Partial Differential Equations/Distributions

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Partial Differential Equations
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Distributions and tempered distributions[edit]

Definition 4.1:

Let be open, and let be a function. We call a distribution iff

  • is linear ()
  • is sequentially continuous (if in the notion of convergence of bump functions, then in the reals)

The set of all distributions for we denote by

Definition 4.2:

Let be a function. We call a tempered distribution iff

  • is linear ()
  • is sequentially continuous (if in the notion of convergence of Schwartz functions, then in the reals)

The set of all tempered distributions we denote by .

Theorem 4.3:

Let be a tempered distribution. Then the restriction of to bump functions is a distribution.

Proof:

Let be a tempered distribution, and let be open.

1.

We show that has a well-defined value for .

Due to theorem 3.9, every bump function is a Schwartz function, which is why the expression

makes sense for every .

2.

We show that the restriction is linear.

Let and . Since due to theorem 3.9 and are Schwartz functions as well, we have

due to the linearity of for all Schwartz functions. Thus is also linear for bump functions.

3.

We show that the restriction of to is sequentially continuous. Let in the notion of convergence of bump functions. Due to theorem 3.11, in the notion of convergence of Schwartz functions. Since as a tempered distribution is sequentially continuous, .

The convolution[edit]

Definition 4.4:

Let . The integral

is called convolution of and and denoted by if it exists.

The convolution of two functions may not always exist, but there are sufficient conditions for it to exist:

Theorem 4.5:

Let such that and let and . Then for all , the integral

has a well-defined real value.

Proof:

Due to Hölder's inequality,

.

We shall now prove that the convolution is commutative, i. e. .

Theorem 4.6:

Let such that (where ) and let and . Then for all :

Proof:

We apply multi-dimensional integration by substitution using the diffeomorphism to obtain

.

Lemma 4.7:

Let be open and let . Then .

Proof:

Let be arbitrary. Then, since for all

and further

,

Leibniz' integral rule (theorem 2.2) is applicable, and by repeated application of Leibniz' integral rule we obtain

.

Regular distributions[edit]

In this section, we shortly study a class of distributions which we call regular distributions. In particular, we will see that for certain kinds of functions there exist corresponding distributions.

Definition 4.8:

Let be an open set and let . If for all can be written as

for a function which is independent of , then we call a regular distribution.

Definition 4.9:

Let . If for all can be written as

for a function which is independent of , then we call a regular tempered distribution.

Two questions related to this definition could be asked: Given a function , is for open given by

well-defined and a distribution? Or is given by

well-defined and a tempered distribution? In general, the answer to these two questions is no, but both questions can be answered with yes if the respective function has the respectively right properties, as the following two theorems show. But before we state the first theorem, we have to define what local integrability means, because in the case of bump functions, local integrability will be exactly the property which needs in order to define a corresponding regular distribution:

Definition 4.10:

Let be open, be a function. We say that is locally integrable iff for all compact subsets of

We write .

Now we are ready to give some sufficient conditions on to define a corresponding regular distribution or regular tempered distribution by the way of

or

:

Theorem 4.11:

Let be open, and let be a function. Then

is a regular distribution iff .

Proof:

1.

We show that if , then is a distribution.

Well-definedness follows from the triangle inequality of the integral and the monotony of the integral:

In order to have an absolute value strictly less than infinity, the first integral must have a well-defined value in the first place. Therefore, really maps to and well-definedness is proven.

Continuity follows similarly due to

, where is the compact set in which all the supports of and are contained (remember: The existence of a compact set such that all the supports of are contained in it is a part of the definition of convergence in , see the last chapter. As in the proof of theorem 3.11, we also conclude that the support of is also contained in ).

Linearity follows due to the linearity of the integral.

2.

We show that is a distribution, then (in fact, we even show that if has a well-defined real value for every , then . Therefore, by part 1 of this proof, which showed that if it follows that is a distribution in , we have that if is a well-defined real number for every , is a distribution in .

Let be an arbitrary compact set. We define

is continuous, even Lipschitz continuous with Lipschitz constant : Let . Due to the triangle inequality, both

and

, which can be seen by applying the triangle inequality twice.

We choose sequences and in such that and and consider two cases. First, we consider what happens if . Then we have

.

Second, we consider what happens if :

Since always either or , we have proven Lipschitz continuity and thus continuity. By the extreme value theorem, therefore has a minimum . Since would mean that for a sequence in which is a contradiction as is closed and , we have .

