= Solution
Write the perturbation <Stokes streamfunction> as $\psi=\phi(r)e^{ikz-i\omega t}$, with $\omega=kc$ and nonzero $k$. Its radial and axial <velocity> <amplitudes> are $u=-ik\phi/r$ and $w=\phi'/r$. These automatically obey <incompressibility>. Linearizing the radial and axial <Euler equations>, with <pressure> <amplitude> $P$, gives
$$
ik(U-c)u=-P'/\rho_0,\qquad ik(U-c)w+U'u=-ikP/\rho_0.
$$
Thus
$$
P/\rho_0=-(U-c)\phi'/r+U'\phi/r,\qquad P'/\rho_0=-k^2(U-c)\phi/r.
$$
Differentiating the first expression, the terms involving $U'\phi'$ cancel. Equating the two expressions for $P'$ proves the <axisymmetric inviscid pipe stability equation>:
$$
\boxed{(U-c)\left[r\left(\frac{\phi'}r\right)'-k^2\phi\right]-r\left(\frac{U'}r\right)'\phi=0}.
$$
At the axis, finite smooth axisymmetric <velocity> requires $\phi=O(r^2)$, after choosing the irrelevant constant in the <streamfunction>. At the wall, impermeability gives $\phi(a)=0$. 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 $c=c_r+ic_i$ with $c_i\ne0$, so division by $U-c$ is legitimate. Multiply the <ordinary differential equation> by $\phi^*/[r(U-c)]$, integrate, and apply <integration by parts>. The endpoint terms vanish by the preceding regularity and wall conditions, leaving
$$
\int_0^a\frac{|\phi'|^2+k^2|\phi|^2}{r}\,dr+
\int_0^a\frac{(U'/r)'|\phi|^2}{U-c}\,dr=0.
$$
Taking its imaginary part yields
$$
c_i\int_0^a\frac{(U'/r)'|\phi|^2}{|U-c|^2}\,dr=0.
$$
\b[A growing temporal mode therefore requires a sign change of $(U'/r)'$, rather than a sign change of $U''$ 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>, $U'/r=-2U_0/a^2$ is constant. Under the specified exclusion of critical levels, division gives
$$
\phi''-\frac{\phi'}r-k^2\phi=0.
$$
Setting $\phi=rf$, or equivalently choosing $n=1$ in $r^{-n}\phi$, turns this into
$$
r^2f''+rf'-(k^2r^2+1)f=0.
$$
It is the <Modified Bessel differential equation> of order one. Hence $\phi=r[A I_1(kr)+B K_1(kr)]$. Since $rK_1(kr)$ tends to a nonzero constant at the axis, it gives a singular radial <velocity> and must be excluded. The admissible <mode shape> is therefore
$$
\boxed{\phi=A rI_1(kr),\qquad I_1(ka)=0}.
$$
For $k=0$, direct solution gives $\phi=C r^2$ 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> are
$$
\boxed{k=\pm i j_{1,l}/a,\qquad \phi\propto rJ_1(j_{1,l}r/a)}.
$$
Their perturbation <vorticity> vanishes: $\partial_z u-\partial_r w=k^2\phi/r-(\phi'/r)'=0$. 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 $\phi(a)=0$ and $\phi'(a)=0$, 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.
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