= Solution
The reduced first-order equation cannot satisfy both endpoint values. The given <outer expansion> satisfies $y(1)=2$ but has $y_0(0)=e$, so the <boundary layer> lies at $x=0$ and has stretched coordinate $X=x/\epsilon$.
Write the <inner expansion> as $Y=Y_0+\epsilon Y_1+\cdots$. After multiplying the differential equation by $\epsilon$, it becomes
$$
Y_{XX}+(1+\epsilon X)Y_X+\epsilon^2XY=0.
$$
At leading order,
$$
Y_0''+Y_0'=0.
$$
The boundary condition $Y_0(0)=0$ and matching to $y_0(0)=e$ give
$$
\boxed{Y_0=e(1-e^{-X}).}
$$
At the next order,
$$
Y_1''+Y_1'=-XY_0'=-eXe^{-X}.
$$
Matching to $y_1(0)=e(\ln2-1)$ and imposing $Y_1(0)=0$ gives
$$
\boxed{
Y_1=e(\ln2-1)
+ee^{-X}\left(\frac{X^2}{2}+X+1-\ln2\right).}
$$
The <additive composite expansion> is outer plus inner minus their common part. With $X=x/\epsilon$, it is
$$
\boxed{
\begin{aligned}
y_{\rm comp}(x)={}&(1+x)e^{1-x}\\
&+\epsilon[-(1+x)\ln(1+x)+(1+x)\ln2+(x-1)]e^{1-x}\\
&+ee^{-x/\epsilon}\left[-1+\epsilon\left(
\frac12\left(\frac x\epsilon\right)^2
+\frac x\epsilon+1-\ln2\right)\right].
\end{aligned}}
$$
It satisfies $y_{\rm comp}(0)=0$ exactly through the retained order, satisfies the right boundary condition up to exponentially small terms, and is uniformly accurate to $O(\epsilon)$.
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