Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 74 / 3 / b / Solution

At fixed positive $x$ the leading outer expansion satisfies $z_0^{\prime }+z_0=e^{-x}/x$. Integrating and imposing the value at one yieldsIt diverges negatively at zero, and the neglected nonlinear shift in the derivative coefficient becomes important when $x\sim \epsilon |\log x|$. Thus the relevant logarithmically enhanced nonlinear boundary layer is larger than a plain $O(\epsilon )$ layer. Write $L=\log (1/\epsilon )$, $x=\epsilon LX$ andFor 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 givesIn particular the exact equation at zero is $-\epsilon z(0)z^{\prime }(0)=1$, so

An implicit inner form also checks the matching constants. With $x=\epsilon \xi$, retain $z$ without assigning it a bounded size. The leading equation is $(\xi -z)z_{\xi }=1$. Inverting it gives $d\xi /dz-\xi =-z$, henceIts large-$\xi$ match to $z\sim 1+\log \epsilon +\log \xi$ fixes $B\sim e^{-1}/\epsilon$. Taking that matched leading value at zero giveswhere $W_0$ 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.