= Solution
Let $s=a_{\min}$ be a small neck radius in the preceding sinusoidal shape. Expanding around a minimum gives $a\simeq s+(gk^2/\sqrt2)(z-z_*)^2$, so its local axial length is $\ell\sim[s/(gk^2)]^{1/2}$. The uniform-<pressure> model has $|a_t|\sim\gamma/\mu$ near the neck. Local <conservation of mass> therefore demands axial inner flux $Q\sim s(\gamma/\mu)\ell$. Internal <Hagen-Poiseuille flow> of viscosity $\lambda\mu$ would require a <pressure> difference
$$
\Delta P\sim\frac{\lambda\mu Q\ell}{s^4}\sim\frac{\lambda\gamma\ell^2}{s^3},\qquad
\frac{\Delta P}{\gamma/s}\sim\frac{\lambda}{gk^2s}.
$$
If $sk\ll\lambda$, while $gk\ll1$ by slenderness, this ratio is very large. Hence \b[internal <pressure> variation certainly cannot be neglected when $a_{\min}k\ll\lambda$]. This is a sufficient breakdown condition; appreciable variation can arise earlier, especially because the radial viscous and capillary terms nearly cancel in the uniform-<pressure> regime.
For the inner flow, the leading axial <pressure> is uniform across a section. Neglect the much smaller interfacial axial <velocity> relative to the internal <pressure>-driven motion. The <lubrication approximation> solves $\lambda\mu r^{-1}(ru_{z,r})_r=P_z$, with regularity at zero and $u_z(a)=0$ at leading order. Thus
$$
u_z=\frac{P_z}{4\lambda\mu}(r^2-a^2),\qquad Q=2\pi\int_0^a ru_z\,dr=-\frac{\pi a^4}{8\lambda\mu}P_z.
$$
The area balance $(\pi a^2)_t+Q_z=0$, combined with the exterior normal-stress relation, gives the <lubrication equation for a bubble with viscous exterior>:
$$
\boxed{2aa_t=\frac1{8\lambda\mu}\partial_z\left[a^4\partial_z\left(\frac{2\mu a_t}a+\frac\gamma a\right)\right].}
$$
The neglected inner viscous <normal stress> is smaller by $\lambda$ in the similarity scales obtained next; the external axial interfacial speed is likewise smaller than the large <pressure>-driven inner speed.
Put $\tau=t_*-t$, $a\sim\tau^p$ and axial length $\ell\sim\tau^q$. The radial viscous <normal stress> scales as $\mu a_t/a\sim\tau^{-1}$ and capillary <pressure> as $\gamma/a\sim\tau^{-p}$. Retaining both requires $p=1$. The left side $aa_t$ scales as $\tau^{2p-1}$, while the axial-flux divergence scales as $a^4P/\ell^2\sim\tau^{4p-1-2q}$. Equating them gives $q=p=1$. Thus \b[both local radius and axial length shrink linearly in the remaining time].
To obtain the specified dimensionless coefficients, choose the <linear pinch-off scaling of a slender viscous bubble>
$$
\boxed{a(z,t)=\frac\gamma\mu\tau A(\zeta),\qquad\zeta=\frac{\mu\sqrt\lambda(z-z_*)}{\gamma\tau}.}
$$
Then $a_t=(\gamma/\mu)(\zeta A'-A)$ and
$$
P=\frac\mu\tau\left[\frac{2\zeta A'+1}{A}-2\right].
$$
The constant $-2$ has no axial derivative. Substituting into the flux equation and cancelling its dimensional factor gives
$$
\boxed{\zeta AA'-A^2=\frac1{16}\left\{A^4\left[\frac{2\zeta A'+1}{A}\right]'\right\}'.}
$$
Both axial derivatives on the right are essential; the outer derivative is present in the PDF but lost in the converted TeX expression. The physical slope is $a_z=\sqrt\lambda A'$, so an order-one similarity profile remains slender for $\lambda\ll1$. No solution of this similarity ordinary differential equation is required.
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