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

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.