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Complex Variables Chapter 20 Conformal Mappings
May 21, 2013 Lecturer: Shih-Yuan Chen
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Except where otherwise noted, content is licensed under a CC BY-NC-SA 3.0 TW License.
Contents Complex functions as mappings Conformal mappings Linear fractional transformations Schwarz-Christoffel transformations
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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.
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Complex Functions as Mappings
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.
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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.
Complex Functions as Mappings
Complex Functions as Mappings
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.
Complex Functions as Mappings
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Complex Functions as Mappings
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
Complex Functions as Mappings
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.
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Complex Functions as Mappings
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
Complex Functions as Mappings
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 =
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Complex Functions as Mappings
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.
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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′.
Complex Functions as Mappings
( ) 41zzf =∴
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Complex Functions as Mappings
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
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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
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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
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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.
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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.
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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′.
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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
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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
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( )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.
Linear Fractional Transformations
( ) ( )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
Linear Fractional Transformations
(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
Linear Fractional Transformations
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
Linear Fractional Transformations
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.
Linear Fractional Transformations
( )[ ] ( )( )
( )( )
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.
Schwarz-Christoffel Transformations
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
Schwarz-Christoffel Transformations
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.
Schwarz-Christoffel Transformations
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
Schwarz-Christoffel Transformations
( ) ( ) 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
Schwarz-Christoffel Transformations
( ) 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.
Schwarz-Christoffel Transformations
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.
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