Exponential-gap slider bearing ($h=e^{-\alpha x},x_m=-{\alpha }^{-1}\log \frac{3(e^{2\alpha }-1)}{2(e^{3\alpha }-1)}$)

In a dimensionless thin bearing with lower-wall speed one, stationary upper wall and gap $h(x)$, lubrication theory gives $u=1-Y/h+(p^{\prime }/2)Y(Y-h)$ and constant volume flux $q=h/2-h^3{p}^{\prime }/12$. Equal endpoint pressures, chosen as zero at $x=0,1$, imply $(h^3{p}^{\prime }-6h{)}^{\prime }=0$.

For a converging exponential gap $h=e^{-\alpha x}$, $\alpha >0$, integration gives $p^{\prime }=6e^{3\alpha x}(e^{-\alpha x}-M)$ with $M=3(e^{2\alpha }-1)/(2(e^{3\alpha }-1))$. This constant is the positive weighted average $\int h{h}^{-3}dx/\int h^{-3}dx$, so $e^{-\alpha }<M<1$. It follows that the pressure first rises, reaches a unique maximum at the displayed $x_m$, and falls back to zero. The interior pressure is positive and supports a load. Expansion gives $x_m=1/2+5\alpha /24+O({\alpha }^3)$ as $\alpha \to 0$.