0

In $e^{-\omega/\epsilon}\leq10^{-9}$ why is $\omega=O(\epsilon)$ and in $e^{-\omega/\epsilon}\leq \epsilon$ why is $\omega=O(\epsilon\ln1/\epsilon)$ where $\epsilon$ is a very small parameter.

Artem
  • 14,414
Vaolter
  • 1,711
  • Are you sure that the direction of the inequalities are correct? It seems to me that for the first one $\omega\in\Omega(\epsilon)$ as $\omega \geq (9\log 10)\epsilon$. – Lord Soth Jul 01 '13 at 17:45
  • @Lord Soth, yes am sure. – Vaolter Jul 01 '13 at 17:52
  • LordSoth was trying to be polite: the first inequality does not imply $\omega=O(\epsilon)$ and the second inequality does not imply $\omega=O(\epsilon\log1/\epsilon)$. What is your source? – Did Jul 01 '13 at 18:41
  • @Did here is the book, http://www.amazon.com/Numerical-Methods-Singular-Perturbation-Problems/dp/9814390739 Chapter 2 page 7 – Vaolter Jul 02 '13 at 12:06
  • 2
    Vaulter, if you read closely it says "the width of the boundary layer [...] is the smallest value $\omega$ such that $e^{-\omega/\epsilon} \leq 10^{-9}$", which is $\omega = (9\log 10)\epsilon$. This is indeed $O(\epsilon)$. Similarly for the second inequality. It's not true that all $\omega$ which satisfy $e^{-\omega/\epsilon} \leq 10^{-9}$ are $O(\epsilon)$ (for example take $\omega = (9\log 10)\sqrt{\epsilon}$), but the smallest one, the one you care about, certainly is. – Antonio Vargas Jul 02 '13 at 14:29
  • Problem solved? – Did Jul 02 '13 at 20:26

1 Answers1

1

(I'll post my comment above as an answer since it seems to address what the OP was looking for.)

In the reference given for the source of this question it states

The width of the boundary layer [...] could be taken to be the smallest value $\omega$ such that $e^{-\omega/\epsilon} \leq 10^{-9}$, say. In this case it is clear that $\omega = O(\epsilon)$.

As shown in this related answer, the smallest such $\omega$ is $\omega = (9\log 10)\epsilon$, which is indeed $O(\epsilon)$.

It's not true that all $\omega$ which satisfy $e^{-\omega/\epsilon} \leq 10^{-9}$ are $O(\epsilon)$ (for example take $\omega = (9\log 10)\sqrt{\epsilon}$), but the smallest one, the one you care about, certainly is. Dealing with the other inequality is similar.

Community
  • 1
Antonio Vargas
  • 24,993