 Solution

At every equilibrium the store balance gives $f=0$, and adding $\gamma$ times that balance to the cytoplasmic one gives $u=\mu$. The unique-root result therefore gives a single positive equilibrium $(\mu ,V(\mu ))$. Its Jacobian iswith derivatives evaluated at that equilibrium. At small $\mu$, $f_u>0$ and the trace is negative; sufficiently large $\mu$ also has $f_u>0$, since $J^{\prime }$ is of order $u^{-3}$ and $K^{\prime }$ of order $u^{-5}$. Thus both extreme regimes have a linearly stable unique equilibrium.

The fast variable is not $v$ alone. Using total-calcium fast fibres, introduce total scaled calcium $w=u+\gamma v$. Then $\dot{w}=\mu -u$, while on fast time $\sigma =\tau /ϵ$ the leading equations give $du/d\sigma =-\gamma f$, $dv/d\sigma =f$ and $w$ constant. Fast paths are therefore straight lines of slope $-1/\gamma$ in the $(u,v)$ plane. Attracting portions of the critical manifold satisfy $f_v-\gamma f_u<0$, rather than merely $f_v<0$. Along $v=V(u)$, slow motion obeysStarting with large $u$ and $v=0$, $f>0$ drives a rapid move upward and left, loading the store until the trajectory approaches the right attracting branch. For large $\mu$ in the right stable regime, slow motion on that branch carries it toward $u=\mu$; the direction reverses appropriately if the starting value lies below that equilibrium. For sufficiently small $\mu$, slow motion travels left along the right branch until its fold, then a fast approximately constant-$w$ jump carries the trajectory to the left attracting branch. Subsequent slow motion reaches the small-$u$ equilibrium. At a fold the slow approximation is singular, so the full equations resolve the transition. At very small $u$ an additional loss of time-scale separation is possible because the rates themselves vanish; the fast/slow drawing describes the separated portions, not a uniform estimate arbitrarily near the origin.