\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{\textsc{Advanced Analysis}} \centerline{\textbf{Math 121, Fall 2004}} \centerline{\textbf{Solutions, Midterm 2}} \bigskip\bigskip\noindent \textbf{1.} [20\%] Use Laplace transforms to find the solution $y(t)$ of the following initial value problem \beann &&y^{\prime\prime} + 9 y = f(t),\\ &&y(t) = 0,\quad y^\prime(0) = 0, \eeann where $f(t)$ is an arbitrary function. Express your answer as a convolution. \bigskip\noindent \textbf{Solution.} \begin{itemize} \item Let $Y(p) = \mathcal{L}[y(t)]$ and $F(p) = \mathcal{L}[f(t)]$, where $\mathcal{L}$ denotes the Laplace transform. Then \[ p^2 Y + 9 Y = F(p), \] so \[ Y(p) = \frac{F(p)}{p^2 + 9}. \] From the tables, \[ \mathcal{L}^{-1}\left[\frac{1}{p^2 + 9}\right] = \frac{1}{3}\sin(3t). \] Therefore, using the convolution theorem, we have \[ y(t) = \frac{1}{3} \int_0^t \sin [3(t-s)] f(s)\, ds. \] \end{itemize} \newpage \noindent \textbf{2.} [20\%] (a) Say what jump conditions the solution $y(t)$ of the following initial value problem satisfies at $t=0$, and find the solution directly (do not use Laplace transforms): \beann &&y^{\prime} + 2 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} + 2 y = f(t),\\ &&y(0) = 0. \eeann \bigskip\noindent \textbf{Solution.} \begin{itemize} \item (a) The solution $y(t)$ has a jump discontinuity of size one at $t=0$, so that $y(t)$ for $t>0$ satisfies the IVP \beann &&y^\prime + 2 y = 0,\\ &&y(0) = 1. \eeann The solution of the ODE is \[ y(t) = c e^{-2t} \] where $c$ is a constant of integration, and the initial condition implies that $c=1$. Hence, \[ y(t) = \cases{e^{-2t} & for $t > 0$,\cr 0 & for $t < 0$.} \] \item (b) The solution is given by \[ y(t) = \int_0^t e^{-2(t-s)}f(s)\,ds. \] \end{itemize} \newpage \noindent \textbf{3.} [20\%] Use Fourier series to find the solution $u(x,y)$ of the following boundary value problem for Laplace's equation in the strip $00$: \beann &&\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0,\\ &&\frac{\partial u}{\partial x} (0,y) = \frac{\partial u}{\partial x}(1,y) = 0,\\ &&u(x,0) = x,\qquad \mbox{$u(x,y)$ is bounded as $y\to\infty$}. \eeann What does the solution $u(x,y)$ approach as $y\to \infty$? \bigskip\noindent \textbf{Solution.} \begin{itemize} \item The appropriate linear combination of separated solutions of Laplace's equation that satisfy the boundary conditions at $x=0,1$ and as $y\to\infty$ is \[ u(x,y) = \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n \cos (n\pi x) e^{-n\pi y}, \] where the $a_n$ are constants. Imposing the boundary condition at $y=0$, we obtain that \[ x = \frac{1}{2}a_0 + \sum_{n=1}^\infty a_n \cos (n\pi x), \] so \[ a_n = 2\int_0^1 x\cos(n\pi x) \, dx. \] Evaluating this integral, we find that $a_0 = 1$, and for $n\ge 1$ \[ a_n = \cases{-4/(n\pi)^2 & for $n$ odd,\cr 0 & for $n$ even.} \] Hence the solution is \beann u(x,y) &=& \frac{1}{2} - \frac{4}{\pi^2}\left\{\cos(\pi x) e^{-\pi y} + \frac{1}{3^2} \cos (3\pi x) e^{-3\pi y}\right.\\ &&\qquad\qquad\qquad\left.+ \frac{1}{5^2} \cos (5\pi x) e^{-5\pi y} +\dots\right\}. \eeann It follows that $u(x,y) \to 1/2$ as $y\to \infty$. Thus, the temperature approaches a constant value equal to the mean value of the boundary data at $y=0$. \end{itemize} \newpage \noindent \textbf{4.