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{\bf Question}

Use the method of matching to find the first terms in the outer
and inner solutions of

$$\varepsilon y''+y'+y=0,\ y(0)=1,\ y(1)=1,$$

given that a boundary layer of width $O(\varepsilon)$ exists near
the origin. Hence write down the one-term composite expansion.
Compare thus with the exact solution.

\vspace{.25in}

{\bf Answer}

$\varepsilon y''+y'+y=0,\ y(0)=1,\ y(1)=1$

Boundary layer of $O(\varepsilon)$ exists at origin

\un{OUTER} $y=y_0(x)+\varepsilon y_1(x)+O(\varepsilon^2)$

Substitute into equations:

$$\varepsilon y_0+(\varepsilon^2 y_1)+y_0'+\varepsilon
y_1'+y_0+\varepsilon y_1'=O(\varepsilon^2)$$

Balance at

$\un{O(\varepsilon^0)}: +y_0'+y_0=0 \Rightarrow y_0=Ae^{-x}$

Boundary condition: $y_0(0)=1$ (Note in outer region. Therefore
irrelevant), $y_0(1)=1$ (Only this one is relevant)

$\Rightarrow \un{y_0=e^{-x+1}}$

\un{INNER}

Given boundary layers is $O(\varepsilon)$ .

Use inner variable $X=\df{x}{\varepsilon}:\ \pl_x=\df{\pl X}{\pl
x}\pl_X=\df{1}{\varepsilon}\pl_x$ and set $y(\varepsilon
x;\varepsilon)=Y(X;\varepsilon)$

Equation becomes:

$\df{1}{\varepsilon}Y;;+\df{1}{\varepsilon}Y'+Y=0,\ Y(0)=1$

$\Rightarrow Y''+Y'+\varepsilon Y=0,\ Y(0)=1$

2nd order equation and only one boundary condition $\Rightarrow$
matching is needed.

Try regular ansatz: $Y(X;\varepsilon)=Y_0(X)+\varepsilon
Y_1(X)+\varepsilon^2Y_2(X)+O(\varepsilon^3)$

$\Rightarrow Y_0''+\varepsilon Y_1''+Y_0'-\varepsilon
Y_1'+\varepsilon Y_0+(\varepsilon^2 Y_1)=O(\varepsilon^2)$

$\un{O(\varepsilon^0)}: \begin{array}{cl} & Y_0''+Y_0'=0;
Y_0(0)=1\\ \Rightarrow & Y_o'+Y_0=const=A\ \rm{say}\\ \Rightarrow
& Y_0=+A+Ce^{-x}\ \mbox{where $A$ and $C$ are arbitary
constants}\end{array}$

Use \un{1} boundary condition $Y_0(0)=1 \Rightarrow 1=A+C$,

only one constant can be determined, say $A=1-C$

Therefore $Y_0=(1-C)+Ce^{-X}$

Match up to get value of $C$ using Van Dyke.

$\begin{array}{lcl} \mbox{One term outer expansion} & = &
e^{1-x}\\ \mbox{Rewritten in inner variable} & = &
e^{1-\varepsilon X}\\ \mbox{Expanded for small $\varepsilon$} & =
& e(1-\varepsilon x+O(\varepsilon^2))\\ \mbox{One term
$O(\varepsilon^0)$} & = & e+O(\varepsilon)\ (\star)\\ \mbox{One
term inner expansion} & = & -(c-1)+ce^{-X}\\ \mbox{Rewritten in
outer variable} & = & -(c-1)+ce^{-\frac{X}{\varepsilon}}\\
\mbox{Expanded for small $\varepsilon$} & = & -(c-1)\\ & &
+\mbox{exp. small term in} \varepsilon \\ & & \mbox{(no +ve power
series in $t$)}\\ \mbox{One term $O(\varepsilon^0)$} & = & -c+1\
(\star\star)\end{array}$

$(\star) and (\star\star)$ must be equal:

$\Rightarrow e=-c+1$ or $c=1-e$

Therefore outer 1-term is:
$y(x;\varepsilon)=e^{1-x}+O(\varepsilon)$

Inner 1-term is $Y(X;\varepsilon)=e+(1-e)e^{-X}$

Composite is:

\begin{eqnarray*} y^{comp} & = & y^{outer}+y^{inner}-\mbox{outer
limit of $y^{inner}$ or inner limit of $y^{outer}$}\\ & = &
e^{1-x}+e+(1-e)e^{-X}-e\ \ (+O(\varepsilon))\\ & = &
e^{1-x}+(1-e)e^{-X}+\cdots\\ & = &
e^{1-x}+(1-e)e^{-\frac{x}{\varepsilon}}+\cdots,\ \varepsilon \to
o^+ \end{eqnarray*}

Exact solution is given by substituting $y=Ae^{kx}$

$\Rightarrow \varepsilon k^2+k+1=0 \Rightarrow k=\df{-1\pm
\sqrt{1-4\varepsilon}}{2\varepsilon}$

General solution is $y=Ae^{\undb{(\frac{-1 +
\sqrt{1-4\varepsilon}}{2\varepsilon})}x}+Be^{\undb{(\frac{-1 -
\sqrt{1-4\varepsilon}}{2\varepsilon})}x}$

\hspace{2.3in} $k_1$ \hspace{1.5in} $k_2$

$y(0)=1 \Rightarrow A+B=1$

$y(1)=1 \Rightarrow Ae^{k_1}+Be^{k_2}=1$

so $A=\left(\df{e^{k_2}-1}{e^{k_2}-e^{k_1}}\right),\
B=\left(\df{1-e^{k_1}}{e^{k_2}-e^{k_1}}\right)$ so exact solution
is

$$y=\df{(e^{k_2}-1)e^{k_1 x}+(1-e^{k_1})e^{k_2
x}}{e^{k_2}-e^{k_1}}$$

Small $\varepsilon$ expansion gives $k_1=-1+O(\varepsilon),\
k_2=-\df{1}{\varepsilon}+1+O(\varepsilon). e^{k_2}$ is therefore
exp. small.

$$\Rightarrow y \sim +e^{k_1(x-1)}+(1-e^{-k_1})e^{k_2 x} \sim
e^{1-x}+(1-e)e^{-\frac{x}{\varepsilon}}\ \surd\surd$$



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