Environmental Engineering Reference
In-Depth Information
continuity and the boundary conditions of (2.31), we obtain
u
(
2
,
s
1
2
,
b
1
=
b
2
=
−
=
−
2
b
1
−
b
2
[
u
(
0
)+
u
(
l
)]
a
1
=
−
b
1
−
(
a
2
+
b
2
)
u
(
0
)
,
a
2
.
u
(
0
)
−
u
(
l
)
Therefore, by the meaning of the Green function, (2.29) leads to an eigenvalue prob-
lem
l
X
(
x
)=
−
λ
G
(
x
,
s
)
X
(
s
)
d
s
.
(2.32)
0
2. Method of Resolving (2.32)
∞
n
=
1
a
n
sin
n
π
x
(
,
)=
(
)
Let
G
x
s
s
; substituting it into (2.32) yields
l
l
∞
n
=
1
X
n
sin
n
π
x
X
(
x
)=
−
λ
,
X
n
=
−
λ
X
(
s
)
a
n
(
s
)
d
s
.
(2.33)
l
0
Multiplying (2.33) by
a
m
(
x
)
and integrating over
[
0
,
l
]
lead to
l
∞
n
=
1
a
mn
X
n
,
sin
n
π
x
X
m
=
−
λ
a
mn
=
a
m
(
x
)
d
x
,
m
=
1
,
2
, ···
l
0
Therefore we obtain an eigenvalue problem
T
=
−
λ
,
=(
)
∞
×
∞
,
=(
,
, ···
)
.
X
AX
A
a
mn
X
X
1
X
2
Let
X
1
,
X
2
, ···
be the eigenvectors corresponding to the eigenvalues
λ
,
λ
, ···
. Sub-
1
2
stituting into (2.33) yields the eigenvector group of (2.32)
X
1
(
x
)
,
X
2
(
x
)
, ··· ,
X
n
(
x
)
, ···
which is complete and orthogonal in
[
0
,
l
]
by the theory of integral equations. Fur-
thermore,
λ
i
>
0,
i
=
1
,
2
, ···
.
3. Analytical Solution of the Original PDS
T
n
(
t
)
corresponding to
λ
=
λ
n
can easily be obtained by Eq. (2.30) so that
n
=
1
D
n
X
n
(
x
)
sin
λ
n
a
(
t
−
τ
)
,
∞
n
=
1
T
n
(
t
)
X
n
(
x
)=
∞
v
(
x
,
t
,
τ
)=
l
λ
n
a
l
0
d
x
X
n
(
D
n
=
X
n
(
x
)
f
(
x
,
τ
)
x
)
d
x
.
0
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