III Analysis of Partial Differential Equations - Elliptic boundary value problems

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

is a bounded open set with smooth boundary.

Suppose

∂U

is a perfect conductor and

U → R

is the charge density inside

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

be open and bounded with

boundary, and for

u ∈ C

(

U), we define

Lu = −

i,j=1

(x)u

)

i=1

(x)u

+ c(x)u,

where

and

are given functions defined on

. Typically, we will assume

they are at least L

∞

, but sometimes we will require more.

If a

∈ C

(U), then we can rewrite this as

Lu = −

i,j=1

(x)u

i=1

(x)u

+ c(x)u

for some

, 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.

i,j

(x)ξ

≥ 0

for all x ∈ U and ξ ∈ R

It turns out this is not quite strong enough, because this condition allows

the a

’s to be degenerate, or vanish at the boundary.

Definition (Uniform ellipticity). An operator

Lu = −

i,j=1

(x)u

)

i=1

(x)u

+ c(x)u

is uniformly elliptic if

i,j=1

(x)ξ

≥ θ|ξ|

for some θ > 0 and all x ∈ U, ξ ∈ R

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

(

) is a solution, and suppose

v ∈ C

(

) also satisfies

∂U

= 0. Multiply the equation

and integrate by parts. Then we

get

vf dx =





v + cuv





dx ≡ B[u, v]. (2)

Conversely, suppose

u ∈ C

(

) and

∂U

= 0. If

[

u, v

] for all

v ∈ C

(

) such that

∂U

= 0, then we claim

in fact solves the original

equation.

Indeed, undoing the integration by parts, we conclude that

vLu dx =

vf dx

for all

v ∈ C

(

) with

∂U

= 0. But if this is true for all

, then it must be

that Lu = f.

Thus, the PDE problem we started with is equivalent to finding

that solves

B[u, v] =

vf dx for all suitable v, provided u is regular enough.

But the point is that (2) makes sense for

u, v ∈ H

(

). So our strategy is

to first show that we can find

u ∈ H

(

) that solves (2), and then hope that

under reasonable assumptions, we can show that any such solution must in fact

be C

(

U).

Definition (Weak solution). We say u ∈ H

(U) is a weak solution of

Lu = f on U

u = 0 on ∂U

for f ∈ L

(U) if

B[u, v] = (f, v)

(U)

for all v ∈ H

(U).

We’ll exploit the Hilbert space structure of H

(U) to find weak solutions.

Theorem

(Lax–Milgram theorem)

Let

be a real Hilbert space with inner

product (

·, ·

). Suppose

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

≤ B[u, u] (coercivity)

Then if

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

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

. 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

– Since B is bilinear, it is immediate that A : H → H is linear.

– A is bounded, since we have

kAuk

= (Au, Au) = B[u, Au] ≤ αkukkAuk.

– A is injective and has closed image. Indeed, by coercivity, we know

βkuk

≤ B[u, u] = (Au, u) ≤ kAukkuk.

Dividing by

kuk

, we see that

is bounded below, hence is injective and

has closed image (since H is complete).

(Indeed, injectivity is clear, and if

→ v

for some

, then

−u

k ≤

kAu

− Au

k →

0 as

m, n → ∞

. So (

) 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

≤ 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

satisfy boundedness and coercivity. Unfortunately, this is not always

true.

Theorem

(Energy estimates for

)

Suppose

, b

, c ∈ L

∞

(

), and

there exists θ > 0 such that

i,j=1

(x)ξ

≥ θ|ξ|

for almost every x ∈ U and ξ ∈ R

. Then if B is defined by

B[u, v] =





v + cuv





dx,

then there exists α, β > 0 and γ ≥ 0 such that

(i) |B[u, v]| ≤ αkuk

(U)

kvk

(U)

for all u, v ∈ H

(U)

(ii) βkuk

(U)

≤ B[u, u] + γkuk

(U)

Moreover, if b

≡ 0 and c ≥ 0, then we can take γ.

Proof.

(i) We estimate

|B[u, v]| ≤

i,j

∞

(U)

|Du||Dv| dx

kbk

∞

(U)

|Du||v| dx

+ kck

∞

(U)

|u||v| dx

≤ c

kDuk

(U)

kDvk

(u)

+ c

kDuk

(U)

kvk

(U)

+ c

kuk

(U)

kvk

(u)

≤ αkuk

(U)

kvk

(U)

for some α.

(ii) We start from uniform ellipticity. This implies

|Du|

dx ≤

i,j=1

(x)u

= B[u, u] −

i=1

u + cu

≤ B[u, u] +

i=1

∞

(U)

|Du||u| dx

+ kck

∞

(U)

|u|

dx.

Now by Young’s inequality, we have

|Du||u| dx ≤ ε

|Du|

dx +

4ε

|u|

for any ε > 0. We choose ε small enough so that

i=1

∞

(U)

≤

So we have

|Du|

dx ≤ B[u, u] +

|Du|

dx + γ

|u|

for some γ. This implies

kDuk

(U)

≤ B[u, u] + γkuk

(U)

We can add

kuk

(U)

on both sides to get the desired bound on

kuk

(U)

To get the “moreover” statement, we see that under these conditions, we have

|Du|

dx ≤ B[u, u].

Then we apply the Poincar´e’s inequality, which tells us there is some

C >

such that for all u ∈ H

(U), we have

kuk

(U)

≤ CkDuk

(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

(U), there exists a unique weak solution to

Lu + µu = f on U

u = 0 on ∂U .

Moreover, we have

kuk

(U)

≤ Ckf k

(U)

for some C = C(L, U) ≥ 0.

Again, if b

≡ 0 and c ≥ 0, then we may take γ = 0.

Proof.

Take

from the previous theorem when applied to

. Then if

µ ≥ γ

and

we set

[u, v] = B[u, v] + µ(u, v)

(U)

This is the bilinear form corresponding to the operator

= L + µ.

Then by the previous theorem,

satisfies boundedness and coercivity. So if we

fix any f ∈ L

, and think of it as an element of H

(U)

∗

hf, vi = (f, u)

(U)

fv dx,

then we can apply Lax–Milgram to find a unique

u ∈ H

(

) satisfying

[

u, v

] =

hf, vi

= (

f, v

)

(U)

for all

v ∈ H

(

). This is precisely the condition for

to be

a weak solution.

Finally, the G˚arding inequality tells us

βkuk

(U)

≤ B

[u, u] = (f, u)

(U)

≤ kf k

(U)

kuk

(U)

So we know that

βkuk

(U)

≤ kf k

(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

is better

behaved, then

is also better behaved, and in general, if

f ∈ H

, then

u ∈ H

k+2

This is known as elliptic regularity. Together Sobolev inequalities, this tells us

is genuinely a classical solution.

Another problem is the presence of the

. We noted that if

is, say, Laplace’s

equation, then we can take

= 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.