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1 \documentclass[a4paper]{article} 2 3 4 \usepackage[T1]{fontenc} 5 \usepackage[utf8]{inputenc} 6 \usepackage{mlmodern} 7 8 %\usepackage{ngerman} % Sprachanpassung Deutsch 9 10 \usepackage{graphicx} 11 \usepackage{geometry} 12 \geometry{a4paper, top=15mm} 13 14 \usepackage{subcaption} 15 \usepackage[shortlabels]{enumitem} 16 \usepackage{amssymb} 17 \usepackage{amsthm} 18 \usepackage{mathtools} 19 \usepackage{braket} 20 \usepackage{bbm} 21 \usepackage{graphicx} 22 \usepackage{float} 23 \usepackage{yhmath} 24 \usepackage{tikz} 25 \usetikzlibrary{patterns,decorations.pathmorphing,positioning} 26 \usetikzlibrary{calc,decorations.markings} 27 28 %\usepackage[backend=biber, sorting=none]{biblatex} 29 %\addbibresource{uni.bib} 30 31 \usepackage[framemethod=TikZ]{mdframed} 32 33 \tikzstyle{titlered} = 34 [draw=black, thick, fill=white,% 35 text=black, rectangle, 36 right, minimum height=.7cm] 37 38 39 \usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref} 40 \usepackage[parfill]{parskip} 41 \usepackage{lipsum} 42 43 44 \usepackage{tcolorbox} 45 \tcbuselibrary{skins,breakable} 46 \newcommand{\eps}{\varepsilon} 47 \pagestyle{myheadings} 48 49 \markright{Popović\hfill Applied Analysis\hfill} 50 51 52 53 \title{Applied Analysis Problems} 54 \author{Milutin Popović} 55 56 \begin{document} 57 \maketitle 58 \tableofcontents 59 60 \section{Sheet 2} 61 \subsection{Problem 4} 62 We consider a quadratic equation with two ways to perturb it by $\eps$: 63 \begin{align} 64 x^2 + 2\eps x -1 = 0, \label{eq: (1)}\\ 65 \nonumber \\ 66 \eps x^2 + 2x - 1 = 0.\label{eq: (2)} 67 \end{align} 68 Equation \ref{eq: (2)} is singular, because the reduced problem ($\eps 69 \rightarrow 0$) has only one solution at $x = \frac{1}{2}$. While the reduced 70 problem in \ref{eq: (1)} has two solutions for $x = \pm 1$, which is the case 71 for this non reduced equation. Let us thereby calculate the asymptotic 72 expansion of the regular case up to $O(\eps^2)$, we take the ansatz for the 73 asymptotic expansion 74 \begin{align}\label{eq: p4 ansatz} 75 x_\eps = x_0 + \eps x_1 + \eps^2 x_2 + O(\eps^3). 76 \end{align} 77 By substituting $x_\eps$ into \ref{eq: (1)} and factoring out the orders of 78 $\eps$ we get 79 \begin{align} 80 \eps^0 (x_0^2 - 1) + \eps^1(2x_0 + 2x_0x_1) + \eps^2(x_1^2+2x_2x_0 81 +2x_1) + O(\eps^3) = 0 82 \end{align} 83 By solving the equations in order of $\eps$, for the coefficients 84 $x_0$, $x_1$ and $x_2$ we get 85 \begin{align} 86 x_0 = \pm 1, \;\;\;\; x_1 = -1, \;\;\;\; x_2 = \pm \frac{1}{2}. 87 \end{align} 88 By substituting into the equation \ref{eq: p4 ansatz} we get 89 \begin{align} 90 x_\eps = \pm 1 - \eps \pm \frac{1}{2} \eps + O(\eps^3). 91 \end{align} 92 For $\eps = 0.001$ we get 93 \begin{align} 94 &x_\eps = -1.0010005 + O(\eps^3), &x_\eps = 0.9990005 + O(\eps^3),\\ 95 &x_\eps = -1.001 + O(\eps^2), &x_\eps = 0.999 + O(\eps^2). 96 \end{align} 97 \subsection{Problem 5} 98 Consider the following equations 99 \begin{align} 100 \label{eq: p5 1}\eps y' + y = x \;\;\;\;\;\; y(0) = 1\\ 101 \label{eq: p5 2}\eps y' + y = x \;\;\;\;\;\; y(0) = 0\\ 102 \nonumber\\ 103 \label{eq: p5 3}\eps y' + y = x \;\;\;\;\;\; y(0) = \eps\\ 104 \label{eq: p5 4}\eps^2 y' + y = x \;\;\;\;\;\; y(0) = \eps\\ 105 \nonumber\\ 106 \label{eq: p5 5}y' + \eps y = x \;\;\;\;\;\; y(0) = 1\\ 107 \label{eq: p5 6}y' + y = \eps x \;\;\;\;\;\; y(0) = 1 108 \end{align} 109 We will go through the equations and elaborate on if the perturbation is 110 regular or singular, if regular we will compute the asymptotic expansion up 111 to second order. 