\documentclass{article} %\pagestyle{empty} \usepackage{graphicx, amsmath, psfrag} \setlength{\textheight}{9in} \setlength{\topmargin}{0.0in} \setlength{\textwidth}{6in} \setlength{\oddsidemargin}{+0.5in} \newcommand{\dis}{\displaystyle} \begin{document} \begin{center}{\bf Lecture Notes for Ch 10\\Fourier Series and Partial Differential Equations}\end{center} {\bf Part I.} Outline \begin{itemize} \item [Page 2.] Review even and odd functions. Very important. \item [Page 3.] Some Damn Useful Integral Formulas. \item [Page 4.] Definition of Fourier series and a Theorem. \item [Pages 5-8.] Verification of parts of the Theorem. \item [Pages 9-10.] Fourier series of a Square Wave. \item [Pages 11-12.] Fourier series of a Triangle Wave. \end{itemize} \vfill \pagebreak \begin{center} {\bf Review of Even and Odd Functions} \end{center} \begin{center} \fbox{ Even: $f(-x) = f(x)$ \hspace{0.6in} Odd: $f(-x)=-f(x)$.} \end{center} \begin{itemize} \item [Even Examples:] $5x^6-3x^4+2$, $\frac{x^4+1}{x^2-7}$, $|x|$, $\cos(x)$, $\cos(x^5)$, $\sin^4(x)$. \item [Odd Examples:] $x^{1/3}$, $x^7-6x^3$, $x|x|$, $\sin(x)$, $\sin^5(x^7)$. \item [Neither:] $x^2+x$, $x+\cos(x)$, $\frac{1}{x+1}$. \begin{center} \psfrag{E}{Three Even Functions} \psfrag{O}{Three Odd Functions} \hspace*{-1in}\includegraphics[height=2in]{Figs/evenodd.eps} \end{center} \item [Problem:] Let $f(x)$ be an odd function and suppose it is defined at $x=0$. What is $f(0)$? Prove this! \item [Problem:] Find a function with domain the whole real line that is both even and odd. (There is only one correct answer.) \item [Fact:] If $f(x)$ is even, then $\dis \int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx$. \item [Fact:] If $f(x)$ is odd, then $\dis \int_{-a}^a f(x)\,dx = 0$. Draw pictures to see intuitively why these two facts hold. \item [Fact:] If $f(x)$ is even and differentiable, then $f'(x)$ is odd, and vice versa. This can be proved with the Chain Rule. Suppose $f(x)$ is even and differentiable. Then $(f(-x))' = f'(-x)(-1)$. But $f(-x)= f(x)$, so $(f(-x))'= (f(x))' = f'(x)$. Thus, $f'(-x)(-1) = f'(x)$, or $f'(-x) = -f'(x)$. You can also draw pictures of tangent lines to curves to see this intuitively. %\end{itemize} \item [Exercises:] Suppose that $f(x)$ and $g(x)$ are even and the $h(x)$ and $k(x)$ are odd. Then show that: \begin{center} \begin{tabular}{ll} (a) $f(x)g(x)$ is even. & (f) $f(g(x))$ is even. \\ (b) $f(x)h(x)$ is odd. & (g) $f(h(x))$ is even. \\ (c) $h(x)k(x)$ is even. & (h) $h(f(x))$ is even. \\ (d) $f(x)+g(x)$ is even. & (i) $h(k(x))$ is odd. \\ (e) $h(x)+k(x)$ is odd. & (j) $f(x) + h(x)$ need not be even or odd. \end{tabular} \end{center} \item [Example:] Let $p(x) = f(x)h(x)$. Then $p(-x) = f(-x)h(-x)= f(x)(-h(x)) = -f(x)h(x) = -p(x)$. Hence we have an odd function. \end{itemize} \vfill \begin{center} \fbox{\bf You may be tested on this!