2 Equations of motion

We consider the motion of the edge of a semi-infinite planar fluid sheet with constant cross-section, which is being stretched in the direction parallel to the edge, as in figure 1. We take the fluid parameters $\unicode[STIX]{x1D707}$ and $\unicode[STIX]{x1D6FE}$ and the strain rate $E$ along the edge all to be uniform, but we allow the strain rate to vary with time $\hat{t}$ . We consider a sheet with uniform thickness $2h_{\infty }(\hat{t})$ in the far field.

We define a coordinate system $(\hat{x},{\hat{y}},\hat{z})$ oriented as shown in figure 1, with the $\hat{z}$ axis parallel to the edge. The $(\hat{x},{\hat{y}})$ -plane is normal to the edge, with the mid-plane of the sheet lying in the $(\hat{x},\hat{z})$ -plane. The corresponding three-dimensional velocity $\hat{\boldsymbol{u}}=(\hat{u} ,\hat{v},{\hat{w}})$ satisfies the Stokes equations

(2.1a,b ) $$\begin{eqnarray}\displaystyle & \displaystyle \hat{\unicode[STIX]{x1D735}}\boldsymbol{\cdot }\hat{\boldsymbol{u}}=0,\quad -\hat{\unicode[STIX]{x1D735}}\hat{p}+\unicode[STIX]{x1D707}\hat{\unicode[STIX]{x1D6FB}}^{2}\hat{\boldsymbol{u}}=\mathbf{0}. & \displaystyle\end{eqnarray}$$

Since the stretching is uniform, we can choose the origin of $\hat{z}$ such that ${\hat{w}}=E(\hat{t})\hat{z}$ , where $E(\hat{t})$ is the strain rate. We parametrise the surface shape by the arclength ${\hat{s}}$ and time $\hat{t}$ as $\hat{\boldsymbol{X}}=(\hat{X}({\hat{s}},\hat{t}),{\hat{Y}}({\hat{s}},\hat{t}))$ , so that ${\hat{Y}}\rightarrow h_{\infty }$ as ${\hat{s}}\rightarrow \infty$ .

Figure 1.

Sketch of the sheet, with the $\hat{z}$ -axis parallel to the edge. The far-field thickness $2h_{\infty }$ , the longitudinal strain rate $E$ and the fluid parameters $\unicode[STIX]{x1D707}$ and $\unicode[STIX]{x1D6FE}$ are uniform.

Since $\hat{u}$ , $\hat{v}$ , $\hat{p}$ and $E$ are independent of $\hat{z}$ , the local mass-conservation equation and the momentum equation for the two-dimensional in-section velocity $\hat{\boldsymbol{u}}_{H}=(\hat{u} ,\hat{v})$ reduce to

(2.2a,b ) $$\begin{eqnarray}\displaystyle & \displaystyle \hat{\unicode[STIX]{x1D735}}\boldsymbol{\cdot }\hat{\boldsymbol{u}}_{H}=-E,\quad -\hat{\unicode[STIX]{x1D735}}\hat{p}+\unicode[STIX]{x1D707}\hat{\unicode[STIX]{x1D6FB}}^{2}\boldsymbol{u}_{H}=\mathbf{0}, & \displaystyle\end{eqnarray}$$

with kinematic and dynamic boundary conditions on the surface

(2.3a,b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\displaystyle \frac{\unicode[STIX]{x2202}\hat{\boldsymbol{X}}}{\unicode[STIX]{x2202}\hat{t}}}\boldsymbol{\cdot }\boldsymbol{n}=\hat{\boldsymbol{u}}_{H}\boldsymbol{\cdot }\boldsymbol{n},\quad \text{and}\quad -\hat{p}\boldsymbol{n}+\unicode[STIX]{x1D707}(\hat{\unicode[STIX]{x1D735}}\hat{\boldsymbol{u}}_{H}+(\hat{\unicode[STIX]{x1D735}}\hat{\boldsymbol{u}}_{H})^{\text{T}})\boldsymbol{\cdot }\boldsymbol{n}=-\unicode[STIX]{x1D6FE}\hat{\unicode[STIX]{x1D705}}\boldsymbol{n}, & \displaystyle\end{eqnarray}$$

where $\hat{\unicode[STIX]{x1D705}}=\hat{\unicode[STIX]{x1D735}}\boldsymbol{\cdot }\boldsymbol{n}$ is the curvature of the surface and $\boldsymbol{n}$ the surface normal.

