Partial Differential Equations/Distributions

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Distributions and tempered distributionsEdit

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 convolutionEdit

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 distributionsEdit

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.

 

EquicontinuityEdit

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 ruleEdit

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 DistributionsEdit

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).

 

ExercisesEdit

  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  .

SourcesEdit