Green’s function notes 2017
1
Introduction
Back in the "formal" notes, we derived the potential in terms of the Green’s
function.
Dirichlet problem: Equation (7) in "formal" notes is
Φ (x) =
1
4πε0
V
GD (x, x ) ρ (x ) d3 x −
1
4π
Φ (x )
S
∂GD (x, x )
dA
∂n
(1)
which is also Jackson 1.44. The Dirichlet Green’s function is symmetric in x
and x (see Lea §C.7.1).
GD (x, x ) = GD (x , x)
Neumann problem: Writing <Φ>S = A1 S Φ dA, the average value of
the potential over the surface S, equation (9) in "formal" notes is
Φ (x) − < Φ >S =
1
4πε0
GN (x, x ) ρ (x ) d3 x +
V
1
4π
GN
S
∂Φ
dA
∂n
(2)
In order to make use of these expressions, we need a form for the Green’s
function that is relatively easy to integrate. That suggests that we expand
the Green’s function in orthogonal functions. The Dirichlet Green’s function
is symmetric in the two sets of coordinates x and x . (See Lea pg 514 and J
problem 1.14.) In the Neumann case it is possible to impose the symmetry as
an additional condition. (In problems 1.14 and 3.27 you can see why this does
not affect the computed potential). So it does not matter whether we compute
G (x, x ) or G (x , x) . I find it easier to use x as the variable and x as fixed while
finding G to reduce the number of primes I have to write. Then the differential
equation satisfied by G is ("formal" notes eqn 3):
∇2 G (x, x ) = −4πδ (x − x )
(3)
This means that G represents the potential at x due to a unit point charge1
at x , multiplied by 4πε0 . This shows that the physical dimensions of G are
1/length.
The boundary conditions are
Dirichlet case:
(4)
GD (x, x ) = 0 for x on S
and
Neumann case:
4π
∂GN
= n̂ · ∇GN = −
for x on S
∂n
A
where A is the total area of the surface S bounding the volume V.
1 In
2 dimensions it is a line charge.
1
(5)
2
Division of region method
The 10-step method is outlined in Lea (pg 515). We begin by (Step 1) drawing
the region under consideration and (Step 2) choosing a coordinate system so
that the boundaries are represented by constant values of one or more of the
coordinates. In this method (Step 3) we place the unit point charge at an
arbitrary point x within the volume, and (Step 5) use this point to divide the
region into two separate regions, I and II, with x on the boundary between
them. Then G satisifies the simpler differential equation
∇2 G (x, x ) = 0
within each of regions I and II, (but not on the boundary between them). This
means that we can (Step 4) make use of the eigenfunctions of Laplace’s equation.
2.1
Dirichlet Green’s Function in Spherical Coordinates
Suppose our volume is the interior of a sphere of radius a. We place a unit
point charge at an arbitrary (but, for the moment, fixed) point in the region
with coordinates r , θ , φ . This point then divides the volume into two regions:
Region I: 0 ≤ r < r
Region II: r < r ≤ a
2
(Step 6) Then in region I the appropriate solution to Laplace’s equation is
GI (x, x ) =
∞
+l
Alm rl Ylm (θ, φ)
(6)
l=0 m=−l
where we have used only the functions rl and excluded r−(l+1) because the
potential should be finite at r = 0. The dependence on the coordinates r , θ
and φ is contained in the coefficient Alm .
Region II contains neither r = 0 nor r → ∞, so we need both functions of
r:
+l
∞
Clm
Blm rl + l+1 Ylm (θ, φ)
GII (x, x ) =
r
l=0 m=−l
However, here we have to deal with the boundary at r = a where G has to be
zero:
∞
+l
Clm
Blm al + l+1 Ylm (θ, φ) = 0
GII (xon S , x ) =
a
l=0 m=−l
We make use of the orthogonality of the Ylm to argue that each term must
separately equal zero:
Clm
Blm al + l+1 = 0 ⇒ Clm = −Blm a2l+1
a
Then
+l
∞
a2l+1
Blm rl − l+1 Ylm (θ, φ)
(7)
GII (x, x ) =
r
l=0 m=−l
We still have two sets of unknown constants: the Alm and Blm . But we have
one more boundary to consider at r = r . (Step 7) The first boundary condition
we need is continuity of the potential (G) at r = r .
GI (x, x )r=r
=
∞
+l
l
Alm (r ) Ylm (θ, φ)
l=0 m=−l
=
∞
+l
l
l=0 m=−l
Blm (r ) −
a2l+1
Ylm (θ, φ) = GII (x, x )r=r
(r )l+1
Again we make use of the orthogonality of the Ylm to argue that
l
a2l+1
l
Alm (r ) = Blm (r ) −
l+1
(r )
for each l, m. Thus
GI (x, x ) =
∞
+l
l=0 m=−l
GII (x, x ) =
∞
(r )2l+1
Blm rl 1 −
a2l+1
r2l+1
+l
l=0 m=−l
a2l+1
Blm rl 1 −
3
Ylm (θ, φ)
Ylm (θ, φ)
We’d like to display the symmetry in r and r more obviously, and we can do
that by relabeling
l
Blm = β lm (r )
(Remember that for the moment r is a constant.) Then
∞
GI (x, x ) =
+l
l=0 m=−l
∞
GII (x, x ) =
+l
l
l=0 m=−l
or
∞
G (x, x ) =
a2l+1
l
β lm (r ) rl 1 −
β lm (r ) rl 1 −
+l
l
l=0 m=−l
β lm (r ) rl 1 −
Ylm (θ, φ)
(r )2l+1
a2l+1
r2l+1
a2l+1
2l+1
r>
Ylm (θ, φ)
Ylm (θ, φ)
(8)
where r> = max (r, r ) and the same expression (8) holds in both regions.
