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 Incoming links: Uniform-pressure evolution of a slender viscous bubble

When pressure is uniform along the bubble, its normal-stress relation is

2 μ a_{t} = P a - γ

. Use the PDF's amplitude convention

a = f + 2 g sin k z

; the square root acts on

2

only, not on

2 g

. Comparing constant and sinusoidal terms gives

2 μ \dot{f} = P f - γ, 2 μ \overset{g}{˙} = P g .

The mean cross-sectional area of an incompressible periodic segment is proportional to

⟨ a^{2} ⟩ = f^{2} + g^{2}

and is conserved. Thus

α_{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 α_{0} = γ f

P = \frac{γ f}{α _{0}}, \dot{f} = - \frac{γ g ^{2}}{2 μ α _{0}}, \overset{g}{˙} = \frac{γ f g}{2 μ α _{0}} .

For a small disturbance to radius

a_{0}

f = a_{0} + O (g^{2})

and

α_{0} = a_{0}^{2} + O (g^{2})

, so its amplitude grows as

g \propto e^{s t}

with

s = \frac{γ}{2 μ 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 > 2 g > 0

, the minimum radius is

a_{m i n} = f - 2 g

\overset{a}{˙}_{m i n} = - \frac{γ}{2 μ α _{0}} (g^{2} + 2 f g) < 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.