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For the shallow-water approximation, let the vertical scale be much smaller than the horizontal scale, so vertical acceleration is negligible and pressure is hydrostatic. Assume constant density, a flat impermeable bottom, a free surface under constant atmospheric pressure, and a horizontally moving column with no leading-order vertical shear. Neglect friction. Taking the bottom at $z=0$, hydrostatic pressure is $p=p_a+\rho g(h-z)$, so its horizontal gradient is independent of $z$. The horizontal momentum equations are thereforeIntegrating incompressibility from the bottom to the free surface, using zero bottom vertical velocity and the surface kinematic condition $w=h_t+u{h}_x+v{h}_y$, givesTogether these are the nonrotating shallow water equations.

Rotation with angular velocity $f\hat{\mathbf{z}}/2$ adds Coriolis force terms $-fv$ and $+fu$ to the left sides of the $u$ and $v$ equations, respectively. The centrifugal potential is absorbed into the background pressure and effective gravity. ThusWrite $\zeta =v_x-u_y$ and $d=u_x+v_y$. Taking the horizontal curl, including the derivatives of the advecting velocity, yieldsFor example the curl of the Coriolis force contribution is $fd$, and the curl of nonlinear advection supplies $\zeta d$. Since $q_a=f+\zeta$ and $D_{H}h=-hd$, division givesThis is exact shallow-water potential vorticity conservation within the rotating shallow water equations.

For the quasi-geostrophic approximation, take a horizontal scale $L$, speed $U$, mean depth $H$, and slow time $L/U$. Require small Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$ and small surface departure $\eta =h-H$. Introduce the Rossby deformation radius $L_R=\sqrt{gH}/|f|$ and Burger number $\mathrm{Bu}=L_R^2/L^2=O(1)$. Geostrophic balance gives $g\eta /L\sim |f|U$, hence $\eta /H\sim \mathrm{Ro}/\mathrm{Bu}\ll 1$. The Froude number is $U/\sqrt{gH}=\mathrm{Ro}/\sqrt{\mathrm{Bu}}\ll 1$. These assumptions select slow balanced motion, with fast gravity-wave transients absent at leading order.

Define the quasi-geostrophic streamfunction by $\psi =g\eta /f$. The leading geostrophic flow is $(u_g,v_g)=(-{\psi }_y,{\psi }_x)$, which is horizontally nondivergent. Expanding the exact shallow-water potential vorticity givesThe correction is smaller than $f/H$ by $O(\mathrm{Ro})$. Advecting it by the leading geostrophic flow, while the constant leading term has zero gradient, givesHere potential-vorticity conservation means that $Q$ is carried unchanged by the leading horizontal trajectories. Potential-vorticity inversion means solving the elliptic equation $({\nabla }_H^2-L_R^{-2})\psi =Q-f$, with appropriate boundary or decay conditions, to recover $\psi$ and thus the velocity. Conservation evolves the scalar field; inversion obtains its velocity field.

For a mean current $U(y)=-{\bar{\psi }}_y$, substitute $Q=\bar{Q}+Q^{\prime }$ and $\psi =\bar{\psi }+{\psi }^{\prime }$ into potential-vorticity conservation. Retaining first-order terms givesLet $\Delta Q=\bar{Q}(0^{+})-\bar{Q}(0^{-})>0$. Differentiating the mean inversion relation gives ${\bar{Q}}_y=-U^{\prime \prime }+L_R^{-2}U$. Away from $y=0$ this is zero, so decay at both infinities gives $U=U_0{e}^{-|y|/L_R}$. A velocity jump would introduce a derivative of a delta function, absent from ${\bar{Q}}_y$, so $U$ is continuous. Integrating through zero now gives $-[U^{\prime }]=\Delta Q$, and therefore

