Environmental Engineering Reference
In-Depth Information
The general solution of Eq. (2.16) is
⎧
⎨
A
e
−
√
−
λ
x
B
e
√
−
λ
x
+
,
λ
<
0
,
A
+
Bx
,
λ
=
0
,
X
(
x
)=
A
cos
√
B
sin
√
⎩
λ
x
+
λ
x
,
λ
>
0
,
where
A
and
B
are constants. By applying the boundary conditions
X
(
0
)=
0and
X
(
l
)=
0, we know that both
λ
<
0and
λ
=
0 lead to
X
(
x
)
≡
0. Therefore
λ
>
0.
Applying the boundary condition
X
(
0
)=
0
yield
s
A
=
0, thus
B
=
0. Applying the
0 leads to
B
√
λ
cos
√
λ
0orcos
√
λ
boundary condition
X
(
l
)=
l
=
l
=
0. Therefore
we obtain the eigenvalues
λ
k
and the eigenfunctions
X
k
(
x
)
:
(
2
+
)
π
sin
(
+
)
π
2
k
1
2
k
1
x
λ
k
=
,
X
k
(
x
)=
,
k
=
0
,
1
,
2
, ··· .
2
l
2
l
It is easy to show both the completeness and the orthogonality of the eigenfunction
group
{
X
k
(
x
)
}
in
[
0
,
l
]
.
Substituting
λ
k
into Eq. (2.17) leads to
a
k
cos
(
2
k
+
1
)
π
at
b
k
sin
(
2
k
+
1
)
π
at
T
k
(
t
)=
+
,
k
=
0
,
1
,
2
, ···
2
l
2
l
where
a
k
and
b
k
are undetermined constants. A superposition of
u
k
=
)
yields the solution satisfying the equation and the boundary conditions, whatever
the constants
a
k
and
b
k
,
X
k
(
x
)
T
k
(
t
a
k
cos
(
sin
(
∞
k
=
0
+
)
π
b
k
sin
(
+
)
π
+
)
π
2
k
1
at
2
k
1
at
2
k
1
x
u
(
x
,
t
)=
+
.
2
l
2
l
2
l
To satisfy the initial conditions
a
k
and
b
k
must be determined such that
∞
k
=
0
a
k
sin
(
2
k
+
1
)
π
x
=
0
,
2
l
∞
k
=
0
(
2
k
+
1
)
π
a
b
k
sin
(
2
k
+
1
)
π
x
=
ψ
(
)
.
x
2
l
2
l
Thus,
l
0
ψ
(
4
sin
(
2
k
+
1
)
π
x
a
k
=
0
,
b
k
=
x
)
d
x
,
(
2
k
+
1
)
π
a
2
l
where we have used the completeness and the orthogonality of the eigenfunction
group and the normal square of the eigenfunction group
l
sin
2
(
2
k
+
1
)
π
x
l
2
.
M
k
=
d
x
=
2
l
0
2
Also,
√
M
k
=
is called the
normal of the eigenfunction group
.
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