Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 2 / 8C (Differential Equations)

The chain rule gives ${\partial }_x={\partial }_u+{\partial }_v$ and ${\partial }_t={\partial }_u-{\partial }_v$. Hence ${\partial }_x^2-{\partial }_t^2=4{\partial }_u{\partial }_v$, and the forced wave equation becomes $y_{uv}=1$. Integrating twice gives $y=uv+F(u)+G(v)$. At $t=0$ the initial conditions areThus $F-G$ is constant, which cancels from the final expression. Combining the remaining sum yieldsDifferentiation checks the forcing and both initial conditions.

Apply differentiation under the integral sign on the moving drop:The moving-boundary term vanishes because $h(R,t)=0$. At the origin, symmetry and regularity give $h_r(0,t)=0$. At the front, the stated $h\propto (R-r{)}^{1/3}$ gives $h^3{h}_r=O((R-r{)}^{1/3})\to 0$. Thus $\boxed{dM/dt=0}$, expressing volume conservation for this axisymmetric viscous gravity current.

With $\eta =r/t^{\alpha }$ and $R={\eta }_0{t}^{\alpha }$, substitute the self-similar solution into the conserved integral:For a nonzero finite drop volume, the power of $t$ must vanish, so $\boxed{\alpha =1/8}$. Balancing powers in the evolution equation gives the same result: its left side scales as $t^{-5/4}$ and its right side as $t^{-1-2\alpha }$. The front-edge exponent alone would not determine this spreading exponent without the conserved volume.