4Elliptic boundary value problems
III Analysis of Partial Differential Equations
4.4 Elliptic regularity
We can finally turn to the problem of regularity. We previously saw that when
solving
Lu
=
f
, if
f ∈ L
2
(
U
), then by definition of a weak solution, we have
u ∈ H
1
0
(
U
), so we have gained some regularity when solving the differential
equation. However, it is not clear that
u ∈ H
2
(
U
), so we cannot actually say
u
solves
Lu
=
f
. Even if
u ∈ H
2
(
U
), it may not be classically differentiable, so
Lu
=
f
isn’t still holding in the strongest possible sense. So we might hope that
under reasonable circumstances,
u
is in fact twice continuously differentiable.
But human desires are unlimited. If
f
is smooth, we might hope further that
u
is also smooth. All of these will be true.
Let’s think about how regularity may fail. It could be that the individual
derivatives of
u
are quite singular, but in
Lu
all these singularities happen to
cancel with each other. Thus, the content of elliptic regularity is that this doesn’t
happen.
To see why we should expect this to be true, suppose for convenience that
u, f ∈ C
∞
c
(R
n
) and
−∆u = f.
Using integration by parts, we compute
Z
R
n
f
2
dx =
Z
R
n
(∆u)
2
dx
=
X
i,j
Z
R
n
(D
i
D
i
u)(D
j
D
j
u) dx
=
X
i,j
Z
R
n
(D
i
D
j
u)(D
i
D
j
u) dx
= kD
2
uk
L
2
(R
n
)
.
So we have deduced that
kD
2
uk
L
2
(R
n
)
= k∆uk
L
2
(R
n
)
.
This is of course not a very useful result, because we have a priori assumed
that
u
and
f
are
C
∞
, while what we want to prove that
u
is, for example, in
H
2
(u). However, the fact that we can control the H
2
norm if we assumed that
u ∈ H
2
(
U
) gives us some strong indication that we should be able to show that
u must always be in H
2
(U).
The idea is to run essentially the same argument for weak solutions, without
mentioning the word “second derivative”. This involves the use of difference
quotients.
Definition
(Difference quotient)
.
Suppose
U ⊆ R
n
is open and
V b U
. For
0 < |h| < dist(V, ∂U), we define
∆
h
i
u(x) =
u(x + he
i
) − u(x)
h
∆
k
u(x) = (∆
h
1
u, . . . , ∆
h
n
u).
Observe that if
u ∈ L
2
(
U
), then ∆
h
u ∈ L
2
(
V
). If further
u ∈ H
1
(
U
), then
∆
h
u ∈ H
1
(V ) and D∆
h
u = ∆
h
Du.
What makes difference quotients useful is the following lemma:
Lemma. If u ∈ L
2
(U), then u ∈ H
1
(V ) iff
k∆
h
uk
L
2
(V )
≤ C
for some C and all 0 < |h| <
1
2
dist(V, ∂U ). In this case, we have
1
˜
C
kDuk
L
2
(V )
≤ k∆
h
uk
L
2
(V )
≤
˜
CkDuk
L
2
(V )
.
Proof. See example sheet.
Thus, if we are able to establish the bounds we had for the Laplacian using
difference quotients, then this tells us u is in H
2
loc
(U).
Lemma. If w, v and compactly supported in U, then
Z
U
w∆
−h
k
v dx =
Z
U
(∆
h
k
w)v dx
∆
h
k
(wv) = (τ
h
k
w)∆
h
k
v + (∆
h
k
w)v,
where τ
h
k
w(x) = w(x + he
k
).
Theorem
(Interior regularity)
.
Suppose
L
is uniformly elliptic on an open set
U ⊆ R
n
, and assume
a
ij
∈ C
1
(
U
),
b
i
, c ∈ L
∞
(
U
) and
f ∈ L
2
(
U
). Suppose
further that u ∈ H
1
(U) is such that
B[u, v] = (f, v)
L
2
(U)
(†)
for all v ∈ H
1
0
(U). Then u ∈ H
2
loc
(U), and for each V b U , we have
kuk
H
2
(V )
≤ C(kf k
L
2
(U)
+ kuk
L
2
(U)
),
with C depending on L, V, U , but not f or u.
Note that we don’t require
u ∈ H
1
0
(
U
), so we don’t require
u
to satisfy the
boundary conditions. In this case, there may be multiple solutions, so we need
the
u
on the right. Also, observe that we don’t actually need uniform ellipticity,
as the property of being in
H
2
loc
(
U
) can be checked locally, and
L
is always
locally uniformly elliptic.
The proof is essentially what we did for the Laplacian just now, except this
time it is much messier since we need to use difference quotients instead of
derivatives, and there are lots of derivatives of
a
ij
’s that have to be kept track
of.
When using regularity results, it is often convenient to not think about it in
terms of “solving equations”, but as something that (roughly) says “if
u
is such
that Lu happens to be in L
2
(say), then u is in H
2
loc
(U)”.
