Environmental Engineering Reference
In-Depth Information
Remark 5. If
ψ (
r
)=
1 in Eq. (5.94), Eq. (5.97) yields
I 0 b
2
0 r + At
r
t
1
2 Ar e
u
(
r
,
t
)=
(
At
)
2
(
r
ρ )
ρ
d
ρ
At
I 0 b
2
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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