Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2007 / B (Mathematical physiology)

For an experiment beginning by mixing unbound enzyme with substrate, take $\boxed{[E](0)=e_0>0,[S](0)=s_0>0,[X](0)=[Y](0)=[E_i](0)=[P](0)=0}$. The enzyme-substrate complexes, inactivated enzyme and product are initially absent. Nonzero pre-existing product could be allowed without changing the dynamics here, since it has no feedback reaction in the specified scheme.

Write concentrations as $E,S,X,Y,E_i,P$ in this calculation. The law of mass action assigns the forward bimolecular rate $k_{1}ES$ and the first-order rates $k_{-1}X,k_{2}X,k_{3}Y,k_{4}Y$. Each species gains the rates of reactions producing it and loses those consuming it:Adding the equations for all enzyme-bearing species cancels every reaction rate, so $\boxed{E+X+Y+E_i=e_0}$. This is conservation of the total enzyme-derived material, including the inactivated form. It does not assert conservation of active enzyme: mechanism-based inhibition transfers some of that conserved material irreversibly into $E_i$.

The conserved total is $e_0=[E](0)$ under the chosen initial conditions. Substitute $E=e_0-X-Y-E_i$ into the equations from the previous part. This closes the subsystem for $S,X,Y,E_i$; product can be recovered afterward by integrating $\dot{P}=k_{3}Y$.

Let $k_{\Sigma }=k_{-1}+k_2$ and $K_m=k_{\Sigma }/k_1$. The required nondimensionalization is completed byIndeed $X=e_0\sigma x/(1+\sigma )$ and the free-enzyme fraction is $1-\sigma x/(1+\sigma )-y-e_i$. Substitution and division by the indicated time scale giveFor example $k_2/k_{\Sigma }=1/(1+\rho )$ explains the coefficient in the third equation. The small multiplier on the intermediate-species derivatives identifies their potentially fast adjustment relative to substrate depletion.

For quasi-steady enzyme kinetics at low substrate saturation, the quasi-steady-state approximation is controlled here by $\epsilon =e_0/(s_0+K_m)$, rather than by $e_0/s_0$ alone. Even with $e_0\approx s_0$, taking $K_m\gg s_0$ makes $\epsilon \ll 1$ and $\sigma =s_0/K_m=O(\epsilon )$.

After the initial fast layer $T=O(\epsilon )$, the third equation givesSince $\varphi =O(\epsilon )$, the last equation gives $(e_i{)}_T=(\varphi /\epsilon )y=O(\epsilon )$. Starting from $e_i(0)=0$, this implies $e_i=O(\epsilon )$ on bounded intervals of the slow time. The quasi-steady second equation then gives $x=s+O(\epsilon )$. In the first equation,Matching to the initial substrate value $s(0)=1$ therefore yields $\boxed{s(T)\sim e^{-T/(1+\rho )}}$. The relation $x\sim s$ is an outer approximation: it does not replace the initial layer needed to reconcile $x(0)=0$ with $s(0)=1$. Likewise the smallness of accumulated inactivated enzyme has been justified for bounded slow times, not arbitrarily long times.

For the travelling wave coordinate $\xi =x+st$, the chain rule givesFast-slow matching for a calcium pulse joins two fast transitions, $B\to C$ and $D\to A$, joined by slow portions $C\to D$ and $A\to B$. Write the nullcline as $V=g(U)$. The resting point has $U_B=\mu$, $V_B=g(\mu )$ in the leading matched description.

Put $\xi =\epsilon X$. Multiplying the first equation by $\epsilon$ gives $U_{XX}=s(U_X+\gamma V_X)+\epsilon (U-\mu )$. At leading order, integrate and use $U_X\to 0$, $(U,V)\to (U_B,V_B)$ at the start of the front. Together with the second equation this givesThe fast equilibria are intersections of $V=g(U)$ and $V=V_B-(U-U_B)/\gamma$. As the slope $-1/\gamma$ changes, there may be one or three simple intersections, with two distinct intersections at a tangency. At a simple point of the nullcline, $f_U=-f_V{g}^{\prime }$, and the sign convention above/below it gives $f_V<0$. The Jacobian matrix of the fast system hasFor the usual three-intersection configuration, the two outer points have $1+\gamma g^{\prime }>0$, hence negative determinant and one positive and one negative eigenvalue: they are saddle points. The middle point has positive determinant; it is linearly stable if $s^2<-f_V$ and unstable if $s^2>-f_V$. Zero trace is a possible oscillatory stability threshold, with a nonlinear classification requiring further information about $f$.

To select the front speed, follow the appropriate branch of the unstable manifold of $B$ and vary $s$ until it meets the stable manifold of $C$. This is a shooting method for a heteroclinic orbit. Follow the high-$U$ slow branch from $C$ to $D$; the falling front obeys the same fast equations with the integration constant anchored at $D$. Choose $D$ so that its descending saddle connection reaches a low-$U$ point $A$ at that same speed. The remaining slow branch returns toward the resting point, completing the matched spike. The construction requires the indicated connections; a nullcline sketch alone does not specify their numerical speed or prove their existence for every possible kinetic function $f$.

On the slow portions, $f(U,V)=0$ at leading order, so $V=g(U)$ and $V^{\prime }=g^{\prime }(U)U^{\prime }$. Neglecting the small diffusion term yieldsOn the outer rising branches the coefficient on the left is positive. Recovery from $C$ toward $D$ requires decreasing $U$, hence $\mu <U_D$; the low branch from $A$ back toward rest requires increasing $U$, hence $U_A<\mu$. Therefore $\boxed{U_A<\mu <U_D}$. These inequalities distinguish the recovery branches from a trajectory that would move away from the required matching points.