We still need to find the β lm and to do this (Step 8) we make use of the
differential equation (3). First we insert expression (8) for G and evaluate the
derivatives in θ and φ. We express the delta function in terms of the spherical
coordinates using the result of Jackson Problem 1.2.
∇2 G =
=
1 ∂
r2 ∂r
∞
r2
∂
G + ∇2ang
∂r
+l
β lm
l=0 m=−l
−
= −
1
r2
∞
∞
+l
l
l=0 m=−l
β lm (r ) rl 1 −
1 ∂
a2l+1
l l
2 ∂
(r
)
r
1
−
r
2l+1
r2 ∂r
∂r
r>
+l
l=0 m=−l
l
β lm (r ) rl 1 −
a2l+1
2l+1
r>
a2l+1
2l+1
r>
Ylm (θ, φ)
Ylm (θ, φ)
l (l + 1) Ylm (θ, φ)
4π
δ (r − r ) δ (µ − µ ) δ φ − φ
r2
Now we multiply both sides of the equation by Yl∗m (θ, φ) and integrate over
the whole solid angle of the sphere. On the LHS we use the orthogonality of the
Ylm . On the RHS we use the sifting property.
βl m ∂
∂
l
r2 (r r)
r2 ∂r
∂r
= −
1−
a2l +1
2l +1
r>
− βl m
4π
δ (r − r ) Yl∗m θ , φ
r2
(r r)l
r2
1−
a2l +1
2l +1
r>
l (l + 1)
(9)
∗
Thus each β lm contains a factor Ylm
θ ,φ .
∗
β lm = γ lm Ylm
θ ,φ
4
and eqn (8) becomes
∞
G (x, x ) =
+l
γ lm (r r)
l
l=0 m=−l
1−
a2l+1
2l+1
r>
∗
θ ,φ
Ylm (θ, φ) Ylm
Now we have symmetry in all the coordinates, and the remaining set of constants
γ lm should be independent of all coordinates. The remaining equation (9) now
takes the form
l
a2l+1
γ lm ∂
l
2 ∂
(r
r)
1
−
r
2l+1
r2 ∂r
∂r
r>
−γ lm
(r r)
r2
1−
a2l+1
2l+1
r>
l (l + 1) = −
4π
δ (r − r )
r2
(Step 9) To make use of this equation we multiply both sides by r2 , then integrate across the internal boundary from r = r − ε to r = r + ε.
r +ε
∂
∂
l
r2 (r r)
∂r
∂r
γ lm
r −ε
1−
a2l+1
2l+1
r>
− (r r)
l
1−
a2l+1
2l+1
r>
l (l + 1) dr
r +ε
= −4π
δ (r − r ) dr
r −ε
The integral on the RHS equals 1. The second term on the LHS → 0 as ε → 0
because the integrand is continuous by construction and has no singularities in
the range of integration. Thus we are left with
lim γ lm r2 (r )
ε→0
l
r +ε
∂
a2l+1
rl 1 − 2l+1
∂r
r>
r −ε
= −4π
At the upper limit, r = r + ε is > r , so r> = r. At the lower limit, r = r − ε
is < r , so r> = r . Thus
l
γ lm (r )
r2
l+2
γ lm (r )
a2l+1
∂
rl − l+1
∂r
r
l−1
l (r )
+ (l + 1)
− r2
r
a2l+1
l+2
(r )
γ lm (r )
a2l+1
∂
rl 1 −
∂r
(r )2l+1
l−1
− l (r )
l+2
(l + 1)
1−
a2l+1
(r )l+2
a2l+1
(r )2l+1
+l
a2l+1
(r )l+2
r
= −4π
= −4π
= −4π
4π
1
2l + 1 a2l+1
is independent of all coordinates, as expected, and thus
γ lm = −
γ lm
G (x, x ) =
∞
+l
l=0 m=−l
4π (r )l rl
2l + 1 a2l+1
a2l+1
∗
θ ,φ
− 1 Ylm (θ, φ) Ylm
2l+1
r>
5
(10)
(Step 10) G has dimensions of 1/length, as required, and displays the necessary
symmetry. It is also positive, since a ≥ r> throughout the region. In the limit
a → ∞ we get
G →
=
∞
+l
l=0 m=−l
∞
+l
l=0 m=−l
=
∞
+l
l=0 m=−l
4π (r )l rl a2l+1
∗
θ ,φ
Y (θ, φ) Ylm
2l+1 lm
2l + 1 a2l+1 r>
4π (r )l rl
∗
θ ,φ
Y (θ, φ) Ylm
2l+1 lm
2l + 1 r>
l
1
4π r<
∗
θ ,φ =
Ylm (θ, φ) Ylm
l+1
2l + 1 r>
|x − x |
which is the correct result for a point charge in infinite space.
2.2
Use of the result
Suppose we have a sphere of radius a with potential V on the top and −V on the
bottom. Inside the sphere, a line charge of length 2b (b < a) with uniform line
charge density λ runs along the diameter of the dividing plane and is centered
at the center of the sphere. Find the potential inside.
