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For a top-hat planar jet in stationary fluid of the same mass density, odd and even velocity potentials have the same temporal growth rate. Put $s=\tanh (|k|L)$. Their dispersion relations use $s$ and $1/s$, but $\sqrt{s}/(1+s)=\sqrt{1/s}/(1+1/s)$. Their real phase velocities are respectively $Vs/(1+s)$ and $V/(1+s)$. The common growth rate is strictly below $|kV|/2$ for finite $|k|L>0$, approaching the isolated vortex sheet value when the sheets decouple.

An initial current $U{1}_{|y|<a}$ with flat surface has two oppositely signed vortex sheets. On an f-plane, its local final geostrophic balance hasThe Rossby deformation radius controls the edge layers. The height is continuous and odd; the along-strip velocity is $-g{\eta }_f^{\prime }/f$ and has the sheet jumps. Outgoing inertia-gravity waves carry the energy not retained in this balanced component.

The total velocity fields on the two sides areEach base flow is divergence-free. Incompressibility therefore makes both disturbance velocity potentials satisfy the Laplace equation:At the fixed horizontal walls, impermeability requires zero vertical velocity:The discontinuity in tangential base velocity at the interface is a vortex sheet. The assumption of irrotational flow is understood separately in each open fluid region; it cannot hold distributionally across that sheet when $U\ne 0$.

Each layer has constant mass density, initially negligible velocity and hence zero vorticity. Its Euler equations for an inviscid fluid have a conservative gravitational body force, so Kelvin's circulation theorem preserves zero bulk vorticity. In a simply connected layer, the irrotational flow therefore has ${\mathbf{v}}_i=\nabla {\varphi }_i$. The incompressible flow condition givesThis applies while the ideal layerwise solution is smooth. It allows an interfacial vortex sheet; it does not assert that the whole two-fluid flow has zero vorticity or that a smooth interface must exist for arbitrarily long times.

The rigid wall no-penetration boundary conditions arewith boundedness and ${\partial }_r{\varphi }_i=0$ at $r=0$ for regular axisymmetric fields. For a small interface displacement and small slope, linearization of the kinematic boundary condition about $z=0$ givesThe small-slope condition is an additional requirement for high-wavenumber modes: $|\alpha \eta |\ll 1$, not just $|\eta |\ll h,a$.

Choose the unperturbed hydrostatic pressures as ${\bar{p}}_i(z)=p_{\ast }-{\rho }_{i}gz$. The linearized Bernoulli equation gives the pressure perturbation $p_i^{\prime }=-{\rho }_i{\partial }_t{\varphi }_i$, after absorbing spatially constant gauges into ${\varphi }_i$. Since surface tension and viscous normal stress vanish, pressure continuity on $z=\eta$ givesorThe initial velocity is zero, so ${\eta }_t(r,0)=0$. Separate conservation of mass in each closed layer also requires ${\int }_0^a\eta (r,t)rdr=0$ relative to the initially horizontal interface.

For an inviscid flow, the force per unit volume is $\mathbf{F}=-\nabla p-\rho g\hat{\mathbf{z}}$. The gravitational acceleration has zero curl. Using the curl product rule,The supplied two-dimensional vorticity equation therefore becomesThe out-of-plane vorticity has no stretching term in this two-dimensional setting. The source is baroclinic vorticity generation: differently directed mass density and pressure gradients generate rotation even from zero initial vorticity.

For an explicit sign check, set ${\omega }_y={\partial }_{z}u-{\partial }_{x}w$, with $z$ upward. A thin smoothed interface has $\rho \simeq \bar{\rho }(z-\eta (x,t))$, with ${\bar{\rho }}^{\prime }>0$. If $P(x,t)$ denotes the total pressure evaluated on the interface, then within the thin layer $p_x+{\eta }_x{p}_z\simeq P_x$ andIntegrating through that layer and taking its thickness to zero gives the vortex sheet strength $\gamma =u_2-u_1$ and its linear evolution:For a growing Fourier mode, the potentials from the preceding calculation with $\alpha$ replaced by $k$ giveThus the right-hand descending slope of a light-fluid crest produces positive ${\omega }_y$, while the left-hand ascending slope produces negative ${\omega }_y$. Their induced velocity raises the light-fluid crest and lowers the heavy-fluid trough, increasing the slope and generating more vorticity. This is the positive feedback underlying Rayleigh-Taylor instability.

