Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 5 4 b iii Solution Created 2026-10-03 Updated 2026-10-06
Fix and the Rankine-Hugoniot condition speed from the preceding part. The new hypothesis is a uniformly convex scalar flux, ; it replaces the globally bounded- hypothesis of part (a). Choose . Thensince a strictly convex function lies below the chord between its two endpoint values. Also andThe zeros at the endpoints are simple, so the separated integral diverges logarithmically there. Thus the travelling wave is a decreasing connection defined for all , unique up to translation.
To specify a limit, fix a number independently of and normalize . If and , uniqueness gives . ConsequentlyAt the normalized profile equals for every . This single-line value is immaterial to the weak solution. Convergence holds pointwise off the line and in by bounded convergence; it cannot be uniform across a nonzero jump. The transition has thickness of order .
This is the vanishing viscosity approximation to a compressive entropy shock. The Rankine-Hugoniot condition makes the step a weak solution of the inviscid scalar conservation law, while means characteristic curves enter the shock from both sides. For every smooth convex function used as an entropy, with entropy flux for a scalar conservation law , the viscous equation givesAgainst compactly supported tests the right side tends to zero, since stays in . Passing to the limit yields the entropy inequality, explaining the direction selected by positive viscosity.
A translation must be fixed to obtain this particular limit. An -dependent translate can converge to a shock at a different location, to a constant if its center escapes, or fail to converge if the centers oscillate. Thus existence of profiles alone does not specify a single vanishing-viscosity limit without a phase normalization.