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Take a steady moving point source in uniform resting gas, with constant sound speed $c$ and source velocity $U{\mathbf{e}}_x$. Place the source at the origin at the current time $t=0$. In the linear acoustics model, an emission at an earlier time $\tau <0$ has a spherical wavefront centered at $(U\tau ,0,0)$ with radius $-c\tau$. An axial cross-section is the circle in the displayed equation. It is centered at the old release position, not the source's present position. The circles describe the geometry of weak acoustic waves, not a measured shock wave or its pressure jump.

For $M=0.5$, every old circle contains the present source because its center is $U|\tau |$ from the origin while its radius is the larger distance $c|\tau |$. It reaches ahead of the source. For $M=2$, the rightmost point of every old circle is $(U-c)\tau <0$: the source outruns all of its old wavefronts. The continuously emitted wavefronts have a common geometric envelope, the Mach cone. The plotted discrete release times sample this continuous family; rotating the axial section about the motion axis recovers the three-dimensional construction.

The normal acoustic impedance of a progressive acoustic wave at a planar boundary is the pressure amplitude divided by the normal-velocity amplitude. For propagation angle $\theta$ from the normal,The distinction from the acoustic impedance $\rho c$ matters at oblique incidence: interface reflection depends on the ratio of normal impedances. For an evanescent outgoing wave $k_n$ is imaginary and the normal impedance is reactive, giving no mean normal acoustic energy flux.

A finite-amplitude compression can steepen because different parts of the wave travel at different characteristic speeds. For a right-moving simple wave in a perfect gas, one may write $dx/dt=c_0+(\gamma +1)u/2$ in terms of the local velocity perturbation $u$. Faster compressive characteristics catch slower ones. When the smooth inviscid description would become multivalued, a physical compression shock replaces it. A rarefaction normally spreads rather than forming an entropy-violating expansion shock.^[ref]

Across a locally planar shock, use velocities relative to its moving front. The Rankine-Hugoniot conditions for a perfect gas conserve mass, momentum and total energy flux:where $h=\gamma p/[(\gamma -1)\rho ]$ is the specific enthalpy. For upstream normal Mach number $M_1>1$,Thus a normal compression shock increases pressure, density and temperature, reduces the normal velocity, and changes a supersonic upstream flow to subsonic downstream flow in the shock frame. In the strong-shock limit, compression approaches $(\gamma +1)/(\gamma -1)$, which is four for a monatomic gas. A shock conserves total energy but generates entropy; an adiabatic shock is not an isentropic compression. For a calorically perfect gas, stagnation temperature is conserved while stagnation pressure falls.^[ref]

Real dissipative processes resolve the narrow front; the jump conditions do not require its detailed internal profile. In a continuum-gas experiment a shock tube can produce a moving shock by releasing a pressure discontinuity. pressure probes separated along the tube measure arrival speed and the pressure jump; distinguishing the shock from the following contact surface is essential. A contact surface can have a density/entropy jump with continuous pressure and normal velocity, whereas a shock carries nonzero mass flux through its front.

Take incidence from A into B and first specify pressure amplitude. For a plane acoustic wave traveling normally to the interface, the acoustic impedance is $Z=\rho c_s$. A right-moving wave has pressure amplitude $P=ZU$, where $U$ is its fluid-velocity amplitude; a left-moving reflected wave has $P=-ZU$.

Let $P_i,P_r,P_t$ denote incident, reflected, and transmitted pressure amplitudes. Continuity of pressure and normal velocity at the interface givesSolving the two linear equations gives the acoustic reflection and transmission at an interface coefficientsFor the same perfect gas, take a common adiabatic index $\gamma$. The adiabatic sound speed satisfies $c_s^2=\gamma p/\rho$. The equal equilibrium pressure therefore implies $Z_B/Z_A=\sqrt{{\rho }_B/{\rho }_A}=\sqrt{\delta }$, givingThe word amplitude is otherwise ambiguous. For fluid velocity, or displacement at the same frequency, the corresponding answer isThese are consistent because pressure and velocity amplitudes differ by the impedance. As an energy check, the time-averaged acoustic intensity is $|P{|}^2/(2Z)$, so its transmission coefficient is $\mathcal{T}=4\sqrt{\delta }/(1+\sqrt{\delta }{)}^2\le 1$, even when the transmitted pressure amplitude exceeds the incident one.

