The Burgers Godunov flux is continuous and consistent

On each open region cut out by $a=b$, $a=0$, and $b=0$, the displayed formula in [F1] is either a polynomial, zero, or $(a^2+b^2+\mathrm{\mid }{a}^2-b^2\mathrm{\mid })\mathrm{/}4$, so it is continuous there.

At a boundary with $a=b>0$ both adjacent formulas tend to $a^2\mathrm{/}2$; at $a=b<0$ both tend to $a^2\mathrm{/}2$; at $a=0\le b$ the left and right limits are zero; at $a\le 0=b$ the limits are zero. On the remaining axis portions $a=0>b$ and $b=0<a$, a neighborhood lies in the single shock branch $b<a$, whose formula is continuous there. At $(a,b)=(0,0)$ every branch has absolute value at most $\max \{a^2,b^2\}\mathrm{/}2$, which tends to zero. These checks cover every branch boundary and prove continuity on all of ${\mathbb{R}}^2$.

On the diagonal, if $a\ge 0$ the first branch gives $G(a,a)=a^2\mathrm{/}2$; if $a<0$ the second branch gives $G(a,a)=a^2\mathrm{/}2$. Thus $G(a,a)=f_B(a)$ for all $a$, and [F2] makes $G$ a consistent numerical flux.