Consider the stress $\hat{\unicode[STIX]{x1D748}}=-\hat{p}\unicode[STIX]{x1D644}+\unicode[STIX]{x1D707}(\hat{\unicode[STIX]{x1D735}}\hat{\boldsymbol{u}}_{H}+(\hat{\unicode[STIX]{x1D735}}\hat{\boldsymbol{u}}_{H})^{\text{T}})$ in the far field. The sheet is almost flat and uniform so, by translational symmetry, the far-field stress is independent of $\hat{x}$ at leading order. In the far field, we have ${\hat{Y}}\sim h_{\infty }$ , so $\boldsymbol{n}\sim (0,1)$ and the curvature is negligible. Hence the stress boundary condition (2.3b ) gives $\hat{\unicode[STIX]{x1D70E}}_{xy}=0$ and $\hat{\unicode[STIX]{x1D70E}}_{yy}=0$ in the far field. In addition, the viscous force over the sheet in the far field, $2h_{\infty }\unicode[STIX]{x1D70E}_{xx}$ , must balance the net force from surface tension, $2\unicode[STIX]{x1D6FE}$ . Together, these conditions specify $\hat{\unicode[STIX]{x1D748}}$ in the far field, and then local mass conservation (2.2) gives the far-field velocity and pressure as

(2.4a,b ) $$\begin{eqnarray}\displaystyle & \displaystyle \hat{\boldsymbol{u}}_{H}\sim {\displaystyle \frac{\unicode[STIX]{x1D6FE}}{4\unicode[STIX]{x1D707}h_{\infty }}}(-\hat{x},{\hat{y}})-{\displaystyle \frac{E}{2}}(\hat{x},{\hat{y}}),\quad \hat{p}\sim {\displaystyle \frac{\unicode[STIX]{x1D6FE}}{2h_{\infty }}}-\unicode[STIX]{x1D707}E\quad \text{as }\hat{x}\rightarrow \infty . & \displaystyle\end{eqnarray}$$

Here, we have chosen the origin so that the velocity grows linearly in the far field, without a constant term. Since the Stokes equations are linear, we may add any solid-body motion to our solution below.

The kinematic boundary condition (2.3a ) with the far-field velocity (2.4) gives

(2.5) $$\begin{eqnarray}\displaystyle & \displaystyle {\displaystyle \frac{\text{d}h_{\infty }}{\text{d}\hat{t}}}={\displaystyle \frac{\unicode[STIX]{x1D6FE}}{4\unicode[STIX]{x1D707}}}-{\displaystyle \frac{Eh_{\infty }}{2}}, & \displaystyle\end{eqnarray}$$

and so we have found the effect of the stretching and capillarity on the far-field thickness.

Note that the two-dimensional velocity $\boldsymbol{u}_{H}$ does not satisfy the two-dimensional Stokes equations when $E\neq 0$ because $\unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}_{H}=-E$ and so mass is not conserved. However, if subtract the far-field velocity and non-dimensionalise by setting

(2.6) $$\begin{eqnarray}\displaystyle & \displaystyle {\displaystyle \frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D707}}}(u,v)=(\hat{u} ,\hat{v})-{\displaystyle \frac{\unicode[STIX]{x1D6FE}}{4\unicode[STIX]{x1D707}h_{\infty }}}(-\hat{x},{\hat{y}})+{\displaystyle \frac{E}{2}}(\hat{x},{\hat{y}}), & \displaystyle\end{eqnarray}$$

then $\boldsymbol{u}=(u,v)$ satisfies the Stokes equations. We also subtract the far-field pressure, and non-dimensionalise lengths with respect to $h_{\infty }$ by setting

