analytic function

maps the region

outside the two open disks |z| < 1 & |z − 5/2| < 1/2 onto the circular region r0 ≤ |w| ≤ 1, where

( )5

627 where,1

+=

−−

= aaz

azzf

6250 −=r



Conformal Mappings One can show that U(w) = (loger)/(loger0) is the

solution to the new Dirichlet problem. The solution to the original BVP is

( ) ( )( ) ( )( )

( ) 156275627log

625log1,

−++−

−==

zzzfUyxu e

e

33

Conformal Mappings A favorite R′ for a simply connected R is the upper

half-plane v ≥ 0. For any real a, Ln(w−a) = loge|w−a| + iArg(w−a) is analytic in R′. Therefore, Arg(w − a) is harmonic in R′ and is a solution to the Dirichlet problem shown in the figure.

It follows that the solution in R′ to the Dirichlet problem with

is the harmonic function

( ) 0 ,0

0 otherwisec a u b

U u< <

=

( ) ( ) ( )0, Arg ArgcU u v w b w aπ

= − − − 34


Linear Fractional Transformations The complex function is called a

linear fractional transformation if a, b, c, d are complex constants with ad − bc ≠ 0. Since , T is conformal at z if

ad − bc ≠ 0 & z ≠ −d/c. When c ≠ 0, T(z) has a simple pole at z0 = −d/c,

Write T(z0) = ∞ as shorthand for this limit.

( )dczbazzT

++

=

( )( )2dcz

bcadzT+−

=′

( ) ∞=⇒→

zTzz 0

lim

35

Linear Fractional Transformations If c ≠ 0, then

Write T(∞) = a/c.

Ex. If T(z) = (2z+1)/(z−i), compute T(0), T(∞), T(i). T(0) = 1/(−i) = i T(∞) = lim|z|→∞T(z) = 2 limz→i|T(z)| = ∞ ⇒ T(i) = ∞

( )ca

zdczbazT

zz=

++

=∞→∞→

limlim

36

( )4 ,1 , 21

21 BAzwz

zdczz +==+=

Linear Fractional Transformations

ca

dczcadbc

dczbazw +

+−

=++

=1

Circle-preserving property If c = 0, the linear fractional transformation

reduces to T(z) = Az + B. A linear function will map a circle in the z-plane to

a circle in the w-plane. If c ≠ 0, we write

If we let A = (bc − ad)/c & B = a/c, T(z) can be written as the composite of transformations:

37

Linear Fractional Transformations If |z − z0| = r & w = 1/z, then

The set of all points w satisfying |w − w1| = λ|w − w2|

is a line when λ = 1 & is a circle when λ > 0 & λ ≠ 1. From (5), the image of the circle |z − z0| = r under

the inversion w = 1/z is a circle except when r = 1/|w0| = |z0| .

( )5 0or 1100

0

0

0

−=−=−

=− wwrwwrwwww

ww

38

Linear Fractional Transformations Thm. A LFT maps a circle in the z-plane to either a line or a circle in the w-plane. The image is a line if & only if the original circle passes through a pole of the LFT. (proof)

From (4), a circle in the z-plane will be mapped to either a circle or a line in the w-plane. If the original circle passes through a pole z0 ⇒ T(z0) = ∞, & so the image is unbounded. Therefore, image of such a circle must be a line. If the original circle does not pass through z0, then the image is bounded & must be a circle.

39

Linear Fractional Transformations Ex. Find the images of the circle |z| = 1 & |z| = 2 & their interiors under T(z) = (z + 2)/(z − 1). (sol) The circle |z| = 1 pass through the pole z0 = 1 of T(z)

& so the image is a line. Since T(−1) = −1/2 & T(i) = −1/2 − i3/2, the image is the line u = −1/2.

The image of |z| < 1 is either the half plane u < −1/2 or u > −1/2. Use z = 0 as a test point, T(0) = −2, & so the image is the half-plane u < −1/2.

The circle |z| = 2 does not pass through the pole & so the image is a circle. 40

For |z| = 2,

Therefore, is a point on the image circle & so the image circle is symmetric to the u-axis. Since T(−2) = 0, T(2) = 4, the image is the circle |w − 2| = 2.

The image of the interior |z| < 2 is either the interior or exterior of the circle |w − 2| = 2. Since T(0) = −2, the image is |w − 2| > 2.


( ) ( )zTzz

zzzTz =

−+

=

−+

==⇒12

12 and 2

( )zT


Linear Fractional Transformations We must construct special functions that map

a given circular region R to a target region R′ where the Dirichlet problem is solvable.

