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Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

Work in ${\mathbb{C}}^N$, the domain of the matrix $A$; the printed ${\mathbb{C}}^n$ in the introduction is a dimension typo. For a fixed nonempty support of a vector $S$, write $u$ for the coordinates of $x$ in $S$. ThenThe Gram matrix $A_S^{\ast }{A}_S$ is a Hermitian matrix, so its difference from the identity matrix is also a Hermitian matrix. By the finite-dimensional spectral theorem, its matrix 2-norm is the largest absolute eigenvalue, equivalentlyIndeed, an expansion in an orthonormal eigenbasis bounds every Rayleigh quotient by the largest absolute eigenvalue, and an appropriate eigenvector attains that bound. The restricted isometry constant must bound this quantity for every $S$ of size at most $s$, and the maximum of these quantities suffices for all sparse vectors of order $s$. There are finitely many sets, so the maximum exists. The empty set contributes zero. Therefore

Let $a_{\xi }=\mathbf{\xi }\cdot \nabla \Psi$, $d_{\xi }=\nabla \cdot \mathbf{\xi }$ and ${\rho }_{\Psi }=d\rho /d\Psi$. Since $\nabla p=-\rho \nabla \Psi$, the pressure and mass density perturbations are $\delta p_{\xi }=\rho a_{\xi }-\gamma p{d}_{\xi }$ and $\delta {\rho }_{\xi }=-\rho d_{\xi }-{\rho }_{\Psi }{a}_{\xi }$. Integrating the pressure-gradient term by parts in the mass density-weighted inner product givesThe boundary term vanishes for regular admissible displacements because $p,\rho$ vanish there and $\delta p=\rho a-\gamma pd$. Completing the pressure square giveswhereThis effective-potential stratification coefficient is real. All coefficients of the bilinear form are real, so $⟨\eta ,\mathcal{F}\xi ⟩=⟨\mathcal{F}\eta ,\xi ⟩$: the operator is symmetric on the stated boundary domain, giving the usual self-adjoint realization of the stellar normal-mode problem.

Set $\eta =\xi$ and use $\mathcal{F}\xi =-{\omega }^2\xi$. The Cowling energy principle for a rotating barotropic star isThe Rayleigh quotient $Q[\xi ]/\int \rho |\xi {|}^2$ tests stability. If $Q\ge 0$ for every admissible displacement, no mode has ${\omega }^2<0$, so there is no exponentially growing mode. If an admissible trial displacement has $Q<0$, the Rayleigh-Ritz variational principle puts negative spectrum below zero; in the usual discrete stellar mode problem this gives a mode with $\omega =i\sigma$, $\sigma >0$. Equality allows neutral modes, rather than establishing strictly positive frequencies. A locally negative coefficient alone is not a complete instability proof: a trial function must also control the pressure and other positive terms.

QMA is the class of promise problems for which a uniform polynomial-size quantum circuit checks a polynomial-size quantum witness. A YES instance has some quantum witness accepted with probability at least $2/3$; on a NO instance every quantum witness is accepted with probability at most $1/3$. The verifier initializes its own ancilla qubits to zero. Mixed quantum witnesses do not improve the maximum because acceptance is linear in their density operator.

For a verifier $U_x$ and quantum witness embedding $J|w⟩=|w⟩|0\cdots 0⟩$, define the witness acceptance operatorThe Rayleigh quotient gives ${\max }_{\Vert w\Vert =1}⟨w|F|w⟩={\lambda }_{\max }(F)$. Every entry of $F$ can be recomputed by the polynomial-storage path summation of part (b); no full quantum witness matrix needs to be stored.

Let the quantum witness have $m$ qubits, so its dimension is $D=2^m$. For this positive operator, trace-power witness optimization usesThis is the useful exponentiated form of the supplied logarithmic inequality. Positivity is required; such a logarithmic statement is not true for an arbitrary operator. No logarithms or roots need to be calculated.

