3.1 Region I: thin-film region

The width of this region is of order one, yet the thickness $\unicode[STIX]{x1D70F}(t)$ of the film shrinks to zero, so we use the ansatz

(3.1a-c ) $$\begin{eqnarray}h(x,t)=\unicode[STIX]{x1D70F}(t)h_{f}(x),\quad u(x,t)=u_{f}(x),\quad \unicode[STIX]{x1D703}(x,t)=\unicode[STIX]{x1D703}_{l},\end{eqnarray}$$

where the temperature is assumed constant, in accordance with our earlier observations. Inserting (3.1) into the equations of motion, (2.2a ) yields

(3.2) $$\begin{eqnarray}\dot{\unicode[STIX]{x1D70F}}h_{f}=-\unicode[STIX]{x1D70F}(h_{f}u_{f})^{\prime }\end{eqnarray}$$

and, at leading order $\unicode[STIX]{x1D70F}^{0}$ , (2.2b ) results in

(3.3) $$\begin{eqnarray}u_{f}u_{f}^{\prime }=4O\frac{(h_{f}u_{f}^{\prime })^{\prime }}{h_{f}},\end{eqnarray}$$

where the surface tension term is of order $\unicode[STIX]{x1D70F}$ , and thus drops out in the limit $\unicode[STIX]{x1D70F}\rightarrow 0$ . Here and in the remainder of the paper, a dot denotes a derivative with respect to time, and a prime denotes a derivative with respect to the spatial variable. Note that (3.3) represents a balance between inertia and viscosity, while surface tension and Marangoni forces drop out. Dividing (3.2) by $\unicode[STIX]{x1D70F}h_{f}u_{f}$ and (3.3) by $4Ou_{f}^{\prime }$ , the term $h_{f}^{\prime }/h_{f}$ can be eliminated between the two equations, and one obtains an equation for the velocity $u_{f}$ alone:

(3.4) $$\begin{eqnarray}\bar{u}+\frac{\dot{\unicode[STIX]{x1D70F}}}{\unicode[STIX]{x1D70F}4O\bar{u}}=\frac{\bar{u}^{\prime \prime }}{\bar{u}^{\prime }}-\frac{\bar{u}^{\prime }}{\bar{u}},\end{eqnarray}$$

where we have rescaled the velocity according to $u_{f}=4O\bar{u}$ . In (3.4) only the second term on the left-hand side depends on time. Therefore, necessarily one has $-\dot{\unicode[STIX]{x1D70F}}/\unicode[STIX]{x1D70F}=4aO$ , where $a>0$ is a constant, which depends on initial conditions, as we will see. This implies

(3.5) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D70F}(t)=\unicode[STIX]{x1D70F}_{0}\exp \{-4aOt\}, & \displaystyle\end{eqnarray}$$

(3.6) $$\begin{eqnarray}\displaystyle & \displaystyle \bar{u}-\frac{a}{\bar{u}}=\frac{\bar{u}^{\prime \prime }}{\bar{u}^{\prime }}-\frac{\bar{u}^{\prime }}{\bar{u}}, & \displaystyle\end{eqnarray}$$

where $\unicode[STIX]{x1D70F}_{0}$ is an arbitrary normalization factor, which depends on the choice of origin for the time coordinate.

As shown in appendix A, (3.6) can be integrated and possesses a one-parameter family of ‘blow-up’ solutions of the form

(3.7) $$\begin{eqnarray}\bar{u}_{A}(x)=A+\tan \left[(x-\bar{x})\sqrt{a-A^{2}}\right]\sqrt{a-A^{2}},\quad \text{for}~A\in (-\sqrt{a},\sqrt{a}).\end{eqnarray}$$

The boundary conditions $\bar{u}(0)=\bar{u}^{\prime \prime }(0)=0$ , which follow from (2.3), together with (3.3), require that $A=\bar{x}=0$ , and thus

(3.8) $$\begin{eqnarray}u_{f}=4O\sqrt{a}\tan (x\sqrt{a}).\end{eqnarray}$$

The flux $j_{f}(x)=h_{f}(x)u_{f}(x)$ is calculated from (3.2) as

(3.9) $$\begin{eqnarray}\ln j_{f}=\int \frac{4Oa}{u_{f}}\,\text{d}x,\end{eqnarray}$$

and then it follows from (3.8) that

(3.10a,b ) $$\begin{eqnarray}j_{f}(x)=j_{0}\sin (\sqrt{a}x),\quad h_{f}(x)=\frac{j_{0}}{4O\sqrt{a}}\cos (\sqrt{a}x),\end{eqnarray}$$

where $j_{0}$ is a positive constant. It is clear from the first equation of (3.1) that, by adjusting $\unicode[STIX]{x1D70F}_{0}$ , we can make $h_{f}(0)=j_{0}/4O\sqrt{a}$ attain any value, which means that $j_{0}$ can be chosen arbitrarily. In the numerical results reported below, we will make the particular choice $j_{0}=4O\sqrt{a}$ .

