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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