Matrix methods

dczbazzT

dcba

++

=

= )( with Associate A

dczbazzTzTT

dzcbzazT

dzcbzazT

++

==

++

=++

=

)())((then

,)( and )( If

12

22

222

11

111

42


(7) adj ismatrix

associated theand ,)( is,that

then ,)( If

)6( where

1

11

11

22

22

A=

−

−+−−

=

+−−

=++

==

=

−

acbd

acwbdwwT

acwbdwz

dczbazzTw

dcba

dcba

dcba

43

Linear Fractional Transformations Ex. If & , find S−1(T(z))

(sol) From (6) & (7), we have , where

( )212

+−

=zzzT ( )

1−−

=iz

izzS

( )( )dczbazzTS

++

=−1

( )( ) ( )( ) izi

izizTS

iiii

ii

ii

dcba

++−+++−

=⇒

+−++−

=

−

−−

=

−

−−

=

−

221212

221212

2112

11

2112

1

1 adj

144

Linear Fractional Transformations Triples to triples Linear fractional transform

has a zero at z = z1, a pole at z = z3, & T(z2) = 1. Thus, T(z) maps three distinct complex numbers z1, z2, z3 to 0, 1, ∞, respectively.

The term is called the cross-ratio of z, z1, z2, z3.

Similarly, maps w1, w2, w3

to 0, 1, ∞. So S−1 maps 0, 1, ∞ to w1, w2, w3.

( )12

32

3

1

zzzz

zzzzzT

−−

−−

=

12

32

3

1

zzzz

zzzz

−−

−−

( )12

32

3

1

wwww

wwwwwS

−−

−−

=

45

( )8 12

32

3

1

12

32

3

1

zzzz

zzzz

wwww

wwww

−−

−−

=−−

−−

Linear Fractional Transformations It follows that the LFT w = S−1(T(z)) maps the triple

z1, z2, z3 to w1, w2, w3. From w = S−1(T(z)), we have S(w) = T(z) & thus

Ex. Construct a LFT that maps the points 1, i, −1 on the circle |z| = 1 to −1, 0, 1 on the real axis. From (8),

iziziw

zzi

ww

ii

zz

ww

+−

−=∴

+−

−=−+

−⇒−+

+−

=+−

−+

11

11

11

11

1010

11

46


Ex. Construct a LFT that maps the points ∞, 0, 1 on the real axis to 1, i, −1 on the circle |w| = 1. Since z1 = ∞, the terms z − z1 & z2 − z1 in the cross-

product are replaced by 1. Then

Or use the matrix method to find w = S−1(T(z)),

)(1

111)(or

110

11

11

11 zT

zwwiwS

zii

ww

=−−

=+−

−=−

−=

−+

+−

.11

11

11

1110

11

adj

iziz

iiziizw

iiiiii

dcba

+−−−

=++−+−−

=∴

+−+−−

=

−−

−=

47


Ex. Solve the Dirichlet problem using conformal mapping by constructing a LFT that maps the given region into a horizontal strip. Boundary circles |z| = 1 & |z − 1/2| = 1/2 each pass

through z = 1. We can map each boundary circle to a line by selecting a LFT that has z = 1

as a pole. If we require T(i) = 0 & T(−1) = 1, then

( ) ( )1

11

111 −

−−=

−−−−

−−

=z

iziiz

izzT

48


Since T(0) = 1 + i & T(1/2 + i/2) = −1 + i, T maps the interior of the circle |z| = 1 onto the upper half-plane & the circle |z − 1/2| = 1/2 onto the line v = 1.

The harmonic fx U(u, v) = v is the solution to the simplified Dirichlet problem in the w-plane, and so u(x, y) = U(T(z)) is the solution to the original Dirichlet problem in the z-plane.


( )[ ] ( )( )

( )( )

solution. theis 1

1,

11

11ImIm

22

22

22

22

yxyxyxu

yxyx

zizizT

+−−−

=∴

+−−−

=

−−

−=

49


Linear Fractional Transformations The level curves u(x, y) = c can be written as

and are circles that pass through z = 1.

22

2

11

1

+

=+

+−

cy

ccx

50


There are analytic functions that map the upper half-plane onto bounded or unbounded polygonal regions.

Riemann mapping thm. If D′ is a simply connected domain, there exists an analytic function g that conformally maps the unit open disk |z| < 1 onto D′.

Schwarz-Christoffel Transformations

51





Special cases First examine the mapping f (z) = (z – x1)α/π, 0 < α <

2π, on the upper half-plane y ≥ 0. It is the composite of ζ = z – x1 & w = ζα/π. Since w = ζα/π

changes the angle in a wedge by a factor of α/π, the interior angle in the image is (α/π)π = α.


52



Schwarz-Christoffel Transformations Note that f ′(z) = A1(z − x1)(α/π)−1 for A1 = α/π. Assume that f (z) is a function that is analytic in the

upper half-plane & that has the derivative

where x1 < x2. Use the fact that a curve w(t) = f (z=t+i0) = f (t) is a

line segment when the argument of its tangent vector f ′(t) is constant. From (9), we get

( ) ( )( ) ( )( ) ( )9 12

11

21

−− −−=′ παπα xzxzAzf

( ) ( ) ( )22

11 Arg1Arg1Argarg xtxtAtf −

−+−

−+=′

πα

πα

53

Schwarz-Christoffel Transformations Since Arg(t – x) = π for t < x,

we can find the variation of f ′(t) along the x-axis.

