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For a jet of velocity $U$ occupying $|y|<h$ in stationary fluid of the same density, disturbances proportional to $e^{ikx+\sigma t}$ satisfyfor a varicose mode of a planar jet, andfor a sinuous mode of a planar jet. Both have unstable growth rate

Decaying normal modes of this continuous piecewise-linear profile satisfy ${\varphi }^{\prime \prime }-k^2\varphi =0$ between the corners. Pressure matching at a piecewise-linear shear interface yields an odd varicose mode of a planar jet with real speed $c=(1-e^{-2k})/(2k)$ and an even sinuous mode of a planar jet whose speeds satisfyIts reduced discriminant factors as $(2k-3-e^{-2k}-4e^{-k})(2k-3-e^{-2k}+4e^{-k})$. The second factor is positive for $k>0$, while the first is strictly increasing and crosses zero once. Thus exactly the long-wave even branch is exponentially unstable, for $0<k<k_c$, where $2k_c-3-e^{-2k_c}-4e^{-k_c}=0$.

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 $1/2$.

For an odd velocity potential, chooseAt $z=L$ the two kinematic conditions give $A=ic{B}_{+}$ and $C\cosh (kL)=i(V-c)B_{+}$. The dynamic condition is $-cA=(V-c)C\sinh (kL)$. Eliminating $A,C,B_{+}$ gives, with $s=\tanh (kL)$,The lower interface gives the same equation by odd parity of the velocity potential. The interior vertical velocity is even, so $B_{-}=B_{+}$: the displacement is a sinuous mode of a planar jet even though the potential is odd.

For $L>0$, $V\ne 0$ and $k>0$, $0<s<1$ and one root has $c_i>0$. Thus every nonzero wavenumber is unstable in this ideal top-hat planar jet. The phrase “all choices” is understood with these assumptions: $k=0$ has no exponential wave growth, and $V=0$ has no shear.

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.

For a normal mode of an inviscid parallel flow, the perturbation pressure is $p=-(U-c){\varphi }^{\prime }+U^{\prime }\varphi$. At an interface where $U$ is continuous but $U^{\prime }$ jumps, continuity of normal velocity and pressure gives continuity of $\varphi$ and the displayed flux. Consequently $(U-c)[{\varphi }^{\prime }]=[U^{\prime }]\varphi$. A sinuous mode of a planar jet may therefore have a cusp at a symmetric velocity-profile corner: even parity alone does not force its one-sided derivative to vanish.