2.1 Standard long-wavelength equations

We consider threads with small aspect ratio ${\it\epsilon}=\sqrt{A_{min}/({\rm\pi}L_{0}^{2})}\ll 1$ , where $A_{min}$ is the initial minimum cross-sectional area. To leading order in ${\it\epsilon}$ , it has been shown (Fitt et al. Reference Fitt, Furusawa, Monro and Please2001) that, the axial fluid velocity, $u(x,t)$ , is independent of the radial coordinate and the long-wavelength Navier–Stokes and continuity equations are given by

and

respectively. Here, $A(x,t)$ is the cross-sectional area that is defined for $0\leqslant x\leqslant L(t)$ . Using the symmetry condition at $x=0$ , the boundary conditions are given by

The initial conditions are given by

We now adopt non-dimensional variables (denoted by bars) defined by

After substituting into the governing equations and dropping the bars, we obtain

where

is the Reynolds number based on the initial length of the thread, and

is a dimensionless capillary number that measures the surface tension force relative to the viscous force. Using (2.8), the Navier–Stokes equation (2.7) may be rewritten in the form

where

denotes the material time derivative. The initial and boundary conditions (2.4)–(2.5) remain unchanged under the non-dimensionalisation. Note that $A_{0}(x)\geqslant 1$ since the cross-sectional area is scaled by the minimum cross-sectional area.

We now consider the case $R\ll 1$ and ${\it\Gamma}\ll 1$ , so that, initially, inertia and surface tension can be neglected. However, we note that the dimensionless estimates of $R$ and ${\it\Gamma}$ are based on the initial length and cross-sectional area of the thread. If the thread becomes sufficiently thin and heavily extended, it is possible that these estimates do not adequately represent the importance of inertial and surface tension effects. This is the case for a thread pulled by a constant force, for which Wylie et al. (Reference Wylie, Huang and Miura2011) showed that even though $R\ll 1$ , inertia must ultimately become important. On the other hand, they were able to use non-trivial asymptotic estimates to show that if ${\it\Gamma}\ll 1$ , then surface tension forces remain negligible throughout the entire stretching process. We therefore proceed by setting ${\it\Gamma}=0$ . The conditions which justify this assumption are discussed in the closing remarks.

Before continuing, we briefly comment on the nature of the initial conditions. To do this we need to address the question of how, in a physical context, the shape of the thread immediately prior to the stretching process arises. The fluid thread must first be attached to the two ends at which the pulling is applied. Following the attachment process, there may be a period of time before the pulling commences. If this time is sufficiently short compared with the time scale for surface tension to significantly deform the thread, then the shape will remain undeformed by surface tension. On the other hand, if the waiting time is sufficiently long, then surface tension will modify the thread shape until eventually a minimal energy surface is achieved. For materials like polymers or glasses that are effectively solid at room temperature, external heating must be applied until the viscosity of the material decreases sufficiently to allow stretching with reasonable extensional forces. After heating, these materials typically have a viscosity of the order $10^{4}~\text{Pa}~\text{s}$ (and possibly many orders of magnitude larger) and surface tension coefficients of order $10^{-1}~\text{N}~\text{m}^{-1}$ . Using these estimates and considering threads with radius of order $10^{-2}~\text{m}$ , the time scale required for surface tension to modify the shape is of order $10^{3}~\text{s}$ . So, as long as the waiting time is limited to a few minutes, surface tension will play a negligible role in determining the initial condition and a wide range of initial shapes are feasible.

2.2 Transformed equations

We now introduce a transformation that significantly simplifies the form of the equations. Following Bradshaw-Hajek et al. (Reference Bradshaw-Hajek, Stokes and Tuck2007), we introduce the variable

Since the thread is symmetric about $x=0$ , $I$ represents half of the total volume of the fluid (and remains constant throughout the stretching process). The variable $Z(x,t)$ is the fraction of the volume between 0 and $x$ . Given $Z(x,t)$ , one can readily calculate the cross-sectional area $A(x,t)$ via

which is obtained by partially differentiating (2.13a ) with respect to  $x$ .

Using conservation of mass (2.3), one can readily check that

This shows that $Z$ is constant when following any given fluid element. From a physical viewpoint this is completely natural, since the amount of fluid between the symmetry axis $x=0$ and a given fluid element must be conserved. In some sense, $Z$ can therefore be considered as a Lagrangian variable, although it is important to note that, in terms of the equations we will derive, $Z$ will be a dependent, rather than an independent variable. (However, for the asymptotics of § 4.2 it will prove useful to use $Z$ as a Lagrangian independent variable.)

Using (2.15), the velocity $u$ can be written as

where the subscript denotes partial differentiation. Substituting (2.16) and (2.14) into (2.11) with ${\it\Gamma}=0$ , we obtain

After some straightforward algebra, this can be expressed as

Integration of (2.18) with respect to the material time introduces a general function of the Lagrangian variable $Z$ . The function is obtained using the initial condition (2.5) and (2.13b ) to give

where $s_{0}(Z)$ is the initial shape of the thread expressed as a function of $Z$ . Since $Z(x,t)$ is the fraction of the volume between 0 and $x$ , the boundary conditions are

