Chapter 2
Boundary-Value Problems: Part I
Problem Set #2: 2.2, 2.11, 2.15 (Due Monday Feb. 18th)
2.1
Image charges
There are four major analytical methods for solving boundary values problems:
(1) Image charges
(2) Green functions,
(3) Expansion in orthogonal functions,
(4) Conformal mappings.
In this course we will only deal with the first three methods.
The method of images is used when there is a boundary condition to the
volume of interest (e.g. conductor held at a given potential). The key idea
is to place the so-called image charges in the region outside of the volume
of interest to match the overall potential at the boundary. This way one
enlarges the considered region, but does not change the solution inside of the
original region of interest. The potentials resulting from the image charges
are the solutions of the Laplace equation, when the potentials from other
charges are the solutions of the Poisson equation. For example one can use
the method of images to solve for a point charge in the presence of an infinite
conducting plane at zero potential (i.e. grounded).
Another less trivial example is a point charge in the presence of a grounded
(i.e. Φ = 0) conducting sphere. Let the sphere be of radius a with center in
origin and the real charge q be placed at y = yŷ with an observation point
at x = xx̂. Now instead of solving this problem let us solve another problem
which is equivalent to the original in the region outside of sphere. Instead of
the conducting sphere we place an image charge q ! at y! = y ! ŷ where q ! and
12
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
13
y ! are so far a free parameter with y ! < a. To fix it we solve for the correct
boundary conditions, i.e.
q
1
q
1
+
=
4π"0 |x − y| 4π"0 |x − y! |
1
q
1
q!
=
+
4π"0 |ax̂ − yŷ| 4π"0 |ax̂ − y ! ŷ|
0 = Φ(x = ax̂) =
or
(2.1)
q
q
=−
|ax̂ − yŷ|
|ax̂ − y !ŷ|
(2.2)
q
q!
= !
.
y−a
y −a
(2.3)
for any direction of x̂. If x̂ = ŷ, then
and if x̂ · ŷ = 0, then
q
or
!
a2 − y 2 − 2ayx̂ · ŷ
= −!
q!
a2 − y !2 − 2ay !x̂ · ŷ
q !2
q2
=
.
a2 + y 2
a2 + y !2
(2.4)
(2.5)
As a result we obtain two equations (2.3) and (2.5) with two unknowns which
can be solved for. From (2.3)
q! =
q(y ! − a)
y−a
(2.6)
and (2.5) can be rewritten as
"
q(y ! −a)
y−a
#2
q
= 2
a2 + y 2
a + y !2
or
!2
y −
$
a2 + y 2
y
%
y ! + a2 = 0.
Solutions of the quadratic equations are given by
&"
#2
' 2
a2 +y 2
a2 +y 2
±
− 4a2
a
y
y
inside sphere
!
y =
= y
2
y outside sphere.
(2.7)
(2.8)
(2.9)
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
14
Of course, the relevant solution is the one corresponding to a charge
#
" 2
q ay − a
qa
=−
(2.10)
q! =
y−a
y
inside sphere at
y! =
a2
.
y
(2.11)
Then the solution for the potential is
)
(
1
1
q
.
−
Φ(x) =
4π"0 |x − y| | ya x − ya y|
(2.12)
Now that we know the potential everywhere outside of the sphere (2.12)
and the potential inside of a grounded sphere (Φ = 0) we can find the surface
charge density in the original problem. For a spherical distribution of charges
the Gauss law implies
+
*
σ
.
(2.13)
(E2 − E1 ) · r̂ =
"0 x=ar̂
and for a perfect conductor
*
σ
E2 · r̂ =
"0
+
.
(2.14)
x=ar̂
In terms of the electric potential in spherical coordinates such that y is placed
at z axis
+
*
∂
Φ
σ = −"0
∂r r=a
 

q ∂ 
1
1

!
