Godunov scheme captures a stationary Burgers entropy shock

The initial cell averages are $U_j^0=1$ for $j<0$ and $U_j^0=-1$ for $j\ge 0$. Same-state faces have flux $1\mathrm{/}2$ by [F1], and the only mixed face has pair $(1,-1)$ and flux $\max \{1,1\}\mathrm{/}2=1\mathrm{/}2$. Thus every face flux equals $1\mathrm{/}2$.

Every cell's left and right fluxes are both $1\mathrm{/}2$, so the conservative update leaves every $U_j^n$ unchanged. Induction in $n$ proves that both reconstructions equal the stationary step function almost everywhere for every $m$; in particular they converge strongly locally to it.

The stationary profile is a bounded weak solution: for every $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$, its time term satisfies ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathbb{R}}u{\phi }_{t}dxdt=-{\int }_{\mathbb{R}}{u}_0(x)\phi (x,0)dx$, while $f_B(u)=1\mathrm{/}2$ almost everywhere and hence ${\int }_{\mathbb{R}}{f}_B(u){\phi }_{x}dx=0$ at each time. This is exactly the weak identity of [F3].

Fix $k\in \mathbb{R}$. The entropy flux traces are $q_k(1)$ on the left and $q_k(-1)$ on the right. If $k\le -1$ or $k\ge 1$, both states lie on the same side of $k$ and these fluxes are equal because $f_B(1)=f_B(-1)$. If $-1<k<1$, then $q_k(1)=\frac{1-k^2}{2},q_k(-1)=-\frac{1-k^2}{2},q_k(1)-q_k(-1)=1-k^2\ge 0.$ For any nonnegative interior test $\phi$, the time term $\int \mathrm{\mid }u-k\mathrm{\mid }{\phi }_t$ vanishes because $u$ is stationary. Integration on the two half-lines gives ${\int }_0^T\text{\hspace{0.17em}⁣}{\int }_{\mathbb{R}}{q}_k(u){\phi }_{x}dxdt=(q_k(1)-q_k(-1)){\int }_0^T\phi (0,t)dt\ge 0.$ The initial trace is exactly $u_0$, so [F5] shows the stationary shock is a Kruzhkov entropy solution.

The left and right traces are $u_L=1$ and $u_R=-1$. By [F4], the shock speed is $s=\frac{f_B(-1)-f_B(1)}{-1-1}=\frac{1\mathrm{/}2-1\mathrm{/}2}{-2}=0,$ which agrees with the stationary interface. The Rankine–Hugoniot supplier applies because step 3.1 verified the weak identity and the traces are distinct.

The translation variation of the datum is $2$: for $h>0$, $u_0(x+h)-u_0(x)$ is nonzero only on $(-h,0)$, where its absolute value is $2$, so $\int \mathrm{\mid }{u}_0(x+h)-u_0(x)\mathrm{\mid }dx=2h$. The ratio is ${\tau }_m\mathrm{/}{h}_m=1\mathrm{/}4$, and the flux bounds $L_1=L_2=1$ from [F2] give $\lambda (L_1+L_2)=1\mathrm{/}2\le 1$. Thus the hypotheses of the main convergence theorem in [F2] hold; since the exact reconstruction is the entropy solution from step 3.2, it is the unique limit. No global $L^1$ assertion is made because the two tails are nonzero.