Geology Reference
In-Depth Information
In addition to the free constants
A
1,1
,
A
4,0
and
A
6,1
already encountered, this system
appears to introduce a fourth free constant,
A
4,2
. This fourth free constant generates
the solution
⎝
⎠
⎝
⎠
A
1,3
A
2,2
A
3,3
A
4,2
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
4,2
,
(3.181)
for α
=
2. Replacing the free constant
A
4,2
with
A
4,0
, this is identical to the
second of the eigenvectors (3.166) found for α
=
n
−
n
. Thus, no new solution arises
from the eigenvalueη
=
n
givingα
=
n
−
2forν
=
3, and we may set
A
4,2
=
0 without
loss of generality.
Using relations (3.178) through (3.180), a third eigenvector,
x
3
,forα
=
n
−
2,
associated with the free constant
A
6,1
, is found to be
⎝
⎠
⎝
⎠
A
1,3
A
2,2
A
3,3
A
4,2
−
n
−
q
1
(
n
)
1
0
ρ
0
p
1
(
n
)
=
A
6,1
.
(3.182)
Another eigenvector,
x
3
, associated with the free constant
A
1,1
, also appears, hav-
ing the form, determined by relations (3.178) through (3.180),
⎝
⎠
⎝
⎠
A
1,3
A
2,2
A
3,3
A
4,2
−
n
p
1
(
n
)
(3
2
m
ω
Ω
/
n
−
q
1
(
n
)
1
0
ρ
0
2
2
=
−
n
)γ
+
ω
+
2
Ω
−
A
1,1
.
(3.183)
This is just the next member of the sequence of eigenvectors beginning with (3.165).
To complete preparation for the next recurrence, we consider the system (3.143) for
ν
=
4, η
=
α
+
3. With α
=
n
−
2andη
=
n
+
1, the system (3.143) becomes
⎝
⎠
⎝
⎠
=
4π
G
ρ
0
⎝
⎠
.
n
+
2
−
1
A
5,4
A
6,3
A
1,3
(3.184)
−
n
1
n
+
3
−
n
1
A
3,3
This can easily be solved to give
3)
(
n
1)
A
3,3
,
4π
G
ρ
0
2 (2
n
+
A
5,4
=
+
3)
A
1,3
−
n
(
n
+
(3.185)
and
A
6,3
=
(
n
+
2)
A
5,4
−
4π
G
ρ
0
A
1,3
.
(3.186)
Search WWH ::
Custom Search