Proof.
We first show that we may in fact assume
b
i
=
c
= 0. Indeed, if we know
the theorem for such L, then given a general L, we write
L
0
u = −
X
(a
ij
u
x
j
)
x
i
, Ru =
X
b
i
u
x
i
+ cu.
Then if
u
is a weak solution to
Lu
=
f
, then it is also a weak solution to
L
0
u
=
f − Ru
. Noting that
Ru ∈ L
2
(
U
), this tells us
u ∈ H
2
loc
(
U
). Moreover,
on V b U,
– We can control kuk
H
2
(V )
by kf − Ruk
L
2
(V )
and kuk
L
2
(V )
(by theorem).
– We can control kf − Ruk
L
2
(V )
by kfk
L
2
(V )
, kuk
L
2
(V )
and kDuk
L
2
(V )
.
–
By G˚arding’s inequality, we can control
k
D
uk
L
2
(V )
by
kuk
L
2
(V )
and
B[u, u] = (f, u)
L
2
(V )
.
– By H¨older, we can control (f, u)
L
2
(V )
by kfk
L
2
(V )
and kuk
L
2
(V )
.
So it suffices to consider the case where
L
only has second derivatives. Fix
V b U
and choose
W
such that
V b W b U
. Take
ξ ∈ C
∞
c
(
W
) such that
ζ ≡
1
on V .
Recall that our example of Laplace’s equation, we considered the integral
R
f
2
d
x
and did some integration by parts. Essentially, what we did was to
apply the definition of a weak solution to ∆
u
. There we was lucky, and we could
obtain the result in one go. In general, we should consider the second derivatives
one by one.
For k ∈ {1, . . . n}, we consider the function
v = −∆
−h
k
(ζ
2
∆
h
k
u).
As we shall see, this is the correct way to express
u
x
k
x
k
in terms of difference
quotients (the
−h
in the first ∆
−h
k
comes from the fact that we want to integrate
by parts). We shall put this into the definition of a weak solution to say
B
[
u, v
] = (
f, v
). The plan is to isolate a
k
∆
h
k
D
uk
2
term on the left and then
bound it.
We first compute
B[u, v] = −
X
i,j
Z
U
a
ij
u
x
i
∆
−h
k
(ζ
2
∆
h
k
u)
x
j
dx
=
X
i,j
Z
U
∆
h
k
(a
ij
u
x
i
)(ζ
2
∆
h
k
u)
x
j
dx
=
X
i,j
Z
U
(τ
h
k
a
ij
∆
h
k
u
x
i
+ (∆
h
k
a
ij
)u
x
i
)(ζ
2
∆
h
k
u
x
j
+ 2ζζ
x
j
∆
h
k
u) dx
≡ A
1
+ A
2
,
where
A
1
=
X
i,j
Z
U
ξ
2
(τ
h
k
a
ij
)(∆
h
k
u
x
i
)(∆
h
k
u
x
j
) dx
A
2
=
X
i,j
Z
U
h
(∆
h
k
a
ij
)u
x
i
ζ
2
∆
h
k
u
x
j
+ 2ζζ
x
j
∆
h
k
u(τ
h
k
a
ij
∆
h
k
u
x
i
+ (∆
h
k
a
ij
)u
x
i
)
i
dx.
By uniform ellipticity, we can bound
A
1
≥ θ
Z
U
ξ
2
|∆
h
k
Du|
2
dx.
This is what we want to be small.
Note that
A
2
looks scary, but every term either only involves “first derivatives”
of
u
, or a product of a second derivative of
u
with a first derivative. Thus, applying
Young’s inequality, we can bound
|A
2
|
by a linear combination of
|
∆
h
k
D
u|
2
and
|Du|
2
, and we can make the coefficient of |∆
h
k
Du|
2
as small as possible.
In detail, since
a
ij
∈ C
1
(
U
) and
ζ
is supported in
W
, we can uniformly
bound a
ij
, ∆
h
k
a
ij
, ζ
x
j
, and we have
|A
2
| ≤ C
Z
W
h
ζ|∆
h
k
Du||Du| + ζ|Du||∆
h
k
u| + ζ|∆
h
k
Du||∆
h
k
u|
i
dx.
Now recall that
k
∆
h
k
uk
is bounded by
k
D
uk
. So applying Young’s inequality, we
may bound (for a different C)
|A
2
| ≤ ε
Z
W
ζ
2
|∆
h
k
Du|
2
+ C
Z
W
|Du|
2
dx.
Thus, taking ε =
θ
2
, it follows that
(f, v) = B[u, v] ≥
θ
2
Z
U
ζ
2
|∆
h
k
Du|
2
dx − C
Z
W
|Du|
2
dx.
This is promising.
