Past exam of the differential equations 2 course of the University of Oxford / 2021 / 3 / c / ii

The reduced equation and the left boundary condition give the leading outer solutionIntroduce the right inner variable $X=(2-x)/ϵ\ge 0$ and write $Y(X)=y(2-ϵX)$. The chain rule gives $y^{\prime }=-Y_X/ϵ$ and $y^{\prime \prime }=Y_{XX}/{ϵ}^2$. Multiplying the rescaled differential equation by $ϵ$ gives the exact equationAt leading order, $Y_{0,XX}+2Y_{0,X}=0$, with general solution $Y_0=A+B{e}^{-2X}$. The right boundary condition gives $Y_0(0)=0$. The asymptotic matching condition in the overlap region $ϵ\ll 2-x\ll 1$ givesThus $A=e$ and $B=-e$, yielding the leading inner expansionEquivalently, the inner variable $\xi =2(2-x)/ϵ$ gives $Y_0=e(1-e^{-\xi })$. Both normalizations describe the same boundary layer.