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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.