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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2019

Paper 4

Again put $x=1+u$. Then
$$
u'(t)=\alpha\bigl(1+u(t-T)\bigr)(-u(t))
=-\alpha u(t)-\alpha u(t-T)u(t).
$$
The linearized equation is therefore
$$
u'=-\alpha u,
$$
whose sole characteristic exponent is $-\alpha<0$ and is independent of $T$. By <delay-independent stability from an undelayed linearization>,
$$
\boxed{x(t)=1\text{ is locally asymptotically stable for every }T\geq0.}
$$

Solution

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

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

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

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