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For the polynomial stellar-mode coefficient reduction, write the highest coefficients asSet $x={\omega }^2/{\omega }_d^2$, $L=n(2l+n+1)$ and $k=l+n+1$. The highest powers in the equations of (b) giveThe factor $3$ follows from $4\pi G\rho =3{\omega }_d^2$; the positive $\gamma d/2$ term comes from the highest, negative $r^2$ coefficient of the equilibrium pressure.

For $l>0$, the final spectrum found below has $x\ne 0$. Eliminating $u,v$ first givesThe remaining pressure relation isA nontrivial leading coefficient has $d\ne 0$, soThusor, in dimensional form,

For the printed stability continuation, define $H=\gamma L/2-4$. The two branches for nonradial stellar oscillations areFor every $l>0$, their product is $-l(l+1)<0$. Therefore one branch oscillates and the other grows exponentially, for every positive $\gamma$. With the chosen time dependence, $x_{-}<0$ gives $\omega =i{\omega }_d\sqrt{-x_{-}}$ and growth rate ${\omega }_d\sqrt{-x_{-}}$. Increasing $\gamma$ does not remove this nonradial instability. At large $n$ or large $\gamma$, $x_{+}\sim \gamma L/2-4$ is a stiff compressive branch, while $x_{-}\sim -l(l+1)/(\gamma L/2-4)$ is a slower unstable buoyancy branch.

The buoyancy frequency supplies the physical explanation. The background density has zero gradient, but pressure decreases outward:The specific entropy of the perfect gas decreases outward, because $p/{\rho }^{\gamma }$ does. An adiabatically displaced fluid parcel therefore experiences destabilizing buoyancy. The model has convective instability of a uniform-density star, even when its radial compression modes are stable. The divergence of this expression at the zero-pressure surface also reflects the artificial equilibrium; the interior negative sign is already decisive.

For $l=0$, the formal quadratic factors asThe physical radial displacement has no $V\nabla F$ component. Using its equations directly givesIndeed $d=(n+1)u$, $j=-3u/n$, and $h=u+\gamma (n+1)u/2$ yield this result directly in the radial momentum equation. The extra identically zero factor in the two-component determinant need not represent an independent radial mode, since $V$ is redundant when $l=0$.

The lowest allowed degree, $n=2$, has ${\omega }_{\mathrm{radial}}^2=(3\gamma -4){\omega }_d^2$. Hence all these radial modes are stable for $\gamma >4/3$; the fundamental radial mode is marginal at $\gamma =4/3$ and unstable below it. Higher $n$ have their individual thresholds $\gamma =8/[n(n+1)]$. These radial stability of a uniform-density star thresholds coexist with nonradial convective instability. They are conclusions within the polynomial interior family assumed in the question; no boundary-condition quantization is being asserted.

Use the inviscid, nonrotating Boussinesq approximation, with a stable background mass density $\bar{\rho }(z)$ and reference density ${\rho }_{\ast }$. Write $B=-g{\rho }^{\prime }/{\rho }_{\ast }$ for the perturbation buoyancy and $\pi =p^{\prime }/{\rho }_{\ast }$ for the kinematic pressure. The buoyancy frequency is $N_0^2=-(g/{\rho }_{\ast }){\bar{\rho }}_z>0$. Dropping products of perturbations in the Boussinesq equations gives the Linearized Boussinesq equationsThe last equation expresses incompressible flow; the second follows by advecting the background mass density gradient. Neglecting rotation and viscosity is part of this internal gravity wave model.

For a plane internal gravity wave proportional to $\exp [i({\mathbf{k}}_h\cdot {\mathbf{x}}_h+mz-\omega t)]$, put $\kappa =|{\mathbf{k}}_h|>0$ and $K^2={\kappa }^2+m^2$. Eliminating the horizontal velocity, pressure and buoyancy from the linear equations givesFor example, the horizontal momentum and continuity equations give $\pi =-\omega mw/{\kappa }^2$; substituting $B=-i{N}_0^{2}w/\omega$ into vertical momentum yields the displayed dispersion relation. Thus an IGW has $0<|\omega |\le N_0$ in this model.

