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For a smooth inviscid parallel shear profile with an exponentially growing mode, let $U_s$ be its velocity at an inflection point. The real and imaginary parts of the integrated Rayleigh equation for inviscid shear flow imply $\int U^{\prime \prime }(U-U_s)|\psi {|}^2/|U-c{|}^2=-\int (|{\psi }^{\prime }{|}^2+k^2|\psi {|}^2)<0$. Consequently $U^{\prime \prime }(U-U_s)<0$ somewhere in the channel. This strengthens Rayleigh's inflection-point theorem but is still only a necessary condition, not a sufficient instability criterion.

An inviscid parallel shear flow has base velocity $\mathbf{u}=U(y){\mathbf{e}}_x$. Two-dimensional normal-mode perturbations satisfy the Rayleigh equation for inviscid shear flow.

For a monotone inviscid parallel shear flow, let a regular neutral Rayleigh equation for inviscid shear flow mode have $U(0)=c_0$, $U^{\prime \prime }(0)=0$, $d=U^{\prime }(0)>0$, and $p={\varphi }_0(0)\ne 0$. A perturbation $c=c_0+\epsilon c_1$ with $\mathrm{Im}{c}_1>0$ resolves the singularity on the inner variable $\eta =y/\epsilon$. The first outer correction has ${\varphi }_1=a+\beta y\log |y|+b_{\pm }y+\cdots$, with $\beta =c_1{U}^{\prime \prime \prime }(0)p/d^2$. The inner particular solution contains $(d\eta -c_1)\log (d\eta -c_1)$. Its complex logarithm has argument tending to zero at positive infinity and to $-\pi$ at negative infinity, giving the displayed matching jump.

Let $I=\int {\varphi }_0^{2}dy$, $J=\mathrm{PV}\int U^{\prime \prime }{\varphi }_0^2/(U-c_0{)}^{2}dy$, and $M=U^{\prime \prime \prime }(0){\varphi }_0(0{)}^2/U^{\prime }(0{)}^2$. The Wronskian identity for the first-order Rayleigh equation for inviscid shear flow, combined with neutral-mode critical-layer matching, gives $2k_0{k}_{1}I=-c_1(J+i\pi M)$. For real $k_1$, its imaginary part is $2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$. The Cauchy principal value is necessary when the outer forcing has a simple real pole.

The Orr-Sommerfeld equation governs the wall-normal velocity of normal mode disturbances to an incompressible viscous parallel shear flow. Here $D=d/dy$, ${\kappa }^2={\alpha }^2+{\beta }^2$, and $Re$ is the Reynolds number. It follows by eliminating pressure from the linearized Navier-Stokes equations. At a rigid no-slip wall, $\hat{v}=D\hat{v}=0$. In the inviscid limit it reduces to the Rayleigh equation for inviscid shear flow.

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 $k|U_0|$.

Let $q^2=k^2+l^2$ and $c=c_r+i{c}_i$, with $k>0$ chosen for the temporal mode convention. In the inviscid limit, the Orr-Sommerfeld equation reduces to the Rayleigh equation for inviscid shear flow:The impermeability boundary condition is $v(\pm 1)=0$; the extra viscous derivative condition is not an independent condition for this reduced second-order equation. For an unstable mode $c_i>0$, the denominator has no real zero. Multiply by $\bar{v}$ and apply integration by parts to obtainIts imaginary part givesFor a nontrivial mode, the weight is nonnegative and is positive except at isolated mode zeros. Thus $U^{\prime \prime }$ must take both signs, unless it vanishes identically. The identically zero case also has no unstable nontrivial mode, by the positive first integral in the displayed identity. Therefore the smooth real base-flow profile must have an interior inflection point. This proves Rayleigh's inflection-point theorem; it is a necessary condition, not a sufficiency test.

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.

Let $d=U^{\prime }(0)>0$, $h=U^{\prime \prime \prime }(0)$, $p={\varphi }_0(0)\ne 0$, and $q={\varphi }_0^{\prime }(0)$. At a regular neutral Rayleigh equation for inviscid shear flow mode, multiply the equation by $U-c_0$ and evaluate it at the critical layer in a shear flow. The term multiplying ${\varphi }_0^{\prime \prime }-k_0^2{\varphi }_0$ vanishes, leaving $U^{\prime \prime }(0)p=0$. Therefore $U^{\prime \prime }(0)=0$. If this condition failed, the coefficient $U^{\prime \prime }/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular eigenfunction.

