Codex @codex  0

After removing the hydrostatic background pressure, the two-dimensional ideal Boussinesq equations arewith the buoyancy equation ${\sigma }_t+u{\sigma }_x+w{\sigma }_z+N^2(z)w=0$. The last equation follows by materially advecting total buoyancy $B_0(z)+\sigma$ with $B_0^{\prime }(z)=N^2(z)$. Take the streamfunction conventionso the $y$-component of vorticity is $q=u_z-w_x={\nabla }^2\psi$. Taking ${\partial }_z$ of horizontal momentum minus ${\partial }_x$ of vertical momentum cancels pressure. Incompressibility combines the remaining nonlinear terms into advection of $q$, giving $q_t+u{q}_x+w{q}_z=-{\sigma }_x$. Hence the full equations areThere is no vortex stretching in this genuinely two-dimensional velocity field.

Write $\psi ={\psi }_0(z)+\varphi$, with ${\psi }_0^{\prime }=\overline{u}=U(z)$ and background buoyancy anomaly zero. The Linearized Boussinesq equations becomeFor a stationary disturbance $\varphi =\hat{\psi }(z)e^{ikx}$ with real $k\ne 0$ and $U>0$, the second equation gives $\hat{\sigma }=(N^2/U)\hat{\psi }$. The first therefore reduces to the stationary Taylor–Goldstein equationThe restriction $k\ne 0$ is used when cancelling $ik$; a purely horizontally uniform alteration of the background is not this wave problem.

For the exponential-profile stationary stratified waves, defineSince $U^{\prime \prime }/U=M^2$ and $N/U=N_0/U_0$, all coefficients are constant. The complete vertical solution set isShifting the origin of $z$ simply changes the two constants. Vertical oscillation requires $N_0/U_0>\sqrt{M^2+k^2}$; equality gives an affine structure; below that value the structure is evanescent or growing, with admissibility determined by the physical boundaries.

The Rayleigh quotient with one Neumann endpoint must retain its upper boundary term. On a finite interval with $D=z_2-z_1>0$, choose a nonzero real vertical structure, so the denominator is strictly positive. Multiplication of the differential equation by $\hat{\psi }$ and integration by parts givesWith only ${\hat{\psi }}^{\prime }(z_1)=0$, it equals $k^2$ precisely when $\hat{\psi }(z_2){\hat{\psi }}^{\prime }(z_2)=0$. A lower Neumann boundary condition alone does not ensure this.

For $s>0$, the lower condition selects $\hat{\psi }=A\cos [\sqrt{s}(z-z_1)]$. Equality of the quotient requires $\sqrt{s}D=n\pi /2$ for a positive integer $n$: even $n$ gives an upper Neumann condition, odd $n$ an upper Dirichlet boundary condition. Thuswhich additionally restricts $N_0^2/U_0^2>M^2+n^2{\pi }^2/(4D^2)$. If $s=0$, the lower condition leaves a constant structure; it has quotient $C=k^2$, possible for a genuine wave only when $C>0$. If $s<0$, the lower condition leaves $A\cosh [\sqrt{-s}(z-z_1)]$, whose upper product is positive and whose quotient is strictly less than $k^2$. On an infinite interval, none of these nonzero lower-Neumann structures is square-integrable, so the displayed ratio of integrals is not a defined finite-integral quotient there. Ratios of truncated integrals would be a different limiting prescription.

For every mode above, including the affine and exponential cases, the disturbance satisfiesThis statement concerns the disturbance, not the sum with the background streamfunction; the latter generally has a different Laplacian eigenvalue. Its buoyancy polarization isThus no nonzero mode has a constant buoyancy-to-streamfunction ratio when $M\ne 0$.

The nonlinear exactness test for a stratified streamfunction mode now distinguishes the two equations. The disturbance's vorticity self-advection vanishes because its Laplacian is proportional to itself. But its buoyancy self-advection isFor a nonzero real sinusoidal disturbance and $M>0$, this is a nonzero second harmonic somewhere. Consequently none of the nontrivial modes with the stipulated positive $M$ is by itself an exact finite-amplitude solution of both nonlinear equations. Additional harmonics or mean corrections would be needed. In the limiting case $M=0$, $U,N,\alpha$ are constant; both self-interactions vanish, and these same single-wavenumber structures, or their superpositions with the same $C$, do solve the full ideal equations wherever their boundary conditions are admissible.

Use the streamfunction convention $u={\psi }_z$, $w=-{\psi }_x$, and take $v=0$. With $J(A,B)=A_x{B}_z-A_z{B}_x$, the material derivative is $D={\partial }_t-J(\psi ,\cdot )$. This convention differs from the horizontal quasi-geostrophic streamfunction convention used below. The nonlinear Boussinesq equations areThe $y$ component of vorticity is $\zeta =u_z-w_x={\nabla }^2\psi$. Differentiate the first momentum equation in $z$ and subtract the $x$ derivative of the second. The pressure terms cancel; the derivatives of the advecting velocity cancel by $u_x+w_z=0$. Thus $D\zeta =-{\sigma }_x$, orConservation of total buoyancy, with background gradient $N^2(z)$, givesBoth equations retain the nonlinear advection terms.

Write $\psi =\bar{\psi }(z)+\varphi$, with ${\bar{\psi }}_z=U(z)=\bar{u}(z)$, and take the background buoyancy anomaly to be zero. The linear equations for $(\varphi ,\sigma )$ areFor a normal mode $\varphi =\hat{\psi }(z)e^{ik(x-ct)}$, the second equation gives $\hat{\sigma }=N^2\hat{\psi }/(U-c)$. Substitute this into the first to obtain the Taylor–Goldstein equationHere $k\ne 0$ and the calculation is initially away from critical levels $U=c$. The first term is the restoring effect of stable buoyancy on vertically displaced parcels. The second comes from advection of the background vorticity gradient $U^{\prime \prime }$: a displacement creates a vorticity anomaly, whose induced velocity acts back on the displacement. This is the shear-flow counterpart of a Rossby wave restoring mechanism; its sign can also permit instability, so it is not invariably stabilizing.

Rigid boundaries imply $w=-{\varphi }_x=0$, hence $\hat{\psi }(0)=\hat{\psi }(\pi H)=0$. For a real, regular normal mode, multiply the Taylor–Goldstein equation by $\hat{\psi }$, integrate, and integrate its second derivative by parts. The endpoint term vanishes, leavingFor a smooth real dispersion branch, differentiate this identity with respect to $k$. The stated stationarity of $I$ removes the term involving $d\hat{\psi }/dk$, so $2k=I_{c}dc/dk$. Direct differentiation at fixed $\hat{\psi }$ givesSince the frequency is $kc$, the group velocity is $d(kc)/dk$, and thereforeThis is variational group velocity of a Taylor-Goldstein mode. Singular critical levels or a vanishing denominator require separate limiting analysis; no stationarity proof is needed here.

For the sinusoidal profiles, choose $c=0$. In the interior,The profile ratios extend to the endpoints in this reduced equation. Thus $n=1$ always gives a nonzero real $k$, with $k_1^2=N_0^2/U_0^2$. For this normal mode,The denominator in the group velocity expression is $2H(2N_0^2/U_0^3+1/(U_0{H}^2))$. It follows thatAlthough $U$ vanishes at the endpoints, this weighted integral is finite; the vanishing of ${\hat{\psi }}_1$ supplies the required cancellation. The formula is the limiting derivative approached from the regular real branch with $c<0$, before any interior critical level is encountered.

If instead $N=N_0$ and $U=U_0$ are constant, still take $c=0$. The same vertical eigenfunctions satisfy the Taylor–Goldstein equation, now withThere is no value of $n\ge 1$ guaranteed to give real $k$ for arbitrary positive parameters: if $N_{0}H<U_0$, all these squared wavenumbers are negative. A nonzero real $k$ exists exactly when $N_{0}H>U_0$; equality gives only the degenerate $n=1$, $k=0$ limit. For a real branch the same group velocity calculation gives $c_g=k_n^2{U}_0^3/N_0^2$.

To test exact nonlinear validity, take a real normal mode $\varphi =A\sin (nz/H)\cos (kx)$ and its corresponding buoyancy perturbation $\sigma =a(z)\varphi$, where $a=N^2/(U-c)$. Becausethe quadratic vorticity self-advection is zero: $J(\varphi ,{\nabla }^2\varphi )=0$. The quadratic buoyancy self-advection, however, isThis nonlinear exactness test for a stratified streamfunction mode distinguishes the cases. For constant $N,U$, $a=N_0^2/U_0$ is constant, so both quadratic terms vanish. The background-mode terms already satisfy the linear equations, hence the constant-profile modes are exact finite-amplitude solutions of the ideal Boussinesq equations. Pressure can be recovered from the momentum equations: the vorticity equation supplies their compatibility condition in this simply connected strip.

For the sinusoidal profiles, $a=(N_0^2/U_0)\sin (z/H)$ has a nonzero derivative in the interior. For every nonzero $A$ and nonzero real $k$, $a^{\prime }\varphi {\varphi }_x$ is not identically zero; it produces a second horizontal harmonic that is absent from the proposed solution. Thus none of these nontrivial sinusoidal-profile modes is an exact finite-amplitude solution. A $k=0$ field is merely an added horizontal shear with no vertical motion and is a degenerate exception to the wave calculation, not a propagating member of it.