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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.

A Riemann problem for a one-dimensional system of conservation laws ${\partial }_{t}U+{\partial }_{x}f(U)=0$ has two constant initial states, $U_L$ for $x<0$ and $U_R$ for $x>0$. The data and equation are invariant under $(x,t)\mapsto (\lambda x,\lambda t)$. When the physically admissible entropy solution is unique, this invariance makes it a self-similar solution depending on $x/t$. For ideal-gas compressible flow, the characteristic speeds are $u-c,u,u+c$. The two acoustic families carry shock waves or rarefaction waves, while the middle family carries a contact discontinuity. Sufficiently strong separating velocities can instead create a vacuum region.

For $\gamma >1$ and positive mass density and pressure, let $\xi =x/t$ in one-dimensional adiabatic compressible flow. The smooth equations are $(u-\xi ){\rho }^{\prime }+\rho u^{\prime }=0$, $(u-\xi )u^{\prime }+p^{\prime }/\rho =0$ and $(u-\xi )p^{\prime }+\gamma p{u}^{\prime }=0$. Away from a constant branch, their coefficient determinant forces $\xi =u+\sigma c$ with $\sigma =\pm 1$ and adiabatic sound speed $c=\sqrt{\gamma p/\rho }$. The entropy equation gives $p=K{\rho }^{\gamma }$, and combining continuity with $c^{\prime }/c=(\gamma -1){\rho }^{\prime }/(2\rho )$ givesThus $u=[2\xi +(\gamma -1)J]/(\gamma +1)$, $c=\sigma (\gamma -1)(\xi -J)/(\gamma +1)$, $\rho =(c^2/(\gamma K){)}^{1/(\gamma -1)}$ and $p=K{\rho }^{\gamma }$, on intervals where $c>0$. The specific entropy is constant. Together with uniform branches, these exhaust smooth positive-gas self-similar solutions; a contact discontinuity is not a smooth fan.