2.1 Base state

We will study stability of the axisymmetric uniform (i.e. with pure radial growth) cylinder base state, which does not depend on the $z$ and $\unicode[STIX]{x1D717}$ coordinates: $\unicode[STIX]{x1D719}=\unicode[STIX]{x1D6F7}(t,r)$ , $p=P(t,r)$ , $f=F(t)$ . It is straightforward to show from the structure of equations (2.1) that this is possible only if there is no flow in the $z$ -direction and $g=0$ . Since we are going to consider the case $Bo=\unicode[STIX]{x1D70C}gF^{2}/\unicode[STIX]{x1D70E}\ll 1$ , then deviation from this base state due to acceleration can be treated as a perturbation (§ 4.1). The Laplace equation (2.1a ) admits a non-trivial solution of the above type only when there is a line source at the axis of symmetry providing a mass flux $Q$ , treated here as constant,

(2.2) $$\begin{eqnarray}\unicode[STIX]{x1D6F7}(t,r)=(Q/2\unicode[STIX]{x03C0})\ln r,\end{eqnarray}$$

producing an axisymmetric flow field; cf. figure 1(b). The logarithmic singularity can be removed by considering a mass source on the cylinder axis of a small radius $\unicode[STIX]{x1D716}>0$ . The time-dependent cylinder radius $F(t)$ is found from the kinematic boundary condition (2.1c ), yielding $F(t)=\sqrt{F_{0}^{2}+Qt/\unicode[STIX]{x03C0}}$ , the same as the rim growth on a retracting soap film in the Taylor–Culick theory (Taylor Reference Taylor1959; Culick Reference Culick1960). We will focus, without loss of generality, on the asymptotic case,

(2.3) $$\begin{eqnarray}F(t)\simeq \sqrt{Qt/\unicode[STIX]{x03C0}},\quad \text{valid for}~t\gg \unicode[STIX]{x03C0}F_{0}^{2}/Q\equiv t_{c}.\end{eqnarray}$$

For a shrinking cylinder, i.e. when $t^{\prime }=t_{c}-t\ll t_{c}$ , $F(t)\simeq \sqrt{Qt^{\prime }/\unicode[STIX]{x03C0}}$ , and since $\unicode[STIX]{x2202}_{t}\rightarrow -\unicode[STIX]{x2202}_{t^{\prime }}$ , the stability results would be the exact opposite of the case considered below. The base state pressure is $P(t,r)=p_{\infty }+\unicode[STIX]{x1D70E}\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{n}=p_{\infty }+\unicode[STIX]{x1D70E}/r$ , so that the constant $C(t)$ determined by evaluating the Cauchy–Lagrange integral (2.1b ) at the cylinder surface $r=F(t)$ becomes $C(t)=\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C}F(t)+(Q/2\unicode[STIX]{x03C0}F)^{2}/2$ . Then, deviations from this base state due to instability development and lateral acceleration can be treated as perturbations in §§ 3 and 4, respectively.

2.2 Perturbation equation

In the analysis to follow, first, we are going to superimpose general perturbations of the form $f(t,z)=F(t)+f^{\prime }(t,\unicode[STIX]{x1D717},z)$ , $\unicode[STIX]{x1D719}(t,r)=\unicode[STIX]{x1D6F7}(t,r)+\unicode[STIX]{x1D719}^{\prime }(t,\unicode[STIX]{x1D717},r)$ , $p(t,r,\unicode[STIX]{x1D717})=P(t,r)+p^{\prime }(t,r,\unicode[STIX]{x1D717})$ . Due to the linearity of inviscid incompressible potential flows, the perturbation of the velocity potential $\unicode[STIX]{x1D719}^{\prime }$ obeys the Laplace equation (2.1a ) as well. Following the same procedure as in the classical case (Drazin & Reid Reference Drazin and Reid2004), modulo the time-dependent boundary $r=F(t)$ , the linearization of the kinematic condition (2.1c ), after projecting onto the undisturbed interface $r=F(t)$ and taking into account that $\unicode[STIX]{x1D6F7}_{r}|_{r=F+f^{\prime }}=\unicode[STIX]{x1D6F7}_{r}|_{F}+\unicode[STIX]{x1D6F7}_{rr}|_{F}\,f^{\prime }\,+\,$ higher-order terms, furnishes

(2.4) $$\begin{eqnarray}r=F(t):\frac{\unicode[STIX]{x2202}f^{\prime }}{\unicode[STIX]{x2202}t}=\left.\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D6F7}}{\unicode[STIX]{x2202}r^{2}}\right|_{F}f^{\prime }+\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}^{\prime }}{\unicode[STIX]{x2202}r}.\end{eqnarray}$$

Similarly, we can linearize the dynamic condition (2.1b ) around $r=F(t)$ ,

(2.5) $$\begin{eqnarray}\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}^{\prime }}{\unicode[STIX]{x2202}t}+\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}}{\unicode[STIX]{x2202}r}\left(\frac{\unicode[STIX]{x2202}^{2}\unicode[STIX]{x1D6F7}}{\unicode[STIX]{x2202}r^{2}}f^{\prime }+\frac{\unicode[STIX]{x2202}\unicode[STIX]{x1D719}^{\prime }}{\unicode[STIX]{x2202}r}\right)=-\frac{1}{\unicode[STIX]{x1D70C}}[p|_{r=F+f^{\prime }}-P|_{r=F}]+g\cos \unicode[STIX]{x1D717}(F(t)+f^{\prime }),\end{eqnarray}$$

where the expression in square brackets simplifies given the pressure perturbation at the interface $p^{\prime }=-\unicode[STIX]{x1D70E}(f_{zz}+f_{\unicode[STIX]{x1D717}\unicode[STIX]{x1D717}}/F^{2}(t))$ related to the perturbed part of the interfacial curvature $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{n}$ ,

(2.6) $$\begin{eqnarray}[\cdots ]=\left[\left.\frac{\unicode[STIX]{x2202}P}{\unicode[STIX]{x2202}r}\right|_{r=F}f^{\prime }+p^{\prime }\right]=-\left.\frac{\unicode[STIX]{x1D70E}}{r^{2}}\right|_{r=F}f^{\prime }+p^{\prime }=-\unicode[STIX]{x1D70E}\left(\frac{f^{\prime }+\unicode[STIX]{x2202}^{2}f^{\prime }/\unicode[STIX]{x2202}\unicode[STIX]{x1D717}^{2}}{F^{2}(t)}+\frac{\unicode[STIX]{x2202}^{2}f^{\prime }}{\unicode[STIX]{x2202}z^{2}}\right).\end{eqnarray}$$

It should be noted that when calculating $\unicode[STIX]{x2202}\unicode[STIX]{x1D6F7}/\unicode[STIX]{x2202}t$ in (2.1b ), we first differentiate (with respect to time  $t$ ) and then evaluate at $r=F(t)$ , not vice versa – these two operations do not commute. The same result (2.5) can be arrived at directly from Euler’s equations of fluid motion.

