= Solution
Take $z$ positive upward, and write $b=\widetilde\rho g$. For a straight steady <viscous buoyant conduit>, the inner and outer axial equations are
$$
\lambda\mu\frac1r(rw_i')'=P_i'-b,\qquad
\mu\frac1r(rw_o')'=P_o'.
$$
Regularity gives $w_i'(0)=0$, the wall has $w_o(R)=0$, and <velocity> and tangential traction are continuous at $r=a_0$: $w_i=w_o$, $\lambda\mu w_i'=\mu w_o'$. With no interfacial tension specified, normal traction is continuous; for this straight steady flow that sets $P_i=P_o$ at the interface. Thus the modified gradients are common in this case.
When their value is zero, integration gives
$$
w_o(r)=\frac{ba_0^2}{2\mu}\ln(R/r),\qquad
w_i(r)=w_o(a_0)+\frac b{4\lambda\mu}(a_0^2-r^2).
$$
Consequently
$$
\boxed{\frac{w_i(0)}{w_i(a_0)}=1+\frac1{2\lambda\ln(R/a_0)}.}
$$
The large center-to-interface speed ratio explains why the inner parabolic component dominates at small <viscosity> ratio.
For a slowly varying radius, set $B=b-P_{i,z}$. The local inner solution has $w_i=w_a+B(a^2-r^2)/(4\lambda\mu)$, so
$$
q=\pi a^2w_a+\frac{\pi a^4B}{8\lambda\mu}.
$$
The shear-driven outer interface speed has scale $a^2B\ln(R/a)/\mu$. The pressure-driven return flow needed for zero total flux adds a <velocity> of order $q/R^2$ and a shear-driven contribution of order $a^2B/\mu$. Relative to the inner parabolic <velocity> these terms are $O(\lambda\ln(R/a))$ and $O((a/R)^2)$, which are small under the stated assumptions. Axial viscous derivatives are also smaller because the plume is slender. Therefore
$$
\boxed{q=\frac{\pi a^4}{8\lambda\mu}(b-P_{i,z})\quad\text{to leading order}.}
$$
For clarity, the outer pressure-gradient estimate must include both return flux and transmitted interface shear. Their flux scales are $R^4P_{o,z}/\mu$, $a^2R^2B/\mu$ and $q$. Zero total flux hence implies
$$
\frac{|P_{o,z}|}{|B|}
=O\left((a/R)^2+\frac{(a/R)^4}{\lambda}\right)\ll1.
$$
This justifies the negligible outer <pressure> gradient. We subsequently choose its negligible <pressure> reference as zero, as in the reduced model.
Near the moving interface the outer <radial source flow> is $u_r=aa_t/r$: the interface's axial advection is smaller because its axial <velocity> is small compared with the inner core speed. Its radial strain is $\partial_ru_r=-aa_t/r^2$. Continuity of normal traction at $r=a$, neglecting the much smaller inner normal viscous stress, gives
$$
-P_i=-P_o-2\mu a_t/a,
\qquad\boxed{P_i-P_o=2\mu a_t/a.}
$$
With $A=a^2/a_0^2$, this is $P_i=\mu A_t/A$ after dropping $P_o$. Conservation of plume volume is $\pi(a^2)_t+q_z=0$.
Let the axial and <velocity> scales be
$$
\ell=\frac{a_0}{\sqrt{8\lambda}},\qquad
U_0=\frac{ba_0^2}{8\lambda\mu},\qquad
\boxed{Z=z/\ell,\quad T=tU_0/\ell.}
$$
Since $\mu U_0/(b\ell^2)=1$, substitution gives the <conduit equation>
$$
\boxed{A_T+\partial_Z\left\{A^2\left[1-\partial_Z(A_T/A)\right]\right\}=0.}
$$
The buoyant flux and the viscous normal-stress <pressure> are both retained at this scaling.
Linearizing about unit area gives $a_T+2a_Z-a_{TZZ}=0$. Thus
$$
\boxed{\omega=\frac{2k}{1+k^2},\qquad\omega/k=\frac2{1+k^2}>0.}
$$
Wave crests propagate upward. The <group velocity> is $2(1-k^2)/(1+k^2)^2$: it is downward for $|k|>1$, so the phase direction is not the direction of every wave packet.
For a <travelling wave> $f(\zeta)$, integrating once and using its uniform far field gives
$$
-cf+f^2+c(ff''-f'^2)=f_0^2-cf_0=:K.
$$
Divide by $f^2$, multiply by $f'/f$ and integrate again. A convenient potential, defined up to an additive constant, is
$$
\boxed{\frac c2\frac{f'^2}{f^2}+V(f)=V(f_0),\qquad
V(f)=\ln f+\frac c f+\frac{f_0^2-cf_0}{2f^2}.}
$$
At the positive crest $f=\alpha f_0$, the derivative is zero. Subtracting the far-field potential yields
$$
\boxed{c(1-2/\alpha+\alpha^{-2})
=f_0(2\ln\alpha-1+\alpha^{-2}).}
$$
This is the <solitary-wave amplitude-speed relation for the conduit equation>. A nontrivial elevation wave has $\alpha>1$ and $c>2f_0$; as $\alpha\downarrow1$, the speed tends to the long-wave speed $2f_0$. Positive area is required throughout the derivation.
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