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When $U^{\prime }/r$ is constant and $U\ne c$, the axisymmetric inviscid pipe stability equation reduces to ${\varphi }^{\prime \prime }-{\varphi }^{\prime }/r-k^2\varphi =0$. Axis regularity selects $r{I}_1(kr)$ and impermeability gives $I_1(ka)=0$. These imaginary-wavenumber disturbances are spatially evanescent potential fields, not bounded whole-pipe temporal normal modes; their radial equation supplies no eigenvalue condition on $c$.

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.