= Solution
Let $L$ be a core-scale length, $B_*$ a representative strength of the prescribed <poloidal magnetic field>, and $v_A=B_*/\sqrt{\mu_0\rho}$ the <Alfvén speed>. A consistent weakly damped, small-amplitude ordering is
$$
\frac{v_A}{\Omega L}\ll1,\qquad U\ll v_A,\qquad
\frac{|\mathbf u_p|}{L}\ll\sigma\sim\frac{v_A}{L},\qquad
\frac{\nu}{L^2},\frac{\eta}{L^2}\ll\sigma.
$$
The first ratio is the <Lehnert number>. It makes the wave slow relative to rotation and allows almost <geostrophic flow>, with the azimuthal velocity approximately constant along a rotation-axis cylinder. The small wave amplitude gives $|B_\phi|\sim\sqrt{\mu_0\rho}|U|\ll B_*$. Poloidal advection then changes neither the background field nor the wave substantially over one wave period. If there are no-slip boundary layers, also require their damping rate $\sqrt{\nu\Omega}/L\ll\sigma$, or otherwise retain their torque; a small bulk viscosity alone does not remove that boundary torque.
In the rotating <magnetohydrodynamic momentum equation>, axisymmetry removes the azimuthal pressure derivative. The neglected nonlinear azimuthal advection has size $|\mathbf u_p|U/L$, and the neglected viscous term has size $\nu U/L^2$. The retained azimuthal <Coriolis force> is $2\Omega u_r$. A typical compatible local balance has $|u_r|\sim\sigma U/\Omega$, small compared with $U$ when the <Lehnert number> is small. The azimuthal <Lorentz force density> can be written using $\nabla\cdot\mathbf B_p=0$:
$$
\frac{1}{\mu_0\rho}\left(\mathbf B_p\cdot\nabla B_\phi+\frac{B_rB_\phi}{r}\right)
=\frac{1}{\mu_0\rho r}\nabla\cdot(r\mathbf B_pB_\phi).
$$
The physical prefactor is $1/(\mu_0\rho)$: the extra $i$ printed inside the first reduced momentum equation is a typographical error, also contradicted by the subsequent real-valued system.
In the <resistive induction equation>, neglecting poloidal advection of the perturbation and magnetic diffusion leaves the winding of the prescribed <poloidal magnetic field> by differential rotation. Thus
$$
U_t+2\Omega u_r=\frac{1}{\mu_0\rho r}\nabla\cdot(r\mathbf B_pB_\phi),\qquad
(B_\phi)_t=r\mathbf B_p\cdot\nabla(U/r).
$$
This is the small-amplitude <torsional Alfvén wave> approximation, with a time-independent background and no external azimuthal forcing.
We now average over the full height of a rotation-axis cylinder. The sphere is impenetrable. Integrating <incompressible flow> over the volume inside any such cylinder gives zero flux through its cylindrical side, since there is no flux through its spherical caps. Therefore
$$
\int_{-z_b}^{z_b}u_r\,dz=0.
$$
This is why the <Coriolis force> disappears from the cylinder average; it need not vanish locally. An axisymmetric field in the insulating exterior is a potential field with $B_\phi=0$, so continuity gives $B_\phi=0$ on the sphere. Consequently both the end flux terms and the moving-endpoint terms in the magnetic-stress integral vanish. Expanding the divergence gives
$$
\frac1r\nabla\cdot(r\mathbf B_pB_\phi)
=\frac1{r^2}\partial_r(r^2B_rB_\phi)+\partial_z(B_zB_\phi).
$$
Since $U$ is independent of $z$, its integrated time derivative is $2z_bU_t$. Defining $T=\int_{-z_b}^{z_b}B_rB_\phi\,dz$ gives
$$
\boxed{U_t=\frac{1}{2z_b\mu_0\rho r^2}\partial_r(r^2T).}
$$
The stationary background and the fact that $U/r$ depends only on $r,t$ also give
$$
T_t=\int_{-z_b}^{z_b}B_r(B_\phi)_t\,dz
=r\partial_r(U/r)\int_{-z_b}^{z_b}B_r^2\,dz,
$$
so
$$
\boxed{T_t=H(r)r\partial_r(U/r),\qquad H(r)=\int_{-z_b}^{z_b}B_r^2\,dz.}
$$
Together these form the <cylinder-averaged torsional Alfvén wave> system.