= −
−0
4π ∂r
y2 2
r 2 + y 2 − 2ry cos θ
r + a2 − 2ry cos θ
a2
$ %
q
a
= −
"
2
4πa
y
1+
1−
a2
y2
a2
y2
− 2 ya cos θ
#3/2
r=a
(2.15)
Consider a bit more complicated problem of a conducting sphere with a
total charge Q and a point charge q. The problem can be reformulated in
terms of three point charges:
- one real charge q at y,
15
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
2
at y! = ay and
- one image charge q ! = − qa
y
- one more image charge Q − q ! = Q + qa
at the origin.
y
The overall potential outside of sphere is given by
(
)
Q + aq
q
1
1
y
Φ(x) =
+
−
4π"0 |x − y| | ay x − ay y|
|x|
or in spherical coordinates

q 
1
1
!
Φ(x) =
−0
+
4π"0
y2 2
r 2 + y 2 − 2ry cos θ
2 − 2ry cos θ
r
+
a
a2
(2.16)
Q
q

+
r
a
y
.
(2.17)
Thus,
+
∂
Φ
σ = −"0
∂r
 r=a
q
∂
1
1
+
= −  !
−0
2
2
2
4π ∂r
y 2
r + y − 2ry cos θ
2 − 2ry cos θ
r
+
a
a2
"
#
2
$ %
1 − ay2
Q + ay q
q
a
= −
"
#3/2 +
4πa2 y
4πa2
2
1 + ya2 − 2 ay cos θ
*
Q
q
+
r

a
y 
Clearly, the correction due to the total charge Q − q ! does not depend on the
angle and is spread uniformly over the unit sphere.
It is also straightforward to calculate the force acting of the point charge
q using the image charges, i.e.
+
*
+
*
!
q
y
1 q
qa3 (2y 2 − a2 )
! (y − y )
!
F =
ŷ
q
− (Q − q ) 3 =
Q−
4π"0
|y − y! |3
|y|
4π"0 y 2
y (y 2 − a2 )2
(2.19)
or equivalently by integrating over the surface charge density on the sphere.
How would you go about solving for a conducting sphere in a uniform
electric field?
2.2
Green functions
Consider a sphere with an arbitrary potential Φ(a, θ, φ) at the surface. The
appropriate conditions for a specified Φ at the boundary are the Dirichlet
r=a
(2.18)
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
16
boundary conditions. Thus we must choose F such that
G(x, x! ) =
1
+ F (x, x! )
|x − x! |
(2.20)
solves the Poisson equation for a unit charge at x! and G(x, x! ) = 0 at the
surface of our sphere. This problem was already solved and the answer is
given by the potential due to a unit charge and its image, i.e.
G(x, x! ) =
1
1
− x!
!
|x − x | | a x −
a !
x|
x!
.
(2.21)
In spherical coordinates
1
1
G(x, x! ) = !
−0
2
!2
!
!2
x
x + x − 2xx cos γ
2
2
!
2 x + a − 2xx cos γ
(2.22)
a
where γ is the angle between two spheres, i.e.
cos γ = cos θ cos θ! + sin θ sin θ! cos(φ − φ! ).
(2.23)
Note that this Green function is good for both external and internal regions
of the sphere (with real and image charges interchanged in the derivation).
The derivative of the Green function (with respect to x’) in the direction
normal to the surface (also the induced charge density) is given by
*
+
x2 − a2
∂G
=
−
.
(2.24)
∂n! x! =a
a (x2 + a2 − 2ax cos γ)3/2
Now use (1.35) to derive an expression for the electric potential outside of
sphere
5
1
a(x2 − a2 )
Φ(x) =
dΩ! .
(2.25)
Φ(a, θ! , φ! )
4π
(x2 + a2 − 2ax cos γ)3/2
6 ∂G 7
Note that a similar expression (with an opposite sign due to ∂n
! x! =a ) would
also be valid for the interior of the sphere. The above expression suggests
that one can think of Green functions as propagators which propagate the
effect of the boundary at x! to the point of observation at x.
