Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 83 2 d Solution Created 2026-10-03 Updated 2026-10-14
Initially the same expansion fan and constant-speed front apply. Its left edge reaches the closed rear wall at . After that, the infinite-reservoir solution would give a nonzero velocity at the wall and violates the no-through-flow condition. A reflected disturbance adjusts the interior while the finite initial volume is progressively spread over a longer, shallower current.
The nose does not respond instantaneously when the fan reaches the wall. Its first return signal follows a characteristic from through the undisturbed fan. There , so integrating givesIt reaches the uniform nose-adjacent region at , where . Crossing that region at speed , it catches the front at . Thus the initial nose remains uninformed about the rear wall until this return time. Subsequently its speed must adjust rather than remain constant indefinitely.
For late times use a prismatic triangular-channel gravity-current box model with uniform representative depth over length . The initial volume is . With no entrainment or loss,Apply the same front closure using that representative depth:Starting from a matching late-time state , integration gives finite-volume inertial spreading in a triangular channel:Hence , and the nose speed decays like . A simple initialized box approximation can instead set and , but it then deliberately replaces the exact early rarefaction rather than reproducing it.
The box's volume equation is compatible with a velocity increasing linearly from zero at the rear wall to the nose speed: . A flat-depth box with this velocity is not an exact solution of the full momentum equation during deceleration. It is an integral approximation that leaves the interior shape and nose dynamics unresolved while predicting late-time bulk scaling.