Equal mean molecular weight alone does not logically guarantee equal $\gamma$. If the gases have different values of the adiabatic index, replace $\sqrt{\delta }$ in these formulas by $\sqrt{{\gamma }_B\delta /{\gamma }_A}$; a density-only answer requires the common-$\gamma$ assumption.

Supersonic communication. A point source moving at speed $V>c_s$ outruns its small acoustic waves. After a time $t$, the source has moved $Vt$, while a disturbance emitted at its old position has radius $c_{s}t$. The envelope of these spherical wavefronts is the Mach cone, with half-angle $\mu$ satisfyingInformation is restricted to the downstream cone, so upstream gas cannot smoothly anticipate a supersonic obstacle. Finite compressions can also steepen because their characteristics converge. Such disturbances form a shock wave, across which the fluid variables change rapidly. A perfectly uniform supersonic flow need not contain a shock: the cone expresses the communication constraint, while a disturbance or compression supplies the reason for a discontinuity.

Conserved fluxes. Work in the shock frame, take velocities normal to the front, and integrate the conservation laws through a thin steady control volume of unit area. Mass conservation gives constant flux $j=\rho u$, soThe normal momentum flux is the advected part $\rho u^2$ plus the pressure stress $p$. With negligible body-force impulse across the thin front,For an ideal gas with constant adiabatic index $\gamma$, the specific internal energy is $e=p/[(\gamma -1)\rho ]$. The energy flux iswhere $h_{\mathrm{sp}}$ is the specific enthalpy. No radiation or other heat flux escapes in this adiabatic shock model. Conservation of the energy flux, divided by the nonzero mass flux, givesThese are the Rankine-Hugoniot conditions for a perfect gas. The shock is adiabatic but irreversible: its entropy increases, so an isentropic relation cannot be imposed across the discontinuity.

Strong-shock compression. Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$ and the upstream enthalpy compared with ${\rho }_1{u}_1^2$ and $u_1^2$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/\chi )$. Substitute this into energy conservation:ThusThe root $\chi =1$ describes no shock. The compressive branch of the Strong-shock Rankine-Hugoniot conditions isIt also gives $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. A monatomic gas has $\gamma =5/3$, hence $\chi =4$.

The finite slab in its initial rest frame. Denote the shock's lab speed by $V_s$. The upstream gas is initially stationary, so $u_1=V_s$ and the downstream shock-frame speed is $u_2=V_s/4$. Transforming to the initial slab frame givesThis is planar shock compression of a finite slab. In the uniform post-shock approximation, breakout occurs at $t_b=L/V_s$. The original front face has then moved $(3V_s/4)t_b=3L/4$, while the shock reaches the originally stationary far face at $L$, leaving thickness $L/4$. Equivalently, the fixed mass per area $\rho L$ becomes $4\rho L_2$. The result assumes a maintained approximately constant shock speed and no important boundary rarefaction during the crossing; an unconfined finite cloud need not remain uniformly compressed.

