Past exam of the mathematical physiology course of the University of Oxford / 2011 (C8.1b, DMAC 2B81, DMCO 2B81, DMSU 2B81)

In the Hodgkin-Huxley model, $V$ is the membrane potential, the voltage inside the axon relative to outside. The space clamp assumption makes it spatially uniform. The gating variables lie between zero and one: $m$ describes sodium activation, $h$ describes availability against sodium inactivation, and $n$ describes potassium activation. Thus large $h$ means channels remain available, not that inactivation is strong.

The left side $C_m\dot{V}$ is capacitive electric current per unit membrane area. The imposed $I$ is inward-positive applied electric current. The other terms are outward-positive ionic electric currents: sodium with electrical conductance $g_{Na}{m}^{3}h$, potassium with conductance $g_K{n}^4$, and a passive leak with conductance $g_L$. Each current is its conductance times the difference between $V$ and the corresponding reversal potential $V_{Na}$, $V_K$ or $V_L$. The coefficients $g_{Na},g_K$ are maximal conductances; the gate factors represent independent permissive gates in an ion channel.

For each gate, $k_{\infty }(V)$ is the equilibrium opening/availability fraction at fixed voltage and ${\tau }_k(V)>0$ its exponential relaxation time. The equation follows from two-state ion-channel gating: if $\dot{k}={\alpha }_k(V)(1-k)-{\beta }_k(V)k$, thenVoltage feedback couples these otherwise elementary gating equations to the membrane potential and produces action potentials.

Write $M(v)=m^3(v)$. In the physically relevant strip $-v_K<v<1$, $0<n<H$, the zero-current nullcline satisfiesThe left side increases strictly from zero to infinity, so there is exactly one nullcline height $n=N(v)$ for each interior voltage. If $M$ is positive at the endpoints, its curve joins $(-v_K,H)$ to $(1,0)$; when low-voltage activation is idealized as exactly zero, the limiting left branch can lie on $v=-v_K$.

The signs of the derivatives explain its folds:At low voltage, potassium's approximately linear contribution dominates and $g_v>0$. During rapid sodium activation the last negative term can dominate, giving $g_v<0$. Once activation saturates, $M^{\prime }$ becomes small and $g_v>0$ again. This folded conductance-model nullcline has the familiar shape: the curve has descending outer branches and an ascending middle branch in the $(v,n)$ plane. Equivalently, an intermediate fixed $n$ can give three zeros of $g$ as voltage rises, with positive, negative and positive $g_v$ at those crossings.

The rapid-activation description permits this folded shape; exact fold positions depend on $M,{\gamma }_K,H,v_K$. The figure supplies an explicit admissible sigmoid example. Its right panel shows $g=0$ as well as the recovery nullcline and an oscillating trajectory.

The voltage nullcline is $g(v,n)=I^{\ast }$ and the recovery nullcline is $n=n_{\infty }(v)$. In the relevant region $g_n>0$, so below the voltage curve $\dot{v}>0$ and above it $\dot{v}<0$. Below the recovery curve $\dot{n}>0$, while above it $\dot{n}<0$.

For small $ϵ$, the outer voltage branches with $g_v>0$ attract fast voltage motion; the middle branch with $g_v<0$ repels it. Arrange the intersections so the equilibrium is on that middle branch, with recovery increasing along the high-voltage attracting branch and decreasing along the low-voltage attracting branch. The trajectory then drifts towards a fold, jumps rapidly to the other attracting branch at nearly constant $n$, and repeats. This is a relaxation oscillation.

The local linear stability calculation supports the construction. At an equilibrium,A sufficiently steep recovery curve gives positive determinant, while $g_v<-ϵ$ gives positive trace. Hence the equilibrium is a repeller. The bounded physical rectangle can be chosen forward invariant, and a small neighborhood of the repeller removed to give a trapping annulus. The Poincaré-Bendixson theorem then supplies a periodic orbit. These conditions can be realized because the recovery function and the applied current control the intersection and its slope; oscillation is possible, not forced for every choice.

For the illustration, ${\gamma }_K=0.1$, $H=0.8$, $v_K=0.2$, $ϵ=0.01$, $I^{\ast }=0$, $M(v)=(1+e^{-25(v-0.2)}{)}^{-1}$ and $n_{\infty }(v)=0.8/(1+e^{-20v})$. The rectangle $[-0.2,1]\times [0,0.8]$ has inward-pointing flow, and the displayed numerical trajectory realizes the folded-branch mechanism.

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.

Let $p_a,p_v,p_{LV}$ be pressures in the arterial reservoir, venous reservoir and left ventricle. Constant vascular compliances $C_a,C_v$ relate volume changes to pressure changes. The ventricular elastance relation iswhere $V_0$ is the unstressed ventricular volume and $E_{LV}(t)>0$ is the time-dependent inverse compliance. Define the positive part $[x{]}_{+}=\max (x,0)$. The systemic, aortic-valve and mitral-valve flows areEach $R$ is a positive hydraulic resistance. The aortic valve permits outflow only when ventricular pressure exceeds arterial pressure, and the mitral valve permits filling only when venous pressure exceeds ventricular pressure.

Mass conservation, with blood treated as incompressible, givesSubstitution and multiplication of the first two equations by $R_c$ give exactlyAdding the three original balances gives zero total volume change. This is a three-compartment ventricular circulation model.

