Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 336 / 2

For the matched asymptotic expansion, first determine the outer expansion. Setting $\epsilon =0$ away from $x=0$ gives $y_0^{\prime }=y_0$, and the right boundary condition sets $y_0=e^x$. At first order,soThere is no outer $\sqrt{\epsilon }$ term: its homogeneous coefficient would be forced to zero by the right boundary condition. The logarithmic overlap forces a switchback term in the inner expansion.

Near zero, balance $\epsilon y^{\prime \prime }$ against $2x{y}^{\prime }$ to get the square-root boundary layer at a vanishing drift:For fixed $z$ expand $Y=Y_0+s{Y}_1+s^2{Y}_2+\cdots$, allowing logarithmic dependence on $s$ in $Y_2$. With ${\mathcal{L}}_0={\partial }_z^2+2z{\partial }_z$, the leading equation is ${\mathcal{L}}_0{Y}_0=0$. Its bounded solutions are a constant plus an error function. The left boundary condition and the overlap value one fixThe following two solution sections complete the half-order and first-order matching; their headings retain the scaffold's organization of the supplied hints.

Write $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. The first correction in the inner expansion satisfies ${\mathcal{L}}_0{Y}_1=2zE$. Direct differentiation verifies that $zE+bg$ is a particular solution. Adding the constant and error function homogeneous solutions gives $Y_1=zE+bg+A+BE$.

The left boundary condition forces $A=-b$. Re-expanding the outer expansion at $x=sz$ givesTherefore $Y_1\sim z$, with no constant term in its large-$z$ overlap. Since $E\to 1$ and $g\to 0$, this sets $B=b$. Thus the required half-order correction isIt vanishes at zero and has the correct linear overlap. Keeping this $O(\sqrt{\epsilon })$ term is essential even though the outer solution has no correction of that order.

Define the nested integralThe order-$s^2$ equation in the inner expansion is ${\mathcal{L}}_0{Y}_2=2z{Y}_1$. The preceding result decomposes its forcing into $2z^{2}E+2bzE-2bz+2bzg$. The particular solutions combine toFor example, ${\mathcal{L}}_{0}H=E$ follows directly by differentiating the nested integral, so the equation can also be verified without relying on the supplied particular-solution formulas. At zero $Y_{2,p}=b^2$; add $-b^2+DE$ to impose the left boundary condition. As $z\to \infty$ this givesThe outer expansion in the overlap, derived in the previous section, requires the remaining constant to be $-\frac{1}{2}\log s$. Thus $D=C+b^2-\frac{1}{2}\log s$. The logarithmic matching constant is an error-function logarithmic switchback.

The complete answer through order $\epsilon$ in the inner expansion isHere $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. This expression is zero at $z=0$, and its large-$z$ expansion agrees with $1+x+x^2/2-(\epsilon /2)\log x$. Together with $y_{\mathrm{out}}=e^x[1-(\epsilon /2)\log x]$, it covers both requested regions. The overlap uses $1\ll z\ll {\epsilon }^{-1/2}$, with the usual termwise ordering restrictions when matching higher terms.

If a single continuous approximation is desired, an additive composite expansion isThe apparent logarithmic divergence at $x=0$ cancels: its residual is $-(\epsilon /2)(e^x-1)\log x\to 0$. The composite therefore extends continuously to the left endpoint and satisfies the right boundary condition to the requested order.

For the outer expansion, the first-order reduced ordinary differential equation gives $w_0=e^{-x^2}$ after imposing the right boundary condition. Its first correction satisfiesAn integrating factor givesThe left boundary is at $x=\epsilon$, where the highest-derivative coefficient is already ${\epsilon }^2$. Balancing that coefficient against $w^{\prime }$ gives a layer of width ${\epsilon }^2$, not $\epsilon$. This is a boundary layer around a moving endpoint:At leading order $W_0^{\prime \prime }+W_0^{\prime }=0$, and matching to the outer limit one while setting $W_0(0)=0$ gives $W_0=1-e^{-z}$. The first correction satisfiesDirect differentiation of $-z(z+2)e^{-z}/2$ verifies the forcing. It vanishes both at zero and at infinity, so no homogeneous term is needed. The requested inner expansion isIts order-$\epsilon$ overlap constant is zero, agreeing with the outer expansion at the moving endpoint. The overlap can be chosen with $1\ll z\ll {\epsilon }^{-1}$.

An independent scaling check drops the reaction term only in the layer and integrates $\epsilon x{w}^{\prime \prime }+w^{\prime }=0$. Its decaying fast solution is proportional to $(x/\epsilon {)}^{1-1/\epsilon }$. Sinceit gives exactly the same first correction. A useful additive composite expansion, with the left boundary value made exact, isThe exponentially small remainder at the right endpoint is beyond all retained algebraic orders.