= Solution
For <Hopf thresholds in a two-pool calcium model>, with the unique <equilibrium> parametrized by $u=\mu$, locate changes of stability by solving the two <equations>
$$
\boxed{f(\mu,v)=0,\qquad f_v(\mu,v)-\gamma f_u(\mu,v)=\epsilon.}
$$
If there are two simple roots $\mu_-<\mu_+$ with positive <matrix trace>[trace] between them, the <equilibrium> is a <unstable equilibrium>[repeller] throughout that interval, since its <determinant> stays positive. At a nondegenerate crossing a <Hopf bifurcation> is possible, with linear <frequency> $[-f_v/\epsilon]^{1/2}$. A complete classification of criticality requires a nonlinear coefficient, not merely the <matrix trace>[trace] calculation. At leading singular order the thresholds approach the folds determined by $f=0$ and $f_v-\gamma f_u=0$, equivalently $V'=-1/\gamma$.
Periodic solutions are also supported globally in the positive quadrant. The axes point inward for positive input, $v'\le(\beta-\delta v)/\epsilon$ bounds $v$, and the $u$ <equation> then bounds $u$ by a linear restoring term plus a bounded source. A large rectangle can thus be chosen <invariant set>[forward invariant]. For an initial point distinct from the <unstable equilibrium>[repelling] unique <equilibrium>, its <omega-limit set> cannot contain that <equilibrium> or a boundary <equilibrium>. The <Poincaré-Bendixson theorem> supplies a <periodic orbit>. In the strong separation regime it follows the two attracting branches and jumps between them, a <relaxation oscillation>.
The interval requires the <maximum> of $f_v-\gamma f_u$ along the <critical manifold> to exceed $\epsilon$; the listed scale hierarchy alone is not a substitute for that test for arbitrary remaining constants. As an explicit example, $\beta=0.1$, $\delta=10^{-5}$, $\epsilon=5\times10^{-4}$ and $\gamma=0.1$ give approximately $\mu_-=0.22618$ and $\mu_+=0.37350$. At $\mu=0.3$ the <matrix trace>[trace] is positive, whereas at $0.03$ and $1$ it is negative. The figures use this example and label their singular-limit portions accordingly.
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