Environmental Engineering Reference
In-Depth Information
⎡
br
I
1
b
r
2
⎤
At
(
)
2
−
At
1
2
A
⎣
r
r
)
d
r
+
⎦
¯
2
At
¯
=
(
ψ
(
ψ
(
At
)
At
)
2
−
r
2
−
At
⎛
⎝
⎞
⎠
bI
1
b
(
r
2
At
At
)
2
−
1
M
)
d
r
=
ψ
(
d
S
8
π
A
r
2
(
At
)
2
−
−
At
S
M
r
1
M
)
+
ψ
(
d
S
4
π
A
2
t
S
r
⎡
⎛
⎞
bI
1
b
(
r
2
At
At
)
2
−
1
1
2
⎣
⎝
⎠
M
)
d
r
=
ψ
(
d
S
4
π
A
2
r
2
(
At
)
−
−
At
S
r
⎤
1
At
⎦
.
M
)
+
ψ
(
d
S
S
r
Thus the solution of
⎧
⎨
u
t
τ
0
+
A
2
R
3
u
tt
=
Δ
u
,
×
(
0
,
+
∞
)
,
(5.119)
⎩
u
(
M
,
0
)=
0
,
u
t
(
M
,
0
)=
ψ
(
M
)
is
⎡
1
1
At
t
2τ
0
⎣
A
e
−
M
)
u
(
M
,
t
)=
W
ψ
(
M
,
t
)=
ψ
(
d
S
4
π
S
At
(5.120)
⎛
⎝
⎞
⎠
⎤
⎦
.
bI
1
b
(
r
2
At
At
)
2
−
1
2
M
)
d
r
+
ψ
(
d
S
r
2
(
At
)
2
−
−
At
S
M
r
Finally, the solution of PDS (5.115) comes from the solution structure theorem,
1
W
ϕ
(
t
τ
0
+
∂
u
(
M
,
t
)=
M
,
t
)+
W
ψ
(
M
,
t
)+
W
f
τ
(
M
,
t
−
τ
)
d
τ
,
(5.121)
∂
t
0
where
f
τ
=
f
(
M
,
τ
)
.
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