Geology Reference
In-Depth Information
For ν
=
3, we have η
=
α
+
2 and the eigenvalue η
=
n
results in a singular
coe
cient matrix in the system (3.158) for α
=
n
−
2. For η
=
n
,wehave
2
n
n
(
n
n
2
1
μ
,
p
1
(
n
)
=
+
2)λ
+
+
2
n
−
(3.171)
2
n
(
n
+
n
2
1
μ
,
r
1
(
n
)
=
2)λ
+
+
2
n
−
(3.172)
p
2
(
n
)
=
n
(
n
+
5)
+
n
(
n
+
3)λ/μ,
(3.173)
r
2
(
n
)
=
n
+
5
+
(
n
+
3)λ/μ.
(3.174)
Thus, the left side of the second equation in the system (3.158) is just
n
times the
left side of the fourth equation. Again,
A
3,1
2 and, substituting
from (3.170) for
A
6,1
, the right side of the second equation becomes
ρ
0
n
γ
−
ω
=
A
1,1
/
n
for α
=
n
−
2
m
ω
Ω
/
n
A
1,1
nA
5,2
,
2
+
−
(3.175)
while the right side of the fourth equation becomes
ρ
0
n
γ
−
ω
2
m
ω
Ω
/
n
A
1,1
/
n
A
5,2
.
2
+
−
(3.176)
Thus, the whole of the second equation in the system (3.158) is just
n
times the
fourth. Dropping the fourth equation, and replacing
A
3,1
by
A
1,1
/
n
,forα
=
n
−
2,
the system can be rearranged to
⎝
⎠
⎝
⎠
01
q
1
(
n
)
00
A
1,3
A
2,2
A
3,3
p
1
(
n
)/ρ
0
10
n
−
⎝
⎠
q
2
(
n
)
A
4,2
(
3
2
m
ω
Ω
/
n
A
1,1
−
A
6,1
−
p
2
(
n
)
A
4,2
/ρ
0
A
4,2
/μ
2
2
=
−
−
n
)
γ
+
ω
+
Ω
−
. (3.177)
2
Given the three free constants
A
1,1
and
A
6,1
,forα
=
n
−
2and
A
4,2
,forα
=
n
,the
second equation of this system yields,
p
2
(
n
)
p
1
(
n
)
A
4,2
+
A
3,3
=
p
1
(
n
)
(3
A
6,1
.
2
m
ω
Ω
/
n
A
1,1
ρ
0
2
2
−
n
)γ
+
ω
+
2
Ω
−
+
(3.178)
Then, the first equation of the system gives
A
2,2
=−
q
1
(
n
)
A
3,3
+
q
2
(
n
)
A
4,2
,
(3.179)
and the third equation gives
1
μ
A
1,3
=−
nA
3,3
+
A
4,2
.
(3.180)
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