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Choose the shock wave normal to point from $S<0$ to $S>0$:The latter identity follows by differentiating $S({\mathbf{x}}_s(t),t)=0$ along the moving surface. Write $[a]=a^{+}-a^{-}$ and $[\mathbf{b}]={\mathbf{b}}^{+}-{\mathbf{b}}^{-}$. The distributional derivative of the Heaviside step function is ${\partial }_{t}H(S)=S_t\delta (S)$ and $\nabla H(S)=\nabla S\delta (S)$. The regular terms cancel using the conservation equations on each side, leavingThus the explicitly requested vector isOnly its normal component matters; adding tangential velocity to the surface parametrization changes neither result. The invariant surface delta distribution is ${\delta }_s=|\nabla S|\delta (S)$, so the formula is the moving-interface conservation jump identity ${\partial }_{t}a+\nabla \cdot \mathbf{b}=([\mathbf{b}]\cdot \mathbf{n}-v_n[a]){\delta }_s$.

The distributional derivative of the Heaviside step function gives $H^{\prime }={\delta }_0$, while $((x-1)H(x-1){)}^{\prime \prime }={\delta }_1$. A particular solution is therefore $H(x)-2(x-1)H(x-1)$.

The difference from any other solution has second derivative zero. Applying a distribution with zero derivative is constant twice shows that difference is affine. ThusThe ramp function is continuous at $x=1$, and its first derivative jumps there, producing the indicated delta source in its second derivative.

The distributional derivative of the Heaviside step function is $H^{\prime }={\delta }_0$. The Leibniz rule and $x{\delta }_0=0$ therefore give

For a compactly supported test function $f$,Thus this is the distributional derivative of the Heaviside step function: $H^{\prime }=\delta$.