\documentclass[12pt]{article} \usepackage{amsfonts} %new commands % reference for equation labels \newcommand{\eq}[1]{(\ref{#1})} % abbreviations for equation environments \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\bea}{\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}} \newcommand{\beann}{\begin{eqnarray*}} \newcommand{\eeann}{\end{eqnarray*}} % abbreviation for reals \def\Cx{\mathbb{C}} \def\Nl{\mathbb{N}} \def\Ra{\mathbb{Q}} \def\Rl{\mathbb{R}} \def\Ts{\mathbb{T}} \def\Ir{\mathbb{Z}} \def\DD{\mathcal{D}} % abbreviations for \varepsilon and \varphi \newcommand{\eps}{\varepsilon} \newcommand{\vphi}{\varphi} \pagestyle{empty} \begin{document} \centerline{\textbf{Solutions to Sample Midterm 2}} \centerline{\textbf{Math 121}, \textbf{Fall 2004}} \bigskip\bigskip \noindent \textbf{1.} Use Fourier series to find the solution $u(x,y)$ of the following boundary value problem for Laplace's equation in the semi-infinite strip $00$: \beann &&\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0,\\ &&u(0,y) = u(1,y) = 0,\\ &&u(x,0) = 1,\\ &&u(x,y) \to 0\quad \mbox{as $y\to\infty$}. \eeann \bigskip\noindent \textbf{Solution.} \begin{itemize} \item The separated solutions of Laplace's equation that satisfy the boundary conditions at $x=0,1$ and as $y\to\infty$ are $\sin (n\pi x) e^{-n\pi y}$, where $n$ is a positive integer. We therefore look for a solution of the form \[ u(x,y) = \sum_{n=1}^\infty b_n\sin (n\pi x) e^{-n\pi y}. \] Imposing the boundary condition at $y=0$, we obtain \[ \sum_{n=1}^\infty b_n\sin (n\pi x) = 1, \] so \beann b_n &=& 2\int_0^1 \sin (n\pi x)\, dx\\ &=& 2\left[\frac{-\cos (n\pi x)}{n\pi}\right]_0^1\\ &=& 2\left[\frac{(-1)^{n+1}}{n\pi} + \frac{1}{n\pi} \right]_0^1\\ &=& \cases{4/(n\pi) & for $n$ odd,\cr 0 & for $n$ even.} \eeann The solution is therefore \[ u(x,y) = \frac{4}{\pi}\left\{\sin (\pi x) e^{-\pi y} + \frac{1}{3}\sin (3\pi x) e^{-3\pi y} + \frac{1}{5}\sin (5\pi x) e^{-5\pi y} + \dots\right\}. \] \end{itemize} \newpage \noindent \textbf{2.} Use Fourier series to find the solution $u(x,t)$ of the following initial-boundary value problem for the wave equation in $00$: \beann &&\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} = 0,\\ &&\frac{\partial u}{\partial x}(0,t) = \frac{\partial u}{\partial x}(1,t) = 0,\\ &&u(x,0) = 0,\\ &&\frac{\partial u}{\partial t}(x,0) = x. \eeann \bigskip\noindent \textbf{Solution.} \begin{itemize} \item The separated solutions of the wave equation that are zero at $t=0$ and satisfy the boundary conditions at $x=0,1$ are $t$ and $\cos (n\pi x) \sin (n\pi t)$, where $n= 1,2,\dots$. We therefore look for a solution of the form \[ u(x,t) = \frac{1}{2} a_0 t + \sum_{n=1}^\infty a_n \cos (n\pi x) \sin (n\pi t). \] Differentiating this series with respect to $t$, we find that \[ \frac{\partial u}{\partial t}(x,t) = \frac{1}{2}a_0 + \sum_{n=1}^\infty n\pi a_n \cos (n\pi x) \cos (n\pi t). \] Imposing the initial condition for $\partial u/\partial t$ at $t=0$, we get the Fourier cosine expansion: \[ \frac{1}{2} a_0 + \sum_{n=1}^\infty n\pi a_n \cos (n\pi x) = x. \] Hence, for $n\ge 1$ we have \beann n \pi a_n &=& 2\int_0^1 x\cos (n\pi x)\, dx\\ &=& 2\left[\frac{x\sin (n\pi x)}{n\pi} + \frac{\cos (n\pi x)}{(n\pi)^2}\right]_0^1\\ &=& 2\left[\frac{(-1)^{n}}{(n\pi)^2} - \frac{1}{(n\pi)^2} \right]_0^1\\ &=& \cases{-4/(n\pi)^2 & for $n$ odd,\cr 0 & for $n$ even,} \eeann and \[ a_n = \cases{-4/(n\pi)^3 & for $n$ odd,\cr 0 & for $n$ even.