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A plane-parallel atmosphere with isothermal sound speed $c_s$ in gravity $-{\Omega }^{2}z$ has mass density ${\rho }_0{e}^{-z^2/(2H^2)}$, $H=c_s/\Omega$. Above a rigid base, vertically directed adiabatic fluid perturbations satisfy $w^{\prime \prime }-z{w}^{\prime }/H^2+({\sigma }^2/{\Omega }^2-1)w/(\gamma H^2)=0$ and $w(0)=0$. Set $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The substitution $y=e^{-x^2/4}w$ converts finite Gaussian-weighted kinetic energy into square integrability for $-y^{\prime \prime }+x^{2}y/4=(\lambda +1/2)y$. Odd extension across the base and the oscillator lowering operator force $\lambda$ to be an odd positive integer. Thus $w_n$ is proportional to the Probabilists' Hermite polynomial ${\mathrm{He}}_{2n+1}(z/H)$. These are vertical acoustic waves; a general stably stratified atmosphere also admits internal gravity waves with horizontal structure.

On polynomials in $L^2$ of standard Gaussian measure, Gaussian integration by parts makes $D$ and $x-D$ adjoints, with commutator $[a^{-},a^{+}]=I$. The Probabilists' Hermite polynomials are obtained by applying $a^{+}$ repeatedly to $1$, and $a^{-}{\mathrm{He}}_n=n{\mathrm{He}}_{n-1}$. These give a Gaussian-space realization of creation and annihilation operators.

The normalized Probabilists' Hermite polynomials diagonalize the Gaussian number operator with eigenvalues $n\ge 0$. Its self-adjoint closure has domain $\sum n^2|c_n{|}^2<\infty$. Polynomial truncations are dense for its graph norm. The semigroup $e^{-tN}$ is the Ornstein-Uhlenbeck semigroup and its energy form is the Gaussian Dirichlet energy.

The first Gaussian Sobolev space consists of functions and their weak first derivatives square-integrable for standard Gaussian measure. If $f=\sum c_n{h}_n$ in the normalized Probabilists' Hermite polynomial basis, Gaussian integration by parts gives $⟨f^{\prime },h_{n-1}⟩=\sqrt{n}{c}_n$. Thus Parseval identity identifies it with $\sum (1+n)|c_n{|}^2<\infty$, and polynomial truncations converge in its norm. It is the form domain of the Gaussian number operator, with Gaussian Dirichlet energy $\sum n|c_n{|}^2$.

In an inviscid locally isothermal Keplerian disk, a radial normal mode of wavenumber $k$ has pressure variable $\xi$ proportional to a Probabilists' Hermite polynomial in $x=z/H$. Gaussian-weighted finite energy excludes the exponentially growing vertical branch. Polynomial termination gives $n=0,1,\dots$ and the local dispersion relation $({\hat{\omega }}^2-n{\Omega }_K^2)({\hat{\omega }}^2-{\Omega }_K^2)=k^2{c}_s^2{\hat{\omega }}^2$, where $\hat{\omega }=\omega -m{\Omega }_K$. At singular values introduced by eliminating velocity components, the original equations must also be checked. The $n=0$ propagating branch is an inertial-acoustic wave; $n=1$ has vertically uniform vertical velocity and odd pressure perturbation, permitting bending disturbances.

The one-dimensional semigroup associated with the standard Ornstein-Uhlenbeck process has invariant standard Gaussian measure and generator $Lf=f^{\prime \prime }-x{f}^{\prime }$. Its Mehler formula for the Ornstein-Uhlenbeck semigroup makes positivity and invariance explicit. The normalized Probabilists' Hermite polynomials diagonalize it, with $P_t{h}_n=e^{-nt}{h}_n$. It has spectral gap one and satisfies the sharp Gaussian Poincaré inequality.

