The Rankine--Hugoniot jump condition in space--time normal form

Side extensions solve the equation classically. On each side of $\mathrm{\Gamma }$ the function $u$ agrees with a $C^1$ extension ${\tilde{u}}^{\pm }$; testing away from $t=0$ and against bumps supported in a single side, the weak identity [F1] shows that ${\mathrm{div}}_{t,x}F({\tilde{u}}^{\pm })$ vanishes as a distribution on that side. Since $F({\tilde{u}}^{\pm })$ is $C^1$, its divergence is continuous, and by [F2] it vanishes pointwise; consequently, for smooth compactly supported $\phi$ supported in the side, ${\mathrm{div}}_{t,x}(F(u^{\pm })\phi )=F(u^{\pm })\cdot D\phi$.

Graph computation on one side. After a permutation of coordinates, write the graph locally as $z_k=\gamma (z^{\mathrm{\prime }})$ and take $\phi$ supported in a box $B^{\mathrm{\prime }}\times (a,b)$ in which the graph stays in $(a,b)$. With $G_j=F_j(u^L)\phi$ on the lower side, [F3] gives ${\int }_{\text{lower}}F(u^L)\cdot D\phi ={\int }_{\text{lower}}\mathrm{div}(F(u^L)\phi )={\int }_{B^{\mathrm{\prime }}}{G}_k^L(z^{\mathrm{\prime }},\gamma (z^{\mathrm{\prime }}))d{z}^{\mathrm{\prime }}-{\sum }_{j\mathrm{\ne }k}{\int }_{B^{\mathrm{\prime }}}{G}_j^L(z^{\mathrm{\prime }},\gamma (z^{\mathrm{\prime }})){\mathrm{\partial }}_j\gamma (z^{\mathrm{\prime }})d{z}^{\mathrm{\prime }}={\int }_{B^{\mathrm{\prime }}}{G}^L(z^{\mathrm{\prime }},\gamma (z^{\mathrm{\prime }}))\cdot N(z^{\mathrm{\prime }})d{z}^{\mathrm{\prime }}$, because the $k$-derivative integrates to the trace at the graph and each tangential derivative of the moving-endpoint integral $A_j(z^{\mathrm{\prime }})={\int }_a^{\gamma (z^{\mathrm{\prime }})}{G}_j(z^{\mathrm{\prime }},r)dr$ contributes $-{\mathrm{\partial }}_j\gamma G_j$ at the graph, the integral of ${\mathrm{\partial }}_j{A}_j$ vanishing by compact support.

Upper side and the interface term. The same computation on the upper side, with the graph as its lower boundary, gives ${\int }_{\text{upper}}F(u^R)\cdot D\phi =-{\int }_{B^{\mathrm{\prime }}}{G}^R(z^{\mathrm{\prime }},\gamma (z^{\mathrm{\prime }}))\cdot N(z^{\mathrm{\prime }})d{z}^{\mathrm{\prime }}$; adding with step 2.1, and noting that $F(u^{+})-F(u^{-})=([u],[f])$ with $u^{\pm }$ the traces from the plus and minus sides, the weak identity becomes $0={\int }_{{\mathrm{\Pi }}_T}F(u)\cdot D\phi =-\sigma {\int }_{B^{\mathrm{\prime }}}\phi (z^{\mathrm{\prime }},\gamma (z^{\mathrm{\prime }}))[F](z^{\mathrm{\prime }},\gamma (z^{\mathrm{\prime }}))\cdot N(z^{\mathrm{\prime }})d{z}^{\mathrm{\prime }}$ for every $\phi$ supported in the box.

Continuity and vanishing of the bracket. The function $z^{\mathrm{\prime }}\mapsto [F](z^{\mathrm{\prime }},\gamma (z^{\mathrm{\prime }}))\cdot N(z^{\mathrm{\prime }})$ is continuous, being a composition of continuous data; if it were nonzero at the point corresponding to ${\zeta }_0$, it would keep one sign on a smaller patch, and a nonnegative smooth bump supported there and positive at ${\zeta }_0$ would make the integral of step 3.1 nonzero, a contradiction. Hence $[F]\cdot N=0$ at ${\zeta }_0$.

Normal form and the one-dimensional case. Since $\nu =\sigma N\mathrm{/}\mathrm{\mid }N\mathrm{\mid }$ with $\sigma =\pm 1$ by [F2], $0=[F]\cdot N=[F]\cdot \nu \sigma \mathrm{\mid }N\mathrm{\mid }$, and $\mathrm{\mid }N\mathrm{\mid }=\mathrm{\mid }{e}_k-{\sum }_{j\mathrm{\ne }k}{\mathrm{\partial }}_j\gamma e_j\mathrm{\mid }>0$, so $[F]\cdot \nu =[u]{\nu }_t+[f]\cdot {\nu }_x=0$ at every point of $\mathrm{\Gamma }$. For a one-dimensional graph $x=s(t)$ with minus side $x<s(t)$, the graph function is $\gamma (t)=s(t)$, so $N=(-s^{\mathrm{\prime }}(t),1)$ and $\nu =(-s^{\mathrm{\prime }},1)\mathrm{/}\sqrt{1+s^{\mathrm{\prime }2}}$; the condition becomes $(-s^{\mathrm{\prime }}[u]+[f])\mathrm{/}\sqrt{1+s^{\mathrm{\prime }2}}=0$, that is, $s^{\mathrm{\prime }}[u]=[f]$. If $[u]\mathrm{\ne }0$ this determines $s^{\mathrm{\prime }}=[f]\mathrm{/}[u]$, and if $[u]=0$ the relation reads $0=[f]$, so $[f]=0$ and no speed is constrained.