4Elliptic boundary value problems

III Analysis of Partial Differential Equations



4.1 Existence of weak solutions
In this chapter, we are going to study second-order elliptic boundary value
problems. The canonical example to keep in mind is the following:
Example.
Suppose
U R
n
is a bounded open set with smooth boundary.
Suppose
U
is a perfect conductor and
ρ
:
U R
is the charge density inside
U
.
The electrostatic field φ satisfies
φ = ρ on U
φ|
U
= 0.
This is an example of an elliptic boundary value problem. Note that we cannot
tackle this with the Cauchy–Kovalevskaya theorem, since we don’t even have
enough boundary conditions, and also because we want an everywhere-defined
solution.
In general, let
U R
n
be open and bounded with
C
1
boundary, and for
u C
2
(
¯
U), we define
Lu =
n
X
i,j=1
(a
ij
(x)u
x
j
)
x
i
+
n
X
i=1
b
i
(x)u
x
i
+ c(x)u,
where
a
ij
,
b
i
and
c
are given functions defined on
U
. Typically, we will assume
they are at least L
, but sometimes we will require more.
If a
ij
C
1
(U), then we can rewrite this as
Lu =
n
X
i,j=1
a
ij
(x)u
x
i
x
j
+
n
X
i=1
˜
b
i
(x)u
x
i
+ c(x)u
for some
˜
b
i
, using the product rule.
We will mostly use the first form, called the divergence form, which is suitable
for the energy method, while the second (non-divergence form) is suited to the
maximum principle. Essentially, what makes the divergence form convenient for
us is that it’s easy to integrate by parts.
Of course, given the title of the chapter, we assume that L is elliptic, i.e.
X
i,j
a
ij
(x)ξ
i
ξ
j
0
for all x U and ξ R
n
.
It turns out this is not quite strong enough, because this condition allows
the a
ij
’s to be degenerate, or vanish at the boundary.
Definition (Uniform ellipticity). An operator
Lu =
n
X
i,j=1
(a
ij
(x)u
j
)
x
i
+
n
X
i=1
b
i
(x)u
x
i
+ c(x)u
is uniformly elliptic if
n
X
i,j=1
a
ij
(x)ξ
i
ξ
j
θ|ξ|
2
for some θ > 0 and all x U, ξ R
n
.
We shall consider the boundary value problem
Lu = f on U
u = 0 on U.
This form of the equation is not very amenable to study by functional analytic
methods. Similar to what we did in the proof of Picard–Lindel¨of, we want to
write this in a weak formulation.
Let’s suppose
u C
2
(
¯
U
) is a solution, and suppose
v C
2
(
¯
U
) also satisfies
v|
U
= 0. Multiply the equation
Lu
=
f
by
v
and integrate by parts. Then we
get
Z
U
vf dx =
Z
U
X
ij
v
x
i
a
ij
u
x
j
+
X
i
b
i
u
x
i
v + cuv
dx B[u, v]. (2)
Conversely, suppose
u C
2
(
¯
U
) and
u|
U
= 0. If
R
U
vf
d
x
=
B
[
u, v
] for all
v C
2
(
¯
U
) such that
v|
U
= 0, then we claim
u
in fact solves the original
equation.
Indeed, undoing the integration by parts, we conclude that
Z
vLu dx =
Z
vf dx
for all
v C
2
(
¯
U
) with
v|
U
= 0. But if this is true for all
v
, then it must be
that Lu = f.
Thus, the PDE problem we started with is equivalent to finding
u
that solves
B[u, v] =
R
U
vf dx for all suitable v, provided u is regular enough.
But the point is that (2) makes sense for
u, v H
1
0
(
U
). So our strategy is
to first show that we can find
u H
1
0
(
U
) that solves (2), and then hope that
under reasonable assumptions, we can show that any such solution must in fact
be C
2
(
¯
U).
Definition (Weak solution). We say u H
1
0
(U) is a weak solution of
Lu = f on U
u = 0 on U
for f L
2
(U) if
B[u, v] = (f, v)
L
2
(U)
for all v H
1
0
(U).
We’ll exploit the Hilbert space structure of H
1
0
(U) to find weak solutions.
Theorem
(Lax–Milgram theorem)
.
Let
H
be a real Hilbert space with inner
product (
·, ·
). Suppose
B
:
H × H R
is a bilinear mapping such that there
exists constants α, β > 0 so that
|B[u, v]| αkukkvk for all u, v H (boundedness)
βkuk
2
B[u, u] (coercivity)
Then if
f
:
H R
is a bounded linear map, then there exists a unique
u H
such that
B[u, v] = hf, vi
for all v H.
Note that if
B
is just the inner product, then this is the Riesz representation
theorem.
Proof.
By the Riesz representation theorem, we may assume that there is some
w such that
hf, vi = (u, v).
For each fixed u H, the map
v 7→ B[u, v]
is a bounded linear functional on
H
. So by the Riesz representation theorem,
we can find some Au such that
B[u, v] = (Au, v).
It then suffices to show that A is invertible, for then we can take u = A
1
w.
Since B is bilinear, it is immediate that A : H H is linear.
A is bounded, since we have
kAuk
2
= (Au, Au) = B[u, Au] αkukkAuk.
A is injective and has closed image. Indeed, by coercivity, we know
βkuk
2
B[u, u] = (Au, u) kAukkuk.
Dividing by
kuk
, we see that
A
is bounded below, hence is injective and
has closed image (since H is complete).
