sea f una función analítica que transforma un dominio d en un dominio d. si u es armonica en d,...
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Sea f una función analítica que transforma un dominio
D en un dominio D. Si U es armonica en D, entonces
la función real u(x, y) = U(f(z)) es armonica en D.
Teorema de transformación de funciones armónicas
Proof We will give a special proof for the special case in which D is simply connected. If U has a harmonic conjugate V in D, then H = U + iV is analytic in D, and so the composite function H(f(z)) = U(f(z)) + iV(f(z)) is analytic in D. It follow that the real part U(f(z)) is harmonic in D.
Solving Dirichlet Problems Using Conformal Mapping
1. Find a conformal mapping w = f(z) that transform s the original region R onto the image R. The region R may be a region for which many explicit solutions to Dirichlet problems are known.
2. Transfer the boundary conditions from the R to the boundary conditions of R. The value of u at a boundary point of R is assigned as the value of U at the corresponding boundary point f().
3. Solve the Dirichlet problem in R. The solution may be apparent from the simplicity of the problem in R or may be found using Fourier or integral transform methods.
4. The solution to the original Dirichlet problems is u(x, y) = U(f(z)).
Ejemplo: The function U(u, v) = (1/) Arg w is harmonic in the upper half-plane v > 0 since it is the imaginary part of the analytic function g(w) = (1/) Ln w. Use this function to solve the Dirichlet problem en la figura superior (a).
Solution The analytic function f(z) = sin z maps the original region to the upper half-plane v 0 and maps the boundary segments to the segments shown in Fig 20.14(b). The harmonic function U(u, v) = (1/) Arg w satisfies the transferred boundary conditions U(u, 0) = 0 for u > 0 and U(u, 0) = 1 for u < 0.
yxyx
yxucoshsinsinhcos
tan1
),( 1
Example 7
From C-1 in Appendix IV, the analytic function
maps the region outside the two open disks |z| < 1 and |z – 5/2| < ½ onto the circular region r0 |w| 1,
. See Fig 20.15(a) and (b).
5627
,1
)(
a
azaz
zf
625 where 0 r
Fig 20.15
Example 7 (2)
In problem 10 of Exercise 14.1, we found
is the solution to the new problem. From Theorem 20.2 we conclude that the solution to the original problem is
))/(log)(log) ,( 0 zrrrU ee
15/)627(5/)627(
log)625(log
1),(
zz
yxu ee
• A favorite image region R for a simply connected region R is the upper half-plane y 0. For any real number a, the complex function Ln(z – a) = loge|z – a| + i Arg (z – a)is analytic in R and is a solution to the Dirichlet problem shown in Fig 20.16.
Fig 20.16
• It follows that the solution in R to the Dirichlet problem with
is the harmonic function U(x, y) = (c0/)(Arg(z – b) – Arg(z –
a))
otherwise
bxacxU
,0
,)0 ,( 0
Fig 20.35(a)
Example 6
Solve the Dirichlet problem in Fig 20.35(a) using conformal mapping by constructing a linear fractional transformation that maps the given region into the upper half-plane.
Example 6 (2)
Solution The boundary circles |z| = 1 and |z – ½| = ½ each pass through z = 1. We can map each boundary circle to a line by selecting a linear fractional transformation that has a pole at z = 1. If we require T(i) = 0 and T(-1) = 1, then
Since , T maps the interior of |z| = 1 onto the upper half-plane and maps |z – ½| = ½ onto the line v = 1. See Fig 20.35(b).
1)1(
111
1)(
ziz
iiz
izzT
iiTiT 1)( ,1)0( 21
21
Example 6 (3)
The harmonic function 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.
22
22
22
22
)1(
1),( issolution the,
)1(
1
is 1
)1()( ofpart imaginary theSince
yx
yxyxu
yx
yx
ziz
izT
Example 6 (4)
The level curves u(x, y) = c can be written as
and are therefore circles that pass through z = 1. See Fig 20.36.
22
2
11
1
cy
cc
x
Fig 20.36
20.4 Schwarz-Christoffel Transformations
• Special Cases First we examine the effect of f(z) = (z – x1)/, 0 < < 2, on the upper half-plane y 0 shown in Fig 20.40(a).
• The mapping is the composite of = z – x1 and w = /. Since w = / changes the angle in a wedge by a factor of /, the interior angle in the image is (/) = . See Fig 20.40(b).
