= Solution
In a slender symmetric <viscous sheet>, the two broad faces have zero tangential traction. A cross-thickness variation of $w$ on the thickness scale would produce leading shear $\mu w_x$, whereas $\mu u_z$ is smaller by the square of the aspect ratio. The leading transverse momentum balance and the shear-free conditions therefore give $w_x=0$: write $w=w(z,t)$.
<Incompressibility> gives $u_x=-w_z$. Symmetry about $x=0$ sets the integration constant to zero, so
$$
\boxed{u=-xw_z.}
$$
The normal traction on either nearly vertical face matches the ambient <hydrostatic pressure>, giving $\sigma_{xx}=-p_a$ to leading order. The <Newtonian fluid stress tensor> then yields
$$
\sigma_{xx}=-p+2\mu u_x=-p-2\mu w_z=-p_a,
\qquad p=p_a-2\mu w_z,
$$
$$
\boxed{\sigma_{zz}=-p+2\mu w_z=4\mu w_z-p_a.}
$$
The factor four is the planar extensional resistance, as in the <planar viscous-sheet stretching equations>, rather than the factor three for uniaxial extension with two contracting transverse directions.
The following slice diagram shows all axial-force contributions per unit $y$-width; the end arrows indicate positive tensile stress and reverse if that signed stress is negative.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2001/iii/paper-44-sheet-forces.png]
{title=Signed end tractions, ambient side pressure and weight on a widening viscous-sheet slice}
The two end tractions have net $z$-component $\partial_z(h\sigma_{zz})\delta z$. The normal <pressure> on the inclined side faces contributes $p_a h_z\delta z$, and <gravitational acceleration> gives the weight $\rho gh\delta z$. Hence <force balance> is
$$
\partial_z(h\sigma_{zz})+p_a h_z+\rho gh=0.
$$
Substitute the stress expression and $p_a'=\rho_a g$:
$$
\partial_z(4\mu h w_z)-h p_a'+\rho gh=0,
$$
so
$$
\boxed{\frac{4\mu}{h}\partial_z(hw_z)+(\rho-\rho_a)g=0.}
$$
Keeping the side-pressure component is essential: dropping it would retain an incorrect dependence on the arbitrary ambient <pressure> level. The remaining equation from <conservation of mass> is
$$
\boxed{h_t+(hw)_z=0,\qquad D_th=-hw_z.}
$$
It follows either by integrating <incompressibility> across the moving faces or directly from the face <kinematic boundary condition>.
When effective <gravitational acceleration> vanishes, $4\mu h w_z$ is independent of $z$. It equals the applied excess tensile <force> $F(t)$, with ambient <pressure> removed from the end loading. Along a material element labeled by its initial position $z_0$,
$$
\frac{Dh}{Dt}=-hw_z=-\frac{F(t)}{4\mu}.
$$
Integration gives <uniform material thinning under planar-sheet tension>:
$$
\boxed{h(z,t)=h_0(z_0)-\Delta(t),\qquad
\Delta(t)=\frac1{4\mu}\int_0^tF(s)\,ds.}
$$
A material interval conserves its area: $h(z,t)\,dz=h_0(z_0)\,dz_0$. Thus
$$
\boxed{\frac{\partial z_0}{\partial z}=\frac{h(z,t)}{h_0(z_0)},\qquad
\frac{\partial z}{\partial z_0}=\frac{h_0(z_0)}{h_0(z_0)-\Delta}.}
$$
With the specified fixed material origin, integrate the latter relation to obtain
$$
\boxed{z=z_0+\int_0^{z_0}\frac{\Delta\,ds}{h_0(s)-\Delta}.}
$$
These formulas hold while the material thickness remains positive.
For the quadratic initial profile, put $q=H-\Delta>0$. If $k\ne0$, evaluating the integral at $z_0=L_0$ gives
$$
\boxed{L(\Delta)=L_0+
\frac{\Delta}{|k|\sqrt{Hq}}
\tan^{-1}\left(|k|L_0\sqrt{\frac Hq}\right).}
$$
For $k=0$ it instead gives
$$
\boxed{L(\Delta)=\frac{HL_0}{H-\Delta}.}
$$
For a constant positive pulling <force>, $\Delta=Ft/(4\mu)$ first reaches the minimum initial thickness $H$ at
$$
\boxed{t^*=\frac{4\mu H}{F}.}
$$
Both expressions for $L$ diverge there. Write $q=F(t^*-t)/(4\mu)$. If $k\ne0$, the inverse tangent tends to $\pi/2$ and $\Delta\to H$, giving
$$
\boxed{\alpha=-\tfrac12,\qquad
A=\frac\pi{|k|}\sqrt{\frac{\mu H}{F}},\qquad
L\sim A(t^*-t)^{-1/2}.}
$$
For $k=0$,
$$
\boxed{\alpha=-1,\qquad A=\frac{4\mu HL_0}{F},\qquad
L\sim A(t^*-t)^{-1}.}
$$
This is <finite-time extension of a quadratically thickened sheet>. With nonzero $k$, the dominant extension comes from the small material region $|z_0|=O(\sqrt{q/H}/|k|)$ near the thickness minimum, where the integrand is proportional to $1/(q+Hk^2z_0^2)$. Its width shrinks like $\sqrt q$, giving an integrated <divergence> $q^{-1/2}$. The fixed ends are asymptotically far away in those local coordinates, so $A$ does not depend on $L_0$. For a uniform sheet the entire material interval has denominator $q$, giving the stronger $q^{-1}$ <divergence> and retaining its length in $A$. The singularity is a prediction of the ideal slender-sheet model as its minimum thickness tends to zero.
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