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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