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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 3} 61 \subsection{Problem 8} 62 Let us look at functions $f: \mathcal{D} \mapsto \mathbb{R}$ that show 63 boundary layer behavior at the following manifolds. 64 65 The \textbf{first} for $\mathcal{D} = \mathbb{R}^2$ and $S = \{0\}$ we have a 66 function e.g. 67 \begin{align} 68 f_{\eps}(x, y) = e^{-\frac{x}{\eps}} + y, 69 \end{align} 70 with the reduced equation 71 \begin{align} 72 \lim_{\eps \rightarrow 0} f_{\eps}(x, y) = 73 \begin{cases} 74 y \;\;\;\;\;\;\;\;\;\; x > 0\\ 75 1+y \;\;\;\; x = 0\\ 76 \end{cases} 77 \end{align} 78 79 The \textbf{second} example is $\mathcal{D} = \mathbb{R}^n$ and $S = \{|x| = 1\}$. 80 \begin{align} 81 f_\eps(x_1,\dots,x_n) = \tanh\left(\frac{|x| - 1}{\eps} \right), 82 \end{align} 83 with the reduced equation 84 \begin{align} 85 \lim_{\eps \rightarrow 0} f_{\eps}(x_1,\dots, x_n) = 86 \begin{cases} 87 -1 \;\;\;\; |x| < 0\\ 88 1 \;\;\;\;\;\;\; |x| > 0\\ 89 \end{cases} 90 \end{align} 91 92 The \textbf{third} example is $\mathcal{D} = \mathbb{R}^3$ and $S = \{x_1 = 93 1\}$ 94 \begin{align} 95 f_\eps(x_1, x_2, x_3) = \tanh\left(\frac{x_1 - 1}{\eps}\right)+x_2x_3 96 \end{align} 97 with the reduced equation 98 \begin{align} 99 \lim_{\eps \rightarrow 0} f_{\eps}(x_1,x_2,x_3) = 100 \begin{cases} 101 -1 + x_2x_3 \;\;\;\; x_1 < 0\\ 102 1 + x_2x_3 \;\;\;\;\;\;\; x_1 > 0\\ 103 \end{cases} 104 \end{align} 105 \subsection{Problem 9} 106 Consider a linear BVP 107 \begin{align} 108 Lu := -\eps u'' + b(x)u' + c(x)u = f(x),\\ 109 u(0) = u(1) = 0, 110 \end{align} 111 for $0 < \eps \ll \eps_0$ and $b, c, f \in C([0,1])$ with the conditions 112 \begin{align} 113 c(x) \geq 0, \qquad b(x) \geq \beta > 0 \qquad x\in[0, 1] 114 \end{align} 115 We are to show that for all $x\in[0, 1)$ the reduced solution $u_0$ of the 116 above BVP satisfies 117 \begin{align} 118 \lim_{\eps \rightarrow 0} u_\eps(x) = u_0(x)))), 119 \end{align} 120 where the reduced solution $u_0$ is the solution to the following 121 differential equation 122 \begin{align} 123 b(x)u' + c(x)u = f(x), \quad u(0) = 0. 124 \end{align} 125 The hint was given: Set 126 \begin{align} 127 w_1(x) = e^{\beta x} \quad w_2(x) = e^{-\beta\frac{1-x}{\eps}}, 128 \end{align} 129 such that $Lw_1 \geq \gamma > 0$ for some suitable $\gamma$ and $Lw_2 \geq 130 0$. Then for 131 \begin{align} 132 v = \pm (u_\eps - u_0), \qquad w = A\eps w_1 + B\eps w_2, 133 \end{align} 134 for some suitable $A, B$. The following comparison principal is applicable: 135 IF 136 \begin{align} 137 &Lv(x) \leq Lw(x) \quad \forall x \in (0, 1) \label{eq:cond1}\\ 138 &v(0) \leq w(0) \label{eq:cond2}\\ 139 &v(1) \leq w(1) \label{eq:cond3}\\ 140 \end{align} 141 then 142 \begin{align} 143 \Longrightarrow v(x) \leq w(x) \quad \forall x\in(0, 1) 144 \end{align} 145 which holds for $u, v \in C^2((0, 1)) \cap C([0, 1])$. Thus a boundary layer 146 is possible only at $x=1$. On the other hand, for $b(x) \leq \beta < 0$ it 147 follows that the boundary layer is possible only at $x=0$. 