Hence, if we define , then . Further, the function

has support contained in , is equal to within and further is contained in due to lemma 4.7. Hence, it is also contained in . Since therefore, by the monotonicity of the integral

, is indeed locally integrable.

Theorem 4.12:

Let , i. e.

Then

is a regular tempered distribution.

Proof:

From Hölder's inequality we obtain

.

Hence, is well-defined.

Due to the triangle inequality for integrals and Hölder's inequality, we have

Furthermore

.

If in the notion of convergence of the Schwartz function space, then this expression goes to zero. Therefore, continuity is verified.

Linearity follows from the linearity of the integral.

Equicontinuity[edit]

We now introduce the concept of equicontinuity.

Definition 4.13:

Let be a metric space equipped with a metric which we shall denote by here, let be a set in , and let be a set of continuous functions mapping from to the real numbers . We call this set equicontinuous if and only if

.

So equicontinuity is in fact defined for sets of continuous functions mapping from (a set in a metric space) to the real numbers .

Theorem 4.14:

Let be a metric space equipped with a metric which we shall denote by , let be a sequentially compact set in , and let be an equicontinuous set of continuous functions from to the real numbers . Then follows: If is a sequence in such that has a limit for each , then for the function , which maps from to , it follows uniformly.

Proof:

In order to prove uniform convergence, by definition we must prove that for all , there exists an such that for all .

So let's assume the contrary, which equals by negating the logical statement

.

We choose a sequence in . We take in such that for an arbitrarily chosen and if we have already chosen and for all , we choose such that , where is greater than .

As is sequentially compact, there is a convergent subsequence of . Let us call the limit of that subsequence sequence .

As is equicontinuous, we can choose such that

.

Further, since (if of course), we may choose such that

.

But then follows for and the reverse triangle inequality:

Since we had , the reverse triangle inequality and the definition of t

, we obtain:

Thus we have a contradiction to .

Theorem 4.15:

Let be a set of differentiable functions, mapping from the convex set to . If we have, that there exists a constant such that for all functions in , (the exists for each function in because all functions there were required to be differentiable), then is equicontinuous.

Proof: We have to prove equicontinuity, so we have to prove

.

Let be arbitrary.

We choose .

Let such that , and let be arbitrary. By the mean-value theorem in multiple dimensions, we obtain that there exists a such that:

The element is inside , because is convex. From the Cauchy-Schwarz inequality then follows:

The generalised product rule[edit]

Definition 4.16:

If are two -dimensional multiindices, we define the binomial coefficient of over as

.

We also define less or equal relation on the set of multi-indices.

Definition 4.17:

Let be two -dimensional multiindices. We define to be less or equal than if and only if

.

For , there are vectors such that neither nor . For , the following two vectors are examples for this:

This example can be generalised to higher dimensions (see exercise 6).

With these multiindex definitions, we are able to write down a more general version of the product rule. But in order to prove it, we need another lemma.

Lemma 4.18:

If and , where the is at the -th place, we have

for arbitrary multiindices .

Proof:

For the ordinary binomial coefficients for natural numbers, we had the formula

.

Therefore,

This is the general product rule:

Theorem 4.19:

Let and let . Then


Proof:

We prove the claim by induction over .

1.

We start with the induction base . Then the formula just reads

, and this is true. Therefore, we have completed the induction base.

2.

Next, we do the induction step. Let's assume the claim is true for all such that . Let now such that . Let's choose such that (we may do this because ). We define again , where the is at the -th place. Due to Schwarz' theorem and the ordinary product rule, we have

.

By linearity of derivatives and induction hypothesis, we have

.

Since

and

,

we are allowed to shift indices in the first of the two above sums, and furthermore we have by definition

.

With this, we obtain

Due to lemma 4.18,

.

Further, we have

where in ,

and

(these two rules may be checked from the definition of ). It follows

.

Operations on Distributions[edit]

For there are operations such as the differentiation of , the convolution of and and the multiplication of and . In the following section, we want to define these three operations (differentiation, convolution with and multiplication with ) for a distribution instead of .

Lemma 4.20:

Let be open sets and let be a linear function. If there is a linear and sequentially continuous (in the sense of definition 4.1) function such that

, then for every distribution , the function is a distribution. Therefore, the function

really maps to . This function has the property

Proof:

We have to prove two claims: First, that the function is a distribution, and second that as defined above has the property

1.

We show that the function is a distribution.

has a well-defined value in as maps to , which is exactly the preimage of . The function is continuous since it is the composition of two continuous functions, and it is linear for the same reason (see exercise 2).