} \] For $n=0$, we get \beann a_0 &=& 2\int_0^1 x\, dx\\ &=& 2\left[\frac{1}{2} x^2\right]_0^1\\ &=& 1. \eeann Hence, the solution is \[ u(x,t) = \frac{1}{2} t -\frac{4}{\pi^3}\left\{ \cos (\pi x) + \frac{1}{3^3} \cos (3\pi x) + \frac{1}{5^3} \cos (5\pi x) + \dots\right\}. \] \end{itemize} \newpage \noindent \textbf{3.} Use Fourier transforms to solve the following initial value problem for $u(x,t)$ in $-\infty < x < \infty$, $t >0$: \beann &&\frac{\partial u}{\partial t} = -\frac{\partial^4 u}{\partial x^4},\\ &&u(x,0) = f(x). \eeann Write the solution for $u(x,t)$ as a convolution, but do not compute any inverse transforms explicitly. How smooth is the solution for $t>0$? \bigskip\noindent \textbf{Solution.} \begin{itemize} \item Let \[ \widehat{u} (k,t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} u(x,t) e^{-ikx}\, dx \] be the Fourier transform of $u$ with respect to $x$. Then, taking the Fourier transform of the initial value problem, we get \beann &&\frac{\partial \widehat{u}}{\partial t} = -(-ik)^4\widehat{u},\\ &&\widehat{u} (k,0) = \widehat{f} (k), \eeann where $\widehat{f}$ is the Fourier transform of $f$. It follows that \beann &&\frac{\partial \widehat{u}}{\partial t} = -k^4\widehat{u},\\ &&\widehat{u} (k,0) = \widehat{f} (k), \eeann which has the solution \[ \widehat{u} (k,t) = \widehat{f} (k) e^{-k^4 t}. \] According to the convolution theorem, if $f$, $g$ have Fourier transforms $\widehat{f}$, $\widehat{g}$ respectively then $\widehat{f}\cdot\widehat{g}$ is the Fourier transform of $\frac{1}{2\pi} f\ast g$. It follows that \[ u(x,t) = \int_{-\infty}^\infty G(x-y,t) f(y)\, dy \] where \[ \widehat{G}(k,t) = \frac{1}{2\pi} e^{-k^4 t}. \] The solution is smooth (infinitely differentiable with respect to $x$) for $t >0$ since its Fourier transform decays exponentially quickly as $k\to \infty$ (assuming, for example, that $\widehat{f} (k)$ is a bounded function of $k$). \end{itemize} \newpage \noindent \textbf{4.} (a) Give the formulas for the Fourier transform $\widehat{f}(k)$ of a function $f(x)$ and the inverse Fourier transform. \smallskip\noindent (b) Compute the Fourier transform of $e^{-|x|}$. \smallskip\noindent (c) State Parseval's theorem, and use it to evaluate \[ \int_0^\infty\frac{1}{(1+k^2)^2}\, dk. \] \bigskip\noindent \textbf{Solution.} \begin{itemize} \item (a) A function $f(x)$ and its Fourier transform $\widehat{f}(k)$ are related by \beann \widehat{f} (k) &=& \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x) e^{-ikx}\, dx,\\ {f} (x) &=& \int_{-\infty}^{\infty} \widehat{f}(k) e^{ikx}\, dk, \eeann \item (b) If $f(x) = e^{-|x|}$, then using \[ |x| = \cases{x & for $x\ge 0$,\cr -x & for $x\le 0$,} \] and changing $x\to -x$ in the integral for $-\infty < x < 0$, we find that \beann \widehat{f} (k) &=& \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-|x|} e^{-ikx}\, dx,\\ &=& \frac{1}{2\pi}\left\{ \int_{-\infty}^{0} e^{(1-ik)x}\, dx + \int_0^{\infty} e^{-(1+ik)x}\, dx\right\}\\ &=& \frac{1}{2\pi} \int_0^{\infty} \left\{ e^{-(1-ik)x} + e^{-(1+ik)x}\right\}\, dx\\ &=& - \frac{1}{2\pi} \left[\frac{e^{-(1-ik)x}}{1-ik} + \frac{e^{-(1+ik)x}}{1+ik}\right]_0^{\infty}\\ &=& \frac{1}{2\pi} \left[\frac{1}{1-ik} + \frac{1}{1+ik}\right]_0^{\infty}\\ &=& \frac{1}{\pi} \frac{1}{1+k^2}. \eeann \item (c) Parseval's theorem states