3.1. Unconfined extrusion of a Newtonian fluid

If the injected fluid is Newtonian then the viscosity $\mu$ is constant and can be factored out of (2.5a ). We introduce the scalings

(3.1a–d) \begin{equation} h \sim h_0, \ \ \ r \sim r_0, \ \ \ u \sim U_0 \equiv \frac {Q}{4 \pi r_0 h_0}, \ \ \ t \sim \frac {r_0}{U_0}. \end{equation}

The governing equations and boundary conditions are unchanged by these scalings, barring (2.6) and (2.11), which, in dimensionless form, become

(3.2a,b) \begin{equation} h = 1, \quad u = 1 \quad \hbox{at} \quad r = 1, \end{equation}

(3.2c) \begin{equation} r_{\!N}=1 \quad \hbox{at} \quad t=0. \end{equation}

As an aside, note that in two dimensions, with the velocity $u$ in the $x$ direction, there are no hoop stresses and the solution of the equivalent dimensionless equations is simply $u\equiv 1$ , $h\equiv 1$ , with the nose position at $x_N(t) = 1+t$ (assuming extrusion from $x_0=1$ ).

We start here by determining full numerical solutions to the extrusion problem. We map the domain $[1, r_{\!N}]$ onto $[0, 1]$ using the independent variable

(3.3) \begin{equation} X(r,t) = {r - 1\over L(t)}, \quad \hbox{where} \quad L(t) = r_{\!N}(t) - 1. \end{equation}

We also introduce the dependent variable

(3.4) \begin{equation} y(r, t) = \textit{ru}(r, t), \end{equation}

which remains bounded as $r\to 0$ . Expressing the dependent variables as $y=y(X,t)$ and $h=h(X,t)$ , the governing equations then become

(3.5a) \begin{equation} 2h\!\left ({y_{XX}\over L} - {y_X\over 1+\textit{LX}}\right ) + h_X\!\left ( {2y_X\over L} - {y\over 1+\textit{LX}}\right ) = 0, \end{equation}

(3.5b) \begin{equation} h_t - {\dot L\over L} X h_X + {(hy)_X\over L(1+\textit{LX})} = 0, \end{equation}

subject to boundary and initial conditions

(3.6a,b) \begin{equation} h = 1, \quad y = 1 \quad \hbox{at} \quad X = 0, \end{equation}

(3.6c,d) \begin{equation} {2y_X\over L} - {y\over 1+L} = 0, \quad \dot L = {y\over 1+L} \quad \hbox{at} \quad X = 1, \end{equation}

(3.6e) \begin{equation} L = 0 \quad \hbox{at} \quad t = 0. \end{equation}

These equations are singular in the limit $L\to 0$ , when the initial distribution $h(X,0)$ is condensed into the point $r=1$ . For consistency with the boundary condition (3.6a ), we define the initial distribution

(3.7) \begin{equation} h(X,0)=1, \end{equation}

to avoid the function $h(r,t)$ being multivalued at $t=0$ .

To avoid this singularity when integrating the equations numerically, we find the early-time asymptotic solution. Letting

(3.8a,b) \begin{equation} y = 1 + L(t) Y(X,t) + O(L^2), \quad h = 1+ L(t) H(X,t)+ O(L^2), \end{equation}

we find that, after expanding in powers of $L$ , the momentum equation, (3.5a ), has leading order $Y_{XX}=0$ , the evolution equation, (3.5b ), has leading order $\dot LH-\dot LXH_X+Y_X+H_X=0$ and the boundary conditions (3.6a – d ) have leading order $H=Y=0$ at $X=0$ , $Y_X= {1}/{2}$ at $X=1$ and $\dot L=1$ , respectively. Solving these equations, along with $L=0$ at $t=0$ , gives

(3.9a–c) \begin{equation} y \sim 1 + \frac {\textit{LX}}{2}, \quad h \sim 1 - \frac {\textit{LX}}{2}, \quad L \sim t. \end{equation}

The small-time evolution of $L(t)$ is illustrated in the figure 2( $b$ ) inset. Ball & Balmforth (Reference Ball and Balmforth2021) derive an equivalent small-time solution for a floating Newtonian extrusion, with an additional term in each of $Y$ and $H$ corresponding to a hydrostatic pressure term in the force balance at the nose (3.6c ). The effect of buoyancy on the solution is discussed further in § 3.2.1.

Figure 2.

The solution to (3.5)–(3.6) for the extrusion of a Newtonian fluid. ( $a$ ) Plots of the current half-thickness $h$ at dimensionless times $t=$ 5, 20, 100, 500 and 2000, from left to right. The dashed curve is $h_N={r_{\!N}}^{-1/2}$ , as predicted by (3.11b ) for the thickness of the intruding fluid at the nose. ( $b$ ) The same results shown in ( $a$ ) plotted in similarity space, according to the scalings defined in (3.12). The similarity solution to (3.13) with $\eta _N\approx 1.71$ is shown in dashed red. The black dots indicate the asymptotic results (3.18) and (3.20). The inset shows a log–log plot of the evolution of the current length $L(t)\equiv r_{\!N}(t)-1$ , illustrating the transition from the small-time solution (3.9) to the similarity solution $L=\eta _Nt^{1/2}-1\sim \eta _Nt^{1/2}$ , with the asymptotic expressions plotted in dashed red. ( $c$ ) Plots of the similarity solution profiles for $y\equiv ru$ and $u$ .

