Geology Reference
In-Depth Information
Now let us assume that the potentials
A
p
and
A
s
have the same direction of
n
:
A
p
D
A
p
n
,
A
s
D
A
s
n
. In this instance, the equations (
10.28
) reduce
to just two standard Poisson's equations:
(
r
2
r
A .r/
D
u
.r/
(10.22)
In fact, (
10.22
) is a classic
Poisson's equation
,
which has a unique solution when
u
decreases
with sufficient rapidity (at least as 1/
r
)for
r
!
1
:
2
A
p
D
1
4 r.œ
C
2/
(10.29)
2
A
s
D
1
4 r
r
Z
u
.r
0
/
j
r
r
0
j
1
4
dV
0
A .r/
D
(10.23)
These equations are easily integrated taking
into account that
r
R
3
2
r
D
2/
r
. Therefore, the so-
lutions have the form:
(
A
p
D
Therefore, using the vector identity (
10.16
)for
the field
A
gives:
r
8 .œC2/
(10.30)
2
r
8
u
Dr
A
Dr
.
r
A/
r
.
r
A/
(10.24)
A
s
D
To determine the displacement, we must mul-
tiply (
10.30
)by
n
and insert the vector potentials
A
p
and
A
s
into (
10.21
). It is convenient to set
n
D
e
j
and determine the
i
-th component of the
displacement field associated with a force at the
origin in the direction
e
j
. We shall indicate this
quantity by
u
i
:
Now, setting:
¥
Dr
A
I
‰
Dr
A
(10.25)
we obtain:
u
Dr
¥
Cr
‰
(10.26)
•
ij
.2
/
1
(10.31)
1
8
r
C
x
i
x
j
u
i
D
This equation implies that a vector field
u
D
u
(
r
) always has an irrotational component
(
r
¥
) and a solenoidal component (
r
‰ ).
The fields ¥ and ‰ are termed respectively
the
scalar potential
and the
vector potential
of
the displacement field, and (
10.26
) is known
as
Helmholtz's decomposition theorem
.Now,
let us consider two vector fields,
A
p
and
A
s
,
such that
A
D
A
p
C
A
s
. In order to satisfy
(
10.21
), we must have that:
r
A
p
D
0
and
r
A
s
D
0.
r
3
where:
œ
C
œ
C
2
(10.32)
The quantities
u
i
form a symmetric tensor that
is known as the
Somigliana tensor
. To visualize
the pattern of deformation, it is usually conve-
nient to express the components of displacement
in spherical coordinates applying the following
transformation (see Appendix I):
Therefore, by
(
10.16
)wehave:
2
A
s
Dr
(
r
A
s
).
Substituting (
10.21
)into(
10.20
) and taking into
account that
r
2
A
p
Dr
(
r
A
p
) d
r
r
2
A
Dr
2
A
p
Cr
2
A
s
leads to:
2
4
3
5
D
2
4
3
5
2
3
u
r
u
™
u
¥
u
1
u
2
u
j
3
n
h
A
p
io
sin ™ cos
¥
sin ™ sin
¥
cos™
cos ™ cos¥ cos™ sin ¥
sin ™
sin ¥
n
2
r
r
4 r
C
.œ
C
2/
r
4
5
Crr
n
A
s
D
0 (10.27)
cos ¥
0
2
4 r
r
(10.33)
This equation is clearly satisfied when:
(
r
For example, for a force applied in the
x
1
direction, it is easy to calculate the components of
the displacement field in the plane
x
1
x
3
(¥
D
0).
2
n
4 r
.
œ
A
p
D
2
/
C
(10.28)
2
A
s
D
n
4 r
r
