 Solution

When internal pressure gradients matter, axial lubrication flow in the low-viscosity interior has volume fluxConservation of cross-sectional area, $(\pi a^2{)}_t+Q_z=0$, and the normal-stress expression for $P$ give

Let $\tau =t^{\ast }-t$. Balancing $2\mu a_t/a$ with $\gamma /a$ gives $a\sim \tau$, so $p=1$. Balancing either pressure contribution after two axial derivatives with $a{a}_t$ then gives $q=1$. DefineAt fixed $z$, $a_t=C(\zeta A^{\prime }-A)$, andSubstitution leaves the parameter-free third-order equationThus the similarity exponents are $\boxed{p=q=1}$; boundary and matching conditions would select a particular profile $A$.