 Solution

Put $x(t)=1+u(t)$. Retaining terms linear in $u$ givesThe exponential ansatz $u=e^{\lambda t}$ gives the characteristic equation of a delay differential equationAt $T=0$ its root is $\lambda =-\alpha$, and the remaining delay roots lie far in the left half-plane, so the equilibrium is stable for sufficiently small $T$.

A stability change can first occur when $\lambda =i\omega$. Separating real and imaginary parts givesFor the first positive crossing, $\omega T=\pi /3$ and $\omega =\sqrt{3}\alpha$. ThusAll positive-frequency crossings have $\omega T=\pi /3+2\pi k$, so there is no earlier imaginary-axis crossing. Implicit differentiation ofgives, with $q=\alpha T_c$,so the conjugate pair crosses into the right half-plane as $T$ increases. Therefore the stability threshold for delayed quadratic crowding is