Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 336 / 2

Introduce $T=ϵt$ and write $x_0=A(T)\cos [t+\theta (T)]$. Averagingover one fast period gives $A{A}_T=-3A^4/8$ and ${\theta }_T=0$. The initial data therefore give the method of multiple scales resultthrough $t=O({ϵ}^{-1})$.

For the replacement damping, $\sin (\dot{x})\dot{x}={\dot{x}}^2-{\dot{x}}^4/6+\cdots$ is even in $\dot{x}$. Every term has zero resonant projection onto the fundamental over a complete orbit, so the $O(ϵ)$ slow amplitude and phase equations vanish. Thusthrough $t=O({ϵ}^{-1})$: there is no first-order secular damping, although bounded mean and higher-harmonic corrections occur.

For constant width, separation of variables with $\varphi =X(x)\sin (n\pi y/h_0)$ givesHence $X=A_n{e}^{i{k}_{n}x}+B_n{e}^{-i{k}_{n}x}$, with imaginary $k_n$ representing an evanescent mode.

For $X=ϵx$ and varying width,Projection of the next-order equation onto $\sin (n\pi y/h)$, whose squared norm is proportional to $h$, givesThus the WKB amplitude in a slowly varying duct isThis applies for $n\ge 1$ away from turning points $k_n=0$.