Differential Equations - Fourier Series: Sine - ii

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

First note that the function we’re working with is in fact an odd function and so this is something we can do. There really isn’t much to do here other than to compute the coefficients for $f\left(x\right)=x$.
Here is that work and note that we’re going to leave the integration by parts details to you to verify. Don’t forget that $n$, $L$, and $\pi$ are constants!
$$\begin{array}{cc}{B}_n& =\frac{2}{L}{\int }_0^{L}x\sin \left(\frac{n\pi x}{L}\right)dx=\frac{2}{L}{\left(\frac{L}{n^2{\pi }^2}\right)\left(L\sin \left(\frac{n\pi x}{L}\right)-n\pi x\cos \left(\frac{n\pi x}{L}\right)\right)|}_0^L\\ & =\frac{2}{n^2{\pi }^2}\left(L\sin \left(n\pi \right)-n\pi L\cos \left(n\pi \right)\right)\end{array}$$These integrals can, on occasion, be somewhat messy especially when we use a general $L$ for the endpoints of the interval instead of a specific number.
Now, taking advantage of the fact that $n$ is an integer we know that $\sin \left(n\pi \right)=0$ and that $\cos \left(n\pi \right)={\left(-1\right)}^n$. We therefore have,
$$B_n=\frac{2}{n^2{\pi }^2}\left(-n\pi L{\left(-1\right)}^n\right)=\frac{{\left(-1\right)}^{n+1}2L}{n\pi }n=1,2,3\dots$$The Fourier sine series is then,
$$x=\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}2L}{n\pi }\sin \left(\frac{n\pi x}{L}\right)=\frac{2L}{\pi }\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}\sin \left(\frac{n\pi x}{L}\right)$$

At this point we should probably point out that we’ll be doing most, if not all, of our work here on a general interval ($-L\le x\le L$ or $0\le x\le L$) instead of intervals with specific numbers for the endpoints. There are a couple of reasons for this. First, it gives a much more general formula that will work for any interval of that form which is always nice. Secondly, when we run into this kind of work in the next chapter it will also be on general intervals so we may as well get used to them now.

Now, finding the Fourier sine series of an odd function is fine and good but what if, for some reason, we wanted to find the Fourier sine series for a function that is not odd? To see how to do this we’re going to have to make a change. The above work was done on the interval $-L\le x\le L$. In the case of a function that is not odd we’ll be working on the interval $0\le x\le L$. The reason for this will be made apparent in a bit.

So, we are now going to do is to try to find a series representation for $f\left(x\right)$ on the interval $0\le x\le L$ that is in the form,

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

or in other words,

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

As we did with the Fourier sine series on $-L\le x\le L$ we are going to assume that the series will in fact converge to $f\left(x\right)$ and we’ll hold off discussing the convergence of the series for a later section.

There are two methods of generating formulas for the coefficients, $B_n$, although we’ll see in a bit that they really the same way, just looked at from different perspectives.

The first method is to just ignore the fact that $f\left(x\right)$ is not odd and proceed in the same manner that we did above only this time we’ll take advantage of the fact that we proved in the previous section that ${\left\{\sin \left(\frac{n\pi x}{L}\right)\right\}}_{n=1}^{\infty }$ also forms an orthogonal set on $0\le x\le L$ and that,

$${\int }_0^L\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx=\{\begin{array}{cc}\frac{L}{2}& \text{if}n=m\\ 0& \text{if}n\ne m\end{array}$$

So, if we do this then all we need to do is multiply both sides of our series by $\sin \left(\frac{m\pi x}{L}\right)$, integrate from 0 to $L$ and interchange the integral and series to get,

$${\int }_0^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dx=\sum _{n=1}^{\infty }{B}_n{\int }_0^L\sin \left(\frac{n\pi x}{L}\right)\sin \left(\frac{m\pi x}{L}\right)dx$$

Now, plugging in for the integral we arrive at,

$${\int }_0^{L}f\left(x\right)\sin \left(\frac{m\pi x}{L}\right)dx=B_m\left(\frac{L}{2}\right)$$

Upon solving for the coefficient we arrive at,

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

Note that this is identical to the second form of the coefficients that we arrived at above by assuming $f\left(x\right)$ was odd and working on the interval $-L\le x\le L$. The fact that we arrived at essentially the same coefficients is not actually all the surprising as we’ll see once we’ve looked the second method of generating the coefficients.

