Geoscience Reference
In-Depth Information
+
∞
1
w
L
(
x
,
z
,
t
)=
−
d
k
2
πτ
−
∞
exp
{−
ikx
}
(sinh(
kz
)
−
[sinh(
kz
)+tanh(
kH
) cosh(
kz
)] cos(
tp
0
))
sinh(
kH
)
×
X
(
k
)
,
(2.70)
η
0
2sin(
ka
)
k
.
In the case of a running displacement of the basin floor the solution is given by
the following formulae:
where
p
0
=(g
k
tanh(
kH
))
1
/
2
,
X
(
k
)=
+
∞
−
η
0
4
d
k
exp
}
k
cosh(
kH
)
{−
ikx
F
R
(
x
,
z
,
t
)=
π
−
∞
exp
ib
k
+
p
0
v
⎛
g
k
cosh(
kz
)+
p
0
sinh(
kz
)
p
0
−
1
⎝
exp
×
{−
ip
0
t
}
k
+
p
0
v
exp
ib
k
⎞
p
0
v
−
−
1
⎠
,
−
exp
{
ip
0
t
}
(2.71)
p
0
v
k
−
exp
ib
k
+
p
0
v
⎛
+
∞
−
1
η
d
k
exp
}
cosh(
kH
)
{−
ikx
0
⎝
exp
ξ
R
(
x
,
t
)=
{−
ip
0
t
}
k
+
p
0
v
4
π
i
−
∞
exp
ib
k
⎞
⎠
,
p
0
v
−
−
1
+ exp
{
ip
0
t
}
(2.72)
p
0
v
k
−
+
∞
u
R
(
x
,
z
,
t
)=
η
}
cosh(
kH
)
{−
0
i
d
k
exp
ikx
4
π
−
∞
exp
ib
k
+
p
0
v
⎛
⎝
exp
−
1
(g
k
cosh(
kz
)+
p
0
sinh(
kz
))
p
0
×
{−
ip
0
t
}
k
+
p
0
v
exp
ib
k
⎞
p
0
v
−
−
1
⎠
,
−
exp
{
ip
0
t
}
(2.73)
p
0
v
k
−