} [20\%] Reduce the boundary conditions to homogeneous boundary conditions and use Fourier series to find the solution $u(x,t)$ of the following initial-boundary value problem for the heat equation in $00$: \beann &&\frac{\partial u}{\partial t} = \frac{\partial^2 u}{\partial x^2},\\ &&u(0,t) = 0, \qquad u(1,t) = 1,\\ &&u(x,0) = 0. \eeann What does the solution $u(x,t)$ approach as $t\to \infty$? \bigskip\noindent \textbf{Solution.} \begin{itemize} \item The steady-state solution $u=x$ satisfies the non-homogeneous boundary conditions and the PDE. We write \[ u(x,t) = x + v(x,t). \] Then $v(x,t)$ satisfies \beann &&\frac{\partial v}{\partial t} = \frac{\partial^2 v}{\partial x^2},\\ &&v(0,t) = 0, \qquad v(1,t) = 0,\\ &&v(x,0) = -x. \eeann We look for a solution for $v$ of the form \[ v(x,t) = \sum_{n=1}^\infty b_n \sin(n\pi x) e^{-n^2\pi^2 t}. \] This function satisfies the PDE and the boundary conditions at $x=0,1$ for any choice of the constants $b_n$. Imposing the initial condition at $t=0$ we get that \[ - x = \sum_{n=1}^\infty b_n \sin(n\pi x). \] Using the expression for the Fourier sine coefficients, we get \beann b_n &=& - 2\int_0^1 x \sin (n\pi x)\, dx\\ &=& \frac{2(-1)^n}{n\pi}. \eeann The solution of the original problem is therefore \[ u(x,t) = x + \frac{2}{\pi} \sum_{n=1}^\infty \frac{(-1)^n}{n} \sin (n\pi x) e^{-n^2\pi^2 t}. \] As $t\to\infty$, we have $u(x,t) \to x$, so the temperature approaches the steady-state temperature distribution associated with the given non-homogeneous boundary conditions. \end{itemize} \newpage \noindent \textbf{5.} [20\%] (a) Consider the following initial value problem for $u(x,t)$ \beann &&\frac{\partial u}{\partial t} = \frac{\partial^3 u}{\partial x^3},\\ &&u(x,0) = f(x), \eeann where $-\infty < x < \infty$, $t >0$. Solve for the Fourier transform of $u(x,t)$ with respect to $x$, and write the solution for $u(x,t)$ as a Fourier integral, but do not attempt to invert the transform explicitly. \smallskip \noindent (b) State Parseval's theorem for the Fourier transform. Use your solution from (a) to show that \[ \int_{-\infty}^\infty u^2(x,t)\, dx = \int_{-\infty}^\infty f^2(x)\, dx \] for all times $t$. (This result expresses `conservation of energy'.) \bigskip\noindent \textbf{Solution.} \begin{itemize} \item (a) Let \[ \widehat{u}(k,t) = \frac{1}{2\pi} \int_{-\infty}^\infty u(x,t) e^{-ikx}\, dx \] be the Fourier transform of $u(x,t)$ with respect to $x$. Then, Fourier tranforming the PDE, we find that \beann &&\frac{\partial \widehat{u}}{\partial t} = (i k)^3\widehat{u},\\ &&\widehat{u}(k,0) = \widehat{f}(k), \eeann where $\widehat{f}$ is the Fourier transform of $f$. Solving this ODE, we get \[ \widehat{u}(k,t) = \widehat{f}(k) e^{(ik)^3 t}. \] Using this expression in the Fourier inversion formula, \[ u(x,t) = \int_{-\infty}^\infty \widehat{u}(k,t) e^{ikx}\, dk, \] and writing $(ik)^3 = -ik^3$, we obtain the solution \[ u(x,t) = \int_{-\infty}^\infty \widehat{f}(k)e^{ikx - i k^3 t}\, dk. \] \item (b) Parseval's theorem states that \[ \frac{1}{2\pi} \int_{-\infty}^\infty |f(x)|^2\, dx = \int_{-\infty}^\infty \left|\widehat{f}(k)\right|^2\, dk. \] Since $|e^{i\theta}| = 1$ for any real number $\theta$, we have that \[ \left|e^{-ik^3 t}\right| = 1. \] Hence, for any $t >0$, we have \[ \left|\widehat{u}(k,t)\right| = \left|\widehat{f}(k) e^{-ik^3 t}\right| = \left|\widehat{f}(k)\right|. \] It then follows from Parseval's theorem and this equation that \beann \int_{-\infty}^\infty u^2(x,t)\, dx &=& 2\pi \int_{-\infty}^\infty \left|\widehat{u}(k,t)\right|^2\, dk\\ &=& 2\pi \int_{-\infty}^\infty \left|\widehat{f}(k)\right|^2\, dk\\ &=& \int_{-\infty}^\infty f^2(x)\, dx. \eeann \end{itemize} \end{document}