Environmental Engineering Reference
In-Depth Information
Substituting it into the Laplace equation in PDS (7.54) yields the ordinary differen-
tial equation of
Y
m
(
y
)
, whose solution can be readily obtained. We thus obtain the
v
that satisfies the Laplace equation in PDS (7.54) and
x
-direction boundary
conditions
(
x
,
y
)
m
=
0
a
m
ch
m
π
y
cos
m
∞
b
m
sh
m
π
y
π
x
(
,
)=
+
,
v
x
y
(7.56)
a
a
a
where
a
m
and
b
m
are constants. Applying the boundary condition
v
y
y
=
0
=
0 yields
b
m
=
0
.
Thus
∞
m
=
0
a
m
ch
m
π
y
∞
m
=
1
cos
m
x
a
,
π
m
π
a
m
a
sh
m
π
y
cos
m
x
a
.
π
v
=
v
y
=
a
a
Note that the normal square of
cos
m
is
a
π
x
2
. Applying the boundary condition
a
v
y
y
=
b
=
g
(
x
)
yields
a
m
π
a
m
a
sh
m
π
b
a
=
2
a
cos
m
π
x
g
(
x
)
d
x
a
0
c
0
−
2
a
abq
kc
cos
m
π
x
=
d
x
a
2
bq
kc
a
m
sin
m
π
x
c
=
−
π
a
0
2
abq
m
kc
sin
m
c
a
,
π
=
−
π
so that
m
2
sh
m
2
a
2
bq
k
sin
m
π
c
b
a
.
π
a
m
=
−
(7.57)
π
2
c
a
Finally, we obtain the solution of PDS (7.53)
y
2
2
+
∞
m
=
1
sin
m
π
c
a
sh
m
π
b
a
2
a
2
b
π
q
k
1
m
2
ch
m
π
y
cos
m
π
x
(
,
)=
−
.
T
x
y
(7.58)
2
c
a
a
Example 2
. Consider the distribution of electric potential in a hollow sphere of inner
radius
R
1
and outer radius
R
2
(
. The electric potential is kept at a constant
v
0
on the inner surface and at zero on the outer surface. Find the steady distribution
of the electric potential inside of the hollow sphere.
Solution
. The electric potential
v
satisfies the Laplace equation at a steady state (see
Section 1.2). By the given boundary conditions,
v
cannot depend on
R
1
<
R
2
)
θ
and
ϕ
in a
sphere coordinate system
(
r
,
θ
,
ϕ
)
, so it varies only along the radial direction. Thus
the
v
(
r
)
must satisfy
⎧
⎨
d
2
v
d
r
2
+
2
r
d
v
d
r
=
0
,
R
1
<
r
<
R
2
,
(7.59)
⎩
v
(
R
1
)=
v
0
,
v
(
R
2
)=
0
.
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