Geoscience Reference
In-Depth Information
v(
x
,
y
,
z
,
t
)
≡ {
u
(
x
,
y
,
z
,
t
)
,
v
(
x
,
y
,
z
,
t
)
,
w
(
x
,
y
,
z
,
t
)
}
=
∇
F
(
x
,
y
,
z
,
t
)
.
(2.33)
The Laplace equation (2.29) is resolved by the standard method of separation of
variables. We shall omit elementary calculations and write out the general solution
of the problem in the form of Laplace and Fourier expansions over the time and
space coordinates:
s
+
i
∞
+
∞
+
∞
F
(
x
,
y
,
z
,
t
)=
{
−
−
}
d
p
d
m
d
n
exp
pt
imx
iny
−
∞
−
∞
s
−
i
∞
×
(
A
(
p
,
m
,
n
) cosh(
kz
)+
B
(
p
,
m
,
n
) sinh(
kz
))
,
(2.34)
where
k
2
=
m
2
+
n
2
.
Substitution of the general solution (2.34) into the boundary condition on the sur-
face (2.30) yields the relationship between the coefficients:
A
(
p
,
m
,
n
)
p
2
B
(
p
,
m
,
n
)=
−
g
k
.
(2.35)
Applying the formulae for the direct and inverse Laplace and Fourier transfor-
mations we obtain the integral representation for the laws, satisfied by motion of
the basin floor:
s
+
i
∞
+
∞
+
∞
1
η
(
x
,
y
,
t
)=
d
p
d
m
d
n
exp
{
pt
−
imx
−
iny
}
H
(
p
,
m
,
n
)
,
(2.36)
π
3
i
8
−
∞
−
∞
s
−
i
∞
where
∞
+
∞
+
∞
H
(
p
,
m
,
n
)=
d
t
d
x
dy
exp
{−
pt
+
imx
+
iny
}
η
(
x
,
y
,
t
)
.
(2.37)
0
−
∞
−
∞
Substituting expression (2.34), written with the aid of formula (2.35), into
the boundary condition on the basin floor (2.31), one can calculate the coefficient
A
(
p
,
m
,
n
). As a result, one obtains the following expression for the potential of
the flow velocity, which corresponds to motions of the basin floor, satisfying the law
η
(
x
,
y
,
t
).
s
+
i
∞
+
∞
+
∞
1
F
(
x
,
y
,
z
,
t
)=
−
d
p
d
m
d
n
8
π
3
i
s
−
i
∞
−
∞
−
∞
p
2
tanh(
kz
)
k
cosh(
kH
)(g
k
tanh(
kH
)+
p
2
)
cosh(
kz
)
g
k
p
exp
{
pt
−
imx
−
iny
}
−
×
H
(
p
,
m
,
n
)
.
(2.38)
Applying formulae (2.32) and (2.33), we obtain expressions describing the be-
haviour of the free surface,
