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t
τ 0
P ( 2 j + 2 )
N
m − 1
j = 1
1
2 A
(
x
)
t
t + τ
2
e −
e −
= τ 0 P N (
x
)
τ 0 − ( τ 0 +
t
)
+
τ 0 G 2 j (
t
− τ )
d
τ ,
(
2 j
)
!
0
(6.191)
1
2 A
d
t
x + A ( t − τ )
P ( 2 k + 2 )
N
m
1
k = 1
−
( ξ )
1
2 A
t
− τ
2τ 0 d
e −
u 12 =
τ
I 0 ·
g 2 k ( τ )
ξ
(
2 k
)
!
x − A ( t − τ )
0
⎡
i = 0 C n u i x n − i ( 2 k + 2 ) d u ⎤
N
4 A 2 t
A ( t − τ )
m
1
k = 1
−
n = 0 a n N
1
g 2 k ( τ )
(
t
− τ
2τ 0 d
e −
⎣
⎦
=
τ
I 0 ·
2 k
)
!
0
−
A
(
t
− τ )
t
1
τ 0 d
P ( 2 k + 2 )
N
m
1
k = 1
−
(
x
)
= τ
t
− τ
0
2 A
e −
g 2 k ( τ )
−
τ
(
2 k
)
!
0
m − 2
k = 1
u i d u
4 A 2 t
A ( t − τ )
P ( 2 k + 2 + i )
N
m
2
i = 1
−
(
x
)
1
g 2 k ( τ )
(
t − τ
2
e −
+
τ 0 d
τ
I 0
·
2 k
)
!
i !
0
− A ( t − τ )
t
0 τ 0 g 2 k ( τ )
1
τ 0 d
P ( 2 k + 2 )
N
m
1
k = 1
−
1
2 A
(
x
)
t − τ
e −
=
−
τ
(
)
2 k
!
t
P ( 2 i + 2 k + 2 )
N
m − 2
∑
1
4 A 2
(
x
)
t − τ
2
e −
+
τ 0 G 2 i (
t
− τ )
g 2 k ( τ )
d
τ .
(6.192)
(
2 i
)
!
(
2 k
)
!
0
i
,
k
=
1
(
i
+
k
≤
m
−
1
)
Remark 1. In the second term of the right side of Eq. (6.192), define a matrix of
order ( m
P ( 2 i + 2 k + 2 )
N
.
−
)
=(
a ik )=
(
)
(
−
)(
−
) /
2elementsof A in the
lower triangular matrix under the auxiliary diagonal are zero so that the sum in the
second term of the right side of Eq. (6.192) is actually over
2
: A
x
m
2
m
3
2terms.
While Eq. (6.192) appears complicated, it is often very simple for specific problems.
If P N (
(
m
−
1
)(
m
−
2
) /
is a polynomial of order not larger than 5, for example, the second term of
the right side of Eq. (6.192) is zero.
x
)
(
,
)
Remark 2. Eq. (6.176) shows that the u 0
x
t
has the same form for both even and
1
n = 0 a n x n ,the u 1 ( x , t ) can be obtained by
adding the solution for an even N and that from the initial value a 2 m + 1 x 2 m + 1 .Let
u 00 and u 10 be the u 0 and u 1 for an odd N . By Eq. (6.176), we have
2 m
+
odd N . For an odd N such that P N
(
x
)=
u 00 = τ 0 a 2 m + 1 x 2 m + 1 1
τ 0
[ N /
]
2
a 2 m + 1 x 2 m + 1
) ( 2 k )
1
2 A
(
t
t
e −
e −
k = 1
−
+
G 2 k (
t
)
2
τ 0
.
(
)
2 k
!
 
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