Figure 2( $a$ , $b$ ) shows the numerical solution of (3.5)–(3.6) starting from this asymptotic solution. The transformation (3.3) means that numerical solution of the equations was performed on a fixed grid. We used central differences for spatial derivatives and a second-order, implicit midpoint scheme in time.

We see that $h(r_{\!N}, t) \ne 0$ but approaches zero on a similar time scale to that taken for the extent of the intruding flow to exceed the radius of injection significantly. To understand the evolution of $h_N = h(r_{\!N}(t), t)$ , we evaluate (2.5b ) at $r=r_{\!N}(t)$ and use the result to write

(3.10) \begin{equation} \dot h_N \equiv \frac {D h}{D t}(r_{\!N},t) = -h_N\! \left (\frac {u(r_{\!N},t)}{r_{\!N}} + \frac {\partial u}{\partial r}(r_{\!N},t) \right )\!. \end{equation}

From (2.8) and (2.10) we have

(3.11a,b) \begin{equation} 2 \dot h_N +h_N \frac {\dot r_{\!N}}{r_{\!N}} =0, \quad \hbox{whence} \quad h_N = r_{\!N}^{-1/2}, \end{equation}

having used the boundary condition (3.2a ) and the initial condition (3.2c ) to determine that $h_N=1$ when $r_{\!N}=1$ . Note that (2.8) and (2.10), and hence also this result, are independent of the rheology of the fluid, and can be seen for different power-law exponents in figures 3 and 4. The relation (3.11b ) is plotted for a Newtonian fluid as the dashed curve in figure 2( $a$ ).

Now, $L(t)=r_{\!N}(t)-1$ , and on short time scales, we found that $L \sim t$ (see (3.9c )), giving $r_{\!N} \sim 1+t$ and therefore $h_N \sim (1+t)^{-1/2} \sim 1-t/2$ , which is consistent with our approximation for $h$ in (3.9b ) when evaluated at the endpoint $X=1$ . We will soon show that at late times, $r_{\!N} \sim t^{1/2}$ so that $h_N \sim t^{-1/4}$ , which represents a slow, algebraic decay towards $h_N=0$ .

3.2. Similarity solution for extrusion of a Newtonian fluid

On long time scales, formally as $r_{\!N}\to \infty$ , the fluid can be considered to be extruded from a line source occupying $r=0$ , $-1\leqslant z\leqslant 1$ . This case was considered by Pegler & Worster (Reference Pegler and Worster2012), where they found that the equations admitted a similarity solution at early times in their scenario, when the effects of buoyancy were negligible. They then used this similarity solution as an initial condition for numerical calculations of buoyancy-driven extension. The physical relevance of this similarity solution to their flow is discussed in § 3.2.1. We can recover their similarity solution by considering a radial similarity variable that is independent of $r_0$ . From (3.1), $t^{-1/2} r$ is the only combination of the dimensionless variables $t$ and $r$ that is independent of $r_0$ , and so we let

(3.12a,b) \begin{equation} h(r, t) = h(\eta ), \quad y(r, t) = y(\eta ), \quad \hbox{where} \quad \eta = {r\over t^{1/2}}. \end{equation}

By plotting $h$ against $r/t^{1/2}$ for the transient problem (3.5)–(3.6), we can verify this scaling and approximate the similarity solution by observing convergence at late time, as shown in figure 2( $b$ ).

For completeness and ease of reference, we present the similarity equations for $y(\eta )$ and $h(\eta )$ here as

(3.13a) \begin{equation} 2 h\! \left (y'' - {y'\over \eta }\right ) + h'\! \left (2 y' - {y\over \eta }\right ) = 0, \end{equation}

(3.13b) \begin{equation} \left (y - {\eta ^2\over 2}\right ) h' + h y' = 0, \end{equation}

(3.13c,d) \begin{equation} y = 1, \quad h = 1 \quad \hbox{at} \quad \eta = 0, \end{equation}

(3.13e, f) \begin{equation} y = {\eta _N^2\over 2}, \quad y' = {\eta _N\over 4} \quad \hbox{at} \quad \eta = \eta _N, \end{equation}

where $\eta _N = t^{-1/2}r_{\!N}$ , and primes denote $\text{d}/\text{d}\eta$ .

We eliminate $h$ between (3.13a ) and (3.13b ) to give

(3.14) \begin{equation} 2\!\left (y - {\eta ^2\over 2}\right ) \left (y'' - {y'\over \eta }\right ) - \left (2 y' - {y\over \eta }\right )y' = 0, \end{equation}

which we solved in Matlab using the in-built integrator ode45, shooting backwards from the nose to find the value $\eta _N\approx 1.71$ that gives $y = 1$ at $\eta =0$ , in agreement with the value found by Pegler & Worster (Reference Pegler and Worster2012). Given the boundary condition (3.13e ), it is clear that this differential equation is singular at $\eta = \eta _N$ . However, it has a removable singularity, and we use a second-order Taylor expansion to show that $y''(\eta _N) = 5/32$ . This expression allows us to avoid the singularity at $\eta _N$ , by writing (3.14) as

(3.15) \begin{equation} y''=\left \{\begin{matrix}\left (\frac {2\eta y'+y-\eta ^2}{2\eta (y-\eta ^2/2)}\right )y'&\text{for }\eta \lt \eta _N\\[10pt]5/32&\text{for }\eta =\eta _N,\end{matrix}\right . \end{equation}

and integrating this new equation backwards from $\eta _N$ , applying (3.13e , f ) as the initial condition, and ending at $\epsilon _0=10^{-8}$ .

