Transcript 3.5 Conformal Mapping - Helmholtz-Zentrum Dresden

3.5 Two dimensional problems
• Cylindrical symmetry
• Conformal mapping
Laplace operator in polar
coordinates
1  V
1  2V
(r
) 2
0
2
r r r
r 
V  R( r )( )
( )  Cn cosn  Dn sin n
R(r)  An r n  Bn r n
n  0,1,2,....
n  1,2,... R(r)  A  B ln(r)
V ( r, )  A  B ln(r )

  ( An r n  Bn r n )(Cn cosn  Dn sin n )
n 1
Example:
Two half pipes
Conformal Mapping
Is there a simple
solution?
For two-dimensional problems complex analytical function
are a powerful tool of much elegance.
z  x  iy  ei
iy
F ( z )  g ( x, y )  ih( x, y )
w  u  iv  F ( z )
Maps (x,y) plane onto (u,v) plane.
For analytical functions the derivative exists.
Examples:
z n , z1/ n , sin z, ez , ln z
x
Analytical functions obey the Cauchy-Riemann equations
g h
g
h

and
 ,
x y
y
x
which imply that g and h obey the Laplace equation,
2 g 2 g
 2h  2h
 2  0 and
 2  0.
2
2
x
y
x
y
If g(x,y) fulfills the boundary condition it is the potential.
If h(x,y) fulfills the boundary condition it is the potential.
g and h are conjugate. If g=V then g=const gives the equipotentials
and h=const gives the field lines, or vice versa.
If F(z) is analytical it defines a conformal mapping.
A conformal transformation maps a rectangular grid onto a curved
grid, where the coordinate lines remain perpendicular.
Example
Cartesian onto polar coordinates:
w  ez , z  x  iy, w  e xeiy .
Polar onto Cartesian coordinates:
w  ln z, z  ei , ln z  ln   i .
z
2 i
w
Full
plane
A corner of conductors
F ( z)  Az2  iVo
potential: V ( x, y)  2 Axy  Vo
field lines: g ( x, y)  A( x 2  y 2 )  const
chargedensityon thehorizontalplate: ( x)  2 A o x
Edge of a conducting plane
F ( z)  Az1/ 2  iVo
A
V ( x, y ) 
( x 2  y 2  x)1/ 2  Vo
2
A
g ( x, y ) 
( x 2  y 2  x)1/ 2
2
equipotentials
field lines
Parallel Plate Capacitor
Vo
F ( z )  ln w
i
d
1
z  [ln w  (1  w2 )]

2
map w  F
map w  z
w  ei , 0     , 0    

Vo
V  Vo , h   ln 


d
1
x  [ln   (1   2 cos2 )]

2
d
1 2
y  [ln    sin 2 ]

2