2.1. Solutions for large $d$

When we formally set $d = \infty$, then (2.4) becomes $\phi (x,y,0)= y^2-x^2$ and we can write

(2.6)\begin{equation} \phi= A(t)(y^2-x^2)+B(t),\quad f=F(t), \end{equation}

where $A(0)=1, B(0)=0$ and $F(0)=1$. Thus, from (2.2) and (2.3),

(2.7a–c)\begin{equation} \frac{{\rm d}A}{{\rm d}t}=2A^2,\quad \frac{{\rm d}F}{{\rm d}t}=2AF \quad \text{and}\quad \frac{{\rm d}B}{{\rm d}t}={-}4A^2 F^2, \end{equation}

and we find that $A=F={1}/({1-2t}), B=-{2}/({3(1-2t)^3})+\tfrac {2}{3}$, which is one case of the family of explicit solutions described by Longuet-Higgins (Reference Longuet-Higgins1972). The pressure $p$ is given by

(2.8)\begin{equation} p=\frac{4(F^2-y^2)}{(1-2t)^2}, \end{equation}

having jumped from $-2(x^2 + y^2)$ to $4(1-y^2)$ at $t=0$. Thus, not only does blow-up occur when $t=\tfrac {1}{2}$ but the pressure has a maximum that increases with time on the whole $x$-axis. We also note that when the stagnation-point flow is reversed so that $A(0)=-1$, the pressure still has a maximum on $y=0$ but $F(t)=-A(t)={1}/({1+2t})$ and thus the free boundary moves towards $y=0$ as $t\rightarrow \infty$.

2.1.1. The case of $d$ large and $0< t<\frac {1}{2}$

A limit that leads to tractable asymptotic solutions occurs when $d$ is large. In this case we write $\epsilon =1/d$ and, when $x,y$ are $O(1)$, the solution (2.6) will hold to lowest order. This is no longer true as $|x|\rightarrow \infty$ and so we need to introduce an outer region which is of width $O(\epsilon ^{-1})$ in the $x$-direction while $y$ remains of $O(1)$. Thus with $X=\epsilon x, \varPhi ={\epsilon }^2 \phi$, the problem becomes

(2.9)\begin{equation} \epsilon^2 \frac{\partial^2\varPhi}{\partial X^2} + \frac{\partial^2\varPhi}{\partial y^2} = 0 \end{equation}

with, on the free boundary, now denoted by $y=F(X,t)$,

(2.10)\begin{equation} \epsilon^2\left(\frac{\partial \varPhi}{\partial t}+\frac{1}{2}\left(\frac{\partial \varPhi}{\partial X}\right)^2\right)+ \frac{1}{2}\left(\frac{\partial \varPhi}{\partial y}\right)^2 = 0, \end{equation}

and

(2.11)\begin{equation} \epsilon^2\left(\frac{\partial F}{\partial t}+\frac{\partial \varPhi}{\partial X}\frac{\partial F}{\partial X}\right)= \frac{\partial \varPhi}{\partial y}. \end{equation}

Also, at $t=0$, expanding the term in brackets in (2.4) for large $d$ and using (2.5) gives

(2.12a,b)\begin{equation} F=1 \quad\text{and}\quad \varPhi \sim{-}\frac{X^2}{1+X^2}+O(\epsilon^2). \end{equation}

Finally, matching with the solution (2.6) gives that, as $X\rightarrow 0$,

(2.13a,b)\begin{equation} \varPhi\sim{-}\frac{X^2}{1-2t},\quad F\sim\frac{1}{1-2t}. \end{equation}

When we now expand in the form

(2.14a,b)\begin{equation} \varPhi\sim{\varPhi}_0 + {\epsilon}^2{\varPhi}_1 +\cdots,\quad F \sim F_0 + {\epsilon}^2F_1 + \cdots , \end{equation}

we find that, for some functions $\alpha$ and $\beta$,

(2.15)\begin{equation} {\varPhi}_0 = \alpha(X,t)y + \beta(X,t), \end{equation}

where, from (2.12b), $\alpha (X,0)=0$ and $\beta (X,0)=-({X^2}/({1+X^2}))$. Then conditions (2.10) and (2.11) imply that $\alpha (X,t)=0$ and so $\varPhi _0$ is independent of $y$. In order to determine $\varPhi _0$, we need to proceed to the second-order terms in (2.10) which reveals that

(2.16)\begin{equation} \frac{\partial \varPhi_0}{\partial t}+\frac{1}{2}\left(\frac{\partial \varPhi_0}{\partial X}\right)^2=0, \end{equation}

with the initial condition from (2.12b) that

(2.17)\begin{equation} \varPhi_0={-}\frac{X^2}{1+X^2}\quad \text{at}\ t=0. \end{equation}

