Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 6C / b / Solution

For $x'(t)=\alpha x(t-T)[1-x(t)]$, linearization about $x=1$ is $u'=-\alpha u$ because the delayed perturbation appears only in a quadratic term. The equilibrium is therefore locally asymptotically stable for every delay $T\geq0$.


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 Incoming links: Delay-independent stability from an undelayed linearization