Environmental Engineering Reference
In-Depth Information
Fig. 2.1
Wave shapes
x
x
0
ψ
(
ξ
)
1
4
2. By (2.65),
Ψ
(
x
)=
d
ξ
,
x
0
≤
x
1
. Thus
⎧
⎨
0
,
x
<
x
1
,
1
Ψ
(
)=
4
(
x
−
x
1
)
ψ
0
,
x
1
<
x
<
x
2
,
x
⎩
1
4
(
x
2
−
x
1
)
ψ
0
,
x
2
≤
x
.
The solution of PDS (2.66) reads
x
+
2
t
1
4
u
(
x
,
t
)=
ψ
(
ξ
)
d
ξ
=
Ψ
(
x
+
2
t
)
−
Ψ
(
x
−
2
t
)
.
x
−
2
t
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