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I'm reading "Introduction to Perturbation Methods", Second Edition, by Mark H. Holmes.

In ch 2.2.5 "Matching Revisit" it explains the approach to match the outer and boundary layer solutions, and I got lost in the Example of (2.20) $$y=\sqrt{1+x+\frac{\varepsilon}{\varepsilon+x}}$$ , where $0\le x \le 1$.

First the outer expansion is found for $0<x\le 1$ as (2.21) $$y\sim \sqrt{1+x} + \frac{\varepsilon}{2x\sqrt{1+x}} + \cdots$$ , and the boundary solution is found by setting $\bar x = x/\varepsilon$ and expanding to obtain (2.22) $$Y\sim \sqrt{\frac{2+\bar x}{1+ \bar x}} + \frac12 \varepsilon \bar x \sqrt{\frac{2+\bar x}{1+ \bar x}} + \cdots $$

The usual matching approach would be (2.14) $$\lim_{\bar x\rightarrow\infty}Y_0 = \lim_{x\rightarrow 0} y_0$$ , but now it does not work as both solutions become unbounded, so it's necessary to use a more refined matching method.

To do that, introduce the intermediate variable $x_\eta$ defined as (2.17) $$x_\eta = \frac{x}{\varepsilon^\beta}$$ where $0<\beta<1$. This interval for $\beta$ comes from the requirement that the scaling for the intermediate variable must lie between the outer scale $O(1)$, and the inner scale $O(\varepsilon)$.

Substitute (2.17) into (2.21) gets $$y_{outer} \sim \sqrt{1+\varepsilon^\beta x_\eta} + \frac{\varepsilon^{1-\beta}}{2x_\eta\sqrt{1+\varepsilon^\beta x_\eta}} + \cdots \sim 1 + \frac12 \varepsilon^\beta x_\eta + \cdots + \varepsilon^{1-\beta} \frac{1}{2x_\eta} + \cdots$$ Similarly, substitute (2.17) into (2.22) gets $$y_{inner} \sim 1 + \varepsilon^{1-\beta} \frac{1}{2x_\eta} + \cdots + \frac12\varepsilon^\beta x_\eta + \cdots$$

Then it says :

Comparing these two expansions it is evident that they match.

and stopped just before giving the final result!

I'm confused here:

Though $y_{outer}$ and $y_{inner}$ match, both would be unbounded at $x_\eta=0$ or $x_\eta=\infty$. What value can $\beta$ take? Any most importantly, what is the final expansion of $$y=\sqrt{1+x+\frac{\varepsilon}{\varepsilon+x}}$$ ?

athos
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  • Not an answer, but it looks like the expansion is about $\epsilon$, so if $f(\epsilon) = \sqrt{1+x+\frac{\epsilon}{\epsilon+x}}$, then $f(\epsilon) \approx f(0) + f'(0) \epsilon = \sqrt{1+x}+ {1 \over 2x \sqrt{1+x}}\epsilon$. – copper.hat Nov 25 '23 at 23:43
  • I must confess that, to me, the steps are not very clear – Claude Leibovici Nov 26 '23 at 02:46
  • If I understand your notation correctly, $y_0(x)=\sqrt{1+x}$ and $Y_0(\bar{x})=\sqrt{\frac{2+\bar{x}}{1+\bar{x}}}$. I don't see how "both solutions become unbounded" as $x\to 0$ and $\bar{x}\to\infty$. – Gonçalo Nov 26 '23 at 05:55
  • It’s not a question on calculus but on perturbation theory, seeking an approximation for $0\le x \le 1$ and small $\varepsilon$. So @copper.hat $f(x,\varepsilon) \approx \sqrt{1+x} + \frac{1}{2x \sqrt{1+x}} \varepsilon$ is not perfect as it cannot handle $x=0$. – athos Nov 26 '23 at 10:52
  • @Gonçalo Indeed. I misread the book and ignore the subscription 0 of $y_0$ and $Y_0$. You are right, thank you for pointing that out! – athos Nov 26 '23 at 21:22

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