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

For fixed $x>0$, set $y=y_0+\epsilon y_1+{\epsilon }^2{y}_2+\cdots$ and impose the data at one at each order. The leading equation gives $y_0^{\prime }=0$, hence $y_0=1$. At the next two orders,The boundary conditions $y_1(1)=y_2(1)=0$ give the three-term outer expansionIts ordering fails when $\epsilon /x=O(1)$, locating a boundary layer of thickness $O(\epsilon )$ at zero. Put $x=\epsilon \xi$ and $y=Y_0(\xi )+\epsilon Y_1(\xi )+\cdots$. The rescaled equation isIts leading equation is ${\xi }^2{Y}_0^{\prime }=Y_0^3$. Matching to the positive outer value one fixes the sign and integration constant:At the next order the bracket $\xi (Y_0^2-1)+2Y_0^2$ vanishes identically, leavingThe general solution is $Y_1=(1+k\xi ){\xi }^{1/2}/(\xi +2{)}^{3/2}$. In the overlap its expansion isThe outer solution re-expressed on this scale has the corresponding terms $1-1/\xi +3/(2{\xi }^2)+\epsilon (1-2/\xi )+\cdots$. Thus matched asymptotic expansion determines $k=1$, consistently in both displayed orders, and the two-term inner expansion isThis positive branch tends continuously to zero at $x=0$, with a square-root cusp. An infinite endpoint derivative is compatible with the degeneracy of the original equation's $x^2{y}^{\prime }$ coefficient.