We conclude this section with a brief note regarding the relationship between $A_{0}(x)$ and $s_{0}(Z)$ that represent the initial shape of the thread specified in Eulerian and Lagrangian coordinates, respectively. It is more physically natural to specify $A_{0}(x)$ , but since $s_{0}(Z)$ appears in the governing equation for $Z$ (2.19), for the purpose of analysis, it is often more convenient to specify $s_{0}(Z)$ . On the one hand, if $A_{0}(x)$ is specified, one can use (2.13) with $t=0$ to obtain $Z\equiv Z(x)$ . Inverting this function allows us to express $x$ as a function of $Z$ , $x\equiv x(Z)$ . Substituting this into $A_{0}(x)$ we obtain $s_{0}(Z)$ . On the other hand, if $s_{0}(Z)$ is specified, one can calculate $I$ from (2.14) with $t=0$ , writing $1/I=Z_{x}/s_{0}(Z(x))$ and integrating to give

Then, since $Z=1$ corresponds to $x=1$ at $t=0$ , we have

One can then use (2.21) to obtain $x\equiv x(Z)$ . Inverting this function allows us to express $Z$ as a function of $x$ , $Z\equiv Z(x)$ . Substituting this into $s_{0}(Z)$ we obtain $A_{0}(x)$ .

3.1 Case 1: initially cylindrical thread

In this case, $s_{0}(Z)=A_{0}(x)=1$ , $I=1$ , and $Z(x,0)=x$ , $0\leqslant x\leqslant 1$ (via (2.13a )), so that (2.19) reduces to

to be solved subject to boundary conditions

The leading-order solution satisfies $Z_{0xx}=0$ and, using the boundary conditions, we find

Substituting for $Z_{0}$ in (3.4) gives

to be solved subject to boundary conditions $Z_{1}(0,t)=Z_{1}(L(t),t)=0$ . We find

Hence, to first order in $R$ , we have the solution

from which we can find the cross-sectional area, $A=IZ_{x}$ .

We must, however, ensure that the series for $A$ remains asymptotically valid, i.e. that the second term in the series for $Z_{x}$ is smaller than the first. The ratio

has largest magnitude at the end of the thread ( $x=L$ ), so that in order for the series to remain valid, we require

At early times, this quantity is small and the series is asymptotically valid. However, for large times, the validity depends on the asymptotic behaviour of $L$ . If $L\sim t^{{\it\alpha}}$ as $t\rightarrow \infty$ , then we require $RL\dot{L}\sim Rt^{2{\it\alpha}-1}\ll 1$ for large $t$ . This quantity gets smaller as $t$ increases for ${\it\alpha}<1/2$ , and for ${\it\alpha}=1/2$ it will be a small constant of order $R$ . Hence, this small-inertia solution will be valid for all time if ${\it\alpha}\leqslant 1/2$ . However, if ${\it\alpha}>1/2$ then the second term in the expansion will eventually become larger than the first term, meaning that at large times, inertia becomes important. We will consider the solution when inertia is non-negligible in § 4.

Figure 2.

Comparison between the numerical solution (solid) and the small-inertia asymptotic solution (3.13) (dashed), with $L(t)=(1+t)^{{\it\alpha}}$ and $R=0.2$ for an initially cylindrical thread (dash-dot). The radius $\sqrt{A(x,t)/{\rm\pi}}$ is plotted against the physical coordinate $x$ for (a)  ${\it\alpha}=0.4$ , ${\it\lambda}=200$ , $k=5,6,\ldots ,12$ , (b)  ${\it\alpha}=0.8$ , ${\it\lambda}=100$ , $k=4,5,\ldots ,12$ . The extension of the thread doubles between subsequent curves (3.16). A better visualisation of these results is obtained by plotting cross-sectional area $s(Z,t)$ as a function of the Lagrangian variable $Z$ for the same parameters in (c,d). The asymptotic and numerical solutions are indistinguishable in (a,c), while the small-inertia solution can be seen to fail near the pulled end of the thread in (b,d).

Figure 2 compares the small-inertia solution (3.13) (dashed) with the numerical solution of (2.19) and (2.20) (solid) for $L(t)=(1+t)^{{\it\alpha}}$ for two different values of ${\it\alpha}>0$ . The numerical results were obtained using an implicit backward-time-centred-space (BTCS) finite difference scheme. The details are given in appendix A. In some sense, it is most physically natural to plot the the radius $\sqrt{A/{\rm\pi}}$ as a function of $x$ as done in figure 2(a,b). As is apparent, such plots have the problem that it becomes difficult to see the curves that represent early times when the thread becomes highly extended. We therefore plot the cross-sectional area $s(Z,t)$ as a function of the Lagrangian coordinate $Z$ ; see (c,d). (Note that here, and throughout the remainder of the paper, $s(Z,t)\equiv A(x(Z,t),t)$ .) This has the advantage of fixing the domain between $Z=0$ and $Z=1$ at all times. The thread also becomes very thin during the stretching and we therefore use a logarithmic scale to plot $s(Z,t)$ . Curves are plotted at times $t$ when the thread extension is given by

where ${\it\lambda}^{-1}$ is a chosen fraction of the initial length. Thus, the extension doubles between consecutive curves. The initial thread shape, $s_{0}(Z)=1$ is shown with a dash-dot line. In figure 2(a,c), we set ${\it\alpha}=0.4$ . As expected, the numerical solution and the small-inertia solution are indistinguishable from each other for all time. In figure 2(b,d), we set ${\it\alpha}=0.8$ . In this case the small-inertia solution fails near the pulled end of the thread at large time, as expected from our analysis. The final time in figure 2(a,c) is much greater than that in figure 2(b,d). The time corresponding to curve $k$ is $t_{k}=(1+2^{k}/{\it\lambda})^{1/{\it\alpha}}-1$ .