Interpret the ordered Iterated Itô integral recursively as $I_0(t)=1$ andThe diagonal boundary of the ordered region does not affect this recursive definition. For $n=1$, $I_1(t)=B_t$. For $n=2$, part (a)(iii) gives $I_2(t)=\frac{1}{2}(B_t^2-t)$. For $n=3$, use parts (b)(ii) and (b)(iii):Thus the three Probabilists' Hermite polynomials giveThe right sides are interpreted as their polynomial extensions at $t=0$. These space-time Hermite polynomials arise here directly from ordered Iterated Itô integrals.

Use the Probabilists' Hermite polynomials, rather than the differently scaled physicists' convention:Thus $H_2=x^2-1$, $H_3=x^3-3x$, and $H_5=x^5-10x^3+15x$. The cumulant-generating function of an exponential variable of mean $\theta$ is $-\log (1-\theta t)$ near zero, so its $r$th cumulant is $(r-1)!{\theta }^r$. Hence the standardized third and fourth cumulants are ${\rho }_3=2$ and ${\rho }_4=6$; the fourth is a cumulant, not the fourth raw central moment.

Write $F_n$ for the cumulative distribution function of the pivot in part (i). Its Edgeworth expansion, valid here because the exponential law is smooth, nonlattice and has exponential moments, isat the fixed endpoints $t=\pm z$. The coverage is exactly $F_n(z)-F_n(-z)$. Since $H_2$ and $\varphi$ are even, the $n^{-1/2}$ terms cancel; the two odd Hermite terms double. Consequently the symmetric Edgeworth coverage cancellation givesIn particular its coverage probability error is $\boxed{O(n^{-1})}$. This is a fixed-confidence-level expansion; it is not a uniform statement for normal cutoffs increasing with $n$.

Let the summands have mean mu, variance ${\sigma }^2>0$, and cumulants ${\kappa }_j$. For the standardized sum $Z_n=(S_n-n\mu )/(\sigma \sqrt{n})$, expansion of its log-characteristic function gives a normal leading term and successively smaller cumulant terms. Fourier inversion turns these terms into Probabilists' Hermite polynomials. Under sufficient moments and a smooth nonlattice condition, the Edgeworth expansion beginsfor bounded z. Higher orders include the fourth cumulant and products of lower cumulants. This is a central expansion in powers of $n^{-1/2}$; it need not remain positive, normalized, or accurate in far tails after truncation.

For a saddlepoint density approximation, require the summand cumulant-generating function K to be finite on an interval about zero. At a sum value $s=ny$, find the interior saddle $\hat{t}$ satisfying $K^{\prime }(\hat{t})=y$. Exponential tilting makes the tilted mean y, so the tilted sum is approximated locally by its normal probability density at its mean. Undoing the tilt givesEquivalently this follows by deforming the transform-inversion contour through its stationary point. Suitable analytic and nonlattice regularity gives relative error $O(n^{-1})$ at fixed interior y. This approximation incorporates a nonquadratic tail exponent and is positive, while an Edgeworth truncation uses a normal probability density times a polynomial. It uses more information from the distribution and stronger exponential-moment assumptions. Its integral may differ from one by a higher-order amount, so positivity does not imply exact normalization.

For the centered Laplace distribution, $K(t)=t^2+t^4/2+\cdots$, hence ${\kappa }_2=2$ and ${\kappa }_3=0$. Its skewness vanishes, so the order-$n^{-1/2}$ Edgeworth correction is zero. With $z=s/\sqrt{2n}$, the requested expansion, omitting terms of order $n^{-1}$ in the standardized correction, isFor bounded z the exact probability density function is this expression times $1+O(n^{-1})$. The prefactor is $n^{-1/2}$ because this is the probability density function of the unstandardized sum; that change of units should not be confused with the orders inside the Edgeworth expansion.

For the saddlepoint, $K^{\prime }(t)=2t/(1-t^2)$. Write $y=s/n$ and $d=\sqrt{1+y^2}$. The root in $(-1,1)$ isThese formulas include $s=0$ without a division by y. Substitution yieldsThis is the leading saddlepoint approximation, without its order-$n^{-1}$ relative correction. Expanding d near one recovers the same central normal approximation, while retaining d captures the Laplace-sum large-deviation exponent away from the center.

