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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