For example 2.25 can be used to solve an electrostatic problem for a
conducting sphere with hemispheres at different potentials (e.g. +V and
−V ):
(5
)
5 π/2
5 2π
5 π
2π
V
a(x2 − a2 )
Φ(x) =
dφ!
dθ! sin θ! −
dφ!
.
dθ! sin θ!
4π
(x2 + a2 − 2ax cos γ)3/2
0
0
0
π/2
(2.26)
17
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
By changing variables of the second integral θ! → π − θ! and φ! → φ! + π (so
that cos γ → − cos γ) we get
V a(x2 − a2 )
Φ(x) =
4π
5
2π
dφ
!
0
5
π/2
dθ! sin θ!
0
"8
x2 + a2 − 2ax cos γ
9− 23
8
9− 3 #
− x2 + a2 + 2ax cos γ 2 .
9− 23
8
9− 3 #
− z 2 + a2 + 2az cos θ! 2
(2.27)
Then on the z axis (i.e. θ = 0)
V a(z 2 − a2 )
Φ(x) =
4π
5
2π
dφ
!
0
5
π/2
dθ! sin θ!
0
"8
z 2 + a2 − 2az cos θ!
(2.28)
or
5
9− 3 8
9− 3 #
V a(z 2 − a2 ) 1 "8 2
Φ(x) =
du z + a2 − 2azu 2 − z 2 + a2 + 2azu 2
2
0
$
%
z 2 − a2
√
= V 1−
(2.29)
z z 2 + a2
where u = cos θ! and du = − sin θ! dθ! .
2.3
Expansion in orthogonal functions
Consider a set of functions Un (x) (real or complex) defined on the interval
(a, b) . Then the two functions Un (x) and Um (x) are called orthogonal if
5
b
a
Un∗ (x)Um (x) dx = 0.
The set of functions is called orthonormal if
5 b
Un∗ (x)Um (x) dx = δmn .
(2.30)
(2.31)
a
Then an arbitrary square integrable function f (x) can be expanded as
f (x) =
∞
:
an Un (x)
(2.32)
n=1
but sometimes a finite number of terms gives good estimate for the original
function
N
:
f (x) ≈
an Un (x).
(2.33)
n=1
18
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
Then one can find the coefficients
∗
f (x)Um
(x) =
b
5
a
∗
f (x)Um
(x)dx =
Thus,
f (x) =
∞
:
∞
:
∗
an Un (x)Um
(x)
n=1
5 b:
∞
a n=1
∗
an Un (x)Um
(x) dx = am .
an Un (x) where an =
5
(2.35)
b
f (x)Un∗ (x)dx
a
n=1
(2.34)
(2.36)
For an infinite interval the total number of orthogonal is not countable
5 +∞
Uk∗ (x)Uk! (x) = δ(k − k ! )
(2.37)
−∞
and the summation in (2.36) is replaced by integration,
5 ∞
5 b
f (x) =
A(k)Uk (x) dk where A(k) =
f (x)Uk∗ (x)dx.
−∞
(2.38)
a
The most famous are sines and cosines for which the expansion (2.38) is
called Fourier series expansion. For an interval from −L/2 to L/2 the set of
orthonormal functions is given by
&
$
%
2
2πnx
sin
(2.39)
L
L
and
&
2
cos
L
$
%
2πnx
.
L
(2.40)
Then an arbitrary square integrable function can be expanded as
$
%
$
%+
∞ *
2πnx
2πnx
A0 :
+
An cos
+ Bn sin
f (x) =
2
L
L
n=1
(2.41)
where
2
An =
L
L/2
$
2πnx
f (x) cos
L
−L/2
5
%
2
dx and Bn =
L
Using Euler’s formula
eix = cos x + i sin x
5
L/2
−L/2
f (x) sin
$
%
2πnx
dx.
