Past exam of the mathematical physiology course of the University of Oxford / 2011 / 1 / c

The nullclines are $w=g(v)$ and $v=0$. The two outer pieces of $g$ have slope $-1$; the middle piece has positive slopeThe curve is continuous, with folds $(\alpha ,-\alpha )$ and $(\beta ,1-\beta )$. The unique equilibrium is $(0,-\alpha /(\beta -\alpha ))$. Its Jacobian matrix has trace $\mu /ϵ>0$ and determinant $1/ϵ>0$, so it is unstable. The outer branches attract the fast voltage motion.

Let $w_L=-\alpha$ and $w_R=1-\beta$. The hypothesis $\beta -\alpha <1$ gives $0<w_L<w_R<1$. The singular relaxation oscillation follows this route:

The jumps are almost horizontal in the phase plane, because $w$ changes negligibly on the fast $O(ϵ)$ time. Below $w=g(v)$, voltage increases; above it, voltage decreases; $w$ increases on the right of $v=0$ and decreases on the left. These signs provide a closed trapping region around the indicated route for sufficiently small positive $ϵ$, with the unstable equilibrium excluded, giving a periodic orbit by the planar return-map or Poincaré-Bendixson theorem argument. The continuous piecewise linear field is locally Lipschitz, so solution uniqueness is available at the corners.

On the attracting right branch, the leading fast balance is $w=1-v$, so $\dot{w}=v=1-w$. On the left branch it is $w=-v$, so $\dot{w}=-w$. The slow travel times are thereforeBoth are positive because $w_R>w_L$. The fast jumps contribute vanishing time as $ϵ\to 0$, giving the piecewise linear relaxation-oscillator periodFor the illustrated $\alpha =-0.25$, $\beta =0.4$, the leading period is $1.5041$ in the given nondimensional time units.