Geoscience Reference
In-Depth Information
Division of the second of Equations (5.52) by
c
0
followed by integration and subsequent
combination with the first of (5.52), yields the following form for the two functions
x
m
1
2
1
2
c
0
F
1
(
x
m
)
=
f
(
x
m
)
−
g
(
s
)
ds
x
m0
(5.53)
x
m
1
2
1
2
c
0
F
2
(
x
m
)
=
f
(
x
m
)
+
g
(
s
)
ds
x
m0
in which
s
is the dummy integration variable. Thus the solution (5.51) can be written for
this case as
x
m
+
c
0
t
m
1
2
1
2
c
0
y
=
[
f
(
x
m
−
c
0
t
m
)
+
f
(
x
m
+
c
0
t
m
)]
+
g
(
s
)
ds
(5.54)
x
m
−
c
0
t
m
which is commonly attributed to d'Alembert. This result (5.54) is equally applicable to the
water surface depth
h
p
, the velocity
V
p
and the rate of flow
q
p
. However, in each case the
functions
f
(
x
m
) and
g
(
x
m
) should represent the initial conditions of the intended variable.
Example 5.4. Infinitely long channel with zero initial time derivative
If the value of the time derivative of the variable in question is initially equal to zero, or
g
(
x
m
)
=
0, d'Alembert's solution (5.54) becomes simply
1
2
1
2
y
=
f
(
x
m
−
c
0
t
m
)
+
f
(
x
m
+
c
0
t
m
)
(5.55)
As an illustration, consider the following function to describe the initial disturbance,
f
(
x
m
)
=
α
1
+
10
x
m
−
1
(5.56)
In this expression
α
is a constant, which should be large to ensure that the perturbation
is small, compared to the undisturbed part of the variable, i.e.
h
0
,
V
0
or
q
0
, describing the
steady uniform flow. With this initial condition, the solution is, in accordance with Equation
(5.55),
1
10(
x
m
+
c
0
t
m
)
2
−
1
10(
x
m
−
c
0
t
m
)
2
−
1
+
1
1
2
α
y
(
x
m
,
t
m
)
=
+
+
(5.57)
This solution is shown in Figure 5.7 for the values
c
0
t
m
=
0
,
0.2, 0
.
4
,
0.8 and 1.2, and
illustrates the propagation of the disturbance with time, relative to the reference point
x
m
=
0, which is moving with the velocity
V
0
.
If the initial disturbance is a unit impulse
y
=
δ
(
x
m
) (see Appendix) (and
∂
y
/∂
t
=
0),
Equation (5.55) yields the unit response function for an infinitely long channel as
1
2
u
=
[
δ
(
x
m
−
c
0
t
m
)
+
δ
(
x
m
+
c
0
t
m
)]
or, in the original coordinate system
1
2
u
=
[
δ
(
x
−
(
V
0
+
c
0
)
t
)
+
δ
(
x
−
(
V
0
−
c
0
)
t
)]
(5.58)
This describes a translatory motion of two delta functions, one with, and one against the
flow
V
0
. Equation (5.56) is not a delta function, but Figure 5.7 gives an idea of how the two
unit impulses in (5.58) proceed.