} \end{center} \vfill \pagebreak \begin{center} {\bf Some Damn Useful Integral Formulas} \end{center} \begin{enumerate} \item $\dis \int_{-L}^L \cos \left( \frac{m\pi x}{L}\right) \cos \left(\frac{n\pi x}{L} \right) \,\, dx = \left\{ \begin{array}{rrr} 0 & \mbox{ for } & m\neq n, \\ L & \mbox{ for } & m = n. \end{array}\right. $ \vspace{.25in} \item $\dis \int_{-L}^L \cos \left( \frac{m\pi x}{L}\right) \sin \left(\frac{n\pi x}{L} \right) \,\, dx = 0,\: \mbox{for all integers $m$ and $n$} $ \vspace{.25in} \item $\dis \int_{-L}^L \sin \left( \frac{m\pi x}{L}\right) \sin \left(\frac{n\pi x}{L} \right) \,\, dx = \left\{ \begin{array}{rrr} 0 & \mbox{ for } & m\neq n, \\ L & \mbox{ for } & m = n. \end{array}\right. $ \end{enumerate} \vspace{.25in} Below and to the left are the overlaid plots of $\cos 4\pi x$ and $\cos 2\pi x$. Below and to the right is the graph of their product. Study this. Make some similar plots on your own until the formulas above make sense. \begin{center} \includegraphics[height=2in]{Figs/cos4xandcos2x.eps} \hspace{0.5in} \includegraphics[height=2in]{Figs/cos4xXcos2x.eps} \end{center} We will prove only a special case of \#1. Let $L=\pi$, $m=3$ and $n=2$. The proof uses the trig identities \[ \cos (\theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi \] and its corollary \[ \cos (\theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi \] Thus, \[ \cos \theta \cos \phi = \frac{1}{2}\left( \cos (\theta + \phi) + \cos (\theta - \phi) \right). \] Applying this to our case gives \[ \cos 3x \cos 2x = \frac{1}{2}\left( \cos 5x + \cos x\right). \] Thus, \[ \int_{-\pi}^{\pi} \cos 3x \cos 2x \, dx = \frac{1}{2}\int_{-\pi}^{\pi} \cos 5x + \cos x \, dx = 0 + 0 = 0. \] Let's consider one more special case: $L=\pi$, $m=n=2$. Then \[ \int_{-\pi}^{\pi} \cos^2 2x \, dx = \frac{1}{2}\int_{-\pi}^{\pi} \cos 4x + \cos 0 \,dx = \frac{0+2\pi}{2} = \pi. \] From these ideas you should be able to derive the three integrals formulas. If you have had linear algebra, you might notice that the integral of a product of two functions is a kind of inner product and so the cosine and sine functions used above are mutually orthogonal. \pagebreak \begin{center}{\bf Definition of Fourier Series}\end{center} Let $f(x)$ be a piecewise continuous periodic function with period $2L$.\footnote{Technical note: It is also required that $f(x)$ is bounded and that in each period of $f(x)$ there are only finitely many extrema.} The Fourier Series of $f(x)$ is, \[ \frac{a_0}{2} + \sum_{n=1}^\infty a_n \cos \left( \frac{n\pi x}{L} \right) + \sum_{n=1}^\infty b_n \sin \left( \frac{n\pi x}{L} \right), \] where \[ a_n = \frac{1}{L} \int_{-L}^L \cos \left( \frac{n\pi x}{L} \right) f(x) \, dx, \:\:\: n = 0,1,2,3,\dots, \] and \[ b_n = \frac{1}{L} \int_{-L}^L \sin \left( \frac{n\pi x}{L} \right) f(x) \, dx, \:\:\: n = 1,2,3,\dots. \] {\bf Theorem.} The Fourier series of $f(x)$ converges to $f(x)$ if $f$ is continuous at $x$. If $f$ is discontinuous at $x$ then this must be a jump discontinuity. Let \[ f(x^+) = \lim_{c \to x^+} f(c) \hspace{.4in}\mbox{and} \hspace{.4in} f(x^-) = \lim_{c \to x^-} f(c). \] Then the Fourier series of $f$ converges to the average of these two limits, \[ \frac{f(x^+) + f(x^-)}{2}. \] If $f$ is an even function then $b_n = 0$, for $n=1, 2, 3, \dots.