Force-free pressure drop of an annular lubrication plug ($\Delta p=\frac{2\mu }{a}\int (4U/h-6Q/h^2)dx$)

The pressure force on a plug-and-film control volume is $\pi a^2\Delta p$. Its opposing wall traction is $2\pi a\mu \int w_y(0)dx$, where $w_y(0)=4U/h-6Q/h^2$. Equating them for a freely moving body proves the displayed formula. Writing $s=2Q/U$, $h=sH$ and $x=\sqrt{s/\kappa }X$ converts it into $H(-\tilde{L})-H(\tilde{L})=C\lambda \int [4/(3H)-1/H^2]dX$, with $C=s/a$ and $\lambda =6\mu U/(\alpha \sqrt{\kappa }{s}^{5/2})$.

For $H_X+\lambda (H^{-2}-H^{-3})=X$, small $\lambda$ gives $H_0=(b^2+X^2)/2$. In the large-domain limit its inverse-power integrals are $I_1=2\pi /b$, $I_2=2\pi /b^3$ and $I_3=3\pi /b^5$. The force-free condition is $I_2-I_3=C(4I_1/3-I_2)$, or $2b^2-3=C(8b^4/3-2b^2)$. Expanding the root near $b^{-2}=2/3$ proves the displayed minus sign and leading pressure drop. The absolute downstream pressure relative to the elastic reference pressure fixes $s$ and therefore closes the speed-pressure relation.