Godunov finite volumes preserve the Burgers shock speed

The cell averages are $1$ for $j<0$ and $-1$ for $j\ge 0$. The only face pairs are $(1,1)$, $(1,-1)$, and $(-1,-1)$; [F3] gives flux $1\mathrm{/}2$ in all three cases.

Every cell therefore has equal left and right fluxes at every step, so the update leaves every $U_j^{m,n}$ unchanged. Induction on $n$ gives exactly the stationary step profile displayed in the Example.

By [F2] and step 2.1, every reconstruction equals $u(x,t)=1$ for $x<0$ and $u(x,t)=-1$ for $x\ge 0$ almost everywhere, so ${\mathcal{U}}_m\to u$ strongly in $L_{\mathrm{l}\mathrm{o}\mathrm{c}}^1$ and $\mathrm{\mid }u\mathrm{\mid }\le 1$. For any $\phi \in C_c^{\mathrm{\infty }}(\mathbb{R}\times (-\mathrm{\infty },T))$, compact support makes $\phi$ vanish near $t=T$, and $u(x,t)=u_0(x)$ for $t>0$; hence ${\int }_0^T{\int }_{\mathbb{R}}u{\phi }_{t}dxdt=-{\int }_{\mathbb{R}}{u}_0(x)\phi (x,0)dx$. Also $f_B(u)=1\mathrm{/}2$ almost everywhere, so ${\int }_0^T{\int }_{\mathbb{R}}{f}_B(u){\phi }_{x}dxdt=0$. Thus the weak identity holds, and [F4] gives the claimed local weak solution without an additional choice axiom.

The traces are $u^{-}=1$ and $u^{+}=-1$. Thus [F5] gives $s=(f_B(-1)-f_B(1))\mathrm{/}(-1-1)=(1\mathrm{/}2-1\mathrm{/}2)\mathrm{/}(-2)=0$, agreeing with the stationary interface.