(2.7a-d ) $$\begin{eqnarray}\displaystyle & \displaystyle p={\displaystyle \frac{h_{\infty }}{\unicode[STIX]{x1D6FE}}}\hat{p}-{\displaystyle \frac{1}{2}}+Ca,\quad \boldsymbol{x}={\displaystyle \frac{\hat{\boldsymbol{x}}}{h_{\infty }}},\quad \boldsymbol{X}={\displaystyle \frac{\hat{\boldsymbol{X}}}{h_{\infty }}}\quad \text{and}\quad \unicode[STIX]{x1D705}=h_{\infty }\hat{\unicode[STIX]{x1D705}}, & \displaystyle\end{eqnarray}$$

where $Ca=\unicode[STIX]{x1D707}Eh_{\infty }/\unicode[STIX]{x1D6FE}$ is the capillary number. Then the two-dimensional Stokes equations (2.2) become

(2.8a,b ) $$\begin{eqnarray}\displaystyle & \displaystyle -\unicode[STIX]{x1D735}p+\unicode[STIX]{x1D6FB}^{2}\boldsymbol{u}=\mathbf{0},\quad \unicode[STIX]{x1D735}\boldsymbol{\cdot }\boldsymbol{u}=0, & \displaystyle\end{eqnarray}$$

where the derivatives are taken with respect to the scaled variables. The stress boundary condition (2.3b ) on the surface becomes

(2.9) $$\begin{eqnarray}\displaystyle & \displaystyle -p\boldsymbol{n}+(\unicode[STIX]{x1D735}\boldsymbol{u}+\unicode[STIX]{x1D735}\boldsymbol{u}^{\text{T}})\boldsymbol{\cdot }\boldsymbol{n}=-\unicode[STIX]{x1D705}\boldsymbol{n}+\left(\begin{array}{@{}cc@{}}1 & 0\\ 0 & 0\end{array}\right)\boldsymbol{\cdot }\boldsymbol{n}, & \displaystyle\end{eqnarray}$$

and in the far field

(2.10a-d ) $$\begin{eqnarray}\displaystyle & \displaystyle X\rightarrow \infty ,\quad Y\rightarrow \pm 1,\quad \boldsymbol{u}\rightarrow \mathbf{0}\quad \text{as }s\rightarrow \pm \infty . & \displaystyle\end{eqnarray}$$

We note that, by subtracting the far-field velocity in the definition of $\boldsymbol{u}$ , we have removed $E$ from the Stokes problem entirely at any given time. Moreover, if we rescale the time variable appropriately, then the kinematic boundary condition (2.3a ) becomes

(2.11a,b ) $$\begin{eqnarray}\displaystyle & \displaystyle {\displaystyle \frac{\unicode[STIX]{x2202}\boldsymbol{X}}{\unicode[STIX]{x2202}t}}\boldsymbol{\cdot }\boldsymbol{n}=\boldsymbol{u}\boldsymbol{\cdot }\boldsymbol{n}-{\displaystyle \frac{1}{2}}(x,0)\boldsymbol{\cdot }\boldsymbol{n},\quad \text{where}~{\displaystyle \frac{\text{d}t}{\text{d}\hat{t}}}={\displaystyle \frac{\unicode[STIX]{x1D6FE}}{\unicode[STIX]{x1D707}h_{\infty }(\hat{t})}} & \displaystyle\end{eqnarray}$$

and $h_{\infty }(\hat{t})$ is given by the solution to (2.5). We have therefore removed the time-dependent stretching from the problem entirely, and rather than solving a family of problems parametrised by $Ca$ , we need only solve a single Stokes problem (2.8)–(2.11). Independent of the initial conditions, we expect that the solution of (2.8)–(2.11) will tend to a steady state, $\boldsymbol{n}\boldsymbol{\cdot }\unicode[STIX]{x2202}\boldsymbol{X}/\unicode[STIX]{x2202}t=0$ , in these scaled variables, which corresponds to a similarity solution with length scale $h_{\infty }(\hat{t})$ in the original variables. We solve for this similarity solution numerically below.