Taking an infinite Fourier transform in the $z$ -direction and a finite Fourier transform in the $\unicode[STIX]{x1D717}$ -direction, the Laplace equation (2.1a ) reduces to $r^{-1}\unicode[STIX]{x2202}_{r}(r\widehat{\unicode[STIX]{x1D719}}_{r}^{\prime })-(k^{2}+n^{2}/r^{2})\widehat{\unicode[STIX]{x1D719}}^{\prime }=0$ , with the solution expressed in terms of the modified Bessel functions $\widehat{\unicode[STIX]{x1D719}}_{n}^{\prime }=C_{n}I_{n}(kr)+D_{n}K_{n}(kr)$ , where $k$ and $n$ are continuous axial and discrete (i.e. quantized due to boundedness of the $\unicode[STIX]{x1D717}$ domain) azimuthal wavenumbers, respectively. Thereby, the conditions (2.4) and (2.5) become

(2.7) $$\begin{eqnarray}\displaystyle r & = & \displaystyle F(t):\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}t}\left(\begin{array}{@{}c@{}}\widehat{f}_{n}^{\prime }\\ \widehat{\unicode[STIX]{x1D719}}_{n}^{\prime }\end{array}\right)=\left(\begin{array}{@{}cc@{}}\unicode[STIX]{x1D6F7}_{rr} & \displaystyle \frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}r}\\ \displaystyle -\unicode[STIX]{x1D6F7}_{r}\unicode[STIX]{x1D6F7}_{rr}+\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}\left(\frac{1-n^{2}}{F^{2}(t)}-k^{2}\right) & \displaystyle -\unicode[STIX]{x1D6F7}_{r}\frac{\unicode[STIX]{x2202}}{\unicode[STIX]{x2202}r}\end{array}\right)\left(\begin{array}{@{}c@{}}\widehat{f}_{n}^{\prime }\\ \widehat{\unicode[STIX]{x1D719}}_{n}^{\prime }\end{array}\right)\nonumber\\ \displaystyle & & \displaystyle +\,\frac{g}{2}\left(\begin{array}{@{}c@{}}0\\ \widehat{f}_{n-1}^{\prime }+\widehat{f}_{n+1}^{\prime }\end{array}\right)+\frac{gF}{2}\left(\begin{array}{@{}c@{}}0\\ \unicode[STIX]{x1D6FF}_{n,-1}+\unicode[STIX]{x1D6FF}_{n,1}\end{array}\right),\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}_{n,m}$ is the Kronecker delta function. As we can see from the above system, the base state is affected by acceleration $g$ only at the mode $n=1$ . Moreover, since the system (2.7) is symmetric with respect to the transformation $n\rightarrow -n$ , in the subsequent stability analysis we can focus on $n\geqslant 0$ . The system (2.7) is universal in the sense that it allows us to compute both distortion of the base state from a circular cylinder state and the time-dependent perturbation responsible for instability (or stability); in the former case, the amplitude of the distortion is limited by the magnitude $g$ of acceleration, while in the latter case, the amplitude can grow unboundedly.

Modes corresponding to the modified Bessel functions of the first kind $I_{n}$ play the dominant role in the instability, since the $K_{n}$ are singular at $r=0$ and decay towards the cylinder surface, where the instability is driven by the surface tension forces. Hence, we can put $D_{n}=0$ as in the classical case of a time-independent cylinder (Drazin & Reid Reference Drazin and Reid2004). If the line source is substituted (to remove the singularity at $r=0$ ) by a finite-size mass source $r\leqslant \unicode[STIX]{x1D716}$ , then the $K_{n}$ would need to be taken into account, but for $\unicode[STIX]{x1D716}\ll 1$ their contribution should be negligible for the above reasons. With these considerations, we can reduce the system (2.7) to a single second-order equation. From the first equation of (2.7), we determine $C_{n}=[\text{d}\widehat{f}_{n}^{\prime }/\text{d}t-\unicode[STIX]{x1D6F7}_{rr}(F)\widehat{f}_{n}^{\prime }]/kI_{n}^{\prime }(kF)$ , where we divided by $I_{n}^{\prime }(kF)$ since it does not have zeros for $kF\geqslant 0$ , and substituting into the second of equations (2.7) we arrive at a second-order equation for $\widehat{f}_{n}^{\prime }$ ,

(2.8) $$\begin{eqnarray}L\,\widehat{f}_{n}^{\prime }=\frac{\text{d}^{2}\widehat{f}_{n}^{\prime }}{\text{d}t^{2}}+a(t)\frac{\text{d}\widehat{f}_{n}^{\prime }}{\text{d}t}+b(t)\widehat{f}_{n}^{\prime }=\frac{g}{2}(\widehat{f}_{n-1}^{\prime }+\widehat{f}_{n+1}^{\prime })+\frac{gF}{2}(\unicode[STIX]{x1D6FF}_{n,-1}+\unicode[STIX]{x1D6FF}_{n,1}),\end{eqnarray}$$

with the coefficients defined by

(2.9a ) $$\begin{eqnarray}\displaystyle & \displaystyle a(t)=-\unicode[STIX]{x1D6F7}_{rr}(F)-k\frac{I_{n}^{\prime \prime }(kF)}{I_{n}^{\prime }(kF)}\frac{\text{d}F}{\text{d}t}+k\frac{I_{n}^{\prime }(kF)}{I_{n}(kF)}\unicode[STIX]{x1D6F7}_{r}(F), & \displaystyle\end{eqnarray}$$

(2.9b ) $$\begin{eqnarray}\displaystyle & \displaystyle b(t)=-\unicode[STIX]{x1D6F7}_{rrr}(F)\frac{\text{d}F}{\text{d}t}+k\frac{I_{n}^{\prime \prime }(kF)}{I_{n}^{\prime }(kF)}\frac{\text{d}F}{\text{d}t}\unicode[STIX]{x1D6F7}_{rr}(F)-k\frac{I_{n}^{\prime }(kF)}{I_{n}(kF)}\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}\left(\frac{1-n^{2}}{F^{2}}-k^{2}\right). & \displaystyle\end{eqnarray}$$

