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Let $x$ measure distance down the incline and $y$ distance perpendicular to it, with the wall at $y=0$ and the free surface at $y=h$. A uniform steady falling film flow has $\mathbf{u}=u(y){\mathbf{e}}_x$. Its convective acceleration vanishes because $\mathbf{u}\cdot \nabla =u{\partial }_x$ and ${\partial }_{x}u=0$. A flat surface exposed to constant atmospheric pressure hasThus ${\partial }_{x}p=0$, and the downslope component of the Navier-Stokes equations reduces to $0=g\sin \alpha +(\eta /\rho )u^{\prime \prime }$. The printed pressure-free equation must be understood as this downslope projection: its full vector version would incorrectly omit the normal hydrostatic pressure gradient that balances $g\cos \alpha$.

The no-slip boundary condition gives $u(0)=0$. Neglecting air shear stress makes the surface shear-free, so $u^{\prime }(h)=0$. Two integrations give the uniform gravity-driven Newtonian film profileThe velocity increases up to the free surface, whereThis assumes a uniform laminar Newtonian fluid film, a stationary wall and negligible surface shear; a varying thickness or externally imposed downslope pressure gradient would alter the balance.

Meaning and microscopic origin. For a Newtonian fluid, the dynamic viscosity $\eta$ measures tangential momentum transfer between fluid layers moving at different speeds. For the Newtonian shear considered here, ${\tau }_{xz}=\eta d{v}_z/dx$, with units $\mathrm{Pa}\mathrm{s}$ for $\eta$. Molecular motion and collisions transport momentum across layers in a gas; intermolecular interactions contribute strongly in liquids. The resulting viscous forces oppose relative shear motion and convert organized kinetic energy to heat. The constitutive law is Newtonian and isotropic; incompressibility alone does not establish it.

Vertical force balance. Choose $z$ downward, as in the original diagram, and $x$ outward normal to the plate. In a uniform steady film, $\mathbf{v}=v_z(x)\hat{\mathbf{z}}$ has zero advective acceleration and no along-plate pressure gradient. On a slice of thickness $dx$, the net shear force per unit face area is $\tau (x+dx)-\tau (x)$, while its weight per unit area is $\rho gdx$. ThereforeHere $\tau =\eta d{v}_z/dx$ denotes the downward traction on a face with positive $x$ normal. The opposite face has the opposite traction orientation, accounting for the sign in the force difference.

Boundary conditions. At the fixed solid wall, the no-slip boundary condition is $v_z(0)=0$, with no normal penetration. At the free surface $x=d$, neglect air shear, so $\tau (d)=0$ and hence $v_z^{\prime }(d)=0$. The flat free surface has normal pressure $p_{\mathrm{atm}}$, neglecting surface-curvature effects. These are a velocity condition at the wall and a stress condition at the free surface, not zero velocity at both boundaries.

The parabolic speed profile. Integrating ${\tau }^{\prime }=-\rho g$ from the shear-free surface gives $\tau (x)=\rho g(d-x)$. Thusand the wall condition yieldsIt is zero at the wall and largest at the free surface, where its gradient vanishes. The wall shear $\rho gd$ supports the downslope weight of the film per unit plate area.

Volume flux per unit width. Integrate through the depth:This is volume flow per unit length in $y$, with units ${\mathrm{m}}^2{\mathrm{s}}^{-1}$.

Inclining the plate. Keep $x$ normal outward from the plate and use $s$ downslope. The gravitational components are $g\sin \theta$ downslope and $-g\cos \theta$ normal to the plate. For a uniform film of normal thickness $d$, the steady slice equations areThe free-surface pressure condition givesUniform $d$ makes ${\partial }_{s}p=0$. The same no-slip and zero-surface-shear conditions then give the uniform gravity-driven Newtonian film,The component normal to the plate changes the hydrostatic pressure, while the tangential component drives the flow. The thickness is measured normal to the tilted plate.

The inclined-film volume flux. Integrating the new profile givesIt recovers the vertical result at $\theta =\pi /2$ and vanishes on a horizontal plate at fixed uniform depth. A varying depth, imposed pressure gradient or significant air shear would require additional terms; the solution here uses the uniform film specified by the simple draining-plate model.

Take $z$ positive away from the wall into the liquid, and write $p_a$ for atmospheric pressure. For a constant-mass density Newtonian fluid, the two-dimensional Navier-Stokes equation and continuity equation areThe stationary wall imposes the no-slip boundary condition and no-penetration boundary condition, $u=w=0$ at $z=0$. For a flat free surface with negligible air shear stress, the stress boundary condition is $u_z=0$, $p=p_a$ at $z=h_0$; the kinematic boundary condition is $w=0$ there.

For steady flow independent of $x$, these equations give $w=0$, $\mu u_{zz}=-\rho g\sin \alpha$ and $p_z=-\rho g\cos \alpha$. Two integrations and the boundary conditions give the uniform gravity-driven Newtonian film:Its surface velocity is $U_0=\rho g{h}_0^2\sin \alpha /(2\mu )$, while its mean velocity is $2U_0/3$.