Environmental Engineering Reference
In-Depth Information
so that
1
τ 0 +( λ m + λ n + λ k )
B 2
D mnk = α mnk
β mnk
C mnk =
2
1
M mnk β mnk
·
ϕ (
x
,
y
,
z
)
X m (
x
)
Y n (
y
)
Z k (
z
)
d x d y d z
.
(6.86)
Ω
By Eq. (6.84), we have
= m , n , k D mnk e α mnk t α mnk sin β mnk t + β mnk cos β mnk t
·
W ϕ (
x
,
y
,
z
,
t
)
t
X m (
x
)
Y n (
y
)
Z k (
z
) ,
1
M mnk β mnk
D mnk =
ϕ (
x
,
y
,
z
)
X m (
x
)
Y n (
y
)
Z k (
z
)
d x d y d z
.
Ω
Thus
m , n , k C mnk e α mnk t cos β mnk t · X m ( x ) Y n ( y ) Z k ( z )
1
τ 0 +( λ m + λ n + λ k )
B 2
=
1
M mnk β mnk
)+ m , n , k
t W ϕ (
,
,
,
x
y
z
t
2
e α mnk t sin
·
ϕ (
x
,
y
,
z
)
X m (
x
)
Y n (
y
)
Z k (
z
)
d x d y d z
β mnk t
·
X m (
x
)
Y n (
y
)
Z k (
z
) .
(6.87)
Ω
Substituting Eqs. (6.86) and (6.84) into Eq. (6.85) and using the structure of
W ψ (
x
,
y
,
z
,
t
)
in Eq. (6.84) yields the solution of PDS (6.81)
1
W ϕ (
τ 0 +
B 2 W
u
(
x
,
y
,
z
,
t
)=
x
,
y
,
z
,
t
)+
( λ m + λ n + λ k ) ϕ (
x
,
y
,
z
,
t
) .
t
Example 3 .Solve
u t
τ 0 +
B 2
A 2
u tt =
Δ
u
+
t Δ
u
+
f
(
x
,
y
,
z
,
t
) ,
0
<
x
<
l
,
0
<
y
<
a
,
0
<
z
<
b
,
0
<
t
,
u
(
0
,
y
,
z
,
t
)=
u
(
l
,
y
,
z
,
t
)=
0
,
(6.88)
u
(
x
,
0
,
z
,
t
)=
u y
(
x
,
a
,
z
,
t
)+
h 1 u
(
x
,
a
,
z
,
t
)=
0
,
(
,
,
,
)
(
,
,
,
)=
(
,
,
,
)=
,
u z
x
y
0
t
h 2 u
x
y
0
t
u
x
y
b
t
0
u
(
x
,
y
,
z
,
0
)= ϕ (
x
,
y
,
z
) ,
u t
(
x
,
y
,
z
,
0
)= ψ (
x
,
y
,
z
) .
 
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