$ If $f$ is an odd function then $a_n = 0$, for $n=0, 1, 2, 3, \dots.$ In all cases $a_0/2$ is the average value of $f(x)$ over one period. \vspace{1in} We will verify parts of this theorem. The full proof is covered in Math 407. \vfill \pagebreak \begin{center} {\bf We verify the formula for $\mathbf a_0$ and show it is the average value of $\mathbf f(x)$.} \end{center} Let $c_n = \cos \left(\frac{n\pi x}{L}\right)$ and $s_n = \sin \left(\frac{n\pi x}{L}\right)$. \vspace{,25in} Suppose $$f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty a_n c_n + \sum_{n=1}^\infty b_n s_n.$$ We will show that \[ a_0 = \frac{1}{L} \int_{-L}^L \cos (0) f(x) \, dx. \] \begin{eqnarray*} \frac{1}{L} \int_{-L}^L \cos (0) f(x) \, dx &=& \frac{1}{L} \int_{-L}^L 1\cdot f(x) \, dx \\ && \\ && \\ &=& \frac{1}{L} \int_{-L}^L \left( \frac{a_0}{2} + \sum_{n=1}^\infty a_n c_n + \sum_{n=1}^\infty b_n s_n\right)\, dx \\ && \\ && \\ &=& \frac{1}{L}\int_{-L}^L \frac{a_0}{2}\, dx + \frac{1}{L} \sum_{n=1}^\infty a_n \int_{-L}^L c_n \, dx + \frac{1}{L} \sum_{n=1}^\infty b_n \int_{-L}^L s_n \, dx\\ && \\ && \\ &=& a_0 + \frac{1}{L} \sum_{n=1}^\infty a_n\cdot 0 + \frac{1}{L} \sum_{n=1}^\infty b_n \cdot 0 = a_0 \end{eqnarray*} since, \vspace{.2in} $\dis \int_{-L}^L \cos \left(\frac{n\pi x}{L} \right) \, dx = \frac{L}{n\pi} \sin \left( \frac{n\pi x}{L} \right) \bigg|_{-L}^{L} = 0 - 0 = 0, $ \vspace{.2in} $\dis \int_{-L}^L \sin \left(\frac{n\pi x}{L} \right) \, dx = \frac{-L}{n\pi} \cos \left( \frac{n\pi x}{L} \right) \bigg|_{-L}^{L} = \frac{-L}{n\pi} \left( \cos(n\pi) - \cos(-n\pi) \right) = 0, $ \vspace{.2in} and \vspace{.2in} $\dis \frac{1}{L} \int_{-L}^L \frac{a_0}{2}\, dx = \frac{a_0}{2L} \int_{-L}^L 1 \, dx = \frac{a_0}{2L} \cdot 2L = a_0. $ \vspace{.2in} Finally, recall that the average value of $f(x)$ over a cycle is $\dis \frac{1}{2L} \int_{-L}^L f(x) \, dx$. Thus, $a_0/2$ is the average value of $f(x)$ over a cycle. \vfill \pagebreak \begin{center} {\bf We check the formula for $\mathbf a_7$, \\but the method is the same for all n $\mathbf \geq$ 1.} \end{center} \vspace{.2in} \begin{eqnarray*} a_7 &\stackrel{?}{=} & \frac{1}{L} \int_{-L}^L \cos \left( \frac{7\pi x}{L}\right) f(x) \, dx \\ &&\\ &=& \frac{1}{L} \int_{-L}^L c_7\left( \frac{a_0}{2} + \sum_{n=1}^\infty a_n c_n + \sum_{n=1}^\infty b_n s_n\right) \, dx \\ &&\\ &=& \frac{a_0}{2L} \int_{-L}^L c_7 \, dx + \frac{1}{L} \sum_{n=1}^\infty a_n \int_{-L}^L c_n c_7 \, dx + \frac{1}{L} \sum_{n=1}^\infty b_n \int_{-L}^L s_n c_7 \, dx \\ &&\\ &=& \frac{a_0}{2L}\cdot 0 + \frac{1}{L} (0+0+0+0+0+0+a_7L+0+0+0+\cdots ) + \frac{1}{L}(0+0+0+\cdots ) \\ &&\\ &=& a_7. \end{eqnarray*} \vspace{0.5in} %\pagebreak \begin{center} {\bf Next we check $\mathbf b_4$}. \end{center} \vspace{.2in} \begin{eqnarray*} b_4 &\stackrel{?