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For a vector $a$ described in axes rotating at constant angular velocity $\Omega$, the rotating-axes theorem is $(da/dt{)}_{\mathrm{inertial}}=(da/dt{)}_{\mathrm{rotating}}+\Omega \times a$. Inertial particle velocity is $u+\Omega \times x$. Applying the theorem once more gives inertial acceleration $Du/Dt+2\Omega \times u+\Omega \times (\Omega \times x)$. Thus the rotating momentum equation isIf $g=-\nabla \Phi$ and density is constant, the centrifugal term can be absorbed into the reduced pressure in a rotating fluid:For small disturbances with $\Omega =\Omega e_z$, linearize to $u_t-2\Omega v=-P_x/\rho$, $v_t+2\Omega u=-P_y/\rho$, $w_t=-P_z/\rho$, together with incompressibility. Put $\zeta =v_x-u_y$. Taking the vertical curl gives ${\zeta }_t=2\Omega w_z$. Taking divergence gives ${\nabla }^{2}P=2\rho \Omega \zeta$. Applying the Laplacian to the vertical momentum equation therefore gives ${\partial }_t{\nabla }^{2}w=-2\Omega {\zeta }_z$. Another time derivative yieldsAll three velocity components are retained, even for fields independent of $y$; rotation produces the transverse component.

Set $m=\pi /h$ and seek $w=a{e}^{-i\omega t}W(x)\sin mz$. The rigid-wall conditions are then automatic and the scalar equation givesFor $0<|\omega |<2|\Omega |$, define $q=m\sqrt{4{\Omega }^2/{\omega }^2-1}$. The two possible oscillatory factors are distinguished by energy propagation. With horizontal wavenumber $k$, the dispersion relation is ${\omega }^2=4{\Omega }^2{m}^2/(k^2+m^2)$ and $c_{g,x}=-\omega k/(k^2+m^2)$. An outgoing wave in $x>0$ must have $\omega k<0$, givingFor $|\omega |>2|\Omega |$, define $\kappa =m\sqrt{1-4{\Omega }^2/{\omega }^2}$. Reject the exponentially growing branch, leaving the evanescent responseThe real part is the physical velocity. The radiation or bounded-decay condition is needed in addition to the wavemaker boundary value to pick a unique half-channel response. Steady forcing ($\omega =0$) is a singular limit: with nonzero rotation a steady geostrophic balance and impermeable top/bottom require $w=0$, incompatible with a nonzero imposed sine profile. At the cutoff equality, the bounded outgoing polarization also degenerates. These exceptional cases cannot be obtained by blindly substituting into the two strict-inequality formulas. The propagating sign is the radiation sign for an inertial wave.

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.