3.1 Numerical integration of the evolution equation

In order to get an initial insight into the solution of (2.8), let us non-dimensionalize it for the convenience of numerical integration only. Letting $t=t_{0}\widetilde{t}$ with $t_{0}=\sqrt{\unicode[STIX]{x1D70C}F_{0}^{3}/\unicode[STIX]{x1D70E}}$ (surface-tension-driven breakup time for a cylinder of radius $F_{0}$ ), (2.8) reduces to

(3.1a ) $$\begin{eqnarray}\displaystyle & \displaystyle \frac{\text{d}^{2}\widehat{f}_{n}^{\prime }}{\text{d}\widetilde{t}^{2}}+\widetilde{a}(\widetilde{t})\frac{\text{d}\widehat{f}_{n}^{\prime }}{\text{d}\widetilde{t}}+\widetilde{b}(\widetilde{t})\widehat{f}_{n}^{\prime }=0, & \displaystyle\end{eqnarray}$$

where

(3.1b,c ) $$\begin{eqnarray}\widetilde{a}(\widetilde{t})=\frac{\widehat{a}(\unicode[STIX]{x1D705})}{2\widetilde{t}},\quad \widetilde{b}(\widetilde{t})=-\frac{1}{4\widetilde{t}^{2}}[2+\widehat{b}_{2}(\unicode[STIX]{x1D705})]-\widehat{b}_{1}(\unicode[STIX]{x1D705})\left(\frac{t_{\ast }}{t_{0}}\right)^{3/2}\frac{1-n^{2}-\unicode[STIX]{x1D705}^{2}}{\widetilde{t}^{3/2}},\end{eqnarray}$$

with the abbreviations $\widehat{b}_{1}(\unicode[STIX]{x1D705})=\unicode[STIX]{x1D705}I_{n}^{\prime \prime }(\unicode[STIX]{x1D705})/I_{n}^{\prime }(\unicode[STIX]{x1D705})$ , $\widehat{b}_{2}(\unicode[STIX]{x1D705})=\unicode[STIX]{x1D705}I_{n}^{\prime }(\unicode[STIX]{x1D705})/I_{n}(\widehat{k})$ , $\widehat{a}(\unicode[STIX]{x1D705})=1+\widehat{b}_{2}(\unicode[STIX]{x1D705})-\widehat{b}_{1}(\unicode[STIX]{x1D705})$ and the non-dimensional wavenumber $\unicode[STIX]{x1D705}=\widehat{k}\,\widetilde{t}^{\,1/2}$ . Here, $\widehat{k}=\widetilde{k}(t_{0}/t_{\ast })^{1/2}$ , $\widetilde{k}=kF_{0}$ and the inertial time scale $t_{\ast }=\unicode[STIX]{x03C0}F_{0}^{2}/Q\equiv t_{1}$ (which can be interpreted as the time to grow from $F=0$ to $F_{0}$ ). Thus, the solution $\widehat{f}_{n}^{\prime }$ of (3.1a ) can be treated as a function of $\widetilde{t}$ , $n$ and $\widehat{k}$ .

Figure 2. Plots of $\widehat{f}_{n}^{\prime }$ at different wavenumbers $\widehat{k}\in [0,0.1]$ for $n=0$ and $t_{\ast }/t_{0}=1$ : (a) short time interval; (b) longer time interval; (c) ultra-long time interval with ultra-short wavenumbers $\widehat{k}=O(10^{-6})$ ; (d) solution for $\widehat{k}=2\times 10^{-2}$ from figure 2(b) for three values of $t_{\ast }/t_{0}$ .

If one naively applies the stability results for a time-independent cylinder to a time-dependent one, it seems to be intuitive to hypothesize that the most unstable wavenumber should still scale as $k\sim F^{-1}$ , now being time-dependent. Indeed, after analysing equations (3.1) with $\unicode[STIX]{x1D705}=\text{const.}$ , we conclude that for large time, $\widetilde{t}\rightarrow \infty$ , the solution behaves as

(3.2a,b ) $$\begin{eqnarray}\widehat{f}_{0}^{\prime }\sim \widetilde{t}^{-\widehat{a}(\unicode[STIX]{x1D705})/4-(1/8)}\text{e}^{4\sqrt{c}\,\,\widetilde{t}^{\,1/4}},\quad c=\widehat{b}_{2}(\unicode[STIX]{x1D705})(1-\unicode[STIX]{x1D705}^{2})(t_{\ast }/t_{0})^{3/2}.\end{eqnarray}$$

For truly $\widetilde{t}\rightarrow +\infty$ , the growth is dominated by the exponential only, so that the maximum growth rate is achieved when $c(\unicode[STIX]{x1D705})$ is at maximum, which occurs at some finite $\unicode[STIX]{x1D705}=O(1)$ similar to the classical dispersion relation (3.5); cf. also figure 1.5 of Drazin & Reid (Reference Drazin and Reid2004). For finite times, the optimum $\unicode[STIX]{x1D705}$ is affected by $a(\unicode[STIX]{x1D705})$ as well. The problem with the above hypothesis $k\sim F^{-1}$ is that it implicitly requires the Fourier transform with time-dependent wavenumber to commute with the operation of the time derivative, whereas they clearly do not. Hence, it can be surmised that the hypothesis is not viable. In fact, by solving an initial value problem for (3.1a ), we can validate this suspicion. Before that, let us first make a few observations from numerical solution of (3.1), which can be constructed accurately even for long times using the conservative nature of the problem at hand and hence appealing to symplectic integrator methods.

From figure 2(a), the mode entanglement makes it clear that the growth is not monotonic with the wavenumber $k$ and time $t$ . Moreover, what appears to be a ‘winning’ mode (bold) on a short time interval becomes a decaying oscillating mode on a longer time interval; cf. figure 2(b). It also appears, cf. figure 2(c), that there is no short-wavenumber cutoff – the limit $k\rightarrow 0$ indicates that the perturbation starts to grow at a very long time and brings up larger maxima of the solution $\widehat{f}_{0}^{\prime }$ compared with shorter wavelengths. The latter observation suggests that there is no optimal, i.e. most amplified, wavenumber, as the maximum achieved increases with decreasing $k$ . Any wavenumber $k$ , no matter how small, is eventually amplified and then damped; cf. figure 2(c). Finally, as observed from figure 2(d), increase of the inertial over the surface tension time scale brings up the decaying oscillation stage closer in time. Overall, figure 2 suggests that there is no single most amplified wavenumber $\widehat{k}$ , and hence the stability should be interpreted in terms different from those in standard eigenvalue studies. The purpose of the subsequent discussion is to disentangle the observed dynamics.

