Past exam of the theoretical mechanics course of the University of Oxford / 2008 / A / 4

Let $X=x_{\dim }/L$, $Y=y/h_0$, scale horizontal and vertical velocities by $U$ and $U{h}_0/L$, and define $P=(p_{\dim }-p_0)h_0^2/(\mu UL)$. The small aspect ratio makes $P$ independent of $Y$ at leading order. The second small parameter, $U{h}_0^2/(\nu L)$, makes inertia negligible compared with transverse viscous stress. Thus the lubrication equations are $u_{YY}=P_X$ and $u_X+V_Y=0$.

No slip gives $u(X,0)=1$, $u(X,h)=0$, soNo penetration at the two solid walls and continuity make $q$ constant. Therefore, writing the dimensionless variables again as $x,p$,The pressure scale is set by balancing the longitudinal pressure gradient with shear across the thin gap, not by the inertial dynamic-pressure scale.

Integrate once and use $h=e^{-\alpha x}$ to obtain $p^{\prime }=6e^{2\alpha x}+C{e}^{3\alpha x}$. The two zero endpoint pressures determineConsequentlyThe constant $M={\int }_0^1{h}^{-2}dx/{\int }_0^1{h}^{-3}dx$ is a weighted average of $h$ with positive weight $h^{-3}$, so $e^{-\alpha }<M<1$. Since $e^{-\alpha x}$ decreases strictly, $p^{\prime }$ changes sign exactly once, from positive to negative. Thus the unique maximum is atFor small $\alpha$, $\log M=-\alpha /2-5{\alpha }^2/24+O({\alpha }^4)$, so $\boxed{x_m=\frac{1}{2}+\frac{5}{24}\alpha +O({\alpha }^3)\sim \frac{1}{2}}$. The exponential-gap slider bearing provides a particularly simple example where a converging gap produces positive load-bearing pressure.