Environmental Engineering Reference
In-Depth Information
Remark 4. Note that
P ( 2 m )
N
S m
m
m
(
u 0 ) ε
=
(
x
)
q
(
t
) ε
,
(6.194)
where N
=
2 m or 2 m
+
1, m
=
0
,
1
,
2
, ···
, q
(
t
)
is a function to be determined. By
introducing an operator
t = τ
I 0 b A 2
u 2 d
d t
t
A ( t τ )
1
2 A
t
τ
0 d
e
B
=
τ
(
t
τ )
2
d u
0
A
(
t
τ )
t = τ
0 t
0
1
τ 0
d
d t
t τ
e
= τ
d
τ ,
we have
B m
τ 0 1
τ 0
t
e
q
(
t
)=
.
(6.195)
Therefore,
B
τ 0 1
τ 0
= τ 0 t
0
1
τ 0 e τ 0 d
t
t
τ
e
e
τ
= τ 0
τ 0
t
e
τ 0 ( τ 0 +
t
)
,
= τ 0 t
0
B 2
τ 0 1
τ 0
1
τ 0
t
t τ
e τ 0 d
e
e
τ
τ
e
τ 0
t 2
2
t
2
0
= τ
τ 0
τ 0 +
t
+
,
τ 0
= τ 0 t
0
B 3
τ 0 1
τ 0
1
τ 0 τ
2
2! e τ 0 d
t
t τ
e
e
τ
e
τ 0
t 2
2!
t 3
3!
t
3
0
= τ
τ 0
τ 0 +
t
+
τ 0 +
.
2
0
τ
For the real computation, m is normally not larger than 4. By the recurrence method,
we have
= τ 0 t
0
B m
τ 0 1
τ 0
1
τ 0
m 1
τ
t
t τ
! e τ 0 d
e
e
q
(
t
)=
τ
(
m
1
)
e
t 2
2!
t m
t
τ 0
0
= τ
τ 0
τ 0 +
t
+
τ 0 + ··· +
.
m
1
m !
τ
0
L 2
T
m
]= TL 2 m ·
S m
m
T m + 1
The unit of the last term is
[
(
u 0
) ε
·
= Θ
, which is cor-
rect. Therefore, the last term of the solution is independent of x if P N (
x
)
is even. It
 
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