Charpit's equations for (2.16) show that $P={\partial \varPhi _0}/{\partial t}$ and $Q={\partial \varPhi _0}/{\partial X}$ are constant on the characteristics of (2.16) through the point $X=X_0$, $t=0$ and these characteristics are the particle paths given by ${{\rm d}X}/{{\rm d}t}=Q$. Hence

(2.18)\begin{equation} Q=\frac{-2X_0}{(1+{X_0}^2)^2}, \end{equation}

and

(2.19)\begin{equation} \varPhi_0 =\frac{2{X_0}^2 t}{(1+{X_0}^2)^4}-\frac{{X_0}^2}{1+{X_0}^2}, \end{equation}

where

(2.20)\begin{equation} X= Qt+X_0 = X_0 - \frac{2{X_0}t}{(1+{X_0}^2)^2}. \end{equation}

The free surface can now be derived from (2.11) in the form

(2.21)\begin{equation} \frac{\partial F_0}{\partial t}+\frac{\partial \varPhi_0}{\partial X}\frac{\partial F_0}{\partial X} = \frac{\partial\varPhi_1}{\partial y} \quad \text{on} \ y=F_0, \end{equation}

where

(2.22)\begin{equation} \varPhi_1 ={-}\frac{1}{2}y^2 \frac{\partial ^2 \varPhi_0}{\partial X^2} + c_0(X,t)y + c_1(X,t), \end{equation}

and $c_0$ and $c_1$ are functions of integration. However, the fact that the velocity is zero at infinity ensures that $c_0(X,t)=0$.

We note that the characteristics of (2.21) are also the particle paths. Hence it is now convenient to change from Eulerian coordinates $(X,t)$ to Lagrangian coordinates $(X_0,t)$. Using (2.20), we see that, in Lagrangian coordinates with $t$ held constant,

(2.23)\begin{equation} \frac{\partial X}{\partial X_0}=1-\frac{2t(1-3{X_0}^2)}{(1+{X_0}^2)^3}. \end{equation}

Hence, from (2.18),

(2.24)\begin{equation} \frac{\partial ^2 \varPhi_0}{\partial X^2} = \left.\frac{\partial Q}{\partial X_0}\right/\frac{\partial X}{\partial X_0}= \frac{2(3{X_0}^2-1)}{(1+{X_0}^2)^3 - 2t(1-3{X_0}^2)}, \end{equation}

and (2.21) becomes

(2.25)\begin{equation} \left(1-\frac{2t(1-3{X_0}^2)}{(1+{X_0}^2)^3}\right)\frac{\partial F_0}{\partial t} = \frac{2(1-3{X_0}^2)}{(1+{X_0}^2)^3}F_0. \end{equation}

Then, using the condition $F_0=1$ at $t=0$, we obtain

(2.26)\begin{equation} F_0 =\frac{1}{\left(1-\dfrac{2t(1-3{X_0}^2)}{(1+{X_0}^2)^3}\right)}, \end{equation}

where (2.20) relates $X,t$ and $X_0$. Note that (2.26) and (2.19) match with (2.13a,b) as $X\rightarrow 0$ with $t<1/2$. Typical profiles for $F_0$ are shown in figure 2.

Figure 2.

Value of $F_0$ defined by (2.26), plotted as a function of $X$ for $t=0,0.1,0.2,0.3$.

It is relatively easy to show that $F_0$ always has a minimum at $X_0=1$ and then approaches $1$ from below as $|X_0|\rightarrow \infty$, as shown in figure 2. Most importantly, when $X_0=0$ we have $X=0$ and $F_0={1}/({1-2t})$ and so the extreme tip of the jet goes to infinity in finite time in the large $d$ limit. Moreover, using (2.2), (2.16) and (2.22), we find that, to lowest order in $\epsilon$,

(2.27)\begin{equation} -p ={-}\frac{1}{2}y^2 \frac{\partial ^3 \varPhi_0}{\partial X^2\partial t}+ \frac{\partial c_1}{\partial t}+ \frac{\partial \varPhi_0}{\partial X}\left(-\frac{1}{2}y^2\frac{\partial ^3 \varPhi_0}{\partial X^3}+ \frac{\partial c_1}{\partial X}\right)+\frac{1}{2}y^2\left(\frac{\partial ^2\varPhi_0}{\partial X^2}\right)^2. \end{equation}

