= Burgers Riemann problem with positive flux
{c}
{title2=$f(u)=u^2/2,\quad s=(u_L+u_R)/2$}
Consider the <Riemann problem> on $\mathbb R$ with <initial condition> $u_0(x)=u_L$ for $x<0$ and $u_0(x)=u_R$ for $x>0$. No endpoint <boundary conditions> are imposed. When $u_L>u_R$, its <entropy solution> is a <shock wave> with constant speed $s=(u_L+u_R)/2$. In the original example $u_L=2$, $u_R=0$, the <Rankine-Hugoniot condition> gives
$$
s=\frac{f(u_R)-f(u_L)}{u_R-u_L}=\frac{0-2}{0-2}=1,\qquad u(x,t)=\begin{cases}2,&x<t,\\0,&x>t.\end{cases}
$$
The left <characteristic speed> is $2$ and the right <characteristic speed> is $0$, so $2>1>0$: both sides enter the <entropy shock>. The <Viscous Burgers travelling shock> gives an exact smooth family tending to this solution.
Reversing the states to $u_L=0$, $u_R=2$ still gives a <Rankine-Hugoniot condition> speed of $1$, but now <characteristic curves> move away from the discontinuity. That expansion jump is a <weak solution> and fails entropy admissibility. For general $u_L<u_R$, the selected <entropy solution> is the continuous <self-similar solution>
$$
u(x,t)=\begin{cases}u_L,&x\leq u_Lt,\\x/t,&u_Lt<x<u_Rt,\\u_R,&x\geq u_Rt,\end{cases}\qquad t>0.
$$
This <rarefaction wave> fills the expanding gap between the two exterior <characteristic curves>. Inside its fan, $u_t=-x/t^2$ and $uu_x=x/t^2$, verifying the <Inviscid Burgers equation>; its values agree continuously with both exterior states. For $0$ to $2$, the fan occupies $0<x<2t$. If $u_L=u_R$, the solution is constant.<Riemann Problems and Jupyter Solutions by David I. Ketcheson, Randall J. LeVeque, and Mauricio J. del Razo/Burgers' equation>[Rarefaction wave, full piecewise solution; expansion-shock example; The entropy condition]{ref}
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