= Pressure lobes in a symmetric translating lubrication gap
{title2=$p(x)=6\arctan x-\dfrac{3\pi x}{\sqrt{2(1+x^2)}}$}
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 $Q=\pi/(4\sqrt2)$. 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.
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