First we have to choose coordinates. No matter what we do, we do not have
azimuthal symmetry. I’m going to put the polar axis along the line charge,
giving a charge density
ρ (x) =
λ
[δ (µ − 1) + δ (µ + 1)] S (b − r)
2πr2
We can check this by finding the total charge on a differential piece of the line
that runs from r to r + dr with r < b. Then
2π
dq
+1
= 2λdr =
−1
0
=
λ
[δ (µ − 1) + δ (µ + 1)] r2 drdµdφ
2πr2
λ
(2) 2πdr = 2λdr
2π
[
(Do you understand the factor of 2? Draw a picture and see!) The potential
on the surface is then
ΦS
= +V
= −V
if 0 ≤ φ < π
if π ≤ φ < 2π
Thus the potential is (eqn 1)
Φ (x) =
1
4πε0
−
1
4π
2π
+1
a
GD (x, x )
0
2π
−1
+1
0
−1
2
2π (r )
[δ (µ − 1) + δ (µ + 1)] S (b − r ) (r ) dr dµ dφ
∂GD (x, x )
dA
∂n
(11)
0
Φ (x )
λ
6
2
Now the field point x is fixed and x is variable. The first integral gives potential
Φ1 where
2π
a ∞
+1
4πε0 Φ1 (x) =
−1
0
+l
4π (r )l rl
2l + 1 a2l+1
l=0 m=−l
0
×
λ
a2l+1
∗
θ ,φ
− 1 Ylm (θ, φ) Ylm
2l+1
r>
2
2π (r )
2
[δ (µ − 1) + δ (µ + 1)] S (b − r ) (r ) dr dµ dφ
By orthogonality of the eimφ ,the integral over φ gives zero unless m = 0, in
which case we get 2π, so we have
a ∞
+1
−1
0
l=0
4πλ (r )l rl
2l + 1 a2l+1
a2l+1
2
− 1 Nl0
Pl (µ) Pl (µ ) [δ (µ − 1) + δ (µ + 1)] S (b − r ) dr dµ
2l+1
r>
where Nl0 = (2l + 1) /4π. Next we use the sifting property of the deltafunctions to do the integral over µ .
a ∞
0
λ
l=0
(r )l rl
a2l+1
a2l+1
− 1 Pl (µ) [Pl (1) + Pl (−1)] S (b − r ) dr
2l+1
r>
Since Pl (−1) = (−1)l , the result is zero unless l is even, when the square bracket
equals 2. Finally we do the integral over r . The integrand is zero for r > b,
so we have
∞
4πε0 Φ1 (x) =
2λ
l=0, even
b
rl
l
(r )
a2l+1
0
a2l+1
− 1 dr Pl (µ)
2l+1
r>
(12)
Looking at the integral over r by itself, we note that if r < b we have to split
the integral into two parts: r < r and r > r .
b
(r )
0
l
a2l+1
− 1 dr
2l+1
r>
r
=
(r )
a2l+1
− 1 dr +
r2l+1
l
0
=
=
b
l
(r )
r
a2l+1
(r )2l+1
− 1 dr
rl+1 a2l+1
a2l+1 1
1
bl+1 − rl+1
−1 −
− l −
2l+1
l
l+1 r
l
b
r
l+1
2l+1
2l+1
l+1
b
a
(2l + 1) a
−
for r < b, l > 0
−
rl l (l + 1)
lbl
l+1
Note that this fails for l = 0 because of the l in the denominator. For l = 0, we
get
b
0
a
− 1 dr
r>
r
b
a
a
− 1 dr +
− 1 dr
r
r
0
r
b
= a − r + a ln − (b − r)
r
b
= a − b + a ln
r
=
7
Including the factor of 1/4πε0 , the first term in the potential for r < b is
Φ1 (x) =
2λ
4πε0
1
a
a − b + a ln
∞
rl
l=2, even
=
λ
2πε0
1−
b
r
a2l+1
b
b
+ ln
a
r
∞
+
P0 (µ) +


bl+1
a2l+1 (2l + 1) a2l+1
−
−
(µ)
P
l

rl l (l + 1)
lbl
l+1
rl
(2l + 1)
− l
l (l + 1) lb
l=2, even
1+
(13)


Pl (µ)

l b2l+1
l + 1 a2l+1
The log term in (13) is expected near a line charge, and even l indicates the
reflection symmetry about the equatorial (z = 0) plane.
If r > b then r = r> throughout the range of integration, and the integral
over r is
b
l
(r )
0
bl+1
a2l+1
−
1
dr
=
r2l+1
l+1
a2l+1
−1
r2l+1
for
r>b
and the potential (first term) is
∞
Φ1 (x) =
l=0,even
λ
rl bl+1
2l+1
2πε0 a
l+1
a2l+1
− 1 Pl (µ)
r2l+1
(14)
Separating out the l = 0 term, we have
1 1
−
r a
2λb
Φ1 (x) =
4πε0
λb
+
2πε0
∞
rl
l=2, even
bl
a2l+1 l + 1
a2l+1
− 1 Pl (µ)
r2l+1
The first term is the potential due to a point charge q = 2λb at the center of a
grounded sphere of radius a. Again this is what we would expect. Finally we
note that the potential Φ1 is continuous at r = b.
We still need to evaluate the second term in (11). It contains
∂G
∂G
= n̂ · ∇G =
∂n
∂r
where the outward normal from the volume at the surface r = a is n̂ = +r̂.
