Arrested Bubble Rise in a Narrow Tube

Abstract

If a long air bubble is placed inside a vertical tube closed at the top it can rise by displacing the fluid above it. However, Bretherton found that if the tube radius, R, is smaller than a critical value $R_{c}=0.918 \; \ell _c$, where $\ell _c=\sqrt{\gamma /\rho g}$ is the capillary length, there is no solution corresponding to steady rise. Experimentally, the bubble rise appears to have stopped altogether. Here we explain this observation by studying the unsteady bubble motion for $R<R_{c}$. We find that the minimum spacing between the bubble and the tube goes to zero in limit of large t like $t^{-4/5}$, leading to a rapid slow-down of the bubble’s mean speed $U \propto t^{-2}$. As a result, the total bubble rise in infinite time remains very small, giving the appearance of arrested motion.

Appendices

Appendix:1 Asymptotics for F as $\eta \rightarrow 0$

We would like to find the most general solution of (42) consistent with the linear asymptotics (43). To find the structure of the solution in this limit, we put $F = B\eta + \delta$, where $\delta$ is a small correction. Then the leading-order contribution to (42), coming from the first two terms on the left, is linear in $\eta$, while $(F^3)_\eta$ is subdominant as $\eta \rightarrow 0$. Linearizing the remaining term $(F^3F_{\eta \eta \eta })_{\eta }$ in $\delta$, (42) yields:

$$\begin{aligned} -\frac{2B\eta }{5} + \frac{B^3}{3}\left( \eta ^{3} \delta '''\right) ' = 0. \end{aligned}$$

(73)

Here and in the remainder of this appendix, primes denote the derivative with respect to the argument. Integrating (73) we find the general solution

$$\begin{aligned} \delta = -\frac{9}{20 B^{2}}\eta ^{2} + \frac{3\eta ^2}{10B^2}\ln (|\eta |) + \delta _1\ln (|\eta |) + \delta _2 + \delta _3\eta + \delta _4\eta ^2, \end{aligned}$$

(74)

where $\delta _1,\dots ,\delta _4$ are constants of integration. The coefficients $\delta _1$ and $\delta _2$ must vanish, since they would dominate (43), contradicting our assumption, while $\delta _3$ can be included in the constant B. Thus the leading behavior of F as $\eta \rightarrow 0$ is

$$\begin{aligned} F = B\eta -\frac{9}{20 B^{2}}\eta ^{2} + \frac{3\eta ^2}{10B^2}\ln (|\eta |) + P\eta ^2, \end{aligned}$$

with a single free constant P.

The general form of the expansion must contain higher powers of $\ln \eta$ as well, since powers of $\ln \eta$ are generated by the term $F^3$. A closer inspection shows that the most general expansion which balances the number of free coefficients with the number of equations is

$$\begin{aligned} F(\eta ) = \sum _{j=1}^{\infty } \sum _{i=0}^{\max (j-2,1)} a_{ji} \eta ^{j}\left( \log |\eta |\right) ^i, \end{aligned}$$

(75)

where $a_{10} = B$, $a_{11} = 0$, $a_{20}=P-9/(20B^2)$ and $a_{21}=3/(10B^2)$ as shown above. The coefficients $a_{ji}$ are now found recursively as follows. Substituting (75) into (42), at each order $j\ge 3$ we obtain terms of the form

$$\begin{aligned} \eta ^j\left( \log |\eta |\right) ^i, \quad i=0,\dots ,j-1, \end{aligned}$$

whose coefficients must equal zero, and which constitute a linear system of equations for $a_{j+1,0},\dots ,a_{j+1,j-1}$. Starting with $i=j-1$, this system can be solved recursively for $a_{j+1,j-1}$, proceeding down to $i=0$, which is an equation for $a_{j+1,0}$. Implementing this scheme in MAPLE, we find

$$\begin{aligned}&a_{30} = -\frac{4P}{15B^{3}} + \frac{19}{75 B^{5}} + \frac{1}{6}, \quad a_{31} = -\frac{2}{25B^5}, \nonumber \\&a_{40} =\frac{P^2}{8 B^4} - \frac{1}{80B^3}-\frac{131 P}{800B^6} + \frac{1507}{32000 B^8},\nonumber \\&a_{41} = \frac{3}{8000}\frac{200 B^2 P - 131}{B^8}, \quad a_{42} = \frac{9}{800 B^8}, \quad \dots . \end{aligned}$$

