Environmental Engineering Reference
In-Depth Information
⎧
⎨
⎡
⎣
1
t
A
e
−
+
2
τ
0
4
π
A
¯
ψ
(
At
)+
4
π
At
¯
ψ
t
(
At
)
⎩
4
π
⎛
⎝
⎞
⎠
⎤
⎦
bI
1
b
(
r
2
⎡
⎤
At
At
)
2
−
1
2
⎣
⎦
M
)
d
r
(
+
ψ
(
d
S
At
)
2
−
r
2
−
At
S
M
r
⎡
b
2
I
1
b
r
2
(
At
)
2
−
1
2
⎣
M
)
+
b
ψ
(
d
S
|
r
=
At
·
A
2
r
2
(
At
)
−
S
r
⎫
⎬
⎤
⎦
b
2
I
1
b
r
2
(
)
2
−
At
M
)
−
b
(
ψ
(
d
S
|
r
=
−
At
·
(
−
A
)
.
⎭
At
)
2
−
r
2
S
r
From the point view of continuation,
M
)
M
)
ψ
(
d
S
|
r
=
−
At
=
ψ
(
d
S
|
r
=
At
.
S
r
S
r
Also
M
)
lim
t
→
+
0
ψ
(
d
S
=
0
.
S
At
Therefore, the
u
(
M
,
t
)
in Eq. (5.120) also satisfies the initial condition
u
t
(
M
,
0
)=
ψ
(
M
)
.
Comparison with Three-Dimensional Wave Equations
The counterpart of PDS (5.119) in wave equations reads
u
tt
=
A
2
R
3
Δ
u
,
×
(
0
,
+
∞
)
,
(5.122)
u
(
M
,
0
)=
0
,
u
t
(
M
,
0
)=
ψ
(
M
)
.
Its solution is, by the Poisson formula
1
M
)
u
(
M
,
t
)=
ψ
(
d
S
,
(5.123)
4
π
A
2
t
S
At
which is the first term of the right-hand side of Eq. (5.120) without the decaying
factor e
−
t
2
τ
0
. Since the solution is a surface integral of the first kind, there exist a
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