Choose $d=2m+2$. On a NO instance,since $2^m(2/3{)}^{2m+2}=(4/9)(8/9{)}^m<1$. On a YES instance, $\mathrm{Tr}(F^d)\ge (2/3{)}^d>(1/2{)}^d$. Thus comparison of $2^d\mathrm{Tr}(F^d)$ with one distinguishes the cases using only multiplication and a final comparison.

Evaluate the trace byThere are $d=O(m)$ active quantum witness indices, each requiring $m$ bits. Accumulate one product at a time and recompute its matrix entries as required. Together with the verifier path workspace, this uses polynomially many real registers. The potentially enormous time and the numerical precision are unrestricted in $C$. Therefore $\mathrm{QMA}\subseteq C$, without solving an exponentially large eigenvalue problem by storing its matrix.

Take a static, nonrotating, nonmagnetic spherical equilibrium with $\nabla P=-\rho \nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. Use a fluid displacement $\mathbf{\xi }(\mathbf{r})e^{-i\omega t}$. Write $p_1,{\rho }_1,{\varphi }_1$ for Eulerian fluid perturbations. The Lagrangian pressure perturbation and corresponding mass density change obey $\Delta P=p_1+\mathbf{\xi }\cdot \nabla P$, $\Delta \rho ={\rho }_1+\mathbf{\xi }\cdot \nabla \rho$. Linearizing mass conservation, the Euler equations for an inviscid fluid and Poisson equation for Newtonian gravity givesThere is no equilibrium acceleration to multiply a perturbed mass density. The adiabatic process condition, with composition carried by the parcel, closes the system:These are the complete linear adiabatic stellar oscillation equations, including the perturbation of self-gravity. Neglect of heat exchange is appropriate when oscillation periods are short compared with relevant thermal relaxation times; it does not determine nonadiabatic excitation or damping.

For the pressure and buoyancy modes, separate angular dependence using spherical harmonics:The horizontal amplitude ${\xi }_h$ has dimensions of length. Angular differentiation gives $\nabla \cdot \mathbf{\xi }=[r^{-2}(r^2{\xi }_r{)}^{\prime }-\ell (\ell +1){\xi }_h/r]Y_{\ell }^m$, and tangential momentum gives ${\omega }^{2}r{\xi }_h=p_{\ell }/\rho +{\varphi }_{\ell }$. Define the adiabatic sound speed, stellar buoyancy frequency and Lamb frequency byThe adiabatic mass density relation becomes ${\rho }_{\ell }/\rho =p_{\ell }/(\rho c_s^2)+(N^2/g){\xi }_r$. Substitution produces a radial form of the full oscillation equations for nonzero $\omega$:The apparent $N^2/g$ factor is evaluated through its defining gradient at the centre rather than by dividing two zeros.

Regularity at the centre excludes singular solutions. At a free surface the Lagrangian pressure perturbation vanishes, $\Delta P=0$; outside the star the gravitational perturbation decays as $r^{-\ell -1}$. For a model with mass density tending to zero at its surface, continuity of ${\varphi }_{\ell }$ and its radial derivative gives ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R$. If the equilibrium mass density jumps to vacuum, include the displaced-surface mass sheet: the outward-minus-inward derivative jump is $4\pi G\rho (R){\xi }_r(R)$, so the interior condition is ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R-4\pi G\rho (R){\xi }_r(R)$. An atmospheric boundary condition can replace the ideal free surface.

These conditions make ${\omega }^2$ an eigenvalue, not an arbitrary local sound frequency. With conservative boundary conditions the adiabatic operator is self-adjoint, giving real ${\omega }^2$; negative values describe instability. The frequencies depend on $P(r),\rho (r),{\Gamma }_1(r)$, self-gravity, stratification and boundaries, as well as angular degree $\ell$ and radial order. In a spherical nonrotating star they are degenerate in $m$. For homologous equilibrium structures, their scale isthus the typical oscillation time measures inverse square root of mean mass density, while individual frequencies probe the interior adiabatic sound speed and stellar buoyancy frequency profiles.

