Past exam of the theoretical mechanics course of the University of Oxford / 2011 / B / 4 / c

Write $\omega$ for the nonzero forcing frequency, to distinguish it from the vertical velocity $w$. Put $k_z=\pi /h$ and $Z(z)=\cos (\pi (2z+h)/(2h))=-\sin (k_{z}z)$. Then $Z(0)=Z(h)=0$ and $Z^{\prime \prime }=-k_z^{2}Z$, so it meets the rigid-wall conditions. Substituting $w=\mathrm{Re}(f(x)e^{-i\omega t}Z(z))$ into the scalar wave equation givesorThe radiation sign for an inertial wave must exclude any disturbance arriving from $x=+\infty$. This means an radiation condition for propagating waves, or bounded decay for an evanescent wave.

For $0<|\omega |<2|\Omega |$, put $q=k_z\sqrt{4{\Omega }^2/{\omega }^2-1}$. The solutions are oscillatory. The inertial-wave dispersion relation is ${\omega }^2=4{\Omega }^2{k}_z^2/(k_x^2+k_z^2)$; its horizontal group velocity is $c_{g,x}=-\omega k_x/(k_x^2+k_z^2)$. Positive energy transport into the channel therefore requires $k_x=-\mathrm{sgn}(\omega )q$. The solution isFor a positive frequency, the phase travels towards negative $x$ while the group velocity travels towards positive $x$. Choosing the outgoing sign from the phase velocity alone would give the wrong result.

For $|\omega |>2|\Omega |$, set $\kappa =k_z\sqrt{1-4{\Omega }^2/{\omega }^2}$. Excluding the exponentially growing solution givesThis disturbance decays away from the wavemaker and does not propagate horizontally. At the cutoff the ODE degenerates to $f^{\prime \prime }=0$; zero frequency is not a nonzero forced harmonic mode with this vertical shape.

There is a second printed inconsistency: the stated velocity $u\mathbf{i}+w\mathbf{k}$, if taken literally with no $y$ dependence and no transverse velocity, cannot support a nontrivial rotating inertial wave. Its transverse momentum equation would give $2\Omega u=0$, then continuity and the rigid walls would force $w=0$. The scalar solutions above are the intended inertial-wave solutions when two-dimensional means dependence on $x,z$ while retaining the transverse component $v$. For a mode, $-i\omega v+2\Omega u=0$ explicitly supplies the generally nonzero transverse velocity. Thus the scalar equation can be solved as requested, but the full vector field needs this correction to the printed two-component description.