Environmental Engineering Reference
In-Depth Information
Solution.
Let
u
(
x
,
y
,
z
)
be the temperature. It must satisfy
⎧
⎨
Δ
u
=
0
,
0
<
x
,
y
,
z
<
π
,
u
(
0
,
y
,
z
)=
u
(
π
,
y
,
z
)=
0
,
(7.37)
⎩
u
(
x
,
y
,
0
)=
u
(
x
,
y
,
π
)=
0
,
u
(
x
,
0
,
z
)=
0
,
u
(
x
,
π
,
z
)=
u
0
.
Consider a Fourier sine transformation of
u
with respect to
x
π
u
=
u
(
x
,
y
,
z
)
sin
kx
d
x
.
0
The PDS (7.37) is thus transformed into
⎧
⎨
k
2
u
u
yy
+
u
zz
=
,
u
=
u
(
k
,
y
,
z
)
,
0
<
y
,
z
<
π
,
u
|
z
=
0
=
u
|
z
=
π
=
0
,
⎧
⎨
⎩
2
u
0
k
,
for odd
k
,
u
|
y
=
0
=
0
,
u
|
y
=
π
=
⎩
0
,
for even
k
.
Consider a Fourier sine transformation of
u
with respect to
z
π
u
u
sin
k
z
d
z
=
.
0
The PDS (7.37) is hence transformed into
⎧
⎨
k
or
k
is even
0
,
,
d
2
u
d
y
2
=(
k
2
k
2
u
u
u
+
)
,
|
y
=
0
=
0
,
|
y
=
π
=
4
u
0
kk
,
⎩
both
k
and
k
are odd
.
Its general solution is
k
)
k
)
u
=
c
1
(
k
,
sh
ly
+
c
2
(
k
,
ch
ly
,
where
l
2
k
2
k
2
2
2
,
m
,
c
1
and
c
2
are constants
with respect to
y
and can be determined by applying the boundary conditions of
u
.
Finally, we obtain
=
+
=(
2
m
+
1
)
+(
2
n
+
1
)
,
n
=
0
,
1
,
2
, ···
⎧
⎨
k
or
k
is even
0
,
,
u
=
4
u
0
kk
sh
ly
sh
l
⎩
k
=
π
,
2
m
+
1
,
k
=
2
n
+
1
.
Its inverse transformation reads
∞
m
=
0
8
u
0
sh
ly
sh
l
sin
(
2
m
+
1
)
z
u
=
.
(
2
n
+
1
)
π
π
2
m
+
1
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