For the regular interfacial normal mode ${\psi }^{\prime }=\varphi (y)e^{ik(x-ct)}$, the linear equation readsIn each region, choose zero perturbation potential vorticity, giving ${\varphi }^{\prime \prime }-(k^2+L_R^{-2})\varphi =0$. This supplies a regular solution even if $U=c$ at isolated locations away from the interface. Continuity and decay require $\varphi =A{e}^{-\gamma |y|}$, where $\gamma =\sqrt{k^2+L_R^{-2}}$. Integrating the equation across zero givesSince $[{\varphi }^{\prime }]=-2\gamma A$, the nonzero normal mode hasThis potential-vorticity step edge wave is an interfacial Rossby wave. Displacing the potential vorticity jump creates alternating anomalies; their inverted velocity displaces the interface further with an intrinsic propagation opposite to the positive mean current. The mechanism requires a nonzero jump, or more generally a gradient, of the mean potential vorticity. It does not require planetary variation of $f$. Reversing the jump reverses the intrinsic propagation; setting the jump to zero removes this interfacial restoring mechanism.

Use incompressible Euler equations with pressure divided by the constant reference density. Writing $L={\partial }_t+U(y){\partial }_x$, their linearization about the parallel shear flow isTaking the divergence requires the commutator $[{\partial }_y,L]=U^{\prime }{\partial }_x$. Both this commutator and the divergence of $U^{\prime }v$ contribute, so ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying the Laplacian to $Lv=-p_y$ givesHence the wall-normal velocity equation isTaking ${\partial }_z$ of the streamwise momentum equation minus ${\partial }_x$ of the spanwise equation cancels pressure and yields the inviscid Squire equation,This is the vorticity component in the original PDF; the converted TeX corrupts its definition. Impermeable inviscid walls impose $v(\pm 1,t)=0$. They do not impose a viscous no-slip condition on the tangential perturbation velocity.

For a Fourier mode, set $k^2={\alpha }^2+{\beta }^2$, $D=d/dy$ and $q=(D^2-k^2)\tilde{v}$. The equations becomeFor Couette flow $U=\lambda y$, $U^{\prime \prime }=0$, so $q(y,t)=e^{-i\alpha \lambda yt}{q}_0(y)$ with $q_0=(D^2-k^2){\tilde{v}}_0$. Invert this using the Dirichlet Green function of $D^2-k^2$ on $[-1,1]$. If $y_{<}=\min (y,s)$ and $y_{>}=\max (y,s)$, it isIt vanishes at both walls and its first derivative has jump one at $y=s$. ThusAt $t=0$, uniqueness of the Dirichlet inversion recovers the compatible initial velocity. This inviscid Couette initial-value Green function directly solves the initial-value problem without a Laplace transform.

The Squire equation is now a first-order forced equation in time. Its integrating factor givesFor nonzero $\alpha$, the velocity formula contains shearing phases and the inviscid Couette continuous spectrum. Velocity cancellation can occur even though the transported $q$ has constant magnitude; it must not be confused with exponential modal instability.

For the separate case $\alpha =0$, $\beta \ne 0$, the Dirichlet operator $D^2-{\beta }^2$ is invertible and $q_t=0$. Hence $\tilde{v}={\tilde{v}}_0$ andSince $\tilde{\eta }=i\beta \tilde{u}$ in this case, $\tilde{u}={\tilde{u}}_0-\lambda t{\tilde{v}}_0$. This is the lift-up effect: a transverse disturbance transports base-flow momentum and creates a streamwise perturbation growing linearly in time, with perturbation kinetic energy generically growing as $t^2$. The growth is algebraic and is possible without an exponentially unstable normal mode. For nonzero initial normal velocity and nonzero shear, nonlinear effects eventually limit the validity of the linearized calculation. The same streamwise-independent result holds with $\lambda$ replaced by $U^{\prime }(y)$ for a general parallel profile.