Systole is ventricular contraction and ejection; diastole includes ventricular relaxation and filling. Let $V_a,V_v,V_{LV}$ denote arterial, venous and left-ventricular blood volumes. Let $Q^{+}$ flow through the aortic valve into the arteries, $Q_c$ through the peripheral circulation, and $Q^{-}$ through the mitral valve into the ventricle. Conservation of volume givesThe same $Q_c$ leaves arteries and enters veins because the capillary compartment has no changing stored volume in this approximation. Adding the equations conserves total modeled blood volume.

Use constant vascular compliances $C_a,C_v$, with ${\dot{V}}_a=C_a{\dot{p}}_a$ and ${\dot{V}}_v=C_v{\dot{p}}_v$, and ventricular elastance $p_{LV}=E_{LV}(t)(V_{LV}-V_0)$. The hydraulic resistances and ideal one-way valves givewhere $[z{]}^{+}=\max (z,0)$. Substitution into the conservation equations gives the two stated arterial/venous pressure equations. Since $V_{LV}-V_0=p_{LV}/E_{LV}$, differentiating this identity gives the third equation. The positive parts encode valve closure against a reverse pressure difference.

During isovolumetric contraction both valves are closed. Let $V^{+}$ be the end-diastolic ventricular volume, $V_0$ its unstressed reference volume, and $p_a^{+}$ arterial pressure before ejection. The rapidly raised elastance generates $E_s(V^{+}-V_0)$, so the aortic valve opens if $\boxed{E_s(V^{+}-V_0)>p_a^{+}}$.

During ejection, $R_a/R_c\ll 1$ makes the open-valve pressure drop small on the peripheral drainage time scale, so $p_{LV}\approx p_a$. The mitral valve is closed, and adding arterial and ventricular volume balances eliminates $Q^{+}$. With $p_v\ll p_a$ this givesFor approximately constant systolic elastance, set $C_s=1/E_s$ and reset time to the start of the slow ejection phase. ThenHere ${\hat{p}}_a$ is the matched pressure at that start. If peripheral outflow is negligible during the preceding rapid artery-ventricle pressure equilibration, conservation gives the more explicit ${\hat{p}}_a=[C_a{p}_a^{+}+V^{+}-V_0]/(C_a+C_s)$.

After aortic closure the ventricular volume is fixed at the end-systolic value $V^{-}$. Relaxation decreases $E_{LV}$ until the mitral valve opens at $p_{LV}=p_v$, hence $\boxed{E_{LV}\sim p_v/(V^{-}-V_0)}$, with $p_v$ the venous filling pressure at opening. During filling the aortic valve is closed, so $R_c{C}_a{\dot{p}}_a=-(p_a-p_v)\approx -p_a$. With $p_a^{-}$ the arterial pressure at the start of filling and time reset there,The change from $C_a+C_s$ to $C_a$ is physical: the ejecting ventricle shares the arterial pressure during systole, whereas a closed aortic valve disconnects it during filling.

In this delayed carbon-dioxide ventilation feedback model, $p$ is a representative carbon-dioxide partial pressure, $K$ an effective storage capacitance, $M>0$ metabolic production, and $pV$ removal by pulmonary ventilation. The delay represents transport and sensing between the gas-exchange compartment and the ventilatory response. The Hill equation response rises sigmoidally with pressure, reaches half its maximum $\alpha$ at $p=\theta$, and saturates at $\alpha$; $n>1$ permits a steep threshold-like response.

For positive parameters, $pV(p)$ increases strictly from zero to infinity because its derivative is $V+p{V}^{\prime }>0$. Its graph therefore intersects the horizontal line $M$ exactly once: $\boxed{M=p^{\ast }V(p^{\ast }),p^{\ast }>0}$. The figure shows this argument and the first imaginary-root crossing.

For a small perturbation $z=p-p^{\ast }$, linearization givesThe sigmoidal delayed ventilation threshold follows from the characteristic equation of a delay differential equation, $\lambda +A+B{e}^{-\lambda \tau }=0$. At an imaginary root $\lambda =i\omega$, separate real and imaginary parts:For $B>A>0$, the first positive crossing hasAt zero delay the scalar equation is stable. No root can cross the imaginary axis before this first crossing, and zero itself is not a root. To check the crossing direction, implicit differentiation givesAll subsequent imaginary crossings have the same direction. Thus the equilibrium is linearly stable below ${\tau }_c$ and unstable above it. This establishes the oscillatory linear threshold; the full nonlinear Hopf bifurcation additionally requires the usual nondegeneracy conditions. If $B\le A$, no finite imaginary-root crossing occurs and the equilibrium is stable for every finite delay. For the stated response, $B/A=n{\theta }^n/({\theta }^n+(p^{\ast }{)}^n)$, showing explicitly when the gain can exceed instantaneous damping.

If $A\ll B$, then ${\omega }_c\approx B$ and ${\omega }_c{\tau }_c\approx \pi /2$. Consequently the small-amplitude period near onset is $\boxed{2\pi /{\omega }_c\approx 4{\tau }_c}$. The comparison is to the critical delay near the first instability, rather than an arbitrary delay far beyond onset. A longer circulatory transit time can move the physiological system across this delayed-feedback threshold, providing a mechanism for periodic breathing associated with impaired cardiac output.