= Solution
At fixed positive $x$ the leading <outer expansion> satisfies $z_0'+z_0=e^{-x}/x$. Integrating and imposing the value at one yields
$$
z_0=e^{-x}(1+\log x).
$$
It diverges negatively at zero, and the neglected nonlinear shift in the <derivative> coefficient becomes important when $x\sim\varepsilon|\log x|$. Thus the relevant <logarithmically enhanced nonlinear boundary layer> is larger than a plain $O(\varepsilon)$ layer. Write $L=\log(1/\varepsilon)$, $x=\varepsilon L X$ and
$$
z=-L+\log L+1+Z(X)+o(1).
$$
For fixed $X$, the leading inner equation is $(X+1)Z_X=1$. Matching to the outer logarithm, for which $Z\sim\log X$, sets the integration constant and gives
$$
Z=\log(1+X),\qquad z(0)=-L+\log L+1+o(1).
$$
In particular the exact equation at zero is $-\varepsilon z(0)z'(0)=1$, so
$$
\boxed{z'(0)\sim\frac1{\varepsilon\log(1/\varepsilon)}>0}.
$$
An implicit inner form also checks the matching constants. With $x=\varepsilon\xi$, retain $z$ without assigning it a bounded size. The leading equation is $(\xi-z)z_\xi=1$. Inverting it gives $d\xi/dz-\xi=-z$, hence
$$
\xi=z+1+B e^z.
$$
Its large-$\xi$ match to $z\sim1+\log\varepsilon+\log\xi$ fixes $B\sim e^{-1}/\varepsilon$. Taking that matched leading value at zero gives
$$
z(0)\simeq-1-W_0(e^{-2}/\varepsilon),\qquad
z'(0)\simeq\frac1{\varepsilon[1+W_0(e^{-2}/\varepsilon)]},
$$
where $W_0$ is the positive real branch of the <Lambert W function>. Its large-argument expansion reproduces the logarithm and log-log terms above. This refinement is a matched approximation, not an exact solution of the full equation; the leading slope conclusion follows directly from the first inner scaling and the exact endpoint identity.
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