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I'm having problems getting my head around a stretching transformation in the method of matched asymptotic expansions. I'm reading Introduction to Perturbation Methods (by Holmes) and he discusses the introduction of a boundary-layer coordinate to treat the region near $x = 0$ for the differential equation $\epsilon \, y'' + 2 y' + 2y = 0$ for $0 < x < 1$ (the boundary conditions are $y(0) = 0$ and $y(1) = 1$). He treats this boundary layer by using the transformation $\bar{x} = \frac{x}{\epsilon^\alpha}$ on the ODE. He explains this transformation by stating that it stretches the region near $x = 0$. I realize this region, due to the nature of the problem, requires special treatment. However, I'm unsure as to how this transformation actually helps us find a solution in this region. Specifically, how does this transformation help us deal with this special $x = 0$ region and what does he mean when he says it 'stretches the region'?

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up vote 1 down vote accepted

The stretching effect can be see this way: the very small interval $0< x< \epsilon^\alpha$ corresponds to the medium-sized interval $0< \bar x<1$. I do not have access to the book right now (and you did not give the page number anyway), so I am going to guess that $\alpha=1/2$ is going to be used here, because in terms of $\bar x$ the equation becomes $y''+2\epsilon^{-1/2} y+2y=0$. If by "fund a solution" you mean an analytic solution, then this substitution does not help much; the advantage could be in the fact that the author's method of finding the asymptotic behavior relies on the highest order coefficient being normalized.

If this is too vague (which it is), please provide some details. For example, after the book introduces the transformation $\bar x=x/\epsilon^\alpha$, what does it do with it?

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