To get solutions to the first more difficult partial differential equations (like, for example, Poisson's equation, the heat equation and Helmholtz' equation), we will now take a look at distributions.
Distributions and tempered distributionsEdit
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
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 .
Let be a tempered distribution. Then the restriction of to bump functions is a distribution.
Let be a tempered distribution, and let be open.
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 .
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.
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, .
Convolution and approximation of Lp functionsEdit
Let such that (where ) and let and . The convolution of and , is defined as:
The convolution is well-defined:
Let such that and let and .. Then for all , the integral
has a well-defined real value.
We shall now prove that the convolution is commutative, i. e. .
Let such that (where ) and let and . Then for all :
Let be open and let . Then
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 regular distributions. But let's first define what regular distributions are.
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.
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:
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
Let be open, and let be a function. Then
is a regular distribution iff .
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.
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
, 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 .
Let , i. e.
is a regular tempered distribution
Well-definedness follows from Hölder's inequality:
Due to the triangle inequality for integrals and Hölder's inequality, we have
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.
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 .
Let be open sets and let be a linear function. If there is a linear and continuous function such that
, then for every distribution , the function is a distribution. Therefore, we may define a function
This function has the property
Noticing that differentiation, convolution and multiplication are linear, we will define these operations for distributions by taking as the respective of these three operations.
We have to prove two claims: First, that the function is a distribution, and second that as defined above has the property
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).
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.
Using the last lemma 4.13, we define operations like multiplication with a smooth function, differentiation and convolution with a smooth function for distributions.
For the bump functions and the Schwartz functions, we also may define the differentiation of distributions. Let and . Let's now define
Then, for the spaces or , the requirements for the above lemma 1.4 are met and we may define the differentiation of distribution in the following way:
This definition also satisfies .
Proof: By integration by parts, we obtain:
, where is the i-th component of the outward normal vector and is the boundary of . For bump functions, the boundary integral vanishes anyway, because the functions in are zero there. For Schwartz functions, we may use the identity
and the decreasing property of the Schwartz functions to see that the boundary integral goes to zero and therefore
To derive the equation
, we may apply the formula from above several times. This finishes the proof, because this equation was the only non-trivial property of , which we need for applying lemma 1.5.
Let , and let . Let's define
This function () is linear, because the integral is linear. It is called the convolution of and .
We can also define: , and:
By the theorem of Fubini, we can calculate as follows:
Therefore, the first assumption for Lemma 1.5 holds.
Due to the Leibniz integral rule, we obtain that for (i. e. is integrable) and (i. e. the partial derivatives of exist up to order and are also continuous):
With this formula, we can see (due to the monotony of the integral) that
From this follows sequential continuity for Schwartz and bump functions by defining and . Thus, with the help of lemma 1.5, we can define the convolution with a distribution of or as follows:
- Show that endowed with the usual topology is a topological vector space.