Introduction and theoretical background - Earth Dynamics: Deformations and Oscillations of the Rotating Earth - page 56

Solving (1.369) and (1.370) gives

P
2π ,

P
4π(λ + μ) .

A = B =− (1.371)

Substituting these values in (1.345) and (1.352) gives the total displacement field
for the Boussinesq problem as

⎣ ⎦ ,

x 1

x 1 −

P
4πμ

x 3
R 0 −

μ
λ + μ

1

u 1 = (1.372)

( R 0 +

x 3 ) R 0

⎣ ⎦ ,

x 2 x 2 −

P
4πμ

R 0 − μ

x 3 1

u 2 = (1.373)

R 0 λ + μ ( R 0 + x 3 )

⎣ ⎦ .

x 3

P
4πμ

1
R 0

λ + 2μ

λ + μ

u 3 =

R 0 +

(1.374)

The Galerkin vector for the Boussinesq problem can be found by adding the
Galerkin vector G AR 0 e 3 to the Galerkin vector corresponding to the extra dis-

placement field (1.352). The former amounts to
P
2π

=

R 0 e 3 , (1.375)

while the latter is

R 0 log( R 0

x 3 e 3 2μ B λ + μ

λ − μ

− + x 3 ) −

R 0 log( R 0

x 3 e 3 ,

P
2π

μ
λ − μ

= + x 3 ) − (1.376)

as can be shown with the aid of the vector identities

R 0 log( R 0

x 3 e 3 ∇· + x 3 ) − = 3 log( R 0 + x 3 ), (1.377)

and

2 R 0 log( R 0

x 3 e 3

2 R 0 +

R 0 e 3

∇ + x 3 ) − =

x 3 ) .

(1.378)

R 0 ( R 0 +

The total Galerkin vector for the Boussinesq problem is then

R 0 e 3

x 3 e 3

R 0 log( R 0

P
2π

μ
λ − μ

G = + + x 3 ) − . (1.379)

1.4.10 Cerruti's problem
When the concentrated force is applied tangentially at x 1 , x 2 ,0 on the free surface,
x 3 0, of an elastic half-space, the problem is associated with the name of Cerruti,

who first solved it in 1882.

=