Vector Rankine–Hugoniot condition

Fix a test $\phi \in C_c^{\mathrm{\infty }}(I\times \mathrm{\Omega };{\mathbb{R}}^m)$. Choose an orthonormal basis of $n^{\perp }$ and write $x=y+(st+r)n$, with $y\in n^{\perp }$ and $r=n\cdot x-st$. In these coordinates $\mathrm{\Phi }(t,y,r)=(t,y+(st+r)n)$ is a linear bijection with absolute determinant $1$, and $y\mapsto y+stn$ is an isometry from $n^{\perp }$ onto the full spatial plane $\{x\in {\mathbb{R}}^d:n\cdot x=st\}$, whose restriction gives the slice in $\mathrm{\Omega }$. [F4] states scaling for measurable sets. For a nonnegative simple function, [F6] and the set-scaling identity for ${\mathrm{\Phi }}^{-1}$ give equality of the integrals of the function and its pullback by summing over the measurable level sets; since $\mathrm{\Phi }$ is bijective, pullback gives a bijection between the simple minorants of a nonnegative measurable function and those of its pullback, so [F7] gives the same integral identity for nonnegative measurable functions. Apply that identity to the positive and negative parts of each signed integrable real integrand; [F8] then gives equality for the integrable integrands used here. Every test-supported integrand is integrable after extension by zero, since the coefficients are locally integrable and the test derivatives are bounded on compact support. Thus this chart preserves the required volume integrals, and [F5] permits their iterated evaluation. The coordinate measure $dy$ is the pushforward of ${\lambda }_{d-1}$ by the chosen linear isometry $E$. Replacing $E$ by another such isometry changes $z$ by an orthogonal linear map, whose determinant has absolute value $1$, so [F4] and the same simple-function argument prove independence of the basis; when $d=1$ this is the one-point measure specified in the Statement. Let $H(r)=1$ for $r>0$ and $H(r)=0$ for $r<0$. By [F3], one-dimensional integration by parts in $r$ gives, for every $C^1$ coefficient field $a$, ${\mathrm{\partial }}_t(aH)=H{\mathrm{\partial }}_{t}a-sa{\mathrm{\mid }}_{\mathrm{\Sigma }}{\delta }_{\mathrm{\Sigma }}$ and ${\mathrm{\partial }}_{x_j}(aH)=H{\mathrm{\partial }}_{x_j}a+n_{j}a{\mathrm{\mid }}_{\mathrm{\Sigma }}{\delta }_{\mathrm{\Sigma }}$, where ${\delta }_{\mathrm{\Sigma }}$ pairs with $dtdy$ on the spatial plane, with the coordinate Lebesgue measure just defined. The time derivative is at fixed $x$, so in this chart ${\mathrm{\partial }}_t{\mathrm{\mid }}_x={\mathrm{\partial }}_t{\mathrm{\mid }}_{y,r}-s{\mathrm{\partial }}_r$; this produces the factor $-s$ and the stated normalization. Tangential spatial derivatives have no boundary term, while the normal derivative contributes $n_j$.

On a neighborhood of $\mathrm{\Sigma }$, write $U=U_L+[U]H$ and $F_j(U)=F_j(U_L)+[F_j(U)]H$, using the $C^1$ side extensions from [F1]. Apply the derivative formulas of step 1.1. The regular terms are the classical residuals on the two open sides, hence vanish by the given equations; the remaining interface term is $({\sum }_j{n}_j[F_j(U)]-s[U]){\delta }_{\mathrm{\Sigma }}$. A partition of unity over the compact support of $\phi$ gives the same formula throughout $I\times \mathrm{\Omega }$.

If the coefficient vanishes on $\mathrm{\Sigma }$, the residual distribution is zero. Conversely, if it is nonzero at a point of $\mathrm{\Sigma }$, continuity of the traces gives a neighborhood on which one component has a fixed strict sign. A nonnegative compactly supported test in that interface patch, placed in the corresponding vector component, has a nonzero pairing with the residual, contradicting that it is the zero distribution. If $\mathrm{\Sigma }$ is empty in the cylinder, the equivalence is vacuous. Thus the residual vanishes exactly when the vector Rankine–Hugoniot equality holds pointwise on $\mathrm{\Sigma }$.