One can close this mechanism quantitatively. For $\eta =H(t)\cos (kx)$, the kinematic boundary conditions give $\gamma =2\dot{H}\coth (kh)\sin (kx)$. Combining with the sheet-generation equation givesFor $kh\ll 1$, the lids confine motion throughout both layers and horizontal redistribution is slow, yielding $\sigma \sim \sqrt{\mathcal{A}gh}k$. For $kh\gg 1$, motion is localized within depth $k^{-1}$ on each side. Less fluid must accelerate at shorter wavelength, giving $\sigma \sim \sqrt{\mathcal{A}gk}$. Since $k_n=n\pi /b$, the ideal growth rate increases with $n$ with no short-wave cutoff.

The sharp-interface result requires $k\delta \ll 1$ for the smoothed layer. When $k\delta$ becomes order one, it no longer applies. A smooth adverse mass density gradient has a local overturning time of order $[g{\bar{\rho }}^{\prime }/\bar{\rho }{]}^{-1/2}$ rather than the unbounded sharp-interface rate. Finite molecular diffusion also relaxes small-scale mass density perturbations on a time of order $(D{k}^2{)}^{-1}$, and viscosity, if included, damps small-scale velocity. Weak diffusion motivates the thin-layer picture; it does not justify extrapolating the discontinuous dispersion relation to arbitrary $n$.

Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.

Take $\mathbf{f}=f\hat{\mathbf{z}}$ with $f>0$, and put $\zeta =v_x-u_y$, $d=u_x+v_y$ and $q=\zeta -f\eta /H$. Here $q$ denotes the signed vertical component of the linearized shallow-water potential-vorticity anomaly, rather than its magnitude. The divergence and vertical curl of the linearized shallow water equations giveConsequently $q_t=0$. Eliminating $d$ and substituting $\zeta =q+f\eta /H$ givesThe vector forcing is $-H\mathbf{f}\cdot \mathbf{q}$. This is potential-vorticity conservation in its linear, f-plane form: the conserved anomaly forces a stationary balanced part, while the homogeneous equation supports inertial-gravity waves.

For the initial strip, differentiating the discontinuous velocity in the sense of distributions givesThese are two oppositely signed vortex sheets. In the final geostrophic balance, $v_f=0$ and $u_f=-g{\eta }_f^{\prime }/f$. The height therefore solvesThis is the Rossby deformation radius. Decay at infinity fixes the Green function to $G(y)=-\lambda e^{-|y|/\lambda }/2$. Thus the geostrophic adjustment of a finite-width current has the particularly useful representationExpanding the exponentials inside the strip gives ${\eta }_f/H=-(U/c)e^{-a/\lambda }\sinh (y/\lambda )$; above the strip it gives $-(U/c)e^{-y/\lambda }\sinh (a/\lambda )$, and below it gives $(U/c)e^{y/\lambda }\sinh (a/\lambda )$. The height is continuous, odd and exponentially localized. Its derivative has the jumps required by the two Dirac delta functions.

Differentiating the height, rather than assuming a uniform final current, gives the complete velocity:The one-sided velocity jump is $U$ at each edge, with opposite orientations. The value exactly on an idealized vortex sheet is immaterial. Inside the strip the current remains in the original direction; outside it a return current develops.

Let $r=a/\lambda$ and $s=y/\lambda$. For $r\ll 1$, $u_f/U=1-r+O(r^2)$ over the narrow strip, while the outside return flow is approximately $-r{e}^{-|s|}$. The height varies almost linearly across the strip, $c{\eta }_f/(HU)\simeq -s$, and has extrema of magnitude $(1-e^{-2r})/2\simeq r$ at its edges. For $r\gg 1$, the central current is exponentially small: $u_f(0)/U=e^{-r}$. Each edge supports a layer of width $\lambda$, with $u_f/U\simeq \frac{1}{2}{e}^{-(a-|y|)/\lambda }$ on its inner side and $-\frac{1}{2}{e}^{-(|y|-a)/\lambda }$ outside. The corresponding height extrema approach $\pm HU/(2c)$, with almost zero height deep inside and far outside. Both requested profiles are drawn from the exact functions below; the dashed jumps represent one-sided velocity limits.

