Environmental Engineering Reference
In-Depth Information
na
n
−
1
a
n
,
B
n
=
na
n
−
1
b
n
. Equation (7.29) becomes
Let
A
n
=
r
a
n
∞
n
=
1
a
n
u
(
r
,
θ
)=
A
0
+
(
A
n
cos
n
θ
+
B
n
sin
n
θ
)
2π
r
a
n
1
π
∞
n
=
1
a
n
(
θ
)[
θ
cos
n
θ
sin
n
θ
=
A
0
+
f
cos
n
θ
+
sin
n
θ
]
d
0
∞
n
=
1
d
2π
r
a
n
a
π
1
n
(
θ
)
(
θ
−
θ
)
θ
=
A
0
+
f
cos
n
0
2
ln
1
2
d
2π
r
a
a
π
1
2
r
a
cos
(
θ
)
(
θ
−
θ
)+
θ
=
A
0
+
f
−
−
0
a
2
ln
a
2
2π
0
a
2
2π
0
(
θ
)
θ
−
(
θ
)
=
A
0
+
f
d
f
π
π
ln
a
2
r
2
d
(
θ
−
θ
)+
θ
·
−
2
ar
cos
2π
ln
a
2
r
2
d
a
2
(
θ
)
(
θ
−
θ
)+
θ
.
=
A
0
−
f
−
2
ar
cos
(7.31)
π
0
This is called the
integral formula of internal Neumann problems in a circular do-
main
. It can be readily shown that the integral formula of external Neumann prob-
lems in a circular domain is
r
a
−
n
∞
n
=
1
a
n
u
(
r
,
θ
)=
A
0
−
(
A
n
cos
n
θ
+
B
n
sin
n
θ
)
2π
ln
a
2
r
2
d
a
2
(
θ
)
(
θ
−
θ
)+
θ
.
=
A
0
+
f
−
2
ar
cos
(7.32)
π
0
Remark 2.
For the case of steady heat conduction in a domain without an internal
source/sink, the net heat flowing into the domain from the boundary must be zero
by the first law of the thermodynamics. This agrees with the necessary condition for
the existence of solutions of the Neumann problems. Under these conditions, the
system can be at different steady-states (depending on the initial conditions of the
system) with a constant temperature difference for all points. This agrees with the
presence of constant
A
0
in Eq. (7.31).
7.2.2 Fourier Sine/Cosine Transformation in a Finite Region
For a function
f
(
x
)
that satisfies the conditions for Fourier sine expansion in
[
0
,
π
]
,
we have
⎧
⎨
π
f
(
k
)=
f
(
x
)
sin
kx
d
x
,
(7.33a)
0
⎩
∞
k
=
1
2
π
f
f
(
x
)=
(
k
)
sin
kx
.
(7.33b)
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