Environmental Engineering Reference
In-Depth Information
β
n
,
if
β
n
=
0
,
β
n
=
(
,
)=
ψ
(
)
where
Applying the initial condition
u
t
x
0
x
yields
1
,
if
β
n
=
0
.
+
∞
n
=
1
B
n
β
n
sin
n
π
x
l
=
ψ
(
x
)
.
Thus
l
0
ψ
(
ξ
)
2
sin
n
πξ
l
B
n
=
d
ξ
.
l
β
n
Finally, the solution of PDS (6.15) is
⎧
⎨
+
∞
n
=
1
B
n
e
α
n
t
sin
β
n
t
sin
n
π
x
(
,
)=
,
u
x
t
l
(6.18)
l
0
ψ
(
ξ
)
⎩
2
sin
n
πξ
l
=
ξ
.
B
n
d
l
β
n
Solution from Initial Temperature
ϕ
(
x
)
Solve
⎧
⎨
A
2
u
xx
B
2
u
txx
u
t
/
τ
+
u
tt
=
+
,
(
0
,
l
)
×
(
0
,
+
∞
)
,
0
u
(
0
,
t
)=
u
(
l
,
t
)=
0
,
(6.19)
⎩
u
(
x
,
0
)=
ϕ
(
x
)
,
u
t
(
x
,
0
)=
0
.
Solution.
Based on the given boundary conditions, let
+
∞
n
=
1
T
n
(
t
)
sin
n
π
x
u
(
x
,
t
)=
.
l
The
T
n
(
t
)
can be determined by substituting it into the equation in PDS (6.19). Thus
+
∞
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
)
.
(6.20)
l
Applying the initial condition
u
(
x
,
0
)=
ϕ
(
x
)
leads to
l
0
ψ
(
ξ
)
2
l
sin
n
πξ
l
A
n
=
d
ξ
.
By taking the derivative of Eq. (6.20) with respect to
t
, we obtain
n
=
1
α
n
e
α
n
t
+
∞
u
t
(
x
,
t
)=
(
A
n
cos
β
n
t
+
B
n
sin
β
n
t
)
)
sin
n
π
x
e
α
n
t
+
(
−
A
n
β
n
sin
β
n
t
+
B
n
β
n
cos
β
n
t
.
l
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