Environmental Engineering Reference
In-Depth Information
B
2
l
0
ϕ
(
ξ
)
l
0
−
sin
n
πξ
l
sin
n
πξ
l
ϕ
(
ξ
)
=
d
ξ
B
2
l
sin
n
πξ
l
0
ϕ
(
ξ
)
=
−
ξ
(
ϕ
(
)=
ϕ
(
)=
)
.
d
0
l
0
Therefore, the
B
n
in Eq. (6.21) reads
l
0
ϕ
(
ξ
)
l
0
ϕ
(
ξ
)
B
2
1
sin
n
πξ
l
ξ
+
−
sin
n
πξ
l
B
n
=
d
d
ξ
.
τ
0
l
β
n
l
β
n
Also, in Eq. (6.30b)
1
τ
0
+
2
l
0
ϕ
(
ξ
)
n
1
π
B
sin
n
πξ
l
B
(
2
)
=
−
α
d
ξ
n
n
l
β
n
l
l
0
ϕ
(
ξ
)
l
0
ϕ
(
ξ
)
B
2
l
1
τ
0
l
sin
n
πξ
l
sin
n
πξ
l
=
−
d
ξ
+
d
ξ
.
β
n
β
n
Therefore, Eqs. (6.21) and (6.31) are the same although they have different forms.
Solve PDS (6.22) and verify its solution
By Theorem 2 in Section 6.1, the solution of PDS (6.22) is
t
u
(
x
,
t
)=
W
f
τ
(
x
,
t
−
τ
)
d
τ
0
+
∞
n
=
1
2
l
e
α
n
(
t
−
τ
)
sin
t
l
sin
n
πξ
l
sin
n
π
x
=
(
ξ
,
)
(
−
τ
)
f
t
d
ξ
β
t
d
τ
n
β
n
l
0
0
l
t
=
(
,
ξ
,
−
τ
)
(
ξ
,
τ
)
τ
,
d
ξ
G
x
t
f
d
0
0
where
+
∞
n
=
1
2
l
1
β
n
e
α
n
(
t
−
τ
)
sin
n
π
x
sin
n
πξ
l
G
(
x
,
ξ
,
t
−
τ
)=
sin
β
n
(
t
−
τ
)
.
l
It is the same as Eq. (6.28) obtained by the Fourier method of expansion.
6.2.2 Existence
The nominal solution of
⎧
⎨
A
2
u
xx
+
B
2
u
txx
,
u
t
/
τ
0
+
u
tt
=
(
0
,
l
)
×
(
0
,
+
∞
)
,
u
(
0
,
t
)=
u
(
l
,
t
)=
0
,
t
>
0
,
(6.32)
⎩
u
(
x
,
0
)=
ϕ
(
x
)
,
u
t
(
x
,
0
)=
ψ
(
x
)
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