Next: Problems Up: Lecture notes for 18.155, Previous: Sobolev embedding   Contents

Differential operators.

In the last third of the course we will apply what we have learned about distributions, and a little more, to understand properties of differential operators with constant coefficients. Before I start talking about these, I want to prove another density result.

So far we have not defined a topology on - I will leave this as an optional exercise.18 However we shall consider a notion of convergence. Suppose is a sequence in . It is said to converge weakly to if

 (133)

There is no uniformity' assumed here, it is rather like pointwise convergence (except the linearity of the functions makes it seem stronger).

Proof. We can use Schwartz representation theorem to write, for some depending on ,

We know that is dense in , in the sense of metric spaces so we can find , in . The density result then follows from the basic properties of weak convergence.

Proof. This follows by writing everyting in terms of pairings, for example if

This weak density shows that our definition of , and are unique if we require Proposition 10.2 to hold.

We have discussed differentiation as an operator (meaning just a linear map between spaces of function-like objects)

Any polynomial on

defines a differential operator19

 (134)

Before discussing any general theorems let me consider some examples.

 (135) (136) (137) (138) (139)

Functions, or distributions, satisfying are said to be holomorphic, those satisfying are said to be harmonic.

This is quite hard to prove and not as interetsing as it might seem. We will however give lots of examples, starting with . Consider the function

 (140)

Proof. Since is smooth and bounded away from the origin the local integrability follows from the estimate, using polar coordinates,

 (141)

Differentiating directly in the region where it is smooth,

so indeed, in .20

The derivative is really defined by

Here I have cut the space out of the integral and used the local integrability in taking the limit as . Integrating by parts in we find

There is a corrsponding formula for integration by parts in so, recalling that away from ,

assuming that both limits exist. Now, we can write

Replacing by either or in (10.13) both limits are zero. For example

Thus we get the same result in (10.13) by replacing by . Then ,

Let me remind you that we have already discussed the convolution of functions

This makes sense provided is of slow growth and . In fact we can rewrite the definition in terms of pairing

 (143)

where the indicates the variable in the pairing.

Proof. If then for any fixed ,

Indeed the seminorm estimates required are

Since and

we conclude that

The continuity of means that for some

so it follows that

 (144)

The argument above shows that is a continuous function of with values in , so is continuous and satisfies (10.15). It is therefore an element of .

Differentiability follows in the same way since for each , with the th unit vector

is continuous in , . Thus, has continuous partial derivatives and

The same argument then shows that . That follows from the definition of derivative of distributions

Finally consider the support property. Here we are assuming that is compact; we also know that is a closed set. We have to show that

 (145)

implies for near . Now (10.16) just means that

 (146)

Since , so both statements mean that there is no with . This can also be written

and as we showed when discussing supports implies

From (10.17) this is an open condition on , so the support property follows.

Now suppose and . Then

 (147)

This is really Hörmander's Lemma 4.1.3 and Theorem 4.1.2; I ask you to prove it as Problem 35.

We have shown that is if and , i.e., the regularity of follows from the regularity of one of the factors. This makes it reasonable to expect that can be defined when , and one of them has compact support. If and then

where . In fact using Problem 35,

 (148)

Here, are both smooth, but notice

Proof. Since has compact support there exists such that . Then

Thus, for some ,

where is one of our norms on . Since is supported in some large ball,

Thus is bounded for each . The same argument applies to the derivative using Theorem 10.6, so

In fact we get a little more, since we see that for each there exists and (depending on and ) such that

This means that

is a continuous linear map.

Now (10.19) allows us to define when and has compact support by

Using the continuity above, I ask you to check that in Problem 36. For the moment let me assume that this convolution has the same properties as before - I ask you to check the main parts of this in Problem 37.

Recall that is a fundamental situation for , a constant coefficient differential operator, if . We also use a weaker notion.

Since the same must work for nearby points in (10.22), the set is closed. Furthermore

 (149)

As Problem 37 I ask you to show that if and the with in a neighbourhood of such that . In particular

 (150)

Proof. One half of this is true for any differential operator:

Proof. We must show that . Now, if we can find , near , such that . Then

The first term is and , so .

It remains to show the converse of (10.26) where is assumed to be hypoelliptic. Take , a parametrix for with and assume, or rather arrange, that have compact support. In fact if we can arrange that

Now with so

Since it suffices to show that

Take with , but

Then , so

where and . It follows that .

Recall from last time that a differential operator is said to be hypoelliptic if there exists with

 (151)

The second condition here means that if and in for some then . Since we conclude that

and we may well suppose that , replaced now by , has compact support. Last time I showed that

I will remind you of the proof later.

First however I want to discuss the important notion of ellipticity. Remember that is really' just a polynomial, called the characteristic polynomial

It has the property

This shows (if it isn't already obvious) that we can remove from thought of as an operator on .

We can think of inverting by dividing by . This works well provided , for all . An example of this is

However even the Laplacian, , does not satisfy this rather stringent condition.

It is reasonable to expect the top order derivatives to be the most important. We therefore consider

the leading part, or principal symbol, of .

So what I want to show today is

We want to find a parametrix for ; we already know that we might as well suppose that has compact support. Taking the Fourier transform of (10.27) we see that should satisfy

 (152)

Here we use the fact that , so too.

