This chapter will show how you can apply distributions to find solutions for certain PDEs (partial differential equations).
Definition: Solutions in the sense of distributionsEdit
Let . Consider the equation:
We call a distribution a solution in the sense of distributions for , if and only if:
Definition: Fundamendal SolutionsEdit
We call a function a fundamental solution for , if and only if
- it is continuous, and
- , where is the Dirac delta distribution.
Let be a sequence of distributions in , and let be a function space, which is a locally convex space. Let us furthermore have that for all . Then
defines a distribution.
Proof: Very difficult; The appendix will have a section about this.
Let be a function space, which is a locally convex normed space. For , let be a family of distributions on that function space. Let's further assume that for all , the function is continuous on and bounded, and let . Then
defines a distribution.
Proof: Due to the truncation of -functions, we have that there are radii such that
, where is the supremum of the function .
is a compact set, since it is bounded as well as closed. Therefore, we may divide into finitely many (let's say ) squares with diameter at most , such that
. This we may do because continuous functions are uniformly continuous on compact sets. At the border, we just round the squares so that they fit in with the sphere. Furthermore, we choose for each square a inside this square.
We choose now
, which is a finite linear combination of distributions and therefore a distribution. Due to the normal triangle inequality for the absolute value, the triangle inequality for the Lebegue integral, our first calculation and the fundamental integral estimation, we obtain:
This obviously goes to zero, and this lemma follows with Lemma 2.1.
Let's assume that in equation , is integrable. Let be a fundamental solution for with respect to the locally convex normed function space , such that , the function is bounded. Then we can know, that:
is well-defined and solves in the sense of distributions.
Proof: Since by the definition of fundamental solutions, the function is continuous, we may apply lemma 2.2, which gives us that is indeed well-defined.
To show that it really solves in the sense of distributions, we need the following calculation:
, which is what we wanted to show.
Green's functions and Green's kernelsEdit
Definition: Green's functionEdit
Assume that for each , the fundamental solution is a regular distribution, i. e. for each , there is an integrable function such that . Then we call this function a Green's function for .
Definition: Green's kernelEdit
Let's assume that has the Green's function . If there exists a function such that
, then we call a Green's kernel for .
Let be a locally integrable function, and be a domain. Then the family of distributions is well-defined and depends continuously on . Furthermore, for each , the function is bounded.
Proof: Well-definedness follows from Lemma 1.3.
Let , and let . Then we can calculate the following:
for sufficiently large , where the last expression goes to as , since the support of is compact and therefore the function is (even uniformly) continuous.
Furthermore, we have
, which is zero for sufficiently large, which is why the function has compact support. But since the function is also continuous, we know that it obtains a maximum and a minimum and is therefore bounded.
This lemma shows that if we have found a locally integrable function such that , we already know that it is a Green's kernel, and don't need to check the continuity property.
Now this theorem finally shows us why distributions are useful:
Let be a Green's kernel for , and let . If
is sufficiently often differentiable such that is continuous, then it is a solution for in the classical sense.
Proof: From a case of Hölder's inequality (namely , i. e. ), we obtain that is locally integrable, which is why is a distribution in .
Furthermore, due to the theorem of Fubini, we have for , that
, which is why solves in the sense of distributions (this is due to theorem 2.3).
Thus, for all , we can calculate the following:
From this follows that almost everywhere. But since and are both continuous, they must be equal everywhere. This is what we wanted to prove.
Definition: Dirichlet problemsEdit
If one looks at a domain with boundary , then finding a function which satisfies
is called a Dirichlet Problem.
- This means here that as a function from to is continuous for every .