Past exam of the topics in fluid mechanics course of the University of Oxford / 2009 / 1

Hydrostatic pressure varies with height as $p_l-p_{\mathrm{atm}}=-\rho gz$ when the far flat bath is the zero-height reference. Surface tension balances that pressure jump by curvature; with the curvature sign chosen in the equation this becomes $\rho gz=\sigma \kappa$. Here $\rho$ is liquid mass density, $g$ gravitational acceleration, $\sigma$ surface tension and $\kappa$ the signed planar curvature. The capillary length ${\ell }_c=\sqrt{\sigma /(\rho g)}$ balances these effects. Scale both physical $z$ and $h$ by ${\ell }_c$ to obtainFor the wall-rising meniscus $h$ decreases as $z$ rises. At the distant bath, $z\downarrow 0$, $h\to \infty$ and $h_z\to -\infty$, because the free surface becomes horizontal when expressed as $z(x)$. In the first integral this gives $\boxed{A=-1}$.

At zero wall contact angle the interface is tangent to the vertical wall. In the representation $x=h(z)$ this means $h(z_0)=0$ and $h_z(z_0)=0$, so $z_0^2/2-1=0$. The rising contact point is $\boxed{z_0=\sqrt{2}}$. Put $q=z^2/2-1$. Then $h_z=q/\sqrt{1-q^2}=\sqrt{2}(z-z_0)+O((z-z_0{)}^2)$. Integration from the contact point givesThe sign of $h_z$ is negative on the approach from below, while $h$ is positive, as required by the geometry. This local quadratic shape is the zero-contact-angle wall meniscus matching law.

Put $z-z_0=ϵ\bar{z}$, with physical transverse thickness $h=\delta H$. Matching the preceding quadratic meniscus gives $\delta ={ϵ}^2$ when the inner amplitude is normalized as $H\sim {\bar{z}}^2/\sqrt{2}$. After multiplying the transformed film equation by $ϵ$, its two capillary/gravity coefficients are ${\delta }^3\hat{C}{ϵ}^{-3}$ and ${\delta }^2\hat{C}$. Setting the curvature coefficient to one therefore givesThus the inner equation is exactly the stated $ϵH_t+v_0{H}_{\bar{z}}+[H^3(H_{\bar{z}\bar{z}\bar{z}}+ϵ)/3{]}_{\bar{z}}=0$. The leading matching conditions are $H\sim {\bar{z}}^2/\sqrt{2}$ as $\bar{z}\to -\infty$ and a bounded film, with vanishing derivatives, as $\bar{z}\to +\infty$. They connect the static bath-side meniscus to a uniform entrained film.

For positive wall speed varying on the outer time scale, neglect the order-$ϵ$ terms at leading order and setThe parabolic matching coefficient is unchanged because $v_0^{2/3}/v_0^{2/3}=1$. Both remaining differential terms scale as $v_0^{4/3}$, reducing the equation to $3g_{\zeta }+(g^3{g}_{\zeta \zeta \zeta }{)}_{\zeta }=0$. The supplied universal solution therefore yields the quasisteady drag-out matching lawAbove the turn-around region, $H$ and its outer derivatives are order one, while ${\delta }^3\hat{C}={ϵ}^3$ and ${\delta }^2\hat{C}=ϵ$. Hence the leading outer equation and inflow condition areThe boundary value comes from matching, not from setting the film to zero at the wall contact point.

Along a characteristic, $\dot{z}=(1+t{)}^{-2}$ and $\dot{H}=0$. Integration gives $z+1/(1+t)=\xi$, so $\boxed{H=F(\xi )}$. A characteristic entering through $z=z_0$ at time $\tau$ has $\xi =z_0+1/(1+\tau )$ and carries $H=C_0(1+\tau {)}^{-4/3}$. Thus the boundary-fed and initially dry portions giveValues at $\xi \le z_0$ are never sampled in the finite-time physical region and need not be prescribed. The moving coating front is $z=z_0+t/(1+t)$. Explicitly,The incompatible initial corner creates a jump at the leading coating front; a trace convention at the jump does not affect the transported volume. The wall's total stroke is ${\int }_0^{\infty }{v}_{0}dt=1$, so within this leading transport model the final profile isEarlier-deposited fluid is thicker and sits farther up the finite coated interval. Its total scaled volume is $3C_0/7$, agreeing with ${\int }_0^{\infty }{v}_{0}H(z_0,t)dt=C_0{\int }_0^{\infty }(1+t{)}^{-10/3}dt$. This is a finite-stroke coating profile for the leading matched model.