3.2 Long-wave approximation

To get a better sense of the system behaviour, let us consider a long-wave, $kF\ll 1$ , and long-time, $t\gg t_{c}$ , approximation. Introducing $u_{n}=\widehat{f}_{n}^{\prime }\,\text{exp}(\int a\,\text{d}t/2)$ , we can reduce (2.8) to

(3.3) $$\begin{eqnarray}u_{n}^{\prime \prime }-s^{2}(t)u_{n}=0,\quad \text{where}~s^{2}(t)=-b+{\textstyle \frac{1}{2}}a_{t}+{\textstyle \frac{1}{4}}a^{2}.\end{eqnarray}$$

Depending upon the index $n$ of the Bessel function, we obtain

(3.4a ) $$\begin{eqnarray}\displaystyle & \displaystyle n=0:s^{2}(t)=\frac{1}{4t^{2}}+\frac{k^{2}}{2F}\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}(1-k^{2}F^{2})+\frac{1}{4}\left(\frac{Q}{4\unicode[STIX]{x03C0}}k^{2}\right)^{2}, & \displaystyle\end{eqnarray}$$

(3.4b ) $$\begin{eqnarray}\displaystyle & \displaystyle n\neq 0:s^{2}(t)=\frac{n}{4t^{2}}+\frac{n}{F^{3}}\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}(1-n^{2}-k^{2}F^{2}), & \displaystyle\end{eqnarray}$$

$F=\text{const.}$

(3.5) $$\begin{eqnarray}s^{2}=\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}\frac{1}{F^{3}}\frac{kFI_{n}^{\prime }(kF)}{I_{n}(kF)}(1-n^{2}-k^{2}F^{2}),\end{eqnarray}$$

with the asymptotics in the long-wave approximation,

(3.6a,b ) $$\begin{eqnarray}n=0:s^{2}\simeq \frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}\frac{k^{2}}{2F}(1-k^{2}F^{2}),\quad n\neq 0:s^{2}\simeq \frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}\frac{n}{F^{3}}(1-n^{2}-k^{2}F^{2}),\end{eqnarray}$$

where now $s$ has the meaning of a growth rate (eigenvalue), i.e. $\widehat{f}_{n}^{\prime }\sim \text{e}^{st}$ : for $n=0$ (axisymmetric mode), the growth is observed for $k^{2}F^{2}<1$ , while for $n\geqslant 1$ , all modes are stable (not growing). The dispersion relation (3.6) suggests that the growth rate in the classical case vanishes in the limit of $k=0$ (long waves, for which the surface tension $\unicode[STIX]{x1D70E}$ is ineffective) and at $k=F^{-1}$ (short waves, when the surface tension starts to suppress instability). Clearly, the long-wave asymptotics is accurate enough to determine the optimal mode $n$ and wavelength $k$ ; indeed, expression (3.6) for $n=0$ gives $kF=0.707$ , versus the precise 0.697 from (3.5). These results are recovered from the time-dependent case (3.4a ) if one puts $F=\text{const.}$ , $Q=0$ and $t\rightarrow \infty$ .

In the time-dependent case, when $n=0$ and $t\rightarrow \infty$ , we observe that for $k\rightarrow 0$ , $s^{2}=(k^{2}/2F)(\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C})>0$ , i.e. a long-wave instability is expected; for $k\rightarrow \infty$ , the coefficients in (2.8) become

(3.7a,b ) $$\begin{eqnarray}a(t)=\frac{1}{2t},\quad b(t)=-\frac{1}{2t^{2}}-\frac{k}{4}\frac{Q}{\unicode[STIX]{x03C0}t}\sqrt{\frac{\unicode[STIX]{x03C0}}{Qt}}-\frac{k\unicode[STIX]{x03C0}}{Qt}\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}(1-n^{2}-k^{2}F^{2})\simeq k^{3}\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}},\end{eqnarray}$$

so that $s^{2}(t)=-k^{3}\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C}$ , i.e. the surface tension has a stabilizing effect on short wavelengths in the long-time limit. Based on the above, one might expect that for some large fixed time $t$ (independent of $k$ ), there exists a sufficiently large wavenumber $k$ at which $s^{2}$ changes sign from positive to negative and hence, loosely speaking, growth is succeeded by decay in time. While general ordinary differential equation theorems (Bellman Reference Bellman1949; Kamke Reference Kamke1961) applied to (3.3) suggest that instability is plausible, they do not provide any insight into the wavenumber structure of the growing/decaying perturbation. Hence, the rest of this section is devoted to a detailed analysis of the instability phenomena.

3.3 Stability time intervals

Instability develops over several distinct time intervals, defined by the values of $Q$ , $\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C}$ and $k$ . In the following analysis, we discuss the solution of (3.3) for $u_{0}(t)$ , bearing in mind that it is related to the total solution via $\widehat{f}_{0}^{\prime }(t)=\text{e}^{-t/8t_{1}}u_{0}(t)$ . In the classical case (3.6), the exponential factor is absent as $t_{1}=\infty$ . In addition to the mass supply time scale $t_{1}=\unicode[STIX]{x03C0}/Qk^{2}$ previously introduced with $k=F_{0}^{-1}$ , the surface tension time scale $t_{2}=(Q/\unicode[STIX]{x03C0})^{3}(\unicode[STIX]{x1D70C}/\unicode[STIX]{x1D70E})^{2}$ emerges. We focus on axisymmetric disturbances, $n=0$ , as the most unstable ones in order to analyse the time intervals on which different terms in (3.4a ) dominate.

Case 1: $((Q/8\unicode[STIX]{x03C0})k^{2})^{2}$ is dominant if

(3.8a-c ) $$\begin{eqnarray}t\gg 4t_{1}\quad \text{and}\quad t\gg 4^{5}t_{1}^{2}t_{2}^{-1}\quad \text{and}\quad t\ll 4^{-5}t_{2},\end{eqnarray}$$

which is possible provided that $t_{2}$ is large enough and $t_{1}$ is sufficiently small. In this case, we get a linear oscillator equation with the solution

(3.9) $$\begin{eqnarray}u_{0}(t)=C_{1}\text{e}^{(Q/8\unicode[STIX]{x03C0})k^{2}t}+C_{2}\text{e}^{-(Q/8\unicode[STIX]{x03C0})k^{2}t};\end{eqnarray}$$

i.e. exponential growth of $u_{0}(t)$ is observed, with the most unstable wavenumber being $k=\infty$ (short waves), which is expected as surface tension is not present in this limit; of course, the conditions (3.8) are invalidated then. However, $\widehat{f}_{0}^{\prime }(t)$ (and hence the blob) is not prone to a Rayleigh–Taylor instability; in fact, the acceleration of the interface $F_{tt}\sim -t^{-3/2}$ is negative at this stage. The observed behaviour in this case corresponds to the flat part of the graphs in figure 2(c). The presence of the term $(Qk^{2}/4\unicode[STIX]{x03C0})^{2}/4=4^{-3}t_{1}^{-2}$ is due to the radial acceleration of the cylinder interface, as can be gleaned from (2.3).

