prb2.tex (9286B)
1 \include{preamble.tex} 2 3 \begin{document} 4 \maketitle 5 \tableofcontents 6 7 \section{Sheet 2} 8 \subsection{Problem 4} 9 We consider a quadratic equation with two ways to perturb it by $\eps$: 10 \begin{align} 11 x^2 + 2\eps x -1 = 0, \label{eq: (1)}\\ 12 \nonumber \\ 13 \eps x^2 + 2x - 1 = 0.\label{eq: (2)} 14 \end{align} 15 Equation \ref{eq: (2)} is singular, because the reduced problem ($\eps 16 \rightarrow 0$) has only one solution at $x = \frac{1}{2}$. While the reduced 17 problem in \ref{eq: (1)} has two solutions for $x = \pm 1$, which is the case 18 for this non reduced equation. Let us thereby calculate the asymptotic 19 expansion of the regular case up to $O(\eps^2)$, we take the ansatz for the 20 asymptotic expansion 21 \begin{align}\label{eq: p4 ansatz} 22 x_\eps = x_0 + \eps x_1 + \eps^2 x_2 + O(\eps^3). 23 \end{align} 24 By substituting $x_\eps$ into \ref{eq: (1)} and factoring out the orders of 25 $\eps$ we get 26 \begin{align} 27 \eps^0 (x_0^2 - 1) + \eps^1(2x_0 + 2x_0x_1) + \eps^2(x_1^2+2x_2x_0 28 +2x_1) + O(\eps^3) = 0 29 \end{align} 30 By solving the equations in order of $\eps$, for the coefficients 31 $x_0$, $x_1$ and $x_2$ we get 32 \begin{align} 33 x_0 = \pm 1, \;\;\;\; x_1 = -1, \;\;\;\; x_2 = \pm \frac{1}{2}. 34 \end{align} 35 By substituting into the equation \ref{eq: p4 ansatz} we get 36 \begin{align} 37 x_\eps = \pm 1 - \eps \pm \frac{1}{2} \eps + O(\eps^3). 38 \end{align} 39 For $\eps = 0.001$ we get 40 \begin{align} 41 &x_\eps = -1.0010005 + O(\eps^3), &x_\eps = 0.9990005 + O(\eps^3),\\ 42 &x_\eps = -1.001 + O(\eps^2), &x_\eps = 0.999 + O(\eps^2). 43 \end{align} 44 \subsection{Problem 5} 45 Consider the following equations 46 \begin{align} 47 \label{eq: p5 1}\eps y' + y = x \;\;\;\;\;\; y(0) = 1\\ 48 \label{eq: p5 2}\eps y' + y = x \;\;\;\;\;\; y(0) = 0\\ 49 \nonumber\\ 50 \label{eq: p5 3}\eps y' + y = x \;\;\;\;\;\; y(0) = \eps\\ 51 \label{eq: p5 4}\eps^2 y' + y = x \;\;\;\;\;\; y(0) = \eps\\ 52 \nonumber\\ 53 \label{eq: p5 5}y' + \eps y = x \;\;\;\;\;\; y(0) = 1\\ 54 \label{eq: p5 6}y' + y = \eps x \;\;\;\;\;\; y(0) = 1 55 \end{align} 56 We will go through the equations and elaborate on if the perturbation is 57 regular or singular, if regular we will compute the asymptotic expansion up 58 to second order. 59 Let us begin with equation \ref{eq: p5 1}. By the first look, the reduced 60 problem does not agree with the boundary condition 61 \begin{align} 62 y_0 = x \;\;\;\;\; y_0(0) = 1, 63 \end{align} 64 is a contradiction in $y_0(0) = 0 \neq 1$, thereby equation \ref{eq: p5 1} is 65 \textbf{singularly perturbed}. 66 67 The reduced problem of equation \ref{eq: p5 2} on the other hand agrees with 68 the boundary condition, since 69 \begin{align} 70 y_0 = x \;\;\;\;\; y_0(0) = 0. 71 \end{align} 72 But by doing the ansatz for the asymptotic expansion 73 \begin{align} 74 y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3), 75 \end{align} 76 plugging in into \ref{eq: p5 2} and separating coefficients in terms of 77 $\eps$, we get 78 \begin{align} 79 \eps^0 (y_0 -x) + \eps^1(y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) = 0 80 \end{align} 81 The solutions to these equations are 82 \begin{align} 83 y_0 = x, \;\;\;\; y_1 = 1, \;\;\;\; y_2 = 0, 84 \end{align} 85 which is a contradiction to the boundary condition of $y_1(0) = 1 \neq 0$. 