Past exam of the topics in fluid mechanics course of the University of Oxford / 2023 / 1 / a (Long-wave model)

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.

Define the aspect ratio and the thickness-based Reynolds number byUse the nondimensionalizationThe small transverse velocity scale follows from incompressibility. Dropping stars, the base fields become $\bar{u}=2z-z^2$ and $\bar{p}=(1-z)\cot \alpha$. Before subtracting the base state, the dimensionless Navier-Stokes equation readsFor example, division of the downslope momentum equation by $\mu U_0/h_0^2$ makes the inertial coefficient $\rho U_0{h}_0^2/(\mu L)=\mathrm{Re}\epsilon$ and the pressure gradient coefficient $2\epsilon$. The transverse equation has been divided instead by $\mu U_0/h_0^2$ after using its smaller velocity scale; this gives the displayed $\mathrm{Re}{\epsilon }^2$ coefficient.

Now substitute $u=\bar{u}+\hat{u}$, $w=\hat{w}$, $p=\bar{p}+\hat{p}$. Since ${\bar{u}}_{zz}+2=0$, ${\bar{p}}_z=-\cot \alpha$, and the base fields have no $x$ or $t$ dependence, subtraction givesThese are nonlinear equations for the deviations; small amplitude has not been assumed. The later asymptotic expansion takes $\mathrm{Re}=O(1)$ as $\epsilon \to 0$, or more generally requires the retained inertial correction to remain small.

The two wall boundary conditions areDefine the capillary parameter for a falling filmRetain $S=O(1)$ in the long-wave approximation. Thus surface tension is retained in the leading normal stress balance even though geometric slope corrections are small. For $h_x=O(1)$ in the rescaled coordinates, the dimensional surface slope is $\epsilon h_x$.

The three free surface boundary conditions, accurate through $O(\epsilon )$, areThe first is the kinematic boundary condition; the second is the zero tangential traction condition. For the third, the dimensional liquid-side pressure is, to the required order,Division by $\rho g{h}_0\sin \alpha$ gives $p=-3S{h}_{xx}/2+\epsilon w_z$ in dimensionless variables. Subtracting $\bar{p}(h)=(1-h)\cot \alpha$ gives the stated perturbation condition. The additional slope-times-shear contribution to normal stress vanishes at this order because the tangential traction is zero. The omitted geometric and tangential-stress corrections are $O({\epsilon }^2)$ under this ordering.

Use an asymptotic expansion at fixed $z$, evaluating the boundary conditions at the instantaneous $z=h(x,t)$. At leading order, the downslope equation gives ${\hat{u}}_{0,zz}=0$. The wall condition and the surface condition ${\bar{u}}_z(h)+{\hat{u}}_{0,z}(h)=0$ implyHere $u^{(0)}$ denotes the total leading velocity, to distinguish it from the dimensional scale $U_0$. The continuity equation and wall impermeability then giveThe transverse momentum equation gives ${\hat{p}}_{0,z}=0$. The normal stress condition therefore yieldsThe leading volume flux per unit width is $q^{(0)}={\int }_0^h{u}^{(0)}dz=2h^3/3$. Consequently mass conservation gives $h_t=-2h^2{h}_x+O(\epsilon )$. This relation eliminates $h_t$ inside the first inertial correction:Set $G=2h_x\cot \alpha -3S{h}_{xxx}$. The $O(\epsilon )$ downslope equation and boundary conditions areTwo integrations giveThe pressure correction ${\hat{p}}_1$ is not needed for this calculation because its downslope pressure gradient enters only at $O({\epsilon }^2)$.

Integrating the total velocity across the film givesIndeed the three inertial polynomial terms integrate to $\mathrm{Re}{h}^6{h}_x(1/30-1/6+2/3)=8\mathrm{Re}{h}^6{h}_x/15$. Integrating the continuity equation and using the kinematic boundary condition gives $h_t+q_x=0$. Hence the Benney equation isNormal gravity smooths height variations on the upper side of the slope, while the inertial correction can amplify them. Surface tension suppresses sufficiently short wavelengths.