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,


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


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 $\alpha=1$.

  • $\begingroup$ Thanks for this answer, but I guess I don't understand why balancing the first and second term,i.e., $\alpha=1/2$ will give an outer expansion? $\endgroup$
    – Hua
    May 2 '16 at 13:15
  • 1
    $\begingroup$ @Hua If the third and fourth terms are the only leading order terms, then the leading order solution to the equation is exactly the one where the derivative terms are ignored entirely, which is the same as taking $\epsilon=0$ in the original equation (prior to defining the stretching coordinate). This is how you define the outer solution. $\endgroup$
    – Ian
    May 2 '16 at 13:19
  • $\begingroup$ Oh, I see it! Thanks for this! $\endgroup$
    – Hua
    May 2 '16 at 13:33
  • $\begingroup$ @Hua Actually, I'm a bit confused about the second term. How did the coefficient wind up being just $\epsilon$? Is there a typo? It seems that could only happen if the original equation had a $\epsilon^{\alpha}$ there (to cancel the $\epsilon^{-\alpha}$ from the derivative), which makes no sense as $\alpha$ only arises in the course of solving the problem. $\endgroup$
    – Ian
    May 2 '16 at 13:37
  • $\begingroup$ I think so. The original equation is $\epsilon^{2}y''+\epsilon xy'-y=-e^{x}$. Then $dy/dx=dY/d\bar{x}*\epsilon^{-\alpha}$ and $\epsilon\bar{x}=\epsilon^{1+\alpha}$ give the transformed one above. $\endgroup$
    – Hua
    May 2 '16 at 13:44

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