Figure 3.

Comparison between the numerical solution (solid) and the small-inertia asymptotic solution (3.23) (dashed), with $L(t)=(1+t)^{{\it\alpha}}$ and $R=0.2$ for a thread that has its initial minimum at the pulled end (dash-dot). The cross-sectional area $s(Z,t)$ is plotted as a function of the Lagrangian variable $Z$ . The extension of the thread doubles between subsequent curves (3.16). Parameters are (a)  ${\it\alpha}=0.8$ , ${\it\lambda}=100$ , $k=3,4,\ldots ,9$ , (b)  ${\it\alpha}=2$ , ${\it\lambda}=100$ , $k=3,4,\ldots ,9$ . The asymptotic and numerical solutions are indistinguishable in (a), while the small inertia solution can be seen to fail near the pulled end of the thread in (b).

3.2 Case 2: thread with initial minimum at the pulled end

Next we consider a thread with initial shape $s_{0}(Z)=1+K(1-Z)$ . This thread shape has an initial minimum cross-sectional area at $Z=1$ and is shown for $K=1$ in figure 3 (dash-dot curve). Using (2.22) we find

and using (2.14) one can readily show that the initial condition in Eulerian coordinates is given by $A_{0}(x)=(1+K)^{1-x}$ , for $0\leqslant x\leqslant 1$ .

Substituting this initial thread shape into (2.19) yields

and the leading-order solution $Z_{0}$ satisfies

which has solution

The first-order correction satisfies

Integrating twice with respect to $x$ and satisfying the boundary conditions $Z_{1}=0$ at $x=0$ and $x=L(t)$ we find

Hence, to first order in $R$

The cross-sectional area is readily calculated via $A=IZ_{x}$ .

Once again, we find the conditions under which the series for the cross-sectional area remains asymptotically valid by taking the ratio of the second and first terms:

We see that this expression is largest where $x=L(t)$ , and at this location

At early times,

and the series is asymptotically valid. However, as for the cylindrical case, the validity at large times depends on the asymptotic behaviour of $L$ . If $L\sim t^{{\it\alpha}}$ as $t\rightarrow \infty$ , then $R\dot{L}/\mathscr{K}\sim Rt^{{\it\alpha}-1}/\mathscr{K}$ . This ratio gets smaller as $t$ increases for ${\it\alpha}<1$ and for ${\it\alpha}=1$ it will be a small constant of order $R$ . Thus, this small-inertia solution (3.23) will be valid for all time if ${\it\alpha}\leqslant 1$ . However, if ${\it\alpha}>1$ , the second term will, over time, dominate the first and the solution will no longer be asymptotically valid. Physically this is because inertia becomes important. We defer examination of this to § 4.

Figure 3 compares the small-inertia asymptotic solution (3.23) (dashed) with the numerical solution (solid). Once again, we plot the cross-sectional area (on a log scale) as a function of the Lagrangian variable so as to clearly show the change in the geometry of the thread as it becomes extremely long and thin. Curves are shown at times satisfying (3.16) for ${\it\lambda}=100$ , $k=0,1,\ldots ,9$ , and the dash-dot curve shows the initial thread shape.

Figure 3(a) shows the behaviour when ${\it\alpha}=0.8$ . The small-inertia and numerical solutions are almost indistinguishable, even when the extension is very large. In figure 3(b), we plot the behaviour when ${\it\alpha}=2$ . In this case, the small-inertia solution deviates from the numerical solution when the extension becomes large at long times. The failure of the small-inertia solution occurs at the pulled end of the thread. We note that, because of the different values of ${\it\alpha}$ , the final time in figure 3(a) is much greater than that in figure 3(b), although the final thread lengths are the same.

3.3 Case 3: thread with initial minimum away from the pulled end

The last case we consider is $s_{0}(Z)=1+KZ^{2}$ with $K>0$ . This represents an initial thread shape that has the minimum cross-sectional area at $Z=0$ . This initial shape is shown as a dash-dot line in figure 4. Using (2.22) and (2.14), we find that

The corresponding initial thread shape in Eulerian coordinates is $A_{0}(x)=1+\tan ^{2}(x\tan ^{-1}(\sqrt{K}))$ , for $0\leqslant x\leqslant 1$ .

For this initial geometry, (2.19) reduces to

The leading-order solution $Z_{0}$ satisfies

By integrating and making use of the boundary conditions, we find

where $f(t)$ satisfies the algebraic equation

The $O(R)$ correction satisfies

On substituting $Z_{0}$ , integrating twice with respect to $x$ and satisfying the boundary conditions, we find $Z_{1}$ . Hence, the solution for $Z$ to first order in $R$ is

where the dot denotes differentiation with respect to $t$ .

Once again, we ensure the asymptotic validity of the series for the cross-sectional area by examining the ratio

Using (3.31) and the fact that $f$ becomes small and $\mathscr{K}L$ becomes large at large times, one can show that $RA_{1}/A_{0}$ is largest where $x=0$ . At this location

At early times,

and the series is asymptotically valid. However, at large times, the validity depends on the asymptotic behaviour of $L$ . From (3.31), we note that $f=O(1/(\mathscr{K}L))$ as $L\rightarrow \infty$ . If $L\sim t^{{\it\alpha}}$ as $t\rightarrow \infty$ , then $R{\dot{f}}/f^{3}=O(Rt^{2{\it\alpha}-1})$ . This quantity gets smaller for ${\it\alpha}<1/2$ and for ${\it\alpha}=1/2$ it will be a small constant. Thus, this small-inertia solution will be asymptotically valid for all time if ${\it\alpha}\leqslant 1/2$ . For larger values of ${\it\alpha}$ , the second term will eventually become larger than the first and so (3.33) loses validity. We consider ${\it\alpha}>1/2$ at large time in § 4.

