The eigenvalues of $L$ are
$$
s_\pm=-\varepsilon\pm\sqrt{\beta^2-1}.
$$
Since $\beta<1$, the origin is linearly stable exactly when the determinant is positive:
$$
\boxed{1+\varepsilon^2-\beta^2>0},
$$
or $|\beta|<\sqrt{1+\varepsilon^2}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-331/3/a/solution