= Solution
When <pressure> is uniform along the bubble, its normal-stress relation is $2\mu a_t=Pa-\gamma$. Use the PDF's amplitude convention $a=f+\sqrt2\,g\sin kz$; the square root acts on $2$ only, not on $2g$. Comparing constant and sinusoidal terms gives
$$
2\mu\dot f=Pf-\gamma,\qquad 2\mu\dot g=Pg.
$$
The mean cross-sectional area of an incompressible periodic segment is proportional to $\langle a^2\rangle=f^2+g^2$ and is conserved. Thus
$$
\boxed{\alpha_0=f(0)^2+g(0)^2=f(t)^2+g(t)^2.}
$$
Its time derivative vanishes, and substitution of the preceding two equations gives $P\alpha_0=\gamma f$. Therefore the <uniform-pressure evolution of a slender viscous bubble> is
$$
\boxed{P=\frac{\gamma f}{\alpha_0},\qquad\dot f=-\frac{\gamma g^2}{2\mu\alpha_0},\qquad\dot g=\frac{\gamma fg}{2\mu\alpha_0}.}
$$
For a small disturbance to radius $a_0$, $f=a_0+O(g^2)$ and $\alpha_0=a_0^2+O(g^2)$, so its amplitude grows as $g\propto e^{st}$ with
$$
\boxed{s=\frac\gamma{2\mu a_0}.}
$$
The leading long-wave growth rate is independent of $k$ because axial curvature has been neglected; this is not a finite-wavelength dispersion relation.
While $f>\sqrt2g>0$, the minimum radius is $a_{\min}=f-\sqrt2g$. Its derivative is
$$
\boxed{\dot a_{\min}=-\frac\gamma{2\mu\alpha_0}(g^2+\sqrt2fg)<0.}
$$
Thus the minimum keeps decreasing in the nonlinear uniform-<pressure> model, with no need to solve $f,g$ explicitly. Finite internal viscosity eventually invalidates this model close enough to the neck.
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