On the positive-frequency branch, the phase velocity normal to a constant-phase plane and the group velocity areConsequently $({\mathbf{k}}_h,m)\cdot {\mathbf{c}}_g=0$, and phase and group velocity are perpendicular. Equivalently, the dispersion relation is homogeneous of degree zero in the wave vector, so differentiating with respect to its scale proves the same orthogonality. Energy travels with the group velocity, along the phase planes. At the degenerate limit $m=0$, $\omega =N_0$ and the group velocity vanishes; the orthogonality statement then has this limiting interpretation.

For a two-dimensional internal gravity wave with fixed horizontal wave number $k\ne 0$, the vertical velocity amplitude satisfiesThis follows from the Linearized Boussinesq equations just as in part (a), allowing the background buoyancy frequency to vary. The WKB approximation gives locallyIt requires slowly varying stratification compared with the local vertical wavelength, in particular $|m_z|\ll m^2$, and a small-amplitude wave in the inviscid, nonrotating regime. Propagation requires $N>|\omega |$; where $N<|\omega |$, the local vertical wave number is imaginary and the solution is evanescent. Turning levels and abrupt changes need a separate connection calculation. Finite viscosity, background flow, rotation or nonlinear effects require modifications of this model.

The ray tracing equations use the local dispersion relation as a Hamiltonian:The stationary, horizontally uniform background conserves the frequency and horizontal wave number. In two dimensions the group velocity gives $dx/dz=-m/k$. For the specified buoyancy frequency and $\omega =1$, $m^2=256k^2{z}^2$. Hence for $z>0$ the two geometric internal-wave rays through the specified point areEach curve has both directed propagation senses, giving four local directed rays. Their formal intersections with $z=0$ are $x=-8$ and $x=8$.

At $z=0$, $N=\omega$ and $m=0$: this is a double-zero internal-wave turning level, since $m^2$ touches zero without changing sign. The ray becomes vertical geometrically, but its group velocity tends to zero. Taking $k>0$ without loss of generality, the downward branch hasThe formal travel time diverges logarithmically as $z\to 0$. Meanwhile $|m_z|/m^2=1/(16|k|z^2)$ grows, so the WKB approximation fails before that limit; its predicted $|m{|}^{-1/2}$ amplitude cannot be extrapolated to infinity. The spatial ray reaches a finite limiting position, but finite-wavelength behaviour near the zero must be found beyond WKB. There is no evanescent half-space supplied by this particular $N(z)$, so ordinary simple-turning-point reflection cannot be assumed solely from the local ray construction.

Use $(\bar{u},\bar{v})=(-{\bar{\psi }}_y,{\bar{\psi }}_x)$ and the hydrostatic thermal-wind balance convention $\bar{b}=f_0{\bar{\psi }}_z$. Because the basic state is independent of $x$, $\bar{v}=0$. The thermal-wind relation isIntegration from the resting lower boundary givesA compatible quasi-geostrophic streamfunction iswhere $C$ is a spatial constant. Its horizontal Laplacian is $z{A}^{\prime \prime }(y)/f_0$ and its second vertical derivative is $B^{\prime }(z)/f_0$. Substituting into the given three-dimensional quasi-geostrophic potential vorticity yieldsIn particular, the horizontal-curvature term has a plus sign: $\bar{\zeta }=-{\bar{u}}_y=z{A}^{\prime \prime }/f_0$.

The reference-buoyancy convention in quasi-geostrophic potential vorticity should be stated. If $\bar{b}$ includes the full reference profile $N^{2}z$, the formula above includes its constant contribution $f_0$. If instead the quasi-geostrophic streamfunction represents pressure relative to that reference hydrostatic state, use $\bar{b}-N^{2}z=f_0{\bar{\psi }}_z$. Then the quoted $\bar{q}$ is reduced by the constant $f_0$. Both conventions give exactly the same velocity and potential-vorticity gradients, and therefore the same subsequent dynamics. The coefficient $N$ in the stated linear model is a prescribed reference buoyancy frequency; if it is also required to equal the full physical buoyancy frequency everywhere, then $B^{\prime }(z)=N^2$, rather than an arbitrary function. The formal expression above retains the requested arbitrary $B$ with the stated constant coefficient.

