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{\bf Question}

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

$$\varepsilon y''+(2x^2+x+1)y'=4x+1,\ y(0)=1,\ y(1)=1$$

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}

$\varepsilon y''+(2x^2+x+1)y'=4x+1,\ y(0)=1,\ y(1)=1$

Assuming again an $O(\varepsilon)$ boundary layer near $x=0$ we
have

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

Substitute into equation

$$\varepsilon y_0''+(2x^2+x+1)(y_0'+\varepsilon
y_1')=4x+1+O(\varepsilon^2)$$

$\un{O(\varepsilon^0)}$

\begin{eqnarray*} (2x^2+x+1)y_0' & = & 4x+1\\ & & y_0(1)=1
\mbox(is only relevant boundary condition)\\ \rm{Therefore}\ y_0'
& = & \df{4x+1}{2x^2+x+1}\\ \ds\int dy_0 & = & \ds\int
\df{4x+1}{2x^2+x+1}\, dx\\ y_0 & = & \ds\int
\df{d(2x^2+x+1)}{2x^2+x+1}=A+\log(2x^2+x+1) \end{eqnarray*}

or see standard text books.

Apply boundary conditions to get

$\begin{array}{rcl} 1 & = & \log(2+1+1)+A\\ \Rightarrow A & = &
1-\log 4\\ \rm{Therefore}\ y_0 & = &
\un{1+\log\left(\df{2x^2+x+1}{4}\right)} \end{array}$

\un{INNER}

Assuming $O(\varepsilon)$ boundary layer near $x=0$, set
$\df{x}{\varepsilon}=X \Rightarrow \pl_x=\df{1}{\varepsilon}\pl_X$
with $y(\varepsilon X;\varepsilon)=Y(X;\varepsilon)$

Equation becomes

$\df{1}{\varepsilon}Y''+(2\varepsilon^2X^2+\varepsilon
X+1)Y'=4\varepsilon^2 X+\varepsilon\ \ Y(0)=1$

Only 1 boundary condition for 2nd order equation $\Rightarrow$
matching needed for finding 2nd arbitrary constant.

Solve perturbatively using regular constants

$Y(X,\varepsilon)=Y_0(X)+\varepsilon
Y_1(\varepsilon)+\varepsilon^2 Y_2(\varepsilon)+O(\varepsilon^3)$

$\un{O(\varepsilon^0)} Y_0''+Y_0'=0,\ Y_0(0)=1$

This is the same as in question 8.

Hence solution:

$$Y_0=(1-c)+ce^{-X}$$

$c$ determined from matching by Van Dyke.

$\begin{array}{lcl} \mbox{One term outer expansion} & = &
1+\log\left(\df{2x^2+x+1}{4}\right)\\ \mbox{Rewritten in inner
variable} & = & 1+\log\left(\df{2\varepsilon^2 X^2+\varepsilon
X+1}{4}\right)\\ \mbox{Expanded for small $\varepsilon$} & = &
1-\log 4+\varepsilon X+O(\varepsilon^2)\\ \mbox{One term
$O(\varepsilon^0)$} & = & 1-\log 4\ (\star)\\ \mbox{One term inner
expansion} & = & ce^{-X}-(c-1)\\ \mbox{Rewritten in outer
variable} & = & ce^{-\frac{x}{\varepsilon}}-(c-1)\\ \mbox{Expanded
for small $\varepsilon$} & = & 1-c\\ & & +\mbox{exp. small in}
\varepsilon\\ \mbox{One term $O(\varepsilon^0)$} & = & 1-c\
(\star\star)\end{array}$

$(\star) and (\star\star)$ must match, hence: $1-c=1-\log 4
\Rightarrow c=\log 4$

Therefore outer 1-term is:
$y(x;\varepsilon)=1+\log\left(\df{2x^2+x+1}{4}\right)+O(\varepsilon)$

Inner 1-term is $Y(X;\varepsilon)=(1-\log 4)+(\log 4)e^{-X}$

Composite is:

\begin{eqnarray*} y^{comp} & = & y^{outer}+y^{inner}-\mbox{inner
limit of $y^{outer}$}\\ & = &
1+\log\left(\df{2x^2+x+1}{4}\right)+(1-\log 4)\\ & & +(\log
4)e^{-X}-(1-\log 4)\\ y^{comp} & = &
\un{1+\log\left(\df{2x^2+x+1}{4}\right)+(\log
4)e^{-\frac{x}{\varepsilon}}+\cdots}\ \ \varepsilon \to 0^+
\end{eqnarray*}

Exact solution is difficult to find!



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