• Next assume that f(z) is a function that is analytic in the upper half-plane and that jas the derivative
(1)where x1 < x2. We use the fact that a curve w = w(t) is a line segment when the argument of its tangent vector w(t) is constant. From (1) we get
(2)
/,)( that Note 1)/(1 AxzAf'(z)
1)/)(()()( 22
1)/(1
1 xzxzAzf
)(Arg1)(Arg1Arg
)(arg
22
11 xtxtA
tf
• Since Arg(t – x) = for t < x, we can find the variation of f (t) along the x-axis. They are shown in the following table.
See Fig 20.41.
22
1221
211
Arg ) ,(
)( Arg ) ,(
0 )( )( Arg ) ,(
arg.in Change )(' arg Interval
απAx
αππαAxx
παπαAx
tf
Fig 20.41
Let f(z) be a function that is analytic in the upper half-plane y > 0 and that has the derivative
(3)
where x1 < x2 < … < xn and each i satisfies 0 < i < 2. Then f(z) maps the upper half-plane y 0to a polygonal region with interior angles 1, 2, …, n.
THEOREM 20.4Schwarz-Christoffel Formula
1)/(1)/(2
1)/(1 )()()(
)(
21
n
nxzxzxzA
zf
Comments(i) One can select the location of three of the
points xk on the x-axis.
(ii) A general form for f(z) is
(iii) If the polygonal region is bounded only n – 1 of the n interior angles should be included in the Schwarz-Christoffel formula.
. and
)...()()(
of compisite theas considered becan and
)...()()(
1)/(1)/(1
1)/(1)/(1
1
1
BAzw
dzxzxzzg
BdzxzxzAzf
n
n
n
n
Example 1
Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the strip |v| 1, u 0. See Fig 20.42.
Example 1 (2)
Solution We may select x1 = −1, x2 = 1 on the x-axis, and we will construct a mapping f with f(−1) = −i, f(1) = i. Since 1 = 2 = /2, (3) gives
.)1(
1
)1(
1
)1()1()('
2/122/12
2/12/1
ziA
zA
zzAzf
Example 1 (3)
.sin2
)(then
/2,0imply 2
,2
then,)1(,)1( Since
.sin Thus
1
1
zizf
ABBAiiBAii
ifif
BzAif(z)
Example 2
• Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the region shown in Fig 20.43(b).
Example 2 (2)
Solution We may select x1 = −1, x2 = 1 on the x-axis, and we will require f(−1) = ai, f(1) = 0. Since 1 = 3/2 2 = /2, (3) gives
cosh)1()(
then,)1(
1
)1()('
write weIf .)1()1()('
12/12
2/122/12
2/12/1
BzzAzf
zz
zAzf
zzAzf
Example 2 (3)
cosh)1()( Therefore
.)1(0,)()1( so and
,01cosh and )1(cosh that Notice
12/12
11
zza
zf
BfBiAfai
i
Example 3
• Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the region shown in Fig 20.44(b).
Example 3 (2)
Solution Since the region is bounded, only two of the 60 interior angles should be included. If x1 = 0, x2 = 1, we obtain
.)1(
1)(
tiveantideriva
get the to18.8 Theorem usecan We
)1()('
0 3/23/2
3/23/2
Bdsss
Azf
zAzzf
z
Example 3 (3)
zds
sszf
dxxx
A
Bff
0 3/23/2
1
0 3/23/2
)1(
1)3/1(
1)( Thus
)3/1()1(
11
and 0 then ,1)1( ,0)0( require We
Example 4
Use the Schwarz-Christoffel formula to construct a conformal mapping from the upper half-plane to the upper half-plane with the horizontal line v = , u 0, deleted. See Fig 20.45.
Example 4 (2)
Solution The nonpolygonal target region can be approximated by a polygonal region by adjoining a line segment from w = i to a pint u0 on the negative u-axis. See Fig 20.45(b). If we require f(−1) = i, f(0) = u0, then
Note that as u0 approaches −, 1 and 2 approach 2 and 0, respectively.
)1()(' 1)/(1)/( 21 zzAzf
Example 4 (3)
This suggests we examine the mappings that satisfy w = A(z + 1)1z-1 = A(1 + 1/z) or w = A(z + Ln z) + B.First we determine the image of the upper half-plane under g(z) = z + Ln z and then translate the image region if needed. For t real
g(t) = t + loge |t| + i Arg tIf t < 0, Arg t = and u(t) = t + loge |t| varies from − to −1. It follows that w = g(t) moves along the line v = from − to −1.
Example 4 (4)
When t > 0, Arg t = 0 and u(t) varies from − to . Therefore g maps the positive 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 = , u −1, deleted. Therefore w = z + Ln z + 1 maps the upper half-plane onto the original target region.
