Environmental Engineering Reference
In-Depth Information
Since
u
≥
c
and
u
is a harmonic function,
u
−
c
is nonnegative and is also a
harmonic function. Now consider two spheres
V
M
1
R
1
and
V
M
2
R
2
.Here
M
1
and
M
2
are
two arbitrary points in
R
3
,
R
1
=
R
2
+
r
M
1
M
2
,and
r
M
1
M
2
is the distance between
M
1
and
M
2
. Thus
V
M
2
V
M
1
R
1
R
2
⊂
.
This, with
u
−
c
≥
0, yields
[
u
(
M
)
−
c
]
d
Ω
≤
[
u
(
M
)
−
c
]
d
Ω
.
V
M
2
R
2
V
M
1
R
1
By using Eq. (7.81), we obtain
R
1
R
2
3
u
(
M
2
)
−
c
≤
[
u
(
M
1
)
−
c
]
.
R
1
R
2
3
Note that
lim
R
2
→
∞
=
1. Thus
u
(
M
2
)
−
c
≤
u
(
M
1
)
−
c
or
u
(
M
2
)
≤
u
(
M
1
)
.
By interchanging the rule for
M
1
and
M
2
, we can also obtain
u
(
M
1
)
≤
u
(
M
2
)
.
Therefore,
(
)=
(
)
.
u
M
2
u
M
1
The
u
thus must be a constant due to the arbitrariness of
M
1
and
M
2
.
Theorem 4 (Extremum Principle).
Assume that
u
(
M
)
is a harmonic function in
, continuous in a closed domain
¯
and non-constant in
¯
Ω
Ω
=
Ω
∪
∂Ω
Ω
.Then
u
(
M
)
can take its maximum and minimum values only on the boundary
∂Ω
.
Proof.
Since a maximum of
u
is the minimum of
−
u
, we need to prove the theorem
only for the case of maximum value.
In contradiction to the theorem, suppose that
u
takes its maximum value at point
M
0
inside
for all points
M
on
S
M
0
R
Ω
. This assumption would lead to
u
(
M
)
≡
u
(
M
0
)
,
which is the spherical surface of a sphere
V
M
0
R
of center
M
0
and radius
R
inside
Ω
.
for a point
M
1
on
S
M
R
. Because of the
Additionally, suppose that
u
(
M
1
)
<
u
(
M
0
)
continuity of
u
in
¯
around
M
1
on
S
M
0
R
Ω
, there is a region
σ
where
u
(
M
)
<
u
(
M
0
)
for
all
M
on
σ
.Hereweuse
σ
to represent both region and its area. We thus have, by
Theorem 2
⎡
⎣
⎤
⎦
R
2
S
M
0
R
1
1
u
(
M
0
)=
u
d
S
=
u
d
S
+
u
d
S
4
π
4
π
R
2
σ
S
M
0
R
\
σ
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