Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 74 / 5 / a / Solution

The positive drift places the boundary layer at the left endpoint. Set $y_{\mathrm{out}}=y_0+ϵy_1+\cdots$ and impose the right boundary condition order by order. The leading ordinary differential equation is $[(1+x)y_0{]}^{\prime }=0$, so $y_0=2/(1+x)$. At the next order,which integrates toThe outer limit at the left endpoint is two, so it cannot satisfy the left boundary value without a rapidly varying correction.

For the inner expansion use $X=x/ϵ$ and $y=Y_0(X)+ϵY_1(X)+\cdots$. Its equations and boundary values areMatching the leading constant gives $Y_0=2-e^{-X}$. The first-order forcing is then $-2+(1-X)e^{-X}$. A particular solution is $-2X+\frac{1}{2}{X}^2{e}^{-X}$, and matching determines the constant to be $3/2$. Enforcing $Y_1(0)=0$ givesIn the overlap region the common expansion is $2+ϵ(3/2-2X)$. Adding the inner and outer approximations and subtracting that common part gives the uniform asymptotic approximationIt has error $O({ϵ}^2)$ uniformly for $0\le x\le 1$, apart from an exponentially small endpoint adjustment of the same asymptotic irrelevance. At $x=0$ it equals one exactly, and at $x=1$ it differs from one only exponentially. The polynomial multiplying the exponential remains bounded on the inner scale.

There is an independent check from a first integral. The equation is $[ϵy^{\prime }+(1+x)y{]}^{\prime }=0$, so its exact solution has the formwhere $C$ is selected by the right boundary value. Endpoint expansion of this integral gives $C=2-ϵ/2+O({ϵ}^2)$, the same outer coefficients and the same inner expansion. This check also establishes the uniform order of the composite approximation without treating an inner residual as an outer estimate.