For balanced shallow-water energy as a signed potential-vorticity pairing, divide out the constant density and work per unit distance in $x$. The kinetic energy plus surface potential energy isMultiplication of the stationary height equation by ${\eta }_f$ and integration by parts, with vanishing end terms, yieldsThe same expression holds with $dA$ for a finite two-dimensional domain or a finite periodic length in $x$. The infinite strip's total energy is infinite, so both energies and their ratio are understood per unit $x$-length.

There is a sign defect in the printed energy expression: $|\mathbf{q}|$ must be replaced by the signed component $\hat{\mathbf{f}}\cdot \mathbf{q}$. Indeed, $|q_0|=|U|[\delta (y-a)+\delta (y+a)]$, so pairing it with the odd ${\eta }_f$ gives zero, although the balanced state has positive energy. The signed pairing givesSince $E_i=H{U}^{2}a$, the energy retention in finite-width geostrophic adjustment isFor $r\ll 1$ this ratio is $1-r+\frac{2}{3}{r}^2+O(r^3)$: nearly all energy remains in the narrow balanced current. For $r\gg 1$ it is approximately $1/(2r)$: only the edge regions retain balanced energy. Conservation of energy still holds for the inviscid evolution. The missing balanced energy is carried away by inertia-gravity waves; “final state” means the local balanced limit after those waves leave, rather than a loss of the total energy over the entire infinite domain.

To recover the printed dispersion relation, interpret the surface maintained flat as an imposed flat upper boundary for the perturbation, so its normal velocity is zero. This is the rigid-lid approximation. Work with perturbation velocity potentials ${\varphi }_{-}=A{e}^{ky}{e}^{ikx+\sigma t}$ below the interface and ${\varphi }_{+}=B\cosh [k(y-h)]e^{ikx+\sigma t}$ in the upper layer. They solve the Laplace equation, decay as $y\to -\infty$, and obey ${\partial }_y{\varphi }_{+}(h)=0$.

The linear kinematic boundary conditions at the interface give $(\sigma +ikU){\eta }_0=kA$ and $\sigma {\eta }_0=-kB\sinh (kh)$. Equality of pressures, using the linear Bernoulli equation, gives $(\sigma +ikU)A=\sigma B\cosh (kh)$. The equal fluid densities cancel the interface gravitational restoring term. Eliminating $A,B,{\eta }_0$ yieldsWith $q=\tanh (kh)$, the roots areFor every $k>0$ one root has positive real part. Thus the vortex sheet has a Kelvin-Helmholtz instability at every wavelength in this model. If $kh\gg 1$, $q\to 1$ and the growing rate is $Uk/2$. If $kh\ll 1$, $q\sim kh$ and it is $Uk\sqrt{kh}$.

There is a physical qualification to the wording. A dynamically deformable free surface at finite gravity is not exactly a flat lid. Its linear conditions combine to ${\sigma }^2{\varphi }_{+}(h)+g{\partial }_y{\varphi }_{+}(h)=0$. Writing the upper potential as $B\cosh (ky)+C\sinh (ky)$ and using the same interface conditions instead givesThe stated quadratic follows in the strong-gravity flat-surface limit, or with the flat condition imposed exactly. It is not the exact dispersion relation of a freely moving upper surface at arbitrary finite $g$. This makes explicit the boundary interpretation needed for the requested formula.

Write ${\varphi }^{\prime }=\Phi (z)e^{ik(x-ct)}$ and let $B_{\pm }$ be the displacement amplitudes at $z=\pm L$. Use subscripts $I$ for the jet interior and $O$ for the adjoining exterior. Each interface is a material vortex sheet, so its linearized kinematic boundary conditions areThe equal mass densities make the hydrostatic displacement terms cancel. The dynamic boundary condition for an inviscid interface therefore givesThe division form assumes $c\ne 0,V$; the preceding undivided equations remain the proper conditions in those exceptional cases. Potentials themselves need not be continuous, and their vertical derivatives need not be equal: the same material displacement is advected by different tangential base velocities. At infinity the exterior potentials decay. These displacement and pressure conditions, rather than continuity of $\Phi$ and ${\Phi }^{\prime }$, determine the equal-density top-hat planar-jet dispersion relation.