Let $V=U-c$ and suppose a growing normal mode has $\mathrm{Im}c=c_i>0$, taking $\alpha >0$ without loss of generality. Since $V$ never vanishes on the real channel, choose a continuous square-root branch and write $\varphi =V^{1/2}H$. Direct differentiation givesSubstitution into the Taylor–Goldstein equation and division by $V^{1/2}$ givesThus the required transformation isThe impermeable walls give $\varphi (\pm 1)=0$, hence $H(\pm 1)=0$. Multiply by $\overline{H}$, integrate across the channel, and use integration by parts. With the boundary term zero, the resulting power-transformed Taylor–Goldstein energy identity isBoth $U^{\prime \prime }$ and $J$ are real. Since $\mathrm{Im}V=-c_i$ and $\mathrm{Im}(1/V)=c_i/|V{|}^2$, its imaginary part, divided by $-c_i$, isIf $J\ge U^{\prime 2}/4$ everywhere, the integrand is nonnegative and the first two terms have positive integral for any nonzero mode. This contradicts the identity. Therefore instability requiresThis is the necessary instability condition of the Miles–Howard theorem. Where $U^{\prime }\ne 0$, it says that the gradient Richardson number $J/U^{\prime 2}$ must be below $1/4$ somewhere. It is not a sufficient condition for instability, and the conclusion concerns exponential normal-mode growth, not exclusion of the algebraic lift-up effect in part (a). The algebra uses the given nondimensional $J$; in dimensional notation its counterpart is the squared buoyancy frequency.

Let $m(r)$ be the background enclosed mass, so $g=Gm/r^2$ and $m^{\prime }=4\pi r^2\rho$. The background hydrostatic equilibrium is $p^{\prime }=-\rho g$. Use a Lagrangian fluid displacement $\xi ={\xi }_r$ of each spherical mass shell; $\Delta$ denotes its Lagrangian perturbation. Conservation of shell mass givesand the adiabatic pressure law gives the Lagrangian pressure perturbationThis retains perturbed self-gravity: a shell keeps its enclosed mass while its gravitational acceleration changes with its radius.

In mass coordinates, the exact radial shell equation is $\ddot{r}=-4\pi r^2\partial p/\partial m-Gm/r^2$. Its first-order perturbation isHydrostatic equilibrium gives ${\partial }_{m}p=-g/(4\pi r^2)$, and ${\partial }_m\Delta p=(\Delta p{)}^{\prime }/(4\pi r^2\rho )$. Thus the perturbation of the pressure-force geometry contributes another $2g\xi /r$, in addition to the gravity term:Put $\eta =\xi /r$, so ${\xi }^{\prime }+2\xi /r=r{\eta }^{\prime }+3\eta$. Expanding the derivative, using constant $\gamma$ and $p^{\prime }=-\rho g$, yields the radial stellar pulsation equation:The coefficient $4$ includes both geometric pressure and self-gravity effects; dropping the gravitational perturbation would not give this equation.

For a normal mode $\xi (r,t)=\xi (r)e^{-i\omega t}$, multiply by $\rho r^2{\xi }^{\ast }$ and integrate. Since ${\xi }^{\ast }/r={\eta }^{\ast }$, integration by parts givesRegularity at the center requires $\xi =O(r)$ with finite $\eta$, so the lower endpoint vanishes. For an isolated star take $p(R)=0$ and the free-surface condition $\Delta p=0$, with a regular finite-energy displacement; these eliminate the upper boundary term. HenceThese boundary conditions give the self-adjoint radial Sturm-Liouville problem behind the weighted stellar pulsation Rayleigh quotient.

Use the homologous trial fluid displacement ${\xi }_r=r$, or $\eta =1$. It is regular at the center, and $\Delta p=-3\gamma p$ vanishes at the free surface. The derivative term in the Rayleigh quotient is zero, leavingThe ratio is a positive weighted average of $g/r$. Since the central mass density is its maximum,For $\gamma >4/3$ multiplication preserves this inequality, proving the central-density upper bound for a radial stellar frequency:The full positive quadratic form also shows radial stability for constant $\gamma >4/3$ under these assumptions. A uniform-density equilibrium has constant $g/r$, making the homologous displacement an exact mode and saturating the bound.

If $\gamma <4/3$, the same trial value is strictly negative. Therefore ${\omega }_{\min }^2<0$: the star has an exponentially growing radial instability, with growth rate $\sqrt{-{\omega }_{\min }^2}$. One must use this negative trial quotient, rather than carrying the central-density inequality across the sign change unchanged. At the limiting value $\gamma =4/3$, the homologous displacement has zero restoring force and is a neutral radial mode in this constant-exponent model.