Escape estimate and the sound-speed convention. Let $M_c$ be the slab mass, $W>0$ its binding-energy magnitude in the star's field, and $E_{\mathrm{th}}$ its initial internal energy. Treat the potential as nearly uniform across the slab. For the bulk-ballistic escape estimate, requireThe Mach-number convention in shock unbinding estimates matters here. With the usual upstream adiabatic sound speed, $c_{\mathrm{ad}}^2=\gamma p/\rho =5p/(3\rho )$, define ${\mathcal{M}}_{\mathrm{ad}}=V_s/c_{\mathrm{ad}}$. ThenIf instead the shock speed is normalized by the isothermal sound speed, $c_T^2=p/\rho$, then ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$ andThe original PDF's “around five” is consistent with the latter convention or an order-of-magnitude sound-speed estimate. It is not the exact coefficient obtained with the conventional adiabatic Mach number and the stated bulk kinetic-energy criterion. Here $W=-E_{\mathrm{grav}}$ if the potential is zero at infinity; using negative gravitational energy directly on the right would be incorrect. Subsequent conversion of shock-generated internal energy to expansion, external pressure, or radiative losses would change the physical escape threshold, so this numerical factor is not universal.

The background and the Jeans prescription. For a resting fluid, ${\mathbf{u}}_0=0$ and continuity is automatic. The remaining hydrostatic equilibrium equations areAn infinite uniform static medium with constant $p_0$ and ${\rho }_0>0$ would require $\nabla {\psi }_0=0$ by force balance, but then its Laplacian would vanish, contradicting the Poisson equation. An isolated static uniform background is therefore not an exact equilibrium of this system.

The Jeans swindle suppresses the gravitational field of the uniform background and retains only the field sourced by perturbations. Formally it replaces the source by $\rho -{\rho }_0$ in the perturbation problem. This can be regarded as a specified compensating background or a local idealization; it does not prove that the original infinite isolated medium can be static.

Linearized equations. Put $\delta \rho =\Delta \rho$, $\delta p=\Delta p$, $\delta \psi =\Delta \psi$, and $\mathbf{v}=\Delta \mathbf{u}$. For uniform ${\rho }_0,p_0$, neglect products of perturbations. Continuity givesThe advective acceleration is quadratic, and the background pressure gradient vanishes, so the Euler equation becomesWith the Jeans prescription and constant sound speed, the gravitational and pressure closures areThe density correction to $1/\rho$ multiplying $\nabla \delta p$ is second order and is consistently omitted.

The longitudinal dispersion relation. Differentiate continuity in time and substitute the divergence of the linearized Euler equation:For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$, ${\partial }_t^2\mapsto -{\omega }^2$ and ${\nabla }^2\mapsto -k^2$. ThereforeThe physical Jeans length isSince $G{\rho }_0$ has units inverse square seconds, $k_J$ has units ${\mathrm{m}}^{-1}$ and ${\ell }_J$ metres. This is the compressive density branch of the Jeans instability; transverse velocity disturbances have no pressure or gravity restoring force in this static nonrotating model and are neutral at linear order.

Which scales grow. For $0<k<k_J$, or wavelength $\ell >{\ell }_J$, ${\omega }^2<0$ and one temporal solution grows exponentially. The static Jeans growth-rate spectrum isIts derivative is negative for $0<k<k_J$, so longer wavelengths grow faster. The supremum is $s\to \sqrt{4\pi G{\rho }_0}$ as $k\to 0^{+}$: there is no selected finite fastest wavelength in this infinite-medium calculation. The exactly uniform $k=0$ density change is not an ordinary finite-potential Fourier perturbation with the fixed background subtraction. In a finite cloud, boundaries set the longest allowed scale.

For $k>k_J$, the density perturbation oscillates as a stable acoustic wave. At $k=k_J$ the linear equation is ${\partial }_t^2\delta \rho =0$, so the mode is marginal rather than exponentially growing.

Comparing acoustic and gravitational times. A sound-crossing estimate for a region of size $\ell$ is $t_{\mathrm{sound}}\sim \ell /c_s$. Self-gravitational acceleration scales as $G{\rho }_0\ell$, so collapse over a distance $\ell$ takes $t_{\mathrm{grav}}\sim (G{\rho }_0{)}^{-1/2}$. Equating these scales giveswhich agrees with the Jeans length up to its numerical factor $\sqrt{\pi }$. This is the physical competition between pressure communication and gravitational contraction.

