9.3: Fourier Series

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Our solution of the diffusion and wave equations will require use of a Fourier series. A periodic function $f(x)$ with period $2L$, can be represented as a Fourier series in the form $$\label{eq:1}f(x)=\frac{a_0}{2}+\sum\limits_{n=1}^\infty\left(a_n\cos\frac{n\pi x}{L}+b_n\sin\frac{n\pi x}{L}\right).$$

Determination of the coefficients $a_0,\: a_1,\: a_2,\ldots$ and $b_1,\: b_2,\: b_3,\ldots$ makes use of orthogonality relations for sine and cosine. We first define the widely used Kronecker delta $\delta_{nm}$ as $$\delta_{nm}=\left\{\begin{array}{rl}1&\text{if }n=m; \\ 0&\text{otherwise.}\end{array}\right.\nonumber$$

The orthogonality relations for $n$ and $m$ positive integers are then given with compact notation as the integration formulas $$\label{eq:2} \int_{-L}^L\cos\left(\frac{m\pi x}{L}\right)\cos\left(\frac{n\pi x}{L}\right)dx=L\delta_{nm},$$ $$\label{eq:3}\int_{-L}^L\sin\left(\frac{m\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right)dx=L\delta_{nm},$$ $$\label{eq:4}\int_{-L}^L\cos\left(\frac{m\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right)dx=0.$$

To illustrate the integration technique used to obtain these results, we derive $\eqref{eq:3}$ assuming that $n$ and $m$ are positive integers with $n\neq m$. Changing variables to $ξ = \pi x/L$, we obtain $$\begin{aligned}\int_{-L}^L\sin&\left(\frac{m\pi x}{L}\right)\sin\left(\frac{n\pi x}{L}\right)dx \\ &=\frac{L}{\pi}\int_{-\pi}^\pi\sin(m\xi)\sin(n\xi)d\xi \\ &=\frac{L}{2\pi}\int_{-\pi}^\pi [\cos ((m-n)\xi)-\cos((m+n)\xi)]d\xi \\ &=\frac{L}{2\pi}\left[\frac{1}{m-n}\sin((m-n)\xi)-\frac{1}{m+n}\sin((m+n)\xi)\right]_{-\pi}^\pi \\ &=0.\end{aligned}$$

For $m=n$, however, $$\begin{aligned}\int_{-L}^L\sin^2\left(\frac{n\pi x}{L}\right)dx&=\frac{L}{\pi}\int_{-\pi}^\pi\sin^2(n\xi)d\xi \\ &=\frac{L}{2\pi}\int_{-\pi}^\pi(1-\cos(2n\xi))d\xi \\ &=\frac{L}{2\pi}\left[\xi-\frac{1}{2n}\sin 2n\xi\right]_{-\pi}^\pi \\ &=L.\end{aligned}$$

Integration formulas given by $\eqref{eq:2}$ and $\eqref{eq:4}$ can be similarly derived.

To determine the coefficient $a_n$, we multiply both sides of $\eqref{eq:1}$ by $\cos (n\pi x/L)$ with $n$ a nonnegative integer, and change the dummy summation variable from $n$ to $m$. Integrating over $x$ from $−L$ to $L$ and assuming that the integration can be done term by term in the infinite sum, we obtain $$\begin{aligned}\int_{-L}^Lf(x)\cos &\frac{n\pi x}{L}dx=\frac{a_0}{2}\int_{-L}^L\cos\frac{n\pi x}{L}dx \\ &+\sum\limits_{m=1}^\infty\left\{a_m\int_{-L}^L\cos\frac{n\pi x}{L}\cos\frac{m\pi x}{L}dx+b_m\int_{-L}^L\cos\frac{n\pi x}{L}\sin\frac{m\pi x}{L}dx\right\}.\end{aligned}$$

If $n = 0$, then the second and third integrals on the right-hand-side are zero and the first integral is $2L$ so that the right-hand-side becomes $La_0$. If $n$ is a positive integer, then the first and third integrals on the right-hand-side are zero, and the second integral is $L\delta_{nm}$. For this case, we have $$\begin{aligned}\int_{-L}^Lf(x)\cos\frac{n\pi x}{L}dx&=\sum\limits_{m=1}^\infty La_m\delta_{nm} \\ &=La_n,\end{aligned}$$ where all the terms in the summation except $m = n$ are zero by virtue of the Kronecker delta. We therefore obtain for $n = 0, 1, 2,\ldots$ $$\label{eq:5}a_n=\frac{1}{L}\int_{-L}^L f(x)\cos\frac{n\pi x}{L}dx.$$

To determine the coefficients $b_1,\: b_2,\: b_3,\ldots$, we multiply both sides of $\eqref{eq:1}$ by $\sin (n\pi x/L)$, with $n$ a positive integer, and again change the dummy summation variable from $n$ to $m$. Integrating, we obtain $$\begin{aligned}\int_{-L}^L f(x)\sin &\frac{n\pi x}{L}dx=\frac{a_0}{2}\int_{-L}^L\sin\frac{n\pi x}{L}dx \\ &+\sum\limits_{m=1}^\infty\left\{a_m\int_{-L}^L\sin\frac{n\pi x}{L}\cos\frac{m\pi x}{L}dx+b_m\int_{-L}^L\sin\frac{n\pi x}{L}\sin\frac{m\pi x}{L}dx\right\}.\end{aligned}$$

Here, the first and second integrals on the right-hand-side are zero, and the third integral is $L\delta_{nm}$ so that $$\begin{aligned}\int_{-L}^L f(x)\sin\frac{n\pi x}{L}dx&=\sum\limits_{m=1}^\infty Lb_m\delta_{nm} \\ &=Lb_n.\end{aligned}$$

Hence, for $n = 1,\: 2,\: 3,\ldots$, $$\label{eq:6}b_n=\frac{1}{L}\int_{-L}^L f(x)\sin\frac{n\pi x}{L}dx.$$

Our results for the Fourier series of a function $f(x)$ with period $2L$ are thus given by $\eqref{eq:1}$, $\eqref{eq:5}$ and $\eqref{eq:6}$.