Figure 4 compares the numerical solution (solid) with the small-inertia solution (3.33) (dashed). As before, we plot the log of the cross-sectional area as a function of the Lagrangian coordinate to enable better comparison of the curves as the extension becomes large and the thread becomes thin. The extension doubles between successive curves and the initial thread shape is shown by a dash-dot curve.

Figure 4.

Comparison between the numerical solution (solid) and the small-inertia asymptotic solution (3.33) (dashed), with $L(t)=(1+t)^{{\it\alpha}}$ and $R=0.2$ for a thread that has its initial minimum away from the pulled end (dash-dot). The cross-sectional area $s(Z,t)$ is plotted as a function of the Lagrangian variable $Z$ . The extension of the thread doubles between subsequent curves (3.16). Parameters are (a)  ${\it\alpha}=0.4$ , ${\it\lambda}=200$ , $k=5,6,\ldots ,12$ , (b)  ${\it\alpha}=0.8$ , ${\it\lambda}=10$ , $k=4,5,\ldots ,14$ . The asymptotic and numerical solutions are indistinguishable in (a), while the small inertia solution can be seen to fail near the centre ( $Z=x=0$ ) of the thread in (b).

Figure 4(a) shows the thread behaviour for ${\it\alpha}=0.4$ and figure 4(b) shows the behaviour for ${\it\alpha}=0.8$ . For ${\it\alpha}=0.4$ , the small-inertia solution is indistinguishable from the numerical solution for all times. In fact, closer examination of the curves shows that as time increases, the relative difference between the numerical and asymptotic solutions decreases. In contrast, for the case in which ${\it\alpha}=0.8$ , the asymptotic solution is seen to fail at large times where the relative difference between it and the numerical solution becomes significant. This failure occurs near $x=0$ , where the cross-sectional area is initially at a minimum. These results are fully consistent with the theory which states that inertia can only become important for values of ${\it\alpha}>1/2$ .

4.1 Case 1: initially cylindrical thread

In the case of an initially cylindrical thread, the critical value is ${\it\alpha}=1/2$ and we look for solutions of

where $L(t)\sim t^{{\it\alpha}}$ at large times and ${\it\alpha}>1/2$ .

Using heat kernels, the fundamental solution that satisfies the boundary condition $Z=0$ at $x=0$ is (Carslaw & Jaeger Reference Carslaw and Jaeger1959)

where

and the function ${\it\phi}({\it\xi})$ must be chosen to satisfy the initial condition and the condition at the pulled end.

Taking the limit of (4.2) as $t\rightarrow 0$ , we see that the kernel becomes the sum of two delta functions,

and since $Z=x$ for $0\leqslant x\leqslant 1$ at $t=0$ , this allows us to determine that ${\it\phi}({\it\xi})={\it\xi}$ for $0\leqslant {\it\xi}\leqslant 1$ .

To find ${\it\phi}({\it\xi})$ for ${\it\xi}>1$ , we apply the boundary condition at the pulled end $x=L(t)$ , so that we require

which is a Fredholm integral equation of the first kind. It is more convenient to rewrite this equation in terms of $\widetilde{{\it\phi}}$ where ${\it\phi}=1+\widetilde{{\it\phi}}$ . Using the fact that $\int _{0}^{\infty }[\text{e}^{-({\it\xi}-a)^{2}}-\text{e}^{-({\it\xi}-a)^{2}}]\,\text{d}{\it\xi}=\sqrt{{\rm\pi}}\,\text{erf}(a)$ , and $\widetilde{{\it\phi}}={\it\xi}-1$ for $0<{\it\xi}<1$ , we obtain

where $\text{erfc}(x)$ is the complementary error function.

In the following discussion, we will show that $\widetilde{{\it\phi}}({\it\xi})\rightarrow 0$ as ${\it\xi}\rightarrow \infty$ . We then use this, together with (4.2) to obtain an asymptotic solution for $Z(x,t)$ at large times.

At large times, $L(t)$ becomes large and the second exponential term in the kernel can be neglected as it represents a Gaussian centred at ${\it\xi}=-L(t)$ which is far from the area of interest, ${\it\xi}\in [1,\infty )$ . Thus, the kernel is approximately a Gaussian, centred at ${\it\xi}=L(t)$ with width $O(\sqrt{t/R})$ . Since $L\gg t^{1/2}$ , the kernel is such that at large $t$ , the contribution to the integral from values of ${\it\xi}=O(1)$ is small. Whilst the following argument can be couched in terms of operator theory, it is perhaps more straightforward to discuss it in terms of a discretised problem in matrix form. To discretise the integral on the left-hand side of (4.6), we rewrite it as a matrix, $\unicode[STIX]{x1D648}$ , multiplied by a column vector of unknowns, $\widetilde{\unicode[STIX]{x1D731}}$ . The right-hand side of the equation is a known column vector,  $\boldsymbol{B}$ .