86 Thereby we can conclude that equation \ref{eq: p5 2} is \textbf{singularly 87 perturbed}. 88 89 Next up is equation \ref{eq: p5 3}, where by the asymptotic expansion the 90 first order coefficient of $\eps$, $y_2$, has the boundary condition $y_2(0) 91 = 0$. But by applying the ansatz of the asymptotic expansion and plugging 92 into the equation we get 93 \begin{align} 94 \eps^2(y_0 - x) + \eps^1 (y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) = 95 0. 96 \end{align} 97 Solving these equations we get 98 \begin{align} 99 y_0 = 0, \;\;\;\; y_1 = 1 \;\;\;\; y_2 = 0 , 100 \end{align} 101 which is a contradiction $y_1(0) = 1 \neq 0 $, thus the equation \ref{eq: p5 102 3} is \textbf{singularly perturbed}. 103 104 The next equation \ref{eq: p5 4} is also singularly perturbed, we 105 can see this by plugging the asymptotic expansion into the equation 106 \begin{align} 107 \eps^0 ( y_0 - x) + \eps^1(y_1) + \eps^2(y_0' + y_2) = O(\eps^3), 108 \end{align} 109 solving for the coefficients we get 110 \begin{align} 111 y_0 = x, \;\;\;\; y_1 = 0, \;\;\;\;\; y_2 = -1, 112 \end{align} 113 which is contradiction by the boundary condition $y_2(0) = -1 \neq 0$, 114 thereby \ref{eq: p5 4} is \textbf{singularly perturbed}. 115 116 Equation \ref{eq: p5 5} on the first sight does not indicate for any 117 contradictions, we may plug the ansatz of the asymptotic expansion into the 118 equation and see what happens 119 \begin{align} 120 \eps^0(y_0 -x) + \eps^1(y_1' + y_0) + \eps^2(y_2' +y_1) + O(\eps^2) = 0, 121 \end{align} 122 with the initial conditions $y_0(0) = 1$, $y_1(0) = y_2(0) = 0$. 123 \begin{align} 124 y_0 = \frac{x^2}{2} + 1, \;\;\;\; 125 y_1 = -\frac{x^3}{6} + x, \;\;\;\; 126 y_2 = \frac{x^4}{24} + \frac{x^2}{2}. 127 \end{align} 128 Finally we get 129 \begin{align} 130 y_\eps(x) = (\frac{x^2}{2}+1) + \eps(-\frac{x^3}{6} -x) 131 +\eps^2(\frac{x^4}{24} + \frac{x^2}{2}) + O(\eps^3). 132 \end{align} 133 Thereby we can conclude that \ref{eq: p5 5} is \textbf{regularly perturbed}. 134 135 The last equation \ref{eq: p5 6} is also regular, let us do the asymptotic 136 expansion of the equation and order the equation in orders of $\eps$. 137 \begin{align} 138 \eps^0(y_0' + y_0) + \eps^1(y_1' + y_1 -x) + \eps^2(y_2' + y_2)+ 139 O(\eps^3) = 0 . 140 \end{align} 141 by solving these differential equations with the boundary conditions $y_0(0) 142 = 1$, $y_1(0) = y_2(0) = 0$ we get 143 \begin{align} 144 y_0 = e^{-x} \;\;\;\; y_1 = (x-1) + e^{-x} \;\;\;\; y_2 = 0. 145 \end{align} 146 The equation we get 147 \begin{align} 148 y_\eps(x) = e^{-x} + \eps(x-1 +e^{-x}) + O(\eps^3). 149 \end{align} 150 Thereby we can conclude that the last equation \ref{eq: p5 6} is 151 \textbf{regularly perturbed}. 152 \subsection{Problem 6} 153 In this section we will calculate the asymptotic expansion of a regularly 154 perturbed equation in two ways, by doing the regular expansion ansatz and by 155 substituting and expanding in terms of $\eps$. The ordinary differential 156 equation we are dealing with is 157 \begin{align} 158 y' = -y + \eps y^2 \;\;\;\;\; y(0) = 1, 159 \end{align} 160 where $t > 0$ and $0 < \eps \ll 1$. The standard expansion ansatz is 161 \begin{align} 162 y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3). 