Environmental Engineering Reference
In-Depth Information
Thus
m
C
m
e
α
m
t
cos β
m
t
·
X
m
(
x
)=
∂
t
W
ϕ
(
x
,
t
)
∂
d
x
e
α
m
t
sin
l
0
ϕ
(
α
m
M
m
β
m
−
m
x
)
X
m
(
x
)
β
m
t
·
X
m
(
x
)
1
m
B
2
τ
0
+
λ
=
∂
∂
1
M
m
β
m
)+
m
t
W
ϕ
(
x
,
t
2
d
x
e
α
m
t
sin
l
0
ϕ
(
·
x
)
X
m
(
x
)
β
m
t
·
X
m
(
x
)
.
(6.62)
Substituting Eqs. (6.61) and (6.62) into Eq. (6.60) leads to the solution of PDS (6.56),
by using the structure of
W
ψ
(
x
,
t
)
in Eq. (6.59),
1
τ
0
+
λ
d
x
e
α
m
t
sin
l
0
ϕ
(
m
B
2
)=
∂
W
ϕ
∂
1
M
m
β
m
+
m
(
,
)
(
)
·
(
)
u
x
t
x
X
m
x
β
m
t
X
m
x
t
2
1
d
x
e
α
m
t
sin
l
0
ϕ
(
τ
0
+
λ
m
B
2
2
1
M
m
+
m
x
)
X
m
(
x
)
β
m
t
·
X
m
(
x
)
β
m
1
W
ϕ
(
τ
0
+
∂
=
x
,
t
)
∂
t
m
λ
m
d
x
e
α
m
t
sin
l
0
ϕ
(
1
M
m
β
m
B
2
+
x
)
X
m
(
x
)
β
m
t
·
X
m
(
x
)
1
W
ϕ
(
τ
0
+
∂
B
2
W
λ
m
ϕ
(
=
x
,
t
)+
x
,
t
)
.
∂
t
Example 1
.Solve
⎧
⎨
A
2
u
xx
+
B
2
u
txx
+
u
t
/
τ
0
+
u
tt
=
f
(
x
,
t
)
,
(
0
,
l
)
×
(
0
,
+
∞
)
,
u
x
(
0
,
t
)=
0
,
u
(
l
,
t
)=
0
,
(6.63)
⎩
u
(
x
,
0
)=
ϕ
(
x
)
,
u
t
(
x
,
0
)=
ψ
(
x
)
.
Solution.
Based on the given boundary conditions, we have, from Row 4 in Ta-
ble 2.1,
(
2
2
m
+
1
)
π
cos
(
2
m
+
1
)
π
x
l
2
.
λ
m
=
,
X
m
(
x
)=
,
M
m
=(
X
m
(
x
)
,
X
m
(
x
)) =
2
l
2
l
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