Lecture 4: Boundary Value Problems

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Lecture 4: Boundary Value Problems
Instructor:
Dr. Gleb V. Tcheslavski
Contact:
[email protected]
Office Hours:
Room 2030
Class web site:
www.ee.lamar.edu/gleb
/em/Index.htm
ELEN 3371 Electromagnetics
Fall 2008
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What will we learn
So far, we considered fields in an infinite space. In practice,
however, we often encounter situations when fields live in a
finite space consisting of bounded regions with different
electromagnetic properties.
We have learned that an electrostatic field could be created
from a charge distribution. The electric potential can be
obtained in terms of charge distributions via Poisson’s
equation.
Now, we will examine how to solve such equations in general.
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Fall 2008
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Boundary conditions
We restrict our
discussion to a 2D
case and Cartesian
coordinates
With respect to the
interface between two
boundaries, an EM
field can be
separated into a
parallel (tangential)
and a perpendicular
(normal) components
ELEN 3371 Electromagnetics
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Boundary conditions: normal
components
Normal components for
the displacement flux
density D and the
magnetic flux density B
“Pillbox”
Thickness:
z 0
Cross-section area s
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Boundary conditions: normal
components (cont)
We assume that there is a charge
distributed along both sides of the interface
and has a total charge density of s.
Electric field from the Gauss’s law:
 D ds  Q
enc
  s s
(4.5.1)
D2n  D1n  s  r 2 0 E2n   r1 0 E1n  s
Magnetic field:
 B ds  0
(4.5.3)
B2n  B1n  0r 2 0 H2n  r10 H1n  0
ELEN 3371 Electromagnetics
(4.5.2)
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(4.5.4)
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Boundary conditions: tangential
components
Electric field: the total
work is
 E dl  0
(4.6.1)
Assume that z portions can be
neglected; two other edges:
 E2t  E1t 
x0
 E2t  E1t   0
ELEN 3371 Electromagnetics
D2t
r2

D1t
(4.6.2)
 r1
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Boundary conditions: tangential
components (cont)
For the magnetic field, Ampere’s law:
H
dl  I enc  J s x
(4.7.1)
surface current density
Here, we again neglected integration
over the top and bottom edges.
 H2t  H1t  x  J s x
H 2t  H1t  J s 
ELEN 3371 Electromagnetics
B2t
r 2
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
B1t
r1
 0 J s
(4.6.5)
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Boundary conditions: Example
The surface current with a density
Js = 20uy A/m is flowing along the
interface between two homogeneous,
linear, isotropic materials with r1 = 2
and r2 = 5.
H1 = 15ux + 10uy + 25uz A/m. Find H2.
1. Normal component:
r 2 0 H 2 n  r10 H1n H 2 x 
2. Tangential component:
r1
2
H1x  15  6 A / m
r 2
5
H2t  H1t  J s H2 z  H1z  J y  25  20  45 A / m
3. There is no change in the y-component of the magnetic field. WHY?
H2  6ux 10uy  45uz  A / m
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Boundary with ideal conductor
Since the tangential electric field
must be continuous, and
accounting for the Ohm’s law, it
must be a tangential current
approaching infinity!
Therefore, the tangential current
density and tangential component
of E must be zero at the interface
with a perfect conductor.
Ideal conductors are equipotential.
The consequence: if we place a point charge above an ideal conductor, it will
create an electric field that would be entirely in a radial direction. Therefore, the
tangential component of E will be zero just beneath the charge. In order to satisfy
the “zero tangential component requirement” at the other points of the surface, we
assume that so called “image charge” exists inside the conductor.
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Poisson’s and Laplace’s equations
The Gauss’s law in differential form:
 E
v
0
(4.10.1)
Electrostatic field is conservative:
 E  0
(4.10.2)
Therefore, E is a gradient of electric potential:
E  V
(4.10.3)
Combining (4.10.1) and (4.10.3), the Poisson’s equation:
v
V 
0
(4.10.4)
2V  0
(4.10.5)
If the charge density in the region is zero, the
Laplace’s equation:
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Poisson’s and Laplace’s equations
From non-existence of magnetic monopole:
 B0
(4.11.1)
A vector magnetic potential such that:
B   A
(4.11.2)
The Ampere’s law in differential form:
 B  0 J
(4.11.3)
Therefore:
 A  0 J
(4.11.4)
Via the vector identity, the Coulomb’s gauge is:
 A0
(4.11.5)
Similarly to the Poisson’s equation:
2 A  0 J
(4.11.6)
In the CCS:
ELEN 3371 Electromagnetics
2 Ax  0 J x ;2 Ay  0 J y ;2 Az  0 J z (4.11.7)
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Poisson’s and Laplace’s equations
Recall that the Laplacian operator in different CSs:
1. Cartesian:
2
2
2

