Suppose I have an ODE in cylindrical coordinates
$$\dot{r} = g_r(r,z) + \epsilon f_r(r,z,\theta), \quad r(0) = r_0$$ $$\dot{z} = g_z(r,z) + \epsilon f_z(r,z,\theta), \quad z(0) = z_0$$ $$\dot{\theta} = \frac{b(r,z)}{\epsilon} + g_\theta (r,z) + \epsilon f_\theta(r,z,\theta), \quad \theta(0) = \theta_0$$
such that $||f_i||_{L^\infty}, ||g_i||_{L^\infty}$ and $||b||_{L^\infty} <\infty$ for $i\in \{r,\theta,z\}$ and $b >0$ . Furthermore suppose I know that for any $\epsilon \in (0,1]$, I now that the solutions $r(t)$ and $z(t)$ are uniformly bounded with respect to $\epsilon$. Is there any way to justify that for any $t\in [0,T]$, with our solution defined on $[0,T]$, our solution functionally has the form
$$r(t;r_0,z_0,{\color{red}{\theta_0}};\epsilon) = R_1(t,r_0,z_0,\epsilon) + \epsilon R_2(t,r_0,z_0, {\color{red}{\theta_0}},\epsilon ) +\mathcal{O}(\epsilon^2)\\ z(t;r_0,z_0,{\color{red}{\theta_0}};\epsilon) = Z_1(t,r_0,z_0,\epsilon ) + \epsilon Z_2(t,r_0,z_0, {\color{red}{\theta_0}},\epsilon) +\mathcal{O}(\epsilon^2) \\ \theta(t;r_0,z_0,{\color{red}{\theta_0}};\epsilon) = \frac{1}{\epsilon}\Theta_1(t,r_0,z_0,\epsilon) +{\color{red}{\theta_0}} + \Theta_2(t,r_0,z_0,\epsilon) +\epsilon \Theta_3(t,r_0,z_0,{\color{red}{\theta_0}},\epsilon) +\mathcal{O}(\epsilon^2)$$
Where the functions $R_i$, $Z_i$ and $\Theta_i$ are uniformly bounded with respect the $\epsilon$. In other words I have suppressed the dependence of the initial condition $\theta_0$ of our solution into a constant plus something small of size $\epsilon$. I suspect this should be the case since our vector field only has small dependence on $\theta$.