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Again
A
5,4
is given by (3.185),
A
6,3
is given by (3.186), and
A
1,5
,
A
2,4
,
A
3,5
,
A
4,4
are
given by the system (3.139) in terms of
A
1,3
,
A
3,3
,
A
6,3
,
A
5,4
for ν
=
5, η
=
n
+
2,
while
A
5,6
,
A
6,5
are given by the system (3.143) in terms of
A
1,5
,
A
3,5
for ν
=
6,
η
=
3.
The fundamental solution generated by the free constant
A
4,0
is
n
+
1
μ
−
A
4,0
n
p
2
(
n
)
p
1
(
n
)
A
1,3
r
2
A
1,5
r
4
z
1
(
r
)
=
+
+
+···
,
(3.200)
q
2
(
n
)
A
4,0
q
1
(
n
)
p
2
(
n
)
p
1
(
n
)
A
2,2
r
2
A
2,4
r
4
=
−
+
+
+···
,
z
2
(
r
)
(3.201)
p
2
(
n
)
p
1
(
n
)
A
4,0
A
3,3
r
2
A
3,5
r
4
z
3
(
r
)
=
+
+
+···
,
(3.202)
A
4,2
r
2
A
4,4
r
4
z
4
(
r
)
=
A
4,0
+
+
+···
,
(3.203)
A
5,4
r
2
A
5,6
r
4
z
5
(
r
)
=
A
5,2
+
+
+···
,
(3.204)
A
6,3
r
2
A
6,5
r
4
z
6
(
r
)
=
A
6,1
+
+
+···
.
(3.205)
A
5,2
and
A
6,1
are given in terms of
A
1,1
and
A
3,1
by (3.168) and (3.169). Next
A
1,3
,
A
2,2
,
A
3,3
,
A
4,2
are given by the system (3.139) in terms of
A
1,1
,
A
3,1
,
A
6,1
,
A
5,2
for
ν
=
3, η
=
n
+
2, while
A
5,4
,
A
6,3
are given by the system (3.143) in terms of
A
1,3
,
A
3,3
for ν
=
3. Then
A
1,5
,
A
2,4
,
A
3,5
,
A
4,4
are given by the system (3.139)
in terms of
A
1,3
,
A
3,3
,
A
6,3
,
A
5,4
for ν
=
4, η
=
n
+
5, η
=
n
+
4, while
A
5,6
,
A
6,5
are given by
the system (3.143) in terms of
A
1,5
,
A
3,5
for ν
=
6, η
=
n
+
5.
Now consider the special case of
n
=
1. From (3.146) and (3.147),
A
1,0
=
A
3,0
for
n
=
1, α
=
0. For ν
=
1, we have η
=
α
=
0 and from (3.146)
A
5,0
=
0.
=
=
=
Thus, the system (3.139) is homogeneous and non-singular and
A
1,1
A
2,0
A
3,1
A
4,0
=
0. The system (3.143), for ν
=
1, becomes
⎝
⎠
⎝
⎠
=
4π
G
ρ
0
⎝
⎠
.
η
+
1
−
1
A
5,1
A
6,0
A
1,0
(3.206)
−
n
1
η
+
2
−
n
1
A
3,0
The coe
cient matrix is singular for η
=
n
−
1
=
0. For
n
=
1, we have
A
1,0
=
A
3,0
,
and for η
=
0 both equations reduce to
3γ
+
2
A
1,0
.
A
6,0
=
A
5,1
−
4π
G
ρ
0
A
1,0
=
A
5,1
−
2
Ω
(3.207)
For ν
=
2, α
=
0, η
=
1
=
n
, the system (3.158) is singular. With η
=
n
=
1,
=
p
1
(1)
r
1
(1),
(3.208)
p
2
(1)
=
r
2
(1).
(3.209)
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