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Write $v=Ly$ asDifferentiating twice, the moving-limit terms cancel at first order and yield $\boxed{v^{\prime \prime }+v=y}$ at second order. Also $v^{\prime }(0)=0$ and $v(\pi )=0$. Thus this integral operator is the inverse of $D^2+1$ with mixed Neumann boundary condition and Dirichlet boundary condition.

For ${\kappa }_n=n+1/2$, $y_n(x)=\cos ({\kappa }_{n}x)$ satisfies those conditions and $(D^2+1)y_n=(1-{\kappa }_n^2)y_n$. The corresponding mixed homogeneous equation has no nonzero solution, so uniqueness givesThis verifies the asserted eigenvalues and eigenfunctions for every $n\ge 0$.

Write $\eta =H{e}^{\sigma t}{J}_0(\alpha r)$ using the Bessel function of the first kind. The wall boundary conditions at $z=\pm h$ selectBecause $J_0^{\prime }(s)=-J_1(s)$, the sidewall no-penetration boundary condition requireswhere $j_{1,n}$ are the positive zeros of $J_1$. These radial eigenfunctions automatically preserve layer volume becauseThe constant radial mode $\alpha =0$ cannot change the interface height in a closed tank: the Laplace equation and the impermeable lids then give zero vertical velocity, and a uniform displacement would change each layer's volume.

The kinematic boundary condition fixesThus the interface values are ${\varphi }_1(0)=\sigma \eta \coth (\alpha h)/\alpha$ and ${\varphi }_2(0)=-\sigma \eta \coth (\alpha h)/\alpha$. Substitution in the normal stress condition gives the finite-depth Rayleigh-Taylor dispersion relation:soHere $\mathcal{A}$ is the Atwood number. Every allowed mode has a positive growth rate and is subject to Rayleigh-Taylor instability. Since $d[\alpha \tanh (\alpha h)]/d\alpha =\tanh (\alpha h)+\alpha h{\mathrm{sech}}^2(\alpha h)>0$, shorter radial scales grow faster. The limits areThere is no fastest finite wavenumber in the ideal sharp-interface model without surface tension, viscosity or diffusion. The zero-initial-velocity condition uses both exponential branches: an initial mode $H{J}_0({\alpha }_{n}r)$ evolves as $H\cosh ({\sigma }_{n,+}t)J_0({\alpha }_{n}r)$. If the mass density ordering were reversed the same calculation would give oscillatory stable modes; equal mass densities give neutral modes.

For the linear differential operator $L=(D-\alpha )(D-\beta )$, the exponential function is an eigenfunction: $L{e}^{ϵx}=(ϵ-\alpha )(ϵ-\beta )e^{ϵx}$. Distinct $\alpha ,\beta$ give the two homogeneous solutions, while the method of undetermined coefficients supplies a particular solution at the distinct forcing exponent $\gamma$:

Differentiate the Legendre differential equation once and put $R_n=P_n^{\prime }$:Multiplying by $1-x^2$ gives its Sturm-Liouville form,Thus the weight is $w(x)=1-x^2$, not one. For $n\ge 1$, $R_n$ is a nonzero polynomial; $R_0=0$ is a trivial solution rather than a nonzero eigenfunction.

Multiply the equations for $n,m$ by $R_m,R_n$ respectively, subtract and integrate. The resulting boundary term is $[(1-x^2{)}^2(R_n^{\prime }{R}_m-R_m^{\prime }{R}_n){]}_{-1}^1=0$, since all these derivatives are finite at the endpoints. For distinct nonnegative indices, ${\lambda }_n-{\lambda }_m\ne 0$, giving the weighted orthogonality of Legendre polynomial derivatives:For $n=0$ the integral is zero simply because $R_0=0$.

The spatial reflection $x\mapsto -x$ acts on the two complex amplitudes as $(A,B)\mapsto (A,-B)$, because the first spatial eigenfunction is even and the second is odd. The first equation must therefore be even in $B,\overline{B}$, and the second odd. For a nonresonant double Hopf bifurcation, the temporal phases give independent actions $A\mapsto e^{i{\phi }_1}A$, $B\mapsto e^{i{\phi }_2}B$ in the cubic normal form of a dynamical system. A cubic monomial in the first equation must have net $A$ phase weight one and net $B$ phase weight zero. Only $A|A{|}^2,A|B{|}^2$ satisfy this; the second equation similarly has $B|B{|}^2,B|A{|}^2$. Parametrizing the independent cross-saturation coefficients givesAll coefficients can be complex: their real parts determine amplitude growth and their imaginary parts frequency shifts. These are cubic truncations of a family of reduced vector fields on the four-real-dimensional extended centre manifold, with the two unfolding parameters entering ${\nu }_1,{\nu }_2$.