Once $y(\eta )$ has been determined, it is in principle straightforward to integrate (3.13b ) using quadrature to evaluate

(3.16) \begin{align} h = \exp \int _0^\eta {2y'(\xi )\over \xi ^2 - 2y(\xi )}\, \text d\xi . \end{align}

However, it proved expedient to integrate (3.13b ) for $h$ numerically alongside the integration for $y$ , starting from the asymptotic solution near the nose, $h \sim A(\eta _N - \eta )^{1/3}$ (see (3.41) for a derivation of the most general form of this asymptote). We write $\phi \equiv (\eta _N-\eta )^{-1/3}h$ , and substitute this expression into (3.13b ) to find

(3.17) \begin{align} \phi '=\left (\frac {-3(\eta _N-\eta )y'+y-\eta ^2/2}{3(\eta _N-\eta )(y-\eta ^2/2)}\right )\phi . \end{align}

Expanding the term in brackets with a second-order Taylor expansion in $(\eta _N-\eta )$ , and substituting in (3.13e , f ) and the value $y''(\eta _N)=5/32$ found above, we find that $\phi '(\eta _N)=A/(48\eta _N)$ , where $A=\phi (\eta _N)$ . This allows us to integrate (3.13b ) directly from $\eta _N$ , using the technique described in (3.15). We shoot to determine $A\approx 0.904$ , and thus

(3.18) \begin{align} h \sim 0.904(\eta _N - \eta )^{1/3} \quad \hbox{as} \quad \eta \to \eta _N, \end{align}

in agreement with Pegler & Worster (Reference Pegler and Worster2012).

An asymptotic approximation to $h$ near $\eta = 0$ can be determined straightforwardly as follows. Firstly, (3.13b ) shows that $(hy)' = 0$ to leading order in $\eta$ , and from (3.13c , d ), $h y =1$ . Therefore, writing $y = 1+\delta (\eta )$ , with $\delta \to 0$ as $\eta \to 0$ , we can deduce that $h\sim 1 - \delta (\eta )$ as $\eta \to 0$ . With these asymptotic expressions, it is straightforward to show that the dominant balance in (3.13a ) is

(3.19) \begin{equation} 2 \delta '' - {\delta '\over \eta } = 0, \end{equation}

whence $\delta \sim B\eta ^{3/2}$ , for some constant $B$ . We determined $B = 0.190$ by evaluating $B \sim -(2/3)h'(\eta )/\eta ^{1/2}$ at $\eta = \epsilon _0$ where the value of $h'(\epsilon _0)$ is obtained from the full numerical solution to (3.13b ). Given now that

(3.20) \begin{equation} h \sim 1 - 0.190\eta ^{3/2} \quad \hbox{as} \quad \eta \to 0, \end{equation}

the intruding fluid remains very close to $h=1$ near the point of extrusion. The similarity solution for $h$ and these asymptotic approximations are shown in figure 2( $b$ ), and the corresponding solutions for $y$ and the similarity velocity $y/\eta \equiv u/t^{-1/2}$ are plotted in figure 2( $c$ ).

Given that $h\lt 1$ throughout the domain $r\gt 1$ , the solutions in this section also describe the extrusion of fluid into a confining channel containing an inviscid ambient fluid, a geometry that is discussed further in § 4. However, the fact that $h$ is very close to unity near the radius of extrusion means that if the ambient fluid has any viscosity, no matter how small, then the assumption that the intruding fluid feels no tangential stress is invalid in the neighbourhood of the source. This is worth exploring but is beyond the scope of the current paper.

3.2.1. Evolution of a buoyant current

Pegler & Worster (Reference Pegler and Worster2012) consider a buoyant current with density $\rho$ floating on an inviscid fluid with density $\rho _w$ , produced by a line source with height $2h_0$ . They find an early-time similarity solution, before buoyancy effects become significant, which is replaced by a late-time, buoyancy-dominated solution with a steady $h$ profile, and a nose that advances with $r_{\!N}\sim t$ .

Their formulation is equivalent to the set-up in figure 1( $a$ ), with the addition of a buoyancy force, and with $r_0$ set to 0. This reproduces the condition for the scalings in (3.12) to be valid, namely independence from the initial extrusion radius, and hence the late-time similarity solution found above for an extruded Newtonian current is identically their early-time solution, before the effects of buoyancy become significant.

They identify a buoyancy time scale and radial length scale

(3.21a,b) \begin{equation} \mathcal{T}\equiv \frac {\mu }{2\rho g'h_0},\quad \mathcal{R}\equiv \left (\frac {\mu Q}{8\pi \rho g'h_0^2}\right )^{1/2}, \end{equation}

respectively, where $g'\equiv (\rho _w-\rho )g/\rho _w$ is the reduced gravity, and all other parameters are as defined in § 2. They find convergence from the purely extensional similarity solution to the buoyancy-driven similarity solution for dimensionless time $t\gtrsim 100$ , non-dimensionalised using (3.21).

