Environmental Engineering Reference
In-Depth Information
t
t
2
A
2
c
2
r
2
2
2
2
2
c
2
=(
−
A
ω
t
)
+(
i
ct
)
=
ω
−
or
r
=
(
A
ω
)
−
,
d
At
e
i
r
cos ϕ cos θ
=
e
−
i
A
ω
t
cos θ
=
e
−
iωβ
,
sin
θ
d
θ
=
−
,
At
2
1
sin
θ
=
−
cos
2
θ
=
1
−
,
(
sin
θ
≥
0
,
θ
∈
[
0
,
π
])
Substituting these and
J
0
(
i
x
)=
I
0
(
x
)
into Eq. (5.110) yields a very important for-
mula for Fourier transformation
sin
t
c
2
⎛
2
⎞
⎠
e
−
iωβ
d
A
At
(
ω
)
2
−
A
1
2
A
⎝
c
(
t
2
=
I
0
−
β
.
(5.111)
A
ω
)
2
−
c
2
−
At
5.8.2 Fourier Transformation for Three-Dimensional Problems
Consider the PDS
⎧
⎨
u
t
τ
0
+
A
2
R
3
u
tt
=
Δ
u
,
×
(
0
,
+
∞
)
,
(5.112)
⎩
u
(
M
,
0
)=
0
,
u
t
(
M
,
0
)=
ψ
(
M
)
,
R
3
. It is transformed to, by the function transfor-
where
M
stands for point
(
x
,
y
,
z
)
∈
t
2τ
0
,
e
−
mation
u
(
M
,
t
)=
v
(
M
,
t
)
v
tt
=
A
2
c
2
v
1
R
3
Δ
v
+
,
c
=
τ
0
,
×
(
0
,
+
∞
)
,
2
(5.113)
v
(
M
,
0
)=
0
,
v
t
(
M
,
0
)=
ψ
(
M
)
.
Applying a triple Fourier transformation to PDS (5.113) yields an initial-value prob-
lem of the ordinary differential equation
v
tt
(
ω
,
A
2
2
c
2
t
)+(
ω
−
)
v
(
ω
,
t
)=
0
,
v
(
ω
,
0
)=
0
,
v
t
(
ω
,
0
)=
ψ
(
ω
)
,
¯
2
1
2
3
. Its solution is
where ¯
ψ
(
ω
)=
F
[
ψ
(
M
)]
and
ω
=
ω
+
ω
+
ω
sin
t
c
2
2
(
A
ω
)
−
v
(
ω
,
t
)=
ψ
(
ω
)
.
¯
(
A
ω
)
2
−
c
2
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