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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.

First estimate transverse relaxation without assuming a thin cross-section. With both depth and width of order $h_0$, a transverse surface-height variation of order $h_0$ gives a pressure gradient of order $\rho g$. The Stokes equation then gives transverse horizontal and vertical velocities of order $g{h}_0^2/\nu$. Moving the surface through distance $h_0$ takesFor longitudinal variations over $L\gg h_0$, the hydrostatic pressure gradient is of order $\rho g{h}_0/L$, so the longitudinal velocity is $u\sim g{h}_0^3/(\nu L)$. Incompressibility gives a vertical velocity $w\sim h_{0}u/L$, henceThus cross-channel variations relax rapidly compared with the longitudinal evolution. These are viscous estimates, requiring inertial relaxation to be negligible; the slow-flow regime in particular excludes a large transverse inertial response. After that initial relaxation the free surface is nearly horizontal across the channel, so its height is $h_0(x,t)$ and its local vertical depth is $h_0-|y|\cot \alpha$.

Retain the full cross-sectional geometry for a fixed $0<\alpha <\pi /2$; no small-angle approximation is used. With $|h_{0x}|\ll 1$, the leading hydrostatic pressure is $p=p_a+\rho g(h_0-z)$. Axial viscous derivatives are small relative to cross-sectional ones. The axial Stokes equation is thereforeIntroduce $Y=y/h_0$, $Z=z/h_0$, and the triangular domainLet the dimensionless cross-sectional Poisson equation beThe two lower sloping sides impose the no-slip boundary condition; the upper horizontal side imposes the stress-free boundary condition. Symmetry would additionally give ${\varphi }_Y=0$ at $Y=0$ if only half the domain were used. The full domain needs no extra midline boundary condition. Define $Q(\alpha )={\int }_{{\mathcal{D}}_{\alpha }}\varphi dYdZ$. ThenThe cross-sectional area is $A=h_0^2\tan \alpha$. Conservation of mass, $A_t+q_x=0$, gives the V-shaped channel gravity current equationIn particular, neither replacing this triangular Poisson problem by a vertically parabolic film nor taking an extreme angle is necessary.

Put $C=gQ(\alpha )/(\nu \tan \alpha )$. For a symmetric fixed-volume similarity solution, write $h_0=t^{-b}f(x{t}^{-d})$. Volume conservation gives $d=2b$, and balancing $(h_0^2{)}_t=C(h_0^4{h}_{0x}{)}_x$ gives $2b+1=5b+2d$. Hence $b=1/7$, $d=2/7$. Integrating the similarity equation once, with zero flux at the centre, yieldsIntegrating again and imposing $h_0=0$ at $x=\pm X(t)$ givesThe constant-volume similarity in a V-shaped channel has a compact support. If $H(t)=h_0(0,t)$, its volume isEliminating $H$ therefore givesThis describes the large-time spreading from a localized release; the point-release profile is singular at $t=0$, and the idealized front has a narrow region where its divergent slope violates the long-wave approximation. Neither feature changes the leading bulk similarity law.

At a fixed nonzero $x^{\ast }$, the depth is initially zero until $X(t)=|x^{\ast }|$. After arrival it rises and then falls. Because $X^2\propto t^{4/7}$, differentiation of $h_0(x^{\ast },t{)}^3=3(X^2-x^{\ast 2})/(7Ct)$ givesThus its unique maximum occurs at $X^2=7x^{\ast 2}/3$; thereafter the depth tends to zero as $t^{-1/7}$. If $t_a$ is the arrival time, $t_{\max }/t_a=(7/3{)}^{7/4}$. At the singular release point $x^{\ast }=0$, the point-source similarity profile instead decreases from the start; a finite initial release regularizes its early behaviour.

The right-hand sketch shows the triangular bulk at half its previous maximum depth, with thin draining films on the previously wetted portions of both walls. These residual wall films in a receding gravity current are not captured by a perfectly dry-wall triangular section. They drain rapidly compared with bulk longitudinal spreading. For an order-one angle, a wall film of thickness $b$ and wetted length $\ell$ has speed $O(g{b}^2/\nu )$ by falling film flow; its drainage time gives $b\sim (\nu \ell /(gt){)}^{1/2}$. At a fixed nonzero station the previous maximum wetted height is finite, so film area decays as $t^{-1/2}$, faster than the bulk area $h_0^2\sim t^{-2/7}$. In stations whose arrival and peak times are of order the observation time, the typical residual film fraction is $O(\sqrt{{\tau }_y/t})\to 0$. Therefore the films modify the detailed receding cross-section but do not hold a leading-order fraction of the volume or invalidate the large-time nose law. Microscopic contact-line physics is beyond this ideal viscous model.

Take $x$ down the plane and $y$ normally away from it, with the solid at $y=0$ and the free surface at $y=h$. A steady unidirectional falling film flow has velocity $\mathbf{u}=u(y){\mathbf{e}}_x$. The streamwise Navier-Stokes equation reduces towhere $\nu$ is the kinematic viscosity. The lower no-slip boundary condition and upper stress-free boundary condition areIntegrating gives the velocity profileThe volume flux per unit width is

If the upper surface is replaced by a stationary solid plane, the second boundary condition becomes $u(h)=0$. The resulting plane-channel profile and flux areThe free surface exerts no tangential shear stress, whereas the stationary upper wall enforces no slip and exerts a retarding shear. This additional drag accounts for the reduced flux.

A falling free surface does not instantaneously dry previously wetted rigid walls. Thin films remain above the bulk liquid and drain under the component of gravitational acceleration parallel to the walls. For a film of thickness $b$ over a wetted distance $\ell$, falling film flow gives a speed of order $g{b}^2/\nu$, hence $b\sim (\nu \ell /(gt){)}^{1/2}$ at late times for an order-one wall angle. These films can be visible in a cross-section while holding asymptotically negligible volume. In a V-shaped channel gravity current, local bulk depth falls only as $t^{-1/7}$; at a fixed nonzero station with a finite previous maximum wetted height, film volume decays as $t^{-1/2}$ while bulk-sectional volume decays as $t^{-2/7}$. Thus the film does not change the leading spreading law.