Differential Equations - Fourier Series: Sine - i

In this section we are going to start taking a look at Fourier series. We should point out that this is a subject that can span a whole class and what we’ll be doing in this section (as well as the next couple of sections) is intended to be nothing more than a very brief look at the subject. The point here is to do just enough to allow us to do some basic solutions to partial differential equations in the next chapter. There are many topics in the study of Fourier series that we’ll not even touch upon here.

So, with that out of the way let’s get started, although we’re not going to start off with Fourier series. Let’s instead think back to our Calculus class where we looked at Taylor Series. With Taylor Series we wrote a series representation of a function, $f\left(x\right)$, as a series whose terms were powers of $x-a$ for some $x=a$. With some conditions we were able to show that,

$$f\left(x\right)=\sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(a\right)}{n!}{\left(x-a\right)}^n$$

and that the series will converge to $f\left(x\right)$ on $\left|x-a\right|<R$ for some $R$ that will be dependent upon the function itself.

There is nothing wrong with this, but it does require that derivatives of all orders exist at $x=a$. Or in other words $f^{\left(n\right)}\left(a\right)$exists for $n=0,1,2,3,\dots$ Also for some functions the value of $R$ may end up being quite small.

These two issues (along with a couple of others) mean that this is not always the best way of writing a series representation for a function. In many cases it works fine and there will be no reason to need a different kind of series. There are times however where another type of series is either preferable or required.

We’re going to build up an alternative series representation for a function over the course of the next couple of sections. The ultimate goal for the rest of this chapter will be to write down a series representation for a function in terms of sines and cosines.

We’ll start things off by assuming that the function, $f\left(x\right)$, we want to write a series representation for is an odd function (i.e. $f\left(-x\right)=-f\left(x\right)$). Because $f\left(x\right)$ is odd it makes some sense that we should be able to write a series representation for this in terms of sines only (since they are also odd functions).

What we’ll try to do here is write $f\left(x\right)$ as the following series representation, called a Fourier sine series, on $-L\le x\le L$.

$$\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)$$

There are a couple of issues to note here. First, at this point, we are going to assume that the series representation will converge to $f\left(x\right)$ on $-L\le x\le L$. We will be looking into whether or not it will actually converge in a later section. However, assuming that the series does converge to $f\left(x\right)$ it is interesting to note that, unlike Taylor Series, this representation will always converge on the same interval and that the interval does not depend upon the function.

Second, the series representation will not involve powers of sine (again contrasting this with Taylor Series) but instead will involve sines with different arguments.

Finally, the argument of the sines, $\frac{n\pi x}{L}$, may seem like an odd choice that was arbitrarily chosen and in some ways it was. For Fourier sine series the argument doesn’t have to necessarily be this but there are several reasons for the choice here. First, this is the argument that will naturally arise in the next chapter when we use Fourier series (in general and not necessarily Fourier sine series) to help us solve some basic partial differential equations.

The next reason for using this argument is the fact that the set of functions that we chose to work with, ${\left\{\sin \left(\frac{n\pi x}{L}\right)\right\}}_{n=1}^{\infty }$in this case, need to be orthogonal on the given interval, $-L\le x\le L$ in this case, and note that in the last section we showed that in fact they are. In other words, the choice of functions we’re going to be working with and the interval we’re working on will be tied together in some way. We can use a different argument but will need to also choose an interval on which we can prove that the sines (with the different argument) are orthogonal.

So, let’s start off by assuming that given an odd function, $f\left(x\right)$, we can in fact find a Fourier sine series, of the form given above, to represent the function on $-L\le x\le L$. This means we will have,

$$f\left(x\right)=\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)$$

As noted above we’ll discuss whether or not this even can be done and if the series representation does in fact converge to the function in later section. At this point we’re simply going to assume that it can be done. The question now is how to determine the coefficients, $B_n$, in the series.

Let’s start with the series above and multiply both sides by $\sin \left(\frac{m\pi x}{L}\right)$ where $m$ is a fixed integer in the range $\left\{1,2,3,\dots \right\}$. In other words, we multiply both sides by any of the sines in the set of sines that we’re working with here. Doing this gives,

$$f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)=\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)$$

Now, let’s integrate both sides of this from $x=-L$ to $x=L$.

$${\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dx={\int }_{-L}^L\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx$$

At this point we’ve got a small issue to deal with. We know from Calculus that an integral of a finite series (more commonly called a finite sum….) is nothing more than the (finite) sum of the integrals of the pieces. In other words, for finite series we can interchange an integral and a series. For infinite series however, we cannot always do this. For some integrals of infinite series we cannot interchange an integral and a series. Luckily enough for us we actually can interchange the integral and the series in this case. Doing this and factoring the constant, $B_n$, out of the integral gives,

$$\begin{array}{cc}{\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dx& =\sum _{n=1}^{\infty }{\int }_{-L}^L{B}_n\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx\\ & =\sum _{n=1}^{\infty }{B}_n{\int }_{-L}^L\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx\end{array}$$

Now, recall from the last section we proved that ${\left\{\sin \left(\frac{n\pi x}{L}\right)\right\}}_{n=1}^{\infty }$ is orthogonal on $-L\le x\le L$and that,

$${\int }_{-L}^L\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx=\{\begin{array}{cc}L& \text{if}n=m\\ 0& \text{if}n\ne m\end{array}$$

So, what does this mean for us. As we work through the various values of $n$ in the series and compute the value of the integrals all but one of the integrals will be zero. The only non-zero integral will come when we have $n=m$, in which case the integral has the value of $L$. Therefore, the only non-zero term in the series will come when we have $n=m$ and our equation becomes,

$${\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dx=B_{m}L$$

Finally, all we need to do is divide by $L$ and we now have an equation for each of the coefficients.

$$B_m=\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dxm=1,2,3,\dots$$

Next, note that because we’re integrating two odd functions the integrand of this integral is even and so we also know that,

$$B_m=\frac{2}{L}{\int }_0^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dxm=1,2,3,\dots$$

Summarizing all this work up the Fourier sine series of an odd function $f\left(x\right)$ on $-L\le x\le L$ is given by,

$$\begin{array}{cccc}f\left(x\right)& =\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)& B_n& =\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dxn=1,2,3,\dots \\ & & & =\frac{2}{L}{\int }_0^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dxn=1,2,3,\dots \end{array}$$

Let’s take a quick look at an example.