Environmental Engineering Reference
In-Depth Information
Assume
v
=
R
(
r
)
Y
(
θ
,
ϕ
)
. Substituting it into (2.55) yields
⎧
⎨
)+
k
2
r
2
)
R
r
2
R
(
2
rR
(
r
)+
r
−
l
(
l
+
1
(
r
)=
0
, |
R
(
0
)
| <
∞
,
sin
2
Y
∂ϕ
1
sin
∂
∂θ
θ
∂
Y
∂θ
1
sin
2
∂
(2.56)
+
2
+
l
(
l
+
1
)
Y
=
0
,
⎩
θ
θ
Y
(
θ
,
ϕ
+
2
π
)=
Y
(
θ
,
ϕ
)
,
in which the separation constant is customarily denoted by
l
(
l
+
1
)
.
Assume
Y
=
Θ
(
θ
)
Φ
(
ϕ
)
. Substituting it into Eq. (2.56) leads to
⎧
⎨
Φ
+
ηΦ
=
0
,
Φ
(
ϕ
+
2
π
)=
Φ
(
ϕ
)
,
(2.57)
l
η
sin
2
Θ
+(
θ
)
Θ
+
cot
(
l
+
1
)
−
Θ
=
0
,
⎩
θ
0
<
θ
<
π
|
Θ
(
θ
)
| <
∞
,
(2.58)
where
η
is the separation constant.
Solution of Eigenvalue Problems
1. The eigenvalue problem (2.57) can be very easily solved. The results are:
m
2
Eigenvalues
η
=
,
m
=
0
,
1
,
2
, ···
;
c
(
1
m
cos
m
c
(
2
m
sin
m
Eigenfunctions
Φ
(
ϕ
)=
ϕ
+
ϕ
.
m
Here
c
(
1
m
and
c
(
2
m
are arbitrary constants that are not all equal to zero. Also,
c
(
1
)
0
=
0.
Φ
0
(
ϕ
)=
1for
η
=
0.
m
2
, Eq. (2.58) forms an eigenvalue problem of the Legendre equation.
Its solution is available in Appendix A.
When
m
η
=
2. With
=
0,
Eigenvalues
l
(
l
+
1
)=
n
(
n
+
1
)
,
n
=
0
,
1
,
2
, ··· ,
Eigenfunctions
Θ
n
(
θ
)=
P
n
(
cos
θ
)
,
P
n
(
x
)
is the
Legendre polynomial
of degree n
.
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