The pinch point $x_{0}$ is determined by where $h_{f}$ goes to zero, which is at

(3.11) $$\begin{eqnarray}x_{0}=\unicode[STIX]{x03C0}/(2\sqrt{a});\end{eqnarray}$$

the interval $0\leqslant x\leqslant x_{0}$ will be referred to as the ‘thin-film region’. At $x_{0}$ , the flux is $j_{f}(x_{0})=j_{0}$ , which means that according to (3.1) the mass flux through the pinch point and into the drop is $\unicode[STIX]{x1D70F}j_{0}$ . In the neighbourhood of $x_{0}$ , the outer (film) profiles are

(3.12a,b ) $$\begin{eqnarray}h\approx -\frac{j_{0}\unicode[STIX]{x1D70F}(x-x_{0})}{4O},\quad u\approx -\frac{4O}{x-x_{0}}.\end{eqnarray}$$

In figure 4 we present the leading-order film solutions (3.8) and (3.10) in the thin-film region (red dash-dotted curves), superimposed with full numerical solutions, rescaled according to (3.1). Even for times of order one, very convincing convergence towards the asymptotic solutions is found.

Figure 4.

Comparison of a numerical solution of (2.2) (initial data and parameters as in figure 2), with the leading-order solutions (3.10) and (3.8) (dash-dotted curves), in the thin-film region. Five snapshots of (a) the height (linear plot) and (b) the velocity (log-linear plot) solutions to (2.2) are shown at times $t_{1}=2.247,~t_{2}=4.032,~t_{3}=5.138,~t_{4}=5.987,~t_{5}=6.800$ . The thinning rate $a=0.819$ was calculated from the numerical solution using (3.6). The pinch point was found to be $x_{0}\approx 1.7355$ , in very good agreement with (3.11).

3.2 Region IV: drop region

The total mass in the film region is of order $\unicode[STIX]{x1D70F}$ , which means that any change in the volume of the drop region is a subdominant correction. To leading order, the drop volume is constant and the drop thus converges towards a static shape, with no flow, and temperature is constant: $h(x,t)=h_{d}(x)$ and $\unicode[STIX]{x1D703}(x,t)=\unicode[STIX]{x1D703}_{r}$ , while $u(x,t)=0$ . The leading-order solution to (2.2b ) in this region must satisfy $h_{d}^{\prime \prime \prime }=0$ , and thus

(3.13) $$\begin{eqnarray}h_{d}(x)=C_{0}[(\unicode[STIX]{x03C0}-x_{0})^{2}-(x-\unicode[STIX]{x03C0})^{2}],\end{eqnarray}$$

where $x_{0}$ is the pinch point as in (3.11). The constants $C_{0}$ and $\unicode[STIX]{x1D703}_{r}$ are determined uniquely by conservation of mass and heat (2.4), which yield

(3.14a,b ) $$\begin{eqnarray}\int _{x_{0}}^{\unicode[STIX]{x03C0}}h_{d}(x)\,\text{d}x=M,\quad \int _{x_{0}}^{\unicode[STIX]{x03C0}}h_{d}(x)\unicode[STIX]{x1D703}_{r}\,\text{d}x=Q.\end{eqnarray}$$

From this the constants can be computed as

(3.15a,b ) $$\begin{eqnarray}C_{0}=\frac{3M}{2(\unicode[STIX]{x03C0}-x_{0})^{3}},\quad \unicode[STIX]{x1D703}_{r}=\frac{Q}{M}.\end{eqnarray}$$

In particular, we have the following expression for the macroscopic contact angle of the drop:

(3.16) $$\begin{eqnarray}h_{d}^{\prime }(x_{0})=\frac{3M}{2(\unicode[STIX]{x03C0}-x_{0})^{2}}\equiv s,\end{eqnarray}$$

which will be used later to match to the pinch region.

3.3 Region II: pinch region

Since this solution lives on an exponentially small scale set by the film thickness $\unicode[STIX]{x1D70F}$ , we try the similarity solution

(3.17a-d ) $$\begin{eqnarray}h(x,t)=\unicode[STIX]{x1D70F}^{\unicode[STIX]{x1D6FC}_{1}}H\left(\unicode[STIX]{x1D709}\right),\quad u(x,t)=\unicode[STIX]{x1D70F}^{\unicode[STIX]{x1D6FC}_{2}}U\left(\unicode[STIX]{x1D709}\right),\quad \unicode[STIX]{x1D703}(x,t)=\unicode[STIX]{x1D70F}^{\unicode[STIX]{x1D6FC}_{3}}\unicode[STIX]{x1D6E9}\left(\unicode[STIX]{x1D709}\right),\quad \unicode[STIX]{x1D709}=\frac{x-x_{0}}{\unicode[STIX]{x1D70F}^{\unicode[STIX]{x1D6FD}}}.\qquad\end{eqnarray}$$

