Environmental Engineering Reference
In-Depth Information
where x
=
cos
θ
,
1
x
1. Note that
{
P n (
x
) }
is orthogonal in
[
1
,
1
]
. Thus we
have
1
1
v 0
2 .
C 0 P 0 (
)
=
,
=
x
d x
v 0 d x
C 0
1
0
Also
1
2 R n 1
+
2 n
1
v 0
P n + 1 (
P n 1 (
C n =
v 0 P n (
x
)
d x
=
0 [
x
)
x
)]
d x
2 R n
0
v 0
2 R n [
=
P n 1 (
0
)
P n + 1 (
0
)] .
2 R 1
2 =
v 0
3 v 0
1
2 . Therefore C 1 =
1
Note that P 0 (
0
)=
0
,
P 2 (
0
)=
+
4 R .
Since
n 1
·
3
·
5
···· (
2 n
1
)
P 2 n + 1 (
0
)=
0
,
P 2 n (
0
)=(
1
)
,
2
·
4
·
6
·····
2 n
we obtain
C 2 k =
0
,
k
=
1
,
2
, ··· .
When n
=
2 k
+
1
(
k
=
1
,
2
, ··· )
,
v 0
2 R 2 k + 1 [
C 2 k + 1 =
P 2 k (
0
)
P 2 k + 2 (
0
)]
2 R 2 k + 1
v 0
k 1
·
3
·
5
···· (
2 k
1
)
1 1
·
3
·
5
···· (
2 k
+
1
)
k
+
=
(
1
)
2 k (
1
)
2
·
4
·
6
····
2
·
4
·
6
···· (
2 k
+
2
)
k v 0
2 R 2 k + 1
= (
1
)
1
·
3
·
5
···· (
2 k
1
)
4 k
+
3
2 .
2
·
4
·
6
····
2 k
2 k
+
Therefore the solution of PDS (7.61) is
1
2 r
P 1 (
k = 1 ( 1 )
v 0
2
3
v 0
2
k
T
(
r
, θ )=
+
cos
θ )
+
R
r
R
2 k + 1
·
·
···· (
)
+
1
3
5
2 k
1
4 k
3
·
P 2 k + 1 (
cos
θ ) .
2
·
4
·
6
····
2 k
2 k
+
2
Remark . The Laplace operator of two and three dimensions plays an important role
in potential equations. Its form depends on the coordinate system. Table 7.1 lists the
Laplace operator in some typical coordinate systems.
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