Assume $H>0$ on the cylinders carrying the wave, with regular endpoint limits. If $H$ vanishes on an interior cylinder, then $B_r$ and $T$ vanish there and that cylinder has no radial magnetic coupling; the displayed $H^{-1}$ energy term must be interpreted on the coupled region rather than literally dividing by zero. Since $H$ is time independent, differentiation of the proposed <conserved energy of a cylinder-averaged torsional wave> gives
$$
\begin{aligned}
\frac{dE}{dt}
&=\int_0^a\left(2z_b\mu_0\rho U U_t+\frac{T}{H}T_t\right)r\,dr\\
&=\int_0^a\left[\frac Ur\partial_r(r^2T)+r^2T\partial_r(U/r)\right]dr\\
&=[rUT]_0^a=0.
\end{aligned}
$$
Regularity at the rotation axis makes the lower endpoint vanish, while vanishing magnetic torque at the collapsed outer cylinder makes the upper endpoint vanish. This proves
$$
\boxed{E=\frac12\int_0^a\left(2z_b\mu_0\rho U^2+\frac{T^2}{H}\right)r\,dr\quad\text{is conserved}.}
$$
The magnetic term has a direct interpretation. Decompose $B_\phi=(T/H)B_r+B_\phi^\perp$, with $\int B_rB_\phi^\perp\,dz=0$. Then $\int B_\phi^2\,dz=T^2/H+\int(B_\phi^\perp)^2\,dz$, and the induction equation gives $(B_\phi^\perp)_t=0$. Thus, apart from a common dimensional factor and a stationary orthogonal magnetic component, the conserved quantity is the physical kinetic plus <magnetic energy>.
For the frequency estimate, put $V=U/r$ and eliminate $T$:
$$
V_{tt}=\frac{1}{2z_b\mu_0\rho r^3}\partial_r(r^3H V_r).
$$
Locally this is a <wave equation> with speed
$$
\boxed{c_T(r)=\sqrt{\frac{H(r)}{2z_b(r)\mu_0\rho}}=\frac{\sqrt{\langle B_r^2\rangle_z}}{\sqrt{\mu_0\rho}}.}
$$
It is the cylinder-averaged radial field component that matters, not simply the dipole measured at the Earth's surface. Taking a cylinder height of order $6\times10^6\,\mathrm m$, $\rho\sim10^4\,\mathrm{kg\,m^{-3}}$, and an assumed internal cylindrical radial field $B_{r,\mathrm{rms}}\sim0.2$--$2\,\mathrm{mT}$ gives
$$
H\sim0.2\text{--}20\,\mathrm{T^2\,m},\qquad c_T\sim2\times10^{-3}\text{--}2\times10^{-2}\,\mathrm{m\,s^{-1}}.
$$
For a standing wave of radial length $L_c\sim2$--$3.5\times10^6\,\mathrm m$, the cyclic frequency is of order $c_T/(2L_c)$ and the angular frequency is of order $\pi c_T/L_c$. Hence
$$
\boxed{\nu_T\sim10^{-10}\text{--}10^{-9}\,\mathrm{Hz},\qquad\text{periods of order years to a century}.}
$$
Geometry and the assumed internal field can shift the estimate substantially. As a concrete calibration, https://www.nature.com/articles/nature09010[the 2010 torsional-wave study] inferred a cylindrical radial rms field near $2\,\mathrm{mT}$ and a roughly four-year crossing time, with an observed six-year recurrence. A much weaker assumed interior field instead gives the older multi-decadal estimate. The paper supplies no numerical interior field strength, so a unique numerical frequency cannot be deduced without such an assumption.
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