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For nondimensional constant shear $U=y$ and streamwise wavenumber one, put $W=(D^2-1)\hat{v}$. The Orr-Sommerfeld equation gives $W^{\prime \prime }=[1+iRe(y-c)]W$. The displayed substitution transforms it into the Airy ordinary differential equation, because $(e^{i\pi /6}{)}^3=i$. The scale $R{e}^{-1/3}$ balances shear advection against viscous diffusion.

An Orr-Sommerfeld mode has nonzero wall-normal velocity satisfying the Orr-Sommerfeld equation. Its wall-normal vorticity satisfies the accompanying forced Squire equation. The triangular coupling separates the velocity eigenproblem from the vorticity forcing; the vorticity need not vanish in three dimensions.

Write the perturbation Stokes streamfunction as $\psi =\varphi (r)e^{ikz-i\omega t}$, with $\omega =kc$ and nonzero $k$. Its radial and axial velocity amplitudes are $u=-ik\varphi /r$ and $w={\varphi }^{\prime }/r$. These automatically obey incompressibility. Linearizing the radial and axial Euler equations, with pressure amplitude $P$, givesThusDifferentiating the first expression, the terms involving $U^{\prime }{\varphi }^{\prime }$ cancel. Equating the two expressions for $P^{\prime }$ proves the axisymmetric inviscid pipe stability equation:At the axis, finite smooth axisymmetric velocity requires $\varphi =O(r^2)$, after choosing the irrelevant constant in the streamfunction. At the wall, impermeability gives $\varphi (a)=0$. There is no tangential no-slip condition in this inviscid problem. A decoupled azimuthal disturbance satisfies $ik(U-c)u_{\theta }=0$ and does not introduce a growing discrete mode when $U\ne c$.

For real $k$, suppose $c=c_r+i{c}_i$ with $c_i\ne 0$, so division by $U-c$ is legitimate. Multiply the ordinary differential equation by ${\varphi }^{\ast }/[r(U-c)]$, integrate, and apply integration by parts. The endpoint terms vanish by the preceding regularity and wall conditions, leavingTaking its imaginary part yieldsA growing temporal mode therefore requires a sign change of $(U^{\prime }/r{)}^{\prime }$, rather than a sign change of $U^{\prime \prime }$ alone. This is the cylindrical counterpart of Rayleigh's inflection-point theorem. If this derivative vanishes identically, the positive first integral rules out a nontrivial growing real-wavenumber mode as well. Isolated zeros without a sign change do not make the second integral vanish for a nonzero regular mode.

For the parabolic Poiseuille flow, $U^{\prime }/r=-2U_0/a^2$ is constant. Under the specified exclusion of critical levels, division givesSetting $\varphi =rf$, or equivalently choosing $n=1$ in $r^{-n}\varphi$, turns this intoIt is the Modified Bessel differential equation of order one. Hence $\varphi =r[A{I}_1(kr)+B{K}_1(kr)]$. Since $r{K}_1(kr)$ tends to a nonzero constant at the axis, it gives a singular radial velocity and must be excluded. The admissible mode shape is thereforeFor $k=0$, direct solution gives $\varphi =C{r}^2$ after axis regularity, and the wall condition again forces the trivial solution. For nonzero $k$, the nonzero roots of $I_1$ are imaginary. Writing $j_{1,l}>0$ for the zeros of the ordinary Bessel function $J_1$, the evanescent potential modes of inviscid Poiseuille flow areTheir perturbation vorticity vanishes: ${\partial }_{z}u-{\partial }_{r}w=k^2\varphi /r-({\varphi }^{\prime }/r{)}^{\prime }=0$. These modes grow exponentially in one axial direction and decay in the other. They can describe end-forced or spatially localized potential fields, but neither sign gives a bounded normal mode on a whole infinite pipe. Moreover, the reduced radial equation contains no $c$: it supplies no temporal eigenvalue condition. An arbitrary imaginary part assigned to $\omega =kc$ here is consequently not evidence for a physical temporal instability.

The real-wavenumber calculation does exclude growing discrete axisymmetric modes of this inviscid profile. It does not establish every form of stability: singular neutral disturbances with $c$ in the range of $U$, transient dynamics, non-axisymmetric disturbances, and finite-amplitude transition require separate treatment. The exclusion $U\ne c$ specifically removes the inviscid continuous spectrum.

With viscosity, linearize the Navier-Stokes equations instead and eliminate pressure to obtain a fourth-order cylindrical counterpart of the Orr-Sommerfeld equation. Impose both $\varphi (a)=0$ and ${\varphi }^{\prime }(a)=0$, with appropriate regularity at the axis. The resulting eigenvalue problem involves the Reynolds number and determines complex frequencies, rather than leaving $c$ arbitrary. Viscosity supplies diffusion, no-slip boundary layers and critical-layer regularization, so the inviscid sign argument alone no longer decides its spectrum. That observation does not imply that viscous pipe flow must have a growing linear eigenmode; finite-amplitude pipe transition is a different question.

For $k>0$, put $q=\sqrt{k^2+l^2}$ and define the Squire transformationThe two-dimensional Orr-Sommerfeld equation then has the same transverse operator because $k^{\prime 2}=q^2$, and the same viscous coefficient because $k^{\prime }{R}^{\prime }=kR$. Its boundary conditions on $v$ and $v^{\prime }$ are unchanged. Thus it has the same eigenfunction and the same phase-speed eigenvalue.