Case 2: $1/4t^{2}$ is dominant if

(3.10a-c ) $$\begin{eqnarray}t\ll 4t_{1}\quad \text{and}\quad t\ll (4^{-1}t_{1}^{2}t_{2})^{1/3}\quad \text{and}\quad t\ll (4^{-1}t_{1}^{4}t_{2})^{1/5},\end{eqnarray}$$

which is possible provided that both $t_{1}$ and $t_{2}$ are large enough, i.e. the behaviour corresponds to the earlier time growth in figure 2(a). Since in this case $s^{2}(t)=1/(4t^{2})$ , we get Euler’s equation, admitting the solution

(3.11) $$\begin{eqnarray}u_{0}(t)=C_{1}t^{(1/2)+(1/\sqrt{2})}+C_{2}t^{(1/2)-(1/\sqrt{2})},\end{eqnarray}$$

i.e. algebraic growth is observed.

Case 3: $(k^{4}/2)(\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C})F$ is dominant if

(3.12a-c ) $$\begin{eqnarray}t\gg t_{1}\quad \text{and}\quad t\gg 4^{-5}t_{2}\quad \text{and}\quad t\gg (4^{-1}t_{1}^{4}t_{2})^{1/5},\end{eqnarray}$$

which is possible provided that both $t_{1}$ and $t_{2}$ are small enough. In this case, the solution can be expressed in terms of Bessel functions of fractional order $\unicode[STIX]{x1D708}$ , namely

(3.13) $$\begin{eqnarray}u_{0}(t)=\sqrt{t}Z_{1/2q}\left(\frac{\text{i}}{q}\sqrt{c}t^{q}\right),\quad \text{where}~\frac{1}{2q}=\frac{2}{5},\,c=-\frac{k^{4}}{2}\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}\sqrt{\frac{Q}{\unicode[STIX]{x03C0}}},\end{eqnarray}$$

and $Z_{\unicode[STIX]{x1D708}}(z)=C_{1}J_{\unicode[STIX]{x1D708}}+C_{2}J_{-\unicode[STIX]{x1D708}}$ , with the asymptotics for large $z$ (large time  $t$ ),

(3.14) $$\begin{eqnarray}J_{\unicode[STIX]{x1D708}}(z)=\sqrt{2/\unicode[STIX]{x1D708}z}\cos [z-(2\unicode[STIX]{x1D708}+1)\unicode[STIX]{x03C0}/4]+O(z^{-3/2})\quad \text{as}~z\rightarrow \infty .\end{eqnarray}$$

Hence, $u_{0}(t)\sim t^{-1/8}\cos (\unicode[STIX]{x1D6FC}t^{5/4})+\cdots \rightarrow 0$ as $t\rightarrow +\infty$ .

Case 4: $(k^{2}/2F)(\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C})$ is dominant if

(3.15a-c ) $$\begin{eqnarray}t\ll t_{1}\quad \text{and}\quad t\ll 4^{5}t_{1}^{2}t_{2}^{-1}\quad \text{and}\quad t\gg (4^{-1}t_{1}^{2}t_{2})^{1/3},\end{eqnarray}$$

which is possible provided that $t_{1}$ is large enough and $t_{2}$ is sufficiently small. In this case, $s^{2}(t)=ct^{-1/2}$ , with $c=2^{-1}t_{1}^{-1}t_{2}^{-1/2}$ , and thus

(3.16) $$\begin{eqnarray}u_{0}=t^{1/2}Z_{1/3}\left({\textstyle \frac{4}{3}}\text{i}c^{1/2}t^{3/4}\right),\end{eqnarray}$$

where $Z_{\unicode[STIX]{x1D708}}(z)=C_{1}J_{\unicode[STIX]{x1D708}}(z)+C_{2}J_{-\unicode[STIX]{x1D708}}(z)$ , $z=\text{i}\unicode[STIX]{x1D6FC}t^{3/4}$ and $\unicode[STIX]{x1D6FC}=(4/3)c^{1/2}$ , with the asymptotics $u_{0}(t)\sim t^{-3/8}\text{e}^{\unicode[STIX]{x1D6FC}t^{3/4}}$ . Hence, the solution first grows (case 4) and then decays in an oscillating fashion (case 3) in time, as observed in figures 2(a–c). Thus, the combination of cases 3 and 4 is analogous to the classical situation (3.6), and therefore one then might infer that selection of the most amplified wavenumber $k$ should be made possible when two terms – $(k^{2}/2F)(\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C})$ and $(k^{4}/2)(\unicode[STIX]{x1D70E}/\unicode[STIX]{x1D70C})F$ – compete, similarly to the classical Rayleigh–Plateau instability. However, this expectation proves to be unfeasible.

3.4 On non-existence of optimal growth time and wavenumber

Here, we will consider cases 3 and 4 together (omitting index 0), which leads to the equation

(3.17a,b ) $$\begin{eqnarray}u^{\prime \prime }-s^{2}(t)u=0,\quad s^{2}(t)=\frac{\unicode[STIX]{x1D6FC}}{\sqrt{t}}-\unicode[STIX]{x1D6FD}\sqrt{t},\quad \text{with}~\unicode[STIX]{x1D6FC}=\frac{k^{2}}{2}\frac{\unicode[STIX]{x1D70E}}{\unicode[STIX]{x1D70C}}\sqrt{\frac{\unicode[STIX]{x03C0}}{Q}},\,\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D6FC}\frac{k^{2}Q}{\unicode[STIX]{x03C0}}.\end{eqnarray}$$

The sign of $s^{2}(t)$ changes at the ‘turning’ point $t_{\ast }$ , which coincidentally equals the previously introduced inertial time scale $t_{\ast }=\unicode[STIX]{x1D6FC}/\unicode[STIX]{x1D6FD}\equiv t_{1}$ . The solution to this problem is amenable to the Wentzel–Kramers–Brillouin method, namely away from the turning point it renders

(3.18) $$\begin{eqnarray}u(t)=\frac{C_{\pm }}{[s^{2}(t)]^{1/4}}\exp \left[\pm \int _{t_{0}}^{t}\sqrt{s^{2}(\unicode[STIX]{x1D70F})}\,\text{d}\unicode[STIX]{x1D70F}\right],\end{eqnarray}$$

which after Taylor expansion of $s^{2}(t)$ near $t_{\ast }$ , $s^{2}(t_{\ast }+t)=-\unicode[STIX]{x1D6FC}t/t_{\ast }^{3/2}+\cdots$ , with $t$ now measured from the turning point $t_{\ast }$ , becomes

