prb3.tex (6306B)
1 \include{preamble.tex} 2 3 \begin{document} 4 \maketitle 5 \tableofcontents 6 7 \section{Sheet 3} 8 \subsection{Problem 8} 9 Let us look at functions $f: \mathcal{D} \mapsto \mathbb{R}$ that show 10 boundary layer behavior at the following manifolds. 11 12 The \textbf{first} for $\mathcal{D} = \mathbb{R}^2$ and $S = \{0\}$ we have a 13 function e.g. 14 \begin{align} 15 f_{\eps}(x, y) = e^{-\frac{x}{\eps}} + y, 16 \end{align} 17 with the reduced equation 18 \begin{align} 19 \lim_{\eps \rightarrow 0} f_{\eps}(x, y) = 20 \begin{cases} 21 y \;\;\;\;\;\;\;\;\;\; x > 0\\ 22 1+y \;\;\;\; x = 0\\ 23 \end{cases} 24 \end{align} 25 26 The \textbf{second} example is $\mathcal{D} = \mathbb{R}^n$ and $S = \{|x| = 1\}$. 27 \begin{align} 28 f_\eps(x_1,\dots,x_n) = \tanh\left(\frac{|x| - 1}{\eps} \right), 29 \end{align} 30 with the reduced equation 31 \begin{align} 32 \lim_{\eps \rightarrow 0} f_{\eps}(x_1,\dots, x_n) = 33 \begin{cases} 34 -1 \;\;\;\; |x| < 0\\ 35 1 \;\;\;\;\;\;\; |x| > 0\\ 36 \end{cases} 37 \end{align} 38 39 The \textbf{third} example is $\mathcal{D} = \mathbb{R}^3$ and $S = \{x_1 = 40 1\}$ 41 \begin{align} 42 f_\eps(x_1, x_2, x_3) = \tanh\left(\frac{x_1 - 1}{\eps}\right)+x_2x_3 43 \end{align} 44 with the reduced equation 45 \begin{align} 46 \lim_{\eps \rightarrow 0} f_{\eps}(x_1,x_2,x_3) = 47 \begin{cases} 48 -1 + x_2x_3 \;\;\;\; x_1 < 0\\ 49 1 + x_2x_3 \;\;\;\;\;\;\; x_1 > 0\\ 50 \end{cases} 51 \end{align} 52 \subsection{Problem 9} 53 Consider a linear BVP 54 \begin{align} 55 Lu := -\eps u'' + b(x)u' + c(x)u = f(x),\\ 56 u(0) = u(1) = 0, 57 \end{align} 58 for $0 < \eps \ll \eps_0$ and $b, c, f \in C([0,1])$ with the conditions 59 \begin{align} 60 c(x) \geq 0, \qquad b(x) \geq \beta > 0 \qquad x\in[0, 1] 61 \end{align} 62 We are to show that for all $x\in[0, 1)$ the reduced solution $u_0$ of the 63 above BVP satisfies 64 \begin{align} 65 \lim_{\eps \rightarrow 0} u_\eps(x) = u_0(x)))), 66 \end{align} 67 where the reduced solution $u_0$ is the solution to the following 68 differential equation 69 \begin{align} 70 b(x)u' + c(x)u = f(x), \quad u(0) = 0. 71 \end{align} 72 The hint was given: Set 73 \begin{align} 74 w_1(x) = e^{\beta x} \quad w_2(x) = e^{-\beta\frac{1-x}{\eps}}, 75 \end{align} 76 such that $Lw_1 \geq \gamma > 0$ for some suitable $\gamma$ and $Lw_2 \geq 77 0$. Then for 78 \begin{align} 79 v = \pm (u_\eps - u_0), \qquad w = A\eps w_1 + B\eps w_2, 80 \end{align} 81 for some suitable $A, B$. The following comparison principal is applicable: 82 IF 83 \begin{align} 84 &Lv(x) \leq Lw(x) \quad \forall x \in (0, 1) \label{eq:cond1}\\ 85 &v(0) \leq w(0) \label{eq:cond2}\\ 86 &v(1) \leq w(1) \label{eq:cond3}\\ 87 \end{align} 88 then 89 \begin{align} 90 \Longrightarrow v(x) \leq w(x) \quad \forall x\in(0, 1) 91 \end{align} 92 which holds for $u, v \in C^2((0, 1)) \cap C([0, 1])$. Thus a boundary layer 93 is possible only at $x=1$. On the other hand, for $b(x) \leq \beta < 0$ it 94 follows that the boundary layer is possible only at $x=0$. 95 96 We shall go through the chronological order of the conditions\ref{eq:cond1}, 97 \ref{eq:cond2}, \ref{eq:cond3} and check them. So for \ref{eq:cond1} 98 we have that 99 \begin{align} 100 Lw(x) &= A\eps Lw_1(x) + B Lw_2(x) \\ 101 &\geq A\eps Lw_1(x) = A\eps e^{\beta x} \left(-\eps \beta^2 - 102 b(x)\beta+c(x)\right)\\ 103 &\geq A\eps \beta e^{\beta x} \left(1-\eps\right)\\ 104 &\geq \eps A\beta^2 e^{\beta}(1-\eps) = \gamma > 0 105 \end{align} 106 And obviously 107 \begin{align} 108 Lv(x) \leq 0 , 109 \end{align} 110 by that we have that 111 \begin{align} 112 Lv(x) \leq \gamma \leq Lw(x). 