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.

Write $d{\gamma }_1(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}dx$. On the dense polynomial algebra in $L^2({\gamma }_1)$, the Gaussian creation and annihilation operators areGaussian integration by parts gives $⟨a^{-}f,g⟩=⟨f,a^{+}g⟩$. Their closed realizations are adjoints. More explicitly, for the orthonormal basis $h_n={\mathrm{He}}_n/\sqrt{n!}$ of normalized Probabilists' Hermite polynomials,Thus the closed annihilation operator has domain $\{\sum c_n{h}_n:\sum n|c_n{|}^2<\infty \}$, and the closed creation operator has domain $\{\sum c_n{h}_n:\sum (n+1)|c_n{|}^2<\infty \}$. These describe the same set of $L^2$ functions, the Gaussian Sobolev space, though the two operators differ. Directly on polynomials, $[a^{-},a^{+}]=I$, and

Use the standard carré du champ operator convention $\Gamma (f,g)=\frac{1}{2}(L(fg)-fLg-gLf)$. The product rule for this semigroup generator givesso the requested squared-gradient and joint energy formulas, for real functions in the standard algebra, areFor complex functions the joint energy is sesquilinear, with $\overline{g^{\prime }}$ in the second factor. If “squared gradient” is defined without the factor $1/2$ in the product-rule expression, its value is instead $2f^{\prime }{g}^{\prime }$ and the joint energy is one half of its integral; the joint energy and logarithmic Sobolev constant used here remain as displayed.

For the Gaussian logarithmic Sobolev inequality, first take smooth bounded $f$ with bounded first and second derivatives, bounded away from zero. The Mehler formula for the Ornstein-Uhlenbeck semigroup isIt preserves ${\gamma }_1$ and differentiating in $x$ gives the Ornstein-Uhlenbeck gradient commutation identityPut $u_t=P_{t}f$. Differentiate its entropy functional. Since $\int u_{t}d{\gamma }_1=\int fd{\gamma }_1$ is constant, and Gaussian integration by parts applies,For such data, Mehler's formula and dominated convergence give $P_{t}f(x)\to \int fd{\gamma }_1$ as $t\to \infty$. The uniform positive upper and lower bounds justify integrating this limit inside the entropy, whose limit is zero. Hence

Apply the allowed weighted Cauchy-Schwarz inequality with $g=f^{\prime }$:After integration, invariance of ${\gamma }_1$ givesIntegrating in $t$ yieldsFor a real smooth compactly supported $g$, take $f=g^2+ϵ$. ThenLetting $ϵ\downarrow 0$ provesSmooth approximation and cutoff in the Gaussian Sobolev norm extend this to the full form domain. The entropy is lower semicontinuous under the resulting $L^2$ convergence: the square densities converge in $L^1$, their masses converge, and, along an almost-everywhere convergent subsequence, Fatou's lemma applies to $h\log h$ after adding the bound $1/e$. The same inequality for complex $g$ follows by applying it to $|g|$ and using the weak-derivative bound $|(|g|{)}^{\prime }|\le |g^{\prime }|$. Thus the constant is at most $2$.

For sharpness of the Gaussian logarithmic Sobolev constant, take $g_s(x)=e^{sx/2}$ with real $s\ne 0$. These functions belong to the Gaussian Sobolev space. The Gaussian moment-generating function givesThereforeThe entropy-to-energy ratio is exactly $2$, so no smaller constant can work. Combining this lower bound with the proved inequality gives

Let the summands be independent and identically distributed random variables with mean $\mu$, variance ${\sigma }^2>0$, and standardized third cumulant ${\gamma }_1=\mathbb{E}[(Y_1-\mu {)}^3]/{\sigma }^3$. DefineAn Edgeworth expansion refines the central limit theorem using the higher cumulants of one summand. Some smoothness and moment assumptions are essential: finite variance alone does not supply a density expansion. A convenient sufficient set for the expansions here is a smooth density whose derivatives decay faster than every inverse power, together with $\mathbb{E}{e}^{\delta |Y_1|}<\infty$ for some $\delta >0$. These assumptions give all required moments, Fourier smoothing and the nonlattice characteristic-function condition. Less restrictive sufficient conditions are possible, but this set covers the logistic model below.

