Geoscience Reference
In-Depth Information
∞
s
+
i
∞
∞
n
=1
J
n
(
kr
)(
C
3
cos(
n
ϕ
)+
C
4
sin(
n
ϕ
))
+
d
k
d
p
exp
{
pt
}
0
s
−
i
∞
(
C
5
(
p
,
k
) cosh(
kz
)+
C
6
(
p
,
k
) sinh(
kz
))
.
×
(2.48)
Substitution of equation (2.48) into the boundary condition on the surface (2.30)
yields the relationship between the coefficients,
C
5
(
p
,
k
)
p
2
C
6
(
p
,
k
)=
−
gk
.
(2.49)
We shall now write the integral representation for the function describing
the space-time law of motion of the basin floor,
r
(
r
)
η
ϕ
(
t
(
t
)
η
(
r
,
ϕ
,
t
)=
η
ϕ
)
η
,
t
)=
∞
s
+
i
∞
H
0
(
p
,
k
)+
∞
s
+
i
∞
J
0
(
kr
)
k
A
0
2
η
(
r
,
ϕ
d
k
dp
exp
{
pt
}
d
k
dp
0
s
−
i
∞
0
s
−
i
∞
∞
n
=1
J
n
(
kr
)
k
(
A
n
cos(
n
ϕ
)+
B
n
sin(
n
ϕ
))
H
n
(
p
,
k
)
,
×
exp
{
pt
}
(2.50)
where
π
1
π
η
ϕ
(
A
n
=
ϕ
) cos(
n
ϕ
) d
ϕ
,
−
π
π
1
π
η
ϕ
(
ϕ
ϕ
ϕ
,
B
n
=
) sin(
n
) d
−
π
∞
∞
1
2
H
n
(
p
,
k
)=
r
(
r
)
t
(
t
) exp
d
t
d
r
η
η
{−
pt
}
rJ
n
(
kr
)
.
π
i
0
0
Substitution of formulae (2.48) and (2.50) into the boundary condition on the basin
floor (2.31) reveals that equality of the left-hand and right-hand parts is possible,
only when the following three conditions are fulfilled:
C
3
=
A
n
,
C
4
=
B
n
,
pH
n
(
p
,
k
)
k
sinh(
kH
)+
p
2
C
5
(
p
,
k
)=
−
.
g
k
cosh(
kH
)
It is now possible to write out the resultant expression for the potential, which is
the solution of equation (2.43) with the boundary conditions (2.30) and (2.31)