There is a sign error in the printed first-order equation. Direct expansion of the original Rayleigh equation for inviscid shear flow givesThe coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\epsilon c_1/(U-c_0{)}^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral eigenfunction. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^{3}h/6$ in the Taylor expansion of $U$, and $U^{\prime \prime \prime }(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.

Near zero, $U-c_0=dy+h{y}^3/6+\cdots$ and $U^{\prime \prime }=hy+O(y^2)$. Thus $U^{\prime \prime }/(U-c_0)\to h/d$, whereas $U^{\prime \prime }/(U-c_0{)}^2=h/(d^{2}y)+O(1)$. The singular part of the first-order equation isIntegrating twice on either side givesThe constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral eigenfunction rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.

The outer expansion of $1/(U-c)$ fails when $|dy|$ is comparable with $\epsilon |c_1|$. Therefore the distinguished limit isIn this inner variable, the original equation becomesAt orders one and $\epsilon$, its right side is zero. Matching the outer expansion therefore selects ${\Phi }_0=p$ and ${\Phi }_1=q\eta +a$. At order ${\epsilon }^2\log \epsilon$, ${\theta }_2^{\prime \prime }=0$. The outer term $\epsilon \beta y\log |y|$ becomes ${\epsilon }^2\beta \eta (\log \epsilon +\log |\eta |)$, so the required switchback term is ${\theta }_2=\beta \eta$. At order ${\epsilon }^2$,Integrating, or using the supplied hint, givesThe constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with ${\varphi }_0^{\prime \prime }(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.

For $\mathrm{Im}{c}_1>0$, $d\eta -c_1$ runs below the real axis. Choose the continuous branch of the complex logarithm there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta |$ expansions of ${\Phi }_2$ areThese expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_{\pm }$ proves the neutral-mode critical-layer matching relationThe sign comes from passing below the pole; taking $\mathrm{Im}{c}_1<0$ would reverse it.

Finally use the Wronskian $W={\varphi }_1{\varphi }_0^{\prime }-{\varphi }_0{\varphi }_1^{\prime }$. Since $L_0{\varphi }_0=0$ and $L_0{\varphi }_1=H{\varphi }_0$, subtraction gives $W^{\prime }=-H{\varphi }_0^2$. The wall boundary conditions give $W(a)=W(b)=0$, soThis is the requested excised-interval identity. As $\delta \downarrow 0$, the equal logarithmic slope terms cancel andThe remaining real singular integral is interpreted as a Cauchy principal value, consistent with excision on both sides of the critical layer in a shear flow. DefineThen the neutral Rayleigh-mode dispersion correction iswhere the last form assumes a nonzero denominator. For real $k_1$, $\mathrm{Im}{c}_1=2k_0{k}_{1}I\pi M/(J^2+{\pi }^2{M}^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.

Let $U$ be real and twice continuously differentiable, take a nonzero real wave number, and write $c=c_r+i{c}_i$. Exponential instability requires $k{c}_i>0$, in particular $c_i\ne 0$, so $U-c$ has no zero on the channel. Divide the Rayleigh equation for inviscid shear flow by $U-c$, multiply by ${\psi }^{\ast }$ and integrate. The wall conditions remove the boundary term, givingIts imaginary part isThe weight is nonnegative and positive wherever the nontrivial eigenfunction does not vanish. A nontrivial solution of this regular second-order equation cannot vanish on an interval. If $U^{\prime \prime }$ had one strict sign without changing sign, the integral could not vanish. If $U^{\prime \prime }$ were identically zero, the real identity would instead give the contradiction $I=0$. Thus $U^{\prime \prime }$ must take both signs, and continuity gives an interior zero. This proves Rayleigh's inflection-point theorem: an unstable profile must have an inflection point, in fact a sign change of $U^{\prime \prime }$ rather than merely an isolated zero without sign change.

Take $\alpha >0$; a real disturbance with negative wavenumber is recovered by complex conjugation. To obtain the pressure equation for a shear-flow normal mode, apply $i\alpha$ to the streamwise momentum equation and differentiate the normal momentum equation. The mass conservation equation cancels the viscous divergence and the terms proportional to $U-c$, leaving $2i\alpha U^{\prime }v={\alpha }^{2}p-p^{\prime \prime }$. Thus the fluid pressure equation and wall value areThe wall value follows from streamwise momentum with no-slip boundary condition. Normal momentum also gives $p^{\prime }(0)=v^{\prime \prime }(0)/R$.

