¿Cuál es el camino más corto para un cable eléctrico?

Planta eléctrica

ciudad

planta eléctrica

Ejemplo típico que motiva la necesidad de integrales de contorno

Integrales de contorno o de camino

Secciones cónicas

Círculo Elipse Parábola Hipérbola

Ecuación general de una sección cónica:Ax2 + Bxy + Cy2 + Dx + Ey + F = 0

Longitud de una curva

• z’(t) = x’(t) + iy’(t) • Si x’(t) e y’(t) son continuas en el intervalo a <= t <= b entonces: C es un arco diferenciable y su longitud es:

b

a

dttzL |)('|

Si z’(t) no es 0 en ningún punto de a < t < b entonces podemos definir un vector tangente:

|)('|

)('

tz

tzT Y decimos que z(t) es un arco suave.

Contorno = arco suave a trozos

2ty

2txtz )(')(')('

y

x

C: z(t)

Ejemplo: la longitud de camino es independiente de la parametrización

a

b

rx= r cos(t)

y= r sin(t)

0 t π/2

2|)(sin')(cos'|

2/

0

2/

0

2/

0

2/

0

rdtrdterdt

dt

derdttirtrL it

it

1|| ititit eeezzz

x= r cos(2)

y= r sin(2)

0 π

222|)2(sin')2(cos'|

00

2

0

2

0

r

dr

der

dd

derdirrL i

i

Dos parametrizaciones distintas

2

2

and where,

11

aty(t) tx(t) iyxz

xaxy

1

1

21

1

1

1

1

1

)2(1|21||)(')('||)('| dtatdtatidttiytxdttzL

sin Substitute iw

atw 2 Substitute a

dwwa

L2

0

211

dwwI 21Let

didiI 22 cos)cos(sin1

a

-1x

y

2

2sin1cos2

2

2sin

2

iI

)(sin Substitute 1 iw

2

)(1)(2)(sin

2

21 iwiw

iwi

I

1

2412ln4122

1

2

0

21ln12

1

2a

0

2w11

L

2

2

aaaaa

aw

wwwww

a

dwa

21ln)(1-sinh)(1-sin :identity following theUse

wwwiwi

dwwwwwwI 222 111ln2

1

2

)(1)(2)(sin

2

21 iwiw

iwi

I

a

-1x

1222 )5.0(5.0

21 a

La

62.125.10.5)5.0(5.0

48152ln2

15

2

2

2

1

2

1.4121

,1When

22

2

a

.. L

a

a

0.5

axy 2'

Circulation and Net Flux• Let T and N denote the unit tangent vector and the

unit normal vector to a positively oriented simple closed contour C. When we interpret the complex function f(z) as a vector, the line integrals

(6)

(7)

have special interpretations.

C C

dyvxdusdf T

C C

vdyxdusdf N

• The line integral (6) is called the circulation around C and (7) is called the net flux across C. Note that

and socirculation = (8)

net flex = (9)

C C

CC

zdzfdyixdivu

sdfisdf

)())((

NT ．．

C zdzf )(Re

C zdzf )(Im

Given the flow f(z) = (1 + i)z, compute the circulation around and the net flux across the circle C: |z| = 1.

Solution

2 ,2ncirculatio

)1(2)1()1()(

20,)( and )1()( Since2

0

2

0

fluxnet

idtidtieeidzzf

tetzzizf

itit

C

it

The complex function where k = a + ib and z1 are complex numbers, gives rise to a flow in the domain z z1. If C is a simple closed contour containing z = z1 in its interior, then we have

The circulation around C is 2b and the net flux across C is 2a. If z1 were in the exterior of C both of them would be zero.

)/()( 1zzkzf

)(2)(1

ibaizdzziba

zdzfC C

Note that when k is real, the circulation around C is zero but the net flux across C is 2k. The complex number z1 is called a source when k > 0 and is a sink when k < 0.

Evaluate

where C is the circle |z| = 1.

SolutionThis integrand is not analytic at z = 0, −4 but only z = 0 lies within C. Since

We get z0 = 0, n = 2, f(z) = (z + 1)/(z + 4), f (z) = −6/(z + 4)3. By (6):

33441

4

1

zzz

zz

z

zdzz

zC

34 4

1

ifi

zdzz

zC 32

3)0(

!22

4

134

Evaluate zdizz

zC

2

3

)(

3

SolutionThough C is not simple, we can think of it is as the union of two simple closed contours C1 and C2 in Fig 18.27.

21

2

3

2

3

22

3

2

3

21

21

)(

3)(

3

)(

3

)(

3

)(

3

II

zdizz

zz

zdz

izz

z

zdizz

zzd

izz

zzd

izz

z

CC

CCC

For I1 : z0 = 0 , f(z) = (z3 + 3)/(z – i)2 :

For I2 : z0 = i, n = 1, f(z) = (z3 + 3)/z, f ’(z) = (2z3 –3 )/z2:

We get

ifizdz

izz

z

IC

6)0(2)(

3

1

2

3

1

)32(2)23(2)(!1

2

)(

3

22

3

2 iiiifi

zdiz

zz

IC

)31(4)32(26)(

3212

3

iiiIIzdizz

zC