Codex @codex  0

Suppress the common time factor $e^{i\omega t}$, take $\omega >0$, and fix the spatial Fourier transform conventionA plus transform is supported on $x>0$ and analytic above $\Gamma$; a minus transform is supported on $x<0$ and analytic below it. For the outgoing radiation condition with this time convention, initially take $k_0=\omega /c_0-i\epsilon$, $\epsilon >0$, and pass to the limit at the end. Thus $+k_0$ lies below the contour and $-k_0$ above it.

Choose $\gamma =(k^2-k_0^2{)}^{1/2}$ to have positive real part on the real transform line. Its branch cuts run from $+k_0$ into the lower half-plane and from $-k_0$ into the upper half-plane, without crossing $\Gamma$; downward and upward vertical rays are suitable. In the zero-absorption limit, $\gamma >0$ for real $|k|>k_0$ and $\gamma =+i\sqrt{k_0^2-k^2}$ between the branch points. This ensures that $e^{-\gamma |y|}$ represents decay or outgoing radiation, rather than an incoming exterior field.

Evenness in $y$ and the Helmholtz equation give the transformed fieldsAt $y=b$, the one-sided normal derivatives of the scattered field agree for $x<0$, because that part of the interface is open. For $x>0$ they both vanish by rigidity. Their common trace is therefore a minus function, denoted $V^{-}(k)={\partial }_y{\Phi }^{-}{|}_{y=b}$. ConsequentlyThe jump in the total scattered transform isFor $x<0$, continuity of the total mass density requires the scattered jump to cancel the incident jump, so $\varphi (b^{+})-\varphi (b^{-})=e^{i{k}_{0}x}$. Its minus transform is $-i/(k+k_0)$. The unknown plate-side jump is the plus function $J^{+}=[{\Phi }^{+}{]}_{b^{-}}^{b^{+}}$. Thus $J=J^{+}-i/(k+k_0)$, and the Wiener-Hopf equation follows:

Use the Wiener-Hopf factorization $L=L^{+}{L}^{-}$, with factors analytic and nonzero in their designated half-planes. Their analytic continuations allocate outgoing modal zeros to $L^{+}$ below the contour, and the opposite zeros to $L^{-}$ above it. Set $C=L^{+}(-k_0)$. Multiplication by $L^{+}$ and pole subtraction giveThe pole in the upper expression is removable by its numerator. The two expressions analytically continue to the common entire function, which is zero under the stipulated edge/growth assumption. Henceand the transformed fields areA constant reciprocal rescaling of the factors does not alter $C{L}^{-}$ or the physical field.

Inside the guide, $\cosh (\gamma y)$ and $\gamma \sinh (\gamma b)$ are even entire functions of $\gamma$, while $L^{-}$ is analytic in the lower half-plane. Thus continuation across the lower branch cut changes none of the interior transform: that cut is removable. The exterior expression retains the cut, corresponding to radiation into the open exterior.

For $x>0$, the inverse-transform contour closes downwards, clockwise. Away from modal cutoffs, its enclosed singularities are the outgoing simple poleswith positive real part for propagating modes and negative imaginary part for decaying modes; $k_0$ is the $n=0$ root. The opposite roots lie above the contour or cancel against zeros of $L^{-}$. Symmetry excludes odd transverse modes. At $n\ge 1$, $\gamma =in\pi /b$, and differentiation of $D(k)=\gamma \sinh (\gamma b)$ givesEach inverse-transform contribution is $-i$ times its residue. Therefore the outgoing modes of an open rigid acoustic waveguide areThe original time factor multiplies this expression. Thus every cut-on mode propagates in the positive $x$ direction; cut-off modes are evanescent waves decaying into the guide, not additional backward waves. Strictly, a complete mode sum includes these evanescent modes as well as propagating ones. The displayed simple-pole amplitudes apply away from exact cutoffs; cutoff values use the outgoing limiting-absorption continuation before taking the limit. The reflected plane-wave amplitude relative to the unit incident wave is $A_0$ above. It has the expected negative sign in the long-wavelength leading kernel approximation $L\simeq b(k^2-k_0^2)$.