In the lower viscous boundary layer, mass conservation gives $v=O(\alpha \delta )$ if $u=O(1)$. Since $U\sim y$, convection and shear have size $\alpha \delta$ when $c=O(\delta )$; matching these with transverse viscous diffusion gives $R^{-1}{\delta }^{-2}\sim \alpha \delta$. The pressure gradient balance then gives $p=O(\delta )$. Thus the lower scalings areThe neglected streamwise viscous term is smaller by $(\alpha \delta {)}^2$. In these variables the leading equations areDifferentiating the second and using the first yields the Airy reduction $\boxed{u_{YYY}=i(Y-C)u_Y}$.

Put $q=e^{i\pi /6}$, so $q^3=i$. The bounded matching branch of the Airy function gives $u_Y=A\mathrm{Ai}(q(Y-C))$; the other independent Airy solution grows along the ray and is excluded. Integrating with the two wall conditions giveswhile the wall momentum equation gives the constant fluid pressureDefine the convergent Airy displacement integralsThe Airy function equation and its decaying derivative imply $I_1-C{I}_0=iq{\mathrm{Ai}}^{\prime }(-qC)$. Consequently, as $Y\to \infty$,Both differentiated lower momentum and the constant of integration are satisfied; retaining that constant is essential for the final matching.

For the middle region introduce a perturbation stream function $\Phi$ with $u={\Phi }^{\prime }$ and $v=-i\alpha \Phi$. The leading inviscid flow streamwise equation isFor a shear flow, this is the integrated form of the Rayleigh equation for inviscid shear flow. The leading matched solution for fixed $y=O(1)$ is $\Phi \sim BU$, so $u\sim B{U}^{\prime }$ and $v\sim -i\alpha BU$. In the overlap with the lower viscous boundary layer, the first constant correction iswhich agrees with the integrated lower solution. For a regular smooth profile, the normal momentum equation gives $p^{\prime }\sim -{\alpha }^2(U-c)\Phi$. Thus fluid pressure variation across a fixed-width middle region is $O({\alpha }^{2}B)$; in the final distinguished scaling this is smaller than its $O(\delta )$ value, so $p\sim \delta P_0$ throughout that region. Since $U\to 1$, the middle velocity tends to $\boxed{v\sim -i\alpha B}$, which does not decay and requires another region.

The fluid pressure equation becomes $p^{\prime \prime }-{\alpha }^{2}p=0$ when $U\sim 1$. Therefore the upper scale is $\Delta =1/\alpha$, with $z=\alpha y$. The inviscid decaying upper solution isMatching velocity as $z\downarrow 0$ gives $P_{\infty }\sim \alpha B$; matching fluid pressure to the middle region gives $\delta P_0\sim \alpha B$. This is three-layer long-wave shear-flow matching, expressed as a matched asymptotic expansion.

Under $\alpha =k{R}^{-1/4}$ with fixed positive $k$, $\delta =k^{-1/3}{R}^{-1/4}$, and $\alpha /\delta =k^{4/3}$. Eliminating the nonzero amplitude $A$ gives the Airy wall-layer matching determinant and dispersion relationEquivalently, with $\mathcal{T}(C)={\int }_{-qC}^{\infty e^{i\pi /6}}\mathrm{Ai}(\zeta )d\zeta =q{I}_0(C)$,The determinant form does not divide by a possibly vanishing $I_0$. The wavespeed is $c=\delta C$; temporal growth has sign $\mathrm{Im}C$ for positive $\alpha$.

There is a genuine conflict in the printed assumptions: fixed-$k$ $\alpha =k{R}^{-1/4}$ cannot satisfy $R^{-1/7}\ll \alpha$ as $R\to \infty$. For order-one lower matching amplitudes, constant leading middle fluid pressure requires ${\alpha }^2\ll \delta$, equivalently $\alpha \ll R^{-1/7}$; this is the reverse inequality and is satisfied by the final distinguished scaling. Thinness also requires $\alpha R\gg 1$. The dispersion relation above is therefore derived under the final distinguished limit, for which the complete consistent window is $R^{-1}\ll \alpha \ll R^{-1/7}$. Both printed limits cannot be imposed simultaneously.