Let $\zeta$ denote free-surface height and $h=\zeta -b$ the water depth. The inviscid rotating shallow water equations with constant Coriolis parameter $f$ areIt is $\zeta$, not $h$, that supplies the horizontal pressure gradient. For a fixed bottom, mass conservation equivalently gives $D_{H}h=-h(u_x+v_y)$.

Write $\xi =v_x-u_y$ for relative vorticity. Differentiate the second momentum equation in $x$ and the first in $y$, then subtract. The pressure terms cancel. Expanding the differentiated advection terms givesConsequently the absolute vorticity $q_a=\xi +f$ satisfies $D_H{q}_a=-q_a{\nabla }_H\cdot {\mathbf{u}}_H$. Combining this with mass conservation yields exact shallow-water potential vorticity conservation:A sloping bottom changes the depth and hence the stretching, but contributes no curl to the gravitational pressure force.

For the quasi-geostrophic approximation, choose constant reference depth $H_0$, speed $U$, and horizontal length $L$. Require Rossby number $\mathrm{Ro}=U/(|f|L)\ll 1$, slow evolution on the advective time $L/U$, smooth horizontal scales, and $L/L_D=O(1)$ or smaller, where $L_D=\sqrt{g{H}_0}/|f|$ is the Rossby deformation radius. Write $\zeta ={\zeta }_0+\eta$, $b=b_0+\delta b$, and $h=H_0+\eta -\delta b$. The leading momentum balance is geostrophic balance:It gives $\eta /H_0=O(\mathrm{Ro}(L/L_D{)}^2)$, while we require $\delta b/H_0=O(\mathrm{Ro})$ and slopes on the chosen horizontal scale. Relative vorticity divided by $f$ is also $O(\mathrm{Ro})$. ThusThe derivative of the constant first term vanishes. In advecting the first-order anomaly, the ageostrophic flow contributes only at the next order. Therefore the shallow-water quasi-geostrophic potential vorticity equation isThis retains the balanced vortical branch, including Rossby waves driven here by the topographic potential-vorticity gradient. It filters the freely oscillating inertia–gravity branches of the full shallow water equations, and does not retain independent boundary Kelvin wave dynamics. Small Rossby number alone is insufficient for the reduction: fast waves must be absent at leading order, as in well-prepared balanced initial data.

In the channel take $H_0=h_{00}$ and measure the bottom anomaly from mid-channel: $\delta b=\epsilon (y-L/2)$. Define $\beta =\epsilon f/H_0$. The resting potential vorticity is $\beta (y-L/2)$, and the linear Rossby wave equation isFor nonzero zonal wavenumber $k$, wall conditions $\psi =0$ selectThe $k=0$ modes are stationary. Write $K^2=k^2+(n\pi /L{)}^2+L_D^{-2}$ and take real $A$ for the following sketch. The total potential vorticity is $q=\beta (y-L/2)-K^2\psi$. For a contour whose undisturbed position is $y_0$, its small meridional displacement isThus potential-vorticity conservation says a northward displacement in a positive background gradient carries a negative anomaly, $q^{\prime }=-\beta \delta y$. Potential-vorticity inversion turns this into positive $\psi$ and anticyclonic motion around the bulge. At a positive cosine crest, $v={\psi }_x$ is northward on its western flank and southward on its eastern flank. Advection raises the contour on the west and lowers it on the east, so the crest moves west. Algebraically, $\delta y_t=v$ and $\delta y=(K^2/\beta )\psi$ give $\omega /k=-\beta /K^2$, the same phase velocity as above. For $\beta <0$ all propagation directions reverse.

