differentiation - How to pick a boundary layer coordinate or stretching transformation

I am following Introduction to Perturbation Methods by Holmes and am unsure how I to pick the power in my boundary layer coordinate if my governing equation is the Laplace equation given by \begin{equation} \nabla^2 f(r,\theta) = f_{rr} + \dfrac{1}{r}f_r+\dfrac{1}{r^2}f_{\theta\theta} = 0 \end{equation}

If I expand my function as $f = f_0 + \epsilon f_1 + ...$ and use the boundary layer coordinate given by \begin{equation} \bar{r} = \dfrac{r}{\epsilon^\alpha}\\ \dfrac{\partial}{\partial r} = \dfrac{1}{\epsilon^\alpha}\dfrac{\partial}{\partial \bar{r}} \end{equation} then I obtain \begin{equation} \nabla^2 f(\bar{r},\theta) = \left[\dfrac{1}{\epsilon^{2\alpha}}(f_{0,\bar{r}\bar{r}}+ \epsilon f_{1,\bar{r}\bar{r}} + ...)\right] + \left[\dfrac{1}{\bar{r}\epsilon^{2\alpha}}(f_{0,\bar{r}}+ \epsilon f_{1,\bar{r}} + ...)\right]+\left[\dfrac{1}{\bar{r}^2\epsilon^{2\alpha}}(f_{0,\theta\theta}+ \epsilon f_{0,\theta\theta}+...)\right] = 0 \end{equation}

From what I understand I want pick $\alpha$ so that each of the terms in the square brackets are the same order of magnitude. However in this case it doesn't seem to matter what $\alpha$ is. Could I just pick $\alpha = 0$ to make my life easy?