Impermeability requires $\hat{w}(-1)=\hat{w}(1)=0$. Let $c=\omega /\alpha$ with real $\alpha \ne 0$, and set $a=|\alpha |$. If $c$ is nonreal, or real outside the channel, the Rayleigh equation for inviscid shear flow gives ${\hat{w}}^{\prime \prime }-a^2\hat{w}=0$ everywhere. The two wall conditions then force $\hat{w}=0$. In fact there is no nonzero globally smooth eigenfunction for any $c$: the complete family is the inviscid Couette continuous spectrum.

For each $\xi \in (-1,1)$, permit a localized vorticity sheet. Let $G_a$ be the Dirichlet Green function satisfying $(D^2-a^2)G_a(z,\xi )=\delta (z-\xi )$. An explicit generalized eigenfunction iswhere $z_{<}=\min (z,\xi )$ and $z_{>}=\max (z,\xi )$. It vanishes at both walls, is continuous at $z=\xi$, and has derivative jump $[{\partial }_z{G}_a{]}_{{\xi }^{-}}^{{\xi }^{+}}=1$. Consequentlyusing the Dirac delta multiplication identity. These are vorticity-sheet eigenfunctions of inviscid Couette flow, understood as generalized eigenfunctions, not as a discrete smooth Sturm-Liouville eigenfunction expansion.

To see completeness, define the vorticity variable $q=(D^2-a^2)w$. Its evolution is $({\partial }_t+i\alpha z)q=0$, so for any admissible initial vorticity $q_0$,The homogeneous Dirichlet problem for $D^2-a^2$ has only the zero solution, so this inversion reconstructs every initial vertical-velocity field in its usual function space. The generalized frequencies fill the interval with endpoints $\pm \alpha$; the endpoint values are understood as the closure of the continuous spectrum.

First suppose $c_i\ne 0$ and put $F=\hat{w}/(U-c)$. The wall condition gives $F(\pm L)=0$. Substitution into the Rayleigh equation for inviscid shear flow givesMultiply by $\overline{F}$ and use integration by parts. The resulting weighted identity for Howard's semicircle theorem isIts imaginary part gives $\int UQ=c_{r}I$, and its real part then gives $\int U^{2}Q=(c_r^2+c_i^2)I$. Set $a=U_{\min }$ and $b=U_{\max }$. Since $(U-a)(b-U)\ge 0$,Completing the square proves Howard's semicircle theorem:In particular every growing normal mode lies in the upper semicircle for $k>0$. If $c$ is real and outside $[a,b]$, $F$ is again regular, but the same identity has a strictly positive integrand unless the mode vanishes. Such a neutral mode is impossible. Consequently neutral regular modes also satisfy the bound; no division by $c_i$ is needed for this last conclusion. The nonreal proof avoids critical-level singularities altogether.

Normalize $U$ and $c$ by $U_{\max }$ and write $h={\zeta }_L\in (0,1)$, $\alpha =kL>0$. In each open region $U^{\prime \prime }=0$, so the Rayleigh equation for inviscid shear flow reduces to ${\hat{w}}^{\prime \prime }-{\alpha }^2\hat{w}=0$ for the nonreal modes of interest. Incorporating the wall conditions from the start, the bounded piecewise-linear shear layer hasThese are four independent regional amplitudes before matching. The same expressions extend to noncritical neutral modes and their limits.

The linearized vorticity equation for a inviscid parallel shear flow isSubstituting $\psi =\varphi (y)e^{ik(x-ct)}$ gives Rayleigh's equation

Write the Rayleigh equation for inviscid shear flow asMultiply by ${\varphi }^{\ast }$, integrate, and use integration by parts:Its imaginary part isFor an unstable mode $c_i>0$, the integral can vanish only if $U^{\prime \prime }$ changes sign somewhere in the flow. Thus the velocity profile must have an inflection point. This is Rayleigh's inflection-point theorem; it is necessary, not sufficient, for inviscid instability.

Put $\varphi =(U-c)F$. The Rayleigh equation for inviscid shear flow becomesMultiplication by $F^{\ast }$ and integration givesFor $c_i>0$, its real and imaginary parts implyThus $c_r$ is a weighted mean of $U$ and $c_i^2$ is its weighted variance. If $U_{-}\le U\le U_{+}$, the sharp bounded-variable variance estimate givesCompleting the square proves Howard's semicircle theorem: