= Solution
On the descending part of the graph, mark the two tangency points where
$$
\boxed{g'(\mu_-)=-1/\gamma=g'(\mu_+),\qquad \mu_-<\mu_+.}
$$
Choose the crossings that enclose the portion with $g'(u)<-1/\gamma$. For fixed $\mu$ strictly between these values and sufficiently small $\varepsilon$, the unique <equilibrium> is a <repelling equilibrium> by part (e). The endpoints here are leading <singular limit> values; at finite $\varepsilon$ the boundaries instead solve $f_v-\gamma f_u=\varepsilon$ and approach these tangencies when the crossings are nondegenerate.
There is a necessary qualification: such an interval exists only when $\gamma$ is large enough that the descending graph actually attains slopes below $-1/\gamma$. If $M=\max_{u>0}[-g'(u)]$, one needs $\gamma M>1$. The supplied graph has a very steep descending section, but an arbitrarily small positive $\gamma$ need not satisfy this condition. For $\delta=0.001$, numerical evaluation gives $M\simeq511.49$, so the leading threshold is $\gamma\simeq0.0019551$. A graph alone does not remove this parameter restriction.
The fast motion is especially informative. In fast time $s=\tau/\varepsilon$,
$$
\frac{du}{ds}=\varepsilon(\mu-u)-\gamma f(u,v),\qquad
\frac{dv}{ds}=f(u,v).
$$
At leading order $z=u+\gamma v$ is constant, so fast paths are straight lines with slope $dv/du=-1/\gamma$. Along such a line, the fast <eigenvalue> at the <nullcline> is
$$
f_v-\gamma f_u=f_v\bigl(1+\gamma g'(u)\bigr).
$$
It is negative on the two outer attracting segments and positive on the intervening repelling segment. The tangencies defining $\mu_\pm$ are consequently the <fold points of a critical manifold> when it is expressed in the coordinates $(z,v)$. They are not generally the maximum and minimum of $g$ itself.
For $\mu_-<\mu<\mu_+$, slow motion on an attracting segment follows
$$
\boxed{(1+\gamma g'(u))\dot u=\mu-u.}
$$
On the left attracting segment $u<\mu_-$ this moves towards increasing $u$; on the right attracting segment $u>\mu_+$ it moves towards decreasing $u$. Each segment ends at a tangency and is followed by a fast jump along a line of slope $-1/\gamma$ to the other attracting segment. This constructs the limiting <relaxation oscillation>. The figure uses $\gamma=0.01$, for which $\mu_-\simeq0.62970$ and $\mu_+\simeq1.01157$; the marked endpoints change with $\gamma$.
\Image[/past-exam-of-the-mathematical-physiology-course-of-the-university-of-oxford/2023-calcium-nullcline.png]
{title=Calcium nullcline, slope tangencies and the limiting relaxation cycle}
To justify existence of a <periodic orbit> of the full planar system, rather than just draw a formal cycle, first establish a compact <trapping region>. The positive quadrant is forward invariant: at $u=0$, $\dot u=\mu+\gamma\delta v/\varepsilon>0$, and at $v=0$, $\dot v=u^2/[\varepsilon(1+u^2)]\ge0$. Moreover
$$
\dot v\le\frac{1-\delta v}{\varepsilon}.
$$
Choose $V>1/\delta$ above the initial store value. Since $f(u,v)\ge-1-\delta V$ when $0\le v\le V$,
$$
\dot u\le\mu-u+\frac\gamma\varepsilon(1+\delta V).
$$
Thus the rectangle $0\le u\le U$, $0\le v\le V$ is a <positively invariant set> if $U>\mu+\gamma(1+\delta V)/\varepsilon$ and the initial point lies inside it. Every forward solution is bounded. There are no <equilibria> on its axes for $\mu>0$, and its sole interior <equilibrium> is a <hyperbolic equilibrium point> with both <eigenvalues> having positive real parts in the strict unstable range. A sufficiently small neighborhood of this <repelling equilibrium> has outward flow; deleting it leaves a compact <trapping region> with no <equilibria>. The <Poincaré-Bendixson theorem> then supplies a <periodic orbit>.
Hence \b[<periodic solutions> exist throughout the strict unstable interval], interpreted with the finite-$\varepsilon$ correction above. This argument establishes existence, without asserting uniqueness, the criticality of possible endpoint <Hopf bifurcations>, or absence of coexisting <periodic orbits> outside the interval.
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