Past exam of the differential equations and applications course of the University of Oxford / 2009 / A / 3

The quantity inside a fixed interval changes by influx at its left endpoint minus outflux at its right endpoint:In a smooth region the fundamental theorem of calculus turns this into ${\int }_a^b[{\rho }_t+{\partial }_{x}q(\rho )]dx=0$. Since the interval is arbitrary, the local conservation law is $\boxed{{\rho }_t+{\partial }_{x}q(\rho )=0}$.

For a jump at $x=S(t)$, write the two smooth pieces using a step function. The singular part of ${\rho }_t$ is $-S^{\prime }[\rho ]\delta (x-S)$ and that of ${\partial }_{x}q$ is $[q]\delta (x-S)$. Cancellation gives the Rankine-Hugoniot conditionThe same identity follows by applying conservation to a shrinking moving control interval across the shock.

In smooth regions the characteristic equations are $\dot{x}=q^{\prime }(\rho )={\rho }^3$ and $\dot{\rho }=0$. If $\xi$ labels the initial point, their parametrization isFor the two constant states, characteristics from the left have speed one and those from the right speed zero; they would overlap. The entropy solution replaces the overlap by a compressive shock. Its speed isThusThe left parametrization $x=\xi +t$ is retained only for $x<t/4$, and the right parametrization $x=\xi$ only for $x>t/4$. The characteristic speeds obey $0<c<1$, so characteristics enter the shock from both sides.

Before intersection, the characteristic map is $x=\xi +{\rho }_0(\xi {)}^{3}t$, with derivative $1+3{\rho }_0^2{\rho }_0^{\prime }t$. Because the initial profile decreases, higher-density characteristics behind move faster than lower-density characteristics ahead. Compression therefore forms shocks, which absorb more characteristics. The far states remain one and zero, so a shock joining their limiting traces has speed tending to $[q]/[\rho ]=1/4$.

The precise large-scale interpretation uses $\zeta =x/t$. The rescaled initial transition tends to a step, and entropy comparison for the convex flux givesOne way to see the comparison is to choose distant initial points beyond which the profile is within $\delta$ of each far state. Step profiles with those nearby states bracket it; their shock speeds tend to $1/4$ as $\delta \downarrow 0$, while their fixed initial locations disappear after division by $t$.

A fixed translation of the shock can persist. If the deviation from a reference step is integrable, its signed mass fixes a phase shift. Without such a tail assumption one should assert convergence on the $x/t$ scale and $S(t)/t\to 1/4$, rather than claim that the exact line $x=t/4$ is the unshifted asymptote for every smooth initial transition.

Mass preservation of a nonnegative finite-mass profile requires $\alpha +\beta =0$. Balancing ${\rho }_t$ and ${\partial }_x({\rho }^4/4)$ requires $\alpha -1=4\alpha -\beta$. ThereforeThe similarity ODE isFor a source-type profile adjoining vacuum, the integration constant is zero. The positive branch is $f={\eta }^{1/3}$ for positive $\eta$, terminated by a shock into zero. Normalize the mass:Thus a unit-mass similarity solution isAt the moving right edge, $S^{\prime }={\eta }_F{t}^{-3/4}/4$ equals ${\rho }_L^3/4$, so the Rankine-Hugoniot condition is satisfied. The characteristic speeds satisfy $0<S^{\prime }<{\rho }_L^3$, making it an entropy shock. The left edge meets vacuum continuously. As $t\downarrow 0$ the mass concentrates at the origin; this is a source solution, not a claim that every finite-mass initial profile equals it at finite time.