Using (2.16), this reduces to

(2.28)\begin{equation} -p = y^2\left(\frac{\partial ^2\varPhi_0}{\partial X^2}\right)^2 + \frac{\partial c_1}{\partial t} + \frac{\partial \varPhi_0}{\partial X}\frac{\partial c_1}{\partial X}. \end{equation}

Since $p=0$ on $y=F_0$, this means that

(2.29)\begin{equation} \frac{\partial c_1}{\partial t} + \frac{\partial \varPhi_0}{\partial X}\frac{\partial c_1}{\partial X} ={-}{F_0}^2\left(\frac{\partial ^2\varPhi_0}{\partial X^2}\right)^2, \end{equation}

and hence that

(2.30)\begin{equation} p = \left(\frac{\partial ^2\varPhi_0}{\partial X^2}\right)^2({F_0}^2-y^2). \end{equation}

Thus, pressure maxima in this regime can only occur on $y=0$.

Using (2.24), we see that the pressure on $y=0$ is a function of $X$, as shown in figure 3. Moreover, since $X$ is a monotone increasing function of $X_0$, $p$ will now have two maxima in $X\geq 0$, with the maximum at $X=0$ being increasingly dominant as $t\rightarrow \tfrac {1}{2}$. We remark that the weaker maximum is generated by our initial conditions and is not associated with a stagnation-point flow.

Figure 3.

Value of $p(X,0)$ plotted from (2.30) for $t=0.01$.

The behaviour revealed in figure 3 enables us to draw the isobars in $X \geq 0, y \geq 0$ as shown in figure 4. The isobars in $y<0$ are the mirror image of those in $y>0$ until the zero-pressure isobar is reached and the pressure is everywhere negative below this isobar.

Figure 4.

Isobars of $p(X,y)$ for $t=0.3$ when $y>0$; the fluid region is shaded.

2.1.2. The limit as $t\uparrow \frac {1}{2}$

We observe that, when $X_0$ and $\tfrac {1}{2}-t=\tau$ are small, (2.19) and (2.26) give

(2.31a–c)\begin{equation} X\sim2X_0(\tau+{X_0}^2),\quad F_0\sim\frac{1}{2\tau+6{X_0}^2},\quad \varPhi_0\sim{-}{X_0}^2(2\tau+3{X_0}^2). \end{equation}

Hence, the solution is singular as $X_0,\tau \rightarrow 0$ and rapid changes are expected when $\tau$ and ${X_0}^2$ are small and comparable. If we assume that $\tau =O({\hat {\epsilon }}^2)$ and $X_0=O(\hat {\epsilon })$ for some small parameter $\hat {\epsilon }$, then the appropriate scalings in this region are

(2.32a–c)\begin{equation} x=O\left(\frac{{\hat{\epsilon}}^3}{\epsilon}\right),\quad y=O({\hat{\epsilon}}^{{-}2}),\quad \phi=O\left(\frac{{\hat{\epsilon}}^4}{\epsilon^2}\right), \end{equation}

and this leads to a region where Laplace's equation (2.1) holds if $\hat {\epsilon } = \epsilon ^{1/5}$. Thus we write

(2.33a–e)\begin{equation} \tau=\epsilon^{2/5} \hat{\tau},\quad x=\epsilon^{{-}2/5}\hat{x},\quad y=\epsilon^{{-}2/5}\hat{y}, \quad f=\epsilon^{{-}2/5}\hat{f} \quad \text{and} \quad \phi= \epsilon^{{-}6/5}\hat{\phi}. \end{equation}

The resulting free-boundary problem is equivalent to the full problem (2.1) with boundary conditions (2.2) and (2.3) but with the solution matching with (2.31a–c) as $\hat {x}\rightarrow \infty$. If we let $X_0\rightarrow 0$ in (2.31a–c) while keeping $\tau$ constant we see that, for $t<\frac {1}{2}$,

(2.34a,b)\begin{equation} \hat{\phi} \sim{-} \frac{\hat{x}^2}{2\hat{\tau}}\quad \text{and} \quad \hat{f}\sim \frac{1}{2\hat{\tau}}, \end{equation}

as $\hat {x} \rightarrow \infty$. As $\hat {\tau } \rightarrow \infty$, we need to match with (2.6) and hence the solution is

(2.35a,b)\begin{equation} \hat{\phi} = \frac{{\hat{y}}^2 - {\hat{x}}^2}{2\hat{\tau}} - \frac{1}{12 {\hat{\tau}}^3}, \quad \hat{f}=\frac{1}{2\hat{\tau}}, \end{equation}

which is equivalent to the limit of (2.6) as $t\uparrow 1/2$. Thus, to lowest order in $\epsilon$, the jet extends to infinity at $\hat {\tau } = 0$. As noted after (2.8), the lowest-order pressure maintains a maximum along $y=0$.

