Differential Equations - Fourier Series: Examples - i

Okay, in the previous two sections we’ve looked at Fourier sine and Fourier cosine series. It is now time to look at a Fourier series. With a Fourier series we are going to try to write a series representation for $f\left(x\right)$ on $-L\le x\le L$ in the form,

$$f\left(x\right)=\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)$$

So, a Fourier series is, in some way a combination of the Fourier sine and Fourier cosine series. Also, like the Fourier sine/cosine series we’ll not worry about whether or not the series will actually converge to $f\left(x\right)$ or not at this point. For now we’ll just assume that it will converge and we’ll discuss the convergence of the Fourier series in a later section.

Determining formulas for the coefficients, $A_n$ and $B_n$, will be done in exactly the same manner as we did in the previous two sections. We will take advantage of the fact that ${\left\{\cos \left(\frac{n\pi x}{L}\right)\right\}}_{n=0}^{\infty }$ and ${\left\{\sin \left(\frac{n\pi x}{L}\right)\right\}}_{n=1}^{\infty }$ are mutually orthogonal on $-L\le x\le L$ as we proved earlier. We’ll also need the following formulas that we derived when we proved the two sets were mutually orthogonal.

$$\begin{array}{cc}& {\int }_{-L}^L\cos \left(\frac{n\pi x}{L}\right)\cos \left(\frac{m\pi x}{L}\right)dx=\{\begin{array}{cc}2L& \text{if}n=m=0\\ L& \text{if}n=m\ne 0\\ 0& \text{if}n\ne m\end{array}\\ & \\ & {\int }_{-L}^L\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx=\{\begin{array}{cc}L& \text{if}n=m\\ 0& \text{if}n\ne m\end{array}\\ & \\ & {\int }_{-L}^L\sin \left(\frac{n\pi x}{L}\right)\cos \left(\frac{m\pi x}{L}\right)dx=0\end{array}$$

So, let’s start off by multiplying both sides of the series above by $\cos \left(\frac{m\pi x}{L}\right)$ and integrating from –$L$ to $L$. Doing this gives,

$${\int }_{-L}^{L}f\left(x\right)\cos \left(\frac{m\pi x}{L}\right)dx={\int }_{-L}^L\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)\cos \left(\frac{m\pi x}{L}\right)dx+{\int }_{-L}^L\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)\cos \left(\frac{m\pi x}{L}\right)dx$$

Now, just as we’ve been able to do in the last two sections we can interchange the integral and the summation. Doing this gives,

$${\int }_{-L}^{L}f\left(x\right)\cos \left(\frac{m\pi x}{L}\right)dx=\sum _{n=0}^{\infty }{A}_n{\int }_{-L}^L\cos \left(\frac{n\pi x}{L}\right)\cos \left(\frac{m\pi x}{L}\right)dx+\sum _{n=1}^{\infty }{B}_n{\int }_{-L}^L\sin \left(\frac{n\pi x}{L}\right)\cos \left(\frac{m\pi x}{L}\right)dx$$

We can now take advantage of the fact that the sines and cosines are mutually orthogonal. The integral in the second series will always be zero and in the first series the integral will be zero if $n\ne m$ and so this reduces to,

$${\int }_{-L}^{L}f\left(x\right)\cos \left(\frac{m\pi x}{L}\right)dx=\{\begin{array}{cc}{A}_m\left(2L\right)& \text{if}n=m=0\\ A_m\left(L\right)& \text{if}n=m\ne 0\end{array}$$

Solving for $A_m$ gives,

$$\begin{array}{cc}& A_0=\frac{1}{2L}{\int }_{-L}^{L}f\left(x\right)dx\\ & A_m=\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\cos \left(\frac{m\pi x}{L}\right)dxm=1,2,3,\dots \end{array}$$

Now, do it all over again only this time multiply both sides by $\sin \left(\frac{m\pi x}{L}\right)$, integrate both sides from –$L$ to $L$ and interchange the integral and summation to get,

$${\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dx=\sum _{n=0}^{\infty }{A}_n{\int }_{-L}^L\cos \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx+\sum _{n=1}^{\infty }{B}_n{\int }_{-L}^L\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx$$

In this case the integral in the first series will always be zero and the second will be zero if $n\ne m$ and so we get,

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

Finally, solving for $B_m$ gives,

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

In the previous two sections we also took advantage of the fact that the integrand was even to give a second form of the coefficients in terms of an integral from 0 to $L$. However, in this case we don’t know anything about whether $f\left(x\right)$ will be even, odd, or more likely neither even nor odd. Therefore, this is the only form of the coefficients for the Fourier series.