3 Numerical solution

We divide the sheet into two regions: a local region $S_{1}=\{(X,Y):X-X(0,t)\leqslant 4\}$ and a far-field region $S_{2}=\{(X,Y):X-X(0,t)>4\}$ . In $S_{1}$ , we solve for the velocity $\boldsymbol{u}$ with the boundary integral equation for Stokes flow (e.g. Pozrikidis Reference Pozrikidis1992). For a point $\boldsymbol{y}$ on the surface,

(3.1) $$\begin{eqnarray}\displaystyle & \displaystyle \int _{S}\unicode[STIX]{x1D645}(\boldsymbol{y}-\boldsymbol{x})\boldsymbol{\cdot }\unicode[STIX]{x1D748}\boldsymbol{\cdot }\boldsymbol{n}\,\text{d}s_{\boldsymbol{x}}+\int _{S}\boldsymbol{u}\boldsymbol{\cdot }\unicode[STIX]{x1D646}(\boldsymbol{y}-\boldsymbol{x})\boldsymbol{\cdot }\boldsymbol{n}\,\text{d}s_{\boldsymbol{x}}={\textstyle \frac{1}{2}}\boldsymbol{u}(\boldsymbol{y}), & \displaystyle\end{eqnarray}$$

where

(3.2a,b ) $$\begin{eqnarray}\displaystyle & \displaystyle \unicode[STIX]{x1D645}(\boldsymbol{x})={\displaystyle \frac{1}{4\unicode[STIX]{x03C0}}}\left(-\unicode[STIX]{x1D644}\log |\boldsymbol{x}|+{\displaystyle \frac{\boldsymbol{x}\boldsymbol{x}}{|\boldsymbol{x}|^{2}}}\right),\quad \unicode[STIX]{x1D646}(\boldsymbol{x})=-{\displaystyle \frac{\boldsymbol{x}\boldsymbol{x}\boldsymbol{x}}{\unicode[STIX]{x03C0}|\boldsymbol{x}|^{4}}}, & \displaystyle\end{eqnarray}$$

and $\unicode[STIX]{x1D748}\boldsymbol{\cdot }\boldsymbol{n}$ is given by (2.9). We discretise the surface in $S_{1}$ with grid points at $(X(s_{i}),Y(s_{i}))$ for $1\leqslant i\leqslant n$ . Between the grid points, we interpolate $X(s)$ and $Y(s)$ with quintic splines. In appendix A, we derive an asymptotic series for the far-field decay of the shape, which we use here to set $Y^{\prime }(s_{n})$ and $Y^{\prime \prime }(s_{n})$ as boundary conditions on the spline. We also use the asymptotic form of $\boldsymbol{u}$ to integrate (3.1) numerically for $\boldsymbol{x}$ in $S_{2}$ .

For any $\boldsymbol{y}$ on the surface, the first integrand in (3.1) diverges at $s_{\boldsymbol{x}}=s_{\boldsymbol{y}}$ , where $\unicode[STIX]{x1D645}(\boldsymbol{y}-\boldsymbol{x})\boldsymbol{\cdot }\unicode[STIX]{x1D748}\boldsymbol{\cdot }\boldsymbol{n}\sim \log |s_{\boldsymbol{y}}-s_{\boldsymbol{x}}|\,\unicode[STIX]{x1D748}(\boldsymbol{y})\boldsymbol{\cdot }\boldsymbol{n}$ . We subtract $\log |s_{\boldsymbol{y}}-s_{\boldsymbol{x}}|\,\unicode[STIX]{x1D748}(\boldsymbol{y})\boldsymbol{\cdot }\boldsymbol{n}$ from the integrand on panels where the integrand diverges and integrate the remainder with seven-point Gaussian quadrature. The subtracted term can be integrated analytically.

