= Solution
In the <Boussinesq approximation>, replace density by a constant reference density $\rho_0$ in inertia and <incompressible flow>, retaining the small density departure only in the gravitational force. Write $\rho=\bar\rho(z)+\rho'$ and subtract the resting <hydrostatic pressure>. The upward <buoyancy perturbation> acceleration and squared <buoyancy frequency> are
$$
\sigma=-\frac{g\rho'}{\rho_0},\qquad N^2(z)=-\frac{g}{\rho_0}\frac{d\bar\rho}{dz}.
$$
Thus stable <density stratification> has $N^2>0$. Let $p$ henceforth denote the pressure anomaly divided by $\rho_0$. The ideal, nonrotating <Boussinesq equations> are
$$
\frac{D\mathbf u}{Dt}=-\nabla p+\sigma\mathbf e_z,
\qquad \frac{D\sigma}{Dt}+N^2(z)w=0,
\qquad \nabla\cdot\mathbf u=0,
\qquad \frac D{Dt}=\partial_t+u\partial_x+v\partial_y+w\partial_z.
$$
The <buoyancy> equation follows by materially conserving the total buoyancy $b_0(z)+\sigma$, where $b_0'=N^2$; it does not assume constant $N$.
For an $x$-periodic flow of period $P$, define $\bar a=P^{-1}\int_0^P a\,dx$. An infinite-domain average with vanishing end contributions works as well. The necessary pressure condition is \b[no mean zonal pressure gradient], $\overline{p_x}=0$: periodic pressure satisfies it, whereas a pressure containing a term proportional to $x$ does not. Use <incompressible flow> to write the zonal momentum equation in conservative form,
$$
u_t+(u^2)_x+(uv)_y+(uw)_z=-p_x.
$$
Averaging its zonal derivative terms yields the exact <Reynolds stress> relation
$$
\boxed{\bar u_t=-\partial_y\overline{uv}-\partial_z\overline{uw}.}
$$
For two-dimensional small disturbances about rest, the <Linearized Boussinesq equations> are $u_t=-p_x$, $w_t=-p_z+\sigma$, $\sigma_t=-N^2w$, and $u_x+w_z=0$. Differentiate the two momentum equations to eliminate pressure: $(u_z-w_x)_t=-\sigma_x$. Differentiating in $x$ and using $u_{xz}=-w_{zz}$ gives $\nabla^2w_t=\sigma_{xx}$. One further time derivative gives
$$
\boxed{\nabla^2w_{tt}+N^2(z)w_{xx}=0.}
$$
\b[The equation remains valid for $N=N(z)$.] No vertical derivative of $N$ was taken: $\partial_x^2(N^2w)=N^2w_{xx}$.
For constant $N$ and $k\ne0$, a <plane internal gravity wave> has
$$
w=\operatorname{Re}\{a e^{i(kx+mz-\omega t)}\},
\qquad \omega^2(k^2+m^2)=N^2k^2,
\qquad m=\pm |k|\sqrt{N^2/\omega^2-1}.
$$
Both signs give the prescribed lower-boundary velocity. The <radiation condition> excludes energy arriving from infinity. At positive frequency, the vertical <group velocity> is
$$
c_{gz}=\frac{\partial\omega}{\partial m}=-\frac{\omega m}{k^2+m^2}.
$$
For a source below the fluid, energy must propagate upward, so \b[the outgoing solution has $m<0$], even though its vertical <phase velocity> $\omega/m$ is downward.
To obtain this selection causally, let $T=\mu t$ and $Z=\mu z$, and replace the constant amplitude by $a(Z,T)$. The <method of multiple scales> treats $(z,t,Z,T)$ as independent during differentiation: the carrier derivatives hold $Z,T$ fixed, while $a_T$ holds $Z$ fixed and $a_Z$ holds $T$ fixed. Only after differentiating do we restrict to $T=\mu t$, $Z=\mu z$. Acting on the complex wave, the operators become
$$
\partial_t\longmapsto-i\omega+\mu\partial_T,
\qquad \partial_z\longmapsto im+\mu\partial_Z.
$$
With $K^2=k^2+m^2$, the wave equation has residual
$$
(\omega^2K^2-N^2k^2)a
+2i\mu\{\omega K^2a_T-m\omega^2a_Z\}+O(\mu^2).
$$
The zeroth-order term is the <dispersion relation>; the first-order term vanishes precisely when
$$
\boxed{a_T+b a_Z=0,\qquad b=-m\omega/K^2=c_{gz}.}
$$
If $a_0(T)$ is the switched-on boundary amplitude, with $a_0(T)=0$ for $T<0$, the outgoing envelope is $a(Z,T)=a_0(T-Z/b)$ for $b>0$. Its characteristics carry the boundary signal into initially undisturbed fluid. If $b<0$, characteristics instead arrive from the upper half-space; zero initial disturbances and no incoming signal cannot establish the proposed boundary-forced wavetrain. This is the <internal-wave envelope radiation condition>.
The leading <incompressible flow> polarization gives $u'=-(m/k)w'$. Hence, to second order in amplitude and leading order in the slow modulation,
$$
\overline{u'w'}=-\frac{m}{2k}|a(Z,T)|^2,
\qquad
\bar u_t=\frac{m}{2k}\partial_z|a|^2
=\frac{\mu m}{2k}\partial_Z|a|^2.
$$
There is no meridional transport here. Envelope transport gives $\partial_t|a|^2=-b\partial_z|a|^2$, so an initially vanishing mean develops as
$$
\boxed{\bar u=-\frac{m}{2kb}|a|^2=\frac{K^2}{2k\omega}|a|^2.}
$$
These are the leading expressions for <internal-wave momentum deposition>; higher slow-modulation corrections to the stress are beyond the displayed approximation. In particular, a spatially uniform fully established wavetrain has zero stress divergence, but the front has already produced its mean flow.
The mean wave <energy density> per unit mass follows from $\sigma'=-iN^2w'/\omega$:
$$
\mathcal E=\frac12\overline{u'^2+w'^2+\sigma'^2/N^2}
=\frac{K^2}{2k^2}|a|^2,
\qquad \overline{p'w'}=-\frac{\omega m}{2k^2}|a|^2=b\mathcal E.
$$
Thus $\bar u=(k/\omega)\mathcal E$. In the original frame the mean <kinetic energy> is $\bar u^2/2=O(a^4)$. In a frame translating at $c=\omega/k$, the undisturbed velocity is $-c$, and the change in mean <kinetic energy> is $-c\bar u+O(a^4)=O(a^2)$. Indeed $\mathcal E-c\bar u=0$ at this order behind the front. \b[Mean-flow energy is therefore essential at the same order as wave energy in the translating frame.] This follows also from the frame transformation of energy, $E\mapsto E-cP$ up to the fixed background constant, and is why <conservation of momentum> cannot be omitted from the moving-frame <conservation of energy> budget.
Back to article page