Thus we have
2π
−4πΦ2 (x) =
+1
Φ (x )
0
−1
π
= V
0
2π
−
π
∂GD (x, x )
dA
∂n
+1
−1
∂
∂r
∞
+l
l=0 m=−l
4π (r )l rl
2l + 1 a2l+1
a2l+1
−1
2l+1
r>
∗
θ , φ a2 dµ dφ
× Ylm (θ, φ) Ylm
8
r =a
On the outer surface, r = a > r, so r> = r and we have
V
∞
+l
π
4π
Ylm (θ, φ)
2l + 1
l=0 m=−l
0
2π
−
∂ (r )l rl
∂r a2l+1
+1
−1
π
a2l+1
−1
2l+1
(r )
∗
× Ylm
r =a
2
θ , φ a dµ dφ
Doing the integral over φ first, we get zero if m = 0 and for m = 0
π
2π
−
0
−imφ
e
dφ
π
e−imφ
−im
=
π
2π
e−imφ
−
−im
0
π
e−imπ − 1 e−im2π − e−imπ
−
−im
−im
2
m
(1 − (−1) )
im
=
=
So we get zero for m even and 4/im for m odd. Since l ≥ m, the sum over l
now starts at 1. Thus the integral has reduced to
V
∞
+l
16πNlm Ylm (θ, φ) ∂ rl
2l + 1
im
∂r a2l+1
l=1 m=−l, odd
a2l+1
(r )
+1
l
l+1
− (r )
Plm (µ ) a2 dµ
r =a
−1
Plm (µ) is an even function of µ if l + m is even and an odd function if l + m is
odd. Thus for the integral over µ to be non-zero, we need l + m to be even,
and thus l odd. Let’s call the integral Ilm .The derivative is
∂ rl
∂r a2l+1
a2l+1
l+1
(r )
− (r )
l
r =a
= − (l + 1)
rl
al+2
−l
rl
a2l+1
al−1 = − (2l + 1)
rl
al+2
and thus we have
−4πΦ2 (x) = −V
∞
+l
(2l + 1)
l=1,odd m=−l, odd
r
a
l
16πNlm Ylm (θ, φ)
Ilm
2l + 1
im
Finally we should tidy this up by combining the positive and negative m terms.
∗
and so Nl,−m Pl−m = (−1)m Nlm Plm . That
Remember that Yl,−m = (−1)m Ylm
m
makes Nl,−m Il,−m = (−1) Nlm Ilm . Thus
Nlm Ylm (θ, φ)
Ilm Nl,−m Yl,−m
+
Il,−m
im
−im
2
= Nlm
Plm (µ) eimφ
2
= 2Nlm
Plm (µ)
9
2
Plm (µ)
Ilm Nl,m
+
Ilm e−imφ
im
−im
Ilm
sin mφ
m
and so the second term in the potential is
∞
32πV
4π
Φ2 (x) =
= 8V
+l
l=1,odd m=1, odd
∞
+l
l=1,odd m=1, odd
2V
π
=
∞
+l
l=1,odd m=1, odd
r
a
r
a
r
a
l
l
2
Nlm
Plm (µ)
Ilm
sin mφ
m
Ilm
2l + 1 (l − m)! m
Pl (µ)
sin mφ
4π (l + m)!
m
l
2l + 1
m
(l − m)!
Ilm Plm (µ) sin mφ
(15)
(l + m)!
The total potential is the sum of the two terms (13 or 14 and 15):

∞
λ 
b
rl
(2l + 1)
l
b
Φ (x) =
− l 1+
+
1 − + ln
2πε0 
a
r
l (l + 1) lb
l+1
l=2, even
+
2V
π
∞
+l
l=1,odd m=1, odd
(2l + 1) (l − m)!
r
Ilm
m (l + m)!
a
l
b
a
2l+1
Plm (µ) sin mφ ... r < b
and
λb
2πε0
Φ (x) =
+
2V
π
1 1
−
r a
+
∞
+l
λb
2πε0
l=1,odd m=1, odd
∞
l=2, even
rl
a2l+1
bl
l+1
a2l+1
− 1 Pl (µ)
r2l+1
(2l + 1) (l − m)!
r
Ilm
m (l + m)!
a
l
Plm (µ) sin mφ ... r > b
Compare the second term, in method and result, with pages 393-395 in Lea.
Let’s evaluate the first few terms in our result.
π
I11 =
0
(− sin θ) sin θdθ = −
π
2
Thus for r < b
λ
b
r2
r2 b3 1
b
5
− 2− 5
3µ2 − 1
1 − + ln
+
2πε0
a
r
6 2b
a 3 2
2V 1 π r
3
(− sin θ) sin φ + · · ·
−
π 22 a
3 cos2 θ − 1
λ
b
b
b3
r2
=
1 − + ln
+ 5− 2 3−2 3
4πε0
a
r
b
a
12
Φ (x) =
+
3V r
sin θ sin φ + · · · ... r < b
2 a
10
Pl (µ)



and for r > b
Φ (x) =
=
λ r2 b3 a5
3V r
1 1
1
−
+
3µ2 − 1 +
sin θ sin φ + · · ·
−1
r a
2πε0 a5 3 r5
2
2 a
1 1 r2 b2 a5
3V r
− +
sin θ sin φ + · · ·
− 1 3 cos2 θ − 1 +
r a
6a5 r5
2 a
λb
2πε0
λb
2πε0
Lea 8.67 gives the value of Ilm for larger odd values of l and m.
