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

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 $|\beta |$.