(3.19a ) $$\begin{eqnarray}\displaystyle & \displaystyle t>0:u=\frac{t_{\ast }^{3/8}}{(\unicode[STIX]{x1D6FC}t)^{1/4}}[A\text{e}^{\text{i}(2/3)(\unicode[STIX]{x1D6FC}^{1/2}/t_{\ast }^{3/4})t^{3/2}}+B\text{e}^{-\text{i}(2/3)(\unicode[STIX]{x1D6FC}^{1/2}/t_{\ast }^{3/4})t^{3/2}}], & \displaystyle\end{eqnarray}$$

(3.19b ) $$\begin{eqnarray}\displaystyle & \displaystyle t<0:u=\frac{t_{\ast }^{3/8}}{(-\unicode[STIX]{x1D6FC}t)^{1/4}}[C\text{e}^{-(2/3)(\unicode[STIX]{x1D6FC}^{1/2}/t_{\ast }^{3/4})(-t)^{3/2}}+D\text{e}^{(2/3)(\unicode[STIX]{x1D6FC}^{1/2}/t_{\ast }^{3/4})(-t)^{3/2}}], & \displaystyle\end{eqnarray}$$

$D=0$

$t\rightarrow -\infty$

In a neighbourhood of the turning point, (3.17) simplifies to the Airy equation

(3.20) $$\begin{eqnarray}u^{\prime \prime }+\unicode[STIX]{x1D6FC}tt_{\ast }^{-3/2}u=0,\end{eqnarray}$$

the solution of which is expressed in terms of Airy functions $u_{0}(t)=a\,\text{Ai}(z)+b\,\text{Bi}(z)$ , $z=-\unicode[STIX]{x1D6FC}^{1/3}t_{\ast }^{-1/2}t$ . In order to determine the constants $a$ and $b$ , we will need the asymptotics of $\text{Ai}$ and $\text{Bi}$ for $z\gg 1(t\rightarrow -\infty )$ and $-z\gg 1(t\rightarrow +\infty )$ (Abramowitz & Stegun Reference Abramowitz and Stegun1965). Matching these asymptotics with the Wentzel–Kramers–Brillouin solutions (3.19) for both  $t\rightarrow \pm \infty$ , we find

(3.21a-d ) $$\begin{eqnarray}A=\frac{\text{e}^{\text{i}(\unicode[STIX]{x03C0}/4)}}{2\sqrt{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D6FC}^{1/6}}{t_{\ast }^{1/4}}[b-\text{i}a],\quad B=\frac{\text{e}^{-\text{i}(\unicode[STIX]{x03C0}/4)}}{2\sqrt{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D6FC}^{1/6}}{t_{\ast }^{1/4}}[b+\text{i}a],\quad C=\frac{a}{2\sqrt{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D6FC}^{1/6}}{t_{\ast }^{1/4}},\quad D=\frac{b}{\sqrt{\unicode[STIX]{x03C0}}}\frac{\unicode[STIX]{x1D6FC}^{1/6}}{t_{\ast }^{1/4}}.\end{eqnarray}$$

Since $D=0$ , then $b=0$ . As a result, the solution for $t>0$ , i.e. after the turning point $t_{\ast }$ , reads

(3.22) $$\begin{eqnarray}u(t)=C\frac{2t_{\ast }^{3/8}}{\unicode[STIX]{x1D6FC}^{1/4}t^{1/4}}\sin \left(\unicode[STIX]{x1D709}+\frac{\unicode[STIX]{x03C0}}{4}\right),\quad \text{where}~\unicode[STIX]{x1D709}=\frac{2}{3}\frac{\unicode[STIX]{x1D6FC}^{1/2}}{t_{\ast }^{3/4}}t^{3/2},\end{eqnarray}$$

and thus the interfacial deflection becomes $\widehat{f}^{\prime }(t)=\text{e}^{-1/2\int a\,\text{d}t}u(t)=\text{const.}\,\text{e}^{-t/8t_{\ast }}u(t)$ , where again $t$ is defined relative to the turning point $t_{\ast }$ . To determine the maximum achieved during the time evolution, we differentiate the above (total) solution,

(3.23a,b ) $$\begin{eqnarray}\frac{\text{d}}{\text{d}t}[\text{e}^{-t/8t_{\ast }}u(t)]=0\Rightarrow \frac{1}{6}\tan \left(\unicode[STIX]{x1D709}+\frac{\unicode[STIX]{x03C0}}{4}\right)[\unicode[STIX]{x1D6FF}\unicode[STIX]{x1D709}^{2/3}+1]=\unicode[STIX]{x1D709},\quad \unicode[STIX]{x1D6FF}=\frac{3^{2/3}}{2^{4/3}}\left(\frac{t_{2}}{t_{1}}\right)^{1/6}\ll 1,\end{eqnarray}$$

where $\unicode[STIX]{x1D6FF}\ll 1$ since $t_{2}/t_{1}\ll 1$ in the considered case 4. For $\unicode[STIX]{x1D6FF}=0$ , the above extremum condition admits the approximate solution reflecting the presence of infinitely many ‘humps’ (cf. figure 3),

(3.24a,b ) $$\begin{eqnarray}\unicode[STIX]{x1D709}_{0}=\unicode[STIX]{x03C0}\left(m+\frac{1}{4}\right)-\unicode[STIX]{x1D709}^{\prime },\quad \unicode[STIX]{x1D709}^{\prime }=\frac{2}{3\unicode[STIX]{x03C0}}\frac{1}{1+4m},\quad m\in \mathbb{Z}.\end{eqnarray}$$

The corresponding maxima of $u(t)$ scale as $u_{max}\sim m^{-1/6}k^{-5/6}$ for $m\gg 1$ , i.e. the extremum decays with $m$ . For $\unicode[STIX]{x1D6FF}\neq 0$ , the solution of (3.23) is corrected to $\unicode[STIX]{x1D709}=(1-3\unicode[STIX]{x1D6FF}/2)\unicode[STIX]{x1D709}_{0}$ ; that is, the time when the maximum of the solution is reached becomes shorter with increase of the ratio $t_{2}/t_{1}$ .

Figure 3. Solution of (3.17) near the turning point $t_{\ast }$ ; the maximum is achieved after passing through $t_{\ast }$ .