163 \end{align} 164 The ODE then expands to 165 \begin{align} 166 \eps^0(y_0' + y_0) + \eps(y_1' + y_1 - y_0^2) + \eps^2(y_2' + y_2 - 167 2y_0y_1) + O(\eps^3) = 0. 168 \end{align} 169 Equations in order of $\eps$ and $\eps^2$ are non-homogeneous ODE's. The 170 solution to these three coefficients with the boundary conditions $y_0(0) = 171 1$, $y_1(0) = 172 y_2(0) = 0$ we get 173 \begin{align} 174 y_0 = e^{-x}, \;\;\;\; y_1 = -e^{-2x} + e^{-x}, \;\;\;\; y_2 = e^{-3x} - 175 2e^{-2x} + e^{-x}. 176 \end{align} 177 The expansion of $y$ is then 178 \begin{align} 179 y_\eps (x) = e^{-x} + \eps(-e^{-2x} + e^{-x}) + \eps^2(e^{-3x} - 2e^{-2x} 180 + e^{-x}) + O(\eps^3). \end{align} 181 182 The second ansatz, considers the substitution $z = \frac{1}{y}$, by 183 calculating the first derivative and substituting the original problem we 184 get 185 \begin{align} 186 z' &= \frac{-y'}{y^2} = \frac{y-\eps y^2}{y^2} = \frac{1}{y} - \eps = z - 187 \eps. \\ 188 z(0) &= \frac{1}{y(0)} = 1. 189 \end{align} 190 The solution is 191 \begin{align} 192 z(x) = \eps + (1-\eps) e^x. 193 \end{align} 194 By substituting this into $y = \frac{1}{z}$ and expanding we get 195 \begin{align} 196 y(x) &= \frac{1}{\eps+(1-\eps)e^x} = e^{-x} \frac{1}{1 - (1- e^{-x})\eps} 197 \\ 198 &= e^{-x} \sum_{n\geq 0} \eps^n(1-e^{-x})^n. 199 \end{align} 200 which is the geometric series. 201 \subsection{Problem 7} 202 The last problem consists of a perturbation of a partial differential 203 equation (heat equation). 204 \begin{align} 205 &\partial_t u(x, t) + \partial_x^2 u(x,t) - \eps u(x, t)^2 = 0 206 &x\in (0, 1),\; t>0,\\ 207 &u(x, 0) = \tilde{u}_0(x) &x\in(0, 1), \\ 208 &u(0, t) = u(1, t) = 0 & t>0. 209 \end{align} 210 The problem is regular because the reduced solution is the regular heat 211 equation in the one special dimension on $x\in (0, 1)$, we know this is 212 solvable. By doing the expansion ansatz we can derive the first equations 213 for the first three terms, the ansatz is always the same 214 \begin{align} 215 u_\eps = u_0 + \eps u_1 + \eps^2 u_2 + O(\eps^3). 216 \end{align} 217 Plugging this into the perturbed problem problem and factoring out the terms 218 in the order of $\eps$ we get 219 \begin{align} 220 &\eps^0 (\partial_t u_0 + \partial_x^2 u_0) + \\ 221 &\eps^1 (\partial_t u_1 + \partial_x^2 u_1 - u_0^2) +\\ 222 &\eps^2 (\partial_t u_2 + \partial_x^2 u_2 - 2u_1u_0) + O(\eps^3) = 0. 223 \end{align} 224 225 We can solve the reduced problem with the initial condition $\tilde{u}_0 = 226 \sin(\pi x)$ by separation of variables. Setting $u(x, t) = \psi(x) \phi(t)$ 227 and substituting into the equation we get two ordinary differential equation 228 \begin{align} 229 \underbrace{\frac{\psi_{xx}}{\psi}}_{=k} 230 +\underbrace{\frac{\phi_t}{\phi}}_{=-k} = 0, 231 \end{align} 232 for some $k$. Solving these two by the exponential ansatz. 233 \begin{align} 234 \psi(x) &= A_1 e^{\sqrt{k} x} +A_2 e^{-\sqrt{k} x},\\ 235 \phi(t) &= A_3 e^{-kt}. 236 \end{align} 237 With the initial condition we get the conditions that 238 \begin{align} 239 A_1A_3 &= -A_2 A_3,\\ 240 k &= \pi^2 241 \end{align} 242 we choose $A_1 = A_3 = 1$ , $A_2 = -1$. We get the following solution to the 243 PDE 244 \begin{align} 245 u(x, t) = \psi(x)\phi(t) = \sin(\pi x) e^{-\pi^2 t}. 246 \end{align} 247 248 %\printbibliography 249 \end{document}