\documentclass[a4paper,12pt]{article} \newcommand{\ds}{\displaystyle} \newcommand{\pl}{\partial} \newcommand{\un}{\underline} \newcommand{\undb}{\underbrace} \newcommand{\df}{\ds\frac} \parindent=0pt \begin{document} {\bf Question} (More difficult) Consider the problem $$\varepsilon y''=y^2=0,\ y(0)=1,\ y(1)=1,\ \varepsilon \to 0^+.$$ Try to solve this when $0 \leq x \leq 1$ by using a regular perturbation scheme of the type $$y(x;\varepsilon)=y_0(x)+\varepsilon y_1(x)+O(\varepsilon^2)$$ Show that it is not possible for thus outer type of expansion to satisfy the boundary conditions at either end. Hence deduce that there must be a boundary layer near both $x=0$ and $x=1$. You are given that both boundary layers have a width of order $\varepsilon^{\frac{1}{2}}$. Determine one term of the inner solution near to the origin in the standard way. Repeat this for the second inner solution near to $x=1$. match the two inner expansions to the common outer expansion in the relevant regions. {\bf Hint:} The solution $Y''-Y^2=0$ can be obtained by first reducing the order with the substitution $u=Y'$ (hence $Y''=u'=\df{du}{dY}\df{dY}{dx}=u\df{du}{dY}$), solving for $u$, and then resubstituting for $Y$ in the resulting equation. Also use the fact that if $\lim_{x\to \infty} Y=0$ then $\lim_{x\to\infty} Y'=0$. \vspace{.25in} {\bf Answer} This is difficult: it has 2 boundary layers and $O(\varepsilon^{\frac{1}{2}})$ width. Given $$\varepsilon y''-y^2=0;\ y(0)=1,\ y(1)=1;\ \varepsilon \to 0^+$$ we try the usual ansatz anyway: $$y(x;\varepsilon)=y_0(x)+\varepsilon y_1(x)+O(\varepsilon^2)$$ $\Rightarrow \varepsilon y_0''-y_0^2-2\varepsilon y_0y_1=O(\varepsilon^2);\ y_0(1)=1,\ y_r(1)=0\ (r>0)$ $\un{O(\varepsilon^0)} \begin{array}{rcl} -y_0^2 & = & 0\\ \Rightarrow y-0 & = & 0\ \rm{but}\ y_0(1)=1\ \rm{from\ boundary\ data.}\end{array}$ Therefore inconsistency. Hence usual ansatz \un{can't work} at $x=0$. Equally, if we think of using ansatz at $x=0$ we run into the same problem: $y_0=0$ by $y_0(0)=1$. Hence we must have a boundary layer at both ends: \setlength{\unitlength}{.5in} \begin{picture}(4,4) \put(0,0){\line(0,1){1}} \put(1,.1){\makebox(0,0){-}} \put(1,0.3){\makebox(0,0){-}} \put(1,0.5){\makebox(0,0){-}} \put(1,0.7){\makebox(0,0){-}} \put(1,.9){\makebox(0,0){-}} \put(4,.1){\makebox(0,0){-}} \put(4,0.3){\makebox(0,0){-}} \put(4,0.5){\makebox(0,0){-}} \put(4,0.7){\makebox(0,0){-}} \put(4,.9){\makebox(0,0){-}} \put(5,0){\line(0,1){1}} \put(0.5,1.1){\vector(-1,0){.5}} \put(0.5,1.1){\vector(1,0){.5}} \put(4.5,1.1){\vector(-1,0){.5}} \put(4.5,1.1){\vector(1,0){.5}} \put(2.5,.5){\vector(-1,0){1.8}} \put(2.5,.5){\vector(1,0){1.8}} \put(0,0){\makebox(0,0)[t]{$x=0$}} \put(5,0){\makebox(0,0)[t]{$x=1$}} \put(2.5,0.4){\makebox(0,0)[t]{$y_0=0$}} \put(0.5,1.2){\makebox(0,0)[b]{Boundary layer}} \put(4.5,1.2){\makebox(0,0)[b]{Boundary layer}} \end{picture} You are given that layers are $O(\varepsilon^{\frac{1}{2}})$ in width. \un{Near $x=0$} Try inner variable $X=\df{x}{\varepsilon^\frac{1}{2}}$ $$\Rightarrow \pl_x=\df{1}{\varepsilon^\frac{1}{2}}\pl_X$$ and use ansatz $y(\varepsilon^\frac{1}{2} X;\varepsilon)=Y(X;\varepsilon)$ $$Y(X;\varepsilon)=Y_0(X)+\varepsilon^\frac{1}{2}Y_1(X)+O(\varepsilon)$$ Substitute into equation: $$\df{\varepsilon}{\varepsilon}Y_0''-\df{\varepsilon}{\varepsilon} \varepsilon^{\frac{1}{2}}Y_1''(X)-Y_0^2-2\varepsilon^\frac{1}{2}Y_0Y_1=0$$ $$Y_0''-Y_0^2=O(\varepsilon^\frac{1}{2})$$ Boundary date: only $x=0$ is relevant and we get $y(0)=1 \Rightarrow Y(0)=1 \Rightarrow Y_0(0)=1,\ Y_{r>0}(0)=0$ Therefore $\un{O(\varepsilon^0)}$ $Y_0''-Y_0^2=0;\ Y_0(0)=1$ Again 2nd order equation. 