20.5 Poisson Integral Formulas
• Introduction It would be helpful if a general solution could be found for Dirichlet problem in either the upper half-plane y 0 or the unit disk |z| = 1. The Poisson formula fro the upper half-plane provides such a solution expressing the value of a harmonic function u(x, y) at a point in the interior of the upper half-plane in terms of its values on the boundary y = 0.
Formula for the Upper Half-Plane
• Assume that the boundary function is given by u(x, 0) = f(x), where f(x) is the step function indicated in Fig 20.55.
• The solution of the corresponding Dirichlet problem in the upper half-plane is
(1)
Since Arg(z – b) is an exterior angle formed by z, a and b, Arg(z – b) = (z) + Arg(z – a), where 0 < < , and we can write
(2)
)](Arg)(Arg[),( azbzu
yxu i
azbzu
zu
yxu ii Arg)(),(
• The superposition principle can be used to solve the more general Dirichlet problem in Fig 20.56.
• If u(x, 0) = ui for xi-1 x xi, and u(x, 0) = 0 outside the interval [a, b], then from (1)
(3)
Note that Arg(z – t) = tan-1(y/(x – t)), where tan-1 is selected between 0 and , and therefore (d/dt) Arg(z – t) = y/((x – t)2 + y2).
n
iii
n
iii
i
zu
xzxzu
yxu
1
11
)(1
)](Arg)(Arg[),(
(4) )(
)0,(),( have we
],[ interval theoutside 0)0 ,( Since
)(
1
)(Arg1
),(
(3), From
22
122
1
1
1
dtytx
tuyyxu
a,bxu
dtytx
yu
dttzddt
uyxu
n
i
x
xi
n
i
x
x i
i
i
i
i
Let u(x, 0) be a piecewise-continuous function on everyfinite interval and bounded on - < x < . Then the
function defined by
is the solution of the corresponding Dirichlet problem
on the upper half-plane y 0.
THEOREM 20.5Poisson Integral Formula for the Upper Half-Plane
tdytx
tuyyxu
22)(
)0,(),(
Example 1
Find the solution of the Dirichlet problem in the upper half-plane that satisfies the boundary condition u(x, 0) = x, where |x| < 1, and u(x, 0) = 0 otherwise.
Solution By the Poisson integral formula
.)(
),(1
1 22 dt
ytx
tyyxu
Example 1 (2)
yx
yxx
yx
yxy
t
t
ytx
xytxy
yxu
txs
e
e
1tan
1tan
)1(
)1(log
2
1
1tan))((log
21
),(
then, Using
1122
22
122
Example 2
The conformal mapping f(z) = z + 1/z maps the region in the upper half-plane and outside the circle |z| = 1 onto the upper half-plane v 0. Use the mapping and the Poisson integral formula to solve the Dirichlet problem shown in Fig 20.57(a).
Fig 20.57
Example 2 (2)
Solution Using the result of Example 4 of Sec 20.2, we can transfer the boundary conditions to the w-plane. See Fig 20.57(b). Since U(u, 0) is a step function, we will use the integrated solution (3) rather than the Poisson integral. The solution to the new Dirichlet problem is
)22
(Arg1
1
)]2(Arg[1
)2(Arg1
),(
ww
wwvuU
Example 2 (3)
2
11
Arg1
1) ,(
)2121
(Arg1
1)1
() ,(
thereforeand
zz
yxu
zzzz
zzUyxu
Let u(ei) be a bounded and piecewise continuous for- . Then the solution to the corresponding Dirichlet Problem on the open units disk |z| < 1 is given by
(5)
THEOREM 20.6Poisson Integral Formula for the Unit Disk
tdze
zeuyxu it
it
2
2
||
||1)(
21
),(
Geometric Interpretation
• Fig 20.58 shows a thin membrane (as a soap film) that has been stretched across a frame defined by u = u(ei).
Fig 20.58
• The displacement u in the direction perpendicular to the z-plane satisfies the two-dimensional wave equation
and so at equilibrium, the displacement function u = u(x, y) is harmonic. Formula (5) provides an explicit solution for u and has the advantage that the integral is over the finite interval [−, ].
2
2
2
2
2
22
t
u
y
u
x
ua
Example 3
A frame for a membrane is defined by u(ei) = | | for − . Estimate the equilibrium displacement of the membrane at (−0.5, 0), (0, 0) and (0.5, 0).
SolutionUsing (5),
dt
ze
ztyxu
it 2
21
21
) ,(
Example 3 (2)
• When (x, y) = (0, 0), we get
For the other two values of (x, y), the integral is not elementary and must be estimated using a numerical solver. We have u(−0.5, 0) = 2.2269, u(0.5, 0) = 0.9147.
