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
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
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
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
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
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
0 1
τ 0 d
1
2 A
t + τ
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
 
Search WWH ::




Custom Search