2.

We show that has the property

For every , we have

Since equality of two functions is equivalent to equality of these two functions evaluated at every point, this shows the desired property.

We also have a similar lemma for Schwartz distributions:

Lemma 4.21:

Let be a linear function. If there is a linear and sequentially continuous (in the sense of definition 4.2) function such that

, then for every distribution , the function is a distribution. Therefore, we may define a function

This function has the property

The proof is exactly word-for-word the same as the one for lemma 4.20.

Noting that multiplication, differentiation and convolution are linear, we will define these operations for distributions by taking in the two above lemmas as the respective of these three operations.

Theorem and definitions 4.22:

Let , and let be open. Then for all , the pointwise product is contained in , and if further and all of it's derivatives are bounded by polynomials, then for all the pointwise product is contained in . Also, if in the sense of bump functions, then in the sense of bump functions, and if and all of it's derivatives are bounded by polynomials, then in the sense of Schwartz functions implies in the sense of Schwartz functions. Further:

  • Let be a distribution. If we define

    ,

    then the expression on the right hand side is well-defined and for all we have

    ,

    and is a distribution.

  • Assume that and all of it's derivatives are bounded by polynomials. Let be a tempered distribution. If we define

    ,

    then the expression on the right hand side is well-defined and for all we have

    ,

    and is a tempered distribution.

Proof:

The product of two functions is again , and further, if , then also . Hence, .

Also, if in the sense of bump functions, then, if is a compact set such that for all ,

.

Hence, in the sense of bump functions.

Further, also . Let be arbitrary. Then

.

Since all the derivatives of are bounded by polynomials, by the definition of that we obtain

, where are polynomials. Hence,

.

Similarly, if in the sense of Schwartz functions, then by exercise 3.6

and hence in the sense of Schwartz functions.

If we define , from lemmas 4.20 and 4.21 follow the other claims.

Theorem and definitions 4.23:

Let be open. We define

, where such that only finitely many of the are different from the zero function (such a function is also called a linear partial differential operator), and further we define

.
  • Let be a distribution. If we define

    ,

    then the expression on the right hand side is well-defined, for all we have

    ,

    and is a distribution.

  • Assume that all s and all their derivatives are bounded by polynomials. Let be a tempered distribution. If we define

    ,

    then the expression on the right hand side is well-defined, for all we have

    ,

    and is a tempered distribution.

Proof:

We want to apply lemmas 4.20 and 4.21. Hence, we prove that the requirements of these lemmas are met.

Since the derivatives of bump functions are again bump functions, the derivatives of Schwartz functions are again Schwartz functions (see exercise 3.3 for both), and because of theorem 4.22, we have that and map to , and if further all and all their derivatives are bounded by polynomials, then and map to .

The sequential continuity of follows from theorem 4.22.

Further, for all ,

.

Further, if we single out an , by Fubini's theorem and integration by parts we obtain

.

Hence,

and the lemmas are applicable.

Definition 4.24:

Let and let . Then we define the function

.

This function is called the convolution of and .

Theorem 4.25:

Let and let . Then

  1. is continuous,
  2. and
  3. .

Proof:

1.

Let be arbitrary, and let be a sequence converging to and let such that . Then

is compact. Hence, if is arbitrary, then uniformly. But outside , . Hence, uniformly. Further, for all . Hence, in the sense of bump functions. Thus, by continuity of ,

.

2.

We proceed by induction on .

The induction base is obvious, since for all functions by definition.

Let the statement be true for all such that . Let such that . We choose such that (this is possible since otherwise ). Further, we define

.

Then , and hence .

Furthermore, for all ,

.

But due to Schwarz' theorem, in the sense of bump functions, and thus

.

Hence, , since is a bump function (see exercise 3.3).

3.

This follows from 1. and 2., since is a bump function for all (see exercise 3.3).

Exercises[edit]

  1. Let be (tempered) distributions and let . Prove that also is a (tempered) distribution.
  2. Let be essentially bounded. Prove that is a tempered distribution.
  3. Prove that if is a set of differentiable functions which go from to , such that there exists a such that for all it holds , and if is a sequence in for which the pointwise limit exists for all , then converges to a function uniformly on (hint: is sequentially compact; this follows from the Bolzano–Weierstrass theorem).
  4. Let such that is a distribution. Prove that for all .
  5. Prove that for the function is a tempered distribution (this function is called the Dirac delta distribution after Paul Dirac).
  6. For each , find such that neither nor .

Sources[edit]

Partial Differential Equations
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