Finally, in order to determine the wavenumber for which the maximum of the solution is absolute, it should be noticed that this maximum corresponds to $m=1$ and is achieved after passing through the turning point $t_{\ast }$ , cf. figure 3, which is analogous to the bifurcation delay phenomena in dynamical systems if $s^{2}(t)$ is deemed to be a time-dependent bifurcation parameter. Given the general turning point solution with $b=0$ , we can first determine where the maximum is achieved,

(3.25) $$\begin{eqnarray}\frac{\text{d}}{\text{d}t}[\text{e}^{-t/8t_{\ast }}\text{Ai}(\unicode[STIX]{x1D70F})]=0\Rightarrow \frac{\text{Ai}(-\unicode[STIX]{x1D703}\widetilde{\unicode[STIX]{x1D70F}})}{\text{Ai}^{\prime }(-\unicode[STIX]{x1D703}\widetilde{\unicode[STIX]{x1D70F}})}=-8\unicode[STIX]{x1D703},\quad \unicode[STIX]{x1D703}=\frac{1}{2^{1/3}}\left(\frac{t_{1}}{t_{2}}\right)^{1/6},\,\unicode[STIX]{x1D70F}=\frac{t}{t_{1}},\end{eqnarray}$$

with $\unicode[STIX]{x1D703}\gg 1$ in case 4. Graphical analysis indicates that the solution of the above equation is near the first zero of $\text{Ai}^{\prime }(-\unicode[STIX]{x1D703}\,\widetilde{\unicode[STIX]{x1D70F}})$ , denoted by $a_{s}^{\prime }=-1.01879\cdots <0$ (Abramowitz & Stegun Reference Abramowitz and Stegun1965), which allows one to apply asymptotic methods to find the solution of (3.25),

(3.26) $$\begin{eqnarray}\widetilde{\unicode[STIX]{x1D70F}}=-\frac{a_{s}^{\prime }}{\unicode[STIX]{x1D703}}+\widetilde{\unicode[STIX]{x1D70F}}^{\prime },\quad \text{where}~\widetilde{\unicode[STIX]{x1D70F}}^{\prime }=-\frac{1}{8\unicode[STIX]{x1D703}^{2}}\frac{\text{Ai}(a_{s}^{\prime })}{\text{Ai}^{\prime \prime }(a_{s}^{\prime })}+\cdots .\end{eqnarray}$$

By evaluating the solution at $\widetilde{\unicode[STIX]{x1D70F}}$ , i.e. $\widehat{f}^{\prime }(\widetilde{\unicode[STIX]{x1D70F}})=\text{const.}\,t_{2}^{1/2}\unicode[STIX]{x1D703}^{5/6}\text{e}^{-\widetilde{\unicode[STIX]{x1D70F}}/8}\text{Ai}(-\unicode[STIX]{x1D703}\,\widetilde{\unicode[STIX]{x1D70F}})$ , it is straightforward to show that there is no ‘optimum’ wavenumber $k$ . Indeed, as $k\rightarrow 0$ ,

(3.27) $$\begin{eqnarray}\widehat{f}^{\prime }(\widetilde{\unicode[STIX]{x1D70F}})\sim \unicode[STIX]{x1D703}^{5/6}\sim k^{-5/18},\end{eqnarray}$$

because $\text{Ai}(-\unicode[STIX]{x1D703}\widetilde{\unicode[STIX]{x1D70F}})\sim \text{const.}$ and $\text{e}^{-\widetilde{\unicode[STIX]{x1D70F}}/8}\sim 1$ (since $t_{1}\rightarrow +\infty$ for $k\rightarrow 0$ ). As expected from the discussion in § 3.1, there is no critical self-similar wavenumber scaled with $F^{-1}(t)$ and growing exponentially since, based on (3.26) and (3.27), the most amplified wavenumber at a given time $t$ scales as $k\sim t^{-3/5}$ . Thus, the fundamental difference from the classical case $Q=0$ is that no matter how small (but non-zero) $k$ is, there is a large enough time when this $k$ is amplified and then damped (for later times). It should be noted that for $k\equiv 0$ , from (3.4a ) we find that $a(t)=0$ , $b(t)=-1/(4t^{2})$ and $s^{2}(t)=1/(4t^{2})$ , so that the leading-order solution reads as

(3.28a,b ) $$\begin{eqnarray}u(t)=C_{1}t^{(1/2)+(1/\sqrt{2})}+C_{2}t^{(1/2)-(1/\sqrt{2})},\quad \text{and therefore}\quad \widehat{f}^{\prime }\sim t^{(1/2)+(1/\sqrt{2})},\end{eqnarray}$$

i.e. it grows faster than the undisturbed interface $F\sim t^{1/2}$ , in analogy with the Rayleigh–Taylor instability of an expanding sphere (Plesset Reference Plesset1954). The result (3.28) for $k\equiv 0$ contrasts with $\widehat{f}^{\prime }\sim t^{1/6}$ for $k\rightarrow 0$ , i.e. the limits $k\rightarrow 0$ and $t\rightarrow \infty$ are not interchangeable and thus distinguished. Since the limit of $k=0$ corresponds to a 2D problem, this transition from 3D to 2D makes some of the mechanisms (transfer of energy from the base state to the perturbation) disappear. Physically, a growing blob imminently breaks up in a finite time, and thus the range of finite non-zero wavenumbers amplified on this time scale ultimately affects the size of the resulting drops, which might be non-uniform due to the lack of a single preferred wavenumber.

4.1 Base state correction

From (2.8), it is obvious that non-zero acceleration $g$ affects mode $n=1$ only and thus induces a non-axisymmetric flow field in the blob such as in figure 1(c),

(4.1) $$\begin{eqnarray}L\widehat{f}_{1}^{\prime }=gF/2,\end{eqnarray}$$

which after the transformation $u_{1}=\widehat{f}_{1}^{\prime }\,\text{exp}(\int a\,\text{d}t/2)$ reduces to

(4.2) $$\begin{eqnarray}u_{1}^{\prime \prime }-s^{2}(t)u_{1}=Ct^{1/2}gF(t)\equiv h(t),\quad \text{where}~s^{2}(t)=t^{-2}[4^{-1}-(t/t_{2})^{1/2}k^{2}F^{2}].\end{eqnarray}$$

Looking for a long-time correction, we can simplify $s^{2}(t)$ to $s^{2}(t)=-t^{-1/2}t_{2}^{-1/2}t_{-1}<0$ , provided that $t\gg 4^{2/3}t^{2/3}t_{2}^{1/3}$ . The solution for the homogeneous part of (4.2) is

