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for three-dimensional and
ln
1
r
a
i
1
2
g
i
=
F
i
=∇
(9.63)
2
π
G
(
2
for two-dimensional problems, where
r
=
x
k
−
ξ
)
is the distance between
ξ
and
x
.
k
ξ
=
=
Here,
k
and
x
k
,k
1
,
2
,
3 for three-dimensional problems and
k
1
,
2 for two-dimensional
ξ
problems, are the coordinates of the points
and
x
.
Equations (9.62) and (9.63) can be solved for
g
i
. Note that Eqs. (9.62) and (9.63) are func-
tions of
r
only. Hence, the solutions to Eq. (9.62) and (9.63) will be functions of
r
only, i.e.,
they will be spherically symmetric for three-dimensional problems and rotationally sym-
metric for two-dimensional problems. For the three-dimensional problems, the operator
∇
2
is
r
2
d
dr
1
r
2
d
dr
2
∇
=
so that Eq. (9.62) becomes
r
2
dg
i
dr
1
r
2
d
dr
1
2
g
i
=
∇
=
rG
a
i
4
π
Integration leads to
r
8
C
2
a
i
C
1
r
g
i
=
G
+
+
π
0
,
the displacement should be finite, and so should be
g
i
. Hence,
C
1
must be
equal to zero. Furthermore, since only the derivatives of
g
i
At
r
=
appear in Eq. (9.58), and
differentiation of
g
i
will cancel
C
2
, the constant
C
2
can be set to zero here. Thus,
r
g
i
=
G
a
i
=
ga
i
(9.64)
8
π
2
with
g
=
r
/(
8
π
G
)
. For the two-dimensional case, the operator
∇
is
r
1
r
d
dr
d
dr
2
∇
=
Then Eq. (9.63) appears as
r
dg
i
dr
ln
1
r
a
i
1
r
d
dr
1
2
g
i
=
∇
=
(9.65)
2
π
G
Integrate Eq. (9.65), and process the integration constants in the same manner as for the
three-dimensional case to obtain
G
r
2
ln
1
a
i
1
g
i
=
=
ga
i
(9.66)
8
π
r
with
G
r
2
ln
1
1
g
=
(9.67)
8
π
r
σ
ij
of Eq. (9.52), let
u
ji
be the displacement in the
i
direc-
Similar to the definitions of
δ(ξ
)
tion due to the force
,x
a
j
. Thus the resultant displacements in the
i
direction due to
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