that \[ \int_{-\infty}^{\infty} \left|\widehat{f}(k)\right|^2\,dk = \frac{1}{2\pi} \int_{-\infty}^{\infty} \left|f(x)\right|^2\,dx. \] For $f(x) = e^{-|x|}$, we compute that \beann \int_{-\infty}^{\infty} \left|f(x)\right|^2\,dx &=& 2 \int_{0}^{\infty} e^{-2x}\,dx \\ &=& -\left[e^{-2x}\right]_0^{\infty}\\ &=& 1. \eeann It follows from Parsevals theorem and (b) that \[ \frac{2}{\pi^2} \int_0^\infty \frac{1}{(1+k^2)^2}\, dk = \frac{1}{2\pi}\cdot 1, \] so \[ \int_0^\infty \frac{1}{(1+k^2)^2}\, dk = \frac{\pi}{4}. \] \item \textbf{Remark.} The integral in (c) can also be evaluated directly by use of the substitution $k = \tan \theta$, which gives \beann \int_0^\infty \frac{1}{(1+k^2)^2}\, dk &=& \int_0^{\pi/2} \frac{1}{(1+ \tan^2\theta)^2} \sec^2\theta\,d\theta\\ &=& \int_0^{\pi/2} \frac{1}{\sec^4\theta} \sec^2\theta\,d\theta\\ &=& \int_0^{\pi/2} \frac{1}{\sec^2\theta}\,d\theta\\ &=& \int_0^{\pi/2} {\cos^2\theta}\,d\theta\\ &=& \frac{\pi}{2} \cdot \frac{1}{2}\\ &=& \frac{\pi}{4}, \eeann which verifies Parseval's theorem explicitly in this case. \end{itemize} \newpage \noindent \textbf{5.} Use Laplace transforms to solve the following initial value problem: \beann &&y^{\prime\prime} +2 y^\prime + 2 y = 1,\\ &&y(t) = 0,\quad y^\prime(0) = 1. \eeann \bigskip\noindent \textbf{Solution.} \begin{itemize} \item Let $Y(p)$ be the Laplace transform of $y(t)$. Then, taking the Laplace transform of the ODE and using the initial conditions, we get that \[ p^2 Y - 1 + 2p Y +2Y = \frac{1}{p}. \] Solving for $Y$, we get \[ Y(p) = \frac{1}{p^2 + 2p + 2} + \frac{1}{p(p^2 + 2p + 2)}. \] We have $p^2 + 2p + 2 = (p+1)^2 +1$, so (from L13 of the table) \[ L^{-1}\left[\frac{1}{p^2 + 2p + 2}\right] = e^{-t}\sin t. \] Also \beann \frac{1}{p(p^2 + 2p + 2)} &=& \frac{1}{2}\left[\frac{1}{p} - \frac{p+2}{p^2 + 2p + 2} \right]\\ &=& \frac{1}{2}\left[\frac{1}{p} - \frac{p+1}{(p+1)^2 +1} - \frac{1}{(p+1)^2 +1}\right]. \eeann So (from L1, L13, L14) we have \[ L^{-1}\left[\frac{1}{p(p^2 + 2p + 2)}\right] = \frac{1}{2}\left[1 - e^{-t}\cos t - e^{-t}\sin t\right] \] Hence, combining these inverse transforms, we get \[ y(t) = \frac{1}{2}\left[1 - e^{-t}\cos t + e^{-t}\sin t\right]. \] \end{itemize} \bigskip \noindent \textbf{6.} (a) Say what jump conditions the solution of $y(t)$ of the following initial value problem satisfies at $t=0$, and find the solution directly (do not use Laplace transforms): \beann &&y^{\prime\prime} - 4 y = \delta(t),\\ &&y(t) = 0\quad\mbox{for $t<0$}. \eeann \smallskip\noindent (b) Write the solution of the following initial value problem, where $f(t)$ is an arbitrary function, as a convolution (you don't need to derive your answer): \beann &&y^{\prime\prime} - 4 y = f(t),\\ &&y(0) = y^\prime(0) = 0. \eeann \bigskip\noindent \textbf{Solution.} \begin{itemize} \item (a) The derivative of $y$ has a jump discontinuity of size one at $t=0$. The solution is therefore \[ y(t) = \cases{y_+(t) & for $t\ge 0$,\cr 0 & for $t < 0$,} \] where \beann &&y_+^{\prime\prime} - 4 y_+ = 0\qquad\mbox{for $t>0$},\\ &&y_+(0) = 0,\qquad y_+^\prime(0) = 1. \eeann The general solution of the ODE is $y_+(t) = a\cosh 2t + b\sinh 2t$, and the initial conditions imply that $a=0$ and $b=1/2$. Hence, \[ y(t) = \frac{1}{2} \sinh 2t\qquad \mbox{for $t\ge 0$}. \] \item (b) The solution is \[ y(t) = \frac{1}{2}\int_0^t \sinh 2(t-s) f(s)\, ds. \] \end{itemize} \end{document}