Before we look at the second method of generating the coefficients we need to take a brief look at another concept. Given a function, $f\left(x\right)$, we define the odd extension of $f\left(x\right)$ to be the new function,

$$g\left(x\right)=\{\begin{array}{cc}f\left(x\right)& \text{if}0\le x\le L\\ -f\left(-x\right)& \text{if}-L\le x\le 0\end{array}$$

It’s pretty easy to see that this is an odd function.

$$g\left(-x\right)=-f\left(-\left(-x\right)\right)=-f\left(x\right)=-g\left(x\right)\text{for}0<x<L$$

and we can also know that on $0\le x\le L$ we have that $g\left(x\right)=f\left(x\right)$. Also note that if $f\left(x\right)$ is already an odd function then we in fact get $g\left(x\right)=f\left(x\right)$ on $-L\le x\le L$.

Let’s take a quick look at a couple of odd extensions before we proceed any further.

Example 2 Sketch the odd extension of each of the given functions.

1. $f\left(x\right)=L-x$ on $0\le x\le L$
2. $f\left(x\right)=1+x^2$ on $0\le x\le L$
3. $f\left(x\right)=\{\begin{array}{cc}\frac{L}{2}& \text{if}0\le x\le \frac{L}{2}\\ x-\frac{L}{2}& \text{if}\frac{L}{2}\le x\le L\end{array}$

Not much to do with these other than to define the odd extension and then sketch it.

a) $f\left(x\right)=L-x$ on $0\le x\le L$ 

Here is the odd extension of this function.

$$\begin{array}{cc}g\left(x\right)& =\{\begin{array}{cc}f\left(x\right)& \text{if}0\le x\le L\\ -f\left(-x\right)& \text{if}-L\le x\le 0\end{array}\\ & =\{\begin{array}{cc}L-x& \text{if}0\le x\le L\\ -L-x& \text{if}-L\le x\le 0\end{array}\end{array}$$Below is the graph of both the function and its odd extension. Note that we’ve put the “extension” in with a dashed line to make it clear the portion of the function that is being added to allow us to get the odd extension.

b) $f\left(x\right)=1+x^2$ on $0\le x\le L$  

First note that this is clearly an even function. That does not however mean that we can’t define the odd extension for it. The odd extension for this function is,
$$\begin{array}{cc}g\left(x\right)& =\{\begin{array}{cc}f\left(x\right)& \text{if}0\le x\le L\\ -f\left(-x\right)& \text{if}-L\le x\le 0\end{array}\\ & =\{\begin{array}{cc}1+x^2& \text{if}0\le x\le L\\ -1-x^2& \text{if}-L\le x\le 0\end{array}\end{array}$$The sketch of the original function and its odd extension are ,

c) $f\left(x\right)=\{\begin{array}{cc}\frac{L}{2}& \text{if}0\le x\le \frac{L}{2}\\ x-\frac{L}{2}& \text{if}\frac{L}{2}\le x\le L\end{array}$ 

Let’s first write down the odd extension for this function.
$$g\left(x\right)=\{\begin{array}{cc}f\left(x\right)& \text{if}0\le x\le L\\ -f\left(-x\right)& \text{if}-L\le x\le 0\end{array}=\{\begin{array}{cc}x-\frac{L}{2}& \text{if}\frac{L}{2}\le x\le L\\ \frac{L}{2}& \text{if}0\le x\le \frac{L}{2}\\ -\frac{L}{2}& \text{if}-\frac{L}{2}\le x\le 0\\ x+\frac{L}{2}& \text{if}-L\le x\le -\frac{L}{2}\end{array}$$The sketch of the original function and its odd extension are,

With the definition of the odd extension (and a couple of examples) out of the way we can now take a look at the second method for getting formulas for the coefficients of the Fourier sine series for a function $f\left(x\right)$ on $0\le x\le L$. First, given such a function define its odd extension as above. At this point, because $g\left(x\right)$ is an odd function, we know that on $-L\le x\le L$ the Fourier sine series for $g\left(x\right)$ (and NOT $f\left(x\right)$ yet) is,

$$g\left(x\right)=\sum _{n=1}^{\infty }{B}_n\sin \left(\frac{n\pi x}{L}\right)B_n=\frac{2}{L}{\int }_0^{L}g\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dxn=1,2,3,\dots$$

However, because we know that $g\left(x\right)=f\left(x\right)$ on $0\le x\le L$ we can also see that as long as we are on $0\le x\le L$ we have,

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

So, exactly the same formula for the coefficients regardless of how we arrived at the formula and the second method justifies why they are the same here as they were when we derived them for the Fourier sine series for an odd function.

Now, let’s find the Fourier sine series for each of the functions that we looked at in Example 2.

Note that again we are working on general intervals here instead of specific numbers for the right endpoint to get a more general formula for any interval of this form and because again this is the kind of work we’ll be doing in the next chapter.

Also, we’ll again be leaving the actually integration details up to you to verify. In most cases it will involve some fairly simple integration by parts complicated by all the constants ($n$, $L$, $\pi$, etc.) that show up in the integral.