I am trying to show that the approximation on $0\leq x \leq 1$ $$\phi(x,\epsilon) \sim \sin x+ \epsilon \cos x - \epsilon$$ is uniformly asymptotic to the exact solution $$f(x,\epsilon) = \frac{1}{1+\epsilon}\left[\sin x +\epsilon \cos x - \epsilon e^{-\epsilon x}\right]$$. The definition of uniformly asymptotic: Suppose $f(x,\epsilon)$ and $\phi(x,\epsilon)$ are continuous functions for $x \in I$, $0 < \epsilon < \epsilon_0$. Then $\phi$ is a uniformly valid asymptotic approximation of $f$ if for all $x \in I$ and $\delta > 0$, there is $\epsilon_1$ such that $|f(x,\epsilon)-\phi(x,\epsilon)|\leq\delta |\phi(x,\epsilon)|$, $0< \epsilon< \epsilon_1$

So here is what I have:

I subtract the two solutions and get $$\left|\frac{-\epsilon}{1+\epsilon} \sin x + \frac{-\epsilon^2}{1+\epsilon} \cos x - \epsilon(e^{-\epsilon x} -1)\right| \leq \delta \left|(\sin x+ \epsilon \cos x - \epsilon)\right|$$ Now if I take the limit as $\epsilon \to 0$ I get that $$|0 + 0 - 0| \leq \delta |\sin x+ 0 - 0| \Rightarrow 0 \leq \delta |\sin x|$$ But for $0\leq x \leq 1$, $|\sin x| > 0$ and $\delta > 0$ so the inequality holds. Is this sufficient to prove that the approximation is uniformly asymptotic?


HINT No, it's not: there's clearly something going on around $(x,\varepsilon)=(0,0)$ which needs to be resolved, plus a subtle point relating to the zeroes of your RHS. A plan of attack would read like this:

  • Define $$ g(x,\varepsilon) = \frac{ (1+\varepsilon)^{-1} \sin(x) + \varepsilon (1+\varepsilon)^{-1} \cos(x) - (1-{\rm e}^{-\varepsilon x}) }{ \sin(x) + \varepsilon \cos(x) - \varepsilon } . $$ Your inequality reads $\sup_{(x,\varepsilon)\in[0,1]\times(0,\varepsilon_1)} \left\vert\varepsilon \, g(x,\varepsilon) \right\vert < \delta$, where $\delta>0$ is given and $\varepsilon_1$ can be taken as small as needed.

  • Plot the above two-variable function to gain some intuition.

  • Focus on the point $(x,\varepsilon)=(0,0)$, which appears problematic: it makes the denominator zero. You thus have to determine whether $\lim_{(x,\varepsilon)\to(0,0)} \left\vert\varepsilon \, g(x,\varepsilon) \right\vert$ exists. Even worse, the denominator may be expected to have an entire curve of zeroes; it's "improbable" (chuckle...) that the numerator will be zero on that precise same set, and you can't bound something "small" by something zero :).

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  • $\begingroup$ After checking my notes, I realized that I made an error in the definition. At the point $(0,0)$ the inequality should still hold. for all $x$. $\endgroup$ – rioneye Oct 2 '13 at 13:46

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