The matrix $\unicode[STIX]{x1D648}$ is constructed using the kernel, $K({\it\xi},L,t;R)$ . As discussed above, this kernel is approximately Gaussian, centred at $L(t)\gg t^{1/2}$ with a width which broadens like $t^{1/2}$ . As a result, if one chooses an appropriate discretisation for ${\it\xi}$ and $t$ , it can readily be seen that $\unicode[STIX]{x1D648}$ is dominated by the entries on its diagonal. Moreover, the matrix entries decay extremely rapidly as one moves away from the diagonal.

The vector $\boldsymbol{B}$ is constructed using the right-hand side of (4.6). After some straightforward calculation, one can see that the values of $\boldsymbol{B}$ are $O(1)$ at the top of the vector, but decay to zero in a Gaussian manner as one moves down the vector. As mentioned above, $\unicode[STIX]{x1D648}$ is dominated by the entries on its diagonal and has entries that decay rapidly as one moves away from the diagonal. As a consequence, the inverse matrix $\unicode[STIX]{x1D648}^{-1}$ also has the same form (Demko, Moss & Smith Reference Demko, Moss and Smith1984). As a result of these two properties, $\widetilde{\unicode[STIX]{x1D731}}$ will decay to zero as one moves down the vector. In terms of functions, this result corresponds to $\widetilde{{\it\phi}}({\it\xi})\rightarrow 0$ or ${\it\phi}({\it\xi})\rightarrow 1$ as ${\it\xi}\rightarrow \infty$ .

In order to evaluate $Z(x,t)$ , we now examine (4.2). For large times, the kernel broadens like $t^{1/2}$ , so that the contribution to the integral from ${\it\xi}=O(1)$ becomes less significant as $t\rightarrow \infty$ . This means that the integral can be approximated by setting ${\it\phi}=1$ , which gives the leading-order solution

To calculate the cross-sectional area, we use $A=IZ_{x}$ to obtain

Figure 5 compares the error function solution (4.7) (dashed) to the numerical solution (solid) for the initially cylindrical thread at large times, for ${\it\alpha}>1/2$ . As before, we plot the cross-sectional area on a log scale and as a function of the Lagrangian coordinate. The extension doubles between subsequent curves. The figure shows that at very large times when the thread is very thin, the error function solution is an extremely good approximation to the numerical solution. The error function solution has not been plotted at early times when the agreement is poor.

Figure 5.

Comparison between the numerical solution (solid) and the large-time asymptotic solution (4.7) (dashed) with $L(t)\sim t^{{\it\alpha}}$ , ${\it\alpha}=0.8$ and $R=0.2$ . The cross-sectional area $s(Z,t)$ is plotted as a function of the Lagrangian variable $Z$ . Curves are plotted at times given by (3.16) with ${\it\lambda}=20$ . The numerical solution is shown for $k=4,5,\ldots ,16$ while the asymptotic solution is shown for $k=11,12,\ldots ,16$ . The initial condition is shown with a dash-dot line. As time increases, the asymptotic solution better approximates the numerical solution.

It is of interest to examine the form of the asymptotic correction to the solution. To do this, one must determine $\widetilde{{\it\phi}}$ from (4.6). The right-hand side of (4.6) contains two terms: a complementary error function and an integral involving the initial condition. As $t\rightarrow \infty$ ,

and

The ratio of these two terms has the following asymptotic form

In the case $L\gg t$ ( ${\it\alpha}>1$ ), the ratio becomes exponentially large as $t\rightarrow \infty$ . For $L\ll t$ ( ${\it\alpha}<1$ ), the ratio grows like $t^{1-{\it\alpha}}/R$ as $t\rightarrow \infty$ . Finally, in the case $L\sim t$ ( ${\it\alpha}=1$ ), the ratio tends to a large constant of size $O(1/R)$ . In all three cases, at large times, the integral involving the initial condition is larger than the complementary error function and so the right-hand side of (4.6) will eventually be dominated by the integral involving the initial conditions. Thus, even for large ${\it\xi}$ , the asymptotic correction $\widetilde{{\it\phi}}$ will also depend on the initial conditions, and as a result, one cannot find an asymptotic correction that is only determined by the local behaviour near the pulled end. For diffusive problems of this type, the effects of the initial condition decay exponentially quickly as time goes to infinity. However, for this problem, the leading-order asymptotic correction is also exponentially small and so the initial condition remains significant indefinitely.

4.2 Case 2: thread with initial minimum at the pulled end

Next we consider a thread in which the initial minimum occurs at the pulled end, i.e.  $s_{0}(Z)=1+K(1-Z)$ , in the case in which inertia eventually becomes important, i.e.  $L\sim t^{{\it\alpha}}$ for ${\it\alpha}>1$ . Figure 3(b) demonstrates that for values of ${\it\alpha}>1$ , the small-inertia solution fails near the pulled end of the thread. In the following, we show that the small-inertia solution is valid in the main part of the thread, even at large times.

It proves to be convenient to rewrite the governing equation (2.19) in terms of coordinates $(Z,{\it\tau})$ , rather than $(x,t)$ , where ${\it\tau}=t$ and the Lagrangian variable $Z$ is an independent variable. In addition, fluid elements travel with the fluid velocity so that $u=x_{{\it\tau}}$ . Multiplying the governing equation (2.19) by $I$ , using (2.14) and (2.16), we obtain

Partially differentiating with respect to $Z$ and using $u=x_{{\it\tau}}$ and (2.14), we obtain

A more detailed derivation of this equation can be found in Wylie et al. (Reference Wylie, Huang and Miura2011).

