Environmental Engineering Reference
In-Depth Information
1
ε
1
ε
=
−
r
1
·
n
d
S
=
−
d
S
=
−
4
π
,
(7.76)
2
2
r
=
ε
r
=
ε
or
1
4
d
−
Δ
Ω
=
1
,
π
r
R
3
where the
r
1
is the unit normal vector of
r
=
M
0
M
.
1
r
does not change its sign in the neighborhood of
Equation (7.76) shows that
Δ
R
3
M
0
. Therefore we have, for any
ϕ
(
x
,
y
,
z
)
∈
C
(
)
,
1
4
1
4
−
Δ
ϕ
(
,
,
)
Ω
=
−
Δ
ϕ
(
,
,
)
x
y
z
d
lim
ε
→
x
y
z
d
Ω
π
r
π
r
0
R
3
r
≤
ε
1
4
d
=
lim
ε
→
0
ϕ
(
ξ
,
η
,
ζ
)
−
Δ
Ω
=
lim
ε
→
0
ϕ
(
ξ
,
η
,
ζ
)=
ϕ
(
x
0
,
y
0
,
z
0
)
.
π
r
r
≤
ε
1
4
Thus, by the definition of the Dirac function,
−
Δ
=
δ
(
M
−
M
0
)
. Similarly,
π
r
1
2
ln
1
r
1
2
ln
1
for the two-dimensional case,
−
Δ
=
δ
(
M
−
M
0
)
. Therefore,
r
and
π
π
1
r
are also called the fundamental solutions of two- and three-dimensional har-
monic equations, respectively.
4
π
7.4.3 Harmonic Functions
Five Theorems
C
2
∈
(
Ω
)
Theorem 1
.Let
u
.
u
is a harmonic function in
Ω
if and only if, for any
closed sub-domain
¯
Ω
∗
with a smooth boundary surface
∂Ω
∗
,
∂
u
¯
Ω
∗
=
Ω
∗
∪
∂Ω
∗
,
n
d
S
=
0
,
(7.77)
∂
∂Ω
∗
where
∂
u
n
is the normal derivative.
Proof.
With
v
∂
0,
∂
v
∂
=
1 (thus
Δ
v
=
n
=
0), applying the second Green formula yields
∂
u
Δ
u
d
Ω
=
n
d
S
.
(7.78)
∂
Ω
∗
∂Ω
∗
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