However, if we consider a modification of their set-up, with extrusion at a finite radius $r_0$ rather than a line source, we can estimate the evolution of the flow by comparing the onset of buoyancy effects found in their paper with the onset of extensional effects described in § 3.1. Here, in the case of a neutrally buoyant fluid extruded from a finite radius, we find convergence to the purely extensional similarity solution around dimensionless time $t\gtrsim 100$ (see figure 2 $b$ ), but non-dimensionalised now using (3.1). Therefore, in the case of extrusion of a buoyant fluid from a cylindrical source, the purely extensional similarity solution is only expected to develop if the buoyancy time scale given in (3.21) is much larger than the geometric time scale given in (3.1), or, equivalently, the buoyancy radius is much larger than the extrusion radius, expressible as

(3.22) \begin{equation} \frac {8\pi \rho g'r_0^2h_0^2}{\mu Q}\ll 1. \end{equation}

If this is not satisfied, the small-time similarity solution found by Pegler & Worster (Reference Pegler and Worster2012), reproduced here as a late-time similarity solution, will not be realised, with the flow instead following the behaviour derived by Ball & Balmforth (Reference Ball and Balmforth2021) for a sliding fluid film. Their model takes the ambient fluid density $\rho _w=0$ , and is therefore equivalent to the model of Pegler & Worster (Reference Pegler and Worster2012) with $g'=g$ , with the added feature of a finite radius of extrusion $r_0$ . They find a gravity-dependent small-time solution (§ 3.3 in that paper) that reduces to (3.9) in identically the limit (3.22), which can be realised within their dimensionless model as the unexplored limit $H_0\ll 1$ , corresponding to a small cylindrical source height, relative to the typical thickness suggested by the buoyancy–viscosity force balance.

3.3. Unconfined extrusion of a power-law fluid

If the viscosity of the intruding fluid is not uniform but has a power-law rheology, then in terms of the variables $y$ and $X$ defined respectively in (3.3) and (3.4), the momentum equation, (2.5a ), becomes

(3.23) \begin{equation} 2\mu h\!\left ({y_{XX}\over L} - {y_X\over 1 + \textit{LX}}\right ) + (\mu h_X + \mu _Xh)\left ( {2y_X\over L} - {y\over 1+ \textit{LX}}\right ) = 0, \end{equation}

with viscosity

(3.24) \begin{equation} \mu = \left ({y_X^2\over L^2(1 + \textit{LX})^2} - {y y_X\over L(1 + \textit{LX})^3} + {y^2\over (1 + \textit{LX})^4}\right )^{(1-n)/(2n)}, \end{equation}

obtained from (2.4). Here, we have used the same scalings as given in (3.1), and in addition we have scaled the dynamic viscosity with $\tilde \mu (U_0/r_0)^{(1-n)/n}$ , as also derived by Pegler et al. (Reference Pegler, Lister and Worster2012). The other equation, boundary conditions, and initial condition (3.5b )–(3.6e ) remain unchanged from those of an extruded Newtonian fluid.

Note that the viscosity can also be written in the form

(3.25) \begin{equation} \mu = \left [\left ({y_X\over L(1 + \textit{LX})} - {1\over 2}{y \over (1 + \textit{LX})^2}\right )^2 + {3\over 4}{y^2\over (1 + \textit{LX})^4}\right ]^{(1-n)/(2n)}, \end{equation}

which shows clearly that it is positive definite, and this form is therefore convenient to retain in computations.

Similarly to the Newtonian case, these equations are singular in the limit $L\to 0$ , and we can find the asymptotic solution at small times by letting $y=1+L(t)Y(X,t)+O(L^2)$ and $h=1+L(t)H(X,t)+O(L^2)$ as in (3.8). To leading order in $L \ll 1$ , the evolution (3.5b ) and boundary conditions (3.6a – d ) are unchanged from that of the Newtonian case, and the momentum equation, (3.23), has leading order

(3.26) \begin{equation} (Y_X^2-Y_X+1)^{(1-n)/(2n)}\left [2+\frac {(1-n)(2Y_X-1)^2}{2n(Y_X^2-Y_X+1)}\right ]Y_{XX}=0. \end{equation}

The first term $(Y_X^2-Y_X+1)^{(1-n)/(2n)}$ is the leading-order viscosity, and the two terms inside the square brackets, when multiplied by $Y_{XX}$ , represent radial strain and viscosity gradients (corresponding to the terms $2\mu hy_{XX}/L$ and $h\mu _X(2y_X/L-y/(1+\textit{LX}))$ in (3.23)), respectively. There are no roots of the expression in square brackets for real $Y_X$ , and so $Y_{XX}=0$ to leading order. The small-time behaviour is therefore independent of the rheology, and we find the same small-time solution (3.9) as obtained for the Newtonian case. The early time behaviour is purely geometric in the situation we consider because, in the narrow annular region, there is little variation in extension, so the viscosity is essentially uniform, and is determined by the rheology-independent stress condition at the nose (2.10). This is in contrast with the similar case of sliding non-Newtonian films studied by Ball & Balmforth (Reference Ball and Balmforth2021), who also find uniform extension and viscosity at early times, but show that the forms of those uniform values are rheology-dependent. This discrepancy is due to the effect of a hydrostatic pressure term in the stress condition at the nose, meaning that the viscosity does not factor out of the stress condition (2.10), as it does in our neutrally buoyant flow.