During isovolumetric contraction, ventricular pressure lies between the venous and arterial pressures, $p_v<p_{LV}<p_a$. Both valves are closed, so $Q_v=Q_a=0$ and ventricular volume remains at its end-filling value $V^{+}$. ConsequentlyHere $V^{+}$ is end-diastolic ventricular volume and $V_0$ is the zero-pressure/unstressed volume.

The arterial and venous reservoirs still exchange systemic blood:If contraction is short compared with both $R_c{C}_a$ and $R_c{C}_v$, the fractional changes are small andwhere the superscript values refer to the beginning of contraction. Thus pressure can rise rapidly in the left ventricle through increasing elastance while the other pressures are approximately fixed. This stage ends when $p_{LV}$ reaches $p_a$ and the aortic valve opens.

During ejection the mitral valve is closed, so $Q_v=0$. Adding the arterial and ventricular balances cancels aortic flow:where the last approximation assumes $p_v\ll p_a$.

Model the rapid contraction as a switch to approximately constant systolic elastance $E_S$, with $C_s=1/E_S$ the systolic ventricular compliance. A sufficiently small aortic-valve resistance makes $p_{LV}\simeq p_a$ after a short pressure-equalization transient. The systolic pressure must be high enough to open the valve, $E_S(V^{+}-V_0)>p_a^{+}$. During that transient neglect systemic runoff. Conservation of combined arterial and stressed ventricular volume givessoThis is an approximation across a rapid transient, not a discontinuity at finite valve resistance. The subsequent common pressure obeys $R_c(C_a+C_s){\dot{p}}_a\simeq -p_a$, givingThe source's $R_C$ in this expression denotes the same systemic resistance $R_c$.

At the end of firing, the remaining ventricular volume is $V_{\mathrm{end}}\simeq V_0+C_s{p}_a(\Delta t_F)$. The stroke volume is $V^{+}-V_{\mathrm{end}}$, henceThe assumptions are a closed mitral valve throughout ejection, negligible venous pressure, rapid valve equalization relative to systemic runoff, and nearly constant systolic elastance during the interval measured by $\Delta t_F$. These give the ventricular ejection pressure and stroke-volume approximation.

For a constant solution $u(t)\equiv U$, its delayed value is also $U$. The equation becomes $0=U(1-U)$, so the only constant equilibria are $\boxed{U=0\text{ and }U=1}$. The delay differential equation requires the same constant history to realize each equilibrium.

A positive solution stays positive, sinceTherefore the sign of $\dot{u}$ is exactly the sign of $1-u(t-\tau )$. Initially the given history below one makes $\dot{u}>0$. If $u$ crosses one upwards at time $t_0$, it continues growing until $t_0+\tau$, because the feedback still sees earlier values below one. It reaches a maximum when the delayed value crosses one. Similarly, after a downward crossing at $t_1$, it continues decreasing until $t_1+\tau$ and reaches a minimum then.

This delayed reversal makes overshoots and undershoots possible. In a derivative sketch, positive lobes persist until one delay after upward crossings, followed by negative lobes persisting until one delay after downward crossings. Sustained oscillation depends on the delay; a short delay can instead give convergence to one. The example shown has $\tau =2>\pi /2$ and a history identically $1/2$.

Put $u=1+x$. Then $\dot{x}=-x(t-\tau )-x(t)x(t-\tau )$, so the linearization is $\dot{x}=-x(t-\tau )$. A mode $e^{\lambda t}$ gives the characteristic equation of a delay differential equationFor real $\lambda \ge 0$, the left side is nonnegative and the right side negative. The two real graphs therefore never meet there, as drawn in the figure. There is no nonoscillatory exponential instability. If stability is lost, the unstable eigenvalues must be a nonreal conjugate pair, producing oscillatory growth; absence of a positive real root alone does not imply stability.

For a purely imaginary root $\lambda =i\omega$ with $\omega >0$, separating real and imaginary parts givesThus $\omega =1$ and $\tau =\pi /2+2\pi m$, $m\ge 0$. The first crossing isFor $0<\tau <\pi /2$ there is no root with nonnegative real part. Indeed, write $\lambda =a+ib$ with $a\ge 0$. The real equation forces $\cos (b\tau )\le 0$, hence $|b|\tau \ge \pi /2$. But the imaginary equation gives $|b|\le e^{-a\tau }\le 1$, contradicting $\tau <\pi /2$.

Differentiate the characteristic equation along the root:The imaginary roots are simple because $1+i{\tau }_c\ne 0$, and they cross into the unstable half-plane as delay increases.

One can also check the nonlinear Hopf bifurcation directly. With $\theta =\omega t$ and small amplitude $a$, substituteThe constant and second harmonics at order $a^2$ give the displayed coefficients. Put $d=\mu +{\tau }_c\nu$. At order $a^3$, absence of resonant first harmonics requires $d+1/20=0$ and $\nu +3/20=0$. HenceThe nonzero coefficient verifies a branch of small periodic solutions for $\tau >{\tau }_c$, consistent with the transverse Hopf bifurcation in the delayed logistic equation. This is the delayed logistic Hopf threshold.

At the crossing the angular frequency is ${\omega }_c=1$, so the limiting small-amplitude period isThis is the period of the bifurcating infinitesimal oscillation. Finite-amplitude oscillations away from the threshold need not have period exactly $4\tau$.