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Take $\gamma >{\gamma }_c$ and let $\epsilon \to 0^{+}$ with the other parameters fixed. The nullclines are the cubic polynomial graph $w=f(v)$ and the line $w=\gamma v$. In the phase plane, $v$ increases below the cubic polynomial graph and decreases above it; $w$ increases below the line and decreases above it. The fast ordinary differential equation, in time $s=t/\epsilon$, freezes $w$ and gives $dv/ds=f(v)-w$.

At $w=0$, the fast roots are $0,a,1$. Their linear stability follows from $f^{\prime }(0)=-a<0$, $f^{\prime }(a)=a(1-a)>0$ and $f^{\prime }(1)=a-1<0$. Thus a fixed $0<v^{\ast }<a$ falls towards the resting branch, whereas a fixed $a<v^{\ast }<1$ rises towards $v=1$. If $v^{\ast }>1$, the first fast segment instead settles down towards $1$, followed by the same recovery excursion. The threshold $a$ is the leading singular perturbation threshold; finite-$\epsilon$ corrections matter for initial values approaching $a$ with $\epsilon$.

LetThese are the roots of $f^{\prime }(v)=0$. The outer branches $v<v_L$ and $v>v_R$ of the critical manifold are attracting under fast motion; the middle branch is repelling. The full excursion has four stages.

For this cubic polynomial, the landing point is explicit. Since $f(v)-f(v_R)=-(v-v_R{)}^2(v-v_J)$, comparison of the quadratic coefficients givesThus a graph of $v(t)$ has a rapid rise, a slowly falling elevated plateau, a rapid fall through zero to a negative undershoot, and a slow approach back to zero. The figure shows both the phase plane and a computed time trace. The smaller initial perturbation stays near the resting branch and produces no elevated plateau.

The four-stage description concerns the singular limit. Close to a simple fold point of a critical manifold, balancing the local quadratic activation term with the slow drift gives $\Delta v=O({\epsilon }^{1/3})$ and $\Delta t=O({\epsilon }^{2/3})$ for fixed parameters. These layers shrink compared with the slow phases and do not alter the leading time integrals below.

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.

Take electrostatic perturbations proportional to $e^{i(kz-\omega t)}$, with ions immobile on the fast electron time scale. Linearized Vlasov gives $f_1=-i(q{E}_1/m)(\partial f_0/\partial v_z)/(\omega -k{v}_z)$. Integrating over transverse velocities gives $F_0(v_z)=n_0\sqrt{m/(2\pi k_{B}T)}{e}^{-m{v}_z^2/(2k_{B}T)}$. Combine the perturbed charge density with Poisson's equation:Since $F_0^{\prime }=-m{v}_z{F}_0/(k_{B}T)$, this is preciselyThe pole selects particles moving at phase speed. A causal initial-value prescription specifies how the contour passes it and produces resonant energy exchange, including Landau damping for a decreasing Maxwellian. A bare real improper integral without that prescription is incomplete. The unclosed brace in the printed Maxwellian exponent is typographical; the normalized Gaussian above fixes its meaning.

A time scale is a characteristic duration over which a specified variable or process changes appreciably. Comparing time scales helps choose a dimensionless time and identify a singular perturbation. An exponential relaxation $x-x_{\ast }=A{e}^{-t/\tau }$ has relaxation time constant $\tau$; a nonlinear process can have several distinct time scales.