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{\bf Question}

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

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

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}

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

Assume that the boundary layer exists near $x=0$ of width
$O(\varepsilon)\ \leftarrow$ This needs to be justified really but
we follow lead from the question.

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

$$\varepsilon y_0''+y_0'+\varepsilon y_1'=1+O(\varepsilon^2)$$

Balance at

$\un{O(\varepsilon^0)}: y_0'=1 \Rightarrow y_0=x+A$

Boundary data: only relevant one is $y(1)=\beta$. (0 is in the
boundary layer).

$$\Rightarrow y_0(1)=\beta\ \rm{so}\ \un{y_0=\beta-1+x}$$

\un{INNER}

Assuming $O(\varepsilon)$ boundary layer near $x=0$, set
$X=\df{x}{\varepsilon}$ as inner variable.

$\pl_x=\df{\pl X}{\pl x}\pl_X=\df{1}{\varepsilon}\pl_X$ etc. with
$y(\varepsilon X;\varepsilon)=Y(X,\varepsilon)$

Equation becomes:

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

Therefore $Y''+Y'-\varepsilon=0$ with $Y(0)=\alpha$

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

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

$\un{O(\varepsilon^0)} Y_0''+Y_0'=0;\ Y_0(0)=\alpha$

$\rightarrow Y_0^(X)=A+Ce^{-X}$ where $A$ and $C$ are arbitrary
constants.

Therefore $\alpha=A+C$, can only find 1 constant. Pick $A$ say.

$A=\alpha-c$

Therefore $Y_0(X)=(\alpha-c)+ce^{-X}$

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

$\begin{array}{lcl} \mbox{One term outer expansion} & = &
\beta-1+x\\ \mbox{Rewritten in inner variable} & = &
\beta-1+\varepsilon x\\ \mbox{Expanded for $\varepsilon \to o^+$}
& = & \beta-1+\varepsilon x\\ \mbox{One term $O(\varepsilon^0)$} &
= & \beta-1\ (\star)\\ \mbox{One term inner expansion} & = &
\alpha-c+ce^{-X}\\ \mbox{Rewritten in outer variable} & = &
\alpha-c+ce^{-\frac{x}{\varepsilon}}\\ \mbox{Expanded for
$\varepsilon \to 0^+$} & = & \alpha-c\\ & & +\mbox{exp. small term
in} \varepsilon\\ \mbox{One term $O(\varepsilon^0)$} & = &
\alpha-c\ (\star\star)\end{array}$

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

$\Rightarrow \beta-1+\alpha-c$ or \un{$c=\alpha-\beta+1$}

Therefore outer 1-term is: $y(x;\varepsilon)=\beta-1+x$

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

Composite is:

\begin{eqnarray*} y^{comp} & = & y^{outer}+y^{inner}-\mbox{inner
limit of $y^{outer}$}\\ & = &
\beta-1+x+(\alpha-\beta+1)e^{-\frac{x}{\varepsilon}}+(\beta-1)-(\beta-1)+\cdots\\
& = &
(\beta-1)+x+(\alpha-\beta+1)e^{-\frac{x}{\varepsilon}}+\cdots,\ \
\varepsilon \to 0^+\end{eqnarray*}

Exact solution is given by $y=\undb{y^{CF}}+\undb{y^{PI}}$

\hspace{2.2in} (1) \ \ (2)

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

(2) PI of $\varepsilon y''+y'=1 \un{y^{PI}=x}$

After using boundary conditions we get

$$y=\df{x+(\beta-1)-\alpha
e^{-\frac{1}{\varepsilon}}+(\alpha-\beta+1)e^{-\frac{x}{\varepsilon}}}
{1-e^{-\frac{1}{\varepsilon}}}\ \rm{\un{exactly}}$$

so as $\varepsilon \to 0^+,\ e^{-\frac{1}{\varepsilon}}$ is exp.
small with respect to $e^{-\frac{x}{\varepsilon}}\ \ x\in [0,1)$

so $y \sim x+(\beta-1)+(x-\beta+1)e^{-\frac{x}{\varepsilon}}+$
exp. small terms $\surd\surd$



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