The constants can be checked using the free-fall time of a uniform sphere. For a pressureless shell initially at radius $a$, enclosing fixed mass $M_a$, energy conservation gives ${\dot{r}}^2=2G{M}_a(1/r-1/a)$. Thuswhere $M_a=4\pi {\rho }_0{a}^3/3$. Consequently the Jeans length and gravitational time conventions satisfyThey agree as characteristic scales, not as identical exact times. Alternatively, using the length $k_J^{-1}$ gives $1/(c_s{k}_J)=1/\sqrt{4\pi G{\rho }_0}$, exactly the gravitational inverse frequency entering the dispersion relation. The actual growth time of a marginal Jeans mode is infinite because pressure and gravity cancel there; it should not be confused with a pressureless free-fall time.

The exponential factor removes the first spatial derivative. Apply the chain rule explicitly:while ${\partial }_t^2\Delta \rho =e^{-z/(2H)}{\Psi }_{tt}$. Substituting into the supplied density equation, the terms proportional to ${\Psi }_z$ cancel, leavingThis is the atmospheric acoustic cutoff transformation. There is a sign error in the target equation in the original PDF, as well as in the transcription: the last term must be $+{\omega }_0^2\Psi$ for real positive ${\omega }_0$. If the displayed minus sign is retained literally, coefficient matching would require ${\omega }_0^2=-c^2/(4H^2)$, so it would not define a real characteristic oscillation period.

For the corrected equation, a plane wave $\Psi \propto e^{i(kz-\omega t)}$ has dispersion relationThus ${\omega }_0$ is the acoustic cutoff frequency: vertically propagating acoustic waves have $|\omega |\ge {\omega }_0$. The scale height and sound speed have units metres and metres per second, respectively, so ${\omega }_0$ has units ${\mathrm{s}}^{-1}$.

For an order-of-magnitude estimate in the lower terrestrial atmosphere, take $H\simeq 8.0\mathrm{km}$ and $c\simeq 340\mathrm{m}{\mathrm{s}}^{-1}$. These are representative atmospheric values, not additional data specified by the paper. They giveThe isothermal atmosphere approximation is local: a real atmospheric temperature profile and the thermodynamic response of the disturbance modify the precise cutoff and period.

A shock wave forms when a compressive flow steepens or when gas must be decelerated or turned faster than upstream acoustic waves can communicate the change. Astrophysical examples include a supernova remnant sweeping up surrounding gas, a fast stellar wind colliding with ambient material, and supersonic accretion onto an obstacle or onto a dense gas layer. A large Mach number by itself is not sufficient: a uniform supersonic stream need not contain a shock wave.

Choose the shock frame, and denote upstream and downstream normal velocities by $u_1,u_2$. Integrating the local conservation laws over a thin control volume spanning a steady normal shock wave gives equal fluxes on both sides. Mass conservation gives constant mass flux per unit area,Pressure and advected momentum together give the normal momentum flux,For a thin non-radiating layer, with negligible gravitational-potential change across it, conservation of energy gives equal fluxes $j(h+u^2/2)$. A perfect gas has specific enthalpy $h=\gamma p/[(\gamma -1)\rho ]$, soThese are the Rankine-Hugoniot conditions for a perfect gas. Dissipation inside the thin layer produces an entropy increase and selects the physical compressive branch. Efficient radiative cooling would change the energy relation and may instead lead to an isothermal shock.

Interpret “reach” first as remaining an energetic, supersonic blast wave, rather than as the arrival of any acoustic wave. An atmospheric blast-wave range estimate must account for the thin atmosphere: the distance $D=8.5\times 10^6\mathrm{m}$ is much greater than its height $H=10^4\mathrm{m}$.

