Québec, September 2012
vmarceau.libres-penseurs.com
Vincent Marceau
vincent.marceau.2@ulaval.ca
PROBLEM 2.19
From problem 2.17 of Jackson (see the solution for more details), we know that the Green’s function
we are looking for can be expressed as a Fourier series in the azimutal coordinate:
G(ρ, φ; ρ′ , φ′ ) =
∞
1 X
′
gm (ρ, ρ′ )eim(φ−φ ) ,
2π m=−∞
where the radial Green’s function gm (ρ, ρ′ ) satisfies
2
∂
1 ∂
m2
4π
+
− ′2 gm (ρ, ρ′ ) = − δ(ρ − ρ′ ) .
∂ρ′2 ρ′ ∂ρ′
ρ
ρ
(1)
(2)
Since we are dealing with Dirichlet boundary conditions, we shall want gm (ρ, ρ′ ) to vanish at the
boundaries of the annular region, namely gm (ρ, b) = gm (ρ, c) = 0.
To complete the Green’s function series expansion, we must solve for gm (ρ, ρ′ ). We proceed in the
following manner. First, we must obtain a solution to the homogeneous differential equation in
− ) and ρ′ > ρ (we’ll call this solution g + ).
each of the two domains ρ′ < ρ (we’ll call this solution gm
m
′
These solutions must vanish when ρ → ∞ and be well behaved at origin. We must afterwards
− = g + and ∂ ′ g + = ∂ ′ g − − 4π/ρ
connect those two solutions so that they satisfy the conditions gm
ρ m
ρ m
m
′
at ρ = ρ.
Let us first consider equation (2) in its homogeneous form:
2
1 ∂
m2
∂
+
− ′2 gm (ρ, ρ′ ) = 0 .
∂ρ′2 ρ′ ∂ρ′
ρ
(3)
We must distinguish two different cases, namely m = 0 and m 6= 0.
For m = 0, the solution is
(
g0− = C1 + C2 ln ρ′
g0 (ρ, ρ′ ) =
g0+ = C3 + C4 ln ρ′
ρ′ < ρ
.
ρ′ > ρ
(4)
Applying the boundary conditions, we get:
g0− (ρ, b) = C1 + C2 ln b = 0
→ C1 = −C2 ln b ,
(5)
g0+ (ρ, c) = C3 + C4 ln c = 0
→ C3 = −C4 ln b .
(6)
The solution can now be written as
(
g0− = C2 ln(ρ′ /b)
g0 (ρ, ρ′ ) =
g0+ = C4 ln(ρ′ /c)
ρ′ < ρ
ρ′ > ρ
(7)
Applying the continuity and derivative jump conditions at ρ = ρ′ , we get:
C2 ln (ρ/b) = C4 ln (ρ/c)
(8)
C2 − 4π = C4 .
(9)
Solving these two equations for C2 and C4 is pretty straightforward (i.e., express C4 as a function
of C2 using the first equation and then solve the second equation for C2 ) and yields
C2 = −4π
ln(ρ/c)
ln(c/b)
and
C4 = −4π
ln(ρ/b)
.
ln(c/b)
The general solution for g0 (ρ, ρ′ ) is therefore

ln(ρ/c) ln(ρ′ /b)


g0− = −4π
ln(c/b)
g0 (ρ, ρ′ ) =
ln(ρ/b)
ln(ρ′ /c)


g0+ = −4π
ln(c/b)
(10)
ρ′ < ρ
.
(11)
ρ′ > ρ
If we define ρ< ≡ min(ρ, ρ′ ) and ρ> ≡ max(ρ, ρ′ ), this can be written more compactly as
g0 (ρ, ρ′ ) = 2π
ln(c2 /ρ2> ) ln(ρ2< /b2 )
.
ln(c2 /b2 )
(12)
Now consider the case m 6= 0. The general solution for gm (ρ, ρ′ ) is
(
− = C ρ′−m + C ρ′m
ρ′ < ρ
gm
1
2
.
gm (ρ, ρ′ ) =
+ = C ρ′−m + C ρ′m
ρ′ > ρ
gm
3
4
(13)
Applying the boundary conditions, we get
gm (ρ, b) = C1 b−m + C2 bm = 0
→ C1 = −C2 b2m ,
(14)
gm (ρ, c) = C3 c−m + C4 cm = 0
→ C3 = −C4 c2m ,
(15)
which yields
(
− = C (ρ′m − b2m ρ′−m )
gm
2
gm (ρ, ρ′ ) =
+
gm = C4 (ρ′m − c2m ρ′−m )
ρ′ < ρ
.
ρ′ > ρ
(16)
Applying the continuity and derivative jump conditions at ρ = ρ′ , we get:
C2 (ρm − b2m ρ−m ) = C4 (ρm − c2m ρ−m )
4π
C2 (ρm + b2m ρ−m ) −
= C4 (ρm + c2m ρ−m ) .
m
(17)
(18)
Once again, by isolating C4 in the first equation, inserting the obtained expression in the second
equation and solving for C2 , we end up with the following result:
C2 = −
2π (ρm − c2m ρ−m )
m (c2m − b2m )
and
C4 = −
2π (ρm − b2m ρ−m )
.
m (c2m − b2m )
The general solution for gm (ρ, ρ′ ) is therefore

m
2m ρ−m )(ρ′m − b2m ρ′−m )

− = − 2π (ρ − c

gm
m
(c2m − b2m )
gm (ρ, ρ′ ) =
m
2m
ρ−m )(ρ′m − c2m ρ′−m )

+ = − 2π (ρ − b

gm
m
(c2m − b2m )
(19)
ρ′ < ρ
,
(20)
ρ′ > ρ
which can be written more compactly as
gm (ρ, ρ′ ) =
m 2m )(ρm − b2m /ρm )
2π (1/ρm
<
<
> − ρ> /c
.
m
(1 − b2m /c2m )
(21)
Notice that if we define k ≡ |m|, the solution takes exactly the same form indenpendently of the
sign of m:
gk (ρ, ρ′ ) =
2π (1/ρk> − ρk> /c2k )(ρk< − b2k /ρk< )
.
k
(1 − b2k /c2k )
(22)
We now return to our Fourier series expansion of G. Inserting the abovementioned solutions for
gm (ρ, ρ′ ), we finally get
G(ρ, φ; ρ′ , φ′ ) =
=
m 2m )(ρm − b2m /ρm )
ln(c2 /ρ2> ) ln(ρ2< /b2 ) X 1 (1/ρm
< im(φ−φ′ )
<
> − ρ> /c
+
e
ln(c2 /b2 )
m
(1 − b2m /c2m )
ln(c2 /ρ2> ) ln(ρ2< /b2 )
ln(c2 /b2 )
m6=0
∞
X
+2
k=1
cos[k(φ − φ′ )]
(1/ρk> − ρk> /c2k )(ρk< − b2k /ρk< ) . (23)
k(1 − b2k /c2k )