Civil Engineering Reference
In-Depth Information
In particular,
Lu
≤
0 implies
max
x
u(x)
≤
max
{
0
,
max
x
∂
u(x)
}
.
(
2
.
5
)
∈
∈
Proof.
(1) Apply the maximum principle to
v
:
=−
u
.
(2) By construction,
Lw
=
Lv
−
Lu
≥
0 and
w
≥
0on
∂
, where
w
:
=
v
−
u
.
It follows from the minimum principle that inf
w
≥
0, and thus
w(x)
≥
0in
.
(3)
Lw
=
0 for
w
:
=
u
1
−
u
2
. It follows from the maximum principle that
w(x)
≤
sup
z
∈
∂
w(z)
≤
sup
z
∈
∂
|
w(z)
|
. Similarly, the minimum principle implies
w(x)
≥−
.
(4) Suppose
is contained in a circle of radius
R
. Since we are free to choose
the coordinate system, we may assume without loss of generality that the center
of this circle is at the origin. Let
sup
z
∈
∂
|
w(z)
|
w(x)
=
R
2
x
i
.
−
i
≤
w
≤
R
2
Since
w
x
i
x
k
=−
2
δ
ik
, clearly
Lw
≥
2
nα
and 0
in
, where
α
is the
ellipticity constant appearing in Definition 2.2. Let
1
2
nα
v(x)
:
=
z
∈
∂
|
u(z)
|+
w(x)
·
sup
z
∈
∂
|
Lu(z)
|
.
sup
Then by construction,
Lv
≥|
Lu
|
in
, and
v
≥|
u
|
on
∂
. The comparison
−
v(x)
≤
u(x)
≤+
v(x)
in
. Since
w
≤
R
2
, we get (2.3)
principle in (2) implies
with
c
=
R
2
/
2
nα
.
(5) It suffices to give a proof for
x
0
∈
and
u(x
0
)
=
sup
z
∈
u(z) >
0. Then
Lu(x
0
)
−
c(x
0
)u(x
0
)
≤
Lu(x
0
)
≤
0, and moreover, the principal part
Lu
−
cu
defines an elliptic operator. Now the proof proceeds as for Theorem 2.1.
Problem
2.4
For a uniformly elliptic differential operator of the form (2.4), show that the
solution depends continuously on the data.
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