Environmental Engineering Reference
In-Depth Information
To demonstrate this, consider two groups of initial values
⎧
⎨
⎧
⎨
(
,
)=
ϕ
(
)
,
u
1
x
0
x
u
2
(
x
,
0
)=
ϕ
2
(
x
)
,
1
and
∂
∂
∂
∂
⎩
⎩
t
u
1
(
x
,
0
)=
ψ
1
(
x
)
t
u
2
(
x
,
0
)=
ψ
2
(
x
)
,
where, for a small positive constant
δ
,
−
∞
<
x
<
+
∞
|
ϕ
1
(
max
x
)
−
ϕ
2
(
x
)
| <
δ
,
−
∞
<
x
<
+
∞
|
ψ
1
(
max
x
)
−
ψ
2
(
x
)
| <
δ
.
By Eq. (2.62), the difference between the two corresponding solutions
u
1
and
u
2
satisfies
x
+
at
)
|≤
2
+
2
+
1
2
a
|
u
1
(
x
,
t
)
−
u
2
(
x
,
t
|
ψ
1
(
ξ
)
−
ψ
2
(
ξ
)
|
d
ξ
x
−
at
1
2
a
δ
≤
δ
+
2
at
=(
1
+
t
)
δ
.
Therefore,
∀
ε
>
0, we can always find a time instant
t
=
t
0
satisfying
δ
=
ε
/
(
1
+
t
0
)
such that
|
u
1
−
u
2
| <
ε
.
2.7.3 Physical Meaning
To understand each term in Eq. (2.62), take
u
(
x
,
t
)
as the displacement of a vibrating
string. For any fixed time instant
t
0
,
u
represents the spatial distri-
bution of displacement. Its graphic representation shows the string shape at
t
0
,i.e.
the wave shape. For any fixed point
x
0
,
u
=
u
(
x
,
t
0
)=
g
1
(
x
)
=
u
(
x
0
,
t
)=
g
2
(
t
)
illustrates the temporal
distribution of displacement at
x
0
.Let
)=
ϕ
(
2
,
x
)=
ψ
(
)
2
a
.
x
Ψ
(
Φ
(
x
x
(2.65)
Equation (2.62) becomes
u
(
x
,
t
)=[
Φ
(
x
+
at
)+
Φ
(
x
−
at
)]+ [
Ψ
(
x
+
at
)+(
−
Ψ
(
x
−
at
))]
=
u
1
(
x
,
t
)+
u
2
(
x
,
t
)
.
Here
Φ
(
x
±
at
)
and
Ψ
(
x
±
at
)
represent the periodic prolongation of
ϕ
(
x
)
and
ψ
(
x
)
, respectively. Since
at
has the dimension of a displacement, the function
=
ϕ
(
x
−
at
)
=
ϕ
(
x
)
u
represents the
u
(the wave shape at
t
=
0) displaced
at
units
2
2
=
ϕ
(
x
−
at
)
withawavespeed
a
to the right.
u
is thus a forward wave. Similarly
2
ϕ
(
x
+
at
)
represents a backward wave. Therefore we have:
2
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