Environmental Engineering Reference
In-Depth Information
is
1
W
ϕ
(
τ
0
+
∂
B
2
W
λ
m
ϕ
(
(
,
)=
,
)+
,
)
u
x
t
x
t
x
t
(6.54)
∂
t
t
+
W
ψ
(
x
,
t
)+
W
f
τ
(
x
,
t
−
τ
)
d
τ
,
(6.55)
0
where
f
τ
=
λ
m
are the eigenvalues in Table 2.1 corresponding to the bound-
ary conditions in PDS (6.53).
f
(
x
,
τ
)
,
Proof
. By Eq. (6.13) in Section 6.1, we only need to prove that the solution of
⎧
⎨
A
2
u
xx
B
2
u
txx
u
t
/
τ
+
u
tt
=
+
,
(
0
,
l
)
×
(
0
,
+
∞
)
,
0
L
(
u
,
u
n
)
|
∂Ω
=
0
,
(6.56)
⎩
u
(
x
,
0
)=
ϕ
(
x
)
,
u
t
(
x
,
0
)=
0
is
1
W
ϕ
(
τ
0
+
∂
B
2
W
λ
m
ϕ
(
u
=
x
,
t
)+
x
,
t
)
.
(6.57)
∂
t
Let the solution of PDS (6.52) be
=
m
T
m
(
t
)
X
m
(
x
)
,
u
m
=
0
or
∞
∞
m
=
1
depending on
{
X
m
(
x
)
}
. Substituting this into the
equation in PDS (6.52) yields the
T
m
(
m
where
stands for
t
)
-equation
1
τ
0
+
λ
m
B
2
T
m
(
T
(
)+
λ
m
A
2
t
m
(
t
)+
t
t
)=
0
.
Its two characteristic roots are
λ
m
A
2
1
1
τ
0
+
λ
m
B
2
τ
0
+
λ
m
B
2
2
1
2
r
1
,
2
=
−
±
−
4
=
α
m
±
β
m
i
.
Thus the solution of PDS (6.52) reads
)=
m
e
α
m
t
u
(
x
,
t
(
A
m
cos
β
m
t
+
B
m
sin
β
m
t
)
X
m
(
x
)
,
(6.58)
where
A
m
and
B
m
are both constants. Applying the initial condition
u
(
x
,
0
)=
0 yields
A
m
=
0. Clearly we can differentiate the series (6.58) term by term. The
B
m
can be
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