The nonresonance assumption must be made explicit. Merely having ${\omega }_1\ne {\omega }_2$ does not imply this form: for ${\omega }_1=2{\omega }_2$, the quadratic terms $B^2$ in $\dot{A}$ and $A\overline{B}$ in $\dot{B}$ satisfy both temporal resonance and spatial parity. Thus the initial form is the generic nonresonant double-Hopf form, rather than a statement for every possible unequal frequency pair.

At the one-to-one reflection-symmetric one-to-one Hopf resonance, only the common temporal phase action remains. Cubic terms must then have total phase weight one. Spatial reflection retains, in the first equation, the three terms $A|A{|}^2,A|B{|}^2,B^2\overline{A}$; the second retains $B|B{|}^2,B|A{|}^2,A^2\overline{B}$. Common phase covariance excludes all quadratic terms. The additional resonant terms are consequently $-{\gamma }_1{B}^2\overline{A}$ and $-{\gamma }_2{A}^2\overline{B}$, with independent complex coefficients. This is the reflection-symmetric one-to-one Hopf resonance.

In the specified equal-coefficient specialization, put $r=R^2$. A nonzero pure-$A$ periodic branch exists for $\mathrm{Re}{\nu }_1>0$ and hasIts radial eigenvalue is $-2{\alpha }_{R}r<0$, and its phase is neutral. For a transverse $B$ perturbation the linear equation isIts time dependence can be removed exactly by $B=C{e}^{i\Omega t}$. Using ${\nu }_1-\alpha r=i\Omega$ and ${\nu }_2={\nu }_1-d$ givesWriting $C=u+iv$, $d=d_R+i{d}_I$, $\beta ={\beta }_R+i{\beta }_I$ yields the constant real stability matrixLet $Q=\mathrm{Re}(\beta \overline{d})={\beta }_R{d}_R+{\beta }_I{d}_I$. The transverse characteristic polynomial isBoth transverse eigenvalues have negative real part exactly whenThese hold at $r=0$ because $d_R>0$. Thus a sufficiently small pure branch is transversely stable; it is stable modulo its neutral temporal phase until one of the two inequalities fails. The secondary steady and Hopf thresholds of a resonant pure mode are derived in the two parts below. They refer to accessible points on the pure branch; in a multi-parameter problem $d$ may also vary, so a proposed parameter path must actually cross the corresponding surface.

For admissible initial data, for example a square-integrable function on the interval, expand in the sine eigenfunctions of the Dirichlet Laplacian:The expansion follows either from separation of variables or the complete sine orthogonal basis. If $\kappa <{\pi }^2$, every growth rate is negative and Parseval's identity givesThus all finite-energy data decay. The convergence is also uniform in $x$ after any fixed positive time: apply Cauchy-Schwarz inequality to the sine series, using the summable factors $e^{-2(j^2-1){\pi }^{2}t}$, to bound its supremum by a constant times $e^{-({\pi }^2-\kappa )t}$ for $t\ge t_0>0$.

Conversely choose $u(x,0)=\sin \pi x$. Then $u(x,t)=e^{(\kappa -{\pi }^2)t}\sin \pi x$, which is stationary at equality and grows when $\kappa >{\pi }^2$. All initial conditions decay if and only if $\kappa <{\pi }^2$. This is the continuous part of the reaction-diffusion spectral decay threshold.

Use the position-space momentum operator $P_j=-i\hslash {\partial }_j$ and set $F=(x+iy)z$. Direct differentiation givesTo compute the squared orbital angular momentum, write $D_x=y{\partial }_z-z{\partial }_y$, $D_y=z{\partial }_x-x{\partial }_z$, $D_z=x{\partial }_y-y{\partial }_x$, so $L_j=-i\hslash D_j$. Expanding the products, including derivatives of their variable coefficients, givesFor example $D_x^2=y^2{\partial }_z^2+z^2{\partial }_y^2-2yz{\partial }_y{\partial }_z-y{\partial }_y-z{\partial }_z$; summing the three cyclic expressions proves the identity.