L
(2.42)
19
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
we can obtain a much more convenient Fourier integral expansion,
%
$
%
$
5
+∞
:
1 L/2
2πnx
2πnx
where An =
f (x) exp −i
dx
f (x) =
An exp i
L
L
L
−L/2
n=−∞
(2.43)
which in the limit of infinite intervals takes a very “symmetric” form
1
f (x) = √
2π
5
∞
ikx
A(k) e
−∞
1
dk where A(k) = √
2π
5
∞
f (x) e−ikx dx.
−∞
(2.44)
On can check that the exponential functions √12π eikx are in fact orthonormal
5 ∞
5 ∞
1
1
1
!
!
√ eikx √ eik x dx =
ei(k−k )x dx = δ(k − k ! ).
(2.45)
2π
2π
2π
−∞
−∞
Note that the above integral expression is an alternative representation of
the delta function.
2.4
Separation of variables
The Laplace equation in rectangular coordinates
∇2 Φ =
∂2Φ ∂2Φ ∂2Φ
+
+ 2 =0
∂x2
∂y 2
∂z
(2.46)
can be easily solved under assumption that the potential can be decomposed
into a product of function of its variables, i.e.
Φ(x, y, z) = X(x)Y (y)Z(z).
(2.47)
Then,
1 d2 X(x)
1 d2 Y (y)
1 d2 Z(z)
+
+
= 0.
(2.48)
X(x) dx2
Y (y) dy 2
Z(z) dz 2
and the three terms must be separately constant. Without loss of generality
we obtain
1 d2 X(x)
= −α2
X(x) dx2
1 d2 Y (y)
= −β 2
Y (y) dy 2
1 d2 Z(z)
= γ2
Z(z) dz 2
(2.49)
(2.50)
(2.51)
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
20
where
α2 + β 2 = γ 2 .
(2.52)
The non-trivial (i.e. α &= 0 and β &= 0) solutions of (2.49), (2.50) and (2.51)
are given by
X(x) = Aeiαx + Be−iαx
Y (y) = Ceiβy + De−iβy
√ 2 2
√ 2 2
Z(z) = F e α +β z + Ge− α +β z .
(2.53)
Of course, A, B, C, D, E, F and G depend on α and β and might be more
appropriate to write
X(x) = Aαβ eiαx + Bαβ e−iαx
Y (y) = Cαβ eiβy + Dαβ e−iβy
√
√
2
2
2
2
Z(z) = Fαβ e α +β z + Gαβ e− α +β z ,
(2.54)
where the values of α, β, Aαβ , Bαβ , Cαβ , Dαβ , Eαβ , Fαβ and Gαβ are determined from the boundary conditions.
For example, consider a box with corners at (0, 0, 0), (a, 0, 0), (0, b, 0),
(a, b, 0), (0, 0, c), (a, 0, c), (0, b, c) and (a, b, c) and zero potential on all sides
with an exception of z = c which is kept at potential Φ = V (x, y). Our task
is to find a potential everywhere in the box. The solutions which agree with
vanishing potential at z = 0, x = 0, y = 0, x = a and y = b are given by
where
X(x) ∝ sin(αx)
Y (y) ∝ sin(βy)
!
Z(z) ∝ sinh( α2 + β 2 z),
nπ
a
mπ
β =
&b
n2 m2
+ 2.
γ =π
a2
b
α
(2.55)
=
(2.56)
Then the most general solution is a linear combination of these solution
( &
)
∞
" nπx #
" mπy #
:
n2 m2
Φ(x, y, z) =
Anm sin
sin
sinh πz
+ 2 . (2.57)
a
b
a2
b
n,m=1
21
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
where the coefficients Anm are set by
V (x, y) =
∞
:
Anm sin
n! ,m! =1
$
( &
)
%
$ ! %
n! πx
m πy
n!2 m!2
sin
sinh πc
+ 2 .
a
b
a2
b
(2.58)
Then
5
a
0
5
b
V (x, y) sin
0
" nπx #
a
" nπx #
sin
" mπy #
b
dxdy
(
)
2
2
n
m
sin2
sinh πc
+ 2 dxdy.