148 149 We shall go through the chronological order of the conditions\ref{eq:cond1}, 150 \ref{eq:cond2}, \ref{eq:cond3} and check them. So for \ref{eq:cond1} 151 we have that 152 \begin{align} 153 Lw(x) &= A\eps Lw_1(x) + B Lw_2(x) \\ 154 &\geq A\eps Lw_1(x) = A\eps e^{\beta x} \left(-\eps \beta^2 - 155 b(x)\beta+c(x)\right)\\ 156 &\geq A\eps \beta e^{\beta x} \left(1-\eps\right)\\ 157 &\geq \eps A\beta^2 e^{\beta}(1-\eps) = \gamma > 0 158 \end{align} 159 And obviously 160 \begin{align} 161 Lv(x) \leq 0 , 162 \end{align} 163 by that we have that 164 \begin{align} 165 Lv(x) \leq \gamma \leq Lw(x). 166 \end{align} 167 For the condition \ref{eq:cond2} we have 168 \begin{align} 169 w(0) &= A\eps w_1(0) Bw_2(0) = Be^{-\frac{\beta}{\eps}},\\ 170 v(0) &= \pm\left(u_\eps(0) - u_0(0)\right) = 0. 171 \end{align} 172 By the simple choice $B \geq 0$ we satisfy the condition 173 \begin{align} 174 v(0) \leq w(0). 175 \end{align} 176 Now for the last condition \ref{eq:cond3} we have 177 \begin{align} 178 w(1) &= A\eps e^\beta + B \geq A\eps e^\beta,\\ 179 v(1) &= \mp u_0(1) = 0. 180 \end{align} 181 And choose $A = \frac{\pm u(1)}{\eps} e^{-\beta}$, which satisfies the last 182 condition 183 \begin{align} 184 v(1) \leq w(1). 185 \end{align} 186 Thereby we have 187 \begin{align} 188 &v(x) &\leq w(x)\\ 189 &\Rightarrow \lim_{\eps \rightarrow 0} v(x) &\leq \lim_{\eps \rightarrow 190 0} w(x) = 0\\ 191 &\Rightarrow \lim_{\eps \rightarrow 0} v(x) = 0 192 \end{align} 193 uniformly on $(0, 1)$. 194 \subsection{Problem 10} 195 Consider the following BVP 196 \begin{align} 197 -\eps u'' + (1 + x)u' + u = 2, \qquad u(0) = u(1) - 0, 198 \end{align} 199 for $0 < \eps \ll 1$. \textbf{Where can this problem have a boundary layer?} 200 To answer this question we need to look at the reduced problem 201 \begin{align} 202 -(1+x)u' + u = 2. 203 \end{align} 204 The solution to the equation is 205 \begin{align} 206 \bar{u}(x) = 2 + A(x+1). 207 \end{align} 208 According to the boundary conditions it is unclear what the value of the 209 constant is, according to $\bar{u}(0)=0$ we get $A = -2$ or according to 210 $\bar{u}(1)=0$ we get $A = -1$. Ultimately this means that there exists a 211 boundary layer near $x=1$ or $x=0$. We choose $x=0$ and according to this the 212 local variable $\xi = x\eps^{-\alpha}$ ($x = \xi \eps^{-\alpha}$). The 213 derivatives of $u$ are calculated using the chain rule 214 \begin{align} 215 \frac{du}{dx}&= \frac{du}{d\xi}\frac{d\xi}{dx} = \eps^{-\alpha} \dot{u}\\ 216 \frac{d^2u}{dx^2}&= \eps^{-\alpha} \frac{d^2u}{d\xi^2}\frac{d\xi}{dx} = 217 \eps^{-2\alpha} \ddot{u}. 218 \end{align} 219 The BVP transforms as follows 220 \begin{align} 221 -\eps^{1-\alpha}\ddot{u} - \dot{u} + \eps(u - \xi\dot{u} - 2) = 222 \begin{cases} 223 -\ddot{u} - \dot{u} = 0 \;\;\;\;\; \alpha=1\\ 224 -\dot{u} = 0 \;\;\;\;\;\;\;\; 0<\alpha<1 225 \end{cases} 226 \end{align} 227 Choosing $\alpha = 1$ for a reasonable solution 228 \begin{align} 229 \hat{u}(\xi) = Be^{-\xi}, 230 \end{align} 231 which converges in the local limit (!). Thereby we have a asymptotic 232 representation up to the degree of $\eps$ 233 \begin{align} 234 u_\eps(x) &= \bar{u}(x) + \hat{u}(\psi) + O(\eps)\\ 235 &= 2 + A(1+x) + B e^{-\frac{x}{\eps}} + O(\eps) 236 \end{align} 237 And by the boundary conditions 238 \begin{align} 239 u_\eps(0) = 2+A+B=0, \qquad u_\eps(1) = 2+2A+B=0, 240 \end{align} 241 we get that the constants are 242 \begin{align} 243 A = -4, \qquad B = 2. 244 \end{align} 245 The asymptotic representation is thereby 246 \begin{align} 247 u_\eps(x) = 2 - 4(1+x) + 2 e^{-\frac{x}{\eps}} + O(\eps) 248 \end{align} 249 %\printbibliography 250 \end{document}