Write $\varphi$ and $\Phi$ for the standard normal density and standard normal distribution function. The first-order density Edgeworth expansion isThe remainder is uniform on every fixed compact interval under the stated assumptions. To see the origin of the correction, for $Z=(Y_1-\mu )/\sigma$ the cumulant expansion of its characteristic function givesExponentiating and using Fourier inversion yields the displayed correction, since the inverse transform of $(it{)}^3{e}^{-t^2/2}$ is $-{\varphi }^{(3)}(x)=(x^3-3x)\varphi (x)$. The Probabilists' Hermite polynomial convention is being used: $H_2(x)=x^2-1$ and $H_3(x)=x^3-3x$.

The corresponding distribution function expansion isuniformly on fixed compact intervals. The sign follows from $\frac{d}{dx}[H_2(x)\varphi (x)]=-H_3(x)\varphi (x)$. These are expansions of both the density and distribution function; a uniform density error alone would not justify integrating an error bound over the entire real line without the accompanying tail control.

For a fixed $\alpha \in (0,1)$, write $z=z_{\alpha }={\Phi }^{-1}(\alpha )$. At leading order the proposed quantile expansion gives $\Phi (p_0(z))=\Phi (z)$, and strict monotonicity gives $p_0(z)=z$. Substitute $y_{\alpha }=z+p_1(z)n^{-1/2}+O(n^{-1})$ into the distribution function expansion. A Taylor expansion at $z$ givesSince $\varphi (z)>0$, the coefficient of $n^{-1/2}$ must vanish. Thus the first Cornish-Fisher expansion coefficients are

For the logistic distribution, set $Z_i=Y_i-\theta$. Its density $q(z)=e^z/(1+e^z{)}^2$ satisfies $q(-z)=q(z)$ and has exponentially decreasing tails. Symmetry and integrability give $\mathbb{E}{Z}_i=0$, and the supplied integral givesTherefore $\mathbb{E}{Y}_i=\theta$, $\mathrm{Var}(Y_i)={\pi }^2/3$, and the central limit theorem givesWriting $z_{-}=z_{\alpha /2}$ and $z_{+}=z_{1-\alpha /2}$, the condition $z_{-}\le T_n\le z_{+}$ is equivalent toConsequently the confidence intervalhas limiting coverage $\Phi (z_{+})-\Phi (z_{-})=1-\alpha$. Whether the finite endpoints are included makes no probability difference because the sample mean has a density.

The centered logistic distribution is symmetric, so ${\gamma }_1=0$. Its smooth density and exponentially decreasing derivatives satisfy the conditions stated above; in particular all moments exist and Fourier smoothing is available. The first correction in the Edgeworth expansion vanishes, leaving $F_{T_n}(z)=\Phi (z)+O(n^{-1})$ at each of the two fixed quantiles. Subtracting gives the stronger conclusionThe standardized law is independent of $\theta$, so this coverage-error conclusion holds with the same bound for every location parameter.

In a thin disk dominated by a central stellar mass, the vertical gravitational potential is harmonic to leading order, so its vertical acceleration is $-{\Omega }_0^{2}z$. For a locally isothermal background, $p=c_s^2\rho$ with $c_s$ independent of height. Vertical hydrostatic equilibrium therefore givesThis is the isothermal vertical structure in stellar gravity. On the upper half of the disk the downward gravity magnitude is $g={\Omega }_0^{2}z>0$; the density profile extends evenly through the midplane.