It now suffices to bound (f, v) from above. By Young’s inequality,
|(f, v)| ≤
Z
|f||∆
−h
k
(ζ
2
∆
h
k
u)| dx
≤ C
Z
|f||D(ζ
2
∆
h
k
u)| dx
≤ ε
Z
|D(ζ
2
∆
h
k
u)|
2
dx + C
Z
|f|
2
dx
≤ ε
Z
|ζ
2
∆
h
k
Du|
2
dx + C(kfk
2
L
2
(U)
+ kDuk
2
L
2
(U)
)
Setting ε =
θ
4
, we get
Z
U
ζ
2
|∆
h
k
Du|
2
dx ≤ C(kf k
2
L
2
(W )
+ kDuk
2
L
2
(W )
),
and so, in particular, we get a uniform bound on
k
∆
h
k
D
uk
L
2
(V )
. Now as before,
we can use G˚arding to get rid of the kDuk
L
2
(W )
dependence on the right.
Notice that this is a local result. In order to have
u ∈ H
2
(
V
), it is enough
for us to have
f ∈ L
2
(
W
) for some
W
slightly larger than
V
. Thus, singularities
do not propagate either in from the boundary or from regions where
f
is not
well-behaved.
With elliptic regularity, we can understand weak solutions as genuine solutions
to the equation
Lu
=
f
. Indeed, if
u
is a weak solution, then for any
v ∈ C
∞
c
(
U
),
we have
B
[
u, v
] = (
f, v
), hence after integrating by parts, we recover (
Lu−f, v
) =
0 for all v ∈ C
∞
c
(U). So in fact Lu = f almost everywhere.
It is natural to hope that we can get better than
u ∈ H
2
loc
(
U
). This is actually
not hard given our current work. If
Lu
=
f
, and all
a
ij
, b
i
, c, f
are sufficiently
well-behaved, then we can simply differentiate the whole qeuation with respect
to
x
i
, and then observe that
u
x
i
satisfies some second-order elliptic PDE of the
form previously understood, and if we do this for all
i
, then we can conclude
that
u ∈ H
3
loc
(
U
). Of course, some bookkeeping has to be done if we were to do
this properly, since we need to write everything in weak form. However, this is
not particularly hard, and the details are left as an exercise.
Theorem
(Elliptic regularity)
.
If
a
ij
, b
i
and
c
are
C
m+1
(
U
) for some
m ∈ N
,
and f ∈ H
m
(U), then u ∈ H
m+2
loc
(U) and for V b W b U, we can estimate
kuk
H
m+2
(V )
≤ C(kf k
H
m
(W )
+ kuk
L
2
(W )
).
In particular, if
m
is large enough, then
u ∈ C
2
loc
(
U
), and if all
a
ij
, b
i
, c, f
are
smooth, then u is also smooth.
We can similarly obtain a H¨older theory of elliptic regularity, which gives
(roughly) f ∈ C
k,α
(U) implies u ∈ C
k+2,α
(U).
The final loose end is to figure out what happens at the boundary.
Theorem
(Boundary
H
2
regularity)
.
Assume
a
ij
∈ C
1
(
¯
U
),
b
1
, c ∈ L
∞
(
U
), and
f ∈ L
2
(
U
). Suppose
u ∈ H
1
0
(
U
) is a weak solution of
Lu
=
f, u|
∂U
= 0. Finally,
we assume that ∂U is C
2
. Then
kuk
H
2
(U)
≤ C(kf k
L
2
(U)
+ kuk
L
2
(U)
).
If
u
is the unique weak solution, we can drop the
kuk
L
2
(U)
from the right hand
side.
Proof.
Note that we already know that
u
is locally in
H
2
loc
(
U
). So we only have
to show that the second-derivative is well-behaved near the boundary.
By a partition of unity and change of coordinates, we may assume we are in
the case
U = B
1
(0) ∩ {x
n
> 0}.
Let
V
=
B
1/2
(0)
∩ {x
n
>
0
}
. Choose a
ζ ∈ C
∞
c
(
B
1
(0)) with
ζ ≡
1 on
V
and
0 ≤ ζ ≤ 1.
Most of the proof in the previous proof goes through, as long as we restrict
to
v = −∆
−h
k
(ζ
2
∆
h
k
u)
with
k 6
=
n
, since all the translations keep us within
U
, and hence are well-defined.
Thus, we control all second derivatives of the form D
k
D
i
u
, where
k ∈
{
1
, . . . , n −
1
}
and
i ∈ {
1
, . . . , n}
. The only remaining second-derivative to
control is D
n
D
n
u
. To understand this, we go back to the PDE and look at the
PDE itself. Recall that we know it holds pointwise almost everywhere, so
n
X
i,j=1
(a
ij
u
x
i
)
x
j
+
n
X
i=1
b
i
u
x
i
+ cu = f.
So we can write
a
nn
u
x
n
u
x
n
=
F
almost everywhere, where
F
depends on
a, b, c, f
and all (up to) second derivatives of
u
that are not
u
x
n
x
n
. Thus,
F
is controlled
in
L
2
. But uniform ellipticity implies
a
nn
is bounded away from 0. So we are
done.
Similraly, we can reiterate this to obtain higher regularity results.