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Use $p$ for disturbance pressure divided by the constant inertial density. The ideal, nonrotating Linearized Boussinesq equations areHere $N^2>0$ is a prescribed time-independent buoyancy frequency squared. Taking the scalar product of momentum with velocity gives ${\partial }_t(|\mathbf{u}{|}^2/2)=-\nabla \cdot (p\mathbf{u})+\sigma w$, using incompressibility. Multiplying the buoyancy equation by $\sigma /N^2(z)$ gives ${\partial }_t[{\sigma }^2/(2N^2)]=-\sigma w$. No spatial derivative of $N$ is needed in this cancellation. ThusThis is positive disturbance energy per unit inertial density; its flux is pressure work.

For constant $N$, write a plane internal gravity wave with wave vector $\mathbf{K}=({\mathbf{k}}_h,m)$ and phase $\theta ={\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t$. Let $k_h=|{\mathbf{k}}_h|>0$ and $K^2=k_h^2+m^2$. In complex amplitudes, incompressibility and horizontal momentum giveConsequently $\hat{p}=-\omega m\hat{w}/k_h^2$. Substitution into vertical momentum, $-i\omega \hat{w}=-im\hat{p}+\hat{\sigma }$, yieldsThe same dispersion relation follows directly from the displacement equation: its two terms become $K^2{\omega }^2\hat{\zeta }$ and $-N^2{k}_h^2\hat{\zeta }$. The positive-frequency branch is $\omega =N{k}_h/K$; the other branch is its sign reversal.

Choose real vertical-displacement amplitude $A$. The internal-wave polarization can be writtenVelocity and displacement lie along the phase planes, perpendicular to $\mathbf{K}$, and are a quarter cycle out of phase. Buoyancy is opposite in sign to vertical displacement. In the requested two-dimensional sketch take $k<0$, $m<0$ and $\omega >0$: the phase velocity $\omega \mathbf{K}/K^2$ points down and left. Then $\mathbf{e}=(-m/k,1)$ points up and left, and positive $p$ accompanies positive $w$ and negative $u$. Negative-pressure bands have the opposite velocity. Multiplication by pressure therefore transports energy up and left in both signs of band.

Differentiating the positive-frequency dispersion relation gives the group velocityIn particular $\mathbf{K}\cdot {\mathbf{c}}_g=0$. Since $|\mathbf{e}{|}^2=K^2/k_h^2$, the wavelength averages areUsing $N^2={\omega }^2{K}^2/k_h^2$ in the last expression provesFor $k,m<0$ the group velocity points up and left, perpendicular to the down-left phase velocity. The pressure sign pattern, rather than the direction of moving crests, determines the energy-flux direction shown below.

For the moving-boundary problem, first take the fluid to occupy $z>0$ with the boundary near $z=0$. An outgoing boundary-forced internal gravity wave must have $c_{gz}>0$. With the specified $k>0$ and $\omega >0$, the radiation condition therefore selectsIf instead the fluid occupies $z<0$, radiation requires $c_{gz}<0$ and the positive root for $m$. The fluid half-space is needed to select a unique sign; the following formulas use the upper half-space.

For slowly modulated internal-wave boundary forcing, put $\zeta =A(T,Z)e^{i\theta }$, $T=\mu t$, $Z=\mu z$. Acting on $A$ by ${\partial }_t\mapsto -i\omega +\mu {\partial }_T$ and ${\partial }_z\mapsto im+\mu {\partial }_Z$, the displacement equation has leading term $(K^2{\omega }^2-N^2{k}^2)A$. Its order-$\mu$ term isThus characteristics carry the boundary amplitude into the fluid:Real parts are understood. To the same leading order,Slow-envelope corrections to these polarization formulas are order $\mu$. At $T=2$, the amplitude is $ϵ$ for $0\le Z\le c_{gz}$, falls smoothly across $c_{gz}<Z<2c_{gz}$, and is zero beyond $2c_{gz}$. This translating ramp is also shown in the figure.

The mean boundary work per unit horizontal area is $\overline{W}={\rho }_{00}\overline{p(0)w(0)}$ at quadratic disturbance order. The small boundary displacement permits evaluation at the unperturbed plane. From the flux calculation,The finite causal wave train has energy integralDifferentiation, or integration of the local energy law with zero flux beyond the front, proves

Finally, distinguish the oscillatory velocity from the second-order Eulerian mean $\overline{u}$. The displacement polarizations giveThe supplied mean-flow relation therefore yields ${\overline{u}}_t=(k/\omega ){\overline{E}}_t+O({\mu }^2)$ at quadratic wave order. Equivalently, the horizontal momentum flux is $\overline{uw}=(k/\omega )c_{gz}\overline{E}$, and envelope-energy transport gives the same acceleration through ${\overline{u}}_t=-{\partial }_z\overline{uw}$.

In the frame moving horizontally at $c=\omega /k$, the boundary pattern is stationary and the initial mean current is $-c$. Subtract that background's kinetic energy before integrating over the infinite half-space. The internal-wave work in a phase-speed frame is thenIts rate of change isThe mean-current cross term supplies the quadratic energy loss in this moving frame; the lab-frame mean kinetic energy alone is fourth order in wave amplitude. The equality is at the leading slow and quadratic orders used throughout the forcing calculation.