Since the flux through the pinch region is $\unicode[STIX]{x1D70F}j_{0}=hu$ , we must have $\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2}=1$ . We also expect (3.17) to match to the linear $h$ -profile (3.12), which implies that $\unicode[STIX]{x1D6FC}_{1}-\unicode[STIX]{x1D6FD}=1$ . Since the temperature changes over a scale of order unity, we have $\unicode[STIX]{x1D6FC}_{3}=0$ . Finally, Marangoni forces drive the pinch-off and thus must come in at leading order near the pinch point. We expect them to be balanced by viscous forces, which already come in the thin-film region, and thus should also be important on even smaller scales. Then from a balance of the last two terms of (2.2b ) we obtain $\unicode[STIX]{x1D6FC}_{2}-2\unicode[STIX]{x1D6FD}=-\unicode[STIX]{x1D6FD}-\unicode[STIX]{x1D6FC}_{1}$ , and combining all of the above yields $\unicode[STIX]{x1D6FD}=1$ , $\unicode[STIX]{x1D6FC}_{1}=2$ and $\unicode[STIX]{x1D6FC}_{2}=-1$ . Then the leading force balance in (2.2b ) is at $O(\unicode[STIX]{x1D70F}^{3})$ , and inertial, viscous and Marangoni forces come in at leading order.

Thus in the pinch region the similarity solution takes the form

(3.18a-d ) $$\begin{eqnarray}h(x,t)=\unicode[STIX]{x1D70F}^{2}H_{p}\left(\unicode[STIX]{x1D709}\right),\quad u(x,t)=\unicode[STIX]{x1D70F}^{-1}U_{p}\left(\unicode[STIX]{x1D709}\right),\quad \unicode[STIX]{x1D703}(x,t)=\unicode[STIX]{x1D6E9}_{p}\left(\unicode[STIX]{x1D709}\right),\quad \unicode[STIX]{x1D709}=\frac{x-x_{0}}{\unicode[STIX]{x1D70F}},\qquad\end{eqnarray}$$

and the similarity equations are

(3.19a ) $$\begin{eqnarray}\displaystyle & \displaystyle j_{0}=H_{p}U_{p}, & \displaystyle\end{eqnarray}$$

(3.19b ) $$\begin{eqnarray}\displaystyle & \displaystyle U_{p}U_{p}^{\prime }=4O\frac{(H_{p}U_{p}^{\prime })^{\prime }}{H_{p}}-{\mathcal{M}}\frac{\unicode[STIX]{x1D6E9}_{p}^{\prime }}{H_{p}}, & \displaystyle\end{eqnarray}$$

(3.19c ) $$\begin{eqnarray}\displaystyle & \displaystyle U_{p}\unicode[STIX]{x1D6E9}_{p}^{\prime }={\mathcal{D}}\left(\frac{H_{p}\unicode[STIX]{x1D6E9}_{p}^{\prime }}{H_{p}}\right)^{\prime }. & \displaystyle\end{eqnarray}$$

(3.20) $$\begin{eqnarray}\min _{x}h(x,t)\propto \unicode[STIX]{x1D70F}^{2}\propto \exp \{-8aOt\},\end{eqnarray}$$

which is confirmed numerically in figure 5.

Figure 5.

Log-linear plot of the minimum height as a function of time (symbols) for the solution to (2.2), with parameters and initial data as in figures 2 and 4. The thinning rate $a=0.819$ was calculated from the numerical solution using (3.6). The solid line is the theoretical prediction (3.20). Constant $C_{4}=\log (\min _{\unicode[STIX]{x1D709}}H(\unicode[STIX]{x1D709})\unicode[STIX]{x1D70F}_{0}^{2})=\log (4^{(8{\mathcal{P}}-1)/(4{\mathcal{P}}-1)}O\sqrt{a})-\log (C_{+})+2\log (\unicode[STIX]{x1D70F}_{0})$ by (3.10b ), (3.19a ) and (3.32).

The flux condition (3.19a ) can be used to eliminate $H_{p}$ , and we obtain the two equations

(3.21a,b ) $$\begin{eqnarray}U_{p}^{\prime }=4O(U_{p}^{\prime }/U_{p})^{\prime }-{\mathcal{M}}\unicode[STIX]{x1D6E9}_{p}^{\prime }/j_{0},\quad \unicode[STIX]{x1D6E9}_{p}^{\prime }={\mathcal{D}}(\unicode[STIX]{x1D6E9}_{p}^{\prime }/U_{p})^{\prime }.\end{eqnarray}$$

Integrating the first equation in (3.21) one expresses the temperature profile dependence on $U_{p}$ explicitly as

(3.22) $$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{p}=\frac{j_{0}}{{\mathcal{M}}}\left(4O\frac{U_{p}^{\prime }}{U_{p}}-U_{p}\right)+\unicode[STIX]{x1D703}_{l},\end{eqnarray}$$

where we have used the boundary conditions on the similarity profiles, as $\unicode[STIX]{x1D709}\rightarrow -\infty$ ,

(3.23a-c ) $$\begin{eqnarray}H_{p}\approx -\frac{j_{0}\unicode[STIX]{x1D709}}{4O},\quad U_{p}\approx -\frac{4O}{\unicode[STIX]{x1D709}},\quad \unicode[STIX]{x1D6E9}_{p}\approx \unicode[STIX]{x1D703}_{l},\end{eqnarray}$$

which follow from comparison to (3.12).

