Environmental Engineering Reference
In-Depth Information
Solution.
The solution of PDS (7.35) can be obtained by using a Fourier method of
expansion or the method of separation of variables. Here we attempt to use a Fourier
transformation to solve it.
Note that the boundary conditions in PDS (7.35) are all of the first kind along
both the
x
-and
y
-directions, thus we should use the Fourier sine transformation.
Also, the boundary conditions are homogeneous along the
x
-direction. A Fourier
sine transformation appears convenient with respect to
x
.
Let
u
π
=
u
(
x
,
y
)
sin
kx
d
x
. Thus
0
π
π
2
u
2
u
d
2
u
d
y
2
.
∂
∂
k
2
u
x
2
sin
kx
d
x
=
−
,
y
2
sin
kx
d
x
=
∂
∂
0
0
Also
π
1
1
u
0
k
k
+
u
0
sin
kx
d
x
=
+(
−
1
)
.
0
Thus PDS (7.35) reduces to
d
2
u
d
y
2
−
k
2
u
=
0
.
Its general solution can be readily obtained as
u
=
c
1
(
k
)
sh
ky
+
c
2
(
k
)
ch
ky
,
where
c
1
(
k
)
and
c
2
(
k
)
are constants with respect to
y
. Applying the boundary con-
dition
u
|
y
=
0
=
0 yields
c
2
(
k
)=
0.
c
1
(
k
)
can be determined by applying another
k
1
1
and using csch
k
u
0
k
+
boundary condition
u
|
y
=
π
=
+(
−
1
)
π
=
1
/
sh
k
π
. Finally,
we have
⎨
2
u
0
k
(
csch
k
π
)
sh
ky
,
k
=
2
n
+
1
,
n
=
0
,
1
,
2
, ··· ,
u
=
⎩
0
,
k
=
2
n
.
Its inverse Fourier sine transformation yields the solution of PDS (7.35)
∞
n
=
0
4
u
0
π
csch
(
2
n
+
1
)
π
·
sh
(
2
n
+
1
)
y
·
sin
(
2
n
+
1
)
x
u
(
x
,
y
)=
.
(7.36)
2
n
+
1
Example 6
. Consider the steady heat conduction in a cube of side length
.The
temperature is kept at zero on five boundary surfaces. The temperature is kept at
constant
u
0
on the sixth boundary surface. Find the temperature distribution in the
cube.
π
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