Here $\Delta F=0$ and $EF=2F$, so $E^{2}F=4F$. ThereforeThis is the angular momentum of a homogeneous harmonic polynomial, with angular quantum numbers $\ell =2,m=1$. The polynomial is a nonzero angular eigenfunction; by itself it is not a normalizable full-space wavefunction.

Use the weighted inner product $⟨u,v{⟩}_r={\int }_a^{b}r(x)u(x)v(x)dx$. For real eigenfunctions, integration by parts with the zero endpoint values givesThus $({\lambda }_n-{\lambda }_m)⟨y_m,y_n{⟩}_r=0$. The regular separated Sturm-Liouville problem has simple eigenvalues: for a fixed eigenvalue, prescribing $y(a)=0$ leaves only the initial flux $p(a)y^{\prime }(a)$ free, so uniqueness of the associated first-order system makes any two eigenfunctions proportional. Distinct eigenfunctions are therefore orthogonal, and choosing their signs and scales so that $\int r{y}_n^2=1$ givesThe positive energy form $\int (p{y}^{\prime 2}+q{y}^2)$ realizes $r^{-1}\mathcal{L}$ as a positive self-adjoint operator in this weighted space, with compact inverse on the bounded interval. The spectral theorem for compact self-adjoint operators supplies a complete eigenbasis. This is the standard regular Sturm-Liouville eigenfunction expansion, also valid with continuous $p$ by working with the continuous flux $p{y}^{\prime }$.

For a source $F$, expand a zero-endpoint solution of $(\mathcal{L}-\mu r)y=F$ in that basis. Multiplication by $y_n$ and unweighted integration giveSince $\mu$ is not an eigenvalue, every coefficient is determined. The spectral Green function of a regular Sturm-Liouville problem is consequentlyWeighted completeness means $r(x){\sum }_n{y}_n(x)y_n(\xi )=\delta (x-\xi )$ as a distribution, so applying $\mathcal{L}-\mu r$ to this series produces the required delta source. The source integration is with $d\xi$, explaining the absence of an additional factor $r(\xi )$ in the stated kernel.

For the subsequent operator $L=d^2/d{x}^2$ the sign is explicitly reversed: $y_n=\sqrt{2/\pi }\sin (nx)$ and ${\lambda }_n=-n^2$. To cover every real non-eigenvalue $\mu$, define the homogeneous solutionIt satisfies $S_{\mu }(0)=0$, $S_{\mu }^{\prime }(0)=1$ and $S_{\mu }^{\prime \prime }=\mu S_{\mu }$. Matching the two endpoint solutions continuously at $x=\xi$ with derivative jump $G_x({\xi }^{+},\xi )-G_x({\xi }^{-},\xi )=1$ yieldsThe denominator is nonzero precisely off the Dirichlet eigenvalues $\mu =-n^2$. In particular $G_0=-x_{<}(\pi -x_{>})/\pi$. At $x=\xi =\pi /2$, its spectral expansion givesThis odd reciprocal-square sum from a Dirichlet Green function proves $\boxed{\sum _{j=0}^{\infty }(2j+1{)}^{-2}={\pi }^2/8}$.

Let $h$ be layer thickness and $g$ gravitational acceleration. The unforced shallow water equations on the equatorial beta plane areLinearize about $u=v=0$, $h=h_0$. Put $c_0=\sqrt{g{h}_0}$ and $L=\sqrt{c_0/\beta }$, and scale $(x,y)=L(X,Y)$, $t=(L/c_0)T$, $(u,v)=c_0(U,V)$ and $h=h_0(1+\eta )$. Dropping capitals, the linearized shallow water equations becomeThe choice $\beta L^2/c_0=1$ makes both the wave speed and the coefficient of the dimensionless Coriolis parameter equal to one.

