Environmental Engineering Reference
In-Depth Information
V
tt
A
2
R
3
=
Δ
V
,
×
(
0
,
+
∞
)
,
(5.65)
e
A
z
V
|
t
=
0
=
0
,
V
t
|
t
=
0
=
ψ
(
x
,
y
)
.
5.4.2 Solution of PDS (5.62)
The solution of PDS (5.65) is, by the Poisson formula (see Chapter 2)
1
e
A
ζ
d
S
V
(
x
,
y
,
z
,
t
)=
ψ
(
ξ
,
η
)
,
4
π
A
2
t
S
At
where
S
At
stands for the spherical surface of a sphere of center
M
and radius
At
.
Now consider the spherical surface
S
At
of a sphere of center
M
(
x
,
y
,
0
)
and radius
At
. Its projection on
plane
-
Oxy
is a circle of center
M
(
x
,
y
,
0
)
and radius
At
denoted
by
D
At
:
2
2
2
.Let
be the angle between
Oz
-axis and normal
of
S
At
. The surface element d
S
on
S
At
is thus related to the area element in
D
At
by
(
ξ
−
x
)
+(
η
−
y
)
≤
(
At
)
γ
1
At
d
S
=
d
ξ
d
η
=
d
ξ
d
η
.
(5.66)
|
cos
γ
|
2
2
2
(
)
−
(
−
ξ
)
−
(
−
η
)
At
x
y
Note also that the mathematical expression of
S
At
is
2
2
2
ζ
=
±
(
At
)
−
(
x
−
ξ
)
−
(
y
−
η
)
,
is corresponding to points on the upper or the lower half of
S
At
.The
method of descent thus yields the solution of PDS (5.63)
±
where the
⎡
⎣
e
A
√
(
At
)
2
2
2
−
(
x
−
ξ
)
−
(
y
−
η
)
1
ψ
(
ξ
,
η
)
v
(
x
,
y
,
t
)=
d
ξ
d
η
4
π
A
2
2
2
(
At
)
−
(
x
−
ξ
)
−
(
y
−
η
)
D
At
⎤
⎦
A
√
(
At
)
2
2
2
e
−
−
(
x
−
ξ
)
−
(
y
−
η
)
ψ
(
ξ
,
η
)
+
d
ξ
d
η
2
2
2
(
At
)
−
(
x
−
ξ
)
−
(
y
−
η
)
D
At
ch
A
2
2
2
(
At
)
−
(
x
−
ξ
)
−
(
y
−
η
)
1
=
(
ψ
(
ξ
,
η
)
d
ξ
d
η
.
2
π
A
2
2
2
At
)
−
(
x
−
ξ
)
−
(
y
−
η
)
D
At
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