Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 64 / 3

Work in a gas region with $\rho ,p>0$ and physical adiabatic exponent $\gamma >1$. With $\xi =x/t$, each time derivative is $-\xi /t$ times differentiation in $\xi$, and each spatial derivative is $1/t$ times it. The continuity equation, Euler momentum equation and the entropy advection equation for pressure reduce toTheir coefficient determinant is $(u-\xi )[(u-\xi {)}^2-c^2]$, where the adiabatic sound speed is $c=\sqrt{\gamma p/\rho }$. If the determinant is nonzero, all derivatives vanish and the solution is uniform.

The apparent third characteristic $u=\xi$ does not give a smooth fan: on such an interval $u^{\prime }=1$, whereas continuity requires $\rho u^{\prime }=0$, impossible for $\rho >0$. It cannot support an isolated smooth entropy variation either. Subtracting $\gamma$ times the logarithmic density equation from the logarithmic pressure equation gives $(u-\xi )(\ln (p/{\rho }^{\gamma }){)}^{\prime }=0$. A nonzero continuous entropy derivative would force $u=\xi$ on an interval and give the same contradiction. Thus the specific entropy is constant throughout each smooth branch.

Every nonconstant branch therefore lies in one acoustic characteristic family. Let $\sigma =\pm 1$ denote it, with $\xi =u+\sigma c$, so $u-\xi =-\sigma c$. The constant entropy means $p=K{\rho }^{\gamma }$, $K>0$, and $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$. Continuity gives $u^{\prime }=\sigma c{\rho }^{\prime }/\rho =2\sigma c^{\prime }/(\gamma -1)$. Differentiating $\xi =u+\sigma c$ then provesIn particular, $u-2\sigma c/(\gamma -1)=J$ is a constant Riemann invariant. All nonconstant smooth local solutions have the explicit formon an interval where $c>0$. Along the left-going family ($\sigma =-1$) sound speed decreases with $\xi$; along the right-going family ($\sigma =+1$) it increases. These are the two self-similar ideal-gas rarefaction branches. Together with arbitrary uniform positive states they exhaust the smooth local solutions. A global solution can join them piecewise at fan edges, where derivatives generally jump, and can also have discontinuities; those do not constitute additional smooth branches.

The initial-value problem is a Riemann problem. Both the one-dimensional Euler equations and the two constant initial states are unchanged under the simultaneous scaling $(x,t)\mapsto (\lambda x,\lambda t)$ for $\lambda >0$. There is no imposed length or time scale. Rescaling a physically admissible entropy solution therefore gives another solution with exactly the same initial data. Uniqueness of the physical Riemann solution makes the two coincide, soThe smooth pieces are consequently those in part (a), while shocks and contacts lie on rays $x=st$. Across shocks one uses the conservative mass, momentum and total-energy equations and the entropy admissibility condition, rather than extending the smooth entropy-advection equation through the discontinuity. Scale invariance by itself, without selecting the unique admissible solution, would not justify the uniqueness step.

Let $c_L=(\gamma p_L/{\rho }_L{)}^{1/2}$ and $c_R=(\gamma p_R/{\rho }_R{)}^{1/2}$, with positive pressures and densities. A nontrivial single rarefaction wave has constant specific entropy and constant Riemann invariant, so for one family label $\sigma =\pm 1$ the necessary matching conditions areThe final inequality says that the fan opens rather than compresses. Its edge speeds are ${\xi }_L=u_L+\sigma c_L$ and ${\xi }_R=u_R+\sigma c_R$, and their difference is $\sigma (c_R-c_L)(\gamma +1)/(\gamma -1)>0$. The profiles in part (a) then interpolate the two end states and supply a sufficient construction. For the left family, $c_R<c_L$, hence ${\rho }_R<{\rho }_L$ and $p_R<p_L$; for the right family all three inequalities reverse. In both cases $u_R>u_L$. These are the single-rarefaction matching conditions for an ideal-gas Riemann problem.

For a concrete nontrivial example, take $\gamma =5/3$ andThe adiabatic sound speeds are $c_L=1$ and $c_R=1/2$, with common $K=3/5$ and left-family invariant $u+3c=3$. The fan edges are ${\xi }_L=-1$ and ${\xi }_R=1$. The complete self-similar solution is uniform at these stated left and right values outside the fan, while inside $-1<\xi <1$,It matches both states continuously, has $u^{\prime }=3/4=2/(\gamma +1)$, and has constant specific entropy. For fixed $x\ne 0$, its $t\downarrow 0$ limit is the required initial step.

A single rarefaction wave enforces two matching equalities: equal specific entropy and one common Riemann invariant. The inequality $u_R>u_L$ does not enforce either. If the entropies differ, one smooth isentropic fan cannot connect them; even if they agree, arbitrary end velocities and pressures need not satisfy either acoustic-family invariant.

The general positive-gas Riemann problem has a left acoustic wave, a middle contact discontinuity, and a right acoustic wave. Across the contact the two intermediate states have the same velocity $u_{\ast }$ and pressure $p_{\ast }$, but can have different mass densities and entropies. Each acoustic wave is a rarefaction wave or a shock wave, depending on the required intermediate pressure relative to its adjacent initial state. Shocks satisfy the conservative jump conditions and increase entropy. A sufficiently large pressure contrast can produce a shock on one side even when the initial velocities obey $u_R>u_L$.

For sufficiently rapid separation, the two rarefactions instead exhaust the gas and create vacuum between them. Their vacuum-edge velocities are $u_L+2c_L/(\gamma -1)$ and $u_R-2c_R/(\gamma -1)$, so a finite vacuum gap occurs whenEquality is the limiting case in which the vacuum edges meet. Thus the additional features can be two acoustic waves, a contact carrying an entropy jump, shocks, or a vacuum region, rather than one fan joining the two original uniform states.