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\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:

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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:

$\left\{\begin{array} {l} y \sim
\df{1}{(1+\frac{1}{\sqrt{6}}\frac{x}{\varepsilon^\frac{1}{2}})},\
x=o(\varepsilon^\frac{1}{2})\\ y \sim 0,\
o(\varepsilon^\frac{1}{2})<x<O(\varepsilon^\frac{1}{2})\\ y \sim
\df{1}{(1+\frac{1}{\sqrt{6}}\frac{(1-x)}{\varepsilon^\frac{1}{2}})^2},\
x-1=O(\varepsilon^\frac{1}{2}) \end{array} \right.$

Solution

PICTURE \vspace{2in}


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