We have integrated these equations numerically for a shear-thinning fluid with $n=3$ . The resulting profiles of $h(r, t)$ at various times are shown in figure 3( $a$ ). At early times, as in the case of a Newtonian fluid, the radially spreading intruding fluid remains thinner than the cylindrical source height, which can be seen in figure 3( $a$ ) for $t=0.2$ and 1. However, the velocity $u$ and its derivative $u_r$ are decreasing functions of $r$ , and correspondingly the viscosity increases radially (see figure 3 $b$ for plots of $y$ , $u$ and $\mu$ at $t=100$ ), leading to deviations from the Newtonian behaviour, as a relatively immobile ring of fluid develops near the leading edge, which inhibits radial spreading, and the bulk of the fluid builds up behind it. This can be seen for $t\geqslant 20$ in figure 3( $a$ ), where we observe a bulge near the radius of extrusion, which seemingly grows without bound.

Figure 3.

The solution to (3.5b ), (3.6), (3.23) and (3.24) for the extrusion of a power-law fluid with $n=3$ . ( $a$ ) Plots of the current half-thickness $h$ at dimensionless times $t=0.2$ , 1, 5, 20, 50 and 100, with greater time corresponding to greater radial extent $r_{\!N}$ . ( $b$ ) Plots of $y\equiv ru$ , $u$ and $\mu$ at $t=100$ , illustrating the radial decay of the velocity field $u$ , and growth of the viscosity $\mu$ . The viscosity (given by (3.24)) is small but positive at $r=1$ due to the large radial velocity gradient $u_r$ .

Figure 4.

The solution to (3.5b ), (3.6), (3.23) and (3.24) for the extrusion of a power-law fluid with $n=1.25$ . ( $a$ ) Plots of the current half-thickness $h$ at dimensionless times $t=$ 5, 20, 100, 500 and 2000, from left to right. The dashed curve is $h_N={r_{\!N}}^{-1/2}$ , as predicted in (3.11) for the thickness of the intruding fluid at the nose (this derivation is independent of the power-law exponent $n$ ). ( $b$ ) The same results shown in ( $a$ ) plotted in similarity space, according to the scalings defined in (3.27), (3.31) and (3.34). The time exponent for radial growth is $\alpha \approx 0.453$ . The similarity solution as given by the solution to (3.28)–(3.32) is shown in dashed red, with the similarity nose position $\eta _N\approx 2.06$ . ( $c$ ) Plots of the similarity solution profiles for $y\equiv ru$ , $u$ and $\mu$ , illustrating the radial decay of $u$ , and growth of $\mu$ .

Similar bulging is seen in figure 4( $a$ ), which shows the case $n = 1.25$ , but in this case it appears that a self-similar shape emerges. If the propagation is self-similar with a horizontal scale $R(t)$ , then the vertical scale should be $H(t) = t/R(t)^2$ , since $R^2H$ is the scale of the volume of the current, which increases linearly in time given constant input flux. If we rescale the radius by $r=r_{\!N}(t)\hat {r}$ , and the thickness by $h(r,t)=(t/{r_{\!N}}^2)\hat {h}(\hat {r},t)$ , the curve $\hat {h}(\hat {r},t)$ appears to collapse at late time onto a universal shape, given by a similarity solution which we derive in § 3.4. We have not found a corresponding collapse for $n=3$ for as long as we have been able to compute. In addition, $r_{\!N}$ and $h$ do not appear to grow with any power of $t$ (see figure 5 $a$ for log–log plots of $L\equiv r_{\!N}-1$ and $\text{max}(h)$ against $t$ ).

Figure 5.

( $a$ ) Log–log plots of the extent of the current, $L\equiv r_{\!N}-1$ (solid curves), and the maximum height of the current, $\text{max}(h)$ (dashed curves), against $t$ , for $n=1,1.25,1.5$ and $3$ . ( $b$ ) Log–log plots of the aspect ratio $\text{max}(h)/L$ against $t$ , for the same values of $n$ .

3.4. Similarity solutions for extrusion of a power-law fluid

The fact that it takes such a long time to reach self-similarity and that it seems likely that the rigidifying rim of the current will cause it to buckle out of plane, as was observed by Hutchinson & Worster (Reference Hutchinson and Worster2025), the search for similarity solutions may be purely of academic interest. However, we include an analysis here for completeness.

As we have seen above, if the intruding fluid is shear-thinning then the thickness of the current increases with time, which means that its radial extent must increase more slowly than $t^{1/2}$ . At very late times, the radius and width of the extruder are vanishingly small with respect to the radial extent and thickness of the current, respectively. Therefore, the similarity solution will appear to emerge from a point source at $r=0$ , where the differential equation (3.23) is singular in any case.