Before we start examples let’s remind ourselves of a couple of formulas that we’ll make heavy use of here in this section, as we’ve done in the previous two sections as well. Provided $n$ in an integer then,

$$\cos \left(n\pi \right)={\left(-1\right)}^n\sin \left(n\pi \right)=0$$

In all of the work that we’ll be doing here $n$ will be an integer and so we’ll use these without comment in the problems so be prepared for them.

Also, don’t forget that sine is an odd function, i.e. $\sin \left(-x\right)=-\sin \left(x\right)$ and that cosine is an even function, i.e. $\cos \left(-x\right)=\cos \left(x\right)$. We’ll also be making heavy use of these ideas without comment in many of the integral evaluations so be ready for these as well.

Now let’s take a look at an example.

Example 1 Find the Fourier series for $f\left(x\right)=L-x$ on $-L\le x\le L$.

So, let’s go ahead and just run through formulas for the coefficients.
$$A_0=\frac{1}{2L}{\int }_{-L}^{L}f\left(x\right)dx=\frac{1}{2L}{\int }_{-L}^{L}L-xdx=L$$$$\begin{array}{cc}{A}_n& =\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx=\frac{1}{L}{\int }_{-L}^L\left(L-x\right)\cos \left(\frac{n\pi x}{L}\right)dx\\ & =\frac{1}{L}{\left(\frac{L}{n^2{\pi }^2}\right)\left(n\pi \left(L-x\right)\sin \left(\frac{n\pi x}{L}\right)-L\cos \left(\frac{n\pi x}{L}\right)\right)|}_{-L}^L\\ & =\frac{1}{L}\left(\frac{L}{n^2{\pi }^2}\right)\left(-2n\pi L\sin \left(-n\pi \right)\right)=0n=1,2,3,\dots \end{array}$$$$\begin{array}{cc}{B}_n& =\frac{1}{L}{\int }_{-L}^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx=\frac{1}{L}{\int }_{-L}^L\left(L-x\right)\sin \left(\frac{n\pi x}{L}\right)dx\\ & =\frac{1}{L}{\left(-\frac{L}{n^2{\pi }^2}\right)\left[L\sin \left(\frac{n\pi x}{L}\right)-n\pi \left(x-L\right)\cos \left(\frac{n\pi x}{L}\right)\right]|}_{-L}^L\\ & =\frac{1}{L}\left[\frac{L^2}{n^2{\pi }^2}\left(2n\pi \cos \left(n\pi \right)-2\sin \left(n\pi \right)\right)\right]=\frac{2L{\left(-1\right)}^n}{n\pi }n=1,2,3,\dots \end{array}$$Note that in this case we had $A_0\ne 0$ and $A_n=0,n=1,2,3,\dots$ This will happen on occasion so don’t get excited about this kind of thing when it happens.
The Fourier series is then,
$$\begin{array}{cc}f\left(x\right)& =\sum _{n=0}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)\\ & =A_0+\sum _{n=1}^{\infty }{A}_n\cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)=L+\sum _{n=1}^{\infty }\frac{2L{\left(-1\right)}^n}{n\pi }\sin \left(\frac{n\pi x}{L}\right)\end{array}$$

As we saw in the previous example sometimes we’ll get $A_0\ne 0$ and $A_n=0,n=1,2,3,\dots$ Whether or not this will happen will depend upon the function $f\left(x\right)$ and often won’t happen, but when it does don’t get excited about it.

Let’s take a look at another problem.