Simple question on perturbation theory for a function with two small parameters

Question

Suppose I have an analytic function $f(x;\epsilon, \delta)$ that depends on two small parameters that are of the same order, $0 < \epsilon << 1$ and $0< \delta <<1$ . For example in the context of odes, $\dot{x} = f(x; \epsilon, \delta)$, and so expanding $f$ in Taylor series, we have $f(x;\epsilon, \delta) = f(x;0,0) + \epsilon \frac{\partial f}{\partial \epsilon}(x;0,0) + \delta \frac{\partial f}{\partial \delta}(x;0,0) + O(\epsilon \delta)$

In the consequent computations with $f$, and any other related functions that arise that also depend on $\epsilon$ and $\delta$, is it justifiable to denote $$\frac{\epsilon}{\delta} = c$$ for some constant $c$ that is $O(1)$? Or do I need to impose some more restrictions on $\epsilon$, $\delta$?

Answer 1

I'm using the definition of the Landau $\mathcal{O}$ symbols as found in F. Verhulst, Nonlinear Differential Equations and Dynamical Systems (2nd ed), Springer, 2005, page 114:

$\delta_1(\epsilon) = \mathcal{O}(\delta_2(\epsilon))$ as $\epsilon \to 0$ if there exists a constant $k$ such that $\delta_1(\epsilon) \leq k\,\delta_2(\epsilon)$ as $\epsilon \to 0$.

(also see the definition in the Wikipedia article, just replace $x$ with $\frac{1}{\epsilon}$).

From the definition of the $\mathcal{O}$ symbol, you see that '$\epsilon = \mathcal{O}(\delta)$' does not necessarily imply that $\lim_{\epsilon \to 0} \frac{\epsilon}{\delta} = k$. For example, $\delta^2 = \mathcal{O}(\delta)$, as the definition is satisfied for any $k$. However, we can invoke the 'little oh symbol', which is defined as

$\delta_1(\epsilon) = \mathcal{o}(\delta_2(\epsilon))$ as $\epsilon \to 0$ if $\lim_{\epsilon \to 0} \frac{\delta_1(\epsilon)}{\delta_2(\epsilon)} = 0$.

So, if you specify '$\epsilon$ is $\mathcal{O}(\delta)$ but not $\mathcal{o}(\delta)$', by using the above definitions, you can show that there exists a constant $c$ such that $\lim_{\epsilon \to 0} \frac{\epsilon}{\delta} = c$.

To be fair, in practice, the statement '$\epsilon = \mathcal{O}(\delta)$' is understood to include '(but not smaller)', i.e. 'but not $\epsilon = \mathcal{o}(\delta)$'.