Geoscience Reference
In-Depth Information
•
For
|
x
|
a
1
ω
j
=1
Q
exp
{−
k
j
a
}
cosh(
k
j
x
)
∞
F
osc
(
x
,
z
,
t
)
H
= exp
{
i
ω
t
}
2
+
z
−
2
η
0
i
ω
+
P
(exp
{
i
(
ω
t
+
k
0
(
x
−
a
))
}
+ exp
{
i
(
ω
t
−
k
0
(
x
+
a
))
}
)
,
(2.92)
•
For
x
a
2
j
=1
Q
exp
{−
k
j
x
}
sinh(
k
j
a
)
∞
F
osc
(
x
,
z
,
t
)
H
= exp
{
i
ω
t
}
η
0
i
ω
+
P
(
−
exp
{
i
(
ω
t
−
k
0
(
x
−
a
))
}
+ exp
{
i
(
ω
t
−
k
0
(
x
+
a
))
}
)
(2.93)
•
For
x
−
a
2
j
=1
Q
exp
{
k
j
x
}
sinh(
k
j
a
)
∞
F
osc
(
x
,
z
,
t
)
H
= exp
{
i
ω
t
}
η
0
i
ω
+
P
(exp
{
i
(
ω
t
+
k
0
(
x
−
a
))
}−
exp
{
i
(
ω
t
+
k
0
(
x
+
a
))
}
)
,
(2.94)
where
2
sinh(
k
0
z
)
ω
k
0
cosh(
k
0
z
)+
P
=
k
0
cosh(
k
0
))
,
k
0
((
ω
2
−
−
1) sinh(
k
0
)
2
sin(
k
j
z
)
k
j
cos(
k
j
z
)+
ω
Q
=
k
j
cos(
k
j
))
.
k
j
((
ω
2
−
1) sin(
k
j
)
−
With knowledge of the velocity potential of the flow it is not difficult to obtain
expressions for the displacement of a free surface and for the velocity components:
•
|
|
For
x
a
1
j
=1
Q
exp
{−
k
j
a
}
cosh(
k
j
x
)
∞
2
ξ
osc
(
x
,
t
)=
η
0
exp
{
i
ω
t
}
−
2
ω
2
P
(exp
+
η
0
ω
{
i
(
ω
t
+
k
0
(
x
−
a
))
}
+ exp
{
i
(
ω
t
−
k
0
(
x
+
a
))
}
)
,
(2.95)
•
For
x
a
∞
j
=1
Q
exp
{−
k
j
x
}
sinh(
k
j
a
)
η
0
e
i
ω
t
2
2
ξ
osc
(
x
,
t
)=
ω
2
P
(
+
η
0
ω
−
exp
{
i
(
ω
t
−
k
0
(
x
−
a
))
}
+ exp
{
i
(
ω
t
−
k
0
(
x
+
a
))
}
)
,
(2.96)
