= Solution
Three <distinguished limit>[distinguished limits] are needed.
In the origin layer, put $x=\varepsilon X$ and $y(x;\varepsilon)=Y(X)$. Since $\operatorname{sech}(\varepsilon X)=1+O(\varepsilon^2)$, the leading equation is
$$
XY''+(X+2)Y'+Y=0.
$$
It has the first integral
$$
XY'+(X+1)Y=C.
$$
Solving this <first-order linear differential equation> gives
$$
Y=\frac{C+D e^{-X}}{X}.
$$
Regularity at $X=0$ requires $D=-C$, and $Y(0)=1$ then fixes $C=1$. Thus Region I, $x=O(\varepsilon)$, has
$$
\boxed{Y(X)=\frac{1-e^{-X}}{X}.}
$$
Its matching limit is $Y\sim X^{-1}$ as $X\to\infty$.
In Region II, $x=O(1)$ with $x\gg\varepsilon$, setting $\varepsilon=0$ reduces the second-order equation to
$$
xy_0'+(x+1)y_0=0.
$$
Hence $y_0=C e^{-x}/x$. Matching $C/x$ with the inner limit $\varepsilon/x$ fixes $C=\varepsilon$, so
$$
\boxed{y\sim\frac{\varepsilon}{x}e^{-x}.}
$$
The correction generated by $\varepsilon x^3y$ changes the local decay rate when $x=O(\varepsilon^{-1/2})$. For Region III set
$$
Z=\sqrt\varepsilon\,x,
\qquad
y=e^{-F(Z)/\sqrt\varepsilon}G(Z).
$$
The terms of order $\varepsilon^{-1/2}$ give the <eikonal equation>
$$
F'(Z)=1+Z^2,
$$
and the next balance gives $ZG'+G=0$. Matching to Region II selects
$$
F(Z)=Z+\frac{Z^3}{3},
\qquad
G(Z)=\frac{\varepsilon^{3/2}}{Z}.
$$
Therefore
$$
\boxed{
y\sim\frac{\varepsilon^{3/2}}{Z}
\exp\left[-\frac{Z+Z^3/3}{\sqrt\varepsilon}\right]}
\qquad(x=O(\varepsilon^{-1/2})).
$$
This solution is already beyond all algebraic orders in $ε$ in the distant region. A leading <multiplicative composite expansion> that contains all three balances and satisfies the boundary value exactly is
$$
\boxed{
y_{comp}(x;\varepsilon)=
\frac{\varepsilon}{x}
\left(1-e^{-x/\varepsilon}\right)
\exp\left(-x-\frac{\varepsilon x^3}{3}\right),}
$$
with its value at $x=0$ understood by continuity.
Finally, the coefficient of $y''$ vanishes at the origin. Taking the regular limit of the original equation there gives
$$
2\varepsilon y'(0)+y(0)=0,
$$
so the prescribed value also fixes $y'(0)=-1/(2\varepsilon)$. Equivalently, the second inner solution behaves as $1/X$ and is excluded by regularity. This <regular singular point> is why one boundary condition determines the unique regular solution.
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