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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
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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