(76)

Appendix: 2 Asymptotics for F as $\eta \rightarrow -\infty$

Here we seek a description of solutions of (42) consistent with the asymptotics (47) as $\eta \rightarrow -\infty$, so we put

$$\begin{aligned} F=\sqrt{-\eta }+\delta \end{aligned}$$

into (42) and linearize in $\delta$. Contributions coming from surface tension are subdominant at this order, and we obtain to leading order

$$\begin{aligned} \frac{2}{5}\delta + \frac{6}{5}\eta \delta ' = \frac{1}{8\eta ^{2}}, \end{aligned}$$

(77)

where the right-hand side comes from gravity. This has the general solution

$$\begin{aligned} \delta = \frac{Q}{(-\eta )^{1/3}} - \frac{1}{16\eta ^{2}}, \end{aligned}$$

(78)

where Q is a constant of integration.

This suggests a general solution of the form

$$\begin{aligned} F = \sum _{i=0}^{\infty } c_i|\eta |^{1/2-5i/6}, \end{aligned}$$

(79)

where all coefficients $c_i$ can be determined uniquely, and contributions from surface tension come in at higher orders. The first few are,

$$\begin{aligned} c_1 = 1, \quad c_2 = Q, \quad c_3 = -Q^2/6, \quad c_4 = -1/16, \dots , \end{aligned}$$

(80)

determined by a single free constant Q.

Next we wish to find all solutions F to (11), which asymptote to $\bar{F}$ as $\eta \rightarrow -\infty$, by performing a linear stability analysis around $\bar{F}$, using the ansatz:

$$\begin{aligned} F= \bar{F} + \delta (\eta ). \end{aligned}$$

(81)

We are interested in all possible $\delta (\eta )$, which grow exponentially as $\eta \rightarrow - \infty$. Inserting (81) into (42) and linearizing, we find

$$\begin{aligned} -\frac{3}{5}\delta +\frac{1}{5}\eta \delta ' +\big (\bar{F}^{2} \delta \bar{F}''' \big )' + \frac{1}{3}\big (\bar{F}^{3}\delta '''\big )' -(\bar{F}^{2} \delta )' = 0 . \end{aligned}$$

(82)

Since the coefficients depend on $\eta$, we use the WKB ansatz $\delta (\eta ) = e^{\lambda (-\eta )^{n}}$. For $n>0$, the leading order balance as $\eta \rightarrow -\infty$ is

$$\begin{aligned} -\frac{6}{5}(-\eta )\delta ' + \frac{1}{3}(-\eta )^{3/2}\delta ''''\approx 0, \end{aligned}$$

(83)

which requires $n = 5/6$ for the powers to balance, and we obtain the 4th order eigenvalue equation:

$$\begin{aligned} \lambda \left[ 1 +\frac{1}{3}\Bigl (\frac{5}{6}\Big )^{4}\lambda ^{3}\right] = 0. \end{aligned}$$

(84)

We are interested in roots with positive real part only, since negative roots decay exponentially as we integrate toward $-\infty$ and $\lambda = 0$ corresponds to the solution itself. There are two such eigenvectors with positive roots that solve (84):

$$\begin{aligned} \lambda _{3,4} = \frac{3}{25} 450^{1/3} \pm i \frac{3\sqrt{3}}{25} 450^{1/3}, \end{aligned}$$

(85)

leading to unstable solutions of the form

$$\begin{aligned} F = \bar{F} + \exp \bigl (\mathcal {A}\eta ^{5/6}\bigr ) \left[ \epsilon _{1}\sin \left( \sqrt{3} \mathcal {A}\eta ^{5/6}\right) + \epsilon _2\cos \left( \sqrt{3}\mathcal {A} \eta ^{5/6}\right) \right] , \end{aligned}$$

(86)

where $\mathcal {A} = \frac{3}{25} 450^{1/3}$.