For radial modes, write ${\xi }_r=r\eta$. Eliminating the pressure and gravitational perturbations gives the radial stellar pulsation equationThis is a Sturm-Liouville problem. Multiplication by $\eta$ and integration, with vanishing boundary terms, gives its Rayleigh quotientFor constant ${\Gamma }_1>4/3$, $P^{\prime }<0$ makes both numerator contributions nonnegative. At ${\Gamma }_1=4/3$, a homologous displacement is neutral; for constant ${\Gamma }_1<4/3$ the same trial displacement makes the quotient negative. For the uniform-density stellar model, $\eta =$ constant is an exact mode and ${\omega }^2=(3{\Gamma }_1-4)GM/R^3$. With varying ${\Gamma }_1$, the integral criterion, rather than a universal pointwise threshold, controls radial stability.

The Cowling approximation neglects ${\varphi }_{\ell }$ while retaining the equilibrium gravitational field. It is useful for short-wavelength modes, but is not needed for the full derivation above. In a locally slowly varying region, take both remaining amplitudes proportional to $e^{i\int k_{r}dr}$ and retain the leading derivative terms. ThenEliminating either amplitude gives the acoustic-gravity propagation relationPositive $k_r^2$ is oscillatory propagation; negative $k_r^2$ means an evanescent wave. The high-frequency branch, ${\omega }^2>S_{\ell }^2,N^2$, describes stellar acoustic modes, restored chiefly by compressibility and pressure. At frequencies well above $N$, this gives ${\omega }^2\simeq c_s^2[k_r^2+\ell (\ell +1)/r^2]$. The low-frequency propagating branch in stable stratification, ${\omega }^2<N^2,S_{\ell }^2$, instead describes stellar gravity modes, restored by buoyancy. At $\ell =0$ there is no such nonradial gravity-wave cavity.

A stellar acoustic mode is trapped between an inner turning point near $\omega =S_{\ell }$ and an outer reflecting region. Low-$\ell$ modes penetrate deeply; radial modes reach the centre. Higher-degree modes turn farther out. Standing waves require the WKB quantization conditionwhere the phase $\alpha$ depends on the central or turning-point behaviour and surface reflection. For high radial order and small degree, the leading acoustic travel-time result is the large frequency separationThe $\ell /2$ term is the leading central angular phase shift; smaller frequency separations depend on detailed interior gradients. The near-surface mass density stratification sets the acoustic cutoff frequency. In a plane-parallel isothermal atmosphere with mass density scale height $H_{\rho }$, ${\omega }_{\mathrm{ac}}\simeq c_s/(2H_{\rho })$. Modes below this cutoff can reflect and form a cavity; waves sufficiently above it escape and need an outgoing-wave boundary condition. Nonadiabatic damping, driving and rotation alter real-star mode properties, but the adiabatic frequency problem isolates their dependence on the equilibrium structure.

The Rayleigh quotient is $R[\psi ]=⟨\psi ,H\psi ⟩/⟨\psi ,\psi ⟩$, interpreted through the energy quadratic form for trial states in its form domain. Expanding in normalized energy eigenstates gives $R[\psi ]=\sum E_n|c_n{|}^2/\sum |c_n{|}^2\ge E_0$. If the ground state is unique, equality requires all coefficients outside its one-dimensional eigenspace to vanish. Thus equality means the same ground-state ray, allowing normalization and an overall phase.