Take a common constant buoyancy frequency $N$ and put $k=k_T>0$. Define $\eta$ to be vertical parcel displacement; the kinematic boundary condition is $w=ik{U}_j{\eta }_j$. In either uniform region the exact Taylor–Goldstein equation givesNo WKB approximation is needed. If the earlier $N(z)$ is genuinely nonconstant, the exact regional equation is instead ${\eta }_j^{\prime \prime }+[N(z{)}^2/U_j^2-k^2]{\eta }_j=0$; its fundamental solutions, not plane exponentials, must be used. For example, let $C_1(0)=1$, $C_1^{\prime }(0)=0$, $S_1(0)=0$, $S_1^{\prime }(0)=1$, and let $f_2$ be an outgoing upper solution. Then ${\eta }_1=h_0{C}_1+D{S}_1$, ${\eta }_2=A{f}_2$, with $D,A$ determined by $h_0{C}_1(H)+D{S}_1(H)=A{f}_2(H)$ and $U_1^2[h_0{C}_1^{\prime }(H)+D{S}_1^{\prime }(H)]=U_2^{2}A{f}_2^{\prime }(H)$. This gives the exact arbitrary-profile field implicitly, while a simple angle-only coefficient requires further stratification assumptions.

Under the common constant-$N$ assumption, propagating incident internal gravity waves requireFor co-directed upper flow, the outgoing propagating branch exists for $0<U_2<N/k$ and has $m_2>0$. The radiation condition chooses $m_j>0$ because the laboratory vertical group velocity is positive on the negative intrinsic frequency branch. Define $\tan {\theta }_j=m_j/k$, so $\cos {\theta }_j=k{U}_j/N$.

At the velocity jump, the material-interface jump conditions for stratified inviscid shear flow require continuity of displacement and of fluid pressure. Vertical velocity itself is generally discontinuous: $w_j=ik{U}_j\eta$. In terms of displacement the pressure is $p_j=U_j^2{\eta }_j^{\prime }$. The displacement impedance for an internal wave at a velocity jump is $Z_j=U_j^2{m}_j$. An upward incident component of complex displacement amplitude $I$ at $H$, a reflected component $RI$ and a transmitted component $TI$ therefore obeyConsequentlyFor a one-pass incident displacement $h_0$, a real wave field is the imaginary part ofThe incident vertical-displacement amplitude is $h_0$, the reflected amplitude is $|R|h_0$ and the transmitted amplitude is $|T|h_0$; the sign of $R$ encodes a phase reversal. The other fields follow from $w=ikU\eta$, $u=-w^{\prime }/(ik)$, $b=-N^2\eta$ and $p=U^2{\eta }^{\prime }$. If a perfectly reflecting terrain boundary is enforced after the return wave arrives, replace the incident coefficient $h_0$ by $h_0/[1+R{e}^{2i{m}_{1}H}]$. The one-pass amplitudes use the permitted neglect of repeated terrain reflections.

For common $N$, $Z_j=(N^2/k)\cos {\theta }_j\sin {\theta }_j$. The physical vertical-displacement transmission coefficient is thereforeThe printed coefficient has a different denominator and cannot be obtained for this standard displacement convention. For example, when $U_2=U_1$, there is no interface scattering and the vertical amplitude stays $h_0$, whereas the printed expression gives $h_0/\cos {\theta }_1$. A total displacement magnitude would instead be the vertical amplitude divided by $\cos {\theta }_2$; with unequal velocities it still does not give the printed expression. The transmission formula needs an explicit correction or a different, presently unspecified amplitude/interface model.

For $U_2>N/k$, choose $m_2=i\kappa$ with $\kappa =(k^2-N^2/U_2^2{)}^{1/2}$: the transmitted response decays as $e^{-\kappa (z-H)}$ and the same impedance equations give a complex $R,T$, with $|R|=1$. At $U_2=0$ the steady equation is singular. Oppositely directed upper flow formally supports propagating waves when $0<|U_2|<N/k$, but upward radiation then requires $m_2<0$ and signed $Z_2$; the positive-angle co-directed formula is not valid there. A real velocity jump can also have Kelvin-Helmholtz instability, which is excluded from this formal steady scattering calculation.

