Environmental Engineering Reference
In-Depth Information
Here
u
0
is the solution of PDS (6.169). The terms containing powers of
come
from the effect of the term of third order derivative
u
txx
. This property of solutions is
useful for examining heat conduction processes. By Eq. (6.188), the task of finding
solutions is reduced to that of applying the operator
S
to
u
0
. In applications, the
order of polynomials is normally not larger than 5 so that
ε
[
N
/
2
]
≤
2. For
N
=
4or
5, for example, Eq. (6.188) becomes
S
2
2
u
=
u
0
+
S
(
u
0
)
ε
+
(
u
0
)
ε
.
Without loss of generality, consider an even
N
such that
N
=
2
m
and calculate
the terms in the right side of Eq. (6.188). By Eq. (6.176), we have
P
(
2
k
+
2
)
N
m
1
k
=
1
−
(
)
1
2
A
x
t
τ
0
e
−
P
N
(
u
0
txx
=
x
)
+
g
2
k
(
t
)
,
(6.189)
(
2
k
)
!
d
t
e
−
,
k
d
t
2τ
0
G
2
k
(
where
g
2
k
(
t
)=
t
)
=
1
,
2
, ··· ,
m
−
1, which can be readily obtained
by the definition of
G
2
k
(
t
)
. By the operator
S
defined in Eq. (6.187),
t
x
+
A
(
t
−
τ
)
1
2
A
t
−
τ
2τ
0
d
e
−
u
1
=
S
(
u
0
)=
τ
I
0
0
x
−
A
(
t
−
τ
)
P
N
(
ξ
)
d
P
(
2
k
+
2
)
N
m
−
1
k
=
1
1
2
A
(
ξ
)
t
τ
0
e
−
·
+
g
2
k
(
τ
)
ξ
,
(6.190)
(
2
k
)
!
whose integrand is the sum of two parts
u
11
and
u
12
,
t
x
+
A
(
t
−
τ
)
1
2
A
t
−
τ
2τ
0
d
e
−
τ
0
d
e
−
P
N
(
ξ
)
u
11
=
τ
I
0
·
ξ
0
x
−
A
(
t
−
τ
)
N
k
=
2
k
(
k
−
1
)
a
k
(
u
+
x
)
k
+
2
d
u
t
A
(
t
−
τ
)
1
2
A
t
+
τ
2τ
0
d
e
−
=
τ
I
0
·
0
−
A
(
t
−
τ
)
N
k
=
2
k
(
k
−
1
)
i
=
0
a
k
C
k
−
2
u
i
x
k
−
2
−
i
d
u
t
A
(
t
−
τ
)
k
−
2
1
2
A
t
+
τ
2τ
0
d
e
−
=
τ
I
0
·
0
−
A
(
t
−
τ
)
N
k
=
2
k
(
k
−
1
)
a
k
x
k
−
2
2
A
τ
0
e
−
t
−
τ
t
2τ
0
1
τ
0
d
1
2
A
t
+
τ
2τ
0
t
−
τ
e
−
e
−
=
−
τ
0
N
k
=
2
d
t
k
−
2
i
=
1
k
(
k
−
1
)
a
k
(
x
k
−
2
)
(
i
)
1
2
A
t
+
τ
2
e
−
+
τ
0
G
i
(
t
−
τ
)
τ
i
!
0
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