First suppose that is actually homogeneous of degree . Thus

The assumption at ellipticity means that

 (153)

Since is compact and is continuous

 (154)

for some constant . Using homogeneity

 (155)

Now, to get from (10.28) we want to divide by or multiply by . The only problem with defining is at . We shall simply avoid this unfortunate point by choosing as before, with in .

Proof. Clearly is a continuous function and , so . It therefore is the Fourier transform of some . Furthermore

Since , . Thus is a parametrix for . We still need to show the hard part' that

 (156)

We can show (10.33) by considering the distributions . The idea is that for large, vanishes rather rapidly at the origin and this should weaken' the singularity of there. In fact we shall show that

 (157)

If you recall, these Sobolev spaces are defined in terms of the Fourier transform, namely we must show that

Now , so what we need to cinsider is the behaviour of the derivatives of , which is just in (10.32).

Proof. The estimate in (10.36) for is just (10.35). To prove the higher estimates that for each there is a polynomial of degree at most such that

 (158)

Once we know (10.37) we get (10.36) straight away since

We can prove (10.37) by induction, since it is certainly true for . Suppose it is true for . To get the same identity for each with it is enough to differentiate one of the identities with once. Thus

Since is a polynomial of degree at most this proves the lemma.

Going backwards, observe that is smooth in , so (10.36) implies that

which certainly holds if , giving (10.34). Now, by Sobolev's embedding theorem

In particular this means that if we choose with then for every , is smooth and

Thus and this is what we wanted to show, .

So now we have actually proved that is hypoelliptic if it is elliptic. Rather than go through the proof again to make sure, let me go on to the general case and in doing so review it.

Proof. Proof of theorem. We need to show that if is elliptic then has a parametrix as in (10.27). From the discussion above the ellipticity of implies (and is equivalent to)

On the other hand

is a polynomial of degree at most , so

This means that id is large enough then in , , so

This means that itself satisfies the conditions of Lemma 10.14. Thus if is equal to in a large enough ball then in and satisfies (10.36) which can be written

The discussion above now shows that defining by gives a solution to (10.27).

The last step in the proof is to show that if has compact support, and satisfies (10.27), then

Let me refine this result a little bit.

Proof. We need to show that then . Once we can fix , we might as well suppose that has compact support too. Indeed, choose a large ball so that

This is possible by the assumed boundedness of . Then choose with on ; it follows from Theorem L16.2, or rather its extension to distributions, that , so we can replace by , noting that . Now if has compact support we can choose compact neighbourhoods , of and such that . Furthermore we an decompose , so that , and . It follows that

Now, , by the support property of convolution and the three other terms are , since at least one of the factors is . Thus .

The most important example of a differential operator which is hypoelliptic, but not elliptic, is the heat operator

 (160)

In fact the distribution

is a fundamental solution. First we need to check that is a distribution. Certainly is in . Moreover as in it vanishes with all derivatives, so it is except at , . Since it is clearly measurable we will check that it is locally integrable near the origin, i.e.,

 (162)

since . We can change variables, setting , so and the integral becomes

Since is actually bounded near infinity, it follows that ,

As before we want to compute

First we check that in , where it is a function. This is a straightforward computation:

Now we can integrate by parts in (10.42) to get

Making the same change of variables as before, ,

As the integral here is bounded by the integrable function , for some , so by Lebesgue's theorem of dominated convergence, conveys to the integral of the limit. This is

Thus

so is indeed a fundamental solution. Since it vanishes in it is canned a forward fundamrntal solution.

Let's see what we can use it for.

Proof. Naturally we try . That it satisfies (10.43)follows from the properties of convolution. Similarly if is such that then

So we need to show uniqueness. If in two solutions of (10.43) then their difference satisfies the homogeneous' equation . Furthermore, in for some . Given any choose with in , in and consider

where for some , near 0. Thus has comapct support and in fact . I ask you to check this last statement as Problem L18.P1.

Anyway,

Now,

Since , the second tier here is supported in . Thus in , but is arbitrary, so .

Notice that the assumption that is not redundant in the statement of the Proposition, if we allow `large'' solutions they become non-unique. Problem L18.P2 asks you to apply the fundamental solution to solve the initial value problem for the heat operator.

Next we make similar use of the fundamental solution for Laplace's operator. If the

 (164)

is a fundamental solution. You should check that in directly, I will show later that , for the appropriate choice of , but you can do it directly, as in the case .

Proof. Since convolution is defined we certainly get a solution to this way. We need to check that . First we know that is hypoelliptic so we can decompose

and then . In fact we can see from (10.44) that

Now, , as we showed before, and continuing the integral we see that

Since it follows that .

So only the uniqueness remains. If there are two solutions, for a given then satisfies . Since we can take the Fourier transform and see that

an earlier problem was to conclude from this that for some constants . This in turn implies that is a polynomial. However the only polynomials in are identically 0. Thus and uniqueness follows.

Next time I will talk about homogeneous distributions. On the functions

where , is locally integrable (and hence a tempered distribution) precisely when . As a function it is homogeneous of degree . Thus if then

Thinking of as a distribution we can set this as

Thus if we define , for any , we can ask whether a distribution is homogeneous:

Next: Problems Up: Lecture notes for 18.155, Previous: Sobolev embedding   Contents
Richard B. Melrose 2003-02-18