Differential Equations - Fourier Series: Sine - iii

Example 3 Find the Fourier sine series for $f\left(x\right)=L-x$ on $0\le x\le L$.

There really isn’t much to do here other than computing the coefficients so here they are,
$$\begin{array}{cc}{B}_n& =\frac{2}{L}{\int }_0^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx=\frac{2}{L}{\int }_0^L\left(L-x\right)\sin \left(\frac{n\pi x}{L}\right)dx\\ & =\frac{2}{L}{\left(-\frac{L}{n^2{\pi }^2}\right)\left[L\sin \left(\frac{n\pi x}{L}\right)-n\pi \left(x-L\right)\cos \left(\frac{n\pi x}{L}\right)\right]|}_0^L\\ & =\frac{2}{L}\left[\frac{L^2}{n^2{\pi }^2}\left(n\pi -\sin \left(n\pi \right)\right)\right]=\frac{2L}{n\pi }\end{array}$$In the simplification process don’t forget that $n$ is an integer.
So, with the coefficients we get the following Fourier sine series for this function.
$$f\left(x\right)=\sum _{n=1}^{\infty }\frac{2L}{n\pi }\sin \left(\frac{n\pi x}{L}\right)$$

In the next example it is interesting to note that while we started out this section looking only at odd functions we’re now going to be finding the Fourier sine series of an even function on $0\le x\le L$. Recall however that we’re really finding the Fourier sine series of the odd extension of this function and so we’re okay.

Example 4 Find the Fourier sine series for $f\left(x\right)=1+x^2$ on $0\le x\le L$.

In this case the coefficients are liable to be somewhat messy given the fact that the integrals will involve integration by parts twice. Here is the work for the coefficients.
$$\begin{array}{cc}{B}_n& =\frac{2}{L}{\int }_0^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx=\frac{2}{L}{\int }_0^L\left(1+x^2\right)\sin \left(\frac{n\pi x}{L}\right)dx\\ & =\frac{2}{L}\left(\frac{L}{n^3{\pi }^3}\right){\left[\left(2L^2-n^2{\pi }^2\left(1+x^2\right)\right)\cos \left(\frac{n\pi x}{L}\right)+2Ln\pi x\sin \left(\frac{n\pi x}{L}\right)\right]}_0^L\\ & =\frac{2}{L}\left(\frac{L}{n^3{\pi }^3}\right)\left[\left(2L^2-n^2{\pi }^2\left(1+L^2\right)\right)\cos \left(n\pi \right)+2L^{2}n\pi \sin \left(n\pi \right)-\left(2L^2-n^2{\pi }^2\right)\right]\\ & =\frac{2}{n^3{\pi }^3}\left[\left(2L^2-n^2{\pi }^2\left(1+L^2\right)\right){\left(-1\right)}^n-2L^2+n^2{\pi }^2\right]\end{array}$$As noted above the coefficients are not the most pleasant ones, but there they are. The Fourier sine series for this function is then,
$$f\left(x\right)=\sum _{n=1}^{\infty }\frac{2}{n^3{\pi }^3}\left[\left(2L^2-n^2{\pi }^2\left(1+L^2\right)\right){\left(-1\right)}^n-2L^2+n^2{\pi }^2\right]\sin \left(\frac{n\pi x}{L}\right)$$

In the last example of this section we’ll be finding the Fourier sine series of a piecewise function and can definitely complicate the integrals a little but they do show up on occasion and so we need to be able to deal with them.

Example 5 Find the Fourier sine series for $f\left(x\right)=\{\begin{array}{cc}\frac{L}{2}& \text{if}0\le x\le \frac{L}{2}\\ x-\frac{L}{2}& \text{if}\frac{L}{2}\le x\le L\end{array}$ on $0\le x\le L$.

Here is the integral for the coefficients.
$$\begin{array}{cc}{B}_n=\frac{2}{L}{\int }_0^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx& =\frac{2}{L}\left[{\int }_0^{\frac{L}{2}}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx+{\int }_{\frac{L}{2}}^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx\right]\\ & =\frac{2}{L}\left[{\int }_0^{\frac{L}{2}}\frac{L}{2}\sin \left(\frac{n\pi x}{L}\right)dx+{\int }_{\frac{L}{2}}^L\left(x-\frac{L}{2}\right)\sin \left(\frac{n\pi x}{L}\right)dx\right]\end{array}$$Note that we need to split the integral up because of the piecewise nature of the original function. Let’s do the two integrals separately
$${\int }_0^{\frac{L}{2}}\frac{L}{2}\sin \left(\frac{n\pi x}{L}\right)dx=-{\left(\frac{L}{2}\right)\left(\frac{L}{n\pi }\right)\cos \left(\frac{n\pi x}{L}\right)|}_0^{\frac{L}{2}}=\frac{L^2}{2n\pi }\left(1-\cos \left(\frac{n\pi }{2}\right)\right)$$$$\begin{array}{cc}{\int }_{\frac{L}{2}}^L\left(x-\frac{L}{2}\right)\sin \left(\frac{n\pi x}{L}\right)dx& ={\frac{L}{n^2{\pi }^2}\left[L\sin \left(\frac{n\pi x}{L}\right)-n\pi \left(x-\frac{L}{2}\right)\cos \left(\frac{n\pi x}{L}\right)\right]|}_{\frac{L}{2}}^L\\ & =\frac{L}{n^2{\pi }^2}\left[L\sin \left(n\pi \right)-\frac{n\pi L}{2}\cos \left(n\pi \right)-L\sin \left(\frac{n\pi }{2}\right)\right]\\ & =-\frac{L^2}{n^2{\pi }^2}\left[\frac{n\pi {\left(-1\right)}^n}{2}+\sin \left(\frac{n\pi }{2}\right)\right]\end{array}$$Putting all of this together gives,
$$\begin{array}{cc}{B}_n=\frac{2}{L}{\int }_0^{L}f\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx& =\frac{2}{L}\left(\frac{L^2}{2n\pi }\right)\left[1+{\left(-1\right)}^{n+1}-\cos \left(\frac{n\pi }{2}\right)+\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\right]\\ & =\frac{L}{n\pi }\left[1+{\left(-1\right)}^{n+1}-\cos \left(\frac{n\pi }{2}\right)-\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\right]\end{array}$$So, the Fourier sine series for this function is,
$$f\left(x\right)=\sum _{n=1}^{\infty }\frac{L}{n\pi }\left[1+{\left(-1\right)}^{n+1}-\cos \left(\frac{n\pi }{2}\right)-\frac{2}{n\pi }\sin \left(\frac{n\pi }{2}\right)\right]\sin \left(\frac{n\pi x}{L}\right)$$

As the previous two examples has shown the coefficients for these can be quite messy but that will often be the case and so we shouldn’t let that get us too excited.