Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2009 / A / 2 / iii

In the outer expansion, omit $ϵy^{\prime \prime }$ to get $y_0^{\prime }=e^{y_0}$. Integrating and imposing the left boundary condition givesThis is regular throughout $[0,1]$. Its right-end value is $B=-\log (C-1)$, generally different from zero.

At $x=1$ set $X=(x-1)/ϵ$. The equation is $Y^{\prime \prime }-Y^{\prime }+ϵe^Y=0$, so the leading inner expansion is $Y=C_0+C_1{e}^X$. Matching as $X\to -\infty$ gives $C_0=B$, while $Y(0)=0$ gives $C_1=-B$. ThusThe common part is $B$. Subtract it once when forming the additive composite expansion:There is one generic boundary layer, at the right endpoint, of width $ϵ$. A left-end exponential would grow into the interval and cannot supply a decaying matching correction. At $A=-\log 2$ the displayed layer amplitude vanishes; the leading outer solution already meets the right condition, though a smaller correction can still be needed at higher order. The composite meets $y(1)=0$ exactly and $y(0)=A$ up to an exponentially small term.