We look for solutions in which the radial extent scales with $t^\alpha$ , in which case the height must scale with $t^{1 - 2\alpha }$ , given that the volume of the current increases linearly with time. The mass-conservation equation, (2.5b ), then requires that $y$ scales with $r^2/t \sim t^{2\alpha - 1}$ . Therefore, we look for solutions of the form

(3.27a,b) \begin{equation} h(r, t) = t^{1-2\alpha }f(\eta ), \quad y(r, t) = t^{2\alpha - 1} g(\eta ), \quad \hbox{where} \quad \eta = {r\over t^\alpha }. \end{equation}

With these forms, (2.5a ) becomes

(3.28) \begin{equation} 2\!\left (g'' - {g'\over \eta }\right ) + M\!\left (2 g' - {g\over \eta }\right ) = 0, \end{equation}

where

(3.29) \begin{equation} M(f, g, \eta ) = \left [\ln f + \frac {1-n}{2n}\ln \left ({g'^2\over \eta ^2} - {gg'\over \eta ^3} + {g^2\over \eta ^4}\right )\right ]'\!, \end{equation}

while the mass-conservation equation, (2.5b ), gives

(3.30) \begin{equation} (1 - 2\alpha )f - \alpha \eta f' + {1\over \eta }(\textit{fg})' = 0. \end{equation}

Given the self-similar forms (3.27), the boundary conditions (3.2a , b ) at the radius of extrusion $r = 1$ imply that

(3.31a,b) \begin{equation} f \sim \eta ^p, \quad g \sim \eta ^{-p} \quad \hbox{as} \quad \eta \to 0, \quad \hbox{where}\quad p = {1 - 2\alpha \over \alpha } \end{equation}

is written for convenience. These conditions are augmented by the boundary conditions at the nose of the current

(3.32a–c) \begin{equation} f = 0, \quad g = \alpha \eta _N^2, \quad g' = {\alpha \eta _N\over 2} \quad \hbox{at} \quad \eta = \eta _N, \end{equation}

where $r_{\!N} \equiv \eta _N t^\alpha$ . The condition on the half-thickness of the current $f=0$ is derived from (3.11b ), which holds for all values of $n$ , and which states that the half-thickness at the nose is given by $h_N={r_{\!N}}^{-1/2}$ . Using (3.27a ), we must therefore have $t^{1-2\alpha }f(\eta _N)\sim t^{-\alpha /2}{\eta _N}^{-1/2}$ as $t\to \infty$ . We will show that $\alpha \leqslant 1/2$ for all relevant values of $n$ , and so $t^{1-2\alpha }\gg t^{-\alpha /2}$ , implying that $f(\eta _N)$ must be zero for the two terms to balance.

The coupled system of equations consisting of the second-order (3.28) and the first-order (3.30), and five boundary conditions (3.31) and (3.32), allows the exponent $\alpha$ and the dimensionless extent of the current $\eta _N$ to be determined.

Note that this is an example of a similarity solution of the second kind, in which, although the self-similar solution emerges from a point ( $\eta =0$ ), it retains memory of the fact that $u$ and $h$ are simultaneously equal to unity at the finite radius of extrusion. As is common for second-kind similarity solutions, the temporal exponent is not rational and cannot be determined solely by scaling.

We start by using the asymptotic expressions from (3.31) in (3.29) to determine that

(3.33) \begin{equation} M \sim {p -(p+2)(1-n)/n\over \eta } \quad \hbox{as} \quad \eta \to 0. \end{equation}

We then use (3.28) to show that

(3.34) \begin{equation} p = {(5 - 2n) - (1+2n)^{1/2}(9 - 6n)^{1/2} \over 4(n - 1)}, \end{equation}

choosing the branch of solutions that passes through $p=0$ at $n=1$ , corresponding to the Newtonian similarity solution found in § 3.1.

Graphs of $p$ and $\alpha$ are shown in figure 6. From the expression above, considering the range $n\geqslant 1$ , we find that the exponent $p$ is real only if $n\leqslant 3/2$ , and that $p$ varies from $0$ to $1$ as $n$ varies from $1$ to $ {3/2}$ . We also note that the aspect ratio $h/r \propto t^{(p - 1)/(p+2)}$ tends to zero as $t\to \infty$ only if $p\lt 1$ , or equivalently for $n \lt 3/2$ (see figure 5 $b$ for plots of the aspect ratio against time for different values of $n$ ). These observations perhaps signify that if $n\gt {3/2}$ then the current never reaches a self-similar form, which appears to be the case for $n=3$ reported above and shown in figure 5( $b$ ), and may evolve to have an aspect ratio greater than unity, which would violate the assumption of a thin-film flow.

It is worth noting that all the formulations in this section are also valid in the range $0\lt n\lt 1$ , corresponding to shear-thickening fluids. The solutions are characterised by a fluid layer that thins over time, with a viscosity field that decays radially outwards.

Although the geometry provides fixed length scales, namely $h_0$ and $r_0$ , the latter is not a relevant radial scale for the flow at large times when $r_{\!N} \rightarrow \infty$ , and, on the scale of the flow, the fluid appears to emerge from the origin. The thickening of the flow of a shear-thinning fluid or the thinning of the flow of a shear-thickening fluid means that $h_0$ is not an appropriate vertical scale for the flow at large times. Therefore, there are insufficient length scales in the flow to determine a first-kind similarity solution at large times. It was serendipitous that the Newtonian extrusion neither swells nor shrinks, which allowed a first-kind similarity solution to be found in § 3.2.

Figure 6.