2.3
Dirichlet Green’s function in Cylindrical coordinates
Here we will find the Green’s function for the interior of an infinitely long tube
of radius a. Because we are going to use the result to find a potential that does
depend on z, we need the three-dimensional Green’s function. (Step 1: see Lea
Figure C.6) (Step 2) We choose cylindrical coordinates with z-axis along the
axis of the tube, and (Step 3) place our unit point charge at x with (fixed)
coordinates (ρ , φ , z ). (Step 4) The solutions of Laplace’s equation in this
coordinate system are
Jm (kρ)
eimφ
e±kz
Nm (kρ)
or
sin kz
Im (kρ)
eimφ
cos kz
Km (kρ)
The sets of functions {eimφ times Jm (kρ) or Nm (kρ)} and {eimφ times sin kz or
cos kz} form complete orthogonal sets of functions, so we may divide our space
in either z or ρ. (Step 5) Let’s divide in ρ. (The other choice is in Lea pg
521-525.) (Step 6) Then in region I, ρ < ρ , we need a function that is finite at
ρ = 0, which is I. Here we have no boundaries at finite z, so we have no way
to determine specific values for k. So we have
+∞
GI (x, x ) =
∞
Am (k) Im (kρ) eikz eimφ dk
(16)
m=−∞ −∞
In region II, a > ρ > ρ , we have both modified Bessel functions.
+∞
GII (x, x ) =
∞
Bm (k) [Im (kρ) + Cm Km (kρ)] eikz eimφ dk
m=−∞ −∞
At ρ = a, GII = 0, so
+∞
∞
m=−∞
−∞
0=
Bm (k) [Im (ka) + Cm Km (ka)] eikz eimφ dk
By orthogonality of the eikz eimφ in z and φ, we can equate each term separately
to zero, so we get
Im (ka)
Cm = −
Km (ka)
11
and thus
+∞
∞
Im (ka)
Km (kρ) eikz eimφ dk
K
(ka)
m
−∞
m=−∞
(17)
(Step 7) Now we use continuity of the potential at ρ = ρ to get
GII (x, x ) =
+∞
∞
Bm (k) Im (kρ) −
Am (k) Im (kρ ) eikz eimφ dk
m=−∞ −∞
+∞
∞
=
m=−∞ −∞
Bm (k) Im (kρ ) −
Im (ka)
Km (kρ ) eikz eimφ dk
Km (ka)
and again by orthogonality we have
Am (k) = Bm (k) 1 −
Im (ka) Km (kρ )
Im (kρ ) Km (ka)
Thus (16) becomes
+∞
∞
Im (ka) Km (kρ )
Im (kρ) eikz eimφ dk
I
(kρ
)
K
(ka)
m
m
−∞
m=−∞
(18)
We may combine the two results (18) and (17) to get
GI (x, x ) =
Bm (k) 1 −
+∞
G (x, x ) =
∞
Bm (k) gm (ρ, ρ ) eikz eimφ dk
(19)
m=−∞ −∞
where
gmI ≡ 1 −
Im (ka) Km (kρ )
Im (kρ)
Im (kρ ) Km (ka)
and
gmII ≡ Im (kρ) −
Im (ka)
Km (kρ)
Km (ka)
(20)
(21)
(Step 8) Next we make use of the differential equation, expressing the delta
function on the right in cylindrical coordinates (cf Problem 1.2):
∇2 G =
1 ∂ 2G ∂ 2G
1 ∂
4π
(ρG) + 2
+
= − δ (ρ − ρ ) δ φ − φ δ (z − z )
2
2
ρ ∂ρ
ρ ∂φ
∂z
ρ
+∞
m=−∞ 0
= −
∞
Bm (k) eikz eimφ
1 ∂ ∂gm (ρ) m2
ρ
− 2 gm − k2 gm dk
ρ ∂ρ
∂ρ
ρ
4π
δ (ρ − ρ ) δ φ − φ δ (z − z )
ρ
12
Now we multiply both sides by e−im φ and integrate from 0 to 2π in φ. On the
RHS we use the sifting property, and on the left hand side, we use orthogonality
of the eimφ.
∞
2π
Bm (k) eikz
−∞
1 ∂ ∂gm (ρ) (m )2
ρ
−
gm − k2 gm
ρ ∂ρ
∂ρ
ρ2
dk = −
4π
δ (ρ − ρ ) δ (z − z ) e−im φ
ρ
Now we can drop the prime on m. Similarly we multiply by e−ik z and integrate
over all z. We use Lea eqn 6.16 to get
∞
−∞
Bm (k) 2πδ (k − k )
Bm (k ) π
1 ∂ ∂gm (ρ) m2
ρ
− 2 gm − k2 gm dk
ρ ∂ρ
∂ρ
ρ
1 ∂ ∂gm (ρ) m2
2
ρ
− 2 gm − (k ) gm
ρ ∂ρ
∂ρ
ρ
2
= − δ (ρ − ρ ) e−ik z e−imφ
ρ
1
= − δ (ρ − ρ ) e−ik z e−imφ
ρ
Drop the primes on the k , and relabel, remembering that z and φ are fixed
for the moment.