113 \end{align} 114 For the condition \ref{eq:cond2} we have 115 \begin{align} 116 w(0) &= A\eps w_1(0) Bw_2(0) = Be^{-\frac{\beta}{\eps}},\\ 117 v(0) &= \pm\left(u_\eps(0) - u_0(0)\right) = 0. 118 \end{align} 119 By the simple choice $B \geq 0$ we satisfy the condition 120 \begin{align} 121 v(0) \leq w(0). 122 \end{align} 123 Now for the last condition \ref{eq:cond3} we have 124 \begin{align} 125 w(1) &= A\eps e^\beta + B \geq A\eps e^\beta,\\ 126 v(1) &= \mp u_0(1) = 0. 127 \end{align} 128 And choose $A = \frac{\pm u(1)}{\eps} e^{-\beta}$, which satisfies the last 129 condition 130 \begin{align} 131 v(1) \leq w(1). 132 \end{align} 133 Thereby we have 134 \begin{align} 135 &v(x) &\leq w(x)\\ 136 &\Rightarrow \lim_{\eps \rightarrow 0} v(x) &\leq \lim_{\eps \rightarrow 137 0} w(x) = 0\\ 138 &\Rightarrow \lim_{\eps \rightarrow 0} v(x) = 0 139 \end{align} 140 uniformly on $(0, 1)$. 141 \subsection{Problem 10} 142 Consider the following BVP 143 \begin{align} 144 -\eps u'' + (1 + x)u' + u = 2, \qquad u(0) = u(1) - 0, 145 \end{align} 146 for $0 < \eps \ll 1$. \textbf{Where can this problem have a boundary layer?} 147 To answer this question we need to look at the reduced problem 148 \begin{align} 149 -(1+x)u' + u = 2. 150 \end{align} 151 The solution to the equation is 152 \begin{align} 153 \bar{u}(x) = 2 + A(x+1). 154 \end{align} 155 According to the boundary conditions it is unclear what the value of the 156 constant is, according to $\bar{u}(0)=0$ we get $A = -2$ or according to 157 $\bar{u}(1)=0$ we get $A = -1$. Ultimately this means that there exists a 158 boundary layer near $x=1$ or $x=0$. We choose $x=0$ and according to this the 159 local variable $\xi = x\eps^{-\alpha}$ ($x = \xi \eps^{-\alpha}$). The 160 derivatives of $u$ are calculated using the chain rule 161 \begin{align} 162 \frac{du}{dx}&= \frac{du}{d\xi}\frac{d\xi}{dx} = \eps^{-\alpha} \dot{u}\\ 163 \frac{d^2u}{dx^2}&= \eps^{-\alpha} \frac{d^2u}{d\xi^2}\frac{d\xi}{dx} = 164 \eps^{-2\alpha} \ddot{u}. 165 \end{align} 166 The BVP transforms as follows 167 \begin{align} 168 -\eps^{1-\alpha}\ddot{u} - \dot{u} + \eps(u - \xi\dot{u} - 2) = 169 \begin{cases} 170 -\ddot{u} - \dot{u} = 0 \;\;\;\;\; \alpha=1\\ 171 -\dot{u} = 0 \;\;\;\;\;\;\;\; 0<\alpha<1 172 \end{cases} 173 \end{align} 174 Choosing $\alpha = 1$ for a reasonable solution 175 \begin{align} 176 \hat{u}(\xi) = Be^{-\xi}, 177 \end{align} 178 which converges in the local limit (!). Thereby we have a asymptotic 179 representation up to the degree of $\eps$ 180 181 \begin{align} 182 u_\eps(x) &= \bar{u}(x) + \hat{u}(\psi) + O(\eps)\\ 183 &= 2 + A(1+x) + B e^{-\frac{x}{\eps}} + O(\eps) 184 \end{align} 185 And by the boundary conditions 186 \begin{align} 187 u_\eps(0) = 2+A+B=0, \qquad u_\eps(1) = 2+2A+B=0, 188 \end{align} 189 we get that the constants are 190 \begin{align} 191 A = -4, \qquad B = 2. 192 \end{align} 193 The asymptotic representation is thereby 194 \begin{align} 195 u_\eps(x) = 2 - 4(1+x) + 2 e^{-\frac{x}{\eps}} + O(\eps) 196 \end{align} 197 %\printbibliography 198 \end{document}