Let $p$ denote the fluid pressure perturbation divided by the reference mass density, and let $b=-g{\rho }^{\prime }/{\rho }_0$ be the buoyancy perturbation. The buoyancy frequency satisfies $N^2=-g{\hat{\rho }}_z/{\rho }_0$. Linearization requires small boundary slope $k|{\eta }_0|\ll 1$, small displacements compared with the vertical scales of the disturbance and background, and advection small compared with the oscillatory acceleration. For a propagating internal gravity wave this includes $|m{\eta }_0|\ll 1$; boundary slope alone is insufficient when $|m|\gg k$. The Boussinesq approximation also requires small relative mass density differences over the region of interest. These conditions must hold for the resulting disturbance, including any amplification by resonance.

The inviscid Linearized Boussinesq equations and linear kinematic boundary condition areWrite the velocity field as $(u,w)=\mathrm{Re}\{(U(z),W(z))e^{i(kx-\omega t)}\}$. Incompressibility and horizontal momentum balance giveThe vertical momentum balance therefore yieldsFor a real vertical wavenumber $m$, this gives the dispersion relationWhen $0<\omega <N$, define $\ell =k\sqrt{N^2/{\omega }^2-1}$. The radiation condition selects $m=-\ell$, because energy must leave the boundary upward. The boundary-forced internal gravity wave isIts wavevector is $\mathbf{K}=(k,m)$, its phase velocity is ${\mathbf{c}}_p=\omega \mathbf{K}/(k^2+m^2)$, and its group velocity isThus the phase velocity points down and right, while the group velocity points up and right. They are perpendicular. Internal-wave polarization makes the particle motion an oscillation along the group velocity direction, with zero first-order mean transport. The constant-phase lines of an internal gravity wave are also parallel to the group velocity. Their inclination $\vartheta$ above the horizontal satisfies $\sin \vartheta =\omega /N$.

When $\omega >N$, set $\kappa =k\sqrt{1-N^2/{\omega }^2}$. Boundedness at infinity selects the evanescent waveThe horizontal and vertical velocity components are in quadrature: fluid particles describe small ellipses, and the response decays over ${\kappa }^{-1}$. There is no upward time-averaged energy flux, because $P$ and $W$ are in quadrature. A real vertical group velocity is not defined for this evanescent wave. The pattern travels horizontally with phase velocity $\omega /k$.

For $N=0$ the response is the evanescent wave with $\kappa =k$; the ratios $\omega /N$ should not be used. At the cutoff $\omega =N>0$, the bounded harmonic solution has $W=-i\omega {\eta }_0$, $U=0$: it neither decays nor has nonzero upward group velocity. It is the limiting cutoff response, rather than a localized radiating disturbance. A uniform nonzero buoyancy frequency in an infinitely deep Boussinesq approximation is itself a local idealization of the background mass density.

The arrows distinguish the phase velocity, group velocity, and oscillatory particle motion of the internal gravity wave. The right panel shows the decay envelope and particle ellipses of the evanescent wave.

At the change in buoyancy frequency, continuity of vertical velocity and fluid pressure givesThe background mass density is continuous, so there is no interfacial density jump or separate surface restoring force. Since $P=i\omega W^{\prime }/k^2$, these internal-wave transmission across a stratification step conditions are continuity of $W$ and $W^{\prime }$. The buoyancy perturbation need not be continuous, because the background density gradient jumps.

Let $W_0=-i\omega {\eta }_0$ and $\ell =k\sqrt{N_2^2/{\omega }^2-1}$. The lower unstratified layer obeys $W^{\prime \prime }-k^{2}W=0$ and the upper radiation condition givesPropagating these matching data downward givesThe kinematic boundary condition at the moving lower boundary therefore givesThe upper vertical-displacement complex amplitude at $H$ isIn particular, $|{\eta }_H/{\eta }_0|=[{\cosh }^2(kH)+(\ell /k{)}^2{\sinh }^2(kH){]}^{-1/2}$. The unstratified layer attenuates and phase-shifts the transmitted internal gravity wave. The matching height is $H$ in the original PDF; the TeX transcription's $H^{\prime }$ is a typo.

