Environmental Engineering Reference
In-Depth Information
Method of Solving (3.46)
∞
n
=
1
a
n
(
s
)
sin
n
π
x
Expand
G
(
x
,
s
)=
. Substitute it into the integral equation to obtain
l
l
∞
n
=
1
X
n
sin
n
π
x
(
)=
−
λ
,
=
−
λ
(
)
(
)
X
x
X
n
X
s
a
n
s
d
s
(3.47)
l
0
Multiplying the equation of (3.47) by
a
m
(
x
)
and then integrating over
[
0
,
l
]
yields
l
∞
n
=
1
a
mn
X
n
,
sin
n
π
x
=
−
λ
=
(
)
,
=
,
, ···
X
m
a
mn
a
m
x
d
x
m
1
2
(3.48)
l
0
T
. Eq. (3.48) can then be written as an
Let
A
=(
a
mn
)
∞
×
∞
,
X
=(
X
1
,
X
2
, ··· ,
X
n
, ···
)
eigenvalue problem of
A
X
=
−
λ
AX
.
(3.49)
We denote its eigenvalues and the corresponding eigenvectors by
λ
1
,
λ
2
, ··· ,
λ
n
, ··· ,
X
1
,
X
2
, ··· ,
X
n
, ··· .
The eigenfunction set associated with the eigenvalues
λ
n
(
n
=
1
,
2
, ···
)
X
1
(
x
)
,
X
2
(
x
)
, ··· ,
X
n
(
x
)
, ···
can thus be obtained by an substitution into Eq. (3.47). The
T
(
t
)
associated with
λ
=
λ
n
can also be obtained by Eq. (3.43),
e
−
(
λ
n
t
/ ρ
c
)
,
(
)=
=
,
, ··· .
T
n
t
n
1
2
Therefore
∞
n
=
1
C
n
X
n
(
x
)
e
−
(
λ
n
t
/
ρ
c
)
,
which satisfies the equation and the boundary conditions of (3.41). Here
C
n
(
n
∞
n
=
1
C
n
T
n
(
t
)
X
n
(
x
)=
u
(
x
,
t
)=
=
1
,
2
, ···
) are constants and can be determined by applying the initial condition
)
}
n
=
1
in
u
(
x
,
0
)=
f
(
x
)
and using the completeness and orthogonality of
{
X
n
(
x
[
0
,
l
]
.
Analytical solution of PDS (3.41)
The analytical solution of PDS (3.41) follows from the above discussion,
⎧
⎨
∞
n
=
1
C
n
X
n
(
e
−
(
λ
n
t
/ ρ
c
)
,
u
(
x
,
t
)=
x
)
d
x
0
X
n
(
C
n
=
0
X
n
(
⎩
x
)
f
(
x
)
x
)
d
x
.
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