Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 86 / 5 / Solution

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.