Physics 505
Fall 2007
Practice Midterm — Solutions
This midterm will be a two hour open book, open notes exam. Do all three problems.
1. A rectangular box has sides of lengths
a
,
b
and
c
c
x
y
z
a
b
a
) For the Dirichlet problem in the interior of the box, the Green’s function may be
expanded as
G
(
x, y, z
;
x , y , z
) =
∞
m
=1
∞
n
=1
g
mn
(
z, z
) sin
mπx
a
sin
mπx
a
sin
nπy
b
sin
nπy
b
Write down the appropriate differential equation that
g
mn
(
z, z
) must satisfy.
Note that sin
kx
satisfies the completeness relation
∞
m
=1
sin
mπx
a
sin
mπx
a
=
a
2
δ
(
x

x
)
Hence the Green’s function equation
∇
2
x
G
(
x, x
) =

4
πδ
3
(
x

x
)
has an expansion
m,n
∇
2
x
g
mn
(
z, z
) sin
mπx
a
sin
mπx
a
sin
nπy
b
sin
nπy
b
=

4
πδ
(
z

z
)
4
ab
m,n
sin
mπx
a
sin
mπx
a
sin
nπy
b
sin
nπy
b
Working out the
x
and
y
derivatives on the lefthand side yields
m,n
d
2
dz
2

mπ
a
2

nπ
b
2
g
mn
(
z, z
) sin
mπx
a
sin
mπx
a
sin
nπy
b
sin
nπy
b
=

16
π
ab
δ
(
z

z
)
m,n
sin
mπx
a
sin
mπx
a
sin
nπy
b
sin
nπy
b
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
However, since sin
kx
forms an orthogonal basis, each term in this sum must
vanish by itself. This results in the differential equation
d
2
dz
2

γ
2
mn
g
mn
(
z, z
) =

16
π
ab
δ
(
z

z
)
(1)
where
γ
mn
=
π
(
m/a
)
2
+ (
n/b
)
2
is given in part
c
). Note that the Fourier sine
expansion automatically satisfies Dirichlet boundary conditions for
x
and
y
. The
remaining boundary condition is that
g
mn
(
z, z
) vanishes whenever
z
or
z
is equal
to 0 or
c
.
b
) Solve the Green’s function equation for
g
mn
(
z, z
) subject to Dirichlet boundary
conditions and write down the result for
G
(
x, y, z
;
x , y , z
).
We may solve the Green’s function equation (1) by first noting that the homoge
neous equation is of the form
g
(
z
)

γ
2
mn
g
(
z
) = 0
This is a secondorder linear equation with constant coefficients admitting the
familiar solution
g
(
z
) =
Ae
γ
mn
z
+
Be

γ
mn
z
However, we want
g
(
z
) = 0 when
z
= 0 or
z
=
c
. This motivates us to write
out the solutions
u
(
z
) = sinh
γ
mn
z
0
< z < z
v
(
z
) = sinh[
γ
mn
(
c

z
)]
z < z < c
The Green’s function solution is then given by
g
mn
(
z, z
) =
Au
(
z
)
z < z
Bv
(
z
)
z > z
The matching conditions
g
<
=
g
>
,
d
dz
g
<
=
d
dz
g
>
+
16
π
ab
then give the system
A
sinh
γ
mn
z
=
B
sinh[
γ
mn
(
c

z
)]
,
A
cosh
γ
mn
z
=

B
cosh[
γ
mn
(
c

z
)] +
16
π
abγ
mn
which may be solved to yield
A
=
16
π
sinh[
γ
mn
(
c

z
)]
abγ
mn
(cosh
γ
mn
z
sinh[
γ
mn
(
c

z
)] + sinh
γ
mn
z
cosh[
γ
mn
(
c

z
)])
=
16
π
sinh[
γ
mn
(
c

z
)]
abγ
mn
sinh
γ
mn
c
=
16
π
abγ
mn
sinh
γ
mn
c
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 Stephens,P
 Physics, Magnetism, Boundary value problem, Multipole moments, Axial multipole moments, quadrupole, Multipole expansion, sin sin sin

Click to edit the document details