221
) ,(
dttyxu
Fourier Series Form
• Note that un(r,) = rn cos n and vn(r,) = rn sin n are each harmonic, since these functions are the real and imaginary parts of zn. If a0, an, bn are chosen to be the Fourier coefficients of u(ei) for − < < , then
(6) sincos2
) ,(1
0
n
nn
nn nrbnra
aru
1
0
2
2
1
0
sincos2
1)(
21
),(
have we(5) From
)(sincos2
),1(
and harmonic is (6) find We
n
nn
nn
iit
it
i
nnn
nrbnraa
dtree
reuru
eunbnaa
u
Example 4
Find the solution of the Dirichlet problem in the unit disk satisfying the boundary condition u(ei) = sin 4 . Sketch the level curve u = 0.
SolutionRather than using (5), we use (6) which reduces to u(r, ) = r4 sin 4 . Therefore u = 0 if and only if sin 4 = 0. This implies u = 0 on the lines x = 0, y = 0 and y = x.
Example 4 (2)
If we switch to rectangular coordinates, u(x, y) = 4xy(x2 – y2). The surface of u(x, y) = 4xy(x2 – y2), the frame u(ei) = sin 4, and the system of level curves were sketched in Fig 20.59.
Fig 20.59
20.6 Applications
• Vector Fields A vector field F(x, y) = P(x, y)i + Q(x, y)j in a domain D can also be expressed in the complex form
F(x, y) = P(x, y) + iQ(x, y)
Recall that div F = P/x +Q/y and curl F = (Q/x −P/y)k. If we require both of them are zeros, then
(1) and yQ
xP
xQ
yP
(i) Suppose that F(x, y) = P(x, y) + Q(x, y) is a vectorfield in a domain D and P(x, y) and Q(x, y) are continuous and have continuous first partial derivatives in D. If div F = 0 and curl F = 0, then complex function
is analytic in D.
(ii) Conversely, if g(z) is analytic in D, then F(x, y) = defined a vector field in D for which div F = 0and curl F = 0.
THEOREM 20.7Vector Fields and Analyticity
),(),()( yxiQyxPzg
)(zg
THEOREM 20.7
Proof If u(x, y) and v(x, y) denote the real and imaginary parts of g(z), then u = P and v = −Q. Then
Equations in (2) are the Cauchy-Riemann equations for analyticity.
(2) ,
is,that ;)(
,)(
xv
yu
yv
xu
xv
yu
yv
xu
Example 1
The vector field F(x, y) = (−kq/|z − z0|2)(z − z0) may be interpreted as the electric field by a wire that is perpendicular to the z-plane at z = z0 and carries a charge of q coulombs per unit length. The corresponding complex function is
. curl0, div,for analytic is )( Since
)()(
0
002
0
0FF
zzzg
zzkq
zzzz
kqzg
Example 2
The complex function g(z) = Az, A > 0, is analytic in the first quadrant and therefore gives rise to the vector field
. curl,0 div satisfieswhich
)(),(
0VV
V
iAyAxzgyx
Potential Functions
• Suppose that F(x, y) is a vector field in a simply connected domain D with div F = 0 and curl F = 0. By Theorem 18.8, the analytic function g(z) = P(x, y) − iQ(x, y) has an antiderivative
(4)
in D, which is called a complex potential for the vector filed F.
),(),()( yxiyxzG
•
Therefore F = , and the harmonic function is called a (real) potential function of F.
(5) , so and
),(),(
),(),()(')( that Note
Qy
Px
yxy
iyxx
yxx
iyxx
zGzg
Example 3
• The potential in the half-plane x 0 satisfies the boundary conditions (0, y) = 0 and (x, 0) = 1 for x 1. See Fig. 20.68(a). Determine a complex potential, the equipotential lines, and the field F.
Fig. 20.68
Example 3 (2)
Solution We knew the analytic function z = sin w maps the strip 0 u /2 in the w-plane to the region R in question. Therefore f(z) = sin-1z maps R onto the strip, and Fig 20.68(b) shows the transferred boundary conditions. The simplified Dirichlet problem has the solution U(u, v) = (2/)u, and so (x, y) = U(sin-1z) = Re((2/) sin-
1z) is the potential function on D, and G(z) = (2/)u sin-1z is a complex potential for F.
