Let the constant in the integrated Reynolds equation be $C$:Both endpoint pressures are zero, so the integral of $d{p}^{\prime }/d{x}^{\prime }$ on the real line vanishes. DefineThe substitution $x^{\prime }=\sqrt{2}\tan \theta$ givesThe second value follows by integrating the supplied identity for ${\cos }^4\theta$; the nonconstant cosine terms have zero integral over this interval. Hence $12C{I}_3-6I_2=0$, givingThe pressure gradient is nowDirect differentiation showsThe candidate tends to zero at both ends, so the additive constant is zero andTogether with the local Couette-Poiseuille flow profile, this solves the sliding-plate flow through a parabolic gap.

To restore dimensions, integrate $u=U{u}^{\prime }$ across $dz=h_{0}d{z}^{\prime }$. The signed dimensional volume flux per unit width, positive in the $+x$ direction, isThus the magnitude drawn in the direction of the moving plate is $2U{h}_0/3$. For a plate width $B$ large enough to neglect side-wall effects, the total volumetric flow rate is $-2BU{h}_0/3$; without a specified width the answer is per unit width.

The Couette flow and plane Poiseuille flow contributions to the dimensional volume flux per unit width are respectivelyThey point in the same direction precisely when $Q_P<0$, givingAt the endpoints the plane Poiseuille flow contribution is zero. Outside that interval it opposes the Couette flow contribution, while their sum remains the constant negative volume flux per unit width $-2U{h}_0/3$. Thus pressure assists the moving plate near the narrowest gap and drives a compensating return contribution farther away.

Write $H(x^{\prime })=h^{\prime }(x^{\prime })$ and $p_x^{\prime }=d{p}^{\prime }/d{x}^{\prime }$ below, so the subscript $x$ in this shorthand denotes differentiation with respect to $x^{\prime }$. The vertical pressure gradient is zero, and integrating $u_{z^{\prime }{z}^{\prime }}^{\prime }=p_x^{\prime }$ twice with the wall boundary conditions gives the local Couette-Poiseuille flowThe first two terms are Couette flow due to the moving plate; the last is plane Poiseuille flow due to the local pressure gradient. Its dimensionless volume flux per unit width isUse the Leibniz integral rule and incompressibility to differentiate this integral:Thus the flux is independent of position, andThe normal velocity can then be recovered from incompressibility:It is zero on the lower wall by construction and zero on the upper wall by the constant-volume flux relation. This supplies the remaining component of the leading velocity field.

The endpoint pressure conditions are matching conditions, not an assertion that the lubrication approximation remains valid at arbitrarily large dimensional $x$. On an outer length scale $R$, a viscous pressure variation is of size $\mu U/R$, whose ratio to the inner scale isEven an outer inertial pressure scale $\rho U^2$ has ratioConsequently the gap pressure is much larger than either outer variation. Matching to fluid with a common ambient pressure at both ends, and taking that value as zero gauge pressure, givesThere is an overlap region $1\ll |x^{\prime }|\ll {\delta }^{-1}$ before the wall slope becomes order one. The infinite endpoints represent matching through that region. An externally imposed pressure difference comparable with $P$ would change these conditions and the resulting volume flux; equal outer pressure is the physical assumption being used here.

The flat plate moves tangentially and the curved plate is stationary. The no-slip boundary condition and no-penetration boundary condition giveAt the curved plate, impermeability can also be written $w=u{h}_x$; together with zero tangential velocity, this gives both components zero. The native PDF writes $v^{\prime }=0$ in the scaled wall conditions although the normal component is defined as $w^{\prime }$. We use $w^{\prime }$ consistently.

The parabolic lubrication gap changes by order $h_0$ when $x^2/R$ is order $h_0$. Its natural horizontal scale is thereforeThe proposed normal velocity scale $h_{0}U/L$ then makes the two terms of the continuity equation comparable. Balancing the horizontal pressure gradient against the transverse viscous term in the Navier-Stokes equation requires $P/L=\mu U/h_0^2$, soHere pressure is measured from a common reference value, which is chosen at the outer fluid in part (b).

For clarity, define the dimensionless material derivativeSubstitution of all the scalings gives the exact dimensionless equationsFor example, the factor in the last left side is obtained by dividing the vertical acceleration scale $\rho h_0{U}^2/L^2$ by the vertical pressure gradient scale $P/h_0$.

Taking $\delta \ll 1$ and ${\delta }^2\mathrm{Re}\ll 1$ leaves the lubrication approximationThe scaled gap and boundary conditions areThe Reynolds number itself need not be small: it is ${\delta }^2\mathrm{Re}$ that measures inertia relative to transverse viscous resistance in this thin geometry.