For an optimistic estimate, suppose the deposited energy $E$ remains in a horizontally expanding atmospheric layer. If its representative mass density is ${\rho }_a$, the swept mass is of order $\pi D^{2}H{\rho }_a$. Conservation of energy gives the self-similar blast wave scalingUsing ordinary atmospheric estimates ${\rho }_a\sim 1\mathrm{kg}{\mathrm{m}}^{-3}$ and speed of sound $c_a\sim 300\mathrm{m}{\mathrm{s}}^{-1}$, this givesThese are scaling estimates, not an exact shock wave solution. The strong shock wave approximation would already have failed atEquivalently, keeping the disturbance strongly supersonic out to $D$ would require substantially more than the sonic-transition energy scalewhich exceeds $10^{22}\mathrm{J}$ by about a factor of twenty. Reasonable atmospheric averages and order-unity geometrical coefficients do not reverse the comparison. Escape of energy through the top of the atmosphere makes the fully confined estimate optimistic; Earth's curvature changes its area estimate by an order-unity factor.

The strong atmospheric blast will not survive to Lansing in this model. A weakened acoustic wave can nevertheless travel there. The supplied energy and atmospheric height do not determine its detectable amplitude, detailed attenuation or damage: those require an atmospheric propagation model. Thus “no disturbance of any kind reaches Lansing” would be an unjustified stronger conclusion.

Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

The equilibrium isothermal equation of state is $p=c_s^2\rho$. Hydrostatic equilibrium gives $p_z=-\rho {\Omega }^{2}z$, henceHere $\Omega \ne 0$ is assumed. This is the same local harmonic-gravity balance as isothermal vertical structure in stellar gravity.

For $\gamma >1$, this equilibrium is stably stratified under adiabatic fluid perturbations. Its squared buoyancy frequency isA general perturbation can therefore support acoustic waves restored by compression and internal gravity waves restored by buoyancy, with nonoscillatory entropy or vortical disturbances also possible. The prescribed velocity is purely vertical and horizontally uniform. It describes vertical acoustic waves; it does not contain the horizontally varying circulation required for an oscillatory internal-gravity branch. For zero horizontal wavenumber, that branch has zero frequency. An isothermal equilibrium must not be confused with an isothermal perturbation: the latter would remove the buoyancy restoring force.

Introduce the fluid Lagrangian displacement $\xi (z)e^{i\sigma t}$, so $w=i\sigma \xi$. Use $\delta \rho ,\delta p$ for fluid Eulerian perturbations and $\Delta \rho ,\Delta p$ for fluid Lagrangian perturbations. The continuity equation and the adiabatic relation areThe fixed external gravitational potential is not perturbed. The vertical linearized Euler equations consequently giveSubstituting the preceding relations, and using $p_z=-\rho g$ and $p_{zz}=-{\rho }_{z}g-\rho g_z$, reduces its right-hand side toSince $g={\Omega }^{2}z$ and $g_z={\Omega }^2$, the same equation holds for $w$:The rigid fixed base supplies the essential condition $w(0)=0$. At infinity the condition is finite kinetic energy, not bounded velocity amplitude; a polynomial can grow while the Gaussian mass density makes its energy finite.

Put $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The equation becomes the Hermite differential equation in its probabilists' convention,For $w={\sum }_{j\ge 0}{a}_j{x}^j$, its recurrence isThe base condition makes $a_0=0$ and therefore every even coefficient zero. An odd series terminates at degree $2n+1$ when $\lambda =2n+1$. It remains to prove that a nonterminating odd solution cannot have finite energy; termination alone would not establish the requested exhaustion of modes.

For that purpose, set $y=e^{-x^2/4}w$. The finite-energy condition is $y\in L^2(0,\infty )$, and its equation isBecause the coefficients are even and the solution is odd at the base, extend $y$ oddly to the whole real line. Define lowering and raising operatorsIntegration by parts gives $\Vert ay{\Vert }^2=\lambda \Vert y{\Vert }^2$, so $\lambda \ge 0$. These identities apply to the square-integrable solution: multiplying its equation by a compactly cut off conjugate and taking larger cutoffs bounds $y^{\prime }$ and $xy$ in $L^2$, and removes the endpoint terms. The same argument applies to each lowered eigenfunction.