=
Anm sin2
a
b
a2
b
0
0
( &
)
$
%
$
5 a5 b
" nπx #
" mπy #%
1 1
n2 m2
1 1
=
Anm
− cos 2
− cos 2
sinh πc
+ 2 dxdy
2 2
a
2 2
b
a2
b
0
0
( &
)
ab
n2 m2
+ 2
(2.59)
= Anm sinh πc
4
a2
b
5
a
5
b
" mπy #
&
or
Anm
2.5
4
$ 0
=
2
ab sinh πc na2 +
m2
b2
%
5
0
a
5
b
V (x, y) sin
0
" nπx #
a
sin
" mπy #
b
dxdy.
(2.60)
Fields in a Corner
Many system are effectively two dimensional whenever the boundary conditions are uniform in one of the dimensions. Moreover if there is a circular
symmetry in the remaining dimensions then it is convenient to use the polar
coordinates to solve the Laplace equation,
$
%
1 ∂
∂Φ
1 ∂2Φ
ρ
+ 2
= 0.
(2.61)
ρ ∂ρ
∂ρ
ρ ∂φ
Using the separation of variables approach we assume the following ansatz
Φ(ρ, φ) = R (ρ) Ψ(φ)
to get
Ψ(φ) ∂
ρ ∂ρ
$
%
∂R(ρ)
1 ∂ 2 Ψ(φ)
ρ
+ R(ρ) 2
=0
∂ρ
ρ ∂φ2
(2.62)
(2.63)
22
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
or
ρ ∂
R(ρ) ∂ρ
$
%
∂R(ρ)
1 ∂ 2 Ψ(φ)
ρ
=−
.
∂ρ
Ψ(φ) ∂φ2
(2.64)
Since (2.64) must be satisfied for all all values of ρ and φ we get two independent equations
$
%
ρ ∂
∂R(ρ)
ρ
= ν2
(2.65)
R(ρ) ∂ρ
∂ρ
1 ∂ 2 Ψ(φ)
= −ν 2
Ψ(φ) ∂φ2
(2.66)
to be solved for a given constant ν which does not depend on ρ and φ . The
solution for ν &= 0
R(ρ) = aν ρν + bν ρ−ν
Ψ(φ) = Aν eiνφ + Bν e−iνφ
(2.67)
R(ρ) = a0 + b0 log ρ
Ψ(φ) = A0 + B0 φ.
(2.68)
and for ν = 0
Of course, the most general solution is a linear superposition of (2.67) and
(2.68) and the appropriate constants aν ’s, bν ’s, Aν ’s and Bν ’s are to be chosen
to satisfy boundary conditions.
The boundary conditions on small scales can often be approximated by
two planes intersecting at some angle, β. Such systems are effectively two
dimensional and one can easily solve the problem in polar coordinates for a
given set of boundary conditions. For example, let
Φ(ρ, φ = 0) = V
Φ(ρ, φ = β) = V
Φ(ρ = ρ0 , φ) = V0 (φ).
(2.69)
The ansatz which is consistent with these boundary conditions is
%
$
∞
:
mπ
mπφ
β
Φ(ρ, φ) = V +
Cm ρ sin
β
m=1
(2.70)
where Cm ’s are chosen such that
V0 (φ) = Φ(ρ0 , φ) = V +
∞
:
m! =1
m! π
β
C ρ0
m!
sin
$
m! πφ
β
%
.
(2.71)
23
CHAPTER 2. BOUNDARY-VALUE PROBLEMS: PART I
Thus,
5
0
or
β
dφ (V0(φ) − V ) sin
$
2 − mπ
Cm = ρ0 β
β
mπφ
β
5
0
%
mπ
β
= Cm ρ0
5
β
dφ sin
0
β
dφ (V0(φ) − V ) sin
$
2
$
mπφ
β
mπφ
β
%
.
%
(2.72)
(2.73)