22

1221

211

Arg ) ,( )( Arg ) ,(

0 )( )( Arg ) ,(arg.in Change )(' arg Interval

απAxαππαAxx

παπαAxtf

−∞−−+

−+−+−∞

54


( )10 )()()()(11

2

1

1

21 −−−−−−=′ π

απα

πα n

nxzxzxzAzf

Schwarz-Christoffel Transformations Schwarz-Christoffel Formula

Thm. Let f (z) be a function that is analytic in the upper half-plane y > 0 and that has the derivative

where x1 < x2 < ⋅⋅⋅ < xn & each αi satisfies 0 < αi < 2π. Then f (z) maps the upper half-plane y ≥ 0 to a polygonal region with interior angles α1, α2, ⋅⋅⋅, αn. 3 Comments: 1. One can select the location of three of the points

xk on the x-axis. 55


2. A general formula for f (z)

and can be considered as the composite of

and w = A×g(z) + B. 3. If the polygonal region is bounded, only n – 1 of

the n interior angles should be included in the Schwarz-Christoffel formula.

( ) ( ) ( ) BdzxzxzxzAzfn

n +

−−−= ∫ −−− 11

21

1

21

)( πα

πα

πα

( ) ( ) ( )∫ −−− −−−= dzxzxzxzzgn

n11

21

1

21

)( πα

πα

πα

56

Ex. Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the strip: u ≤ 0, |v| ≤ 1. (sol) Select x1 = −1 & x2 = 1 on the x-axis, & construct a

conformal mapping f with f (−1) = −i & f (1) = i.


57



Schwarz-Christoffel Transformations Since α1 = α2 = π/2, (10) gives

Thus, f (z) = −Ai sin−1z + B. Since f (−1) = −i & f (1) = i,

( ) ( ) ( ) ( ) 212212

2121

1111)(

ziA

zAzzAzf

−=

−=−+=′ −−

( ) zizf

ABBAii

BAii

1sin2

2 and 0

2

2

−=∴

−==⇒

+−=

+=−⇒

π

ππ

π

58

Ex. Use the S-C formula to construct a conformal mapping from the upper half-plane to the region shown in the figure.

(sol) Select x1 = −1 & x2 = 1, and require f (−1) = ai & f (1)

= 0. Since α1 = 3π/2 & α2 = π/2, (10) gives


( ) ( ) 2121 11)( −−+=′ zzAzf 59



Schwarz-Christoffel Transformations Write

Note that cosh−1(−1) = πi & cosh−1(1) = 0, and so ai = f (−1) = A(πi) + B and 0 = f (1) = B.

( )( ) ( ) ( )

( ) ( )[ ] BzzAzf

zzzA

zzAzf

++−=⇒

−+

−=

−

+=′

−1212

212212212

cosh1

11

111)(

( ) ( )[ ]zzazf 1212 cosh1 −+−=∴π

60

Ex. Use the S-C formula to construct a conformal mapping from the upper half-plane to the region shown in the figure.

(sol) Since the region is bounded, only two of the 60°

interior angles should be included. If x1 = 0 & x2 = 1, we obtain


( ) 3232 1)( −− −=′ zAzzf 61



Schwarz-Christoffel Transformations Use the Theorem from Ch.18, p.43 to get the

antiderivative

If we require that f (0) = 0 & f (1) = 1,

( )Bds

ssAzf

z+

−= ∫0 3232 1

1)(

( )

( )( )∫

∫

−=∴

×=−

==⇒

zds

sszf

Adxxx

AB

0 3232

1

0 3232

111

111 and 0

α

α

62

Ex. Use the S-C formula to construct a conformal mapping from the upper half-plane to the upper half- plane with the horizontal line v = π, u ≤ 0, deleted. (sol) The non-polygonal target region can be

approximated by a polygon region by adjoining a line segment from w = πi to a point u0 on −u-axis.


63



Schwarz-Christoffel Transformations If we require f (−1) = πi & f (0) = u0, then

Note that as u0 approaches −∞, the interior angles α1 & α2 approach 2π & 0, respectively.

This suggests we examine the mappings that satisfy w′ = A(z + 1)1z−1 = A(1 + 1/z)

⇒ w = A(z + Ln z) + B. Consider g(z) = z + Ln z. For real t, z = t+i0

( ) 1121

1)(−−+=′ π

α

πα

zzAzf

( ) ( )titttg e Arg log ++=64

Schwarz-Christoffel Transformations If t < 0, v(t) = Arg(t) = π & u(t) = t + loge|t| varies

from −∞ to −1. It follows that w = g(t) moves along the line v = π from −∞ to −1.

If t > 0, v(t) = Arg(t) = 0 & u(t) varies from −∞ to ∞. Therefore, g maps the +x-axis onto the u-axis.

We can conclude that g(z) = z + Ln z maps the upper half-plane onto the upper half-plane with the horizontal line v = π, u ≤ −1, deleted. Therefore, w = z + Ln z + 1 maps the upper half-plane onto the original target region.

65

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