Partial Differential Equations/Fundamental solutions, Green's functions and Green's kernels
In the last two chapters, we have studied test function spaces and distributions. In this chapter we will demonstrate a method to obtain solutions to linear partial differential equations which uses test function spaces and distributions.
We now introduce equicontinuity, because we will absolutely need this concept at two very very important places of this wikibook. We will need the concept in the following way: We will show some things which hold for equicontinuity, then we will show that certain things are equicontinuous, and then we can conclude that the things which hold for equicontinuity hold for these things for which we had shown that they are equicontinuous.
Now in the previous sentence I often mentioned the word 'thing'. Let's see for which 'things' equicontinuity is defined:
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 iff:
So the 'things' are in fact sets of continuous functions mapping from (a set in a metric space) to the real numbers .
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.
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 is 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 , which we shall denote by . Let us call the limit of this sequence, very creatively, . As was equicontinuous, we can choose such that . Further, since (if of course), we may choose such that for all , we have . But then follows for and the reverse triangle inequality:
Since we had and further, by the reverse triangle inequality and how the sequence is defined:
, we obtain:
Thus we have a contradiction to .
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 . Note that this choice is independent of ; we get away in this proof with choosing always the same old boring for all the :-)
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:
In the last chapter, we had defined multiplication of a distribution with a smooth function and derivatives of distributions. Therefore, for a distribution , we are able to calculate such expressions as
for a smooth function and a -dimensional multiindex . We therefore observe that in a linear partial differential equation of the form
we could insert any distribution instead of in the left hand side. However, equality would not hold in this case, because on the right hand side we have a function, but the left hand side would give us a distribution (as finite sums of distributions are distributions again due to theorem 4.?; remember that only finitely many are allowed to be nonzero). If we however replace the right hand side by (the regular distribution corresponding to ), then there might be distributions which satisfy the equation. In this case, we speak of a distributional solution. Let's summarise this definition in a box.
Let be open, let
be a linear partial differential equation, and let . is called a distributional solution to the above linear partial differential equation iff
Now we will show how we can obtain distributional solutions to a partial differential equation. The method of choice will be to guess a so-called fundamental solution and then construct solutions with the help of that fundamental solution.
Let be open and let
be a linear homogenous partial differential equation. If has the two properties
, we call a fundamental solution.
Now why we defined this is: Once we have a fundamental solution for the homogenous equation (i. e. ), we can easily construct solutions to the inhomogenous problem. We shall now explain how this works.
Let be a family of distributions, where . Let's further assume that for all , the function is continuous on and bounded, and let . Then
is 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.
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 .
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.
Theorem 5.?: (Fubini's theorem)
Let and , where are arbitrary natural numbers, and let be a function. Then
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.
- 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 can be proved with help of the Bolzano–Weierstrass theorem).