Rarefaction wave

A rarefaction wave is a continuous expanding fan of characteristic curves joining constant states of a hyperbolic system or scalar conservation law. For $u_t+f(u{)}_x=0$ with $f^{\prime \prime }>0$ between $u_L<u_R$, the Riemann problem has the self-similar solutionTo derive the fan, substitute $u(t,x)=V(\xi )$, $\xi =x/t$, into the smooth equation. It gives $(f^{\prime }(V)-\xi )V^{\prime }=0$, so the nonconstant branch has $f^{\prime }(V)=\xi$. Strict convexity makes $f^{\prime }$ invertible on the state interval, and the two fan edges match the exterior states continuously. Derivatives can jump at those edges.

For the Inviscid Burgers equation with $u_L=0$ and $u_R=1$, the solution is $0$ for $x\le 0$, $x/t$ for $0<x<t$, and $1$ for $x\ge t$. The characteristic speed equals $u$, so the gap between the outgoing exterior characteristic curves fills with all intermediate speeds. This is the selected entropy solution, whereas the corresponding expansion discontinuity is only a weak solution.^[ref]

A resting shallow water layer has depth $h_0$ and wave speed $c_0=\sqrt{g{h}_0}$. A vertical wall receding into negative $x$ at speed $0<V<2c_0$ creates a simple wave with invariant $u-2c=-2c_0$. Inside the expansion fan $x/t=u+c$, proving the displayed profiles. The wall-adjacent state is $u=-V$, $c=c_0-V/2$; the fan occupies $c_0-3V/2<x/t<c_0$, with the resting state ahead. The wall lies at $x/t=-V$, strictly to the left of the fan tail because $c_0-V/2>0$. Thus the assumption on wall speed is exactly the condition for positive depth and continued contact between fluid and wall. The depth follows from $h=c^2/g$.

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.