(4.3) $$\begin{eqnarray}u_{1}(t)=\sqrt{t}Z_{2/3}\left({\textstyle \frac{4}{3}}t_{1}^{-1/2}t_{2}^{-1/4}t^{3/4}\right),\end{eqnarray}$$

where $Z_{\unicode[STIX]{x1D708}}(z)=C_{1}J_{\unicode[STIX]{x1D708}}(z)+C_{2}J_{-\unicode[STIX]{x1D708}}(z)$ , $z=\unicode[STIX]{x1D6FE}t^{3/4}$ , $\unicode[STIX]{x1D6FE}=(4/3)t_{1}^{-1/2}t_{2}^{-1/4}$ and $\unicode[STIX]{x1D708}=2/3$ . If $u_{1}=C_{1}v_{1}+C_{2}v_{2}$ , with $v_{1}=t^{1/2}J_{\unicode[STIX]{x1D708}}(z)$ and $v_{2}=t^{1/2}J_{-\unicode[STIX]{x1D708}}(z)$ , then the Wronskian is $w=v_{1}v_{2}^{\prime }-v_{2}v_{1}^{\prime }$ and the solution to the inhomogeneous problem (4.2) reads

(4.4) $$\begin{eqnarray}u_{1}(t)=v_{2}(t)\int ^{t}\frac{v_{1}(\unicode[STIX]{x1D70F})h(\unicode[STIX]{x1D70F})}{w(\unicode[STIX]{x1D70F})}\,\text{d}\unicode[STIX]{x1D70F}-v_{1}(t)\int ^{t}\frac{v_{2}(\unicode[STIX]{x1D70F})h(\unicode[STIX]{x1D70F})}{w(\unicode[STIX]{x1D70F})}\,\text{d}\unicode[STIX]{x1D70F}.\end{eqnarray}$$

Since we are considering the long-time limit, given the asymptotics (3.14), we can choose as independent solutions $v_{1}=t^{1/8}\text{e}^{\text{i}\unicode[STIX]{x1D6FE}t^{3/4}}$ and $v_{2}=t^{1/8}\text{e}^{-\text{i}\unicode[STIX]{x1D6FE}t^{3/4}}$ , so that the Wronskian becomes constant, $w=-3\text{i}\unicode[STIX]{x1D6FE}/2$ , and the long-time limit of the solution (modulo a constant coefficient) is

(4.5) $$\begin{eqnarray}u_{1}(t)\sim t^{1/8}\text{e}^{\text{i}\unicode[STIX]{x1D6FE}t^{3/4}}\int ^{t}\unicode[STIX]{x1D70F}^{9/8}\text{e}^{-\text{i}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F}^{3/4}}\,\text{d}\unicode[STIX]{x1D70F}-t^{1/8}\text{e}^{-\text{i}\unicode[STIX]{x1D6FE}t^{3/4}}\int ^{t}\unicode[STIX]{x1D70F}^{9/8}\text{e}^{\text{i}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F}^{3/4}}\,\text{d}\unicode[STIX]{x1D70F}\sim t^{3/2},\end{eqnarray}$$

where we took into account that the leading-order contribution of the integrals in the above expression can be evaluated through integration by parts,

(4.6) $$\begin{eqnarray}\int ^{t}\unicode[STIX]{x1D70F}^{9/8}\text{e}^{\text{i}\unicode[STIX]{x1D6FE}\unicode[STIX]{x1D70F}^{3/4}}\,\text{d}\unicode[STIX]{x1D70F}=-\text{i}\frac{4}{3\unicode[STIX]{x1D6FE}}t^{11/8}\text{e}^{\text{i}\unicode[STIX]{x1D6FE}t^{3/4}}+O(t^{5/8}).\end{eqnarray}$$

As a result, the correction to the base state scales with time $t$ , acceleration $g$ and mass flux $Q$ as $u_{1}\sim g\sqrt{Q}t^{3/2}$ . For shorter times, i.e. when $1/(4t^{2})$ dominates in $s^{2}(t)$ , it is still the same mode $n=1$ that is excited, but with a different time exponent.

4.2 Contribution to instability

Since lateral acceleration acts at all times, it makes sense to consider its contribution to instability at early times – if it boosts it, the effect will be observable. From (2.8), we see that the axisymmetric mode is affected by mode one through the presence of acceleration  $g$ ,

(4.7) $$\begin{eqnarray}L\widehat{f}_{0}^{\prime }=g(\widehat{f}_{-1}^{\prime }+\widehat{f}_{1}^{\prime })/2=g\widehat{f}_{1}^{\prime },\end{eqnarray}$$

where we took into account the symmetry $n\rightarrow -n$ and $\widehat{f}_{1}^{\prime }$ is determined from the homogeneous equation $L\widehat{f}_{1}^{\prime }=0$ or, after the transformation $\widehat{f}_{1}^{\prime }=\text{exp}(-\!\int a\,\text{d}t/2)u_{1}=t^{-1/2}u_{1}$ corresponding to $a(t)=1/t$ , from (3.3) with $n=1$ . For short times, $s^{2}(t)=1/(4t^{2})$ , and hence the leading-order solution of (3.3) is $u_{1}(t)\sim t^{(1/2)+(1/\sqrt{2})}$ , so that $\widehat{f}_{1}^{\prime }\sim t^{1/\sqrt{2}}$ . To construct the solution to (4.7), we make the transformation $u_{0}=\widehat{f}_{0}^{\prime }\,\text{exp}(\int a\,\text{d}t/2)$ , where $a=1/(4t_{1})$ , to produce

(4.8) $$\begin{eqnarray}u_{0}^{\prime \prime }-s^{2}(t)u_{0}=\text{e}^{-t/8t_{1}}g\,\widehat{f}_{1}^{\prime }/2\equiv h(t)\sim g\text{e}^{-t/8t_{1}}t^{1/\sqrt{2}}.\end{eqnarray}$$

Since for short times $s^{2}(t)=1/(4t^{2})$ , two independent solutions of the homogeneous part are

(4.9) $$\begin{eqnarray}u_{0}(t)=C_{1}v_{1}+C_{2}v_{2},\quad \text{with}~v_{1}(t)=t^{(1/2)+(1/\sqrt{2})},\,v_{2}(t)=t^{(1/2)-(1/\sqrt{2})},\end{eqnarray}$$

i.e. given by case 2, and the Wronskian $w=-\sqrt{2}$ . Using formula (4.4), the corresponding perturbation driven by acceleration $g$ is then to leading order given by $u_{0}(t)\sim g\text{e}^{-t/8t_{1}}t^{1+(1/\sqrt{2})}$ , i.e. $\widehat{f}_{0}^{\prime }=t^{1+(1/\sqrt{2})}$ , and thus the (algebraic) growth is faster than that in the absence of acceleration, which is $t^{(1+\sqrt{2})/2}\text{e}^{t/8t_{1}}$ . Here, the exponential should be discarded as $t\ll t_{1}$ according to the case 2 assumptions. Hence, the presence of lateral acceleration entails amplification of the instability at early times without changing the wavenumber structure of the perturbation.