Geology Reference
In-Depth Information
1, we have η
=
α and
b
1
is a null vector, so the system (3.158) is homo-
geneous. Since α
≥
For ν
=
0, only the eigenvalues η
=
α
=
n
−
2andη
=
α
=
n
are admiss-
ible. The corresponding eigenvectors,
x
1
,are
⎝
⎠
⎝
⎠
A
1,1
A
2,0
A
3,1
A
4,0
1
1)μ
1/
n
2μ(
n
2 (
n
−
=
A
1,1
,
(3.165)
−
1)/
n
and
⎝
⎠
⎝
⎠
A
1,1
A
2,0
A
3,1
A
4,0
1/μ
−
np
2
(
n
)/
p
1
(
n
)
q
2
(
n
)
−
q
1
(
n
)
p
2
(
n
)/
p
1
(
n
)
p
2
(
n
)/
p
1
(
n
)
1
=
A
4,0
,
(3.166)
where
A
1,1
and
A
4,0
are arbitrary free constants, determining the two free solutions
for α
=
−
2andα
=
n
, respectively. For ν
=
1, the system (3.143) is also homogen-
eous. Sinceα
≥
0, only the eigenvalueη
=
n
−
1 is admissible, and thus for η
=
n
−
2
and η
=
n
, only the trivial solution
A
5,1
=
A
6,0
=
n
0 exists.
For ν
=
2,
the system (3.158) is homogeneous and non-singular for both
α
=
n
−
2, η
=
n
−
1andα
=
n
, η
=
n
+
1, and it has only the trivial solution
A
1,2
=
A
2,1
=
A
3,2
=
A
4,1
=
0. The system (3.143) becomes
⎝
⎠
⎝
⎠
=
4π
G
ρ
0
⎝
⎠
.
η
+
1
−
1
A
5,2
A
6,1
A
1,1
(3.167)
−
n
1
η
+
2
−
n
1
A
3,1
For ν
=
2, we haveη
=
α
+
1. Then, for the valueα
=
n
of the free solution (3.166),
the coe
cient matrix of this system is non-singular, giving
A
5,2
and
A
6,1
in terms
of
A
1,1
and
A
3,1
for η
=
n
+
1, allowing continuation of the free solution. Solving
for
A
5,2
,wehave
3
(
n
1)
A
3,1
.
2π
G
ρ
0
2
n
A
5,2
=
+
3)
A
1,1
−
n
(
n
+
(3.168)
+
Then,
A
6,1
=
(
n
+
2)
A
5,2
−
4π
G
ρ
0
A
1,1
.
(3.169)
For the eigenvalue η
=
n
−
1ofthecoe
cient matrix, α
=
n
−
2. From the sys-
=
A
1,1
/
n
for α
=
−
tem (3.165),
A
3,1
n
2, and both equations of the system (3.167)
reduce to
3γ
+
2
A
1,1
,
A
6,1
=
nA
5,2
−
4π
G
ρ
0
A
1,1
=
nA
5,2
−
2
Ω
(3.170)
giving rise to a third free solution by choice of
A
6,1
.
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