When linearsing and expanding in normal modes the Navier-Stokes equations in cylindrical coordinates, two coupled modified Bessel equations are obtained for the radial and the azimuthal velocity,
$\frac{\mathrm{d}^2 u}{\mathrm{d} r^2} + \frac{1}{r} \frac{\mathrm{d} u}{\mathrm{d} r} - \left(\tilde{k}^2 + \frac{m^2 +1}{r^2} \right)u = \frac{2 i m}{r^2} v + A \, k \, I_{m-1}(k r) - A \, \frac{m \, I_m(k r)}{r}$,
$\frac{\mathrm{d}^2 v}{\mathrm{d} r^2} + \frac{1}{r} \frac{\mathrm{d} v}{\mathrm{d} r} - \left(\tilde{k}^2 + \frac{m^2 +1}{r^2} \right)v = - \frac{2 i m}{r^2} u + A \, i \, m \, I_m (kr)$,
where $u$ and $v$ are the radial and the azimuthal velocities respectively, $k$ is the axial wavenumber and $\tilde{k}$ is another axial wavenumber modified by the viscosity, $A$ is a constant, $m \in \mathbb{Z}$ is the azimuthal number and $i$ is the imaginary unit.
Since the $u$ and $v$ have to be bounded, I suppose that the solution is a linear combination of first order modified bessel functions ($K_m(x)$ diverges at $x = 0$), as for the axisymmetric case ($m = 0$), although I do not see an easy way to solve both equations since they are coupled. Any ideas?
Thanks in advance, Alex
Working with the pairs $z^\pm=\pm iu+ v$ should help solving the problem: denoting $L[.]$ the L.H.S Bessel differential operator \begin{align} L[z^+]&=-\frac{2m}{r^2}z^++iA\left(kI_{m-1}(kr)-\frac{mI_m(kr)}{r}+kI_m(kr)\right)\\ L[z^-]&=\frac{2m}{r^2}z^--iA\left(kI_{m-1}(kr)-\frac{mI_m(kr)}{r}-kI_m(kr)\right) \end{align} Then, we have to solve (if I am not wrong) \begin{equation} \frac{d^2z^{\pm}}{dx^2}+\frac{1}{x}\frac{dz^\pm}{dx}-\left( 1+\frac{(m\mp 1)^2}{x^2} \right)z^\pm=\pm iA\frac{k}{\bar{k}^2}\left[I'_m(ax)\pm I_m(ax)\right] \end{equation} where $x=\bar{k}r$ and $a=\tfrac{k}{\bar{k}}$.