Use the normalized trial function $\sqrt{2/a}\cos (\pi x/a)$ on $|x|<a/2$, zero outside. Its kinetic energy is ${\pi }^2{\hslash }^2/(2m{a}^2)$ and its mean square position is $c_0{a}^2$, where $c_0=1/12-1/(2{\pi }^2)$. The energy isMinimization gives $a^4={\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_0)$ andFor the first excited well state use $\sqrt{2/a}\sin (2\pi x/a)$, zero outside. Now the kinetic energy is $2{\pi }^2{\hslash }^2/(m{a}^2)$ and the position factor is $c_1=1/12-1/(8{\pi }^2)$. Minimizing gives $a^4=4{\pi }^2{\hslash }^2/(m^2{\omega }^2{c}_1)$ andThis particular excited-state estimate is an upper bound. The trial function is odd and the oscillator ground state is even, so it is exactly orthogonal to the ground state. Its spectral expansion therefore contains only energies at least $E_1$, giving the same variational bound on that orthogonal subspace. An arbitrary trial state without this orthogonality would bound only $E_0$, not $E_1$. The compactly supported trial functions belong to the quadratic-form domain despite derivative jumps at the well edges; their kinetic energies are correctly computed as ${\hslash }^2\int |{\psi }^{\prime }{|}^2/(2m)$.

For the reality of the eigenvalues, one may initially allow a nonzero complex eigenvector $\mathbf{v}$ with $M\mathbf{v}=\lambda \mathbf{v}$. Since $M$ is real symmetric, it is a Hermitian matrix, andis real: the numerator equals its own complex conjugate and the denominator is positive. Once $\lambda$ is real, a nonzero real or imaginary part of $\mathbf{v}$ supplies a real eigenvector. We work with the real unit eigenvectors of the problem.

For two such eigenvectors, symmetry givesDistinct eigenvalues therefore give zero dot product. Together with the prescribed unit lengths,These $n$ vectors form an orthonormal basis of ${\mathbb{R}}^n$.

Expand a real unit vector as $\mathbf{x}={\sum }_a{c}_a{\mathbf{e}}_a$. Then ${\sum }_a{c}_a^2=1$, and its Rayleigh quotient isMoreover,Every coefficient on the right is strictly positive, so equality forces $c_a=0$ for $a\ge 2$. Equality occurs precisely along the top eigendirection:

To obtain a unit vector in $S$ using just ${\mathbf{e}}_1,{\mathbf{e}}_2$, let $u=({\mathbf{e}}_1{)}_1$ and $v=({\mathbf{e}}_2{)}_1$ denote their first coordinates. If $u^2+v^2>0$, takeThen ${\alpha }_{1}u+{\alpha }_{2}v=0$ and ${\alpha }_1^2+{\alpha }_2^2=1$. Hence ${\alpha }_1{\mathbf{e}}_1+{\alpha }_2{\mathbf{e}}_2$ has first coordinate zero and unit length. If $u=v=0$, simply take ${\alpha }_1=1,{\alpha }_2=0$. In either case the Rayleigh quotient of the constructed vector isThe continuous quadratic form attains its maximum on the compact space $S$, so

For $\mathbf{x}=(0,\mathbf{y}{)}^T$, deletion of the first row and column gives ${\mathbf{x}}^{T}M\mathbf{x}={\mathbf{y}}^{T}A\mathbf{y}$, while $|\mathbf{x}|=|\mathbf{y}|$. The spectral theorem for real symmetric matrices applied to $A$ identifies its largest eigenvalue with this maximum:Combining the upper bound for every unit vector with the constructed lower bound provesThis is largest-eigenvalue interlacing for a principal submatrix: restricting a quadratic form to a coordinate hyperplane cannot exceed its old maximum, but leaves some direction in the span of the top two eigenvectors.

Multiply the Dirichlet Laplacian eigenfunction equation by $u$ and use integration by parts. The zero Dirichlet boundary condition removes the boundary term and givesThe denominator is positive because the eigenfunction is nonzero. The numerator is also positive: if it were zero, the given Sobolev inequality for $u\in W_0^{1,2}(\Omega )$ would force $u=0$. Therefore the Dirichlet Laplacian eigenvalue has the positive Rayleigh quotientEvery nonzero Dirichlet eigenfunction in this problem has a positive eigenvalue.

