Past exam of the mathematics and the environment and mathematical physiology course of the University of Oxford / 2009 / B / 6 / d

Distinguish physical time $t$ from scaled time $T=t/\tau$. PutThen $K(P_s/\tau )p^{\prime }=M-[P_0+P_{s}p](M/P_0)v$, givingA steady state must be positive, since otherwise ventilation vanishes while production remains positive. It solves $\mu p_{\ast }^2+p_{\ast }-1=0$, soThe value $1/(1+\mu )=1-\mu +{\mu }^2+O({\mu }^3)$ printed in the question agrees only through first order. Thus it is a small-$\mu$ approximation with error of order ${\mu }^2$, not an exact equilibrium for nonzero $\mu$. The other quadratic root is negative and inadmissible for this steady ventilation balance.

Since the steady state is strictly positive, the positive-part function is differentiable in a neighbourhood of it. Write $p=p_{\ast }+q$. Retaining first-order terms givesSubstitute the exponential mode $q=e^{\lambda T}$ to obtainHere the last equality uses the exact steady-state equation. For small $\mu$, $\beta =\alpha \mu +O(\alpha {\mu }^2)$ and $\gamma =\alpha (1+\mu )+O(\alpha {\mu }^2)$. Linearizing at the exact root avoids carrying a spurious constant residual from the printed approximate equilibrium.

For real $\lambda \ge 0$, draw $y=\lambda +\beta$ and $y=-\gamma e^{-\lambda }$. The first is nonnegative and increasing, while the second is negative and approaches zero from below. They cannot intersect on the nonnegative real axis. At zero the characteristic function equals $\beta +\gamma >0$, so a stationary zero-root crossing is excluded as well.

Consequently a loss of stability from a stable regime cannot occur by a real eigenvalue crossing through zero. It must involve a conjugate pair crossing the imaginary axis at nonzero frequency, yielding oscillatory instability. This does not mean that every characteristic root is nonreal: negative real roots may exist and are harmless for stability. The graph and imaginary-axis conditions are shown together below.

For the first Hopf boundary of scalar delayed negative feedback, first, if a root has nonnegative real part then $|\lambda +\beta |=\gamma e^{-\mathrm{Re}\lambda }\le \gamma$. For $\beta \ge \gamma$ this is impossible, apart from the potential equality case $\lambda =0$, already excluded. Thus $\beta <\gamma$ is necessary for instability, although not sufficient by itself.

At an imaginary root $\lambda =i\omega$ with $\omega >0$, separate real and imaginary parts:The first crossing uses ${\omega }_c\in (\pi /2,\pi )$ for $\beta >0$, and satisfiesThe period is in scaled time; its physical value is $\tau {\mathcal{T}}_c$. Higher-frequency branches have larger critical gains. For fixed $\beta$, stability at zero delayed gain and absence of any earlier imaginary or zero-root crossing identify this lowest branch as the first stability boundary. Its crossing is transverse: implicit differentiation of $\lambda +\beta +\gamma e^{-\lambda }=0$ gives at the imaginary rootFor very small $\beta$, ${\omega }_c=\pi /2+2\beta /\pi +O({\beta }^2)$, so ${\mathcal{T}}_c\to 4$ and ${\gamma }_c=\pi /2+2\beta /\pi +O({\beta }^2)$. Therefore ${\gamma }_c=\pi /2$ is exact in the limit $\beta =0$ and is the leading approximation for small positive $\beta$. At that limit $\lambda e^{\lambda }=-\gamma$ first has an imaginary pair at $\lambda =\pm i\pi /2$; subsequent pairs occur at $\pi /2+2n\pi$ and cannot be the first loss of stability. A nonlinear Hopf theorem, including its nondegeneracy conditions, is needed to turn the linear imaginary mode into a claim about a particular finite-amplitude periodic branch.