I am studying perturbation theory, and I have a problem when reading the book "Introduction to Perturbation Methods" by M.H. Holmes. This is about boundary layer. We know when seeking inner expansion, we usually need to introduce an inner variable or stretching variable by defining $\bar{x}=\epsilon^{\alpha}x$, where $x$ is original variable and $\bar{x}$ is inner variable. Then the $\alpha$ need to be determined during the later analysis by balancing. In page 62 of the book, there is an example I put here,
The original equation is, $$\epsilon^{2}y''+\epsilon xy'-y=-e^{x}$$
and the stretched one is,
$$\epsilon^{2-2\alpha}\frac{d^{2}Y}{d\bar{x}^{2}}+\epsilon\bar{x}\frac{dY}{d\bar{x}}-Y=-e^{\epsilon^{\alpha}\bar{x}}$$
The book says the balance is between the first, third and fourth terms and so $\alpha$ is $0$. I can't follow this statement. Why we can neglect the second term?
I tried myself by expanding the fourth term by taylor series. Then it becomes
$$\epsilon^{2-2\alpha}\frac{d^{2}Y}{d\bar{x}^{2}}+\epsilon\bar{x}\frac{dY}{d\bar{x}}-Y=-(1+\epsilon^{\alpha}\bar{x}+\dots)$$
Now I don't know how to proceed. Actually $\alpha=1/2$ seems also OK to me. I think I can only handle three terms balancing:-(
You assume that you have chosen $\alpha$ so that all the factors that don't explicitly contain $\epsilon$ are $O(1)$ as $\epsilon \to 0$. Then the first term is $O(\epsilon^{2-2\alpha})$, the second term is $O(\epsilon^1)$, the third term is $O(1)$ and the last term is $O(1)$. So the third and fourth terms can always balance one another in principle...but leaving them to actually balance each other alone would give the outer solution, contradicting the assumption that there is a boundary layer. The only other way to make the $O(1)$ terms vanish is to have $2-2\alpha=0$ so $\alpha=1$.