Environmental Engineering Reference
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0at M 0 Ω . Without loss of generality, consider v
v
(
M 0 ) =
(
M 0 ) >
0. By the con-
dition lim
r
v
=
0, there always exists a spherical surface S R with a sufficiently large
radius R : x 2
y 2
z 2
R 2 such that M 0 is inside the region
∂Ω
+
+
=
Ω
R of boundary
and S R and sup
S R |
v
| <
v
(
M 0 )
. By the extremum principle, the harmonic function in
∂Ω . This requires a strictly pos-
Ω R must take its maximum value at the boundary
v
itive
n somewhere on the boundary by the strong extremum principle. However,
∂Ω =
v
0andsup
S R |
v
| <
v
(
M 0 )
so v does not take its maximum on the boundary
n
of region
Ω R , which is contrary to the extremum principle. Thus we obtain v
0so
the solution is unique.
Theorem 5. If a Robin problem
, ∂Ω ,
=
Δ
u
0
u ∂Ω =
(7.89)
u
n + σ
f
, σ >
0
.
has a solution, the solution must be unique.
=
Proof. Let u 1 and u 2 be two solutions of PDS (7.89). v
u 1
u 2 must thus satisfy
Δ
v
=
0
,
v ∂Ω =
v
n + σ
0
, σ >
0
.
First prove v
. By the extremum
principle, v can take its minimum and maximum values only on the boundary
c (constant). Suppose that v
c (constant) in
Ω
,
say at points M 0 and M 1 , respectively. By the strong extremum principle, we have
∂Ω
v ∂Ω =
0, so that, by the boundary condition
v
(
M 0 )
v
<
n + σ
0, v
(
M 0 )=
n
1
σ
v
M 0 )
(
>
0. Thus the minimum value of the harmonic function v is larger than
n
zero in ¯
(
) <
0, so that the maximum value of v is
smaller than zero. We thus arrive at a contradiction. Therefore, v
Ω
. Similarly, we can show v
M 1
c in ¯
Ω
. Also, by
the boundary condition,
v ∂Ω =
c Ω = σ
v
c
n + σ
n + σ
v
| ∂Ω =
0
,
so that c
=
0and v
0. Therefore the solution is unique.
¯
u in the first Green formula. We can readily use C 2
C 1
Remark .Let v
=
( Ω )
(
Ω )
to
=
show the uniqueness of boundary-value problems of
0ofthefirstandthethird
kinds and the uniqueness up to a constant of the second boundary-value problems
of
Δ
u
=
Δ
u
0.
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