The odd-potential branch uses $s=\tanh (kL)$, whereas the even-potential branch uses $1/s$. The imaginary parts are identical becauseConsequently the equality of top-hat jet parity growth rates givesHere $\gamma =k{c}_i$ is the positive temporal growth rate. The real phase velocities differ: $c_{r,\mathrm{odd}}=\frac{V}{2}(1-e^{-2kL})$ and $c_{r,\mathrm{even}}=\frac{V}{2}(1+e^{-2kL})$. Thus equal amplification does not imply equal propagation. As $kL\to \infty$, both approach the isolated-vortex sheet phase velocity $V/2$ and growth rate $k|V|/2$.

For a single equal-density vortex sheet with stream speeds $0$ and $V$, the gravitational contribution vanishes. The Kelvin-Helmholtz dispersion relation with gravity reduces to $c^2+(V-c{)}^2=0$, with roots $c=V/2\pm i|V|/2$. Its positive temporal growth rate is ${\gamma }_{\mathrm{sheet}}=k|V|/2$.

Both jet parity branches have the common growth rate derived above, henceEquivalently, $2\sqrt{s}\le 1+s$ follows from $(\sqrt{s}-1{)}^2\ge 0$. Equality is approached when the two interfaces are separated by many disturbance decay lengths, $kL\to \infty$. The zero-shear case has both growth rates zero. For small $kL$, the growth rate ratio is approximately $2\sqrt{kL}$: interaction between the sheets suppresses amplification relative to an isolated sheet.

Assume incompressible inviscid layers, gravity $-g{\mathbf{e}}_y$, no surface tension, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla \cdot {\mathbf{u}}_j=0$ and ${\partial }_t{\mathbf{u}}_j+{\mathbf{u}}_j\cdot \nabla {\mathbf{u}}_j=-{\rho }_j^{-1}\nabla p_j-g{\mathbf{e}}_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta (x,t)$ each layer obeys $u_{jy}={\eta }_t+u_{jx}{\eta }_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this vortex sheet.

Write ${\mathbf{u}}_j=(U_j,0)+\nabla {\varphi }_j$, with $U_1=U$, $U_2=-U$. Then ${\nabla }^2{\varphi }_j=0$, and the wall conditions are ${\partial }_y{\varphi }_1(h)={\partial }_y{\varphi }_2(-h)=0$. Linearizing the interface conditions at zero givesThe second follows from linear Bernoulli pressure $p_j^{\prime }=-{\rho }_j({\partial }_t+U_j{\partial }_x){\varphi }_j$ and the hydrostatic pressure shift $-{\rho }_{j}g\eta$.

For $k>0$, use amplitudes ${\varphi }_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and ${\varphi }_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma +ikU)A/[k\sinh kh]$ and $B_2=(\sigma -ikU)A/[k\sinh kh]$. The dynamic condition becomesCompleting the square yieldsA positive real part occurs precisely whenThe function $\tanh (kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its derivative has numerator $kh{\mathrm{sech}}^2(kh)-\tanh (kh)<0$, since $\sinh (2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero wavenumber isAt equality every $k>0$ still satisfies the strict growth inequality, since $\tanh (kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\mathrm{crit}}$.

Take $k>0$; reflection in $x$ supplies the result for negative wavenumber. Since each region has uniform base velocity, the perturbation is an incompressible, irrotational flow with a velocity potential satisfying the Laplace equation. Its decaying outer forms areFor a varicose mode of a planar jet, reflection symmetry makes the inner potential proportional to $\cosh (ky)$; for a sinuous mode of a planar jet, it is proportional to $\sinh (ky)$.

