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For this problem the best way to separate variables is to consider
G
(
r
)
φ
(
θ
) and obtain the equations
r
G
d
dr
±
r
dG
dr
²
=

1
φ
d
2
φ
dθ
2
=

λ.
Considering the
G
(
r
) equation yields the boundary value problem
d
dr
±
r
dG
dr
²
+
λ
1
r
G
= 0
G
(
a
) =
G
(
b
) = 0
We now have a regular SturmLiouville eigenvalue problem, which is why we chose the separation constant
to take the form

λ
. The solution to this problem has the same form as in
§
2
.
5
.
2 of Haberman, only now
we have complex numbers in the exponent
G
(
r
) =
c
1
r
i
√
λ
+
c
2
r

i
√
λ
=
c
1
e
i
√
λ
ln
r
+
c
2
e

i
√
λ
ln
r
where the equality
r
=
e
ln
r
is used in the second equality. It takes some work, but this latter expression
can be rewritten as
G
(
r
) =
c
3
cos
³
√
λ
ln
³
r
a
´´
+
c
4
sin
³
√
λ
ln
³
r
a
´´
.
We now have 0 =
G
(
a
) =
c
3
·
1 +
c
4
·
0 =
c
3
so that
G
(
r
) collapses to
G
(
r
) =
c
4
sin
³
√
λ
ln
³
r
a
´´
which means
0 =
G
(
b
) =
c
4
sin
±
√
λ
ln
±
b
a
²²
⇒
√
λ
ln
±
b
a
²
=
nπ
⇒
λ
=
±
nπ
ln(
b/a
)
²
2
.
Factoring in
φ
(0) = 0 in the
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This note was uploaded on 10/05/2010 for the course EE 5543 taught by Professor Sim during the Spring '10 term at Punjab Engineering College.
 Spring '10
 sim

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