Environmental Engineering Reference
In-Depth Information
If
u
is a harmonic function in
Ω
such that
Δ
u
=
0, Eq. (7.78) leads to Eq. (7.77). If
¯
Ω
∗
, Eq. (7.78) yields
Eq. (7.77) holds for any
∀
Ω
∗
.
Ω
=
,
Δ
u
d
0
Ω
∗
Ω
∗
,
This shows that, by the continuity of
Δ
u
and the arbitrariness of
Δ
u
=
0.
Ω
∗
in Theorem 1 to be
C
2
Remark 1
.Take
Ω
. For a harmonic function
u
∈
(
Ω
)
∩
¯
C
1
(
Ω
)
, an application of Theorem 1 yields
∂
u
n
d
S
=
0
.
∂
∂Ω
Therefore a necessary condition for the existence of solutions of the Neumann
problem
⎧
⎨
Δ
u
=
0
,
Ω
,
∂Ω
=
∂
u
⎩
(7.79)
f
.
∂
n
is
=
.
f
d
S
0
∂Ω
It can be readily proven that
∂Ω
f
d
S
=
0 is also a sufficient condition for the exis-
tence of solutions of PDS (7.79).
For example, the problem
⎨
1
r
z
2
x
2
y
2
Δ
u
=
0
,
r
<
=
+
+
,
r
=
1
=
∂
u
⎩
1
.
∂
n
has no solution because
r
f
d
S
=
d
S
=
4
π
=
0.
=
1
r
=
1
Remark 2.
Take the
u
in Theorem 1 as the steady temperature in a domain
Ω
with-
k
∂Ω
∗
∂
u
out any source or sink inside so that
Δ
u
=
0. The
−
n
d
S
is, by the Fourier
∂
∂Ω
∗
. Theorem 1 shows
that at a steady state, both the necessary and sufficient condition for non-existence
of an internal heat source/sink is zero net heat flux across any closed surface in the
domain. This clearly agrees with the physical reality.
law of heat conduction, the net heat across the boundary
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