Bm (k) ≡ β m (k) e−ikz e−imφ
(Step 9) Finally we multiply both sides by ρ and integrate across the boundary
at ρ = ρ
ρ +ε
πβ m (k)
ρ −ε
∂ ∂gm (ρ) m2
ρ
−
gm − k2 ρgm dρ = −
∂ρ
∂ρ
ρ
ρ +ε
ρ −ε
δ (ρ − ρ ) dρ
Making use of the fact that gm /ρ and ρgm are well behaved in the range of
integration, we get
ρ +ε
∂gm (ρ)
1
=−
lim β m (k) ρ
ε→0
∂ρ
π
ρ −ε
At the upper limit we are in region II where g = gII (21), and at the lower limit
we are in region I where g = gI (20), so
β m (k) kρ
Im (ka)
Im (ka) Km (kρ )
Km (kρ ) − 1 −
I (kρ )
Km (ka)
Im (kρ ) Km (ka) m
Km (kρ ) Im (kρ )
Im (ka)
β m (k) kρ −
Km (kρ ) + Im (ka)
Km (ka)
Km (ka) Im (kρ )
Im (kρ ) −
Thus
β m (k) =
1
Km (ka) Im (kρ )
πkρ Im (ka) [Km (kρ ) Im (kρ ) − Im (kρ ) Km (kρ )]
The denominator is the Wronskian W (kρ ) of the modified Bessel differential equation. To evaluate it, let’s use the large argument form of the functions. We can do this because W (x) is the same function for all x. (Also see
http://www.physics.sfsu.edu/~lea/courses/grad/Besselwronskian.pdf. )
Km (x)
π −x
e ;
2x
13
Im (x)
1
√
ex
2πx
1
π
1
= −
π
= −
W (x) = Km (x) Im (x) − Im (x) Km (x)
1
π −x
1
e
− √ − 3/2
2
x 2x
1
= −
x
1 x
e −
2πx
=
Thus
β m (k) = −
π x
e
2
1
1
√ − 3/2
x 2x
e−x
√
2πx
(22)
1 Km (ka) Im (kρ )
π
Im (ka)
and then (18) becomes
+∞
∞
GI (x, x ) =
−1 Km (ka) Im (kρ ) ik(z−z )
e
π
Im (ka)
m=−∞ −∞
× 1−
1
π
=
+∞
∞
m=−∞ −∞
Im (ka) Km (kρ )
Im (kρ) eim(φ−φ ) dk
Im (kρ ) Km (ka)
eik(z−z ) [Im (ka) Km (kρ ) − Km (ka) Im (kρ )]
Im (kρ) im(φ−φ )
e
dk
Im (ka)
for ρ < ρ and (17)
+∞
∞
GII (x, x ) =
m=−∞ −∞
1
π
=
+∞
Im (ka)
−1 Km (ka) Im (kρ ) ik(z−z )
e
Km (kρ) eim(φ−φ ) dk
Im (kρ) −
π
Im (ka)
Km (ka)
∞
m=−∞ −∞
Im (kρ ) ik(z−z )
[Im (ka) Km (kρ) − Km (ka) Im (kρ)] eim(φ−φ ) dk
e
Im (ka)
for ρ > ρ , or, in both cases, (19)
+∞
∞
Im (kρ< )
[Im (ka) Km (kρ> ) − Km (ka) Im (kρ> )] eik(z−z ) eim(φ−φ ) dk
I
(ka)
m
m=−∞ −∞
(23)
(Step 10) The Green’s function has dimensions 1/length (from the integral
over k) as expected. It also displays the necessary symmetry. As a → ∞,
Km (ka) → 0 and so
G (x, x ) =
G→
1
π
1
π
+∞
∞
Im (kρ< ) Km (kρ> ) eik(z−z ) eim(φ−φ ) dk =
m=−∞ −∞
1
|x − x |
(Jackson 3.148), as required.
Now let’s solve the same problem that is in the math phys book (pg 524)
with V0 on the wall between z = −a and z = +a, and the rest of the cylinder
14
grounded. The outward normal is n̂ = +ρ̂, and at ρ = a, ρ> = ρ , so
Φ (x) = −
1
4π
= −
V0
4π
Φ (x )
S
a
∂G(x, x )
dA
∂n
2π
−a
+∞
−∞
0
×
+∞
1 Im (kρ) ik(z−z ) im(φ−φ )
e
e
π Im (ka)
m=−∞
∂
[Im (ka) Km (kρ ) − Km (ka) Im (kρ )]|ρ =a dkadφ dz
∂ρ
The integral over φ gives zero unless m = 0, so Φ is independent of φ, as
expected.
Φ (x) = −
V0
2π
4π
+∞
−∞
a
−a
1 I0 (kρ) ik(z−z )
k [I0 (ka) K0 (ka) − K0 (ka) I0 (ka)] dkadz
e
π I0 (ka)
Here again we have the Wronskian of the modified Bessel functions W (ka)
(22), so
Φ (x) = +
=
V0
2π
V0
2π
=
V0
2π
=
V0
π
=
2V0
π
+∞
−∞
+∞
−∞
+∞
−∞
+∞
−a
I0 (kρ) ik(z−z ) k
e
dkadz
I0 (ka)
ka
I0 (kρ) eik(z−z )
I0 (ka)
−ik
+a
dk
−a
−ika
− eika
I0 (kρ) eikz e
dk
I0 (ka)
−ik
I0 (kρ) eikz sin ka
dk
I0 (ka)
k
−∞
+∞
0
a
I0 (kρ) sin ka cos kz
dk
I0 (ka)
k
The result is dimensionally correct. Since ρ ≤ a, I0 (kρ) ≤ I0 (ka) , and so the
integral converges. Let’s see how this looks for ρ → a.