Take Fourier modes $\{u(y),v(y),\eta (y)\}{e}^{i(kx-\omega t)}$ that decay as $y\to \pm \infty$. Their equations areFor $v=0$ and nonzero $k$, the first two require ${\omega }^2=k^2$. The root $\omega =k$ gives $u=\eta$, and meridional balance becomes $u^{\prime }=-yu$. Hence the equatorial Kelvin wave isIt travels eastward at speed $c_0$ in dimensional units. The other sign gives $\eta =-u$ and $u^{\prime }=yu$, so $u\propto e^{y^2/2}$ is not trapped.

For $v\ne 0$, $\omega \ne 0$ and ${\omega }^2\ne k^2$, inversion of the first two equations gives the equatorial wave velocity polarizationSubstitute in the remaining momentum equation. Multiplication by $({\omega }^2-k^2)/i$ leavessoWriting $v=e^{-y^2/2}H(y)$ yields $H^{\prime \prime }-2y{H}^{\prime }+(\lambda -1)H=0$, the Hermite differential equation. The trapped solutions have the Hermite oscillator quantization $\lambda =2n+1$ and $H=H_n$.

For completeness, quantization follows without assuming that every acceptable solution is a polynomial. On decaying smooth functions, define $a=d/dy+y$, $a^{†}=-d/dy+y$ and $\mathcal{H}=-d^2/d{y}^2+y^2=a^{†}a+1$. Integration by parts gives $\lambda \ge 1$ and $\Vert av{\Vert }^2=(\lambda -1)\Vert v{\Vert }^2$. If $av\ne 0$, it is an eigenfunction of $\mathcal{H}$ with eigenvalue $\lambda -2$. Repeated lowering must terminate: otherwise it eventually produces an eigenvalue below one. At termination $av=0$, so the terminal eigenfunction is $e^{-y^2/2}$ and has eigenvalue one. Repeated raising gives $H_n(y)e^{-y^2/2}$ and $\lambda =2n+1$; uniqueness of a decaying solution at one end of this second-order equation fixes each mode up to scale.

Thus, away from the exceptional quotients, all fields are determined byInsert this $v$ and its derivative in the polarization formulae to obtain $u$ and $\eta$. For $n\ge 1$, the cubic ${\omega }^3-(k^2+2n+1)\omega -k=0$ supplies the low-frequency equatorial Rossby wave and two equatorial inertia--gravity wave branches. Their meridional structures are Gaussian-weighted Hermite polynomials.

For $n=0$, multiplication by $\omega$ factorizes the equatorial shallow-water dispersion relation asThe physically admissible quadratic gives the mixed Rossby-gravity wave, equivalently the Yanai wave:These fields satisfy all three original equations directly and decay at both infinities, including where the quotient formulae are singular.

To test the other factor, set $\omega =-k\ne 0$. The original equations give $u+\eta =-iyv/k$ and $v^{\prime }=-yv$. Put $\delta =u-\eta$; meridional momentum then givesDecay of $u,\eta$ at both infinities requires the integral of the right side to vanish. Hence $C=0$ unless $k^2=1/2$. Ordinarily the extra factor supplies no trapped mode. At the exceptional root of the mixed Rossby-gravity mode, $k^2=1/2$, it coincides with the quadratic root and gives the same physical fields, not an additional branch. This distinction avoids discarding a regular physical mode at the crossing.

These arguments concern nonzero wave angular frequency. At $k=0$, the oscillatory nonzero-$v$ modes have $\omega =\pm \sqrt{2n+1}$, while zero-frequency $v=0$ geostrophic states must be considered in the original equations rather than in expressions divided by $\omega$.

At the double threshold, $L{\psi }_0=L{\psi }_1=0$. For $j\ge 2$, $q_j^2\ge 9{\pi }^2/4$, so $|K^{\ast }-q_j^2|\ge 13{\pi }^2/8>3{\pi }^2/8$ and every remaining eigenvalue is strictly negative. There is a spectral gap separating this two-dimensional kernel from the stable modes. The centre manifold theorem therefore gives a two-dimensional state centre manifold, withIf the two detuning parameters are adjoined as variables with zero time derivatives, the extended centre manifold has total dimension four; its state fibers still have dimension two.

The only nontrivial spatial symmetry of the bounded interval is the reflection $x\mapsto -x$. Since ${\psi }_0$ is even and ${\psi }_1$ odd, its group action on the amplitudes is $(A,B)\mapsto (A,-B)$. Thus equivariance requires $\dot{A}$ to be even and $\dot{B}$ odd in $B$. There is no sign symmetry $\psi \mapsto -\psi$, because the quadratic nonlinearity breaks it. The general smooth form is $\dot{A}=F(A,B^2)$, $\dot{B}=BG(A,B^2)$, with the zero solution preserved. The quadratic even-odd mode interaction follows on truncation:Writing $m=\mu -{\mu }^{\ast }$ and $\kappa =K-K^{\ast }$, the exact linear growth rates are $m-3{\pi }^2\kappa /4-{\kappa }^2$ and $m+3{\pi }^2\kappa /4-{\kappa }^2$. To first order in detuning,The orthogonal projection of the quadratic term onto the neutral eigenfunctions determines the leading coefficients. Both eigenfunctions have squared $L^2$ norm one, and parity kills the unwanted products. The quadratic correction $h$ contributes no term through $Lh$, since $L$ is self-adjoint and the projected modes lie in its kernel. ConsequentlyThese projections are also a concrete way of computing the normal form. In this normalization they give $a_1=8/(3\pi )$, $a_2=32/(15\pi )$ and $a_3=64/(15\pi )$, consistent with the stated coefficient ratios.

The linearization at the zero solution is the self-adjoint differential operator $L=\mu -({\partial }_x^2+K{)}^2$, with domain satisfying both endpoint conditions. The Dirichlet boundary condition for $-{\partial }_x^2$ on $(-1,1)$ gives the orthogonal eigenfunctions $\cos ((n+\frac{1}{2})\pi x)$ and $\sin ((n+1)\pi x)$. Their second derivatives are multiples of themselves, so the additional condition ${\psi }_{xx}(\pm 1)=0$ is automatic. Completeness of the Dirichlet eigenfunctions also shows that these exhaust the spectrum of the polynomial operator $L$. Set $q_j=(j+1)\pi /2$; each eigenfunction has growth rate ${\sigma }_j=\mu -(K-q_j^2{)}^2$. Hence the stationary bifurcation thresholds areEquality of the first two thresholds requires $K$ to be the midpoint of ${\pi }^2/4$ and ${\pi }^2$, since these numbers are distinct. ThereforeAt this point the even and odd lowest eigenfunctions become neutral simultaneously.

The angular momentum commutation relations are $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$. The commutator product rule givessince the bracketed expression is symmetric in $j,k$ and ${ϵ}_{ijk}$ is antisymmetric. Alsoproving $\boxed{L^2=L_{-}{L}_{+}+L_3^2+\hslash L_3}$.

For orbital angular momentum, $L_i=-i\hslash {ϵ}_{ijk}{x}_j{\partial }_k$. A differentiable radial function has ${\partial }_{k}f(r)=f^{\prime }(r)x_k/r$, so its contraction with the antisymmetric symbol vanishes: $Lf(r)=0$. Put $s=x_1+i{x}_2$. The coordinate differential operators give $L_{3}s=\hslash s$, whileannihilates every power of $s$, because $({\partial }_1+i{\partial }_2)s=0$. Since all $L_i$ annihilate the radial factor, the product rule yieldsSubstitution into the expression for $L^2$ gives the highest-weight complex-coordinate orbital wavefunction resultAs $[L^2,L_{-}]=0$, lowering preserves this eigenvalue whenever the lowered function is nonzero. Explicitly,For $n=0$ the lowered function is zero, so it is not an eigenfunction. For $n\ge 0$ the original nonzero function has the usual regular angular dependence. The PDF's “any integer” differential identities also hold locally for negative $n$ where $s\ne 0$, but those negative powers are singular on the axis and are not square-integrable regular angular eigenstates. The converted TeX incorrectly writes $L_4$ in place of the PDF's $L_{+}$.

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.

Substitution of the power series, with termwise differentiation, gives the coefficient identityThus $a_0$ and $a_1$ are arbitrary, and all power series solutions of the Hermite differential equation areAn empty product is one. For a nonterminating branch the ratio of consecutive terms in its own parity subsequence is asymptotic to $|x{|}^2/j$, hence tends to zero for every fixed $x$. The ratio test proves convergence everywhere; the resulting power series can be differentiated termwise and indeed solve the differential equation. A terminating branch is already a polynomial.