Krdytkyu

Obtain the leading order uniform approximation of the solution to $ epsilon y′′-x^2y′-y=0$.

The boundary conditions are $y(0)=y(1)=1$.

Since $a(x)<0$ the boundary layer is at $x=1$.

The outer solution will be of the form $y(x; epsilon) = y_0(x) + epsilon y_1(x) + ...$

So the leading order problem is $-x^2y_0' - y_0 =0$ and $y_0(0) = 1$.

Hence $y_0 = ke^{1/x}$.

Here's where the issue arises: to work out $k$ I get $y_0(0)=ke^{1/0}=1$ using the boundary condition at $x=0$, but $e^{1/0}$ is obviously undefined. How can I proceed form here?

This is a complicated problem, but if you're careful it does work out. Firstly, let's write out the equation assuming there is a boundary layer or width $epsilon^alpha$ at $x_0$, so let $X=(x-x_0)/epsilon^alpha$ and $Y(X)=y(x)$. We get
$$ epsilon^{1-2alpha}Y_{XX}-epsilon^{-alpha}(epsilon^alpha X+x_0)^2Y_X-Y=0,$$
expanding
$$ epsilon^{1-2alpha}Y_{XX}-epsilon^{alpha}X^2Y_X-2Xx_0Y_X-epsilon^{-alpha}x_0^2Y_X-Y=0.$$

Now for dominant balance. If $x_0neq0$, then we would balance $epsilon^{1-2alpha}$ with $epsilon^{-alpha}$ (you can check that this is the only dominant balance) to give $alpha=1$. You can leave $x_0$ unknown and determine that it must be 1, but since you already know that we'll use the fact and write our leading-order inner equation at $x=1$ as
$$Y_{XX}-Y_X=0,$$
with boundary condition $Y(0)=1$ (since $x=1$ corresponds to $X=0$). The solution is $Y(X)=A+Be^X$ where $A+B=1$.

If $x_0=0$ then we have a different balance, with $alpha=1/2$. Let $mathsf Y(chi)=y(x)$ with $chi=x/sqrt{epsilon}$ and the leading-order inner equation at $x=0$ is
$$mathsf Y_{chichi}-mathsf Y=0,$$
with $mathsf Y(0)=1$. The solution is $mathsf Y=Ce^chi+De^{-chi}$, and since our solution must be bounded as we exit the boundary layer, we need $A=0$, and the boundary condition gives $D=1$.

The leading order outer solution is, as you found, $y=ke^{1/x}$. You can say here that $k=0$ since the solution must be bounded as you enter each boundary layer, or do it through asymptotic matching.

To fix the remaining constants $k$ and $A$ (or $B$), we need to match all the parts of the solution. To do this, we need the outer solution to be bounded, so we need $k=0$. Alternatively, consider matching the outer solution to the inner solution at $x=0$,
$$lim_{chirightarrowinfty}e^{-chi}=lim_{xrightarrow0}ke^{1/x}Rightarrow0=lim_{xrightarrow0}ke^{1/x}Rightarrow k=0.$$ Then the match between the outer layer and the inner layer at $x=1$ gives
$$lim{xrightarrow1}0=lim{Xrightarrow-infty}A+Be^XRightarrow A=0$$
and hence $B=1$.

So the inner solution at $x=0$ is $mathsf Y(chi)=e^{-chi}$, the outer solution is $y(x)=0$ and the inner solution at $x=1$ is $Y(X)=e^X$. We can find a uniformly valid approximation by adding the three equations (writing them all in terms of $x$) and subtracting off the matching constants (all zero) as,
$$y_{unif}(x)=e^{x/sqrt{epsilon}}+e^{(x-1)/epsilon}.$$

This will not be particularly accurate, it's an $O(sqrt{epsilon})$ approximation, but I did verify that numerically. I've made a plot of the solution for $epsilon=0.01$ first and $epsilon=0.02$ second.