Environmental Engineering Reference
In-Depth Information
Applying the initial condition
u
t
(
x
,
0
)=
0 yields
+
∞
n
=
1
(
A
n
α
n
+
B
n
β
n
)
sin
n
π
x
l
=
0
.
Thus
A
n
α
n
β
n
A
n
α
n
+
B
n
β
n
=
0or
B
n
=
−
.
Finally, the solution of PDS (6.19) is
⎧
⎨
+
∞
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
l
0
ϕ
(
ξ
)
2
l
sin
n
πξ
l
A
n
=
d
ξ
,
(6.21)
⎩
l
0
ϕ
(
ξ
)
B
n
=
−
2
α
sin
n
πξ
l
n
d
ξ
.
l
β
n
Solution from Source Term f
(
x
,
t
)
Solve
⎧
⎨
A
2
u
xx
+
B
2
u
txx
+
u
t
/
τ
0
+
u
tt
=
f
(
x
,
t
)
,
(
0
,
l
)
×
(
0
,
+
∞
)
,
u
(
0
,
t
)=
u
(
l
,
t
)=
0
,
(6.22)
⎩
(
,
)=
,
(
,
)=
.
u
x
0
0
u
t
x
0
0
Solution.
Based on the given boundary conditions, let
+
∞
n
=
1
T
n
(
t
)
sin
n
π
x
T
n
(
u
(
x
,
t
)=
,
T
n
(
0
)=
0
)=
0
.
(6.23)
l
l
+
∞
n
=
1
f
n
(
t
)
sin
n
π
x
2
l
sin
n
πξ
l
f
(
x
,
t
)=
,
f
n
(
t
)=
f
(
ξ
,
t
)
d
ξ
.
l
0
Substituting them into the equation of PDS (6.22) yields
1
τ
0
+
2
T
n
(
n
n
2
π
B
π
A
T
n
(
t
)+
t
)+
T
n
(
t
)=
f
n
(
t
)
.
(6.24)
l
l
Its corresponding homogeneous equation is Eq. (6.17). The solution of Eq. (6.24)
subject to
T
n
(
T
n
(
0
)=
0
)=
0 can be written as, by variation of constants,
c
(
n
)
1
c
(
n
)
2
e
α
n
t
cos
e
α
n
t
sin
T
n
(
t
)=
(
t
)
β
n
t
+
(
t
)
β
n
t
,
(6.25)
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