For an adiabatically displaced parcel kept in pressure equilibrium, the parcel's logarithmic density gradient is ${\gamma }^{-1}d\log p/dz$, whereas the environmental density gradient is $d\log \rho /dz$. A small upward displacement $\eta$ consequently produces a density contrast $\delta \rho /\rho =[{\gamma }^{-1}d\log p/dz-d\log \rho /dz]\eta$. Its buoyancy acceleration is $-g\delta \rho /\rho =-N^2\eta$. Thus the Brunt–Väisälä frequency isBoth background gradients equal $-z/H^2$ in the isothermal atmosphere, soAn isothermal background is therefore stably stratified for adiabatic perturbations with $\gamma >1$, but not buoyant for isothermal perturbations. In applying the adiabatic equation of state to a stratified atmosphere, its proportionality relates Lagrangian parcel changes: $\Delta p=(\gamma p/\rho )\Delta \rho$. Eulerian perturbations generally obey $p^{\prime }+\eta p_z=(\gamma p/\rho )({\rho }^{\prime }+\eta {\rho }_z)$ instead. Thus the printed simple proportionality must be interpreted as a parcel relation for the buoyancy calculation. For $\gamma =1$ in this isothermal background, the gradient terms cancel and the Eulerian proportionality holds too.

Now take isothermal perturbations, so $\gamma =1$ and $N^2=0$. The two supplied wave equations reduce toFor the nonsingular oscillatory case $\omega \ne 0$ and ${\omega }^2\ne {\Omega }_0^2$, substitute $u_z={\xi }_z/{\omega }^2$. With $x=z/H$, the resulting Hermite differential equation isThe exceptional frequencies must be treated through the original velocity-pressure equations rather than division by these factors.

The physical vertical condition is finite perturbation energy when integrated with the Gaussian density. In particular the pressure contribution is proportional to $\int e^{-x^2/2}|\xi {|}^{2}dx$ for nonzero frequency. Generic nonterminating solutions have an exponentially growing component at at least one vertical end and fail this condition; acceptable modes are the terminating Hermite polynomials. This is a finite-energy condition, not pointwise boundedness of the velocity in the vanishing-density atmosphere.

To prove the allowed values of $\alpha$, let the nonzero polynomial have degree $n$ and leading coefficient $a_n$. The highest power in ${\xi }_{xx}-x{\xi }_x+\alpha \xi$ has coefficient $(\alpha -n)a_n$. Hence $\alpha =n\in \{0,1,2,\dots \}$. Conversely, writing $\xi ={\sum }_j{a}_j{x}^j$ gives the recurrenceFor $\alpha =n$, choose the parity of $n$ and iterate until the series terminates at degree $n$. Thus all allowed Hermite vertical modes of an isothermal disk havewhere ${\mathrm{He}}_n$ is the Probabilists' Hermite polynomial. In particular, ${\mathrm{He}}_0=1$ and ${\mathrm{He}}_1=x$.

Let $K=kH$ and $s={\omega }^2/{\Omega }_0^2$. Eliminating the denominator for regular modes givesFor $n=0$, the nonzero-frequency solution isIts pressure variable is independent of height and $u_z=0$. These are inertial-acoustic waves, restored by epicyclic motion and pressure; at short radial wavelength they become ordinary acoustic waves. The multiplied dispersion polynomial also has $s=0$. This is not an extra oscillatory solution represented by nonzero $\xi$, since $\xi =i\omega p^{\prime }/\rho$ vanishes at zero frequency for finite pressure perturbation. The full horizontal equations have a zero-frequency balanced mode of an axisymmetric disk, with no radial velocity and azimuthal Coriolis acceleration balancing a radial pressure gradient. It must be recovered without dividing by $\omega$, rather than counted as a propagating wave. At $k=0$, the propagating branch limits to an epicyclic oscillation at $|\omega |={\Omega }_0$, another value at which the eliminated form must be interpreted by a limit.

