III Analysis of Partial Differential Equations - Elliptic boundary value problems

4Elliptic boundary value problems

III Analysis of Partial Differential Equations

4.2 The Fredholm alternative

To understand the second problem, we shall seek to prove the following theorem:

Theorem (Fredholm alternative). Consider the problem

Lu = f, u|

∂U

= 0. (∗)

For

a uniformly elliptic operator on an open bounded set

with

boundary,

either

(i)

For each

f ∈ L

(

), there is a unique weak solution

u ∈ H

(

) to (

∗

); or

(ii)

There exists a non-zero weak solution

u ∈ H

(

) to the homogeneous

problem, i.e. (∗) with f = 0.

This is similar to what we know about solving matrix equations

—

either there is a solution for all

, or there are infinitely many solutions to the

homogneous problem.

Similar to the previous theorem, this follows from some general functional

analytic result. Recall the definition of a compact operator:

Definition

(Compact operator)

A bounded operator

H → H

is compact

if every bounded sequence (

)

∞

m=1

has a subsequence

such that (

)

∞

j=1

converges strongly in H.

Recall (or prove as an exercise) the following theorem regarding compact

operators.

Theorem

(Fredholm alternative)

Let

be a Hilbert space and

H → H

be a compact operator. Then

(i) ker(I − K) is finite-dimensional.

(ii) im(I − K) is finite-dimensional.

(iii) im(I − K) = ker(I − K

†

)

⊥

(iv) ker(I − K) = {0} iff im(I −K) = H.

(v) dim ker(I − K) = dim ker(I −K

†

) = dim coker(I −K).

How do we apply this to our situation? Our previous theorem told us that

is invertible for large

, and we claim that (

)

−1

is compact. We can

then deduce the previous result by applying (iv) of the Fredholm alternative

with K a (scalar multiple of) (L + γ)

−1

(plus some bookkeeping).

So let us show that (

)

−1

is compact. Note that this maps sends

f ∈ L

(

)

u ∈ H

(

). To make it an endomorphism, we have to compose this with the

inclusion

(

)

→ L

(

). The proof that (

)

−1

is compact will not involve

(

)

−1

in any way — we shall show that the inclusion

(

)

→ L

(

) is

compact!

We shall prove this in two steps. First, we need the notion of weak conver-

gence.

Definition

(Weak convergence)

Suppose (

)

∞

n=1

is a sequence in a Hilbert

space H. We say u

converges weakly to u ∈ H if

, w) → (u, w)

for all w ∈ H. We write u

* u.

Of course, we have

Lemma. Weak limits are unique.

Lemma. Strong convergence implies weak convergence.

We shall show that given any bounded sequence in

(

), we can find

a subsequence that is weakly convergent. We then show that every weakly

convergent sequence in H

(U) is strongly convergent in L

(U).

In fact, the first result is completely general:

Theorem

(Weak compactness)

Let

be a separable Hilbert space, and suppose

(

)

∞

m=1

is a bounded sequence in

with

k ≤ K

for all

. Then

admits

a subsequence (u

)

∞

j=1

such that u

* u for some u ∈ H with kuk ≤ K.

One can prove this theorem without assuming

is separable, but it is slightly

messier.

Proof.

Let (

)

∞

i=1

be an orthonormal basis for

. Consider (

, u

). By Cauchy–

Schwarz, we have

|(e

, u

)| ≤ ke

kke

k ≤ K.

So by Bolzano–Weierstrass, there exists a subsequence (

) such that (

, u

)

converges.

Doing this iteratively, we can find a subsequence (

) such that for each

there is some c

such that (e

, v

) → c

as ` → ∞.

We would expect the weak limit to be

. To prove this, we need to first

show it converges. We have

j=1

= lim

k→∞

j=1

|(e

, v

≤ sup

j=1

|(e

, v

≤ sup kv

≤ K

using Bessel’s inequality. So

u =

∞

j=1

converges in H, and kuk ≤ K. We already have

, v

) → (e

, u)

for all

. Since

− uk

is bounded by 2

, it follows that the set of all

such

that

(w, v

) → (v, u) (†)

is closed under finite linear combinations and taking limits, hence is all of

To see that it is closed under limits, suppose

→ w

, and

satisfy (

†

). Then

|(w, v

)−(w, u)| ≤ |(w −w

, v

−u)|+|(w

, v

−u)| ≤ 2Kkw−w

k+|(w

, v

−u)|

So we can first find

large enough such that the first term is small, then pick

such that the second is small.

