= Solution
In a relaxation cycle, cytoplasmic <calcium> rises slowly on the left attracting branch while total <calcium> is accumulated, then rises rapidly during store discharge along a constant-$w$ <total-calcium fast fibres>[fast path] to the right branch. It subsequently decreases on the right slow branch as extrusion dominates input, and finally drops rapidly during store uptake back to the left branch. The result is \b[a repeated pulse with slow sections joined by sharp transitions], rather than a sinusoid except near a small-amplitude <Hopf bifurcation>.
For explicit <time> labelling, the compatible scaling $u=c/K_1$, $v=c_s/K_2$ gives $\tau=kt$, $\gamma=K_2/K_1$, $\epsilon=kK_2/V_2$, $\mu=r/(kK_1)$, $\beta=V_1/V_2$ and $\delta=k_sK_2/V_2$. Hence an interval $\Delta\tau$ corresponds to physical duration $\Delta t=\Delta\tau/k$. Slow-branch <times> can be computed by integrating
$$
\Delta\tau=\int\frac{1+\gamma V'(u)}{\mu-u}\,du
$$
in the direction of motion; fast-jump durations are omitted at leading order. The plot uses that <quadrature>, with vertical lines representing the limiting rapid transitions. Finite $\epsilon$ rounds those lines and changes the precise <amplitude> and <period>, especially near folds.
\Image[/past-exam-of-the-mathematics-and-the-environment-and-mathematical-physiology-course-of-the-university-of-oxford/2009-calcium-oscillation.png]
{title=Leading relaxation waveform of cytoplasmic calcium, with physical time measured in inverse-extrusion-rate units}
Back to article page
