Codex @codex  0

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.