Next, substitution of (3.22) into the second (temperature) equation of (3.21) gives the following ODE for the profile $U_{p}$ :

(3.24) $$\begin{eqnarray}4{\mathcal{D}}O(U_{p}^{\prime \prime }-U_{p}^{\prime 2}/U_{p})-(4O+{\mathcal{D}})U_{p}^{\prime }U_{p}+U_{p}^{3}+C_{2}U_{p}^{2}=0,\end{eqnarray}$$

with $C_{2}$ being a constant of integration. Evaluating the left-hand side of (3.24) for $\unicode[STIX]{x1D709}\rightarrow -\infty$ , once more using (3.23), one concludes that $C_{2}=0$ .

The second-order equation (3.24) can be turned into a first-order equation by putting $U_{p}^{\prime }(\unicode[STIX]{x1D709})=P(U_{p})$ , so that $U_{p}^{\prime \prime }=P^{\prime }P$ , and

(3.25) $$\begin{eqnarray}\frac{\text{d}P}{\text{d}U_{p}}=\frac{P}{U_{p}}+[{\mathcal{D}}^{-1}+(4O)^{-1}]U_{p}-\frac{U_{p}^{3}}{4{\mathcal{D}}OP}.\end{eqnarray}$$

The observation that (3.25) is invariant under $U_{p}\rightarrow C_{+}U_{p}$ and $P\rightarrow C_{+}^{2}P$ suggests the substitution

(3.26) $$\begin{eqnarray}P=\frac{wU_{p}^{2}}{4\sqrt{{\mathcal{D}}O}},\end{eqnarray}$$

which reduces (3.25) to the separable ODE

(3.27) $$\begin{eqnarray}\frac{\text{d}U_{p}}{U_{p}}=\frac{\text{d}w}{4\sqrt{{\mathcal{P}}}+\sqrt{{\mathcal{P}}}^{-1}-4/w-w}.\end{eqnarray}$$

From (3.23) it follows that $w$ must satisfy the boundary condition $w\approx \sqrt{{\mathcal{P}}}^{-1}$ for $U_{p}\rightarrow 0$ . Hence, for each ${\mathcal{P}}>0$ , bounded solutions of (3.27) have the form

(3.28) $$\begin{eqnarray}\displaystyle U_{p}(w)=\left\{\begin{array}{@{}ll@{}}\left.C_{+}\left(\sqrt{{\mathcal{P}}}^{-1}-w\right)^{1/(4{\mathcal{P}}-1)}\right/\left(4\sqrt{{\mathcal{P}}}-w\right)^{4{\mathcal{P}}/(4{\mathcal{P}}-1)}\quad & \text{for }{\mathcal{P}}\not ={\displaystyle \frac{1}{4}},\\ \left.C_{+}\exp \left\{{\displaystyle \frac{2}{w-2}}\right\}\right/(2-w)\quad & \text{for }{\mathcal{P}}={\displaystyle \frac{1}{4}};\end{array}\right. & & \displaystyle\end{eqnarray}$$

they are parametrized by a positive constant $C_{+}$ and are defined in the range $w\in (-\infty ,\sqrt{{\mathcal{P}}}^{-1})$ . A typical plot of $P(U_{p})$ is shown in figure 6. The behaviour near the origin on the upper lobe is $P\approx U_{p}^{2}/(4O)$ , and corresponds to $\unicode[STIX]{x1D709}\rightarrow -\infty$ , where $U_{p}^{\prime }>0$ . This matches the expected asymptotic behaviour (3.23). The lower lobe near the origin, on the other hand, corresponds to $\unicode[STIX]{x1D709}\rightarrow \infty$ , and here $U\approx -C_{+}/w$ , so that

(3.29) $$\begin{eqnarray}P\approx -\frac{C_{+}U}{4O\sqrt{{\mathcal{P}}}^{-1}},\quad \unicode[STIX]{x1D709}\rightarrow \infty .\end{eqnarray}$$

Figure 6.

The homoclinic orbit defined by (3.26)–(3.28), with $C_{+}=2\times 4^{4/3}$ and ${\mathcal{P}}=1$ .

