A mapping of the upper half-plane onto the unit disk.. Let a be the point of the upper half-plane which should be taken to the center w =0 of the disk Fig.. A mapping of the upper half-p
Trang 1Let us present some formulas that allow one to find the images of straight lines and circles for an arbitrary linear-fractional mapping (10.2.1.2)
1 The straight lines Re(λz) = α that do not pass through the point z = –d/c (α ≠
– Re(λd/c)) are taken to the circles|w – w0|= ρ, where
w0= 2aα ¯c + a¯d¯λ + b¯cλ
2α|c|2+2Re(c¯ d ¯ λ), ρ=
a
c – w0.
2 The straight lines Re(λz) = – Re(λd/c) passing through the point z = –d/c are taken
to the straight lines
Read – bc
c2 λ w¯
= Read – bc
c2
λ ¯a
¯c
3 The circles|z – z0|= r that do not pass through the point z = –d/c (r≠ |z0+ d/c|) are taken to the circles|w – w0|= ρ, where
w0= (az0+ b)( ¯c¯z0+ ¯d) – a ¯cr2
|cz0+ d|2–|c|2r2 , ρ=
r|ad – bc|
|cz0+ d|2–|c|2r2
4 The circles|z – z0|=|z0+ d/c|are taken to the straight lines
Re
* ad – bc
c(cz0+ d) w¯
+
= |ad – bc|2+2Re[c(az0+ b)(ad – bc)]
2|c(cz0+ d)|2 .
If a linear-fractional mapping takes four points z1, z2, z3, and z to points w1, w2, w3,
and w, respectively, then the following relation holds:
w – w1
w – w3
w2– w3
w2– w1 =
z – z1
z – z3
z2– z3
z2– z1. (10.2.1.3) THEOREM There exists a unique linear-fractional mapping of the extended z-plane onto the extended w-plane taking three arbitrary distinct points z1, z2, and z3 to three arbitrary
distinct points w1, w2, and w3, respectively
THEOREM Any disk of the extended z-plane can be transformed into any disk of the extended w-plane by a linear-fractional function.
Example 6 A mapping of the upper half-plane onto the unit disk.
Let a be the point of the upper half-plane which should be taken to the center w =0 of the disk (Fig 10.14) Then the problem is solved by the linear-fractional function
w = e iβ z – a
z–¯a,
where β is an arbitrary real number (Changing α means rotating the disk around the center w =0 )
X a
Y
U O
V
Figure 10.14 A mapping of the upper half-plane onto the unit disk.
Trang 2Example 7 A mapping of the unit disk onto the upper half-plane.
Let a = ih be the point of the upper half-plane to which the center z =0 of the disk should be taken (Fig 10.15) Then the problem is solved by the linear-fractional function
w = ih e
iβ + z
e iβ – z, where β is an arbitrary real number (Changing β means rotating the disk around the center w =0 )
U ih
V
X O
Y
Figure 10.15 A mapping of the unit disk onto the upper half-plane.
Example 8 A mapping of the unit disk onto the unit disk.
Let a be the point of the disk|z| < 1 that should be taken to the center of the disk |w| < 1 (Fig 10.16) Then the problem is solved by the linear-fractional function
w = e iβ z – a
where β is an arbitrary real number.
Figure 10.16 A mapping of the unit disk onto the unit disk.
Geometrically, β is the angle of rotation of the mapping (10.2.1.4) at the point a:
β= argdw
dz.
If the radius of the disk in the z-plane is equal to R, then the function w = f (z) mapping this disk onto the disk
|w| < 1and satisfying f (a) =0and arg f z (a) = β has the form
w = e iβ R(z – a)
R2– z ¯a.
