Past exam of the mathematical methods II course of the University of Oxford / 2018 / 11 / Solution

For a plane wave, ${\partial }_t$ acts as $-i\omega$ and ${\partial }_x$ as $ik$. Substitution givesso, for $\omega \ne {\omega }_d$, the dispersion relation isAssume ${\omega }_n>{\omega }_d>0$ and real angular frequency. The ratio is negative exactly on ${\omega }_d<\omega <{\omega }_n$, giving imaginary wavenumbers and an evanescent wave. Thus the propagating frequency gap is $({\omega }_d,{\omega }_n)$. At ${\omega }_d$ there is no finite-$k$ solution, while ${\omega }_n$ has $k=0$; neither endpoint is an ordinary finite-wavelength propagating mode.

Choose the branch $k=(\omega /c)\sqrt{(\omega -{\omega }_n)/(\omega -{\omega }_d)}$, so the phase velocity is positive in the allowed bands. Direct differentiation gives the phase velocity and group velocityThe opposite spatial branch reverses both signs. On the chosen branch, the requested limits areThe band-edge limits are taken from their propagating sides. More precisely, with $ϵ=\omega -{\omega }_n\downarrow 0$ and $\delta ={\omega }_d-\omega \downarrow 0$,

For the final case, set $K=k_{0}c>0$. Substitution of $T(t)e^{i{k}_{0}x}$ at ${\omega }_d=0$ gives $T^{(3)}+(3iK/2)T^{\prime \prime }+K^2{T}^{\prime }=0$. Its characteristic polynomial factors as $r(r+2iK)(r-iK/2)$, henceThe initial data give $A+B+C=1$, $-2B+C/2=0$ and $-4B-C/4=-5$. Thus $B=1$, $C=4$, $A=-4$, andThe time-independent contribution is essential: static modes lost by cancelling frequency would be missed if one divided the ${\omega }_d=0$ dispersion relation by $\omega$ and retained only its two nonzero-frequency roots.