1 boundary condition $\Rightarrow$ matching. Solve this by setting $\left.\begin{array}{rcl} u & = & Y_0'\\ u' & = & Y_0''\end{array}\right\}$ by hint of question $u=\df{dY_0}{dX} \Rightarrow \df{d^2Y_0}{dX^2}=\df{du}{dX}=\df{du}{dY_0} \times \df{dY_0}{dX}=\df{du}{dY_0}Y_0'=u\df{du}{dY_0}$ Therefore $\begin{array}{rcl} u\df{du}{dY_0}-Y_0^2 & = & 0\\ \Rightarrow \ds\int u\, du & = & \ds\int Y_0^2\, dY_0\\ \df{u^2}{2} & = & \df{Y_0^3}{3}+const\\ \rm{or}\ \df{1}{2}\left(\df{dY_0}{dX}\right)^2 & = & \df{Y_0^3}{3}+c\\ \rm{or}\ \df{dY_0}{dX} & = & \pm \sqrt{\df{2}{3}Y_0^3 +2c}\ \ (\star) \end{array}$ This is tricky to solve unless we start imposing boundary conditions. We know $Y_0(0)=1$ but this doesn't tell us about $\df{dY_0}{dX}$. Clearly though if things are to match up, we must have at leading order. $$Y_0 \rightarrow y_0=0\ \rm{as}\ \varepsilon\to 0^+$$ i.e., as $X \to +\infty$ in the outer (middle) region Therefore $lim_{X\to \infty} Y_0=0$. Thus if $\lim_{X\to +\infty} Y_0=0$ the curve $Y_0(X)$ must \un{flatten}. So $Y_0' \to 0$ as $X \to + \infty$ Therefore $0=\pm \sqrt{0+2c}$ as $X\to +\infty$ from $(\star)$. Hence $\df{dY_0}{dX}=\pm \sqrt{\df{2}{3}}Y_0^\frac{3}{2}$ which is variables separable \begin{eqnarray*} \ds\int Y_0^{-\frac{3}{2}}\, dY_0 & = & \pm \ds\int \sqrt{\df{2}{3}}\, dX\\ -2 Y_0^{-\frac{1}{2}}+D & = & \pm \sqrt{\df{2}{3}}X \end{eqnarray*} Now we can use the actual boundary condition: $Y_0(0)=1$ $$\begin{array} {l} 0=D-2 \Rightarrow D=2\\ \rm{Therefore}\ 2-\df{2}{\sqrt{Y_0}} = \pm \sqrt{\df{2}{3}}X\\ \Rightarrow Y_0 = \df{1}{(1 \mp \sqrt{\frac{1}{6}} X)^2}\end{array}$$ Now, if we don't want a singularity for finite $X$, the - sign must be discarded. Hence $$\un{Y_0=\df{1}{(1+\sqrt{\frac{1}{6}}X)^2}}$$ \un{Near $x=1$}: To determine the inner expansion. Near $x=1$ we use layer near $x=1$ $z=\df{\overbrace{1-x}}{\undb{\varepsilon^\frac{1}{2}}}$, say since $O(\varepsilon^\frac{1}{2})$ width is given to you ${}$ Therefore $\pl_x=\df{\pl z}{\pl x}\pl_z=-\df{1}{\varepsilon^\frac{1}{2}}\pl_z$ and $y(1-\varepsilon^\frac{1}{2}z; \varepsilon)=\bar{Y}(z;\varepsilon)$ and $\bar{Y}(z;\varepsilon)=\bar{Y_0}(z)+\varepsilon^\frac{1}{2}\bar{Y_1}(z)+O(\varepsilon)$ Equation becomes: $\df{\varepsilon}{(-\varepsilon^\frac{1}{2})}\df{\pl^2\bar{Y}}{\pl z^2}-\bar{Y^2}=0 \Rightarrow \df{\pl^2 \bar{Y}}{\pl z^2}-\bar{Y}^2=0$ so $\bar{Y_0}^2-\bar{Y_0}^2=O(\varepsilon^\frac{1}{2})$ This is the same equation as at $x=0$ so we expect a solution $$\df{d\bar{Y_0}}{dz}=\pm \sqrt{\df{2}{3}\bar{Y_0}^3+\bar{c}}$$ Again since $y(1)=1 \ \ (x=1 \Rightarrow z=0) \Rightarrow \bar{Y_0}(0)=1$ and since $\bar{Y_0}$ must match onto $y_0=0$ \un{as} $\varepsilon \to +\infty$ i.e., as $z\to +\infty$ Therefore we also have $\bar{Y_0}' \to 0$ as $z \to +\infty$ as before. Hence we get an identical solution: $$\un{\bar{Y_0}=\df{1}{(1+\sqrt{\frac{1}{6}}z)^2}}$$ (negative sign discarded to avoid singularity at finite $z$) Rewrite in original variables and summarise: $\left\{\begin{array} {l} y \sim \df{1}{(1+\frac{1}{\sqrt{6}}\frac{x}{\varepsilon^\frac{1}{2}})},\ x=o(\varepsilon^\frac{1}{2})\\ y \sim 0,\ o(\varepsilon^\frac{1}{2})