For a symmetric matrix $H$ with simple largest eigenvalue and spectral gap $\gamma$, the Davis-Kahan curvature lemma says $\frac{\gamma }{2}\Vert u{u}^{\top }-v{v}^{\top }{\Vert }_F^2\le \mathrm{tr}(H(v{v}^{\top }-u{u}^{\top }))$, where $v$ is its leading unit eigenvector. Applying this to a leading eigenvector $\hat{v}$ of $H+W$, the maximal Rayleigh quotient and duality of the operator norm and trace norm give $\Vert \hat{v}{\hat{v}}^{\top }-v{v}^{\top }{\Vert }_F\le 2\sqrt{2}\Vert W{\Vert }_{\mathrm{op}}/\gamma$; the projector difference has matrix rank at most two. This is a rank-one eigenprojector perturbation bound.

Here $W=M-M_0=A-A_0$ has independent centered bounded upper-triangular entries, apart from symmetry. The spectral norm bound for a centered Bernoulli adjacency matrix gives $\mathbb{E}\Vert W{\Vert }_{\mathrm{op}}\le C_0\sqrt{n}$. Taking $H=M_0$ and $\gamma =t\sqrt{n}/2$ therefore gives, for the largest-eigenvalue estimator,The printed argmin selects the wrong end of the spectrum. For a minimizing unit eigenvector $u$, compare its Rayleigh quotient with any unit vector orthogonal to $v$. This yields $\gamma |u^{\top }v{|}^2\le 2\Vert W{\Vert }_{\mathrm{op}}$, and hence $\mathbb{E}|u^{\top }v{|}^2\le 4C_0/t$. Its mean projector error is at least $\sqrt{2}(1-4C_0/t)$, using $\sqrt{1-x}\ge 1-x$. Along $t=n^{1/4}$ with sufficiently large even $n$, the printed parameter restriction holds and this lower bound tends to $\sqrt{2}$, contradicting an error of order $1/t$. Replace argmin by argmax, or negate the entire matrix before minimizing.

Assume the integrand is sufficiently smooth near the positive admissible profiles. Define the second variation as ${\delta }^{2}F[y,\xi ]={d^{2}F[y+\epsilon \xi ]/d{\epsilon }^2|}_{\epsilon =0}$, so it is twice the quadratic Taylor coefficient. Differentiation under the integral givesIntegrating the middle term by parts, with $\xi (\alpha )=\xi (\beta )=0$, yieldsAll coefficients are evaluated along $y$.

For the surface-area integrand $f=2\pi y\sqrt{1+p^2}$, the Beltrami identity for a stationary profile readsEquivalently, its Euler-Lagrange equation is $y{y}^{\prime \prime }=1+y^{\prime 2}$, solved by the positive catenoid profile $y=E\cosh ((x-x_0)/E)$ with $E>0$. Equal boundary radii at $x=\pm L$, for $L>0$, require $x_0=0$, soDirect substitution also verifies the Euler-Lagrange equation. This is a stationary profile whenever a positive $E$ satisfying the boundary relation exists; such an $E$ does not exist for arbitrary ring separations and radii.

On this profile, with $z=x/E$, the coefficients are $f_{yy}=0$, $f_{pp}=2\pi E{\mathrm{sech}}^{2}z$ and $d{f}_{yp}/dx=(2\pi /E){\mathrm{sech}}^{2}z$. Changing variables therefore givesThe printed strict inequality must mean every nonzero admissible variation: the zero function makes the integral zero, so the literal quantifier including it is false.