(Indeed, injectivity is clear, and if
Au
m
v
for some
v
, then
ku
m
u
n
k
1
β
kAu
m
Au
n
k
0 as
m, n
. So (
u
n
) is Cauchy, and hence has a
limit u. Then by continuity, Au = v, and in particular, v im A)
Since im A is closed, we know
H = im A im A
.
Now let w im A
. Then we can estimate
βkwk
2
B[w, w] = (Aw, w) = 0.
So w = 0. Thus, in fact im A
= {0}, and so A is surjective.
We would like to apply this to our elliptic PDE. To do so, we need to prove
that our
B
satisfy boundedness and coercivity. Unfortunately, this is not always
true.
Theorem
(Energy estimates for
B
)
.
Suppose
a
ij
=
a
ji
, b
i
, c L
(
U
), and
there exists θ > 0 such that
n
X
i,j=1
a
ij
(x)ξ
i
ξ
j
θ|ξ|
2
for almost every x U and ξ R
n
. Then if B is defined by
B[u, v] =
Z
U
X
ij
v
x
i
a
ij
u
x
j
+
X
i
b
i
u
x
i
v + cuv
dx,
then there exists α, β > 0 and γ 0 such that
(i) |B[u, v]| αkuk
H
1
(U)
kvk
H
1
(U)
for all u, v H
1
0
(U)
(ii) βkuk
2
H
1
(U)
B[u, u] + γkuk
2
L
2
(U)
.
Moreover, if b
i
0 and c 0, then we can take γ.
Proof.
(i) We estimate
|B[u, v]|
X
i,j
ka
ij
k
L
(U)
Z
U
|Du||Dv| dx
+
X
i
kbk
C
(U)
Z
U
|Du||v| dx
+ kck
L
(U)
Z
U
|u||v| dx
c
1
kDuk
L
2
(U)
kDvk
L
2
(u)
+ c
2
kDuk
L
2
(U)
kvk
L
2
(U)
+ c
3
kuk
L
2
(U)
kvk
L
2
(u)
αkuk
H
1
(U)
kvk
H
1
(U)
for some α.
(ii) We start from uniform ellipticity. This implies
θ
Z
U
|Du|
2
dx
Z
U
n
X
i,j=1
a
ij
(x)u
x
i
u
x
j
dx
= B[u, u]
Z
U
n
X
i=1
b
i
u
x
i
u + cu
2
dx
B[u, u] +
n
X
i=1
kb
i
k
L
(U)
Z
|Du||u| dx
+ kck
L
(U)
Z
U
|u|
2
dx.
Now by Young’s inequality, we have
Z
U
|Du||u| dx ε
Z
U
|Du|
2
dx +
1
4ε
Z
U
|u|
2
dx
for any ε > 0. We choose ε small enough so that
ε
n
X
i=1
kb
i
k
L
(U)
θ
2
.
So we have
θ
Z
U
|Du|
2
dx B[u, u] +
θ
2
Z
U
|Du|
2
dx + γ
Z
U
|u|
2
dx
for some γ. This implies
θ
2
kDuk
2
L
2
(U)
B[u, u] + γkuk
2
L
2
(U)
We can add
θ
2
kuk
2
L
2
(U)
on both sides to get the desired bound on
kuk
H
1
(U)
.
To get the “moreover” statement, we see that under these conditions, we have
θ
Z
|Du|
2
dx B[u, u].
Then we apply the Poincar´e’s inequality, which tells us there is some
C >
0
such that for all u H
1
0
(U), we have
kuk
L
2
(U)
CkDuk
L
2
(U)
.
The estimate (ii) is sometimes called G˚arding’s inequality.
Theorem.
Let
U, L
be as above. There is a
γ
0 such that for any
µ γ
and
any f L
2
(U), there exists a unique weak solution to
Lu + µu = f on U
u = 0 on U .
Moreover, we have
kuk
H
1
(U)
Ckf k
L
2
(U)
for some C = C(L, U) 0.
Again, if b
i
0 and c 0, then we may take γ = 0.
Proof.
Take
γ
from the previous theorem when applied to
L
. Then if
µ γ
and
we set
B
µ
[u, v] = B[u, v] + µ(u, v)
L
2
(U)
,
This is the bilinear form corresponding to the operator
L
µ
= L + µ.
Then by the previous theorem,
B
µ
satisfies boundedness and coercivity. So if we
fix any f L
2
, and think of it as an element of H
1
0
(U)
by
hf, vi = (f, u)
L
2
(U)
=
Z
U
fv dx,
then we can apply Lax–Milgram to find a unique
u H
1
0
(
U
) satisfying
B
µ
[
u, v
] =
hf, vi
= (
f, v
)
L
2
(U)
for all
v H
1
0
(
U
). This is precisely the condition for
u
to be
a weak solution.
Finally, the G˚arding inequality tells us
βkuk
2
H
1
(U)
B
µ
[u, u] = (f, u)
L
2
(U)
kf k
L
2
(U)
kuk
L
2
(U)
.
So we know that
βkuk
H
1
(U)
kf k
L
2
(U)
.
In some way, this is a magical result. We managed to solve a PDE without
having to actually work with a PDE. There are a few things we might object
to. First of all, we only obtained a weak solution, and not a genuine solution.
We will show that under some reasonable assumptions on
a, b, c
, if
f
is better
behaved, then
u
is also better behaved, and in general, if
f H
k
, then
u H
k+2
.
This is known as elliptic regularity. Together Sobolev inequalities, this tells us
u
is genuinely a classical solution.
Another problem is the presence of the
µ
. We noted that if
L
is, say, Laplace’s
equation, then we can take
γ
= 0, and so we don’t have this problem. But in
general, this theorem requires it, and this is a bit unsatisfactory. We would like
to think a bit more about it.