Past exam of the theoretical mechanics course of the University of Oxford / 2009 / A

For a material volume $V(t)$ moving with velocity $\mathbf{u}$, the Reynolds transport theorem isIt applies componentwise to vector quantities. Cauchy's stress theorem represents traction on a surface with outward unit normal $\mathbf{n}$ as $t_i={\sigma }_{ij}{n}_j$. Conservation of mass gives $d{\int }_{V(t)}\rho dV/dt=0$ and hence ${\partial }_t\rho +\nabla \cdot (\rho \mathbf{u})=0$. Constant density reduces this to $\boxed{\nabla \cdot \mathbf{u}=0}$.

With no body force, conservation of momentum and the divergence theorem giveApplying transport and then mass conservation cancels the terms proportional to $u_i$, leaving $\rho ({\partial }_t{u}_i+u_j{\partial }_j{u}_i)={\partial }_j{\sigma }_{ij}$. For the stated Newtonian stress, constant viscosity gives ${\partial }_j{\sigma }_{ij}=-{\partial }_{i}p+\mu \Delta u_i+\mu {\partial }_i(\nabla \cdot \mathbf{u})$. The last term vanishes. Since the material volume was arbitrary,A body force, if present, would have to be added to the momentum balance; it is absent from the specified model.

For $\mathbf{u}=u(y,t){\mathbf{e}}_x$, incompressibility is automatic and the nonlinear advection term is zero. With no pressure gradient the streamwise momentum equation is $\boxed{\rho u_t=\mu u_{yy}}$. Seek its periodic response as $u=\mathrm{Re}[Uf(y)e^{i\Omega t}]$. The amplitude satisfies $f^{\prime \prime }=i\rho \Omega f/\mu$ with $f(\pm d)=1$. Since $[k(1+i){]}^2=i\rho \Omega /\mu$ for $k=\sqrt{\rho \Omega /(2\mu )}$, even symmetry givesOdd components vanish because the two boundary amplitudes are equal. This is the unique periodic solution. Without a specified initial profile, decaying zero-wall diffusion transients may be added, so the formula should not be read as a claim that every transient is absent.

As $kd\to \infty$, the wall motions penetrate only a distance $k^{-1}=\sqrt{2\mu /(\rho \Omega )}$. Near either wall, with inward distance $s=d-|y|$, the leading velocity isThe two oscillatory viscous layers are separated by an almost motionless core; the phase lag increases inward. This is in-phase oscillating parallel-plate flow.

Let $\delta =R{e}^{-1/2}$ and use the inner variables $\boxed{Y=y/\delta =R{e}^{1/2}y,V=v/\delta =R{e}^{1/2}v}$. Continuity balances $u_x$ with $V_Y$. In the tangential momentum balance, advection is order one while the normal viscous term is $R{e}^{-1}{\delta }^{-2}{u}_{YY}=u_{YY}$. Streamwise viscosity is smaller by $R{e}^{-1}$. The normal momentum balance forces the leading pressure to be independent of $Y$. ThusOn the stationary impermeable plate, no-slip boundary condition and no-penetration boundary condition give $u=V=0$ at $Y=0$, $x>0$. Matching to the uniform outer stream gives $u\to 1$ as $Y\to \infty$, and the leading-edge/upstream data are $u\to 1$ as $x\downarrow 0$ at fixed $Y>0$. There is no requirement that the scaled transverse velocity $V$ vanish at infinity; its physical value $v=\delta V$ matches a smaller outer correction. The outer streamwise Euler equation with constant speed one gives $p_x=0$, and $p_Y=0$ carries this pressure gradient throughout the layer.