For a square basin use a nonzero-frequency normal mode $\psi =\mathrm{Re}\{\Psi (x,y)e^{-i\omega t}\}$, with $\Psi =0$ on all walls. Its equation isSet $\alpha =\beta /(2\omega )$ and $\hat{\Psi }=e^{i\alpha x}\Psi$. Substitution cancels the first derivative and givesThe square's Dirichlet Laplacian eigenfunctions are $\hat{\Psi }=A\sin (m\pi x/L)\sin (n\pi y/L)$, with $m,n$ positive integers. Consequently the square-basin topographic Rossby mode hasFor $\beta >0$, the positive-frequency branch therefore tends to $\beta L/[2\pi \sqrt{m^2+n^2}]$ as $L/L_D\to 0$, as required. The negative-frequency solution is its complex-conjugate representation for a real disturbance; the physical frequency magnitude is $|\beta |/[2\sqrt{{\pi }^2(m^2+n^2)/L^2+L_D^{-2}}]$.

Consider a nonreal phase velocity $c$, so $U-c$ never vanishes. Substitute $\varphi =(U-c)\psi$ into the Rayleigh equation for inviscid shear flow. Differentiating cancels the two terms involving $U^{\prime \prime }$, leavingFor $k>0$, the wall condition on normal velocity gives $\varphi =0$, hence $\psi =0$, at both walls. Multiply by $\overline{\psi }$ and use integration by parts to obtain the weighted identity for Howard's semicircle theorem:For a nonzero normal mode, $N=\int Wdy>0$. The imaginary and real parts of this identity give, respectively,Since $c_i\ne 0$, these implyThe first equation says that $c_r$ is a positive weighted mean of $U$, and consequently $U_{\min }\le c_r\le U_{\max }$. The division by $c_i$ is essential: the weighted-mean assertion concerns nonreal modes. A neutral mode may have a critical level where $U=c$ and the substitution for $\psi$ is singular.

Pointwise, $(U-U_{\max })(U-U_{\min })\le 0$. Integrating against the nonnegative weight $W$ and substituting both identities yieldsCompleting the square proves Howard's semicircle theorem:Thus all nonreal phase velocities lie in the closed disk with the real interval $[U_{\min },U_{\max }]$ as diameter. For the convention $e^{ik(x-ct)}$ and $k>0$, exponentially growing normal modes occupy its upper semicircle, while decaying modes occupy its lower semicircle.

Away from the velocity-profile corners, $U^{\prime \prime }=0$ and an unstable normal mode satisfies ${\varphi }^{\prime \prime }-k^2\varphi =0$. On the upper half of the jet writeThe exterior solution is selected by decay at infinity. From the given velocity components, the perturbation pressure isContinuity of normal velocity and pressure therefore gives continuity of $\varphi$ and pressure matching at a piecewise-linear shear interface:At $y=1$ the slope of $U$ jumps from $-1$ to zero. Using ${\varphi }^{\prime }(1^{+})=-k\varphi (1)$ yieldsThe lower interface then follows by parity.

For the even sinuous mode of a planar jet, the derivative can jump at the center: being an even function does not require the one-sided derivative to vanish. The slope jump at $y=0$ is $-2$, whereas $[{\varphi }^{\prime }]=2{\varphi }^{\prime }(0^{+})$. ThusSubstitute $A=-k(1-c)B$ into the upper-interface relation. With $q=e^{-2k}$, division by $e^k$ and multiplication by $2k$ giveCollecting powers of the phase velocity gives the required dispersion relation:These equations can also be treated as a homogeneous two-by-two system in $A,B$, avoiding division by any potentially zero amplitude or phase-velocity factor.

For the odd varicose mode of a planar jet, $\varphi (0)=0$, so $A=0$; the central pressure matching at a piecewise-linear shear interface is automatic. The upper-interface equation becomes $ck{e}^k=\sinh k$, and thereforeThis phase velocity is real for every $k>0$, so the discrete varicose mode of a planar jet is neutral and there is no exponentially unstable odd mode.