If $\lambda$ is not a nonnegative integer, repeatedly apply $a$. Each application lowers $\lambda$ by one and remains nonzero whenever its current value is positive, by the norm identity. After finitely many applications it would yield a nonzero square-integrable eigenfunction with negative $\lambda$, contradicting that identity. Thus $\lambda =N$ is a nonnegative integer. At $N=0$, $ay=0$ gives $y\propto e^{-x^2/4}$, which is even. Lowering reverses parity, so an initially odd eigenfunction can reach this even ground state only if $N$ is odd. Consequently $N=2n+1$. The recurrence then terminates and uniquely determines the odd solution from $a_1$.

The complete adiabatic vertical modes of a Gaussian atmosphere are thereforeHere ${\mathrm{He}}_j$ is the Probabilists' Hermite polynomial. For example the first shapes are $x$, $x^3-3x$, and $x^5-10x^3+15x$, with frequencies squared ${\Omega }^2(1+\gamma )$, ${\Omega }^2(1+3\gamma )$, and ${\Omega }^2(1+5\gamma )$. Either sign of each real frequency gives a time-harmonic solution. Gaussian moments make every displayed polynomial have finite kinetic energy. Without the fixed-base condition, finite energy alone would also admit even polynomials, which is why that condition is necessary.

Use the Burgers reduction of weakly nonlinear rightgoing acoustics, with $\theta =t-x$, $X=Mx$, ${\partial }_t={\partial }_{\theta }$ and ${\partial }_x=-{\partial }_{\theta }+M{\partial }_X$. The equation of state expands asAt order one, mass conservation and momentum conservation give $({\rho }_0-u_0{)}_{\theta }=0$ and $(u_0-p_0{)}_{\theta }=0$. For the pure rightgoing acoustic wave with no independent leading mean density or velocity shift, ${\rho }_0=u_0=p_0$. This is the background choice implicit in the requested traveling-wave expansion; additional mean components would modify its phase convention. The source's duplicated phrase about expanding $p$ should refer to expanding $\rho$ as well as $u$.

At order $M$, the mass conservation equation becomeswhereas the momentum conservation equation becomesIn the latter equation, the two nonlinear products ${\rho }_0{u}_{0,\theta }$ and $-u_0{u}_{0,\theta }$ cancel. Substitute the equation of state relation into the former equation and add the two equations to eliminate the first corrections. The result isThe same calculation gives $u_1-p_1=-(\gamma +1)p_0^2/4+C(X)$, so the first corrections may be chosen without an explicit secular factor of $\theta$. This is why the retarded-time description can remain uniform for $|\theta |=O(M^{-1})$, provided the leading profile and its relevant derivatives remain bounded over that range and the slow propagation interval.

This Inviscid Burgers equation can develop a shock, so the source's uniformity assertion needs a pre-shock qualification. If $p_0(\theta ,0)=F(\theta )$ and $a=(\gamma +1)/2$, its method of characteristics givesThe smooth construction ends when this denominator vanishes; if $\max F^{\prime }>0$, the first such distance is $X_s=[a\max F^{\prime }{]}^{-1}$.

For viscosity write $\beta =M\nu$ with bounded positive $\nu$. Since $u_{0,xx}=p_{0,\theta \theta }+O(M)$, the order-$M$ momentum conservation equation instead reads $(u_1-p_1{)}_{\theta }+p_{0,X}=\nu p_{0,\theta \theta }$. Addition now yields the viscous Burgers equationIts positive diffusion resolves the inviscid steepening rather than supporting an unqualified smooth inviscid expansion through shock formation.