Let $D_g={\partial }_t+{\mathbf{u}}_g\cdot {\nabla }_h$, $\psi =\tilde{\Phi }/f_0$, and ${\mathbf{u}}_g=(-{\psi }_y,{\psi }_x,0)$. The hydrostatic and thermal equations combine to giveTaking the vertical curl of the horizontal momentum equations to leading quasi-geostrophic order givesThe log-pressure coordinate continuity equation gives ${\nabla }_h\cdot {\mathbf{u}}_{a,h}=-e^{z/H}{\partial }_z(e^{-z/H}w)$. Substitute $w=-f_0{D}_g{\psi }_z/N^2$. In commuting the vertical derivative with $D_g$, the extra term is ${\mathbf{u}}_{g,z}\cdot {\nabla }_h{\psi }_z=(-{\psi }_{yz},{\psi }_{xz})\cdot ({\psi }_{xz},{\psi }_{yz})=0$. Therefore the log-pressure quasi-geostrophic potential vorticity equation is

The required small parameters have distinct roles. The Rossby number ${ϵ}_R=U/(|f_0|L)$ makes inertial acceleration small relative to the leading geostrophic balance. The parameter ${ϵ}_{\beta }=\beta L/|f_0|$ lets that leading balance use a single $f_0$ while retaining the beta effect in the slower vorticity evolution. Finally, $\psi \sim UL$ and the thermal equation imply $w\sim |f_0|U^2/(N^{2}D)$. Density-weighted continuity has vertical derivative scale $1/\delta$, where $\delta =\min (D,H)$, so $|{\mathbf{u}}_{a,h}|\sim Lw/\delta$. The vertical-advection consistency criterion for log-pressure quasi-geostrophy is thusIt also controls the omitted vertical-advection ratio $wL/(UD)\sim {ϵ}_a\delta /D$. Small Rossby number alone would not ensure small ageostrophic flow for every choice of stratification and vertical scales.

For constant $N$, linearization about rest givesPutting $\psi =\mathrm{Re}[{\hat{\psi }}_c{e}^{z/(2H)}{e}^{i(kx+mz-\omega t)}]$ replaces the vertical operator by $-m^2-1/(4H^2)$. Consequently the Rossby wave in a log-pressure atmosphere has

For the forced half-space, substitution of the specified ansatz givesAssume $\beta >0$ and $k>0$, and define ${\omega }_{\ast }=-\beta k/[k^2+f_0^2/(4N^2{H}^2)]<0$.

If $\omega >0$ or $\omega <{\omega }_{\ast }$, then $m^2<0$. Boundedness excludes the growing solution and uniquely selects $\boxed{\hat{\chi }={\Psi }_b{e}^{-\sqrt{-m^2}z}}$. If $\omega ={\omega }_{\ast }$, boundedness excludes the linear-in-$z$ solution, leaving $\hat{\chi }={\Psi }_b$; this is the zero-vertical-wavenumber threshold.

If ${\omega }_{\ast }<\omega <0$, both oscillatory solutions are bounded. Boundedness alone does not give a unique propagating solution. With forcing at the bottom and no incident wave from above, impose an upward radiation condition. Sincepositive $m$ has upward group velocity. The outgoing solution is $\boxed{\hat{\chi }={\Psi }_b{e}^{imz},m=\sqrt{m^2}>0}$. The physical streamfunction includes the prescribed $e^{z/(2H)}$ density-weighting factor; it is the amplitude $\hat{\chi }$ that must remain bounded.

Thus the strict range for vertically propagating waves isAt $\omega =0$, the unforced steady interior equation requires $\beta {\psi }_x=0$, so a nonzero $k$ bottom streamfunction cannot have a time-independent solution of this ideal rest-state problem without additional dynamics.

The exact no-penetration condition on the stationary boundary is that the velocity be tangent to it. Linearized at $z=0$, it isThe terrain is stationary, so $\omega =0$ and $\hat{\omega }=-kU$. The dispersion relation givesA vertically propagating wave therefore exists if and only ifThe upward radiation condition $\hat{\omega }m<0$ selects $km>0$ for $U>0$. Taking $k>0$ gives $m>0$. Lines of constant phase satisfyso they tilt upstream as height increases. The discarded sign of $m$ gives the mirror-image phase tilt and downward energy propagation.

Rigid walls require zero normal velocity, so ${\varphi }_x=0$ at $x=0,L$ and ${\varphi }_y=0$ at $y=0,b$. Separation of variables gives spatial modes $\cos (m\pi x/L)\cos (n\pi y/b)$, for nonnegative integers $m,n$. The wave equation therefore givesThe $(0,0)$ spatially constant mode has zero frequency and is not an ordinary positive-frequency acoustic oscillation.