We next want to show that if

* u

(

), then

→ u

. We

may as well assume that

is some large cube of length

by extension. Notice

that since

is bounded, the constant function 1 is in

(

). So

* u

particular implies

− u) dx → 0.

Recall that the Poincar´e inequality tells us if

u ∈ H

(

), then we can bound

kuk

(Q)

by some multiple of

(U)

. If we try to prove this without the

assumption that

vanishes on the boundary, then we find that we need a

correction term. The resulting lemma is as follows:

Lemma

(Poincar´e revisited)

Suppose

u ∈ H

(

). Let

= [

, ξ

]

×···×

[ξ

, ξ

+ L] be a cube of length L. Then we have

kuk

(Q)

≤

|Q|



u(x) dx



kDuk

(Q)

We can improve this to obtain better bounds by subdividing

into smaller

cubes, and then applying this to each of the cubes individually. By subdividing

enough, this leads to a proof that u

* u in H

implies u

→ u in H

Proof. By approximation, we can assume u ∈ C

∞

(

Q). For x, y ∈ Q, we write

u(x) − u(y) =

u(t, x

, . . . , x

) dt

u(y

, t, x

, . . . , x

) dt

+ ···

u(y

, . . . , y

n−1

, t) dt.

Squaring, and using 2ab ≤ a

+ b

, we have

u(x)

+ u(y)

− 2u(x)u(y) ≤ n



u(t, x

, . . . , x

) dt



+ ···

+ n



u(y

, . . . , y

n−1

, t) dt



Now integrate over x and y. On the left, we get

Q×Q

dx dy (u(x)

+ u(y)

− 2u(x)u(y)) = 2|Q|kuk

(Q)

− 2



u(x) dx



On the right we have



u(t, x

, . . . , x

) dt



≤



u(t, x

, . . . , x

)



dt (Cauchy–Schwarz)

≤ L



u(t, x

, . . . , x

)



dt.

Integrating over all x, y ∈ Q, we get

Q×Q

dx dy I

≤ L

|Q|kD

(Q)

Similarly estimating the terms on the right-hand side, we find that

2|Q|kuk

(Q)

− 2



u(x) dx



≤ n|Q|

i=1

(Q)

= n|Q|L

kDuk

(Q)

It now follows that

Theorem

(Rellich–Kondrachov)

Let

U ⊆ R

be open, bounded with

boundary. Then if (

)

∞

m=1

is a sequence in

(

) with

* u

, then

→ u

in L

In particular, by weak compactness any sequence in

(

) has a subsequence

that is convergent in L

(U).

Note that to obtain the “in particular” part, we need to know that

(

) is

separable. This is an exercise on the example sheet. Alternatively, we can appeal

to a stronger version of weak compactness that does not assume separability.

Proof.

By the extension theorem, we may assume

for some large cube

with U b Q.

We subdivide

into

many cubes of side length

, such that the cubes

only intersect at their faces. Call these {Q

}

a=1

We apply Poincar´e separately to each of these to obtain

− uk

(Q)

a=1

− uk

)

≤

a=1



− u) dx



nδ

kDu

− Duk

)

a=1



− u) dx



nδ

kDu

− Duk

(Q)

Now since

−

(Q)

is fixed, for

small enough, the second term is

Then since u

* u, we in particular have

− u) dx → 0 as i → ∞

for all

, since this is just the inner product with the constant function 1. So for

i large enough, the first term is also <

The same result holds with

(

) replaced by

(

). The proof is in fact

simpler, and we wouldn’t need the assumption that the boundary is C

Corollary.

Suppose

(

)

→ H

(

) is a bounded linear operator. Then

the composition

(U) H

(U) L

(U)

is compact.

The slogan is that we get compactness whenever we improve regularity, which

is something that happens in much more generality.

Proof.

Indeed, if

∈ L

(

) is bounded, then

is also bounded. So by

Rellich–Kondrachov, there exists a subsequence u

→ u in L

(U).

We are now ready to prove the Fredholm alternative for elliptic boundary

value problems. Recall that in our description of the Fredholm alternative, we

had the direct characterizations

(

I − K

) =

ker

(

I − K

†

)

⊥

. We can make the

analogous statement here. To do so, we need to talk about the adjoint of

Since

is not an operator defined on

(

), trying to write down what it means

to be an adjoint is slightly messy. Instead, we shall be content with talking

about “formal adjoints”.

It’s been a while since we’ve met a PDE, so let’s recall the setting we had.

We have a uniformly elliptic operator

Lu = −

i,j=1

(x)u

)

i=1

(x)u

+ c(x)u

on an open bounded set U with C

boundary. The associated bilinear form is

B[u, v] =





i,j

(x)u

i=1

(x)u

v + c(x)uv





dx.