112 Let us begin with equation \ref{eq: p5 1}. By the first look, the reduced 113 problem does not agree with the boundary condition 114 \begin{align} 115 y_0 = x \;\;\;\;\; y_0(0) = 1, 116 \end{align} 117 is a contradiction in $y_0(0) = 0 \neq 1$, thereby equation \ref{eq: p5 1} is 118 \textbf{singularly perturbed}. 119 120 The reduced problem of equation \ref{eq: p5 2} on the other hand agrees with 121 the boundary condition, since 122 \begin{align} 123 y_0 = x \;\;\;\;\; y_0(0) = 0. 124 \end{align} 125 But by doing the ansatz for the asymptotic expansion 126 \begin{align} 127 y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3), 128 \end{align} 129 plugging in into \ref{eq: p5 2} and separating coefficients in terms of 130 $\eps$, we get 131 \begin{align} 132 \eps^0 (y_0 -x) + \eps^1(y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) = 0 133 \end{align} 134 The solutions to these equations are 135 \begin{align} 136 y_0 = x, \;\;\;\; y_1 = 1, \;\;\;\; y_2 = 0, 137 \end{align} 138 which is a contradiction to the boundary condition of $y_1(0) = 1 \neq 0$. 139 Thereby we can conclude that equation \ref{eq: p5 2} is \textbf{singularly 140 perturbed}. 141 142 Next up is equation \ref{eq: p5 3}, where by the asymptotic expansion the 143 first order coefficient of $\eps$, $y_2$, has the boundary condition $y_2(0) 144 = 0$. But by applying the ansatz of the asymptotic expansion and plugging 145 into the equation we get 146 \begin{align} 147 \eps^2(y_0 - x) + \eps^1 (y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) = 148 0. 149 \end{align} 150 Solving these equations we get 151 \begin{align} 152 y_0 = 0, \;\;\;\; y_1 = 1 \;\;\;\; y_2 = 0 , 153 \end{align} 154 which is a contradiction $y_1(0) = 1 \neq 0 $, thus the equation \ref{eq: p5 155 3} is \textbf{singularly perturbed}. 156 157 The next equation \ref{eq: p5 4} is also singularly perturbed, we 158 can see this by plugging the asymptotic expansion into the equation 159 \begin{align} 160 \eps^0 ( y_0 - x) + \eps^1(y_1) + \eps^2(y_0' + y_2) = O(\eps^3), 161 \end{align} 162 solving for the coefficients we get 163 \begin{align} 164 y_0 = x, \;\;\;\; y_1 = 0, \;\;\;\;\; y_2 = -1, 165 \end{align} 166 which is contradiction by the boundary condition $y_2(0) = -1 \neq 0$, 167 thereby \ref{eq: p5 4} is \textbf{singularly perturbed}. 168 169 Equation \ref{eq: p5 5} on the first sight does not indicate for any 170 contradictions, we may plug the ansatz of the asymptotic expansion into the 171 equation and see what happens 172 \begin{align} 173 \eps^0(y_0 -x) + \eps^1(y_1' + y_0) + \eps^2(y_2' +y_1) + O(\eps^2) = 0, 174 \end{align} 175 with the initial conditions $y_0(0) = 1$, $y_1(0) = y_2(0) = 0$. 176 \begin{align} 177 y_0 = \frac{x^2}{2} + 1, \;\;\;\; 178 y_1 = -\frac{x^3}{6} + x, \;\;\;\; 179 y_2 = \frac{x^4}{24} + \frac{x^2}{2}. 180 \end{align} 181 Finally we get 182 \begin{align} 183 y_\eps(x) = (\frac{x^2}{2}+1) + \eps(-\frac{x^3}{6} -x) 184 +\eps^2(\frac{x^4}{24} + \frac{x^2}{2}) + O(\eps^3). 185 \end{align} 186 Thereby we can conclude that \ref{eq: p5 5} is \textbf{regularly perturbed}. 187 188 The last equation \ref{eq: p5 6} is also regular, let us do the asymptotic 189 expansion of the equation and order the equation in orders of $\eps$. 190 \begin{align} 191 \eps^0(y_0' + y_0) + \eps^1(y_1' + y_1 -x) + \eps^2(y_2' + y_2)+ 192 O(\eps^3) = 0 . 193 \end{align} 194 by solving these differential equations with the boundary conditions $y_0(0) 195 = 1$, $y_1(0) = y_2(0) = 0$ we get 196 \begin{align} 197 y_0 = e^{-x} \;\;\;\; y_1 = (x-1) + e^{-x} \;\;\;\; y_2 = 0. 198 \end{align} 199 The equation we get 200 \begin{align} 201 y_\eps(x) = e^{-x} + \eps(x-1 +e^{-x}) + O(\eps^3). 202 \end{align} 203 Thereby we can conclude that the last equation \ref{eq: p5 6} is 204 \textbf{regularly perturbed}. 