Past exam of the mathematical physiology course of the University of Oxford / 2011 / 3 / c

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 $4\tau$.