Example 3 (3)
Note that the equipotential lines = c are the images of the equipotential lines U = c in the w-plane under the inverse mapping z = sin w. We found that the vertical lines u = a is mapped onto a branch of the hyperbola1
cossin 2
2
2
2
a
y
a
x
Example 3 (4)
Since the equipotential lines U = c, 0 < c < 1 is the vertical line u = /2c, it follows that the equipotential lines = c is the right branch of the hyperbola
2/122/12
2/121
2
2
2
2
)1(
12
)1(
12then
,)1/(1sin)/( and )( Since
1)2/(cos)2/(sin
zz
zzdzdzG'
c
y
c
x
F
F
Steady-State Fluid Flow
• The vector V(x, y) = P(x, y) + iQ(x, y) may also be expressed as the velocity vector of a two-dimensional steady-state fluid flow at a point (x, y) in a domain D.If div V = 0 and curl V = 0, V has a complex velocity potential
G(z) = (x, y) + (x, y)that satisfies
V)(zG'
• Here special importance is placed on the level curves (x, y) = c. If z(t) = x(t) + iy(t) is the path of a particle, then
.0),()(
or ),(/),(/ Hence
(6) ),(),,(
dyyxPdxx,yQ
yxPyxQdxdy
yxQdtdy
yxPdtdx
•
The function (x, y) is called a stream function and the level curves (x, y) = c are streamlines for the flow.
.),(satisfy (6) of solutions all and
and
equationsRiemann -Cauchy by the and
)(
implies 0div Since
cyxψ
Pxy
Qyx
xP
yQ
V
Example 4
• The uniform flow in the upper half-plane is defined by V(x, y) = A(1, 0), where A is a fixed positive constant. Note that |V| = A, and so a particle in the fluid moves at a constant speed. A complex potential for the vector field is G(z) = Az = Ax + iAy, and so the streamlines are the horizontal lines Ay = c. See Fig 20.69(a). Note that the boundary y = 0 of the region is itself streamline.
Fig 20.69(a)
Example 5
• The analytic function G(z) = z2 gives rise to the vector field in the first quadrant. Since z2 = x2 − y2 + i(2xy), the stream function is (x, y) = 2xy and the streamlines are the hyperbolas 2xy = c. See Fig 20.69(b).
)2,2()(),( yxzG'yx V
Fig 20.69(b)
Suppose that G(x) = (x, y) + i(x, y) is analytic in a region R and (x, y) is continuous on the boundary of
R. Then V(x, y) = defined an irrotational and incompressible fluid flow in R. Moreover, if a particle
is placed is placed inside R, its path z = z(t) remains in
R.
THEOREM 20.8Streamline
)(zG
Example 6
The analytic function G(z) = z + 1/z maps the region R in the upper half-plane and outside the circle |z| = 1onto the upper half-plane v 0. The boundary of R is mapped onto the u-axis, and so v = (x, y) = y – y/(x2 + y2) is zero on the boundary of R. Fig 20.70 shows the streamlines. The velocity field is given by
ii er
reG'
zzG'
22
2
11)(
so and ,/11)(
Example 6 (2)
It follows that V (1, 0) for large values of r, and so the flow is approximately uniform at large distance from the circle |z| = 1. The resulting flow in R is called flow around a cylinder.
Fig 20.70
Example 7
The analytic function f(w) = w + Ln w + 1 maps the upper half-plane v 0 to the upper half-plane y 0, with the horizontal line y = , x 0, deleted. See Example 4 in Sec 20.4. If G(z) = f -1(z) = (x, y) + i(x, y), then G(z) maps R onto the upper half-plane and maps the boundary of R onto the u-axis. Therefore (x, y) = 0 on the boundary of R.
Example 7 (2)
It is not easy to find an explicit formula for (x, y). The streamlines are the images of the horizontal lines v = c under z = f(w). If we write w = t + ic, c > 0, then the streamlines can be
See Fig 20.71.
)(Arg ),(log21
1
is, that ,1)Ln()(
22 ictcycttx
ictictictfz
e
Fig 20.71
Example 8
The analytic function f(w) = w + ew + 1 maps the strip 0 v onto the region R shown in Fig 20.71. Therefore G(z) = f -1(z) = (x, y) + i(x, y) maps R back to the strip and from M-1 in the Appendix IV, maps the boundary line y = 0 onto the u-axis and maps the boundary line y = onto the horizontal line v = . Therefore (x, y) is constant on the boundary of R.
Example 8 (2)
The streamlines are the images of the horizontal lines v = c, 0 < c < , under z = f(w). If we write w = t + ic, then the streamlines can be
See Fig 20.72.
cecycetx
ictictfztt
ict
sin,cos1
is,that ,1e)( )(
Fig 20.72