}{=} & \frac{1}{L} \int_{-L}^L \sin \left( \frac{4\pi x}{L}\right) f(x) \, dx \\ &&\\ &=& \frac{1}{L} \int_{-L}^L s_4\left( \frac{a_0}{2} + \sum_{n=1}^\infty a_n c_n + \sum_{n=1}^\infty b_n s_n\right) \, dx \\ &&\\ &=& \frac{a_0}{2L} \int_{-L}^L s_4 \, dx + \frac{1}{L} \sum_{n=1}^\infty a_n \int_{-L}^L c_n s_4 \, dx + \frac{1}{L} \sum_{n=1}^\infty b_n \int_{-L}^L s_n s_4 \, dx \\ &&\\ &=& \frac{a_0}{2L}\cdot 0 + \frac{1}{L} (0+0+0+\cdots ) + \frac{1}{L}(0+0+0+b_4L+0+0+0+\cdots ) \\ &&\\ &=& b_4. \end{eqnarray*} \vfil \pagebreak \begin{center}{\bf Even and Odd Properties}\end{center} \vspace{1in} If $f(x)$ is an odd function then \[ a_n = \frac{1}{L} \int_{-L}^L \cos \left( \frac{n\pi x}{L} \right) f(x) \, dx = 0, \] since the product of an even function with an odd function is odd. \vspace{1in} If $f(x)$ is an even function then \[ b_n = \frac{1}{L} \int_{-L}^L \sin \left( \frac{n\pi x}{L} \right) f(x) \, dx = 0, \] since the product of an odd function with an even function is odd. \vfill \pagebreak \begin{center}{\bf An Example: The Square Wave}\end{center} Let \[ f(x) = \left\{ \begin{array}{lll} -1 & \mbox{ for } & -\pi < x < 0 \\ 1 & \mbox{ for } & 0 < x < \pi, \end{array} \right. \] with $f(x+2\pi) = f(x)$ for all $x$. \vspace{.3in} \begin{center}\psfrag{Square}{Square}\psfrag{Wave}{\hspace*{0.3in}Wave} \includegraphics[width=6in]{Figs/sqplot1.eps} \end{center} {\bf Problem.} Find the Fourier series of $f(x)$. \vspace{.3in} {\bf Solution.} Since $f(x)$ is odd, $a_n=0$ for $n=0, 1, 2, \dots$. \vspace{.3in} Next, \[ b_n = \frac{1}{\pi} \int_{-\pi}^\pi \sin(nx) f(x) \\, dx = \frac{2}{\pi} \int_0^\pi \sin(nx)\cdot 1 \, dx = \] \[ \frac{-2}{n\pi} \left( \cos(n\pi) - \cos(0)\right) = \left\{\begin{array}{lll} 0 & \mbox{ for } & n \mbox{ even } \\ \frac{4}{n\pi} & \mbox{ for } & n \mbox{ odd. }\end{array} \right. \] \vspace{.25in} Thus, \[ f(x) = \sum_{k=1}^\infty \frac{4}{(2k-1)\pi} \sin \left( (2k-1)x \right) = \frac{4}{\pi} \sin x + \frac{4}{3\pi} \sin 3x + \frac{4}{5\pi} \sin 5x + \cdots. \] Recall that using $(2k-1)$ is a way to generate only odd numbers. \vspace{.3in} On the next page we plot a few partial sums. \vspace{.3in} Let $f_N(x) = \sum_{k=1}^N \frac{4}{(2k-1)\pi} \sin \left( (2k-1)x \right)$. \vfill \pagebreak \begin{center} {\bf Plots of $f_N(x)$ for $N=1, 2, 3, 4, 10, 100$} \end{center} \begin{center}\psfrag{N}{$N=1$}\psfrag{=}{}\psfrag{1}{} \includegraphics[width=6in]{Figs/sqplot2.eps} \end{center} \begin{center}\psfrag{N}{$N=2$}\psfrag{=}{}\psfrag{2}{} \includegraphics[width=6in]{Figs/sqplot3.eps} \end{center} \begin{center}\psfrag{N}{$N=3$}\psfrag{=}{}\psfrag{3}{} \includegraphics[width=6in]{Figs/sqplot4.eps} \end{center} \begin{center}\psfrag{N}{$N=4$}\psfrag{=}{}\psfrag{4}{} \includegraphics[width=6in]{Figs/sqplot5.eps} \end{center}\psfrag{N}{$N=10$}\psfrag{=}{}\psfrag{10}{} \begin{center} \includegraphics[width=6in]{Figs/sqplot6.eps} \end{center} \begin{center}\psfrag{N}{$N=100$}\psfrag{=}{}\psfrag{100}{} \includegraphics[width=6in]{Figs/sqplot7.eps} \end{center} \vfill \pagebreak \begin{center}{\bf Another Example! A Triangle Wave}\end{center} Problem: Find the Fourier series of the wave depicted below. \begin{center} \includegraphics[width=6in]{Figs/triplots1.eps} \end{center} Solution: First note that since the period is 2, $L=1$. We can see that for $0