2.2. The case of $d$ finite and $t\ll 1$

1. (i) $d>1$ We will consider only the small-time solution of (2.1), (2.2), (2.3) and (2.4) for $d>1$. Assuming a smooth dependence on time, we seek expansions of the form

   (2.36)\begin{equation} \displaystyle\phi\sim\phi(x,y,0)+t{\phi}_1(x,y)+\cdots , \end{equation}
   and
   (2.37)\begin{equation} \displaystyle f\sim1+tf_1(x)+\cdots . \end{equation}
   Then we find that
   (2.38) \begin{equation} f_1=\frac{\partial \phi(x,y,0)}{\partial y}|_{y=1}= \frac{-2d^4(3x^4+2(1+d^2)x^2-(1-d^2)^2)}{(x^2+(1-d)^2)^2(x^2+(1+d)^2)^2}, \end{equation}
   and we see that the free surface immediately adopts the jet profile shown in figure 5. The zeros of $f_1$ are where $3x^2=2\sqrt {1-d^2+d^4}-1-d^2$ and $f_1(0)={2d^4}/{(1-d^2)^2}$. However, the lowest-order pressure can only be calculated by finding $\phi _1$, which involves solving a complicated potential problem which will not be addressed here. We also note that figure 5 has the same geometrical features as were seen in the longer-time solutions shown in figure 2.

2. (ii) $d\downarrow 1$ The profile (2.38) tends to infinity as $x\rightarrow 0$ when $d=1$ and hence, to understand what happens as $\delta = d-1$ tends to zero, we expand (2.4) by writing

   (2.39a,b)\begin{equation} y= 1 + \delta \tilde{y}, \quad x = \delta \tilde{x}, \end{equation}
   and, noting that there is only one effective dipole in this region, (2.38) gives that, at $t=0$,
   (2.40)\begin{equation} \phi \sim{-}\frac{\tilde{y}-1}{2\delta ({\tilde{x}}^2 + (\tilde{y} - 1)^2)} + O(1). \end{equation}
   Hence, for $t$ of $o({\delta }^2)$,
   (2.41)\begin{equation} f_1 \sim{-}\frac{({\tilde{x}}^2 - 1)}{2\delta^2 (1+{\tilde{x}}^2)^2} + O(1). \end{equation}
   In this short time-scale regime, the free boundary is still of the same form as in figure 5 and the length of the jet is ${t}/{2\delta ^2}$ while its width is of $O(\delta )$.
   Figure 5.

   Value of $f_1$ plotted as a function of $x$ from (2.38).

   

   In summary, we have shown that, when $d$ is either near to unity or tends to infinity, the formal asymptotic analysis suggests that the interface tends to infinity in a finite time. In both cases the free surface is focussed into a jet which contains 4 inflection points, as shown in figures 2 and 5.

We now turn to the axisymmetric case, which will be found to lead to similar consequences but with even stronger jet accelerations.

3.1. The solution for small $\epsilon$

When $z=O(1)$ and $r=O(\epsilon ^{-1})$, we write $r=R/\epsilon$ and $\phi =\varPhi /\epsilon ^2$ so that

(3.7)\begin{equation} \frac{\partial^2\varPhi}{\partial z^2}+\epsilon^2 \left(\frac{\partial^2\varPhi}{\partial R^2}+ \frac{1}{R}\frac{\partial \varPhi}{\partial R}\right)=0, \end{equation}

with, on $z=F(R,t)$,

(3.8)\begin{equation} \frac{1}{2}\left(\frac{\partial \varPhi}{\partial z}\right)^2 + \epsilon^2\left( \frac{1}{2}\left(\frac{\partial \varPhi}{\partial R}\right)^2+\frac{\partial \varPhi}{\partial t}\right) = 0, \end{equation}

and

(3.9)\begin{equation} \frac{\partial \varPhi}{\partial z} = \epsilon^2\left(\frac{\partial F}{\partial t}+ \frac{\partial \varPhi}{\partial R}\frac{\partial f}{\partial R}\right). \end{equation}

Writing

(3.10)\begin{equation} \varPhi\sim\varPhi_0 + \epsilon^2\varPhi_1 +\cdots, \end{equation}

and

(3.11)\begin{equation} F \sim F_0 + \epsilon^2 F_1 +\cdots, \end{equation}

the lowest-order solution gives $\varPhi _0$ as a function of $R,t$ only, and such that

(3.12)\begin{equation} \frac{\partial \varPhi_0}{\partial t} + \frac{1}{2}\left(\frac{\partial \varPhi_0}{\partial R}\right)^2 =0. \end{equation}

