Environmental Engineering Reference
In-Depth Information
Since
u
10
=
S
(
u
00
)
,wehave
x
=
ξ
t
=
τ
t
x
+
A
(
t
−
τ
)
3
u
00
1
2
A
∂
t
−
τ
2
e
−
u
10
=
τ
0
d
τ
I
0
·
d
ξ
.
(6.193)
x
2
∂
t
∂
0
x
−
A
(
t
−
τ
)
0,
a
2
m
+
1
x
2
m
+
1
g
l
(
and
a
2
m
x
2
m
g
l
(
Since
G
2
m
+
1
(
share the same form for any
nonnegative integer
l
and any differentiable function
g
l
(
t
)=
t
)
t
)
. Therefore, the right side
of Eq. (6.193) share the same form of
u
1
in Eq. (6.190) after expansion. We can
obtain
u
10
simply by replacing
P
2
m
(
t
)
x
)
in Eqs. (6.191) and (6.192) by
P
2
m
+
1
(
x
)
.
The
P
N
(
x
)
in Eq. (6.190) can be any even or odd polynomial with
m
=[
N
/
2
]
.
This can also be extended to
u
2
,
u
3
,
,
u
[
N
/
2
]
. In finding solutions of PDS (6.181)
by Eq. (6.188), the solutions corresponding to initial values
P
2
(
···
x
)
and
P
3
(
x
)
have
the same form. This is also true for the solutions from
P
4
(
,etc.From
the point of view of approximation, we should use the odd polynomial to approxi-
mate
x
)
and
P
5
(
x
)
, the principle of superposition implies that the solution corresponding to
a higher order polynomial can be formed by adding that for a lower order polyno-
mial and that due to additional terms in the higher order polynomial. Therefore, the
results for a lower order polynomial are still useful even for the case of a higher
order polynomial.
ψ
(
x
)
Remark 3.
It can be shown that the unit
u
1
is correct. For the second term in the
right side of Eqs. (6.191) and (6.192), for example,
L
2
T
·
T
L
·
Θ
TL
2
j
+
2
·
L
2
j
+
1
[
ε
u
11
]=
·
T
=
Θ
,
L
2
T
·
T
2
L
2
Θ
TL
2
i
+
2
k
+
2
·
L
2
i
+
1
L
2
k
+
1
[
ε
u
12
]=
·
·
·
T
=
Θ
.
Thus
[
ε
u
1
]=
Θ
.Since
T
L
·
1
TL
2
·
T
L
2
,
T
L
2
,
[
S
]=
T
·
L
=
[
u
1
]=
we have
u
2
(
2
T
2
L
4
[
u
2
]=[
S
][
u
1
]=
and
x
,
t
)
ε
=
Θ
.
In general,
u
k
(
k
x
,
t
)
ε
=
Θ
,
k
=
0
,
1
,
2
, ···
[
N
/
2
]
.
Therefore, the units of Eq. (6.186) and (6.188) are correct.
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