Differentiating (3.28) with respect to $\unicode[STIX]{x1D709}$ yields

(3.30) $$\begin{eqnarray}\displaystyle \frac{wU_{p}^{2}}{4\sqrt{{\mathcal{P}}}^{-1}O}=\left\{\begin{array}{@{}ll@{}}\!-C_{+}{\displaystyle \frac{\text{d}w}{\text{d}\unicode[STIX]{x1D709}}}w(\sqrt{{\mathcal{P}}}^{-1}-\,w)^{(2-4{\mathcal{P}})/(4{\mathcal{P}}-1)}(4\sqrt{{\mathcal{P}}}\,-\,w)^{(1-8{\mathcal{P}})/(4{\mathcal{P}}-1)}\!\quad & \text{for }{\mathcal{P}}\not ={\displaystyle \frac{1}{4}},\\ \!-C_{+}{\displaystyle \frac{\text{d}w}{\text{d}\unicode[STIX]{x1D709}}}{\displaystyle \frac{w}{(2-w)^{3}}}\exp \left\{{\displaystyle \frac{2}{w-2}}\right\}\!\quad & \text{for }{\mathcal{P}}={\displaystyle \frac{1}{4}},\end{array}\right. & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

which, upon substituting (3.28) back in, can be integrated to give

(3.31) $$\begin{eqnarray}\displaystyle \unicode[STIX]{x1D709}=\left\{\begin{array}{@{}ll@{}}\left.-{\displaystyle \frac{4O\sqrt{{\mathcal{P}}}^{-1}}{C_{+}}}\displaystyle \int _{0}^{w}\left(4\sqrt{{\mathcal{P}}}-s\right)^{1/(4{\mathcal{P}}-1)}\!\right/\!\left(\sqrt{{\mathcal{P}}}^{-1}-s\right)^{4{\mathcal{P}}/(4{\mathcal{P}}-1)}\,\text{d}s\quad & \text{for }{\mathcal{P}}\not ={\displaystyle \frac{1}{4}},\\ \left.-{\displaystyle \frac{8O}{C_{+}}}\displaystyle \int _{0}^{w}\exp \left\{-{\displaystyle \frac{2}{s-2}}\right\}\right/(2-s)\,\text{d}s\quad & \text{for }{\mathcal{P}}={\displaystyle \frac{1}{4}}.\end{array}\right. & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle\end{eqnarray}$$

In (3.31), the origin $\unicode[STIX]{x1D709}=0$ has been chosen arbitrarily as the point with $U_{p}^{\prime }(0)=0$ and $U_{p}(0)=U_{max}$ , with the maximum given by

(3.32) $$\begin{eqnarray}U_{max}=\left\{\begin{array}{@{}ll@{}}C_{+}/4^{4{\mathcal{P}}/(4{\mathcal{P}}-1)}\quad & \text{for }{\mathcal{P}}\not =\frac{1}{4},\\ C_{+}/(2e)\quad & \text{for }{\mathcal{P}}=\frac{1}{4}.\end{array}\right.\end{eqnarray}$$

Combining (3.28) with (3.31) yields a parametric representation of the pinch profile $U_{p}(\unicode[STIX]{x1D709})$ with respect to the parameter $w\in (-\infty ,\sqrt{{\mathcal{P}}}^{-1})$ (for a typical profile, see figure 7). For $w\rightarrow -\infty$ , (3.31) implies that

(3.33) $$\begin{eqnarray}\unicode[STIX]{x1D709}\approx \frac{4O\sqrt{{\mathcal{P}}}^{-1}}{C_{+}}\ln |w|,\quad \forall ~{\mathcal{P}}>0\end{eqnarray}$$

and since according to (3.28) $U_{p}\approx C_{+}/|w|$ in the same limit, we have

(3.34a,b ) $$\begin{eqnarray}H_{p}\approx \frac{j_{0}}{C_{+}}\text{e}^{(C_{+}\sqrt{{\mathcal{P}}}/4O)\unicode[STIX]{x1D709}},\quad U_{P}\approx C_{+}\text{e}^{-(C_{+}\sqrt{{\mathcal{P}}}/4O)\unicode[STIX]{x1D709}},\quad \forall ~{\mathcal{P}}>0.\end{eqnarray}$$

Note that we can write (3.34) in the original variables as

(3.35) $$\begin{eqnarray}\displaystyle u(x,t) & {\approx} & \displaystyle \frac{1}{\unicode[STIX]{x1D70F}(t)}C_{+}\exp \left\{-\frac{C_{+}\sqrt{{\mathcal{P}}}}{4O}\frac{(x-x_{0})}{\unicode[STIX]{x1D70F}(t)}\right\}\nonumber\\ \displaystyle & = & \displaystyle C_{+}\exp \left\{-\frac{(C_{+}\sqrt{{\mathcal{P}}}/4O)(x-x_{0})+\unicode[STIX]{x1D70F}(t)\log [\unicode[STIX]{x1D70F}(t)]}{\unicode[STIX]{x1D70F}(t)}\right\},\end{eqnarray}$$

a representation that will turn out to be useful in the next subsection for matching to the solutions in the transitional region (cf. (3.51)).

Figure 7.

