Environmental Engineering Reference
In-Depth Information
Remark 5.
If
ψ
(
r
)=
1 in Eq. (5.94), Eq. (5.97) yields
I
0
b
2
2τ
0
r
+
At
r
t
1
2
Ar
e
−
u
(
r
,
t
)=
(
At
)
2
−
(
r
−
ρ
)
ρ
d
ρ
−
At
I
0
b
2
2τ
0
At
−
t
1
2
Ar
e
−
2
=
(
At
)
−
ξ
(
r
−
ξ
)
d
ξ
At
I
0
b
2
d
τ
0
At
−
ξ
=
τ
0
1
τ
0
t
2
1
t
2
A
e
−
e
−
=
(
At
)
2
−
ξ
−
,
At
which is the same as Eq. (5.46).
5.7.4 Discussion of Solution (5.98)
1. By Eq. (5.98), the solution of
⎧
⎨
u
t
τ
A
2
+
=
(
,
)
,
<
<
+
∞
,
<
,
u
tt
Δ
u
r
t
0
r
0
t
0
(
,
)=
,
u
r
0
t
0
(5.99)
⎩
u
(
r
,
0
)=
ϕ
(
r
)
,
u
t
(
r
,
0
)=
0
is
1
W
ϕ
(
τ
0
+
∂
u
=
r
,
t
)
∂
t
⎧
⎨
t
(
r
−
At
)
ϕ
(
r
−
At
)+(
r
+
At
)
ϕ
(
r
+
At
)
e
−
2
τ
0
=
⎩
2
r
⎡
τ
0
I
0
b
2
r
+
At
1
2
Ar
1
2
⎣
2
+
(
At
)
−
(
r
−
ρ
)
r
−
At
I
1
b
2
⎤
⎫
⎬
⎭
.
2
(
At
)
−
(
r
−
ρ
)
⎦
ρϕ
(
ρ
)
+
0
b
t
d
ρ
(5.100)
2
4
τ
(
At
)
2
−
(
r
−
ρ
)
2
1. It is straightforward to show that the
u
(
r
,
t
)
in Eq. (5.100) satisfies the initial
conditions
u
(
r
,
0
)=
ϕ
(
r
)
and
u
t
(
r
,
0
)=
0.
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