= Solution
The broad faces have curvature zero to leading order, so their <stress boundary condition> gives $\sigma_{xx}=-p_e$. The semicircular edges have curvature $1/h$, giving $\sigma_{yy}=-p_e-\gamma/h$. <Incompressible flow> gives $u_x+v_y+w_z=0$; uniform transverse normal stresses imply uniform transverse extension rates, and eliminating them from the <Newtonian fluid stress tensor> yields
$$
\boxed{\sigma_{zz}=-p_e-\frac{\gamma}{2h}+3\mu w'.}
$$
At an edge, the <kinematic boundary condition> balances axial advection of $b$, lateral strain, and capillary retraction. This gives
$$
\boxed{wb'+\frac b2w'+\frac{\gamma b}{4\mu h}=0.}
$$
The equation $hbw=Q$ expresses conservation of volume flux, while
$$
3\mu hbw'+\frac{\gamma b}{2}=F
$$
states that the total axial tension is constant. Define
$$
\boxed{\Gamma=\frac\gamma{\mu Q},
\qquad T=\frac F{\mu Q}.}
$$
Then $[\Gamma]=L^{-2}$ and $[T]=L^{-1}$, and substitution of $h=Q/(bw)$ gives
$$
\frac{b'}b+\frac12\frac{w'}w=-\frac{\Gamma b}{4},
\qquad
3\frac{w'}w=-\frac{\Gamma b}{2}+T.
$$
Subtracting these logarithmic-derivative equations gives
$$
\left(\log\frac wb\right)'=\frac T2,
\qquad
\boxed{w(z)=\frac{w(0)}{b_0},b(z)e^{Tz/2}.}
$$
The remaining width equation is
$$
\frac{b'}b=-\frac{\Gamma b+T}{6}.
$$
Thus, for $T\ne0$,
$$
\boxed{
\frac1{b(z)}=\frac{e^{Tz/6}}{b_0}
+\frac\Gamma T\left(e^{Tz/6}-1\right),}
$$
while the continuous $T=0$ limit is $1/b=1/b_0+\Gamma z/6$.
If $\Gamma=0$, then $b/b_0=e^{-Tz/6}$, $w/w(0)=e^{Tz/3}$, and <mass conservation> gives $h/h(0)=e^{-Tz/6}$. Therefore a prescribed thinning ratio $R=h(L)/h(0)$ requires
$$
\boxed{F=\mu QT
=\frac{6\mu Q}{L}\log\frac{h(0)}{h(L)}.}
$$
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