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Let $\rho$ be the liquid density and neglect atmospheric shear. The shallow-channel lubrication approximation makes the surface nearly horizontal in each cross-section, at $z=h(x,t)$, with hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. At fixed $y$, the floor is $z_b=\alpha |y|$ and the local liquid thickness is $H_y=h-z_b$. Axial momentum and boundary conditions areWriting $\zeta =z-z_b$, integration givesThe wetted interval is $|y|<h/\alpha$. Its cross-sectional area and total axial flux areThe shallow V-channel lubrication flux and volume conservation $A_t+Q_x=0$ therefore giveThe slope $\alpha \ll 1$ makes transverse shear small relative to vertical shear; the longitudinal current must also be slender for this reduction.

For typical height $H$ and axial length $L$, fixed volume gives $H^{2}L\sim \alpha V$, while the equation gives $L^2\sim C{H}^{3}t$. Eliminating either scale yieldsThus the current grows in length like $t^{2/7}$ and thins like $t^{-1/7}$. Equivalently, the porous-medium transformation of V-channel spreading sets $w=h^2$ and gives $w_t=(C/5)(w^{5/2}{)}_{xx}$ with constant mass $\alpha V$.

Seek a symmetric solution $h=t^{-1/7}f(\xi )$ with $\xi =x/t^{2/7}$. Substitution into the conservative depth equation givesThe left side is $-(2/7)(\xi f^2{)}^{\prime }$. Symmetry sets the integration constant to zero, so $C{f}^4{f}^{\prime }=-(2/7)\xi f^2$ within the support. Integrating $f^2{f}^{\prime }=-2\xi /(7C)$ gives $f^3=3({\xi }_0^2-{\xi }^2)/(7C)$. In physical variables the constant-volume similarity in a V-shaped channel iswhere $X(t)$ is the half-length of its support and the positive part makes $h=0$ outside it.

Define the positive normalizationThe first substitution is $s=\cos \theta$; the gamma form follows from the beta function integral. Conservation of volume now givesand consequentlyThe full current length is $2X(t)$. Its flux vanishes at the two noses, since the integrated similarity equation gives $C{h}^4{h}_x=-2x{h}^2/(7t)$ there, so no volume is lost at the support boundary. A shifted origin or a positive shift of time gives the corresponding translated similarity family.

In an intended trigonometric $I(\beta )$ notation, $J=I(7/3)$ requires either cosine integrated over $[-\pi /2,\pi /2]$ or absolute cosine over $[0,\pi ]$. The printed ${\cos }^{\beta }\theta$ on $[0,\pi ]$ is not a positive real normalization for the required fractional exponent: with a real cube-root interpretation at $\beta =7/3$, its two halves cancel, while the principal complex power is not real. The similarity normalization uses the positive sine integral $J$ above.

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.

The variable $w=h^2$ converts the shallow V-channel lubrication flux equation into a porous medium equation:Indeed $h^4{h}_x=w^{3/2}{w}_x/2=(w^{5/2}{)}_x/5$. Its fixed-mass source-type solution gives $h\propto t^{-1/7}$ and support length proportional to $t^{2/7}$. Integrating the similarity equation once gives $C{h}^4{h}_x=-2x{h}^2/(7t)$, and hence $h=[3(X(t{)}^2-x^2)/(7Ct){]}_{+}^{1/3}$. The constant-volume similarity in a V-shaped channel fixes $X$ from the positive integral $J={\int }_{-1}^1(1-s^2{)}^{2/3}ds$: $X=(\alpha V/J{)}^{3/7}(7Ct/3{)}^{2/7}$.