I’m going to start this blog off right: a repost from my previous blog of a talk I gave Dec. 2013.

I’d like to post it here as a more well thought out, well formulated (and hopefully much more clear) post that is complementary to the talk I gave.

Before I delve right into it, I’d like to provide some background. The Fourier Transform of a function is very important to solving any sort of differential equation. Similar to Fourier Series, it expands a function to its sum of trigonometric functions (sometimes finite, sometimes infinite). But unlike Fourier Series, your function need not be periodic. In which case, we just say that the period, *T*, tends to infinity. Formally, we write the Fourier transform as:

Really, I could write up an entire post about the Fourier Transform alone (and perhaps I will) but for now, this will do.

Now, we consider a point fixed if, given some operator, *F(X) = X*. The solution of this kind of equation is called a “fixed point”. A good example of commonly used fixed point is the equation

Where *u’* is *u* when the derivative operator is applied.

Essentially, all PDEs can be considered fixed points. So you can have something like,

thus,

Now, for most of my time while working on researching this I worked in L2 space. L2 space is a subset of the real space where the Fourier Transform of a function in this space satisfies the following:

This is called the Parseval Identity, and it is very important with regards to functions in L2 space (there is also much more to be said about L2 space, but this is all that we need for the moment). From here, you can see that the Fourier Transform of a function still lies in L2. Moreover, this implies that the Fourier Transform maps L2 to L2.

Now, L2 space has a lot of interesting properties about it. Some of which are very helpful in our analysis of elliptic PDEs. One of those properties is the norm. On L2, the norm is denoted as:

(You can see how this will be useful later on)

Now, for any operator that maps L2 to L2, it is considered a contraction iff there is a constant *α < 1*, and for any *f1(x)*, *f2(x)* etc element of L2:

A contraction means that after *n* iterations, the function will decrease by a factor of *α*. Thus, the function converges. Because of that, we can conclude that the function must exist. We can prove this computationally as well. Say we have an equation for *U(x)*,

Now we can take an operator, *F*, as it maps the space L2 to L2. Assumming *M(x,y)* satisfies

we can prove that it is a contraction:

This is what I was talking about in my previous post about wanting to go more in depth on contractions, especially how they relate to L2 space. Now, moving on: what happens when we move away from general ODEs? Away from “normal” PDEs? A big problem in solving PDEs stems from the fact that occasionally, our PDEs will be what we like to dub as “disturbed”; we introduce a disturb term, *ϵ*, to our equation. Unfortunately, psychiatric treatment for non-physical entities is not a very lucrative field of study so we must use a different approach in determining if a solution exists. Suppose our disturbed PDE is,

where *a* and *c* are positive integers and *d(x)* is a real-valued function and *ϵ* is assumed to be very small. We can start by taking the Fourier Transform of the equation like usual,

So,

Oh, hey! It’s a fixed point problem now! So, since we’re still on L2 (and always will be…) this becomes,

Remember to exchange *k*‘s for *x*‘s at this point because we are now working in terms of the original equation. *g(x)* is used in place of the large constant involving* d(k)*, since by fixed point theory we know that any operator applied (or “unapplied”) to *d(x)* will equal *d(k)*. Thus the original function is,

That crazy integral you see there is the definition of a convolution.

Now, we want to see if this is a contraction on L2, and we know by definition of the norm on L2 that,

where the right half of the equation is equal to

Now, to test for a contraction, we need not worry about the outer portion of the integral; we can simply work with the dy integral on the inside; we can manipulate it as follows,

Basic properties of integration tells us that this integral can be broken up into two separate integrals. Because of that, we only need work with one of them, and we can choose them arbitrarily. Let’s work with the first one. By Cauchy-Schwartz we know that

Thus, by definition of the norm on L2,

Because this holds true with the first integral that we’ve broken up, in the same vein, it must hold true for the second,

Because both of these hold true, it must hold true for the entire integral. That is, for the outer portion of the original integral we were working with beforehand,

And finally, we can take epsilon out of the integral so it becomes,

Our disturb term acts exactly like *α* does, which we know to be *< 1*. Which means that this is a contraction and thus there exists a solution for this disturbed elliptic PDE.

Hashtag math

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