At large times and in regions away from the pulled end, the right-hand side of (4.13) is $O(1)$ . Since $s$ is also $O(1)$ here, and $R$ is small, the left-hand side is small. From this, it follows that in regions away from the pulled end, $(s_{0}-s)_{ZZ}=0$ approximately describes the evolution of the thread. This precisely corresponds to the small-inertia problem (3.18) of § 3.2. We therefore conclude that the small-inertia solution is valid for a large portion of the thread, even at large times and when ${\it\alpha}>1$ . This can be confirmed by examining figure 3.

In contrast, at large times and near the end of the thread, $s$ becomes small so that the left-hand side of (4.13) becomes significant. In fact, inertia will be important in regions near the end of the thread where $s=O(R)$ . In this case, we must solve (4.13) in its entirety, together with the appropriate initial and boundary conditions.

To examine how the thread shape changes near the pulled end, we look for a large-time asymptotic solution there. Using the transformation $p(x,t)=\text{e}^{\mathscr{K}x/2+\mathscr{K}^{2}t/4R}Z(x,t)$ , the governing equation (3.18) can be rewritten as the heat equation, however the initial and boundary conditions become more complicated. Using the same technique as that presented in § 4.1, we find that, in the original coordinates, the leading-order term at large times near the end of the thread is $Z\sim 1$ . As in the case of the cylinder, we find that the dominant contribution to the first correction term involves the initial condition. Since the influence of the initial condition does not become negligible near the end of the thread as $t\rightarrow \infty$ , it is not possible to find an asymptotic correction that is determined only by the local behaviour at the end.

Therefore, in summary, the majority of the stretching occurs in fluid elements that originated near the pulled end. For these fluid elements, inertia is important. However, for fluid elements that originated in the main portion of the thread, inertia is always negligible so that the small-inertia solution is valid, even at large times.

4.3 Case 3: thread with initial minimum away from the pulled end

For the case $s_{0}(Z)=1+KZ^{2}$ , the governing equation is

where $\mathscr{K}$ is defined in § 3.3. The boundary conditions are $Z(0,t)=0$ and $Z(L(t),t)=1$ , with $L(t)\sim t^{{\it\alpha}}$ at large times. The small-inertia solution given by (3.30) and (3.33), whilst valid for all times when ${\it\alpha}\leqslant 1/2$ , was shown to fail at large times when ${\it\alpha}>1/2$ . We therefore now consider the case $L(t)\gg t^{1/2}$ .

From (3.30) and (3.33), we anticipate that $s(Z,t)$ will become small in the middle of the thread. In this case, the natural scalings of (4.14) suggest that we make the transformation

where

Writing (4.14) in terms of these new variables yields

with boundary conditions

We proceed by neglecting the ${\it\tau}G_{{\it\tau}}$ term in (4.17). A priori it is unclear that this term is negligible, but after obtaining the solution, we will verify that it is indeed the case. Under this assumption, (4.17) becomes an ordinary differential equation in ${\it\eta}$ with ${\it\tau}$ as a parameter. After integrating with respect to ${\it\eta}$ , we obtain

where $c({\it\tau})$ is an arbitrary function of ${\it\tau}$ . The solution to (4.19) can be obtained in terms of Whittaker functions and is given by

where $D({\it\tau})$ is an arbitrary function of ${\it\tau}$ . Applying the boundary condition $G(0)=0$ and making use of the small ${\it\eta}$ asymptotics for the Whittaker function, one obtains

where ${\it\Gamma}$ is the gamma function. Substituting this back into (4.20), we obtain

This has asymptotic form

Since, we require that $G>0$ , we must restrict our attention to $c>-1/4$ . In figure 6(a), we illustrate that the function $G$ must exhibit one of three different types of behaviour for different values of $c$ . For $c<1/4$ , $G$ increases to a maximum value before eventually decaying to zero. For $c>1/4$ , $G$ has a singularity at a finite value of ${\it\eta}={\it\eta}_{\ast }$ and is a monotonic increasing function for $0<{\it\eta}<{\it\eta}_{\ast }$ . For $c=1/4$ , it is easy to check that the solution is given by the straight line $G={\it\eta}$ . The solution $G={\it\eta}$ separates the two solutions types for $c>1/4$ and $c<1/4$ . As $c\rightarrow 1/4^{+}$ , the value of ${\it\eta}_{\ast }\rightarrow \infty$ , whereas as $c\rightarrow 1/4^{-}$ , the location of the maximum tends to infinity. In both cases, for any fixed value of ${\it\eta}$ , $G\rightarrow {\it\eta}$ as $c\rightarrow 1/4$ .

Figure 6.

(a) Behaviour of the scaled Lagrangian variable $G({\it\eta})$ (4.22), for different values of  $c$ . (b) A comparison of the asymptotic approximation (4.27) for $G({\it\eta})$ (dashed) with (4.22) (solid) for values of $c$ close to  $1/4$ .

We need to apply a boundary condition at ${\it\eta}=R^{1/2}L({\it\tau})/(2{\it\tau}^{1/2})$ . Since we are considering the case $L({\it\tau})\gg {\it\tau}^{1/2}$ , this value will become large in the long-time limit. In the case $c>1/4$ , we therefore require that the singularity at ${\it\eta}_{\ast }({\it\tau})$ is not in the domain $0\leqslant {\it\eta}\leqslant R^{1/2}L({\it\tau})/(2{\it\tau}^{1/2})$ . This implies that ${\it\eta}_{\ast }({\it\tau})$ must become large as ${\it\tau}\rightarrow \infty$ . On the other hand, in the case $c<1/4$ , we require that the location of the maximum is also not in the domain. This is because $G$ must be a monotonically increasing function of ${\it\eta}$ since $Z$ is, by definition, a monotonically increasing function of $x$ . These conditions can only be true if $c({\it\tau})\rightarrow 1/4$ as ${\it\tau}\rightarrow \infty$ . We therefore focus our attention on the $c({\it\tau})\rightarrow 1/4$ limit.

