I am reading Bender and Orszag's book page 336, example 2.
I pasted the problem below:
I am wondering why they chose $\epsilon^{-1/5}\ll x\ll\epsilon^{-1/4}$ as the overlapping region. Then they said that it is not the only possible choice, they could choose $\epsilon^{-1/6}\ll x\ll\epsilon^{-1/5}$ or even $\epsilon^{-1/99}\ll x\ll\epsilon^{-32/99}$ as long as they satisfy the general matching criterion that x lie in the asymptotic interval $1\ll x\ll\epsilon^{-1/3}$.
I really have no clue about the general matching criterion that they talked about. I am wondering whether anyone could provide some explanation of why $1\ll x\ll\epsilon^{-1/3}$ is the largest possible matching region.
Many thanks!
I am no expert at regions for asymptotic matching of DEs, but let me see if I can guide to where they got that.
We have:
$$\displaystyle \tag 1 y_L = e^{-x + \frac{1}{x}}$$
$$\displaystyle \tag 2 y_R = a e^{-\epsilon \frac{x^3}{3} - x}$$
Using a Taylor series expansion for $(1)$ with $x$ large, we have:
$$\displaystyle \tag 3 y_L = e^{-x + \frac{1}{x}} \approx e^{-x}\left( 1 + \dfrac{1}{x} + O\left(\dfrac{1}{x^2}\right) \right)$$
Using a Taylor series expansion for $(2)$ with $x$ small, we have:
$$\displaystyle \tag 4 y_R = a e^{-\epsilon \frac{x^3}{3} - x} \approx a e^{-x}\left( 1 - \epsilon \dfrac{x^3}{3} + O\left(\epsilon^2 \dfrac{x^6}{18} \right) \right)$$
Now, we are interested in a region where $y_L = y_R$, so lets equate them and see what we get:
$$\displaystyle \tag 5 e^{-x}\left( 1 + \dfrac{1}{x}\right) = a e^{-x}\left( 1 - \epsilon \dfrac{x^3}{3} \right)$$
On the LHS of $(5)$, what if we have $x \gg 1$ ($x$ large), then we are left with $e^{-x}$.
On the RHS of $(5)$, what if we have $x \ll \epsilon^{-1/3}$ ($x$ small), then we are left with $a e^{-x}$.
What do we need for $a$ to be? Well $a = 1$ will do the trick and that makes the LHS = RHS.
Thus, we have as our asymptotic interval:
$$\tag 6 1 \ll x \ll \epsilon^{-1/3}, a = 1$$
With $(6)$ we have a very wide range of choice for $x$ as the authors describe. As long as we choose values in that region, we are okay. As for the particular values they chose, it might make the solution curves look more appealing to the eyes is as a good a guess as any. I think in these perturbation style problems, it really helps to do what-if scenarios and experiment to see the impacts of varying $x$ and $\epsilon$ in the original DEQ.