Consider the BVP:
$\epsilon \dfrac{d^2y}{dx^2}-(x^2-2)y=-1 \\ \text{where} -1<x<1 \;\text{and} \; y(-1)=y(1)=0, \; 0<\epsilon<<1$
I am trying to show the existence of a boundary layer at $x=-1$ and $x=1$, but am stuck. So far I have shown that the solution $y(x)$ is symmetric, so this should simplify matters to just proving $x=1$ or $x=-1$ is a boundary layer. What is a good way to approach this?
i will give it a try. you get the outer solution by setting $\epsilon = 0.$ so you get $$y_{outer} = \frac1{x^2 - 2}$$ to find the solution valid in a boundary layer near $x = 1,$ you define a stretched variable $$\xi = (x-1)/\sqrt \epsilon, x = \sqrt \epsilon \xi + 1, \frac{d^2y}{dx^2} = \frac{1}{\epsilon}\frac{d^2y}{d\xi^2}$$ so the differential equation becomes $$\frac{d^2y}{d\xi^2}-\left((\sqrt \epsilon\xi + 1)^2 - 2\right)y=-1 $$ at the leading order we get $$ \frac{d^2y}{d\xi^2}+y=-1,\,y(0) = 0. $$ so the inner solution is $$y_{inner}(\xi) = -1 + \cos \xi = -1 + \cos\left(\frac{x-1}{\sqrt \epsilon} \right)$$
i hope you can take it from here.