The second integral in (3.1) depends on the unknown velocity $\boldsymbol{u}$ . We discretise this integral by considering a set of test functions for the velocity. For each component of the velocity at each grid point in $S_{1}$ , we set a test function to 1 at that point and 0 on the other points. We interpolate the velocity with quintic splines and again use Gaussian integration to find the contribution to the integral. The result of this discretisation is a matrix problem for the velocity components in $S_{1}$ , which we invert for the velocity.

We advect the surface with the kinematic boundary condition (2.11), using a time step of order $10^{-2}$ . After each advection step, we rescale the shape so that $Y(s_{n})-1$ matches the asymptotic expression derived in appendix A. We repeat and step forward in time up to $t=30$ to allow any transients to decay.

We initialised the shape with a rectangle of fluid rounded by a semicircular cap. With 51 points spaced so that $\unicode[STIX]{x0394}s\approx 0.1$ , the far end of the rectangle is initially at $X(s_{n})\approx 4.4$ .

The shape relaxes to the steady-state solution shown in figure 2(a). The edge of the sheet is at $X(0)=-0.312$ , and we recall that the origin of $X$ (the frame of reference) was chosen such that the far-field velocity is given by (2.4). This choice implies that the steady-state shape must have the same area as a rectangle of height 2 with its end at $X=0$ . The sheet thickness does not monotonically decay to the far-field thickness; the maximum half-thickness is $Y=1+1.5\times 10^{-3}$ at $X=2.03$ , as shown in figure 2(b). Changing the number of points used to discretise the surface does not affect these constants, so we can be confident that we have resolved the shape accurately.

Figure 2.

The numerically calculated steady-state solution to (2.8)–(2.11) (solid lines) and its far-field behaviour (dashed lines). (a) The self-similar shape of the retracting edge. (b) The decay of the shape to its far-field thickness $Y=1$ . (c) The full curvature $\unicode[STIX]{x1D705}$ of the surface. (d,e) The velocity components $\boldsymbol{u}=(u,v)$ , defined by (2.6). (f) The maximum of the normal velocity at each time step. In (b–e), the asymptotic solution found in appendix A is shown for comparison.

We can compare the shape and velocity with the far-field asymptotic series given in appendix A, calculating the amplitude $A_{1,0}$ of the series from the final velocity components $(u(s_{n}),v(s_{n}))$ . Figure 2(b) shows the asymptotic decay of the surface shape to its far-field thickness, and there is good agreement between the numerically calculated similarity solution and the asymptotic series. The full curvature is compared with $Y_{XX}$ in figure 2(c), showing that the approximation $\unicode[STIX]{x1D705}\sim Y_{XX}$ holds far from the edge. The velocity components are shown in figure 2(d,e). The numerically calculated values agree well with the far-field asymptotic series not only at $s_{n}$ , where the series is set to match the velocity components, but also in some overlap region where $X-X(0)\gtrsim 2$ . This verifies the accuracy of the solution.

The similarity solution is stable; changing the initial shape of the sheet does not affect the resultant steady state. Moreover, our numerical solution shows the decay rate of the slowest-decaying mode. Figure 2(f) shows the maximum absolute value of the normal component of the velocity at each time step. After some shorter-lived transients, the residual normal velocity decays exponentially with time. We fit an exponential to the data between $t=5$ and $t=10$ and find that the slowest mode decays proportional to $\text{e}^{-0.50t}$ (shown as a dashed line in figure 2 f). We can therefore be sure that the solution has converged to the required accuracy by time $t=30$ .