If the original disturbance grows, $\mathrm{Im}c>0$, so the transformed disturbance also grows. Since $R^{\prime }\le R$, every unstable three-dimensional mode maps to an unstable two-dimensional mode at a no larger Reynolds number. It follows that the minimum critical Reynolds number is attained among two-dimensional disturbances. This is the Squire theorem for temporal normal-mode onset, not a statement that three-dimensional transient amplification is impossible. The case $k=0$ is treated separately without dividing by $k$; its streamwise-uniform viscous normal modes decay, although coupling can produce transient growth.

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.

Choose a length $\ell$, velocity $U_s$, time $\ell /U_s$, and pressure $\rho U_s^2$, giving Reynolds number $Re=U_s\ell /\nu$. Write perturbation velocity as $(u,v,w)$ and let $\mathcal{L}={\partial }_t+U{\partial }_x-R{e}^{-1}{\nabla }^2$. Linearizing the incompressible Navier-Stokes equations givesTaking the divergence gives ${\nabla }^{2}p=-2U^{\prime }{v}_x$. Applying ${\nabla }^2$ to the wall-normal momentum equation and usingthen cancels the pressure derivatives and yields $\mathcal{L}{\nabla }^{2}v-U^{\prime \prime }{v}_x=0$. Define wall-normal vorticity $\eta =u_z-w_x$. Applying ${\partial }_z$ to the streamwise equation minus ${\partial }_x$ to the spanwise equation gives $\mathcal{L}\eta =-U^{\prime }{v}_z$.

For the assumed normal modes, set $D=d/dy$ and ${\kappa }^2={\alpha }^2+{\beta }^2$. The two equations become the Orr-Sommerfeld equation and Squire equation:At rigid no-slip walls the conditions are $\hat{v}=D\hat{v}=\hat{\eta }=0$ for nonzero horizontal wavenumber.

An Orr-Sommerfeld mode has $\hat{v}\ne 0$ satisfying the Orr-Sommerfeld equation, with accompanying vorticity satisfying the forced Squire equation. A Squire mode has $\hat{v}=0$ and $\hat{\eta }\ne 0$, so its Squire equation is homogeneous.

For a Squire mode multiply that homogeneous equation by $\overline{\eta }$ and integrate over the channel. Under the no-slip condition $\eta =0$ at the walls, integration by parts givesSince $U,\alpha$ are real, taking the real part provesThe numerator is strictly positive for a nontrivial finite-channel mode, and $Re>0$. Thus Squire modes are exponentially damped in the convention $e^{-i\omega t}$. This does not rule out transient growth from non-normal modes in the coupled system.

With $U=U_0$ and $\beta =0$, the Orr-Sommerfeld equation factors into constant-coefficient operators:For distinct nonzero characteristic roots, the general solution isEither square-root choice for $\gamma$ gives the same solution space. The no-slip boundary condition and impermeability impose $\hat{v}=D\hat{v}=0$ at each wall.

The displayed exponential basis needs the usual repeated-root qualification. If ${\gamma }^2={\alpha }^2\ne 0$, replace it by $(a_1+a_{2}y)e^{\alpha y}+(a_3+a_{4}y)e^{-\alpha y}$. If just one root pair is zero, that pair contributes $1,y$; if both pairs are zero, the basis is $1,y,y^2,y^3$. These are the complete limiting cases, not four independent copies of a repeated exponential.

Put $W=(D^2-1)\hat{v}$ and ${\delta }^3=R{e}^{-1}$. The constant-shear Orr-Sommerfeld equation givesLet $a=e^{i\pi /6}$ and $Z=a(y-c-i{\delta }^3)/\delta$. Since $d/dy=(a/\delta )d/dZ$ and $a^3=i$, the equation becomesThis is the Airy ordinary differential equation, yielding the Airy reduction of the Orr-Sommerfeld equation in constant shear. It is distinct from the evolution PDE represented by the existing Airy equation article.

Take $k^{\ast }>0$ and positive shear scale $S^{\ast }$, with kinematic viscosity ${\nu }^{\ast }$. ChooseThe velocity scale is $S^{\ast }/k^{\ast }$. A reversed shear can be treated by reversing the corresponding coordinate orientation and mode convention. The dimensional Orr-Sommerfeld equation, with $U^{\ast \prime \prime }=0$, isSubstitution of the scales and cancellation gives

The Squire equation governs wall-normal vorticity $\eta ={\partial }_{z}u-{\partial }_{x}w$ in the same normal mode convention as the Orr-Sommerfeld equation. It follows by taking ${\partial }_z$ of the streamwise momentum equation minus ${\partial }_x$ of the spanwise momentum equation. At a rigid no-slip wall, $\hat{\eta }=0$.

For positive streamwise wavenumber $k$, replace a three-dimensional Orr-Sommerfeld equation by a two-dimensional one with the displayed parameters. Both the transverse operator and the coefficient $1/(ikR)$ remain unchanged, preserving the eigenfunction and phase-speed eigenvalue. This proves the modal comparison in Squire's theorem.