10.2.1-4 Mappings determined by the Zhukovskii function
The mapping (see also Paragraph 10.1.2-3)
w= 1 2
z+ 1
z
(10.2.1.5)
is conformal except at the points z = 1
Trang 3The mapping given by the Zhukovskii function is equivalent to the following mappings:
z = w + √
w2–1 or w–1
w+1 =
z–1
z+1
2
(10.2.1.6)
If we denote z = re iϕ and w = u + iv, then the mapping (10.2.1.5) can be written as
u= 1 2
r+ 1
r
cos ϕ, v= 1
2
r– 1
r
sin ϕ. (10.2.1.7) The main properties of the mapping (10.2.1.5) are given in Table 10.2
TABLE 10.2
Properties of mapping w = 1
2
z+ 1
z
, where z = x + iy = re iϕ , w = u + iv = ρe iθ
1 |z|Circle= r
0 < 1
Ellipse
u2
a2 +v
2
b2 = 1 ,
where a = 12( 1/r0+ r0 ),
b=12( 1/r0– r0)
The ellipse has the negative sense The foci are at the points 1
2 |z|Circle= r
0 > 1
Ellipse
u2
a2 +v
2
b2 = 1 ,
where a = 12( 1/r0+ r0 ),
b=12( 1/r0– r0)
The ellipse has the positive sense The foci are at the points 1
3 arg z = ϕRadii
0 ( 0< r <1 )
Hyperbolas
u2
cos2ϕ0 – v
2
sin2ϕ0 = 1 The foci are at the points 1.
4 |z|=Semicircle1, Im z≥ 0 |Re w| ≤ 1Segment, Im w =0 u decreases with increasing ϕ.
5 |z|=Semicircle1, Im z≤ 0 |Re w| ≤ 1Segment, Im w =0 u increases with increasing ϕ.
6 0< Re z <Segment1, Im z =0 Re w >Half-line1, Im w =0 —
7 –1< Re z <Segment0, Im z =0 Re w < –Half-line1, Im w =0 —
8 Re z >Half-line1, Im z =0 Re w >Half-line1, Im w =0 —
9 Re z < –Half-line1, Im z =0 Re w < –Half-line1, Im w =0 —
Example 9 The Zhukovskii function defines the following conformal maps:
1 It maps the interior of the semicircle |z| < 1, Im z >0 , onto the lower half-plane (Fig 10.17) The point
z= 1is taken to the point w =1, and the point z = i is taken to the point w =0
2 It maps the upper half-plane with the disk |z| < 1 deleted onto the upper half-plane (Fig 10.18).
3 It maps the half-annulus 1 < |z|< k, Im z >0 , onto the half-ellipse given by the equation
2ku
k2+ 1
2
+ 2kv
k2– 1
2
= 1
on the w-plane (Fig 10.19).
Trang 41
i Y
1
V
Figure 10.17 Zhukovskii function maps the interior of the semicircle|z| < 1, Im z >0 , onto the lower half-plane.
1
i
U X
V Y
Figure 10.18 Zhukovskii function maps the upper half-plane with the disk|z| < 1 deleted onto the upper half-plane.
X
Y
i
k +1 2k
U V
Figure 10.19 Zhukovskii function maps the half-annulus1 < |z|< k, Im z >0 , onto the half-ellipse. 10.2.1-5 Symmetry principle and mapping of polygons
In a special case, the symmetry principle gives a simple sufficient condition for the existence
of an analytic continuation of a function realizing a conformal mapping
THE RIEMANN–SCHWARZ THEOREM Suppose that a function w = f1(z) realizes a
conformal mapping of a domain D1 onto a domain D1∗ and takes a circular arc C of the boundary of D1to a circular arc C ∗ of the boundary of D1∗ Then the function f1(z)admits
an analytic continuation f2(z) through the arc C into a domain D2symmetric to D1about C, the function w = f2(z) realizes a conformal mapping of the domain D2onto the domain D ∗2 symmetric to D1∗ about C ∗, and the function
w = f (z) =
f
1(z) in D1,
f1(z) = f2(z) on C,
f2(z) in D2 realizes a conformal mapping of the domain D1+ C + D2onto the domain D1∗ + C ∗ + D2∗
An arc C is said to be analytic if it can be described by parametric equations
x = x(t), y = y(t) (α≤t≤β) such that x(t) and y(t) are analytic functions of the real variable t on the interval (α, β).
SCHWARZ’S ANALYTIC CONTINUATION PRINCIPLE Suppose that a function w = f (z) realizes a conformal mapping of a domain D onto a domain D ∗ and takes an analytic arc C
of the boundary of D to an arc C ∗ of the boundary of D ∗ Then the function w = f (z) can
be analytically continued through the arc C.
Trang 5THESCHWARZ–CHRISTOFFEL THEOREM If a function w = f (z) realizes a conformal mapping of the upper half-plane Im z >0onto the interior of a bounded polygon Δ with
angles πα k(0< α k≤ 2, k =1,2, , n) at the vertices and if the points a kof the real axis (–∞ < a1< < a n< ∞) corresponding to the vertices of this polygon are known, then the function f (z) can be represented by the Schwarz–Christoffel integral
f (z) = C
z
z0
(z – a1)α1 – 1(z – a
2)α2 – 1· · · (z – a n)α n– 1dz + C
1,
where z0, C, and C1are some constants
The Schwarz–Christoffel integral is obtained under the assumption that the points a k corresponding to the vertices A kof the polygon are known In practice, only the vertices of
the polygon are given, and the points a k are unknown Determining the points a kis a very difficult task
Table 10.3 presents some conformal mappings of given domains D onto the unit disk.
TABLE 10.3 Conformal mappings onto the unit disk |w| ≤ 1, where z = x + iy = r(cos ϕ + i sin ϕ)
1 Upper half-plane, Im z >0 w = e iβ z – a
z–¯a (β is a real number)
2 Right half-plane, Re z >0 w = e iβ z – a
z–¯a (β is a real number)
3 Disk of radius R, |z|< R w = e iβ R(z – a)
R2–¯az (β is a real number)
4 Strip of width 12π, –14π < Re z < 14π w = tan z
5 |z|Sector of unit disk,<1, 0< arg z < πα w= (1+ z1/α)2– i(1– z1/α)2
( 1+ z1/α) 2+ i(1– z1/α) 2
6 Plane with cut from z =along the positive real axis0to z = ∞ w=
√
z – i
√
z + i
7 Exterior of the ellipse, x
2
a2 +y
2
b2 = 1 z = R
mw+ 1
w
, R= a + b
a – b
a + b
8 Exterior of the parabola, rcos2 ϕ
2
w+ 1
2
9 Interior of the parabola, rcos2 ϕ
π 4
√ z
10 |z|< R,Half-disk,Re z >0 w = i z
2 + 2Rz – R2
z2– 2Rz – R2
Remark.In the items 1, 2, and 3 the point z = a of the domain is taken to the center w =0 of the disk;
β determines the rotation of the disk about the center w =0
10.2.2 Boundary Value Problems
10.2.2-1 Dirichlet problem
Find a function u(z) harmonic in the domain D, continuous in D, and taking given contin-uous values u(ξ) on the boundary of D.
Trang 6Generalized Dirichlet problem Given a function u(ξ) defined on the boundary C of a
domain D and continuous everywhere except for finitely many points ξ1, , ξ n, where it
has jump discontinuities, find a function u(z) harmonic and bounded in D and equal to u(ξ)
at all points of continuity of this function on C.
THEOREM ON THE UNIQUENESS OF A SOLUTION OF THE GENERALIZEDDIRICHLET PROB-LEM. In a given domain for a given boundary function u(ξ), there exists at most one solution
of the generalized Dirichlet problem
THEOREM ON THE EXISTENCE OF A SOLUTION OF THE GENERALIZEDDIRICHLET PROB-LEM. For any simply connected domain D and any piecewise continuous boundary function u(ξ)with jump discontinuities, the generalized Dirichlet problem has a solution
POISSON’S THEOREM The solution of the generalized Dirichlet problem for the unit disk is given by the Poisson integral
u(z) = 1
2π
2π
0 u(e
it) 1– r2
1–2r cos(t – ϕ) + r2 dt (z = re
it (10.2.2.1)
1◦ Let z0be an arbitrary point of a domain D, and let
w = f (z; z0), f (z0; z0) =0 (10.2.2.2)
be a function mapping the domain D onto the unit disk|w|<1 The function
g(z; z0) = ln 1
|f (z; z0)| (10.2.2.3)
is called a Green’s function of the domain D A Green’s function is harmonic everywhere
in D except for the point z0at which it has a pole
The solution of the generalized Dirichlet problem is given by Green’s formula
u(z) = 1
2π
2π
0 u(ξ)gn(ξ, z)ds, (10.2.2.4)
where gnis the inward normal derivative
Green’s formula expresses the solution of the Dirichlet problem for some domain D
in terms of the logarithm of the conformal mapping of the domain D onto the unit disk,
i.e., reduces solving the Dirichlet problem to solving the conformal mapping problem The converse statement is also true If the solution of the Dirichlet problem is known for some
domain D, then a conformal mapping of this domain onto the unit disk can be constructed.
2◦ Suppose that we need to find a function f (z) that is analytic in the disk|z|<1and whose
real part on the circle takes given values u(ξ) at each point of continuity of the function u(ξ) The solution of this problem is given by the Schwarz formula
f (z) = 1
2π
2π
0 u(ξ)
ξ + z
ξ – z dt + iC (ξ = e
it (10.2.2.5)
where C is a real constant The integral on the right-hand side in (10.2.2.5) is called the
Schwarz integral.
Trang 73◦ Suppose that a bounded function u(t) with finitely many points of discontinuity is
given on the real axis and there exist finite limits of u(t) as t → ∞ The solution of the Dirichlet problem for the upper half-plane is given by the following Poisson integral for the
half-plane:
u(z) = 1
π
+∞ –∞ u(t)
y (t – x)2+ y2 dt. (10.2.2.6) Since
y (t – x)2+ y2 = Re
1
i(t – z),
we can also write out the Schwarz integral for the half-plane in the form
f (z) = 1
2π
2π
0 u(t)
dt
t – z + iC, (10.2.2.7)
where C is a real constant.
10.2.2-2 Neumann problem
Find a function u(z) harmonic in the domain D with given normal derivative
un= u x cos α + u y sin α = g(ξ) (10.2.2.8)
on the boundary C and given value u(z0) at a point z0of the domain D.
It is assumed in (10.2.2.8) that the outward normal is considered and α is the angle between this normal and the axis OX The function g(ξ) is allowed to have only finitely many points of jump discontinuity on C; the function u(z) and its first partial derivatives
are assumed to be bounded
Necessary condition for the solvability of Neumann problem:
C g(ξ) ds =0 (10.2.2.9)
If, in addition, we assume that the partial derivatives are continuous in D, then solving
the Neumann problem can be reduced to solving the Dirichlet problem for the conjugate
harmonic function Suppose that v(z) is a harmonic function conjugate to u(z) By the
Cauchy–Riemann conditions written for the curve C in the directions of s and n, we have
vs= un= g(ξ).
If vsalong the curve C is known, then straightforward integration gives
v(ξ) =
ξ
ξ0
vsds=
ξ
ξ0
g(ξ) ds. (10.2.2.10)
Now the problem of determining v(z) in the domain D is reduced to the Dirichlet problem.
If v(z) is known, then the desired function u(z) can also be obtained by integration Now suppose that the domain D is the unit disk If we set f (z) = u+iv, then the function
f (z) satisfies the formula
f(z) = –1
π
2π
0 g(e
it ) ln(e it – z) dt + const. (10.2.2.11) Separating the real part, we obtain the formula for the desired function:
u(z) = –1
π
2π
0 g(e
it) ln|e it – z|dt+ const, (10.2.2.12)
which is called the Dini formula.