Environmental Engineering Reference
In-Depth Information
Solution.
Based on the given boundary conditions, let
+
∞
n
=
1
T
n
(
t
)
sin
n
π
x
u
(
x
,
t
)=
.
(6.16)
l
Substituting it into the equation in PDS (6.15) yields
T
n
(
sin
n
n
2
n
2
+
∞
n
=
1
t
)
π
A
π
B
π
x
l
=
T
n
T
n
(
0
+
(
t
)+
T
n
(
t
)+
t
)
0
.
τ
l
l
Thus
1
τ
0
+
2
T
n
(
n
n
2
π
B
π
A
T
n
(
t
)+
t
)+
T
n
(
t
)=
0
.
(6.17)
l
l
Its general solution reads
e
α
n
t
T
n
(
t
)=
(
A
n
cos
β
n
t
+
B
n
sin
β
n
t
)
,
where
A
n
and
B
n
are constants,
1
τ
0
+
2
n
1
2
π
B
α
n
=
−
,
l
4
n
1
τ
0
+
2
2
2
n
1
2
π
A
π
B
β
=
−
,
n
l
l
sin
β
n
t
,
if
β
n
=
0
,
sin
β
n
t
=
t
,
if
β
n
=
0
.
Substituting
T
n
(
t
)
into Eq. (6.16) leads to
+
∞
n
=
1
e
α
n
t
sin
n
π
x
u
(
x
,
t
)=
(
A
n
cos
β
n
t
+
B
n
sin
β
n
t
)
.
l
(
,
)=
Applying the initial condition
u
x
0
0 yields
+
∞
n
=
1
A
n
sin
n
π
x
l
=
0
.
Thus
A
n
=
0
,
n
=
1
,
2
,...,
+
∞
n
=
1
e
α
n
t
B
n
sin
β
n
t
sin
n
π
x
u
(
x
,
t
)=
,
l
+
∞
n
=
1
B
n
(
α
n
sin
β
n
t
+
β
n
cosβ
n
t
)
e
α
n
t
sin
n
π
x
u
t
(
x
,
t
)=
,
l
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