The three basic modes are radiation, convection and thermal conduction. Radiative transfer dominates transparent atmospheric layers; convection carries heat through sufficiently unstable fluid regions; thermal conduction matters especially in solids or highly conducting dense material.

Consider a chemically homogeneous ideal gas atmosphere and a small, dry parcel displacement, with rapid pressure equilibration and negligible heat exchange during the displacement. The parcel's first law of thermodynamics per unit mass isFor an adiabatic process, $ds=0$. Using hydrostatic equilibrium, $dP/dz=-\rho g$, the parcel followsAfter an upward displacement $\delta z>0$, its temperature relative to its new surroundings isHere $C_p$ is the specific heat capacity at constant pressure, and $N$ is the buoyancy frequency. At the same pressure, a warmer parcel has lower density and accelerates upward. Thus strict instability requires $dT/dz<-g/C_p$. Equivalently, the parcel equation is $\ddot{\delta z}=-N^2\delta z$, withNegative $N^2$ gives growing displacements. The printed non-strict inequality includes the onset boundary: equality is neutrally stable, not a strictly growing instability. This dry parcel buoyancy in a homogeneous atmosphere is the dry Schwarzschild criterion; condensation and composition gradients require modified criteria.

The WKB approximation treats the vertical wavenumber as locally constant over one vertical wavelength. Write $w=\Re \{A(z)e^{ikx+i{\int }^{z}m(s)ds}\}$, with $|m^{\prime }|\ll m^2$ and $|A^{\prime }|\ll |mA|$. For the stationary internal-wave WKB solution, $A\propto m^{-1/2}$ away from singular or turning levels. A small terrain slope and small perturbations are also needed for Linearized Boussinesq equations.

Let $p$ be the perturbation fluid pressure divided by reference mass density, and $b$ the buoyancy perturbation. The steady Linearized Boussinesq equations and incompressibility condition areFor an $e^{ikx}$ component, elimination gives $u=-w^{\prime }/(ik)$, $b=-N^{2}w/(ikU)$ and $p=(U{w}^{\prime }-U^{\prime }w)/(ik)$. Substitution into vertical momentum gives the exact stationary Taylor–Goldstein equationThe leading WKB approximation therefore identifiesThe kinematic boundary condition fixes the horizontal wavenumber to $k=k_T$ for the sinusoidal forcing.

For explicit heights assume a constant positive buoyancy frequency $N$, $0<k{U}_0<N$, and $|S|/N\ll 1$. The latter is a large gradient Richardson number. The linear shear flow has $U^{\prime \prime }=0$, so the propagation condition is $|U|<N/k$.

For $S<0$, the first obstruction is the critical level of an internal gravity wave atAs $z\uparrow z_c$, $m\sim N/U\to \infty$. The critical-height approach of a stationary wave in linear shear has upward group velocity $c_{gz}\sim k{U}^2/N$, so the travel time to $z_c$ diverges: the height is a supremum, approached in the ideal ray model. Here $|m^{\prime }|/m^2\to |S|/N$, which can remain small; it would be incorrect to say that the short-wavelength WKB approximation necessarily fails solely because $m$ diverges. Nevertheless $w\propto U^{1/2}$ and displacement $w/(ikU)\propto U^{-1/2}$, so a finite disturbance eventually violates linearity. Dissipation and overturning also matter near this singular level.

For $S>0$, the turning height of a stationary wave in linear shear isAt $z_t$, $m=0$ and the WKB approximation fails. An Airy function matching region describes reflection and an upper evanescent wave. Thus $z_t$ is the highest height of vertical propagation, rather than a sharp cut-off of every disturbance.

