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

The eigenvalues of $L$ areSince $\beta <1$, the origin is linearly stable exactly when the determinant is positive:or $|\beta |<\sqrt{1+{\epsilon }^2}$.

Direct multiplication shows $L{L}^t\ne L^{t}L$ for $\beta \ne 0$, so $L$ is a non-normal matrix. MeanwhileThe symmetric part has eigenvalues $-\epsilon \pm |\beta |$, so instantaneous growth is possible exactly when $|\beta |>\epsilon$. Under the intended $0\le \beta <1$ regime this is $\boxed{\beta >\epsilon }$.

For $\epsilon =0$, $L^2=-(1-{\beta }^2)I=-{\Gamma }^{2}I$. The matrix exponential therefore gives

The maximum of $E(t)/E(0)$ is the square of the largest singular value of $A(t)$, hence the largest eigenvalue $\lambda$ of $A(t{)}^{t}A(t)$. Its determinant is one and its trace givesFor $0<\beta <1$, this is largest when $|\sin \Gamma t|=1$, namely $\Gamma t=\pi /2$ modulo $\pi$. ThenAt such a time $A(t)$ is off diagonal, and the maximizing initial condition is $\boxed{x(0)=0}$ with $y(0)\ne 0$. For negative $\beta$, the axes interchange and the formula uses $|\beta |$.

When $\epsilon =0$, trajectories conserveso they are closed ellipses around the origin. Ordinary energy measures circular radius rather than this conserved elliptical radius. Starting on the short-energy axis and rotating to the long-energy axis produces the transient amplification from part d; the state later returns, so the growth is transient despite neutral eigenvalues.