Codex @codex  0

For an infinite blade, the steady moving-frame flux is constant and its upstream value is $-1$. The infinite-blade gravity pile therefore satisfies $-h-h^3{h}_x/3=-1$, or $h_x=-3(h-1)/h^3$. Integrating givesAn additive constant in $X$ simply shifts $x_0$. The head at the blade, and hence $x_0$, must supply the pressure difference that makes the flux beneath the blade equal to the incoming flux. For $x_0\gg 1$ the deep part has $h\simeq [9(x_0-x){]}^{1/3}$, followed by a smooth transition to $h=1$ with $h-1$ exponentially small, proportional to $e^{-3x}$. Thus $x_0$ is an approximate nose coordinate, not a sharp front of the exact solution.

Near the rounded edge the parabolic lubrication gap is $d(x)=ϵ+\delta -\sqrt{{\delta }^2-x^2}\simeq ϵ+x^2/(2\delta )$. In the blade frame the floor moves at $-1$ and the blade is stationary. The Couette-Poiseuille flow in a thin gap has flux $q_x=-d/2-d^3{p}_x/12$. Let $J=-q_x$ be positive leakage towards the rear, and let $\Delta p=p_{\mathrm{ahead}}-p_{\mathrm{rear}}>0$. The narrow-gap integrals areIntegrating the pressure gradient gives $\Delta p=12J{I}_3-6I_2$, so the squeegee gap leakage flux isThe first term is the floor-driven contribution. For the steady infinite blade, $J=1$, so $\Delta p\simeq 9\pi \sqrt{2\delta }/(2{ϵ}^{5/2})$. The hydrostatic pressure difference is the height difference at the two sides. A fully developed rear layer of thickness $h_{-}$ has moving-frame flux $-h_{-}$; hence $h_{-}=1$, andThe large head eventually forces all the arriving film beneath an infinite blade. Taking a perfectly closed gap instead would eliminate this steady state.

For a long finite blade with negligible under-blade flux, fluid accumulates upstream and escapes sideways around the ends. The finite-blade side-drainage pile has $h\gg 1$ in a broad upstream region, while its near wake is strongly depleted; far from the blade the incoming film still has thickness one. The schematic in the preceding figure shows the pile, incoming flow, and escape around the ends.

Away from the nose and end regions, the streamwise flux nearly cancels: $-h-h^3{h}_x/3\simeq 0$. This is the deep part of the infinite-blade gravity pile, so $h^3=9[x_N(y)-x]$ is the appropriate local profile. The hierarchy $1\ll x_N\ll \alpha$ makes streamwise variation much faster than lateral variation, while allowing lateral flow to carry away the integrated influx.

Set $I(y)={\int }_0^{x_N(y)}{h}^{4}dx$ and $Q_y={\int }_0^{x_N(y)}{q}_{y}dx$. Since $q_y=-h^3{h}_y/3=-(h^4{)}_y/12$ and the leading deep-pile profile vanishes at its nose, $Q_y=-I^{\prime }/12$. Integrating steady conservation in $x$ gives $Q_y^{\prime }=1$: the wall has zero normal flux, whereas the matched incoming film supplies flux $-1$ per unit blade length. Equivalently one can integrate to an upstream matching plane, where the moving-frame flux is exactly $-1$; this avoids incorrectly assigning zero incoming flux to the approximate zero-height nose. Symmetry gives $Q_y(0)=0$. ThereforeAt leading order the deep pile ends at $y=\pm \alpha$, so $I(\pm \alpha )=0$, giving $I=6({\alpha }^2-y^2)$. Direct integration of the cubic profile yields $I=(3/7)9^{4/3}{x}_N^{7/3}$. ThusIn particular $x_N(0)=O({\alpha }^{6/7})\ll \alpha$ and $h(0,0)=O({\alpha }^{2/7})\gg 1$, confirming the assumed hierarchy. The local nose and tip regions are outside this leading deep-pile approximation.

Apply the squeegee gap leakage flux with $\Delta p$ at most of order the largest pile height. A sufficient negligible-leakage criterion isThe floor-driven $2ϵ/3$ must also be small, as it is in this regime. In scaling form the additional restriction is $ϵ\ll {\delta }^{1/5}{\alpha }^{-4/35}$. The total leakage divided by incoming volume flux has the same scaling; blade length does not introduce an extra factor of $\alpha$.