Codex @codex  0

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.