For completeness, the strict positivity here is sufficient for a strict local minimum in the fixed-endpoint $C^1$ topology, among positive axisymmetric graph profiles. On this finite interval the weight $p(z)={\mathrm{sech}}^{2}z$ has a positive minimum. The regular Sturm-Liouville problem associated with $Q[\xi ]=\int p({\xi }_z^2-{\xi }^2)$ has an attained lowest Dirichlet eigenvalue ${\lambda }_1$ in the Rayleigh quotient. Its first eigenfunction is a nonzero smooth admissible variation; the assumed strict positivity gives ${\lambda }_1>0$. Hence $Q\ge {\lambda }_1\Vert \xi {\Vert }_{L^2}^2$. Combining this with $Q\ge p_{\min }\Vert {\xi }_z{\Vert }_{L^2}^2-p_{\max }\Vert \xi {\Vert }_{L^2}^2$ proves coercivity in the Sobolev space $H_0^1$.

The coefficients of the second variation depend continuously on $(y,y^{\prime })$ while $y>0$. For a sufficiently small $C^1$ perturbation, the quadratic forms along the segment from the stationary profile to the perturbed one retain a uniform positive coercivity bound. Taylor's formula with integral remainder, and the vanishing first variation, then gives $F[y+v]-F[y]\ge c\Vert v{\Vert }_{H^1}^2>0$ for nonzero sufficiently small admissible $v$. The stationary area is therefore a strict local minimum under the stated nonzero-variation condition. This supplies the uniform bound needed in an infinite-dimensional problem, rather than relying on pointwise positivity alone in an arbitrary function space.

Write $c_s^2=p/\rho$ for the constant isothermal sound speed squared and $v_a^2=B^2/({\mu }_0\rho )$ for the constant Alfvén speed squared. The horizontal magnetic field has no vertical magnetic tension; magnetostatic equilibrium isConsequently the magnetized isothermal atmosphere hasThe sign of $B_0$ specifies the orientation of the magnetic field. The adiabatic sound speed is $v_s^2=\gamma p/\rho =\gamma c_s^2$, also constant.

For the proposed normal mode, let $A=\exp (i\mathbf{k}\cdot \mathbf{x}+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\mathbf{\xi }$ for its constant complex amplitude. The divergence amplitude is $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. Using ${\rho }^{\prime }/\rho =-1/H$ and $B^{\prime }/B=-a$, the Eulerian and Lagrangian fluid perturbations giveSince ${\Pi }^{\prime }/\rho =-g$, the magnetohydrodynamic total pressure amplitude isThe vertical derivative of $\delta \Pi =\rho A\psi$ is $\rho A(i{k}_z-a)\psi$. For the magnetic tension perturbation, the terms proportional to $ai{k}_y{\xi }_z{\mathbf{e}}_y$ from $\delta \mathbf{B}\cdot \nabla \mathbf{B}$ and $\mathbf{B}\cdot \nabla \delta \mathbf{B}$ cancel, leavingThus the linearization really reduces to the constant-coefficient eigenvalue problem

For a static ideal magnetohydrodynamics equilibrium with conservative gravitational acceleration, the magnetohydrodynamic energy principle gives a Hermitian operator for the restoring force in the mass density weighted inner product, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential mass density weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then ${\omega }^2\mathbf{\xi }=\mathsf{K}\mathbf{\xi }$, whereThis is a Hermitian matrix for real $\mathbf{k}$ and real positive $s,b$. Therefore every squared frequency ${\omega }^2$ is real; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized plane waves, not finite-total-energy global modes.

At ${\omega }^2=0$ and $k_y\ne 0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta =g{\xi }_z$. For nonzero magnetic field, the $x$ equation also gives $i{k}_x{\xi }_x=k_x^2\psi /(k_y^{2}b)$. Substituting $\psi =-b[\Delta -i{k}_y{\xi }_y]+g{\xi }_z-s\Delta$ and then $s\Delta =g{\xi }_z$ yieldsFor a nontrivial mode ${\xi }_z\ne 0$; otherwise these equations and $k_{y}b\ne 0$ force every component to vanish. Substitution in the vertical equation givesFor example, the determinant provides an independent check: