= Solution
After removing the hydrostatic background pressure, the two-dimensional ideal <Boussinesq equations> are
$$
u_t+uu_x+wu_z=-p_x,\qquad
w_t+uw_x+ww_z=-p_z+\sigma,\qquad u_x+w_z=0,
$$
with 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 $B_0'(z)=N^2(z)$. Take the <streamfunction> convention
$$
u=\psi_z,\qquad w=-\psi_x,
$$
so 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 $q_t+uq_x+wq_z=-\sigma_x$. Hence the full equations are
$$
\boxed{\nabla^2\psi_t+\sigma_x=\psi_x(\nabla^2\psi)_z-\psi_z(\nabla^2\psi)_x,}
$$
$$
\boxed{\sigma_t+\psi_z\sigma_x-\psi_x\sigma_z-N^2(z)\psi_x=0.}
$$
There is no vortex stretching in this genuinely two-dimensional velocity field.
Write $\psi=\psi_0(z)+\phi$, with $\psi_0'=\overline u=U(z)$ and background buoyancy anomaly zero. The <Linearized Boussinesq equations> become
$$
(\partial_t+U\partial_x)\nabla^2\phi-U''\phi_x+\sigma_x=0,\qquad
(\partial_t+U\partial_x)\sigma-N^2\phi_x=0.
$$
For a stationary disturbance $\phi=\widehat\psi(z)e^{ikx}$ with real $k\ne0$ and $U>0$, the second equation gives $\widehat\sigma=(N^2/U)\widehat\psi$. The first therefore reduces to the stationary <Taylor–Goldstein equation>
$$
\boxed{\widehat\psi''+m^2(z)\widehat\psi=0,\qquad
m^2(z)=\frac{N^2}{U^2}-\frac{U''}{U}-k^2.}
$$
The restriction $k\ne0$ 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>, define
$$
C=\frac{N_0^2}{U_0^2}-M^2,\qquad s=C-k^2.
$$
Since $U''/U=M^2$ and $N/U=N_0/U_0$, all coefficients are constant. The complete vertical solution set is
$$
\boxed{\widehat\psi(z)=
\begin{cases}
A\cos(\sqrt s\,z)+B\sin(\sqrt s\,z),&s>0,\\
A+Bz,&s=0,\\
Ae^{\sqrt{-s}\,z}+Be^{-\sqrt{-s}\,z},&s<0.
\end{cases}}
$$
Shifting 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 $\widehat\psi$ and <integration by parts> gives
$$
\mathcal R=\frac{\int_{z_1}^{z_2}(C\widehat\psi^2-\widehat\psi'^2)\,dz}
{\int_{z_1}^{z_2}\widehat\psi^2\,dz}
=k^2-\frac{[\widehat\psi\widehat\psi']_{z_1}^{z_2}}
{\int_{z_1}^{z_2}\widehat\psi^2\,dz}.
$$
With only $\widehat\psi'(z_1)=0$, it equals $k^2$ precisely when $\widehat\psi(z_2)\widehat\psi'(z_2)=0$. A lower <Neumann boundary condition> alone does not ensure this.
For $s>0$, the lower condition selects $\widehat\psi=A\cos[\sqrt s(z-z_1)]$. Equality of the quotient requires $\sqrt sD=n\pi/2$ for a positive integer $n$: even $n$ gives an upper Neumann condition, odd $n$ an upper <Dirichlet boundary condition>. Thus
$$
k^2=C-\frac{n^2\pi^2}{4D^2}>0,
$$
which 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 satisfies
$$
\boxed{\nabla^2\phi=-C\phi=\left(M^2-\frac{N_0^2}{U_0^2}\right)\phi.}
$$
This statement concerns the disturbance, not the sum with the background <streamfunction>; the latter generally has a different Laplacian eigenvalue. Its buoyancy polarization is
$$
\boxed{\sigma=\alpha(z)\phi,\qquad \alpha(z)=\frac{N_0^2}{U_0}e^{Mz}.}
$$
Thus no nonzero mode has a constant buoyancy-to-streamfunction ratio when $M\ne0$.
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 is
$$
\phi_z\sigma_x-\phi_x\sigma_z=-\alpha'(z)\phi\phi_x=-M\alpha(z)\phi\phi_x.
$$
For a nonzero real sinusoidal disturbance and $M>0$, this is a nonzero second harmonic somewhere. Consequently \b[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.
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