Pressure lobes in a symmetric translating lubrication gap ($p(x)=6\arctan x-\frac{3\pi x}{\sqrt{2(1+x^2)}}$)

For a symmetric gap $h(x)=\sqrt{1+x^2}$ on $[-1,1]$, unit lower-wall speed and equal zero gauge pressures at the ends, the flux average for a translating lubrication gap gives $Q=\pi /(4\sqrt{2})$. Integrating $p_x=6(h-2Q)/h^3$ gives the displayed odd pressure. It has a positive maximum at $x=-\sqrt{{\pi }^2/8-1}$ and a negative minimum at the opposite point. The converging and diverging portions of a symmetric gap therefore have opposite pressure lobes even though the geometry itself is even: the directed wall motion breaks reflection symmetry of the flow.