Similarity description of the pinch region. Five snapshots of the velocity (a,b) and temperature (c,d) profiles taken at times $t_{1}=4.032,~t_{2}=5.138,~t_{3}=5.987,~t_{4}=6.800,~t_{5}=7.6410$ (solid lines). (a,c) Numerical solution using the same simulation as in figure 2. (b,d) Profiles have been rescaled according to (3.17), with exact solutions (3.28) and (3.36) (dash-dotted lines) superimposed.

The temperature profile is found from (3.22) and (3.28) to be

(3.36) $$\begin{eqnarray}\unicode[STIX]{x1D6E9}_{p}=\left\{\begin{array}{@{}ll@{}}\unicode[STIX]{x1D703}_{l}-{\displaystyle \frac{j_{0}C_{+}}{{\mathcal{M}}}}\sqrt{{\mathcal{P}}}\left({\displaystyle \frac{\sqrt{{\mathcal{P}}}^{-1}-w}{4\sqrt{{\mathcal{P}}}-w}}\right)^{4{\mathcal{P}}/(4{\mathcal{P}}-1)}\quad & \text{for }{\mathcal{P}}\not ={\displaystyle \frac{1}{4}},\\ \unicode[STIX]{x1D703}_{l}-{\displaystyle \frac{j_{0}C_{+}}{2{\mathcal{M}}}}\exp \left\{{\displaystyle \frac{2}{w-2}}\right\}\quad & \text{for }{\mathcal{P}}={\displaystyle \frac{1}{4}}.\end{array}\right.\end{eqnarray}$$

In figure 7, the similarity description (3.17) of the pinch region is tested against a typical numerical simulation of the original system (2.2). On the left, we show the raw data close to the pinch point $x_{0}$ , while rescaled profiles are shown on the right. One sees very good convergence towards the exact solutions (3.28) and (3.36), which are shown as the red dash-dotted lines. Note that the temperature experiences a finite variation over the pinch region, which is of size $\unicode[STIX]{x1D70F}$ . This length is much larger than the thickness of the sheet, which is of order $\unicode[STIX]{x1D70F}^{2}$ . Hence information from the boundary diffuses quickly into the bulk, and the temperature can effectively be treated as constant across the sheet, so our long-wavelength theory still applies.

By taking the limit $w\rightarrow -\infty$ , which corresponds to $\unicode[STIX]{x1D709}\rightarrow \infty$ , we find the following condition on the jump of the temperature across the pinch-off (see figure 3):

(3.37) $$\begin{eqnarray}\unicode[STIX]{x1D703}_{r}=\unicode[STIX]{x1D703}_{l}-\frac{j_{0}C_{+}}{{\mathcal{M}}}\sqrt{{\mathcal{P}}}<\unicode[STIX]{x1D703}_{l},\quad \forall ~{\mathcal{P}}>0;\end{eqnarray}$$

in particular, it shows that necessarily $\unicode[STIX]{x1D703}_{r}<\unicode[STIX]{x1D703}_{l}$ . It is thus seen from (3.34) that $H_{p}$ grows exponentially, which does not match the drop profile, which has a finite slope (3.16). This means we need another region between the pinch region II and the drop IV, which we call the transition region.

3.4 Region III: transition region

Here we use the same similarity form (3.17) as before, but the balance is different. On account of flux conservation, we have $\unicode[STIX]{x1D6FC}_{1}+\unicode[STIX]{x1D6FC}_{2}=1$ as before. We also require the transitional solution to match onto the linear drop profile for $\unicode[STIX]{x1D709}\rightarrow \infty$ , which implies that $\unicode[STIX]{x1D6FC}_{1}=\unicode[STIX]{x1D6FD}$ . Finally, we expect surface tension to enter the force balance (2.2b ), so that from a balance between surface tension and viscous forces we have $\unicode[STIX]{x1D6FC}_{1}-3\unicode[STIX]{x1D6FD}=\unicode[STIX]{x1D6FC}_{2}-2\unicode[STIX]{x1D6FD}$ . From these conditions we deduce the exponents $\unicode[STIX]{x1D6FC}_{1}=1$ , $\unicode[STIX]{x1D6FC}_{2}=0$ and $\unicode[STIX]{x1D6FD}=1$ , and the similarity solution becomes

(3.38a-d ) $$\begin{eqnarray}h(x,t)=\unicode[STIX]{x1D70F}H_{t}\left(\unicode[STIX]{x1D709}\right),\quad u(x,t)=U_{t}\left(\unicode[STIX]{x1D709}\right),\quad \unicode[STIX]{x1D703}(x,t)=\unicode[STIX]{x1D6E9}_{t}\left(\unicode[STIX]{x1D709}\right),\quad \unicode[STIX]{x1D709}=\frac{x-x_{1}(t)}{\unicode[STIX]{x1D70F}},\quad\end{eqnarray}$$

where $x_{1}(t)$ denotes the centre of the transition region, which will be shown below to be slightly different from the pinch point $x_{0}$ . The similarity equations corresponding to (3.38) are

(3.39a ) $$\begin{eqnarray}\displaystyle & \displaystyle j_{0}=H_{t}U_{t}, & \displaystyle\end{eqnarray}$$