( $a$ ) The exponent of the profile $p$ near the origin, and ( $b$ ) the temporal exponent of the extent of the current $\alpha$ as functions of the rheological exponent $n$ , as given by (3.34) and (3.31), respectively.

For ease of computation, $f$ is readily eliminated from (3.29) using (3.30), which when combined with (3.29) yields

(3.35) \begin{equation} M(g)=-\frac {g'+(1-2\alpha )\eta }{g-\alpha \eta ^2}+\frac {1-n}{2n}\left [\ln\! \left (\frac {{g'}^2}{\eta ^2}-\frac {gg'}{\eta ^3}+\frac {g^2}{\eta ^4}\right )\right ]'\!, \end{equation}

with (3.28) otherwise unchanged.

As in the Newtonian case (3.14), it is possible to remove the singularity at $\eta _N$ using (3.32) to find the quadratic behaviour of $g$ , which takes the most general form

(3.36) \begin{equation} g''(\eta _N)=\frac {\alpha }{8}(9\alpha -2). \end{equation}

Using this expression for $g''$ at the nose, it is possible to integrate (3.28) directly from the nose using the same technique as in (3.15), avoiding the equation singularity at $\eta _N$ . To avoid the singularity at $\eta =0$ , we also restrict our domain of integration to a minimum small value $\eta =\epsilon _0$ , with $\eta _N$ adjusted in a shooting scheme to meet the condition $\epsilon _0^p {} g(\epsilon _0) = 1$ , derived from (3.31).

The similarity half-thickness $f$ can then be determined from (3.30) as

(3.37) \begin{equation} f = {\exp\! \left [\! -\int _0^\eta {\eta \, \text{d}\eta \over g - \alpha \eta ^2}\right ] \over g - \alpha \eta ^2}, \end{equation}

which can, in principle, be evaluated using quadrature. Alternatively, we can find the asymptotic behaviour of $f$ near $\eta _N$ to set up a boundary condition for backwards integration from the nose.

Near $\eta _N$ , we write

(3.38) \begin{equation} g = \alpha \eta _N^2-\frac {\alpha \eta _N}{2}(\eta _N-\eta )+O((\eta _N-\eta )^2), \end{equation}

using the boundary conditions (3.32). The integrand in (3.37) can therefore be written

(3.39) \begin{equation} \frac {\eta }{g-\alpha \eta ^2}=\frac {2}{3\alpha (\eta _N-\eta )}+O(1), \end{equation}

implying that $\ln (f)$ has the leading-order behaviour in $(\eta _N - \eta )$

(3.40) \begin{align} \ln f &\sim -\frac {2}{3\alpha }\int _0^\eta \frac {\text{d}\eta }{\eta _N-\eta }-\ln (\eta _N-\eta ) + C_1\nonumber \\&\sim \frac {2-3\alpha }{3\alpha }\ln (\eta _N-\eta ) + C_2, \end{align}

and hence

(3.41) \begin{equation} f \sim A(\eta _N - \eta )^{(2 - 3\alpha )/(3\alpha )}, \end{equation}

for related integration constants $C_1$ , $C_2$ , $A$ . Writing $f=(\eta _N-\eta )^{(2-3\alpha )/(3\alpha )}\phi (\eta )$ , we use (3.30), (3.32b , c ) and (3.36) to find $\phi$ to first order in $(\eta _n-\eta )$ near the nose, using the same method as in (3.17)

(3.42) \begin{equation} \phi \sim A\!\left (1-\frac {(2-3\alpha )^2}{24\alpha \eta _N}(\eta _N-\eta )\right )\!. \end{equation}

Having removed the singularity in $f$ , we now integrate (3.30) backwards directly from the nose, using the same technique as in (3.15), with $A$ determined by shooting to meet the condition $\epsilon _0^{-p} f(\epsilon _0) = 1$ . For numerical ease, we used this latter approach. An example, for $n = 1.25$ , is shown in figure 4( $b$ ). With this value of $n$ , we find that $p = 0.209$ , $\alpha = 0.453$ , $\eta _N = 2.06$ and $A = 0.639$ , all to three significant figures. The small parameter value $\epsilon _0=10^{-8}$ was used when integrating.

Features of the large-time similarity solution are illustrated in figure 4 $(c)$ . The viscosity of the current becomes very large and increases with radius. Relative to those values, the fluid appears to emerge from the cylindrical source with zero viscosity, although in reality, the viscosity is always positive at the source. The decay in radial velocity seen in the Newtonian case (see figure 2 $c$ ) contributes to, and is itself magnified by, the radial increase in viscosity, leading to greater radial volume flux gradients than seen in the Newtonian case, and hence a faster rate of thickening, allowing the current thickness to exceed the cylindrical source height. The increased thickness of the current additionally contributes to the total resistance of the flow to stretching and thinning, leading to a current whose thickness and viscosity become asymptotically large, and whose velocity becomes asymptotically small, as $t\to \infty$ (note that the depth-integrated radial volume flux $hy$ remains $O(1)$ everywhere). However, the current thickness must still match the cylindrical source height at $r=1$ , and hence there is a transition from a region with $O(1)$ velocity and thickness at the nozzle, to a region with asymptotically small velocities, and large thicknesses and viscosities, in the bulk of the current, which is the origin of the singular behaviour in $f$ and $g$ near $\eta =0$ .