To determine the unstable even modes, factor the reduced discriminant of the quadratic:The second factor vanishes at zero and has derivative $2(1-e^{-k}{)}^2>0$ for $k>0$, so it is positive there. The first factor starts at $-8$, is strictly increasing, and tends to infinity. It therefore has one positive zero $k_c$. The even phase velocities areIt follows thatFor $0<k<k_c$, one member of the complex conjugate pair grows at rate $k{c}_i=\sqrt{-D(k)}/4$ and the other decays. At $k=k_c$ the speeds merge and there is no exponential growth; for $k>k_c$ both speeds are real. This is inviscid instability of a triangular jet. The unstable sinuous mode of a planar jet also satisfies Howard's semicircle theorem with $U_{\min }=0$ and $U_{\max }=1$: its speed lies in the upper half-disk centered at $1/2$ with radius $1/2$.

Varying the Lagrangian gives, with the cyclic angle conventions,All successive angular gaps are $2\pi /N$ in equilibrium, so their two contributions cancel. Because the potential depends only on gaps, a uniform change of all angles costs no energy. Its Hessian therefore annihilates $(1,\dots ,1)$, giving a zero-frequency normal mode. The corresponding motion is rigid rotation, ${\theta }_i(t)={\theta }_i^{(0)}+A+Bt$, including a constant common angular shift.

For two particles write ${\theta }_2-{\theta }_1=\pi +{\delta }_2-{\delta }_1$. Both cyclic gaps contribute, so the quadratic potential is $k({\delta }_2-{\delta }_1{)}^2$. With the specified masses, the mass matrix is $km\mathrm{diag}(1,2)$ and the stiffness matrix is $2k\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$. The generalized eigenvalue problem givesThe finite-frequency normal mode keeps the mass-weighted mean angle fixed and changes the angular separation sinusoidally. Together with the two rigid-rotation constants, it spans all small motions about equilibrium.

For small transverse displacement, the wave equation on a string iswhere $T$ is the tension and $\rho$ is mass per unit length. Using separation of variables, put $y=X(x)F(t)$, the ratios $X^{\prime \prime }/X=F^{\prime \prime }/(c^{2}F)$ equal a constant. The fixed-end conditions $X(0)=X(L)=0$ exclude nontrivial solutions for nonnegative separation constants; for $-{\kappa }^2$ they give $X=\sin (\kappa x)$ and $\kappa =n\pi /L$, $n\ge 1$. The associated angular frequencies are ${\omega }_n=n\pi c/L$. Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so $A_n=0$. Assume the hammer interval lies inside $[0,L]$. The imposed velocity $g(x)=y_t(x,0)$ has Fourier sine series coefficientTherefore $B_n=g_n/{\omega }_n$, giving the amplitude of every excited normal mode.

The string energy is $\frac{\rho }{2}{\int }_0^L(y_t^2+c^2{y}_x^2)dx$. At the impulse it is $E_{\mathrm{total}}=\rho a{v}^2/2$. Orthogonality of the sine modes gives $E_n=\rho L{g}_n^2/4$ initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin (n\pi /3)=0$ or $\sin (n\pi /20)=0$. Hence the unexcited modes are precisely

First suppose $W[\mathbf{\xi }]>0$ for every nonzero admissible displacement. The conserved energy from part (b) then bounds both $T$ and $W$ by $E$. In particular,where $\Vert \mathbf{\xi }{\Vert }_{\rho }^2=⟨\mathbf{\xi },\mathbf{\xi }⟩$. The positive quadratic form $\Vert \dot{\mathbf{\xi }}{\Vert }_{\rho }^2+2W[\mathbf{\xi }]$ defines the perturbation energy norm and is conserved. Small perturbations in this norm remain small, proving energetic linear stability. Also $\Vert \mathbf{\xi }(t){\Vert }_{\rho }\le \Vert \mathbf{\xi }(0){\Vert }_{\rho }+\sqrt{2E}t$, excluding exponential displacement growth. For any admissible normal mode $\mathbf{\xi }=\mathbf{\zeta }{e}^{-i\omega t}$, multiplication of $-{\omega }^2\mathbf{\zeta }=F\mathbf{\zeta }$ by ${\mathbf{\zeta }}^{\ast }$ gives directlyso all such frequencies are real. No variational principle has been assumed.