Earth deformations - Earth Dynamics: Deformations and Oscillations of the Rotating Earth - page 242

From (3.207), A 5,1 = A 6,0 for A 1,0 = 0. The power series expansions for this regular

solution have the forms

= A 1,2 r 2 + A 1,4 r 4 z 1 ( r ) = y 1 ( r ) +··· , (3.221)

= A 2,1 r 2 + A 2,3 r 4 z 2 ( r ) = r y 2 ( r ) +··· , (3.222)

= A 3,2 r 2 + A 3,4 r 4 z 3 ( r ) = y 3 ( r ) +··· , (3.223)

= A 4,3 r 4 z 4 ( r ) = r y 4 ( r ) +··· , (3.224)

= y 5 ( r )/ r = A 6,0 + A 5,3 r 2 + A 5,5 r 4 z 5 ( r ) +··· , (3.225)

= A 6,0 + A 6,2 r 2 + A 6,4 r 4 z 6 ( r ) = y 6 ( r ) +··· . (3.226)

For the regular solution generated by the free constant A 4,1 , from equations
(3.210) through (3.212), we have
⎝ ⎠ ⎝ ⎠

A 1,2
A 2,1
A 3,2
A 4,1

1/μ − p 2 (1)/ p 1 (1)

q 2 (1) − q 1 (1) p 2 (1)/ p 1 (1)

p 2 (1)/ p 1 (1)
1

= A 4,1 . (3.227)

From (3.207), A 5,1 = 0for A 1,0 = A 6,0 = 0. The power series expansions for this

regular solution have the forms

= y 1 ( r )/ r 2 A 1,4 r 2 A 1,6 r 4 z 1 ( r ) = A 1,2 + + +··· , (3.228)

A 2,3 r 2 A 2,5 r 4 z 2 ( r ) = y 2 ( r )/ r = A 2,1 + + +··· , (3.229)

= y 3 ( r )/ r 2 A 3,4 r 2 A 3,6 r 4 z 3 ( r ) = A 3,2 + + +··· , (3.230)

A 4,3 r 2 A 4,5 r 4 z 4 ( r ) = y 4 ( r )/ r = A 4,1 + + +··· , (3.231)

= y 5 ( r )/ r 3 A 5,5 r 2 A 5,7 r 4 z 5 ( r ) = A 5,3 + + +··· , (3.232)

= y 6 ( r )/ r 2 A 6,4 r 2 A 6,6 r 4 z 6 ( r ) = A 6,2 + + +··· . (3.233)

Higher terms in the power series expansions for all three regular solutions are
found from the recurrence relations provided by the systems (3.143) and (3.139).
For ν = 3, η = 2, the system (3.143) gives

⎝

⎠ ⎝ ⎠ =

4π G ρ 0 ⎝

⎠ .

3 − 1 A 5,3

A 6,2

A 1,2

(3.234)

− n 1 4 − n 1 A 3,2

This can be easily solved to give

4π G ρ 0
10

4 A 1,2 −

2 A 3,2 ,

A 5,3 = (3.235)

and

A 6,2 = 3 A 5,3 − 4π G ρ 0 A 1,2 . (3.236)