4.1. Similarity solutions for extrusion of Newtonian and power-law fluids

Noting that there is a fixed scale for $h$ in this scenario, any similarity solution must have  $r\sim t^{1/2}$ by conservation of mass, so we write

(4.2a,b) \begin{equation} h(r, t) = f(\eta ), \quad y(r, t) = g(\eta ), \quad \hbox{where} \quad \eta = {r\over t^{1/2}}, \end{equation}

with

(4.3a,b) \begin{equation} r_G = \eta _G t^{1/2}, \quad r_{\!N} = \eta _N t^{1/2}. \end{equation}

The dimensionless functions $f$ and $g$ satisfy the differential (3.28)–(3.30) with $\alpha = ({1}/{2})$ . We consider cases in which the current just expands to fill the gap, so that the fluid meets the bounding walls tangentially, with $f'(\eta _G)=0$ . Note that similarity solutions can be found for a non-zero contact angle at the grounding line, giving $f'(\eta _G)=\lambda$ for any value of $\lambda \lt 0$ . The solutions we present here, for $f'(\eta _G)=0$ , seem appropriate as an approximation if there is a thin lubricating layer between the extruded fluid and the confining walls, to which the ambient fluid is connected smoothly.

The boundary conditions are therefore

(4.4a–c) \begin{equation} f =1, \quad f' = 0, \quad g = 1 \quad \hbox{at} \quad \eta =\eta _G, \end{equation}

(4.4d–f) \begin{equation} f = 0, \quad g = {\eta _N^2\over 2}, \quad g' = {\eta _N\over 4} \quad \hbox{at} \quad \eta = \eta _N. \end{equation}

It should be noted that the kinematics and dynamics at $\eta _N$ , and hence the boundary conditions, are identical to the unconfined case (3.32), with $\alpha =1/2$ .

As before, shown in (3.41), an asymptotic analysis near the nose of the current is used to determine that $f\sim C(\eta _N - \eta )^{1/3}$ as $\eta \to \eta _N$ . Thus we have six conditions on a coupled system consisting of the second-order momentum equation and first-order evolution equation, allowing us to find the unknowns $\eta _G$ , $\eta _N$ and $C$ . The equations were solved using the Matlab integrator ode45. Note also that $f'(\eta _G) = 0 \Rightarrow g'(\eta _G) = 0$ , which can be seen by substituting the initial conditions (4.4a – c ) into (3.30) with $\alpha ={1}/{2}$ . Example calculations are shown in figure 7( $a$ ) for $n = 1$ (identical to the unconfined case shown in figure 2), 1.25, 2 and $\infty$ (the perfect plastic limit). Note that the power-law index enters the equations only through (3.29), which has a straightforward, finite limit as $n \rightarrow \infty$ .

Figure 7.

( $a$ ) Solutions to (3.28)–(3.30) with boundary conditions (4.4) for the self-similar half-thickness $f(\eta )$ of an axisymmetric extrusion between rigid, horizontal plates (indicated by the dashed line), with power-law exponents $n=1$ , 1.25, 2 and $\infty$ . The fluid fills the gap between the plates until $\eta _G=0$ , 0.435, 0.648, 0.826, and the end of the current is at $\eta _N=1.71$ , 1.62, 1.58, 1.55, all to three significant figures. ( $b$ ) The solid curves show the variation of $\eta _G$ (lower curves) and $\eta _N$ (upper curves) with $n$ , the exponent of the power-law fluid being extruded between parallel plates. The dashed curves show the extent (upper curve at $\eta =\eta _N ={\sqrt {2}}$ ) of a strictly two-dimensional current (i.e. $h\equiv 1$ ) and the position (lower curve at $\eta = \eta _0$ ) where the pressure is zero (given by (4.6)). The pressure, determined by the constraint of two-dimensionality, is negative between the dashed curves, indicating that the fluid would tend to pull away from confining plates were the constraint of two-dimensionality to be relaxed. The diamonds represent the asymptotes as $n\to \infty$ for the solid lines (black diamonds) and dashed lines (white diamonds).

In figure 7( $b$ ), we show how $\eta _G$ and $\eta _N$ vary with $n$ . This result can be compared with the strictly two-dimensional calculations presented by Sayag & Worster (Reference Sayag and Worster2019b ), which is equivalent to imposing that $h$ is constant. In that case, $\eta _N = \sqrt {2}$ , and the pressure field is

(4.5) \begin{equation} p \propto 2(n-1)\left ({\eta \over \eta _N}\right )^{-2/n} - 2n, \end{equation}

which is negative for

(4.6) \begin{equation} \eta \gt \eta _0 \equiv \left (1 - {1\over n}\right )^{n/2}\eta _N. \end{equation}

Note that $\eta _0\to \eta _N/\sqrt {\mathrm{e}}\text{ as }n\to \infty$ . Given that $h$ is constant in this calculation, there is no vertical velocity, and so the normal stress on the confining plates is $p-2\mu \partial w/\partial z= p$ . We can therefore expect the intruding fluid to lose contact with the plates once the pressure is negative, and estimate $\eta _G$ by $\eta _0$ . In figure 7( $b$ ), we show with dashed curves the two-dimensional results of Sayag & Worster (Reference Sayag and Worster2019b ), having $\eta _N = \sqrt {2}$ and $\eta _G = \eta _0$ .