= Logarithmically enhanced nonlinear boundary layer
{title2=$x=O(\varepsilon\log(1/\varepsilon))$}
If a leading <outer expansion> diverges as $\log x$ and a <derivative> coefficient contains $x-\varepsilon z$, balancing that nonlinear correction gives a layer larger than $O(\varepsilon)$. Write $x=\varepsilon L X$ with $L=\log(1/\varepsilon)$ and subtract the large negative constant from $z$. Matching often introduces a $\log L$ correction and yields an endpoint slope of order $1/(\varepsilon L)$. The balance must be derived from the actual equation; logarithmic divergence alone does not fix a universal layer width.
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