For $n=1$, solve $(s-1{)}^2=K^{2}s$. The two branches areBoth signs of frequency occur on each squared-frequency branch. The upper branch is acoustic in the large-$|K|$ limit, $|{\omega }_{+}|\sim |k|c_s$; the lower branch is an inertial wave with $|{\omega }_{-}|\sim {\Omega }_0/|K|$. At $|K|\ll 1$, both frequencies approach ${\Omega }_0$, reflecting the equality of the radial and vertical epicyclic frequencies in a Keplerian disk. Here $\xi \propto z$, so $u_z={\xi }_z/{\omega }^2$ is independent of height: these modes have a coherent vertical displacement of the disk column, rather than a vertical breathing motion.

For a local non-axisymmetric disturbance proportional to $e^{i(kR+m\varphi -\omega t)}$, the material frequency is $\hat{\omega }=\omega -m{\Omega }_0$. In the same local radial-wave approximation, replace $\omega$ by $\hat{\omega }$. For $n=1,m=1$, this givesThe four frequencies can equivalently be written $\omega /{\Omega }_0=1\pm (\sqrt{K^2+4}\pm |K|)/2$, with independent choices of the two signs. This is a local dispersion relation, not a claim that Doppler substitution supplies an exact global non-axisymmetric solution in a differentially rotating disk. The radial-wave ordering neglects azimuthal gradients compared with radial gradients; $|k|R\gg 1$ can coexist with $|k|H\ll 1$ in a thin disk.

Write $w=\omega /{\Omega }_0$. For the two low inertial-frame frequency branches, $|w|\ll 1$, while the material frequency remains near $-{\Omega }_0$. Expanding the dispersion relation givesThereforeFor signed $k$, the two leading branches are conventionally labelled by $\pm kH/2$. These are bending waves of a Keplerian disk: an $m=1$ vertical displacement varies once around an annulus and corresponds to a small tilt of its orbital plane. A radial variation of the tilt is a warp. The waves transport the warp radially at group speeds $c_s/2$ in opposite directions. At zero radial wavenumber their low-frequency limit is a stationary rigid tilt; the apparent slow pattern does not mean that the orbiting gas has slow vertical motion in its own frame.

Consider the operators on all of $C_0(X)$ or $L^2(\mu )$, extending the displayed positive-function formula linearly. Here $C_0(X)$ is the space of continuous functions vanishing at infinity. The multiplier has modulus at most one, so $\Vert P_{t}f\Vert \le \Vert f\Vert$, and multiplication of exponentials gives $P_{s+t}=P_s{P}_t$ and $P_0=I$. For strong continuity on $C_0(X)$, choose a compact $K$ outside which $|f|$ is small. On $K$ the continuous $k$ is bounded, making $e^{-tk}\to 1$ uniform; outside $K$, $|(e^{-tk}-1)f|\le |f|$. For $L^2(\mu )$, use pointwise convergence and the dominated convergence theorem with bound $|(e^{-tk}-1)f{|}^2\le |f{|}^2$. Thus both are contraction semigroups, even when $k$ is unbounded.

The generator of a multiplication semigroup isIndeed pointwise difference quotients converge to $-kf$. A norm limit therefore forces the displayed domain condition, using an almost-everywhere convergent subsequence for the $L^2$ case. Conversely $|(e^{-tk}-1)/t|\le k$, so the dominated convergence theorem proves the $L^2$ limit when $kf\in L^2$. In $C_0(X)$ write the error as $kf$ times $1-(1-e^{-tk})/(tk)$, with the value at $k=0$ supplied by continuity. That second factor tends uniformly to zero where $k$ is bounded and has modulus at most one; the compact-set and small-tail argument applied to $kf$ proves uniform convergence. To prove directly that $L$ is a closed operator, take $f_n\to f$ and $L{f}_n\to g$. Pointwise limits in $C_0$, or a common almost-everywhere convergent subsequence in $L^2$, give $g=-kf$. Thus $f$ is in the appropriate generator domain and $Lf=g$. In $L^2$ the domain is dense because $f{1}_{\{k\le n\}}\to f$; in $C_0$ compactly supported continuous functions are a dense subspace contained in the domain.