Lecture Eight: Maximum principles and gradient estimates
1
The Maximum Principle for more general operators
Let u : Br → Rn be a C 2 function, and let L be a uniformly elliptic differential operator
taking
Lu =
�
Aij
i,j
∂2u
∂xi ∂xj
(1)
for some real n × n symmetric matrix A with continuously differentiable entries. If x is an
internal maximum of u then
�u = 0 and Lu ≤ 0
at x. When A is the identity matrix this is exactly the maximum principle from lecture 1.
If not then we pick coordnates at x so that A(x) is diagonal. Since all the eigenvalues of
A(x) are positive we have
Lu =
�
bi
i
∂2u
|
∂x2i (x)
for some collection of positive constants bi . Since x is a maximum
for all i, and Lu ≤ 0 as expected.
2
(2)
∂2u
|
∂x2i (x)
is non­positive
The gradient estimate for L­harmonic functions
Recall the gradient estimate
sup |�u| ≤
Br
c(n)
sup |u|
r B2r
for harmonic functions u. We will prove a similar estimate for uniformly elliptic operators.
1
Proposition 2.1 Let L be a uniformly elliptic operator as above, with
λ|v|2 ≤ v · (Av) ≤ Λ|v|2
for some real 0 < λ ≤ Λ. There are constants C which depend only on the operator and
the dimension of the space such that
C
sup |�u| ≤ sup |u|
(3)
r B2r
Br
for all L­harmonic functions u on B2r .
Proof This proof follows essentially the same steps as the proof for the earlier gradient
estimate from lecture 2, but each step is now more complicated. When constants are
intrioduced it is important to noctive that they depend only on A, n, λ and Λ. As before
we will first prove the case r = 1, and then extend to general r.
Step 1. One key part of the proof in the harmonic case was the Bochner formula. We
will prove a similar result for L harmonic functions. Calculate
∂2
|�u|2 =
∂xi ∂xj
∂ �
∂ 2 u ∂u
2
∂xi
∂xj ∂xk ∂xk
(4)
k
= 2
�
k
∂3u
∂u
∂2u
∂2u
+
.
∂xi ∂xj ∂xk ∂xk
∂xj ∂xk ∂xi ∂xk
(5)
Therefore
L|�u|2 = 2
�
Aij
ijk
Let vk =
in gives
∂u
∂xk .
∂3u
∂u
+2
∂i ∂xj ∂xk ∂xk
�
k
⎛
⎞
2
2
�
∂ u
∂ u ⎠
⎝
Aij
.
∂xj ∂xk ∂xi ∂xk
(6)
ij
The last term is �vk · A�vk ≥ λ|�vk |2 by uniform ellipticity. Substituting
2
L|�u| ≥ 2
�
ijk
�
∂3u
∂u
Aij
+ 2λ
∂xi ∂xj ∂xk ∂xk
ik
�
∂2u
∂xi ∂xk
�2
.
Now work on the first term.
�
�
� ∂ �
� ∂Aij ∂ 2 u ∂u
∂3u
∂u
∂2u
∂u
Aij
=
Aij
−
∂xk ∂xi ∂xj ∂xk
∂xi ∂xj ∂xk ∂xk
∂xk
∂xi ∂xj ∂xk
ijk
ijk
=
ijk
∂2u
� ∂Aij
� ∂
∂u
∂u
−
(Lu)
∂xk
∂xk
∂xk ∂xi ∂xj ∂xk
k
= −
ijk
� ∂Aij
ijk
∂2u
∂u
∂xk ∂xi ∂xj ∂xk
2
(7)
since Lu = 0. Together with (7) this gives
2
L|�u| ≥ −2
ijk
Let cij =
∂Aij ∂u
k ∂xk ∂xk
�
�
∂ 2 u ∂u
+ 2λ
∂xk ∂xi ∂xj ∂xk
� ∂Aij
and dij =
ik
∂2u
∂xi ∂xj .
�
cij dij + 2λ
�
ij
λd2ij
�2
.
(8)
d2ij .
(9)
ij
c2ij
− 2cij dij +
=
λ
and so
∂2u
∂xi ∂xk
We can re­write (8) as
L|�u|2 ≥ −2
Note that
�
�
√
cij
√ − λdij
λ
�2
≥ 0,
1
λd2ij − 2cij dij ≥ − c2ij .
λ
(10)
(11)
Apply this to (9) to get
L|�u|2 ≥ −
�
1� 2
cij + λ
d2ij ,
λ
ij
(12)
i,j
or, in the old notation,
�
�2
� � ∂ 2 u �2
1 � � ∂Aij ∂u
+λ
≥ −
λ
∂xk ∂xk
∂xi ∂xk
ij
k
ik
� � ∂ 2 u �2
1�
(�Aij · �u)2 + λ
≥ −
λ
∂xi ∂xk
ij
ik
� � ∂ 2 u �2
1�
2
2
≥ −
|�Aij | |�u| + λ
λ
∂xi ∂xk
ij
ik
⎛
⎞
� � ∂ 2 u �2
1 �
≥ − ⎝
.
|�Aij |2 ⎠ |�u|2 + λ
λ
∂xi ∂xk
L|�u|2
ij
Pick c1 = supB2r
1
λ
ik
��
�
2 > 0. We have
|�A
|
ij
ij
� � ∂ 2 u �2
L|�u| ≥ −c1 |�u| + λ
.
