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For $U=-1$ below $z=-1$, $U=z$ between $z=-1$ and $z=1$, and $U=1$ above $z=1$, decaying normal modes solve ${\varphi }^{\prime \prime }-k^2\varphi =0$ in each region. Pressure matching at a piecewise-linear shear interface gives continuity of $\varphi$ and $(U-c){\varphi }^{\prime }-U^{\prime }\varphi$. The two corner conditions yield the displayed dispersion relation. At $k=1/2$ the wave speeds are $c=\pm i/e$, so one mode grows exponentially with rate $1/(2e)$.

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 $2k_c-3-e^{-2k_c}-4e^{-k_c}=0$.

Let the perturbation velocity and pressure be $(u,w,p)(z)e^{ik(x-ct)}$, with real $k>0$. Linearizing the Euler equations about the inviscid parallel shear flow givesUse $u={\varphi }^{\prime }$, $w=-ik\varphi$ to satisfy incompressibility. The first momentum equation gives $p/{\rho }_0=-(U-c){\varphi }^{\prime }+U^{\prime }\varphi$; differentiate it and use the second equation. The Rayleigh equation for inviscid shear flow follows:A growing temporal normal mode has $c_i=\mathrm{Im}c>0$ and growth rate $k{c}_i$.

Rayleigh's inflection-point theorem states that such an instability of a smooth parallel profile requires $U^{\prime \prime }$ to change sign somewhere. To prove it, divide by $U-c$, multiply by ${\varphi }^{\ast }$, integrate over $z$ and integrate the second derivative by parts. Decay removes the endpoint terms, givingTaking the imaginary part yieldsThe weighting is nonnegative. A one-signed $U^{\prime \prime }$ that is not identically zero cannot satisfy this identity for a nontrivial mode; otherwise the mode vanishes on an interval and uniqueness of the differential equation forces it to vanish throughout. If $U^{\prime \prime }\equiv 0$, the real part of the identity is already a strictly positive integral for a nontrivial decaying mode. This proves the necessary sign-change condition.

Fjørtoft's theorem strengthens this: if $U_s$ is the velocity at an inflection point, instability requires $U^{\prime \prime }(U-U_s)<0$ somewhere. This is a necessary condition, not a sufficient criterion. For $U=\tanh z$, $U_s=0$ andThe possibility of instability is therefore consistent with Fjørtoft's theorem.

For the piecewise-linear approximation, $U^{\prime \prime }=0$ within each of the three regions. The decaying solutions outside and the two independent interior solutions can be writtenAt each corner, the normal velocity is continuous, so $\varphi$ is continuous. Pressure matching at a piecewise-linear shear interface additionally gives $[(U-c){\varphi }^{\prime }-U^{\prime }\varphi ]=0$, hence $(U-c)[{\varphi }^{\prime }]=[U^{\prime }]\varphi$. At $z=1$ the slope jumps from $1$ to $0$; at $z=-1$ it jumps from $0$ to $1$. Substituting the decaying outer solutions givesThus the inviscid instability of a piecewise-linear mixing layer has dispersion relationAt $k=1/2$, $c=\pm i/e$, so the root with positive imaginary part is an explicitly growing mode, with growth rate $1/(2e)$. More generally the unstable interval is $0<k<k_c$, where the unique $k_c>1/2$ solves $2k_c-1=e^{-2k_c}$. This direct calculation establishes instability of the approximation; passing a necessary inflection test alone would not have established it.

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 $k|U_0|$.

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 $1/2$.

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 $c=\pm i{U}_0$.