The initial conditions are

(3.13)\begin{equation} \varPhi_0(R,0)=\frac{1}{3(1+R^2)^{3/2}}-\frac{1}{3}, \quad F_0(0) = 1, \end{equation}

and the analogue of (2.21) is

(3.14)\begin{equation} \frac{\partial F_0}{\partial t}+\frac{\partial \varPhi_0}{\partial R}\frac{\partial F_0}{\partial R} = \frac{\partial \varPhi_1}{\partial z}={-}\left(\frac{\partial^2\varPhi_0}{\partial R^2} +\frac{1}{R}\frac{\partial \varPhi_0}{\partial R}\right)F_0 - c_0(R,t), \end{equation}

and again $c_0(R,t)$ has to vanish since there is no mean flow at infinity. The relevant solution of (3.12) is

(3.15)\begin{equation} \varPhi_0= \frac{{{R_0}^2}t}{2(1+{R_0}^2)^5} + \frac{1}{3(1+{R_0}^2)^{3/2}} - \frac{1}{3}, \end{equation}

where

(3.16)\begin{equation} R=R_0 - \frac{{R_0} t}{(1+{R_0}^2)^{5/2}}, \end{equation}

and $R_0$ is a parameter analogous to $X_0$ in two dimensions. Then, $F_0$ is determined in terms of the Lagrangian variables $R_0,t$ from (3.14) in the form

(3.17)\begin{equation} \frac{\partial F_0}{\partial t}={-}\left(\frac{Q}{Qt+R_0}+\frac{Q'}{Q't+ 1}\right)F_0, \end{equation}

where

(3.18)\begin{equation} Q = \frac{\partial {\varPhi}_0}{\partial R} ={-}\frac{R_0}{(1+{R_0}^2)^{5/2}}, \end{equation}

and

(3.19)\begin{equation} Q'=\frac{{\rm d}Q}{{\rm d}R_0}. \end{equation}

Hence, $F_0$ has a time dependence that is different from that in (2.26), namely

(3.20)\begin{equation} F_0=\frac{R_0}{(R_0+Qt)(1+Q't)}, \end{equation}

so that $F_0$ grows like $(1-t)^{-2}$ as $t\uparrow 1$ when $R=0$. A typical profile for $F_0$ is shown in figure 6, which is geometrically similar to figure 2. However, when we calculate $p$ as we did to derive (2.30), we find that

(3.21)\begin{equation} p = \left( \left(\frac{\partial ^2 \varPhi_0}{\partial R^2}\right)^2 + \frac{1}{R}\frac{\partial \varPhi_0}{\partial R}\frac{\partial ^2 \varPhi_0}{\partial R^2} + \frac{1}{R^2}\left(\frac{\partial \varPhi_0}{\partial R}\right)^2\right)({F_0}^2-z^2). \end{equation}

The leading bracket is positive for all $R\geq 0$ and hence the only maximum of $p$ is at $R=0, z=0$.

Figure 6.

Value of $F_0$ plotted from (3.20) as a function of $R$ for $t=0,0.1,0.2,0.3,0.5,0.75$.

As for the two-dimensional case, if we let $R_0\rightarrow 0$ and $t\rightarrow 1$, we find, from (3.15), (3.16) and (3.20) that, the scalings in (2.33a–e) need to be modified. For (3.1), (3.2), (3.3) the appropriate scalings are

(3.22a–e) \begin{align} 1-t =O({\epsilon}^{2/7}),\quad r\!=\! O({\epsilon}^{{-}4/7}),\quad z \!=\! O({\epsilon}^{{-}4/7}),\quad f\!=\!O({\epsilon}^{{-}4/7})\quad \text{and}\quad \phi\!=\!O(\epsilon^{{-}10/7}). \end{align}

This means that $\phi$ still satisfies Laplace's equation and when we apply the matching condition analogous to (2.34a,b), the resulting lowest-order solution in this region emerges as the limit of (3.6) as $t\uparrow 1$, namely

(3.23a,b)\begin{equation} \phi = \frac{z^2-\dfrac{1}{2}r^2}{1-t} - \frac{3}{5(1-t)^5} \quad \text{and} \quad f=\frac{1}{(1-t)^2}. \end{equation}

Finally, we note that a small-time analysis when $d=O(1)$ again leads to a profile similar to that in figure 5 and that if we set $d=1+\delta$ where $\delta$ is small, then, for small times the jet length is of $O({t}/{{\delta }^3})$ as $\delta \rightarrow 0$, which is an order of magnitude longer than that in the two-dimensional case. This is to be expected since there is now flow into the jet from all directions.