Environmental Engineering Reference
In-Depth Information
2τ
0
t
e
−
+
v
1
(
ξ
,
η
,
ζ
,
t
)
ψ
(
x
−
ξ
,
y
−
η
,
z
−
ζ
)
d
ξ
d
η
d
ζ
R
3
⎤
t
t
2τ
0
d
e
−
⎦
.
+
τ
v
1
(
ξ
,
η
,
ζ
,
t
)
f
(
x
−
ξ
,
y
−
η
,
z
−
ζ
,
t
−
τ
)
d
ξ
d
η
d
ζ
0
R
3
(5.116)
Remark.
By the definition of Fourier transformation, we have
L
3
L
−
3
[
d
x
d
y
d
z
]=
,
[
d
ω
1
d
ω
2
d
ω
3
]=
,
Thus, in Eq. (5.114) ,
sin
t
c
2
⎡
⎤
2
(
A
ω
)
−
⎣
⎦
[
TL
−
3
[
v
1
(
M
,
t
)] =
(
d
ω
1
d
ω
2
d
ω
3
]=
.
A
ω
)
2
−
c
2
The unit of the first term in Eq. (5.116) is
1
τ
0
+
∂
[
v
1
][
ϕ
][
d
ξ
d
η
d
ζ
]=
Θ
=[
u
]
.
∂
t
It is straightforward to show that the unit of the second term in Eq. (5.116) is also
Θ
T
−
2
. Note that
[
f
]=
Θ
.
The unit of the third term in Eq. (5.116) is thus
[
d
τ
][
v
1
][
f
][
d
ξ
d
η
d
ζ
]=
Θ
=[
u
]
.
5.8.3 Method of Spherical Means for PDS (5.115)
Since the equation in PDS (5.115) is hyperbolic, its solution has wavelike properties.
We may obtain its solution by the method of spherical means which we have used to
obtain the Poisson formula of three-dimensional wave equations in Section 2.8. By
the solution structure theorem, we can focus our attention on PDS (5.113) to find
the solution of PDS (5.115).
Consider now a sphere
V
r
of center
M
and radius
r
. Its spherical surface is
denoted by
S
r
. The mean value of
V
on
S
r
(
M
,
t
)
is
r
2
S
r
1
M
,
v
(
r
,
t
)=
v
(
t
)
d
S
.
(5.117)
4
π
(
,
)=
(
,
)
.
Also
v
M
t
lim
r
→
+
0
v
r
t
be the infinitesimal spherical area on
S
r
and the corresponding
Let d
S
and d
ω
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