(3.39b ) $$\begin{eqnarray}\displaystyle & \displaystyle 0=H_{t}^{\prime \prime \prime }-4O\frac{(U_{t}^{\prime }H_{t})^{\prime }}{H_{t}}, & \displaystyle\end{eqnarray}$$

(3.39c ) $$\begin{eqnarray}\displaystyle & \displaystyle 0=(H_{t}\unicode[STIX]{x1D6E9}_{t}^{\prime })^{\prime }. & \displaystyle\end{eqnarray}$$

$O(\unicode[STIX]{x1D70F}^{-2})$

Once more, we insert $U_{t}^{\prime }=-j_{0}H_{t}^{\prime }/H_{t}^{2}$ into (3.39b ) and integrate once, to obtain

(3.40) $$\begin{eqnarray}C_{3}=H_{t}^{\prime \prime }H_{t}-\frac{H_{t}^{\prime 2}}{2}-4Oj_{0}\frac{H_{t}^{\prime }}{H_{t}}.\end{eqnarray}$$

For (3.39c ) to be consistent with a general $H_{t}$ -profile, $\unicode[STIX]{x1D6E9}=\unicode[STIX]{x1D703}_{r}$ must be a constant, which is consistent with (3.37), where we matched the temperature profile in the pinch region directly to the constant value $\unicode[STIX]{x1D703}_{r}$ . In order to match to the constant slope (3.16) for $\unicode[STIX]{x1D709}\rightarrow \infty$ , we must have $H_{t}\approx s\unicode[STIX]{x1D709}$ for $\unicode[STIX]{x1D709}\rightarrow \infty$ , and thus $C_{3}=-s^{2}/2$ .

Rescaling (3.40) according to

(3.41a,b ) $$\begin{eqnarray}H_{t}=\frac{8Oj_{0}}{\sqrt{2}s}f\left(\unicode[STIX]{x1D701}\right),\quad \unicode[STIX]{x1D701}=\frac{s^{2}}{8Oj_{0}}\unicode[STIX]{x1D709},\end{eqnarray}$$

it turns into

(3.42) $$\begin{eqnarray}-1=f^{\prime \prime }f-f^{\prime 2}/2-f^{\prime }/f.\end{eqnarray}$$

Putting $p(f)=f^{\prime }$ , the phase-plane representation of (3.42) is

(3.43) $$\begin{eqnarray}p^{\prime }=\frac{p}{2f}+\frac{1}{f^{2}}-\frac{1}{pf}.\end{eqnarray}$$

In order to match to (3.34), $H_{t}$ must behave exponentially for $\unicode[STIX]{x1D709}\rightarrow -\infty$ , which means that $p\sim f$ near the origin of the $p{-}f$ plane. On the other hand, for $\unicode[STIX]{x1D709}\rightarrow \infty$ we have seen that $H_{t}\approx s\unicode[STIX]{x1D709}$ , and so $p\approx \sqrt{2}$ for $f\rightarrow \infty$ .

Figure 8.

Phase-plane portrait of the ODE (3.43). The solid lines correspond to the orbits and the dash-dotted line is the nullcline $p=0$ .

The nullcline $p^{\prime }=0$ is the curve

(3.44) $$\begin{eqnarray}f=\frac{p}{1-p^{2}/2},\end{eqnarray}$$

which is shown in the phase portrait, figure 8, together with some typical solutions of (3.43). An inspection of the phase plane reveals that there is a unique orbit that approaches the nullcline asymptotically as $f\rightarrow \infty$ . As seen from (3.44), this is the solution that has the right asymptotics for $f\rightarrow \infty$ .

For the solutions shown in figure 8, a more careful analysis at the origin of the phase plane is necessary. Assuming a regular expansion yields the series

(3.45) $$\begin{eqnarray}p=f+\frac{f^{3}}{2}+\cdots \equiv p_{0}(f),\end{eqnarray}$$

which has no free parameters. To find the missing degree of freedom, we put $p(f)=p_{0}(f)+\unicode[STIX]{x1D6FF}(f)$ , and linearize in $\unicode[STIX]{x1D6FF}$ to find

(3.46) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}^{\prime }=\unicode[STIX]{x1D6FF}\left(\frac{2}{f}-\frac{p_{0}^{\prime }}{p_{0}}+\frac{1}{p_{0}f^{2}}\right)\approx \frac{\unicode[STIX]{x1D6FF}}{f^{3}},\end{eqnarray}$$

for small $f$ . Making the Wentzel–Kramers–Brillouin (WKB) ansatz

(3.47) $$\begin{eqnarray}\unicode[STIX]{x1D6FF}=\unicode[STIX]{x1D6FF}_{0}\text{e}^{-Af^{\unicode[STIX]{x1D6FC}}+\cdots },\end{eqnarray}$$

a leading-order balance as $f\rightarrow \infty$ yields $\unicode[STIX]{x1D6FC}=-2$ and $A=1/2$ . Thus close to the origin, we arrive at the representation

(3.48) $$\begin{eqnarray}p=f+\frac{f^{3}}{2}+\cdots +\unicode[STIX]{x1D6FF}_{0}\text{e}^{-1/(2f^{2})},\end{eqnarray}$$

where the degree of freedom is in the parameter $\unicode[STIX]{x1D6FF}_{0}$ . Now one can solve (3.43) by shooting from the origin to infinity, as shown in figure 8. The value of $\unicode[STIX]{x1D6FF}$ is varied until the solution asymptotes to the correct value $p=\sqrt{2}$ .

