Environmental Engineering Reference
In-Depth Information
is
I
0
b
2
d
x
+
At
1
2
A
t
e
−
2
u
=
2
τ
0
(
At
)
−
(
x
−
ξ
)
ξ
.
(5.41)
x
−
At
The Initial Conditions
By Eq. (5.41), it is clear that
u
(
x
,
0
)=
0. Since
I
0
b
2
d
τ
0
x
+
At
x
1
2
A
1
2
t
e
−
2
u
t
=
−
2
(
At
)
−
(
x
−
ξ
)
ξ
τ
−
At
0
τ
0
x
+
At
x
∂
∂
t
e
−
+
2
t
I
0
d
ξ
+
I
0
|
ξ
=
x
+
At
A
−
I
0
|
ξ
=
x
−
At
(
−
A
)
,
−
At
we obtain
u
t
(
x
,
0
)=
1. Therefore
u
(
x
,
t
)
in Eq. (5.41) satisfies the two initial condi-
tions of PDS (5.40).
The Equation
1
2
A
By using the series expansion of
I
0
(
x
)
(see Appendix A) and
b
=
τ
0
, we obtain
I
0
b
2
d
x
+
At
1
3
b
2
A
3
t
3
1
60
b
4
A
5
t
5
2
(
At
)
−
(
x
−
ξ
)
ξ
=
2
At
+
+
+
··· .
(5.42)
x
−
At
This shows that the right-hand side of Eq. (5.41) is independent of
x
. Thus we only
need to show that the
u
in Eq. (5.41) satisfies
u
t
τ
0
+
u
tt
=
0
.
(5.43)
)
=
u
v
2
u
v
+
uv
.
This can be achieved by using the formula
(
uv
+
But, the proc-
ess is quite involved. Here we show it by using another method.
By expanding e
−
t
2τ
0
into a series and using Eq. (5.42), we have
+
∞
n
=
1
(
−
1
)
t
n
n
−
1
u
(
x
,
t
)=
.
(5.44)
n
−
1
n
!
τ
0
Thus
+
∞
n
=
1
(
−
1
)
t
n
−
1
+
∞
n
=
2
(
−
1
)
t
n
−
2
n
−
n
−
1
1
u
t
=
,
u
tt
=
−
,
(5.45)
n
−
1
n
−
1
(
n
−
1
)
!
τ
(
n
−
2
)
!
τ
0
0
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