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 + 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
=
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 ,
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
G ρ 0 A 1,3 .
(3.186)
 
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