The Von Mises boundary-layer transformation uses $\Psi$ as transverse coordinate. With $u=U(x,\Psi )$ and ${\Psi }_x=-V$, ${\Psi }_Y=u=U$, the chain rule givesThe streamwise advection terms reduce to $u{u}_x+V{u}_Y=U{U}_x$, so, for $Y>0$ where $U>0$, division by $U$ givesContinuity was already enforced by the stream function. Substitute $U=g(\eta )$ with $\eta =\Psi /\sqrt{x}$. Then $U_x=-\eta g^{\prime }/(2x)$ and ${\partial }_{\Psi }(U{U}_{\Psi })=(g{g}^{\prime }{)}^{\prime }/x$, yieldingThe wall coordinate is zero because $\Psi$ was set to zero there, and the far condition matches unit external speed. This is the transformed Blasius similarity problem. One must not additionally prescribe a finite $g^{\prime }(0)$: the coordinate change degenerates at the no-slip wall, and the regular physical shear corresponds to $g\propto \sqrt{\eta }$ near zero. This similarity reduction is obtained for the attached monotone boundary layer assumed in the question.

Existence of a positive normalized profile can also be seen directly. Put $q=g{g}^{\prime }$ and choose $q(0)=a>0$. In $g$ coordinates the regular initial-value system is $dq/dg=-\eta /2$, $d\eta /dg=g/q$, with $\eta (0)=0$. It starts as $g\sim \sqrt{2a\eta }$. For finite positive $\eta$, $q^{\prime }/q=-\eta /(2g)$ keeps $q$ positive and $q\le a$, while $(g^2{)}^{\prime }=2q$ gives $g\le \sqrt{2a\eta }$. Consequently $q^{\prime }/q\le -\sqrt{\eta }/(2\sqrt{2a})$, so $q$ is integrable at infinity. It follows that $g$ increases to a finite positive limit $M$. The rescaled profile $\tilde{g}(\eta )=M^{-1}g(\sqrt{M}\eta )$ satisfies the same equation and has limit one. This supplies the requested similarity solution with both boundary conditions, rather than only its formal differential equation.

Scale $x,y$ by $L$, $z$ by $\delta L$, tangential velocities by $U$, normal velocity by $\delta U$, and time by $L/U$. Continuity balances all three derivatives. Tangential viscous forces scale as $\mu U/({\delta }^2{L}^2)$, hence their pressure balance requiresThe inertial/normal-viscous ratio is ${\delta }^2\rho LU/\mu$, so it is negligible under the given smallness condition. Tangential viscous derivatives are smaller by ${\delta }^2$; the normal pressure balance gives $p_z=0$ at leading order. In dimensional variables the remaining equations are continuity, $\mu u_{zz}=p_x$, $\mu v_{zz}=p_y$, $p_z=0$.

No-slip boundary condition at each plate imposes $u=v=0$, while their normal velocities are $w(0)=0$ and $w(h)=\dot{h}$. Integrating the two tangential equations twice givesIntegrate continuity through the gap. Since $h$ is independent of in-plane position and the tangential velocity vanishes at the upper wall, the flux divergence is $\nabla \cdot (h\overline{\mathbf{u}})$ and the normal contribution is $\dot{h}$. Therefore

For spatially uniform $h(t)$, the gap equation gives $\Delta p=12\mu \dot{h}/h^3$. Axisymmetry makes this $r^{-1}(r{p}_r{)}_r=12\mu \dot{h}/h^3$. Its general integral contains a logarithmic term; bounded pressure at zero excludes that term. Enforcing zero gauge pressure at the outer edge gives the axisymmetric squeeze-film pressureThe mean radial velocity is ${\overline{u}}_r=-h^2{p}_r/(12\mu )=-r\dot{h}/(2h)$. At the material edge this must equal $\dot{R}$, so $2h\dot{R}+R\dot{h}=0$. HenceIt is conservation of the incompressible liquid volume, with no liquid entering or leaving through the plates or material rim.

When the plates approach, $\dot{h}<0$ and $p>0$ inside: the liquid is squeezed outward. Separation requires $p<0$ relative to the surrounding air and inward motion of the edge. Negative gauge pressure is not intrinsically impossible, but substantial suction can cause cavitation or air invasion, and an advancing air/liquid interface can develop fingering. Thus the same mathematical lubrication solution exists for either sign, whereas its opening-plate physical validity needs these additional failure mechanisms to be absent; the closing case is the robust regime of the stated idealized model.