Geoscience Reference
In-Depth Information
The symbol
G
() denotes Green's function, which can be shown (Brutsaert, 1973) to be in
this case
=−
4
gh
0
−
1
/
2
exp[
d
1
(
x
G
(
ξ,τ
;
x
,
t
)
−
ξ
)
−
d
2
(
t
−
τ
)]
I
0
d
3
t
−
τ
−
1
/
2
⎨
⎩
⎬
⎭
1
/
2
t
−
τ
−
(
x
−
ξ
)
(
x
−
ξ
)
c
01
c
02
×
H
t
−
τ
−
+
(
gh
0
)
1
/
2
|
ξ
−
x
|
c
01
c
02
V
0
(
x
−
ξ
)
c
01
c
02
(5.70)
×
−
I
0
d
3
t
−
τ
−
1
/
2
1
/
2
t
−
τ
−
x
c
01
+
c
02
x
c
02
+
c
01
×
H
t
−
τ
−
x
c
01
+
c
02
The constants in (5.70) are
d
1
=
(
aS
0
/
h
0
);
d
2
=
(
S
0
V
0
/
h
0
)(
a
Fr
0
+
1)
/
Fr
0
; and
d
3
=
(
S
0
V
0
/
h
0
)[(1
−
Fr
0
)(1
−
a
2
Fr
0
)]
1
/
2
/
Fr
0
; the steady uniform Froude number Fr
0
is defined
in Equation (5.63) and the dynamic wave celerities
c
01
and
c
02
in (5.64). The symbol H( )
is the Heaviside step function (see Appendix) and I
0
( ) is the modified Bessel function of
the first kind of order zero.
Example 5.6. Semi-infinite channel with known upstream inflow
In many situations of practical interest, the lateral inflow does not have a large effect on the
solution; therefore, to bring out the most important features of the solution (5.69), in what
follows its simplest form is considered, that is the case
i
=
0. When the lateral flow
i
is
absent, only the second term on the right of (5.69) remains. After carrying out the operations
the result can be given as a simple convolution integral (see Appendix)
t
q
(
x
,
t
)
=
q
0
+
q
u
(
τ
)
u
(
x
,
t
−
τ
)
d
τ
(5.71)
0
As before,
u
(
x
,
t
) denotes the unit response of this channel, that is the flow rate
q
p
(
x
,
t
)at
any time
t
and at any point
x
, resulting from an upstream inflow at
x
=
0 given by a unit
impulse (or Dirac delta function)
q
u
(
t
)
=
δ
(
t
). This can be written as consisting of two parts
u
=
u
1
+
u
2
(5.72)
The first part of
u
is given by
t
−
x
c
01
u
1
=
exp(
−
d
4
x
)
δ
(5.73)
where
(1
−
a
Fr
0
)
Fr
0
+
Fr
0
and the Froude number for steady uniform flow is defined in Equation (5.63). The second
part of
u
in (5.72) is given by
S
0
h
0
d
4
=
x
c
01
−
exp(
d
1
x
d
2
t
)I
1
(
d
3
t
0
)H
t
d
3
2
t
0
x
c
02
x
c
01
u
2
=
−
−
(5.74)