The problem initially allows $N(z)$ to vary. Without the extra constant-$N$ assumption, replace the explicit turning height by the first root of $N(z)=k|U_0+Sz|$ encountered on a propagating branch. A declining $N(z)$ can cause a turning level before the negative-shear critical level. The two closed-form heights require the stated constant-stratification assumption.

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.

Consider the permanent stationary topographic internal gravity wave component of a localized source, with constant buoyancy frequency in each layer and positive $U_1,U_2$. A sufficiently broad obstacle spectrum is assumed. The dispersion relation on the upward branch in a uniform layer isFor a stationary wave, $K=(k^2+m^2{)}^{1/2}=N/U$. Differentiating before imposing stationarity gives the laboratory group velocityThe vertical component is at most $U/2$, attained at $\theta =\pi /4$. The first arrival of a stationary internal wavefront at a layer boundary is thusIf the obstacle spectrum lacks that wavenumber, the first arrival of its actual permanent wave packets is later. This statement concerns the stationary part; arbitrary long-wave startup transients are not bounded by this particular envelope.

In the lower layer an elapsed propagation time $t$ gives $x=U_{1}t{\cos }^2{\theta }_1$, $z=U_{1}t\sin {\theta }_1\cos {\theta }_1$. Elimination gives the causal envelope of stationary internal wavesLater-emitted packets fill the interior. At $t=T$ this semicircle just touches $H$ at $x=H$. At $2T$ its unobstructed continuation would reach $2H$, but rays that hit the interface must instead be refracted or reflected.

For a lower-layer ray labelled by ${\theta }_1$, its interface arrival time and point areThe refraction of a stationary internal-wave causal envelope conserves horizontal wavenumber, so $\cos {\theta }_2=(U_2{N}_1)/(U_1{N}_2)\cos {\theta }_1$. If ${\tau }_1\le t$ and $0<\cos {\theta }_2<1$, its transmitted front is parametrized byThe reflected front instead has $x=U_{1}t{\cos }^2{\theta }_1$ and $z=2H-U_{1}t\sin {\theta }_1\cos {\theta }_1$, until it reaches the ground. At $2T$ that first return just touches the ground. If the $\pi /4$ incident ray is evanescent upstairs, upper-layer propagating arrival is delayed until an allowed ray reaches $H$; this does not change the first lower-layer arrival time $T$.

The stationary phase of topographic internal waves also gives the phase sketch. In a uniform lower layer the Fourier transform integrand has phase $kx+m(k)z$. The stationary phase method requires $x=kz/m$, giving phase $\Phi =(N_1/U_1)(x^2+z^2{)}^{1/2}$. Thus the dominant permanent phase lines are downstream circular arcs centred on the obstacle. Reflected phase arcs have the image source $(0,2H)$. Upstairs the ray arriving at $(x_H,H)$ has phasewhich is continuous at the interface and determines refracted constant-phase curves. Individual monochromatic components have straight phase lines of slope $-k/m_j$ instead; the arcs describe their localized-source superposition.

The schematic uses $H=U_1=N=1$ and $U_2=U_1/2$. It shows the direct front below $H$, the transmitted front above it and the reflected front below it; dotted arcs are permanent phase curves, not material trajectories.

Define $K=N_s^2{z}_s^{-\beta }>0$. The buoyancy frequency definition gives $F_z=-KQ{z}^{\beta }$. The powers in a plume similarity in power-law stratification satisfySolving givesA rising positive-buoyancy turbulent plume entrains, so $Q_z>0$; its momentum flux increases, so $M_z>0$; its buoyancy flux decreases in the stable ambient, so $F_z<0$. With positive coefficients, this requires $q>0$, $m>0$ and $f<0$. The strongest restrictions areIn this interval $1<q<5/3$, $0<m<4/3$ and $-2<f<0$.

Coefficient matching makes the positive similarity solution explicit:soIt is an ideal scaling for $z>0$, often a far-field description. As $z\to 0$, the prescribed ambient stratification and this model's buoyancy flux are singular; a finite physical source would require a regularized near-source region.