= Solution
The radius law integrates to $r(z)=r_0e^{-\beta z}$, so the cross-sectional area is $A(z)=\pi r_0^2e^{-2\beta z}$. Conserving gas volume in a horizontal slice yields
$$
\partial_t(A\phi)+\partial_z\bigl[AV_s\phi(1-\phi)\bigr]=0,
$$
or the <kinematic bubble transport in an exponentially narrowing vessel> equation
$$
\boxed{\phi_t+V_s(1-2\phi)\phi_z=2\beta V_s\phi(1-\phi).}
$$
The geometrical term increases the concentration as a given gas flux enters a smaller area. On concentration <characteristic curves>,
$$
\boxed{\frac{dz}{dt}=V_s(1-2\phi),\qquad
\frac{d\phi}{dt}=2\beta V_s\phi(1-\phi).}
$$
Put $c=2\beta V_s$. A characteristic starting at $(z_0,0)$ with $\phi_i=\phi(z_0,0)$ has
$$
\boxed{\phi(t)=\frac{\phi_i e^{ct}}{1-\phi_i+\phi_i e^{ct}},\qquad
z(t)=z_0+V_st-\frac1\beta\log\left(1-\phi_i+\phi_i e^{ct}\right).}
$$
The first formula follows by integrating $d\phi/[\phi(1-\phi)]=c\,dt$; inserting it into $dz/dt$ gives the second. For initially uniform $\phi_0$, the uninterrupted bubbly region remains spatially uniform, with this time-dependent concentration. These <characteristic curves> carry concentration information; they are not individual bubble trajectories, whose speed is $V_s(1-\phi)$.
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