V

V

V
2V  2  2  2
x
y
z
(4.12.1)
2. Polar:
1   V
V 

   
 1  2V  2V
 2
 2
2
z
  
(4.12.2)
3. Spherical:
2
1


V
1


V
1

V




2V  2  r 2

sin





r r  r  r 2 sin   
  r 2 sin 2   2
(4.12.3)
ELEN 3371 Electromagnetics
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Poisson’s equation: Example
Show that the 2D potential distribution
Satisfies the Poisson’s equation
v 2
V 
x  y2 

4 0
Let us evaluate the Laplacian operator in the Cartesian CS:
v
v
v  2V
 2V
 2V
 2

; 2  
; 2  0
2
x
4 0
2 0
y
2 0 z
Therefore:
2V  
v


 v 0  v
2 0
2 0
0
At this point, we derived the Poisson’s and Laplace’s equations in 3D. Next, we
will attempt to solve them to find a potential.
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Fall 2008
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Analytical solution in 1D – Direct integration
Calculate the potential variation
between two infinite parallel
metal plates in a vacuum.
x=0
x = x0
We assume no resistance,
therefore, no variation in y and z
directions.
V V

0
y z
(4.14.1)
Since no charges exist between plates, we need to solve the Laplace’s equation:
d 2V
0
2
dx
Boundary conditions: V = V0 at x = 0; V = 0 at x = x0
ELEN 3371 Electromagnetics
Fall 2008
(4.14.2)
(4.14.3)
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Analytical solution in 1D – Direct integration
The solution of the Laplace’s equation will be in the form:
V C1 x  C2
(4.15.1)
Here C1 and C2 are integration constant that can be found from boundary cond.
V0  C1  0  C2
C2  V0


0  C1  x0  C2
C1   V0 x0
Finally, the solution:
The electric field is:
ELEN 3371 Electromagnetics

x
V  V0  1  
 x0 
V0
dV
E
ux  ux
dx
x0
Fall 2008
(4.15.2)
(4.15.3)
(4.15.4)
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Analytical solution in 1D – Direct integration
Let’s assume that there is a charge with v uniformly distributed between plates
2
v
d
V
Poisson’s equation:

2
dx
0
v x 2
 C3 x  C4
The solution will be in form: V  
0 2
from the boundary
conditions:
Finally:

v 02
C4  V0
V0    2  C3  0  C4


0
2




x
1


v
0
2
C


V


x
 0

0   v 0  C x  C
 3
x

2
3 0
4
0
0




0 2


v
x
V  V0 1 
x0 x 1  
 2 0V0
 x0 
ELEN 3371 Electromagnetics
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(4.16.1)
(4.16.2)
(4.16.3)
(4.16.4)
V0  v x0 2 
x 
 1 
1  2   ux
x0  2 0V0 
x0  
(4.16.5)
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Analytical solution in 1D – Direct integration
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Fall 2008
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Conclusions
Steps for solving either Poisson’s or Laplace’s equation analytically:
1. Chose the most appropriate representation of
Laplacian based on any symmetry.
2. Perform the integration of the differential equation to
obtain the most general solution for the potential.
3. Let this general solution to satisfy the boundary
conditions to find constants of integration.
Additionally, numerical methods, such as Finite Difference Method, Finite
Element Method, Method of Moments, are developed to solve Boundary
Value problems in situations where analytical solution is hard to find.
ELEN 3371 Electromagnetics
Fall 2008

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