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 Mapping1. 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.

yxyxyxu

coshsinsinhcostan1),( 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

azazzf

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(log1),(

zzyxu e

e

• 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)(

zizi

izizzT

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

yxyxyxu

yxyx

zizizT

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

ccx

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

απAxαππαAxx

παπαAxtf

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 20.4 Schwarz-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

ififBzAif(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

zzzAzf

zzAzf

Example 2 (3)

cosh)1()( Therefore

.)1(0,)()1( so and,01cosh and )1(cosh that Notice

12/12

11

zzazf

BfBiAfaii

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)(

tiveantiderivaget 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[),( azbzuyxu i

azbzuzuyxu 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

xzxzuyxu

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

dttzddtuyxu

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 20.5 Poisson 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

ytxtyyxu

Example 1 (2)

yx

yxx

yxyxy

tt

ytxxytxyyxu

txs

e

e

1tan1tan)1()1(log

2

11

tan))((log2

1),(

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(Arg11

)]2(Arg[1)2(Arg1),(

ww

wwvuU

Example 2 (3)

2

11Arg11) ,(

)2121(Arg11)1() ,(

thereforeand

zzyxu

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 20.6 Poisson Integral Formula for the Unit Disk

tdze

zeuyxu itit

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

tu

yu

xua

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

2121) ,(

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 nrbnraaru

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

eunbnaau

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 20.7 Vector 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

zzkqzz

zzkqzg

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 )(),(

0VVV

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

ya

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'

cy

cx

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,yQyxPyxQdxdy

yxQdtdyyxP

dtdx

•

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 ofR. Then V(x, y) = defined an irrotational and incompressible fluid flow in R. Moreover, if a particleis placed is placed inside R, its path z = z(t) remains inR.

THEOREM 20.8 Streamline

)(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 ),(log211

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