Figure 9.

Similarity description of the transition region. (a) Five snapshots of the spatial derivative of the height profile $h_{x}(x,t)=H_{t}^{\prime }(\unicode[STIX]{x1D709})$ , using the same simulation as in figure 2, for times $t_{1}=4.032$ , $t_{2}=5.138$ , $t_{3}=5.987$ , $t_{4}=6.800$ , $t_{5}=7.6410$ (solid lines). Data from the transition region are shown in self-similar variables (3.38). Each curve is shifted such that $H_{t}^{\prime }(0)=2$ , which defines $x_{1}(t)$ ; the collapsed profiles are compared to the solution of (3.40) (dotted line), obtained by the shooting method. (b) The corresponding values of $\tilde{\unicode[STIX]{x1D709}}_{i}=(x_{1}(t_{i})-x_{0})/\unicode[STIX]{x1D70F}(t_{i}),~i=1,\ldots ,5$ (dots) compared to the prediction (3.54) (dotted line). The value of $a$ is taken from figure 4. Constant $C_{5}=-\log (\unicode[STIX]{x1D70F}_{0})4O/(C_{+}{\mathcal{P}})$ by (3.54).

Once we have obtained $p(f)=f^{\prime }(\unicode[STIX]{x1D701})$ , we find $f(\unicode[STIX]{x1D701})$ by (numerical) integration and thus $H_{t}(\unicode[STIX]{x1D709})$ from (3.41). The derivative $H_{t}^{\prime }(\unicode[STIX]{x1D709})$ is shown on the left of figure 9 as the red dotted line, and compared to $h_{x}$ , as found from a numerical simulation of the full system (2.2). Allowing for a horizontal shift (which determines $x_{1}(t)$ , see below), excellent agreement is found.

For small $f$ ( $\unicode[STIX]{x1D709}\rightarrow -\infty$ ), $f^{\prime }=f$ yields

(3.49) $$\begin{eqnarray}f\approx f_{0}\text{e}^{\unicode[STIX]{x1D701}}=f_{0}\text{e}^{(s^{2}/8Oj_{0})\unicode[STIX]{x1D709}};\end{eqnarray}$$

which transforms to the asymptotics

(3.50a,b ) $$\begin{eqnarray}H_{t}\approx B\text{e}^{(s^{2}/8Oj_{0})\unicode[STIX]{x1D709}}\quad \text{and}\quad U_{t}\approx \frac{j_{0}}{B}\text{e}^{-(s^{2}/8Oj_{0})\unicode[STIX]{x1D709}}\quad \text{as}~\unicode[STIX]{x1D709}\rightarrow -\infty ,\end{eqnarray}$$

as well as

(3.51) $$\begin{eqnarray}u(x,t)\approx \frac{j_{0}}{B}\exp \left\{-\frac{s^{2}}{8Oj_{0}}\frac{x-x_{1}(t)}{\unicode[STIX]{x1D70F}(t)}\right\}\end{eqnarray}$$

in the original variables. Comparing to the asymptotics in the pinch-off region (3.35), we find the following matching conditions:

(3.52) $$\begin{eqnarray}\displaystyle & \displaystyle C_{+}=\frac{s^{2}}{2j_{0}\sqrt{{\mathcal{P}}}}, & \displaystyle\end{eqnarray}$$

(3.53) $$\begin{eqnarray}\displaystyle & \displaystyle B=\frac{j_{0}}{C_{+}}, & \displaystyle\end{eqnarray}$$

(3.54) $$\begin{eqnarray}\displaystyle & \displaystyle \tilde{\unicode[STIX]{x1D709}}(t)=\frac{x_{1}(t)-x_{0}}{\unicode[STIX]{x1D70F}(t)}=-\log (\unicode[STIX]{x1D70F}(t))\frac{4O}{C_{+}{\mathcal{P}}}=-\log (\unicode[STIX]{x1D70F}_{0})\frac{4O}{C_{+}{\mathcal{P}}}+\frac{16O^{2}a}{C_{+}\sqrt{{\mathcal{P}}}}t>0, & \displaystyle\end{eqnarray}$$

where in the last equality in (3.54) we have used (3.5). On the right of figure 9 we test (3.54), shown as the dotted line, against the shift $x_{1}(t)$ , as obtained from a direct numerical simulation (dots). For large times, the dots are seen to approach the theoretical prediction.