Φ (a, z) =
=
=
=
=
V0 a +∞ ikz
V0 +∞ eikz sin ka
dk =
e cos ku dkdu
π −∞
k
π 0 −∞
a
V0
δ (z + u) + δ (z − u)
2π
du
π
2
0
V0 if 0 ≤ z ≤ a (from the second delta function)
V0 if − a ≤ z ≤ 0 (from the first delta function)
0 otherwise
So we get back the correct boundary values.
15
Comparing with the result in the book (pg 525),
Φ (x) = 2V0
∞
J0 (x0n ρ/a)
z
1 − e−x0n cosh x0n
x
J
(x
)
a
0n
n=1 0n 1
,
we see that the main difference is that we have an integral instead of an infinite
sum. We can evaluate the integral numerically. For z = a/2, we have
+∞ I0 (.1k) sin k cos(k/2)
ρ = 0.1a : π2 0
dk = 0.768 66
I0 (k)
k
ρ = 0.2a :
ρ = 0.3a :
ρ = 0.4a :
ρ = 0.5a :
ρ = 0.6a :
ρ = 0.7a :
ρ = 0.8a :
ρ = 0.9a :
2
π
2
π
2
π
2
π
2
π
2
π
2
π
2
π
+∞
0
+∞
0
+∞
0
+∞
0
+∞
0
+∞
0
+∞
0
+∞
0
sin k cos k/2
dk = 0.774 76
k
sin k cos k/2
dk = 0.785 18
k
sin k cos k/2
dk = 0.800 28
k
sin k cos k/2
dk = 0.820 58
k
sin k cos k/2
dk = 0.846 6
k
sin k cos k/2
dk = 0.878 66
k
sin k cos k/2
dk = 0.916 36
k
sin k cos k/2
dk = 0.957 96
k
2 +∞ I0 (.1k) sin k cos(2k)
we have π 0
dk
I0 (k)
k
I0 (.2k)
I0 (k)
I0 (.3k)
I0 (k)
I0 (.4k)
I0 (k)
I0 (.5k)
I0 (k)
I0 (.6k)
I0 (k)
I0 (.7k)
I0 (k)
I0 (.8k)
I0 (k)
I0 (.9k)
I0 (k)
For z/a = 2, ρ = 0.1a
= 6.877 8 × 10−2
Compare with Lea Figure C.7. The results are the same.
3
Expansion in eigenfunctions without division.
This method is outlined in Jackson section 3.12 and Lea sections C.2 and C.7.5.
The result is a single function, valid everywhere in the region, at the expense of
an extra sum. We’ll begin with a one-dimensional problem, and then extend
the result to three dimensions.
We start with the Sturm-Liouville equation
d
dx
f
dy
dx
− g (x) y + λw (x) y = 0
(24)
valid for a ≤ x ≤ b, where we either have boundary conditions (Dirichlet or
Neumann) that guarantee orthogonality of the solutions, or else f (x) = 0 on
the boundary. The solutions are the eigenfunctions yn (x) with eigenvalues λn .
Let’s normalize the eigenfunctions (as we did with the Ylm ) so that they satisfy
the orthonormality relation
b
w (x) yn (x) ym (x) dx = δ nm
(25)
a
Now we look for a Green’s functions that satisfies the differential equation
d
dx
f
dG (x, x )
dx
− g (x) G (x, x ) + λw (x) G (x, x ) = −4πδ (x − x )
16
(26)
where the constant λ does not equal λn for any n. Since G is expected to be
well-behaved, we may expand G (x, x ) in the eigenfunctions yn (x) :
G (x, x ) =
γ n (x ) yn (x)
n
and substitute into the differential equation (26).
γ n (x )
n
d
dyn (x)
f
dx
dx
− g (x)
γ n (x ) yn (x) + λw (x)
n
γ n (x ) yn (x)
n
= −4πδ (x − x )
We use the eigenfunction equation (24) to express the first two terms in terms
of λn yn :
γ n (x ) [−λn w (x) yn (x)] + λw (x)
n
n
γ n (x ) yn (x) = −4πδ (x − x )
Now we multiply both sides by ym (x) and integrate over the range x = a to
b. We use orthonormality (eqn 25) on the LHS and the sifting property on the
RHS.
(λ − λm ) γ m (x ) = −4πym (x )
Thus
ym (x )
(λ − λm )
Now it is clear why λ cannot equal any λn . Then
γ m (x ) = −4π
G (x.x ) = 4π
n
yn (x ) yn (x)
λn − λ
(27)
To extend this result to potential problems in 3-d, we start with equation
(3). The eigenvalue λ in this equation is zero, so we need eigenfunctions that
satisfy an equation with non-zero λ, and that is the Helmholtz equation.
∇2 f (x) + k2 f (x) = 0
(We have solved this equation before: See e.g waveguide notes pages 8 and 9.)