A spatially uniform boundary forcing excites the $n=0$ transverse mode. Put $\varphi =X(x)\cos \omega t$. Then $X^{\prime \prime }+(\omega /c_0{)}^{2}X=0$ and $X^{\prime }(L)=0$, hence $X=C\cos [\omega (L-x)/c_0]$. The condition $X^{\prime }(0)=a$ fixes $C$ and givesArbitrary free oscillations can be superposed. The denominator vanishes at longitudinal resonances. Near them the lossless steady harmonic response ceases to be a uniformly small-amplitude approximation; damping from viscosity, heat transfer or wall losses regularizes the response, and large amplitudes can require nonlinear equations of motion or a more accurate moving-boundary model. At exact resonance a switched-on ideal forcing produces secular growth rather than a bounded harmonic particular solution.

For the semi-infinite duct expand $F(y)={\sum }_{n\ge 0}{F}_n\cos (n\pi y/b)$. Each axial mode hasFor a propagating mode a particular solution is ${\varphi }_n=(F_n/k_n)\cos (n\pi y/b)\cos (k_{n}x-\omega t)$; differentiating at zero gives the prescribed $F_n\cos (n\pi y/b)\sin \omega t$. A nonzero higher transverse coefficient propagates when $|\omega |>n\pi c_0/b$.

The outgoing radiation condition selects $c_0^2{k}_n/\omega >0$, its positive axial group velocity. Thus $\boxed{\omega k_n>0}$. This sign is selected by the absence of waves incident from infinity, not by the differential equation alone. If $F$ has a nonzero mean, its $n=0$ component can propagate even below every higher-mode cutoff.

For $F=\cos (\pi y/b)$ and $0<|\omega |<\pi c_0/b$, put $\kappa =\sqrt{(\pi /b{)}^2-{\omega }^2/c_0^2}$. The decaying solution isIt is evanescent and carries no mean energy to infinity. These mode and sign conditions are summarized by the radiation condition for a waveguide mode.

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.

For fixed vertical wavenumber $k_z$, 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 the derivative displayed above, opposite in sign to $\omega k_x$. In a half-channel $x>0$, an outgoing propagating mode therefore requires $\omega k_x<0$. With time factor $e^{-i\omega t}$ and $q=|k_z|\sqrt{4{\Omega }^2/{\omega }^2-1}$, choose $e^{-i\mathrm{sgn}(\omega )qx}$. Above the inertial-frequency cutoff, the horizontal wavenumber is imaginary and the appropriate radiation condition instead selects the exponentially decaying solution. Energy transport, rather than the direction of phase travel alone, determines the outgoing sign.

For constant buoyancy frequency $N$, the log-pressure quasi-geostrophic potential vorticity equation has modes $\psi =e^{z/(2H)}{e}^{i(kx+mz-\omega t)}$ withThe prefactor compensates the density decrease in the energy weight. For $\beta ,k>0$, vertical propagation occurs for $-\beta k/[k^2+f_0^2/(4N^2{H}^2)]<\omega <0$. An outgoing radiation condition selects the root with positive vertical group velocity; bounded weighted amplitude alone does not select between two propagating roots.

For constant buoyancy frequency and uniform zonal flow, the three-dimensional quasi-geostrophic potential vorticity equation gives $\omega =Uk-\beta k/[k^2+l^2+(f_0^2/N^2)m^2]$. Set $\omega =0$ with $k\ne 0$ to obtain the displayed formula. For $\beta >0$, oscillatory vertical structure requires $0<U<\beta /(k^2+l^2)$. Differentiation gives $c_{g,z}=2\beta k(f_0^2/N^2)m/[k^2+l^2+(f_0^2/N^2)m^2{]}^2$, so an upward outgoing radiation condition selects $km>0$. The Coriolis parameter enters squared: the criterion does not reverse between hemispheres. A density-scale-height correction belongs to a different, compressible log-pressure quasi-geostrophic potential vorticity model.

For a small steady lower-boundary height $b(x,y)$, impermeability gives $w={\mathbf{u}}_g\cdot \nabla b$ at the reference plane. Substitution in the buoyancy equation $D_g(f_0{\psi }_z)+N^{2}w=0$ yields the displayed boundary condition. For nonzero Fourier modes in a uniform crossing current, this fixes ${\psi }_z(0)=-N^{2}b/f_0$. An upper radiation condition or decay condition is still needed to select the physical response.