 Solution

The requested second phase is the second slow recovery phase, beginning after the downward fast jump. Its starting point is $v_3=v_J<0$, fixed by $f(v_J)=f(v_R)$, with the negative root chosen. Along it the same chain rule givesThe approach is to zero from below. Thus, if the printed cutoff $\beta$ is a signed value, it must satisfy $v_J<\beta <0$ and $v_4=\beta$. If $\beta$ denotes a small positive magnitude, the endpoint is instead $v_4=-\beta$. To express both conventions unambiguously, put $b=|\beta |>0$ and writeThere is no positive small endpoint on this leading slow branch. A positive cutoff would require a different event or finite-$\epsilon$ oscillatory recovery.

For bounded negative $v$ and $\gamma \gg 1$,Integration therefore gives the useful more detailed approximationTaking also $b\ll 1$, the leading logarithm isIndeed near zero, $f(v)=-av+O(v^2)$ and the slow equation is $\dot{v}=-(\gamma +a)v/a+O(v^2)$, so at fixed $\gamma$ the exact leading logarithmic coefficient is $a/(\gamma +a)$. Its large-$\gamma$ limit is $a/\gamma$, agreeing with the integral.

The order of approximation matters: first take the singular limit at fixed $\gamma$, then the large-$\gamma$ and small-cutoff limits. For simultaneous limits, a sufficient separation near rest is $\epsilon \gamma \ll a^2$. Otherwise the fast and recovery time scales merge, and these slow-branch formulas need not describe the late-time motion.