Consider Burgers' equation $u_t + f(u)_x = 0$ where $f(v)=\frac{v^2}{2}$, with initial condition $$u_0(x)=\begin{cases} -1 & x<0 \\ 1& x>0 .\end{cases} $$ It's clear that $u_l=-1<u_r=1$, so the discontinuity is not admissible and the entropy solution is a rarefaction wave. Consider the following discrete initial condition $$v_{j}^{0}= \begin{cases} -1 & j\le -1 \\ 1& j\ge 0 \end{cases} $$ Show that using the Murman-Roe scheme that the numerical solution is a stationary shock and is given by $$v_{j}^{n}=v_{j}^{0}= \begin{cases} -1 & j\le -1 \\ 1& j\ge 0 \end{cases} $$
Can you give me some hint pls, or can you solve it.Because I know the Roe Scheme in case of linear function f
What is the solution using Godunov Scheme? Is it by using the Riemann Solver $W_R(\frac{x}{t},u_l,u_r)$ ,where $\frac{x}{t}$ is non-zero?
Roe-type approximate Riemann solvers are known to fail at providing the entropy-satisfying solution in the particular case of transonic rarefaction waves (i.e., rarefaction waves that cross the zero-velocity state $u=0$). The present example illustrates this feature. First, we need to write the Murman-Roe scheme. To do so, consider the mesh size $\Delta x$ and the time step $\Delta t$, such that the coordinates $(x,t)$ are discretized as $x_j = j\Delta x$ and $t^n = n\Delta t$. The scheme reads $$ v_j^{n+1} = v_j^n - \frac{\Delta t}{\Delta x} \left(g(v_{j}^n, v_{j+1}^n) - g(v_{j-1}^n, v_{j}^n)\right) $$ with numerical flux $$ g(u,v) = \tfrac12\left(f(u) + f(v) - |a(u,v)|(v-u)\right) $$ $$ a(u,v) = \left\lbrace\begin{aligned} &\tfrac{f(u) - f(v)}{u-v} &&\text{if}\quad u\neq v\\ &f'(u) &&\text{if}\quad u= v . \end{aligned}\right. $$ Then, we need to show that the unphysical stationary shock solution is obtained by iterating this scheme -- instead of the physical rarefaction solution. Note that this bad feature doesn't show up for the Godunov scheme (i.e., the exact Riemann solver), which numerical solution will be an approximation of the correct rarefaction wave.