Just for reference I am looking at the problem $\epsilon y''(x)+xy'(x)+y(x)=0$ with
$y(1)=0$ and
$y(2)=1$
for which I have found the leading-order outer and inner solutions and composite solution. There's a follow-on question which asks how the method would change if $\epsilon$ were negative and I am having trouble getting my head around this. I read somewhere (it may have been MIT lecture notes but I can't find the link anymore) that then the boundary layer is found at $x=2$ instead of $x=1$ where it is now. Is this true? Is there an intuitive explanation for it?
Thank you for your help!
Your inner solution are simple exponentials $C_0+C_1e^{-x_0(x-x_0)/ϵ}$. To connect them to the outer solution, you need this to be bounded (and convergent) in the connection direction. For these simple exponentials this is only possible in one direction, so the boundary layer has to be, indeed, at a boundary. With $ϵ>0$ this exponential is bounded for increasing $x$, so $x_0=1$ has to be on the left side. If $ϵ<0$, then the exponential is bounded for falling $x$, thus $x_0=2$ has to be on the right side of the interval.