We are interested solving in the boundary value problem

Lu = f, u|

∂u

= 0

with f ∈ L

(U).

The formal adjoint of L is defined by the relation

(Lφ, ψ)

(U)

= (φ, L

†

ψ)

(U)

for all φ, ψ ∈ C

∞

(U). By integration by parts, we know L

†

should be given by

†

v = −

i,j=1

)

−

i=1

(x)v

c −

i=1

Note that here we have to assume that

∈ C

(

). However, what really

interests us is the adjoint bilinear form, which is simply given by

†

[v, u] = B[u, v].

We are actually just interested in

†

, and not

†

, and we can sensibly talk about

†

even if b

is not differentiable.

As usual, we say

v ∈ H

(

) is a weak solution of the adjoint problem

†

v = f, v|

∂U

= 0 if

†

[v, u] = (f, u)

(U)

for all u ∈ H

(U).

Given this set up, we can now state and prove the Fredholm alternative.

Theorem

(Fredholm alternative for elliptic BVP)

Let

be a uniformly elliptic

operator on an open bounded set U with C

boundary. Consider the problem

Lu = f, u|

∂U

= 0. (∗)

Then exactly one of the following are true:

(i) For each f ∈ L

(U), there is a unique weak solution u ∈ H

(U) to (∗)

(ii)

There exists a non-zero weak solution

u ∈ H

(

) to the homogeneous

problem, i.e. (∗) with f = 0.

If this holds, then the dimension of

ker L ⊆ H

(

) is equal to the

dimension of N

∗

= ker L

†

⊆ H

(U).

Finally, (∗) has a solution if and only if (f, v)

(U)

= 0 for all v ∈ N

∗

Proof.

We know that there exists

γ >

0 such that for any

f ∈ L

(

), there is a

unique weak solution u ∈ H

(U) to

u = Lu + γu = f, u|

∂U

= 0.

Moreover, we have the bound

kuk

(U)

≤ Ckf k

(U)

(which gives uniqueness).

Thus, we can set

−1

to be this

, and then

−1

(

)

→ H

(

) is a

bounded linear map. Composing with the inclusion

(

), we get a compact

endomorphism of L

(U).

Now suppose u ∈ H

is a weak solution to (∗). Then

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

(U)

for all v ∈ H

(U)

is true if and only if

[u, v] ≡ B[u, v] + γ(u, v) = (f + γu, v) for all v ∈ H

(U).

Hence, u is a weak solution of (∗) if and only if

u = L

−1

(f + γu) = γL

−1

u + L

−1

In other words, u solves (∗) iff

u − Ku = h,

for

K = γL

−1

, h = L

−1

Since we know that

(

)

→ L

(

) is compact, by the Fredholm alternative

for compact operators, either

(i) u − Ku = h admits a solution u ∈ L

(U) for all h ∈ L

(U); or

(ii)

There exists a non-zero

u ∈ L

(

) such that

u − Ku

= 0. Moreover,

im(I − K) = ker(I −K

†

)

⊥

and dim ker(I − K) = dim im(I − K)

⊥

There is a bit of bookkeeping to show that this corresponds to the two alternatives

in the theorem.

(i) We need to show that u ∈ H

(U). But this is trivial, since we have

u = γL

−1

u + L

−1

and we know that L

−1

maps L

(U) into H

(U).

(ii)

As above, we know that the non-zero solution

. There are two things to

show. First, we have to show that

v − K

†

= 0 iff

is a weak solution to

†

v = 0, v|

∂U

= 0.

Next, we need to show that h = L

−1

f ∈ (N

∗

)

⊥

iff f ∈ (N

∗

)

⊥

For the first part, we want to show that

v ∈ ker

(

I − K

†

) iff

†

[

v, u

] =

B[u, v] = 0 for all u ∈ H

(U).

We are good at evaluating

[

u, v

] when

is of the form

−1

, by definition

of a weak solution. Fortunately,

im L

−1

contains

∞

(

), since

−1

for all

φ ∈ C

∞

(

). In particular,

im L

−1

is dense in

(

). So it

suffices to show that

v ∈ ker

(

I − K

†

) iff

[

−1

w, v

] = 0 for

w ∈ L

(

This is immediate from the computation

B[L

−1

w, v] = B

−1

w, v]−γ(L

−1

w, v) = (w, v)−(Kw, v) = (w, v−K

†

v).

The second is also easy — if v ∈ N

∗

= ker(I −K

†

), then

−1

f, v) =

(Kf, v) =

(f, K

†

v) =

(f, v).