In the case $c({\it\tau})\rightarrow 1/4^{+}$ , we can obtain the location of the singularity ${\it\eta}_{\ast }$ by finding the zero of the denominator of the second fraction in (4.22), i.e. by solving

Taking the limit as $c\rightarrow 1/4$ and using the large ${\it\eta}$ asymptotics of the Whittaker functions, we obtain

We now obtain an approximation for (4.22), by taking the limit as $c({\it\tau})\rightarrow 1/4$ and noting that the expression should be valid for values of ${\it\eta}$ close to ${\it\eta}_{\ast }$ , that is

In the above expression, we have retained the term $(c-1/4)M_{1/4,1/4}({\it\eta}^{2})$ , even though it is $O(c-1/4)$ smaller than the term $W_{5/4,1/4}({\it\eta}^{2})$ for ${\it\eta}=O(1)$ . This is because near ${\it\eta}={\it\eta}_{\ast }$ , the two terms are of similar order and it is the cancelation between these two terms that gives the singularity in (4.22). The Whittaker functions in (4.26) can be expressed in terms of simple functions to give

where $\text{erfi}(x)$ is the imaginary error function. We note that (4.27) is also valid for $c\rightarrow 1/4^{-}$ . The comparison between the asymptotic approximation (4.27) and (4.22) for ${\it\eta}$ near to $1/4$ is shown in figure 6(b). The asymptotic approximation shows good agreement even for relatively moderate values of  ${\it\eta}_{\ast }$ .

In order to determine the value of $c({\it\tau})$ or equivalently ${\it\eta}_{\ast }({\it\tau})$ , we need to satisfy the boundary condition (4.18b ). To satisfy this condition, we require that $G\gg {\it\eta}$ , or

so that $L\ll t/R$ . Thus, we now assume that ${\it\alpha}\leqslant 1$ (so that $t^{1/2}\ll L\ll t$ ) and return to the case where ${\it\alpha}>1$ later. The only solutions with $G>{\it\eta}$ are those with $c({\it\tau})>1/4$ which have a singularity at ${\it\eta}_{\ast }$ . Since $G/{\it\eta}\gg 1$ at ${\it\eta}=LR^{1/2}/(2{\it\tau}^{1/2})$ , we must choose the location of the singularity only slightly larger than ${\it\eta}=LR^{1/2}/(2{\it\tau}^{1/2})$ . To asymptotically satisfy the boundary condition, we first determine the asymptotics of (4.27) near ${\it\eta}={\it\eta}_{\ast }$ . Using (4.25) to eliminate $c$ from (4.27), expanding near ${\it\eta}={\it\eta}_{\ast }$ and assuming that ${\it\eta}_{\ast }$ is large, we obtain

Applying (4.18b ) to (4.29), we obtain an expression for ${\it\eta}_{\ast }$ ,

Using (4.25) and (4.27), we therefore obtain

with ${\it\eta}_{\ast }$ given by (4.30). This represents the uniformly valid solution to the problem for case 3 with $L(t)\ll t/R$ .

We now need to check that the ${\it\tau}G_{{\it\tau}}$ term that we neglected in (4.17) is indeed negligible in comparison to the other terms in the equation. For ${\it\eta}=O(1)$ , this is clearly true since $G({\it\eta})\sim {\it\eta}$ . However, near ${\it\eta}={\it\eta}_{\ast }$ , the situation is more complicated. Using, the asymptotic form (4.29), we obtain

Since ${\it\tau}^{1/2}\ll L({\it\tau})\ll {\it\tau}/R$ , we have ${\it\tau}\,\text{d}{\it\eta}_{\ast }/\text{d}{\it\tau}=O({\it\eta}_{\ast })$ and so

On the other hand, the term $(G^{2})_{{\it\eta}}=O(({\it\eta}-{\it\eta}_{\ast })^{-3})$ and therefore

Using (4.29) near ${\it\eta}=R^{1/2}L({\it\tau})/(2{\it\tau}^{1/2})$ to obtain the scaling for $({\it\eta}-{\it\eta}_{\ast })$ and (4.30) to obtain the scaling for ${\it\eta}_{\ast }$ , we have

since $L\ll t/R$ . Hence, we conclude that the term ${\it\tau}G_{{\it\tau}}$ is negligible.

