separating the quasi-integrable and hard chaos regions. Of course, this phase boundary depends on the precise value of the cutoff $\lambda_c$ we choose. Nevertheless, we checked that on varying the cutoff $\lambda_c$, the phase boundary shifts only slightly and furthermore, the shape of the boundary remains qualitatively the same. Indeed, to a high degree of precision, the phase boundary can be fitted by $$\label{eq:eq18}
{\tilde\omega_c} = C(\lambda_c)/\epsilon,$$ where $C(\lambda_c)$ is a constant which depends on the value of the cutoff. Figure 1 shows the phase diagram for a cutoff $\lambda_c=1$.

We observe from the classical Poincaré sections that there is a symmetry line in the $(r,p_r)$ plane. This arises from the time-reversal invariance present in the problem as follows. Consider a particle kicked at $\tau=0$. The position $r_0$ remains unchanged, while the momentum changes : $p_r^{(+)} = p_r^{(-)} + \eta\ r_0$. Denoting $p_r^{+}$ by $p_0$, then at time $0^{(-)}$ the particle had momentum $p_r^{(-)} = p_0 -\eta\ 
r_0$. Taking into account the fact that the angular momentum is conserved, we see that propagating a particle [*forward*]{} in time from $(r_0,p_0)$ is the same as propagating it [*backward*]{} from $(r_0,\eta\ r_0-p_0)$. Thus, the motion is symmetric about the line $p_r = 