Let’s find the Dirichlet Green’s function for the interior of a rectangular
box measuring a by b by c. We put the origin at one corner, with Cartesian
axes along the three sides, and solve by separation of variables: f (x, y, z) =
X (x) Y (y) Z (z). Then the eigenfunctions we want are the solutions of
Y
Z
X
+
+
+ k2 = 0
X
Y
Z
with X (0) = X (a) = 0, Y (0) = Y (b) = 0 and Z (0) = Z (c) = 0. Then
X = sin nπx/a, Y = sin mπy/b and Z = sin pπz/c, and
2
= π2
knmp
n2 m2 p2
+ 2 + 2
a2
b
c
17
We still need to normalize the eigenfunctions. The orthogonality integral is
a
a
nπx
dx =
a
2
sin2
0
so the normalized function is
nπx
2
sin
a
a
Xn (x) =
and so
nπx 2
mπy 2
pπz
2
sin
sin
sin
a
a
b
b
c
c
Now we just stuff this result into (27) (extended to 3-d) with λ = 0.
fnmp (x, y, z) =
G (x, x ) = 4π
=
2 ∞
8
abc
∞
mπy
pπz
mπy
pπz
nπx
sin nπx
a sin b sin c sin a sin b sin c
π2
n=1 m=1 p=1
∞
∞
∞
∞
sin nπx
32
a sin
πabc n=1 m=1 p=1
mπy
b
n2
a2
+
m2
b2
p2
c2
+
mπy
pπz
nπx
sin pπz
c sin a sin b sin c
n2
a2
+
m2
b2
+
p2
c2
Again check that G has dimensions of [1/length]. The benefit of this approach is that we have a single expression that we can use in the whole region.
The disadvantage is that we have an extra sum.
Suppose the box, with its walls remaining grounded, contains a sheet of
charge with uniform surface charge density σ that extends from x = a/4 to
x = 3a/4 and y = b/4 to y = 3b/4 at z = c/2. Find the potential inside the
box.
In this case we have the volume integral in (1), but the surface integral is
zero. The charge density is
c
a/4 < x < 3a/4, b/4 < y < 3b/4
ρ (x) = σδ z −
2
= 0 otherwise
So
Φ (x) =
=
1
4πε0
σ
4πε0
a
b
c
G (x, x ) ρ (x ) dx dy dz
0
0
3a/4
a/4
0
3b/4
b/4
∞
∞
∞
mπy
b
0
sin nπx
32
a sin
πabc n=1 m=1 p=1
∞
c
dx dy dz
2
mπy
pπz
pπ
sin nπx
a sin b sin c sin 2
c
mπy
pπz
nπx
sin pπz
c sin a sin b sin c
n2
a2
+
m2
b2
+
p2
c2
×δ z −
=
∞
∞
σ 32
4πε0 πabc n=1 m=1 p=1
n2
a2
+
m2
b2
+
p2
c2
3a/4
×
18
sin
a/4
nπx
dx
a
3b/4
sin
b/4
mπy
dy
b
The result is zero if p is even, so
Φ (x) =
∞
∞
8σ
ε0 π2 abc n=1 m=1
∞
mπy
pπz
p−1
sin nπx
a sin b sin c
(−1)( 2 )
2
2
2
p
n
m
p=1,odd
a2 + b2 + c2
×
=
∞
∞
∞
8σ
(−1)(
ε0 π4 c n=1 m=1 p=1,odd
p−1
2
)
nπx
a
cos
nπ
a
3a/4
mπx
b
cos
mπ
b
a/4
mπy
nπx
sin a sin b sin pπz
c
p2
n2
m2
nm a2 + b2 + c2
× cos
nπ
3nπ
− cos
4
4
cos
3b/4
b/4
3mπ
mπ
− cos
4
4
Now we combine the cosines to get
Φ (x) =
=
∞
∞
32σ
ε0 π4 c n=1 m=1
∞
(−1)(p−1)/2
mπy
pπz
sin nπx
a sin b sin c
nm
p=1,odd
n2
a2
+
m2
b2
+
p2
c2
sin
nπ
nπ
mπ
mπ
sin
sin
sin
2
4
2
4
∞ ∞ ∞
mπy
pπz
nπx
mπ
nπ
32σ
(p−1)/2+(n−1)/2+(m−1)/2 sin a sin b sin c
sin
sin
(−1)
p2
n2
m2
ε0 π4 c n=1 m=1 p=1
4
4
nm a2 + b2 + c2
odd odd odd
If n = 2l + 1, then
sin
nπ
4
(2l + 1) π
lπ π
= sin
= sin
+
4
2
4
√
lπ
2
lπ
+ cos
=
sin
2
2
2
√
2
(−1)(l−1)/2 if l is odd
=
2
√
2
=
(−1)l/2 if l is even
2
= sin
π
lπ
π
lπ
cos + cos sin
2
4
2
4
Thus replacing n (odd) with 2n + 1 and similarly for m and p we get
Φ (x) =
∞
∞
∞
σ 16
ε0 π4 c n=0 m=0 p=0
(−1)p+n+m+q
(2n+1)2
a2
+
× sin
where
q = integer part
(2m+1)2
b2
+
(2p+1)2
c2
(2m + 1) πy
(2p + 1) πz
(2n + 1) πx
sin
sin
a
b
c
m
n
+ integer part
2
2
Check the dimensions!
19
At z = c/4 we have
Φ (x, y)
16σa2 / (ε0 π4 c)
=
∞
∞
∞
n=0 m=0 p=0
(−1)p+n+m+q
(2n + 1) (2m + 1) (2n + 1)2 +
× sin
(2m+1)2 a2
b2
+
(2p+1)2 a2
c2
(2m + 1) πy
(2p + 1) π
(2n + 1) πx
sin
sin
a
b
4
The diagram shows the dimensionless potential (28), computed using values of
n, m and p from zero to five, and with a = b = c.
0.15
0.1
z
0.05
1
0.8
0.4
0.6 x
0
20
0.4
y 0.6
0.8
1
(28)