205 \subsection{Problem 6} 206 In this section we will calculate the asymptotic expansion of a regularly 207 perturbed equation in two ways, by doing the regular expansion ansatz and by 208 substituting and expanding in terms of $\eps$. The ordinary differential 209 equation we are dealing with is 210 \begin{align} 211 y' = -y + \eps y^2 \;\;\;\;\; y(0) = 1, 212 \end{align} 213 where $t > 0$ and $0 < \eps \ll 1$. The standard expansion ansatz is 214 \begin{align} 215 y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3). 216 \end{align} 217 The ODE then expands to 218 \begin{align} 219 \eps^0(y_0' + y_0) + \eps(y_1' + y_1 - y_0^2) + \eps^2(y_2' + y_2 - 220 2y_0y_1) + O(\eps^3) = 0. 221 \end{align} 222 Equations in order of $\eps$ and $\eps^2$ are non-homogeneous ODE's. The 223 solution to these three coefficients with the boundary conditions $y_0(0) = 224 1$, $y_1(0) = 225 y_2(0) = 0$ we get 226 \begin{align} 227 y_0 = e^{-x}, \;\;\;\; y_1 = -e^{-2x} + e^{-x}, \;\;\;\; y_2 = e^{-3x} - 228 2e^{-2x} + e^{-x}. 229 \end{align} 230 The expansion of $y$ is then 231 \begin{align} 232 y_\eps (x) = e^{-x} + \eps(-e^{-2x} + e^{-x}) + \eps^2(e^{-3x} - 2e^{-2x} 233 + e^{-x}) + O(\eps^3). \end{align} 234 235 The second ansatz, considers the substitution $z = \frac{1}{y}$, by 236 calculating the first derivative and substituting the original problem we 237 get 238 \begin{align} 239 z' &= \frac{-y'}{y^2} = \frac{y-\eps y^2}{y^2} = \frac{1}{y} - \eps = z - 240 \eps. \\ 241 z(0) &= \frac{1}{y(0)} = 1. 242 \end{align} 243 The solution is 244 \begin{align} 245 z(x) = \eps + (1-\eps) e^x. 246 \end{align} 247 By substituting this into $y = \frac{1}{z}$ and expanding we get 248 \begin{align} 249 y(x) &= \frac{1}{\eps+(1-\eps)e^x} = e^{-x} \frac{1}{1 - (1- e^{-x})\eps} 250 \\ 251 &= e^{-x} \sum_{n\geq 0} \eps^n(1-e^{-x})^n. 252 \end{align} 253 which is the geometric series. 254 \subsection{Problem 7} 255 The last problem consists of a perturbation of a partial differential 256 equation (heat equation). 257 \begin{align} 258 &\partial_t u(x, t) + \partial_x^2 u(x,t) - \eps u(x, t)^2 = 0 259 &x\in (0, 1),\; t>0,\\ 260 &u(x, 0) = \tilde{u}_0(x) &x\in(0, 1), \\ 261 &u(0, t) = u(1, t) = 0 & t>0. 262 \end{align} 263 The problem is regular because the reduced solution is the regular heat 264 equation in the one special dimension on $x\in (0, 1)$, we know this is 265 solvable. By doing the expansion ansatz we can derive the first equations 266 for the first three terms, the ansatz is always the same 267 \begin{align} 268 u_\eps = u_0 + \eps u_1 + \eps^2 u_2 + O(\eps^3). 269 \end{align} 270 Plugging this into the perturbed problem problem and factoring out the terms 271 in the order of $\eps$ we get 272 \begin{align} 273 &\eps^0 (\partial_t u_0 + \partial_x^2 u_0) + \\ 274 &\eps^1 (\partial_t u_1 + \partial_x^2 u_1 - u_0^2) +\\ 275 &\eps^2 (\partial_t u_2 + \partial_x^2 u_2 - 2u_1u_0) + O(\eps^3) = 0. 276 \end{align} 277 278 We can solve the reduced problem with the initial condition $\tilde{u}_0 = 279 \sin(\pi x)$ by separation of variables. Setting $u(x, t) = \psi(x) \phi(t)$ 280 and substituting into the equation we get two ordinary differential equation 281 \begin{align} 282 \underbrace{\frac{\psi_{xx}}{\psi}}_{=k} 283 +\underbrace{\frac{\phi_t}{\phi}}_{=-k} = 0, 284 \end{align} 285 for some $k$. Solving these two by the exponential ansatz. 286 \begin{align} 287 \psi(x) &= A_1 e^{\sqrt{k} x} +A_2 e^{-\sqrt{k} x},\\ 288 \phi(t) &= A_3 e^{-kt}. 289 \end{align} 290 With the initial condition we get the conditions that 291 \begin{align} 292 A_1A_3 &= -A_2 A_3,\\ 293 k &= \pi^2 294 \end{align} 295 we choose $A_1 = A_3 = 1$ , $A_2 = -1$. We get the following solution to the 296 PDE 297 \begin{align} 298 u(x, t) = \psi(x)\phi(t) = \sin(\pi x) e^{-\pi^2 t}. 299 \end{align} 300 301 %\printbibliography 302 \end{document}