∂xi ∂xk
2
2
ik
Step 2. We will also need to estimate
3
(13)
∂u2
∂xi ∂xj
i,j
�
�
∂
∂u
= Aij
2u
∂xi
∂xj
∂2u
∂u ∂u
= 2uAij
+ 2Aij
∂xi ∂xj
∂xj ∂xi
∂u ∂u
= 2Aij
∂xj ∂xi
�
L(u2 ) =
Aij
since Lu = 0. Now apply uniform ellipticity to get
L(u2 ) ≥ 2λ|�u|2 .
(14)
Step 3. Assume r = 1, and pick φ a test function with φ = 0 on ∂B2 , and φ > 0 on
the interior. We need a bound for L(φ2 |�u|2 ). Calculate
∂ 2 (φ2 |�u|2 )
∂xi ∂xj
∂ 2 |�u|2
∂ 2 φ2
∂φ ∂|�u|2
= φ2 Aij
+ |�u|2 Aij
+ 2Aij φ
∂xi ∂xj
∂xi ∂xj
∂xi ∂xj
∂u ∂φ ∂ 2 u
= φ2 L(|�u|2 ) + L(φ2 )|�u|2 + 4Aij φ
.
∂xk ∂xi ∂xj ∂xk
L(φ2 |�u|2 ) = Aij
(15)
(16)
(17)
Note that Lφ2 is bounded on B2 , say Lφ2 ≥ c2 . Applying this bound together with
inequality (13) we get
2
2
2
2
L(φ |�u| ) ≥ (c2 − c1 φ )|�u| + λφ
2
�
∂2u
∂xj ∂xk
�2
+ 4Aij φ
∂u ∂φ ∂ 2 u
.
∂xk ∂xi ∂xj ∂xk
(18)
The function φ2 is also bounded on B2 , so we can pick a positive constant c3 such that
c1 − kφ2 ≥ −c3 . Then
� 2
�2
∂u ∂φ ∂ 2 u
∂ u
2
2
2
2
+ 4Aij φ
.
(19)
L(φ |�u| ) ≥ −c3 |�u| + λφ
∂xj ∂xk
∂xk ∂xi ∂xj ∂xk
2
u
∂u ∂φ
Now set γjk = φ ∂x∂j ∂x
and δjk = Aij ∂x
, and rewrite as
k
k ∂xi
2
L(φ2 |�u|2 ) ≥ −c3 |�u|2 + λγik
+ 4γjk δjk .
Use an absorbing inequality to show that
4
(20)
2
L(φ2 |�u|2 ) ≥ −c3 |�u|2 − c4 δjk
(21)
for some positive constant c4 . Substitute in for δjk and we have
�2
∂u
∂φ
Aij
L(φ2 |�u|2 ) ≥ −c3 |�u|2 − c4
∂xk ∂xi
i
j.k
�
�
��
∂u ∂φ 2
2
≥ −c3 |�u| − c4 n
Aij
∂xk ∂xi
�
� �
j.k
i
since for any collection of real numbers the average of the squares is greater than the
square of the averages. The functions Aij are bounded on B2 , so there is a constant k with
|Aij | ≤ k. Thus
� � � ∂u ∂φ �2
L(φ |�u| ) ≥ −c3 |�u| − c4 nk
∂xk ∂xi
2
2
2
2
2
2 2
j.k
i
≥ −c3 |�u| − c4 n k |�φ|2 |�u|2 ,
and since |�φ|2 is bounded on B2
L(φ2 |�u|2 ) ≥ −c5 |�u|2
(22)
for some constant c5 which depends only on the dimension and L.
Step 4. Apply the maximum principle. By steps 2 and 3
c5 2
u ) ≥ 0,
(23)
2λ
c5 2
so, by the maximum principle, φ2 |�u|2 + 2λ
u has no interior maxima. Therefore it takes
it’s maximum on the boundary. We have shown that
L(φ2 |�u|2 +
sup(φ2 |�u|2 +
B2
c5 2
c5 2
u ) = sup(φ2 |�u|2 +
u ).
2λ
2λ
∂B2
(24)
Recall that φ is zero on the boundary, and choose it to be identically one on B1 . Thus
c5 2
u )
2λ
B1
c5 2
≤ sup(φ2 |�u|2 +
u )
2λ
B2
c5 2
≤ sup(φ2 |�u|2 +
u )
2λ
∂B2
sup |�u|2 ≤ sup(φ2 |�u|2 +
B1
5
≤
≤
c5
sup u2
2λ ∂B2
c5
sup u2 .
2λ B2
Finally take square roots to obtain
sup |�u| ≤
B1
c5
sup |u|
2λ B2
(25)
as expected.
step 5. Scale for general r. Assume u is L­harmonic on B2r , and define u� (x) = u(rx).
Then u� is L harmonic on B2 . Plugging u� into (25) we get
r sup |�u| ≤
Br
as required.
6
c5
sup |u|
2λ B2r
(26)