At the upper interface, the linearized kinematic boundary condition isThe first equation gives ${\varphi }_{+}(h)=-\sigma \eta /k$. The second givesThe linearized Unsteady Bernoulli equation gives $p^{\prime }=-\rho ({\partial }_t+U{\partial }_x)\varphi$. Applying pressure continuity therefore produces the equal-density top-hat planar-jet dispersion relationSolving the quadratic equation givesEquivalently,The plus sign is the unstable Kelvin-Helmholtz instability branch. In particular, the two modes have exactly the same growth rate,Its limiting forms areThus neither type grows faster in this equal-density idealization, although their phase speeds differ. The unbounded large-$k$ growth is the familiar short-wave pathology of an inviscid vortex sheet.

Put $q=|k|$. Decay at infinity and the symmetry of the disturbance suggest the mode potentialsThe middle potential is an even function of $y$ because its vertical derivative must take opposite values at the two interfaces. The upper kinematic conditions giveThe lower conditions give the same two relations. The upper dynamic condition isEliminating $A$, $B$ and ${\eta }_0$ yields the dispersion relationWriting $C=\coth (qh)$, its two roots areFor every real $k\ne 0$, $C>0$ and one root has the positive growth rateThus every nonzero wavenumber is linearly unstable in this inviscid vortex sheet model; $k=0$ is neutral.

When $|k|h\gg 1$, the two shear layers are separated by many disturbance decay lengths. Since $\coth (|k|h)\to 1$, the roots reduce toThis is exactly the Kelvin-Helmholtz instability of one isolated vortex sheet separating equal-density streams of speeds $U$ and $0$: the pattern travels at their mean speed $U/2$, while its unstable amplitude grows at rate $|k|U/2$. The exponentially weak interaction between the two layers disappears in this short-wavelength limit.

A shear layer is a thin region across which the tangential velocity changes rapidly. In an inviscid idealization it can collapse to a vortex sheet.

A top-hat planar jet has one constant streamwise velocity in a layer and another constant velocity outside it. In the inviscid model, the two layer boundaries are parallel vortex sheets.

A uniform-density inviscid shear flow with $U=U_{0}z/d$ for $|z|<d$ and $U=\pm U_0$ outside has two vorticity jumps but no walls. Decaying normal modes satisfy $w^{\prime \prime }-k^{2}w=0$ in each region. Continuity of $w$ and pressure matching at a piecewise-linear shear interface gives a two-by-two determinant whose diagonal entries are $1-2kd\pm 2kdc/U_0$ and off-diagonal entries $e^{-2kd}$. Its vanishing gives the displayed dispersion relation. The Kelvin-Helmholtz instability band is $0<kd<{\alpha }_s$, where $2{\alpha }_s-1=e^{-2{\alpha }_s}$ has its unique positive root above $1/2$, approximately $0.63923227$. Short waves decouple the two edges and remain neutral; long waves approach the vortex sheet instability with $c=\pm i{U}_0$.

A long array of equal co-oriented vortex rings, spaced by $l\ll r$, approximates an azimuthal vortex sheet on a cylinder of radius $r$. The circulation per unit axial length is $\Gamma /l$. The conditions $\nabla \cdot \mathbf{u}=0$ and $\nabla \times \mathbf{u}=\mathbf{\omega }$ are mathematically analogous to the magnetostatic equations for a solenoid. Away from its ends, an array long compared with $r$ produces nearly uniform axial flow $\Gamma /l$ inside and weak flow outside. A finite array has return flow outside connecting its two ends. The uniform-core approximation fails close to individual vortex cores and near the ends, and the field requires a stated circulation orientation and background flow.

Consider an equal-density vortex sheet halfway between rigid walls, with equal layer depths $h$, base velocities $U_1,U_2$ and no restoring interfacial force. The irrotational perturbations have vertical factors $\cosh (k(y-h))$ above and $\cosh (k(y+h))$ below. Their kinematic boundary conditions and pressure continuity giveThe common depth factor cancels, yielding the displayed dispersion relation roots and Kelvin-Helmholtz instability for nonzero shear. The real phase velocity of the pattern is $(U_1+U_2)/2$ and the growing branch has temporal growth rate $|k(U_1-U_2)|/2$. This exact depth cancellation depends on equal depths and equal mass densities; a single nearby wall or unequal layers changes the inertia factors.