We now analyse the solution and show that the majority of the stretching occurs for material that was initially located near the initial minimum. In (4.17), the $G_{{\it\eta}{\it\eta}}$ and $(G^{2})_{{\it\eta}}$ terms correspond to the $Z_{xx}$ and $ZZ_{x}$ terms in (4.14), that represent the viscous forces, whereas the $({\it\eta}G)_{{\it\eta}}$ term corresponds to the $RZ_{t}$ term that represents inertial forces. For ${\it\eta}\ll {\it\eta}_{\ast }$ , $G$ is well approximated by $G\sim {\it\eta}$ . In this case, the main balance is between the $(G^{2})_{{\it\eta}}$ and $({\it\eta}G)_{{\it\eta}}$ terms indicating that the main balance is between viscous and inertial forces. However, close to ${\it\eta}={\it\eta}_{\ast }$ , the solution is well approximated by $G\sim ({\it\eta}_{\ast }-{\it\eta})^{-1}/2$ and the main balance is between the $(G^{2})_{{\it\eta}}$ and the $G_{{\it\eta}{\it\eta}}$ terms. In this case, the inertial term $({\it\eta}G)_{{\it\eta}}$ is negligible. Hence, inertial terms are only significant when $G\sim {\it\eta}$ , which, using (4.15), corresponds to $Z=O(RL/t)\ll 1$ . This implies that inertia only ever becomes important for the material that was initially located near the local minimum. This material becomes stretched into a long thin thread that occupies the vast majority of the length. In this region $G\sim {\it\eta}$ , which corresponds to $Z\sim xR/(2\mathscr{K}t)$ . Since $A=IZ_{x}$ , we see that this region corresponds to a thin thread whose cross-section is $O(R)$ . Moreover, in this region, the cross-section is uniform in space, but decreases inversely with time. On the other hand, near the singularity, $G\sim ({\it\eta}_{\ast }-{\it\eta})^{-1}/2$ , which corresponds to $Z\sim [1-\mathscr{K}(x-L)]^{-1}$ . This shows that the fluid near the pulled end eventually experiences negligible thinning and is just advected along with the pulled boundary at $x=L(t)$ .

In figure 7 we compare the large-time solution given by (4.15), (4.27), for the case where $1/2<{\it\alpha}<1$ , with the numerical solution. Once again, we plot the solutions on a log scale and as a function of the Lagrangian coordinate. The agreement becomes better as time becomes larger and the thread becomes thinner, indicating the validity of the long-time solution. The long-time solution has not been plotted at early times when the agreement is poor. We note that, at later times, the region close to $Z=0$ represents an extremely long and thin filament that forms the vast majority of the length of the thread. As time increases, this filament becomes progressively longer and thinner while the remaining fluid experiences negligible thinning as we commented above.

Figure 7.

Comparison between the numerical solution (solid) and the large-time solution given by (4.15) and (4.27) (dashed) with $L(t)\sim t^{{\it\alpha}}$ and $R=0.2$ for a thread that has its initial minimum away from the pulled end (dash-dot). The cross-sectional area $s(Z,t)$ is plotted as a function of the Lagrangian variable  $Z$ . The extension of the thread doubles between successive curves (3.16). Here, ${\it\alpha}=0.8$ , ${\it\lambda}=40$ . The numerical solution is shown for $k=5,6,\ldots ,15$ , while the asymptotic solution is shown for $k=11,12,\ldots ,15$ . Agreement between the curves becomes better as time becomes larger and the thread becomes thinner.

Finally, we discuss the behaviour of this type of thread in the case where ${\it\alpha}>1$ . In this case, one can readily see that the expression (4.27), that corresponds to a solution in which the majority of the extension is concentrated in those fluid elements that were originally located near the initial minimum, cannot satisfy the boundary condition at $X=L(t)$ at large times.

In figure 8(a), we plot results of a numerical simulation for ${\it\alpha}=2$ . At early times, the solution is well approximated by the small-inertia solution (3.33), but the approximation breaks down when $s$ becomes small (see figure 4 b). Beyond this time, the region in $Z$ in which the thread becomes thin expands as time progresses. This is in direct contrast to the case of $1/2<{\it\alpha}\leqslant 1$ , in which the thread only becomes thin in the vicinity of $Z=0$ . Thus, in the case ${\it\alpha}>1$ , the pulling rate is sufficiently fast that the extension cannot remain localised to fluid elements near the initial minimum.

Figure 8.

Thinning of a thread with the initial minimum away from the pulled end for $R=0.2$ , ${\it\alpha}=2$ . The extension doubles between successive curves (3.16). We have used ${\it\lambda}=100$ . The solutions are shown for $k=3,4,\ldots ,16$ . (a) Cross-sectional area $s(Z,t)$ is plotted as a function of the Lagrangian variable $Z$ . (b) Departure from the cross-sectional area, $s_{0}(Z)-s(Z,t)$ .

Wylie et al. (Reference Wylie, Huang and Miura2011) showed that this type of behaviour must occur for threads pulled with a fixed force. They divided the $Z$ -space into two parts: an ‘outer region’ in which $s=O(1)$ and a ‘necking region’ in which $s$ is small. They considered the quantity $s_{0}-s$ which corresponds to the deviation from the initial shape and showed that, in the necking region, $s_{0}-s$ is small, whereas in the outer region, $s_{0}-s$ must be a linear function of $Z$ . They further showed that the two types of solutions must be patched together at a location at which $s_{0}-s$ and its derivative are continuous. Since the largest value that $s_{0}-s$ can attain is $s_{0}$ , the leading-order problem can be thought of as a purely geometrical obstacle problem with an elastic string whose end moves with time. The necking regions correspond to the regions in which the string is in contact with the object, whereas the outer regions correspond to the regions in which the string is not in contact with the object. In figure 8(b), we replot the data in figure 8(a) for the variable $s_{0}-s$ . The curve, which is initially an approximately horizontal line can clearly be seen to wrap around the obstacle. Using asymptotic methods, Wylie et al. (Reference Wylie, Huang and Miura2011) were able to obtain an explicit solution for the necking region that allowed them to compute the total extension. A similar approach can be used for the current problem, but since the methodology has been developed in detail in their paper, we will not repeat the details here.